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http://math.stackexchange.com/questions/197919/let-p-and-q-be-two-distinct-primes-pick-the-correct-statements-from | # Let $p$ and $q$ be two distinct primes. Pick the correct statements from
Let $p$ and $q$ be two distinct primes. Pick the correct statements from the following:
a. $Q(\sqrt p)$ is isomorphic to $Q(\sqrt q)$ as fields.
b. $Q(\sqrt p)$ is isomorphic to $Q(\sqrt{−q})$ as vector spaces over Q.
c. $[Q(\sqrt p,\sqrt q) : Q] = 4$.
d. $Q(\sqrt p,\sqrt q) = Q(\sqrt p + \sqrt q)$.
(c) & (d) are correct. (a) is not correct for fields but correct for vector spaces. (b) not sure. i think it is correct by the same arguement of (a).am i right?
-
Is this a homework question, or just one for personal development, as this may effect the type of response that people give? – David Ward Sep 17 '12 at 8:16
@DavidWard I don't know why it should affect the type of response. If one should not give a full answer to a homework question, many good questions might have no full answer in this site. I think this is against the policy of this site. – Makoto Kato Sep 17 '12 at 18:13
@MakotoKato My reasoning is that if it is a homework question, then the answer itself is of importance to the person asking the question. However, if the question is not homework, then for a question as above, I would feel happy to give the answer, but then lead someone through the reasoning on how to get there. – David Ward Sep 18 '12 at 8:26
@DavidWard I'm sorry that I don't understand your reasoning. Could you explain why you don't want to give the answer to a homework question? – Makoto Kato Sep 18 '12 at 8:43
@MakotoKato My reasoning is as follows: Firstly suppose that the question asked is from an assessed piece of work. Then as the question is a multiple-choice question, the actual answer will attract some if not all of the marks (depending on the desire of the assessor). Thus knowledge of the answer is of critical importance. On the other hand, if the question asked is purely for personal development, whilst leading the member towards the solution, it can be beneficial to actually know where you are being led. It just gives some motivation for where the reasoning is heading. – David Ward Sep 18 '12 at 15:26
a. Suppose $\mathbb{Q}(\sqrt p)$ is isomorphic to $\mathbb{Q}(\sqrt q)$ as fields. Then $\mathbb{Q}(\sqrt q)$ has an element $\alpha$ such that $\alpha^2 = p$. Let $\alpha = a + b\sqrt q$, where $a, b \in \mathbb{Q}$. We denote the conjugate of $\alpha$ by $\alpha'$. Since $\alpha + \alpha' = 0$, $a = 0$. Since $\alpha^2 = p$, $p = b^2 q$. Hence $p = q$. This is a contradiction. Hence $a$. is not true.
b. Since the both fields have dimension 2 as vector spaces over $\mathbb{Q}$, $b.$ is true.
c. As we see in the above, $\sqrt p$ is not contained in $\mathbb{Q}(\sqrt q)$. Hence $[\mathbb{Q}(\sqrt p,\sqrt q) : \mathbb{Q}(\sqrt q)] = 2$. Hence $[\mathbb{Q}(\sqrt p,\sqrt q) : \mathbb{Q}] = 4$. Thus $c.$ is true.
d. Let $K = \mathbb{Q}(\sqrt p,\sqrt q)$. Clearly $K/\mathbb{Q}$ is Galois. Let $\sigma \in Gal(K/\mathbb{Q})$. Then $\sigma(\sqrt p) = \sqrt p$ or $-\sqrt p$, $\sigma(\sqrt q) = \sqrt q$ or $-\sqrt q$. Since $|Gal(K/\mathbb{Q})| = 4$ by $c.$, $Gal(K/\mathbb{Q}) = \{1, \sigma_1, \sigma_2. \sigma_3\}$, where
$$\sigma_1(\sqrt p) = \sqrt p,\ \ \ \sigma_1(\sqrt q) = -\sqrt q$$
$$\sigma_2(\sqrt p) = -\sqrt p,\ \sigma_2(\sqrt q) = \sqrt q$$
$$\sigma_3(\sqrt p) = -\sqrt p,\ \ \ \sigma_3(\sqrt q) = -\sqrt q$$
Let $\alpha = \sqrt p + \sqrt q$.
Clearly $\alpha, \sigma_1(\alpha) = \sqrt p - \sqrt q, \sigma_2(\alpha) = -\sqrt p + \sqrt q, \sigma_3(\alpha) = -\sqrt p - \sqrt q$ are distinct. Hence $K = \mathbb{Q}(\alpha)$. Thus $d.$ is true.
-
please explain why α+α′=0? In the proof of a ? – GA316 Aug 9 '15 at 4:01 | 2016-05-27T16:28:35 | {
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https://cs.stackexchange.com/questions/106135/how-do-i-describe-formally-complexity-of-2-sum-problem-algorithm?noredirect=1 | # How do I describe formally complexity of 2-sum problem algorithm?
I have algorithm that finds if there are two elements in sorted array that have sum zero.
1.ZeroSumPair(A[1..n]) // A[1..n] <-- sorted
2. l <- 1, r <- n
3. while(l < r)
4. while(l < r or A[l] + A[r] > 0)
5. r--
6. if(A[l] + A[r] = 0)
7. return true
8. l++
9. return false
Intuitively I know that this algorithm is $$O(n)$$, but how do I deduct it using proof with summations like in CLRS book?
I've also saw How to develop an $O(N)$ algorithm solve the 2-sum problem?, but I didn't see any formal proof.
• I would try putting in a counter t. Then increment t by 1 every time you do an operation. Then try to prove a loop invariant for the outer-while loop that claims $t = O(n)$. With the correct loop invariant on $t$, this can be proven by induction relatively easily. Perhaps something like $t \leq c(l + (n - r))$ for some constant $c$ might work. Then at termination (in the worst case) you know $l = r$, thus $t \leq cn \implies t = O(n)$. – ryan Mar 27 at 18:28
Consider this slightly modified algorithm where I've added an "operation counter" t. This will be incremented every time we do a comparison or assignment.
1. ZeroSumPair(A[1..n]) // A[1..n] <-- sorted
2. l <- 1, r <- n
3. t <- 3 // 2 for first two assignments and 1 for initial while check
4. while(l < r)
5. t++ // 1 for initial while check
6. while(l < r or A[l] + A[r] > 0)
7. t++ // 1 for decrementing r
8. r--
9. t++ // 1 for following while check
10. t++ // 1 for if comparison
11. if(A[l] + A[r] = 0)
12. return true
13. t++ // 1 for incrementing l
14. l++
15. t++ // 1 for following while check
16. return false
This is a bit verbose, but it will work. Now we must simply prove that, at termination we have $$t = O(n)$$. We can do this inductively with a loop invariant.
Let's use the following loop invariant for the loop on line 4.
$$t = 3 + 4(l-1) + 2(n-r)$$
### Base Case
Initially $$t = 3$$, $$l = 1$$, and $$r = n$$. Thus we have:
$$3 = 3 + 4(1 - 1) + 2(n - n) = 3$$
### Inductive Case
Let $$t'$$, $$l'$$, and $$r'$$ be the values of $$t$$, $$l$$, and $$r$$ at the end of our previous iteration. At the end of our current iteration we have $$l = l' + 1$$, $$r = r' - k$$ for some $$k$$, and $$t = t' + 4 + 2k$$. Thus we have:
\begin{align*} t & = t' + 4 + 2k\\ & = 3 + 4(l' - 1) + 2(n - r') + 4 + 2k\\ & = 3 + 4(l - 2) + 2(n - (r + k)) + 4 + 2k\\ & = 3 + 4(l - 1) - 4 + 2(n - r) - 2k + 4 + 2k\\ & = 3 + 4(l - 1) + 2(n - r) & \square \end{align*}
Thus, we can conclude the loop invariant holds. At the end of the loop (in the worst case) we have $$l = r \leq n$$. We then have: \begin{align*} t & = 3 + 4(l - 1) + 2(n - r)\\ & = 3 + 4l - 4 + 2n - 2l\\ & = 2(n + l) - 1\\ & \leq 2(n + n) - 1\\ & = 4n - 1\\ & = O(n) & \square \end{align*}
Thus it is linear. There might be an easier way to do this, but this way makes sense to me pretty clearly.
• I see that this looks formal but I feel like it is “hacking”, amending the algorithm in order to prove its complexity, is this real thing in computer science? Not talking about practice here – kuskmen Mar 27 at 21:53
• @kuskmen you don't need $t$ explicitly in your code. I put it there for clarity. You can have an implicit "operation count" which is realistically what you're trying to determine. Then prove (through loop invariants) that the "operation count" is bounded by $n$. That's all I did here with $t$ being the operation counter. – ryan Mar 27 at 22:49
• I see, I have to admit I am bad in mathematical proofs or any sort of proofs as you can tell. Can you tell why $t=t'+ 4 + 2k$ in the inductive step? – kuskmen Mar 28 at 8:15
• Assume $t$ starts at $t'$ just before the loop executes. Now hop into the loop. Increment by 1 for the initial inner while loop check (line 5). Now assume the inner while loop executes $k$ times. There are 2 increments in the inner while loop (line 7 and line 9), so $t$ is then incremented $2k$ times. Next, exit the inner while loop, increment by 1 for the if comparison (line 10). Assume the if condition is false because we do worst case analysis, then increment by 1 for incrementing l (line 13). Increment 1 last time for the following while check (line 15). Add them all up and you get that. – ryan Mar 28 at 16:11
• @kuskmen, this may also help you: cs.stackexchange.com/a/23595/68251 – ryan Mar 28 at 17:45 | 2019-11-19T15:41:32 | {
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https://math.stackexchange.com/questions/2495831/show-that-the-equilibrium-point-0-0-is-asymptotically-stable-and-an-estimate | # Show that the equilibrium point $(0,0)$ is asymptotically stable and an estimate of its basin of attraction
Consider the system \begin{aligned} \dot{x} &=-y-x^3+x^3y^2\\ \dot{y}&=x-y^3+x^2y^3\end{aligned} Show that the equilibrium point $$(0,0)$$ is asymptotically stable and an estimate of its attractiveness basin.
Clearly the point $$(0,0)$$ is a point of equilibrium and also $$D_f(x,y)=\begin{pmatrix}-3x^2+3y^4 & -1+2yx^3\\ 1+2xy^3 &-3y^2+3y^2x^2 \end{pmatrix}$$, so $$D_f(0,0)=\begin{pmatrix}0 & -1\\1 &0 \end{pmatrix}$$ and has eigenvalues $$\pm i$$ So, I can not conclude anything of stability for this point, I need a Liapunov function and I do not know what it is or how to find it, could someone help me please? Thank you very much.
• Always try the simplest one first. For example, quadratic. More precisely, $V(x, y) = x^2+y^2$. Oct 30 '17 at 5:44
In addition to what has been said by @Evgeny and @MrYouMath: the set $$M=\left\{ (x,y)\in\mathbb R^2 :\; x^2+y^2<2 \right\}$$ is a positively invariant set of the considered system since $\forall (x,y)\in M$ $$\dot V=-x^4-y^4+x^2y^2(x^2+y^2)<-x^4-y^4+2x^2y^2=-(x^2-y^2)^2\leq 0;$$ it is also a subset (guaranteed estimation) of the domain of attraction.
• Nice one! This estimate is even better than $\| (x, y) \|_{\infty} < 1$ . Oct 30 '17 at 18:06
• +1: Very nice observation. This might be useful in the future. Oct 31 '17 at 14:03
• How Do I show that $M$ is positively invariant with this fact?
– P.G
Nov 20 '20 at 13:44
• @P.G Just as it is done in the proof of the Lyapunov stability theorem
– AVK
Nov 21 '20 at 7:02
As @Evgeny suggested you could use the Lyapunov function candidate
$$V(x,y)=\dfrac{1}{2}\left[x^2+y^2\right].$$
$V(x,y)$ is clearly positive definite at the origin and radially unbounded (which we would need for assessing global asymptotic stability).
The time derivative of $V(x,y)$ is given by $$\dot{V}=x\dot{x}+y\dot{y}=x(-y-x^3+x^3y^2)+y(x-y^3+x^2y^3)$$ $$\dot{V}=-x^4-y^4+x^2y^2(x^2+y^2)=-(1-y^2)x^4-(1-x^2)y^4.$$
As the lower order terms $-x^4-y^4$ are negative definite we can conclude that the equilibrium point is asymptotically stable in a region around the origin. Using the comment given by @Evgeny we can see that this expression is negative semidefinite if $(x,y)$ lie inside the unit circle $D=\{(x,y)\in \mathbb{R}^2|x^2+y^2<1\}$. This is the basin of attraction (correction due to @Artem).
We cannot say if the origin is globally asymptotically stable because $\dot{V}$ is not negative definite for all regions around the origin. This does not mean that the origin cannot be globally asymptotically stable. It just means we can only show (local) asymptotic stability of the origin with $V(x)$ as our Lyapunov function candidate.
• Actually, you can do better if regroup terms correctly: $\dot{V} = -x^4(1-y^2)-y^4(1-x^2)$. This also gives a rough estimate of basin of attraction since this expression is negative when $\vert x \vert < 1$ and $\vert y \vert < 1$. Oct 30 '17 at 13:18
• @Evgeny: You are right. I felt that I should be able to factor this expression. Oct 30 '17 at 13:19
• Also, speaking about radial unboundedness: it's not a necessary condition really, see recent discussion in comments. Oct 30 '17 at 18:09
• It is not necessary for asymptotic stability but it is necessary for global asymptotic stability (see the comment in the brackets in my answer). I think that is what the example showed. Or do you mean something else? Oct 30 '17 at 18:14
• Even for the global asymptotic stability it is not necessary. You can prove the global asymptotic stability with a Lyapunov function which is not radially unbounded. Moreover, you can transform the image of any Lyapunov function such that it will become bounded but still will prove that something is globally asymptotic stable. Oct 30 '17 at 20:29
This problem can be handled with an optimization procedure, having in mind that generally is a non convex problem. The result depends on the test Lyapunov function used so we will generalize to a quadratic Lyapunov function
$$V(p) = p^{\dagger}\cdot M\cdot p = a x^2+b x y + c y^2,\ \ \ p = (x,y)^{\dagger}$$
and
$$f(p) = \{-y - x^3 + x^3 y^2, x - y^3 + x^2 y^3\}$$ with $$a>0,c>0, a b-b^2 > 0$$ to assure positivity on $$M$$. We will assure a set involving the origin $$Q_{\dot V}$$ such that $$\dot V(Q_{\dot V}) < 0$$. The optimization process will be used to guarantee a maximal $$Q_{\dot V}$$.
After determination of $$\dot V = 2 p^{\dagger}\cdot M\cdot f(p)$$ we follow with a change of variables
$$\cases{ x = r\cos\theta\\ y = r\sin\theta }$$
so $$\dot V = \dot V(a,b,c,r,\theta)$$. The next step is to make a sweep on $$\theta$$ calculating
$$S(a,b,c, r)=\{\dot V(a,b,c,r,k\Delta\theta\},\ \ k = 0,\cdots, \frac{2\pi}{\Delta\theta}$$
and then the optimization formulation follows as
$$\max_{a,b,c,r}r\ \ \ \ \text{s. t.}\ \ \ \ a > 0, c> 0, a c -b^2 > 0, \max S(a,b,c,r) \le -\gamma$$
with $$\gamma > 0$$ a margin control number.
Follows a MATHEMATICA script which implements this procedure in the present case.
f = {-y - x^3 + x^3 y^2, x - y^3 + x^2 y^3};
V = a x^2 + 2 b x y + c y^2;
dV = Grad[V, {x, y}].f /. {x -> r Cos[t], y -> r Sin[t]};
rest = Max[Table[dV, {t, -Pi, Pi, Pi/30}]] < -0.1;
rests = Join[{rest}, {r > 0, a > 0, c > 0, a c - b^2 > 0}];
sols = NMinimize[Join[{-r}, rests], {a, b, c, r}, Method -> "DifferentialEvolution"]
rest /. sols[[2]]
dV0 = Grad[V, p].f /. sols[[2]]
V0 = V /. sols[[2]]
r0 = 2;
rmax = r /. sols[[2]];
gr0 = StreamPlot[f, {x, -r0, r0}, {y, -r0, r0}];
gr1a = ContourPlot[dV0, {x, -r0, r0}, {y, -r0, r0}, ContourShading -> None, Contours -> 80];
gr1b = ContourPlot[dV0 == 0, {x, -r0, r0}, {y, -r0, r0}, ContourStyle -> Blue];
gr2 = ContourPlot[x^2 + y^2 == rmax^2, {x, -r0, r0}, {y, -r0, r0}, ContourStyle -> {Red, Dashed}];
Show[gr0, gr1a, gr1b, gr2]
Follows a plot showing in black the level sets $$Q_{\dot V}$$ an in blue the trace of $$\dot V = 0$$. In dashed red is shown the largest circular set $$\delta = 1.42486$$ defining the maximum attraction basin for the given test Lyapunov function's family. | 2021-11-30T19:30:55 | {
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https://math.stackexchange.com/questions/3147904/if-ah-bh-implies-ha-hb-for-a-subgroup-h-having-finite-index-then-gh-hg-f | # If $aH=bH \implies Ha=Hb$ for a subgroup H having *finite index*, then $gH=Hg$ for all $g \in G$?
Problem 2.5.9 of Herstein's Topics in Algebra asks us to prove that if $$H$$ is a subgroup of a group $$G$$ such that $$Ha \not = Hb \implies aH \not = bH$$, then $$gHg^{-1} \subset H$$ for all $$g \in G$$, which is equivalent to $$gH \subset Hg.$$ Suppose I have proved this (this result is of course true). I want to prove more, namely that, under the same premise, $$gH = Hg$$ for all $$g \in G$$. I think I can prove it if I further assume that the index of $$H$$ in $$G$$ is finite. Is this true? And if so, is this additional hypothesis necessary as well as sufficient? I.e. can you give an example of a subgroup $$H$$ of infinite index such that $$aH \subset Ha$$ but not vice versa? Thanks in advance. [DISCLAIMER: I am not yet familiar with normal subgroups]
Here's my attempt at a proof:
Suppose that $$|G:H|=n \in \mathbb{N}$$ and that $$aH \subset Ha$$ for all $$a \in G$$. Now, suppose that there is an element $$x$$ such that $$x \in Ha$$ but $$x \not \in aH$$. Then $$x$$ must be contained in a different left coset, say $$bH$$, because the left cosets form a partition of $$G$$. Then, by our hypothesis, $$x \in Hb$$, which implies $$Hb = Ha$$, because the right cosets form a partition of $$G$$ as well. So far we have shown that $$aH, bH \subset Ha=Hb$$. But now we can prove that there are more left cosets than right ones! In fact, the remaining $$n-1$$ right cosets distinct from $$Ha$$ must contain at least one left coset, by our hypothesis, but $$Ha$$ contains two left cosets. So there are at least $$n+1$$ left cosets, a contradiction. Therefore $$aH=Ha. \square$$
• Compare with this duplicate and its answers. – Dietrich Burde Mar 14 at 12:00
• @DietrichBurde The answers in your linked question only prove $gHg^{-1} \subset H$ and not equality. So I don't think they address the OP’s problem. – Claudius Mar 14 at 12:15
• @Claudius This is not true. They prove equality, e.g., see Deven's answer. – Dietrich Burde Mar 14 at 12:18
• @DietrichBurde Yes, Deven claims to have shown $gHg^{-1} =H$. But what he really did was to prove $gHg^{-1} \subset H$. At least, I don't see how he proved the other inclusion. – Claudius Mar 14 at 12:20
Your proof seems correct to me.
In fact, you can remove the finite index hypothesis. If $$gHg^{-1} \subset H$$ for all $$g\in G$$, then for each $$g\in G$$ you also have $$H = g^{-1}(gHg^{-1})g \subset g^{-1}Hg = g^{-1}H(g^{-1})^{-1} \subset H,$$ so we must have equality throughout, i. e. $$H = g^{-1}Hg$$ (for each $$g\in G$$).
More generally, for any subgroup $$H$$ of $$G$$ the following holds: the set $$N:= \{ g\in G \mid gHg^{-1}\subset H\}$$ is a subgroup of $$G$$ if and only if $$gHg^{-1} = H$$ for all $$g\in N$$. (In that case $$N$$ is the normalizer of $$H$$ in $$G$$.)
• That's brilliant, thank you. Now I feel stupid for not having come up with that immediately! But why would Herstein keep this result from the reader and only ask to prove one inclusion? And what about this thread math.stackexchange.com/questions/217601/… in which there seems to be a counterexample? – The Footprint Mar 14 at 12:36
• In the answer of Brian M. Scott the set $\{g\in G\mid gHg^{-1} \subset H\}$ is not stable under inversion, hence not a group. It is only closed under composition. – Claudius Mar 14 at 12:40
You don't need any additional hypotheses. Multiplying both sides by $$g^{-1}$$ on the left and $$g$$ on the right yields
$$gHg^{-1} \subset H \implies H \subset g^{-1}Hg$$
Since this is true for all $$g \in G$$, we can substitute $$g$$ for $$g^{-1}$$, concluding
$$gHg^{-1} \subset H \subset gHg^{-1}$$
$$H = gHg^{-1}$$
This is equivalent to
$$gH = Hg$$ | 2019-08-21T15:34:23 | {
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https://brilliant.org/discussions/thread/a-calculus-problem/ | ×
# A Calculus Problem!
The Problem:
Find $$I=\displaystyle\int_{0}^{2} (x^2+1)\,d\left \lfloor x\right \rfloor$$
My Doubt:
There seem two probable approaches to the question but both of them yield different results. Both seem to be mathematically correct and hence the confusion.
Approach 1:
We know that,
$$g(x). f'(x) + f(x). g'(x) = \large \frac{d\left(f(x). g(x)\right)}{dx}$$
Integrating the above expression we get:
$$\displaystyle\int_{a}^{b} g(x)\,df(x) + \displaystyle\int_{a}^{b} f(x)\,dg(x) = f(b).g(b)-f(a).g(a)$$
Using the above property to solve the integration we get the answer as $$\boxed{7}$$
Approach 2:
We can write
$$I=\displaystyle\int_{0}^{1} (x^2+1)\,d\left \lfloor x\right \rfloor +\displaystyle\int_{1}^{2}(x^2+1)\,d\left \lfloor x\right \rfloor$$
But $$\left \lfloor x\right \rfloor$$ assumes constant values of $$\left \lfloor x\right \rfloor=0$$ and $$\left \lfloor x\right \rfloor=1$$ in the respective intervals. And hence in both cases $$d\left \lfloor x\right \rfloor=0$$ and therefore $$I=0$$
I genuinely can't understand what's the correct method. I will really be grateful if someone can explain it to me.
Thank You!
Note by Miraj Shah
1 year, 9 months ago
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In Approach 2, the floor function fails to be constant on the two subintervals; consider the two endpoints! The first integral takes the value of 1 due to the step at 1, and the second one takes a value of 5. With this correction, both approaches work and yield the same result.
- 1 year, 9 months ago
First off, integration by parts doesn't work as the step function isn't differentiable.
Secondly, the integral is 2. I'll provide a proof of this later.
Btw, @Aditya Kumar The integral makes perfect sense and is defined.
- 1 year, 9 months ago
It would be really nice of you to put the proof here if possible! But I would also like to know then what's wrong with the Approach 1&2 illustrated above? It's a bit baffling
- 1 year, 9 months ago
Wait, the integral is 7.
We shall use the definition of integration as given in this book, with $$\alpha(x)=\lfloor x \rfloor$$ and $$f(x)=1+x^2$$.
Consider a partition $$P$$ of $$[0,2]$$ such that $$x_{r-1}\leq 1\leq x_{r}$$ where strict inequality holds in atleast one of the two inequalities. Also, by the definition of a partition, we have $$x_{n-1}<2=x_n$$. Then,
$U(P, f, \alpha)=f(x_r)+f(x_n) \\ L(P, f, \alpha)=f(x_{r-1})+f(x_{n-1}) \\\implies \inf U(P, f, \alpha)=\sup L(P, f, \alpha)=f(1)+f(2) \\\implies \int_0^2 f \, d\alpha = 7$
Note: I've only used the fact that $$f$$ is monotonic.
Plus, in the book mentioned above, in the exercises, a more generalised version of IBP, valid for even discontinuos functions is given. You've made use of this version in approach 1 (luckily, the step function is monotonic).
If the integrals are calculated using the definition in approach 2, you get the right answer, 7.
- 1 year, 9 months ago
- 1 year, 9 months ago
Oh yes! Really helpful @Deeparaj Bhat ! Sorry for replying a bit late. Busy studying for the big one which is to be held on 22nd May! Actually the above problem was from a test paper it self! Had a doubt in the question, so thought of taking help from the brilliant Brilliant Community as I was pretty sure that I will be able to get at least some insight!
Thank you!
- 1 year, 9 months ago
You're writing any other exams? (Except JEE ADVANCED, which is on 22nd)
- 1 year, 9 months ago
Actually yes. BITS and all! You have to give JEE this year?
- 1 year, 9 months ago
Yup. That, BITSAT and CMI in my case.
- 1 year, 9 months ago | 2018-02-21T18:57:59 | {
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https://mathhelpboards.com/threads/eulers-theorem-for-calculating-mod.6407/ | # Number TheoryEuler's theorem for calculating mod
#### ATroelstein
##### New member
I am trying to use Euler's theorem for calculating the following:
$$8^{7} mod 187$$
I have determined that:
$$\phi(187) = \phi(11*17) = 160$$
and
$$8^{7} = 8^{4} * 8^{2} * 8^{1}$$
but am unfortunately confused about how to now proceed beyond this point. Thank you.
#### caffeinemachine
##### Well-known member
MHB Math Scholar
I am trying to use Euler's theorem for calculating the following:
$$8^{7} mod 187$$
I have determined that:
$$\phi(187) = \phi(11*17) = 160$$
and
$$8^{7} = 8^{4} * 8^{2} * 8^{1}$$
but am unfortunately confused about how to now proceed beyond this point. Thank you.
I don't think here Euler's theorem can be fruitfully used. But you can use Fermat's little in conjunction with Chinese Remainder to calculate this nicely.
Note that $8^7=2^{21}=x$ (say). Also, $187=11\times 17$. Now by Fermat we have $2^{10}\equiv 1 \pmod{11}$ and $2^{16}\equiv 1\pmod{17}$. These give $x\equiv 2\pmod{11}$ and $x\equiv 32\equiv -2\pmod{17}$. From here it's routine to use the Chinese remainder and get the remainder $x$ leaves mod $187$.
#### Deveno
##### Well-known member
MHB Math Scholar
As I have stated here many times before, I am pathetically lazy, so I look for the easy way out.
87 = (64)(8)(64)(8)(8) (mod 187)
Next, I calculate 64*8 the old-fashioned way:
(6*8*10) + (4*8) = 480 + 32 = 512.
I'm going to guess that 2 multiples of 187 are all we're going to pack into 512. So:
512 = 512 - 374 = 138 (mod 187).
So 87 = (138)(138)(8) (mod 187).
Did I mention I'm afraid of big numbers? Well, I am. So next I'm going to apply a trick that often works well in modular arithmetic:
Since 138 = -49 (mod 187),
87 = (138)(138)(8) = (-49)(-49)(8) = (49)(49)(8) (mod 187).
Here, I have an unpleasant choice to make: do I tackle 49*49 next, or 49*8?
The small numbers win!
Now 49*8 = (4*8*10) + (9*8) = 320 + 72 = 392. Well, that's pretty close to 374, which makes me happy. So:
87 = (49)(49)(8) = (49)(392) = (49)(18) (mod 187).
I've avoided "hard calculations" as much as possible, but I have no choice but to evaluate 49*18 now:
49*18 = (40+9)(10+8) = 400 + 320 + 90 + 72 = 882.
So I want to know what 882 (mod 187) is. Knowing that 180*5 = 900, I don't think I can get 5 multiples of 187 in 882, so 4 is my best guess:
187*4 = 400 + 320 + 28 = 748. So:
87 = 882 = 882 - 0 = 882 - 748 = 134 (mod 187)
Does this jibe with caffeinemachine's answer?
134 = 132 + 2 = 0 + 2 = 2 = (mod 11). Check.
134 = 119 + 15 = 0 + 15 = 15 = -2 (mod 17). Check.
#### Deveno
##### Well-known member
MHB Math Scholar
I'd like to point out here that caffeinemachine's solution is not so easy to carry out as it might seem.
Given that:
x = 2 (mod 11)
x = -2 (mod 17)
we seek integers k,m such that:
11k + 2 = 17m - 2
or, equivalently:
17m - 11k = 4
If we knew some integers r,s with:
17r - 11s = 1, we could take m = 4r, k = 4s.
Fortunately, the euclidean division algorithm gives us a way to find these two integers r and s:
17 = 11*1 + 6
11 = 6*1 + 5
6 = 5*1 + 1
Therefore:
1 = 6 - 5
5 = 11 - 6
6 = 17 - 11, so:
1 = 6 - 5 = 6 - (11 - 6) = 2*6 - 11 = 2*(17 - 11) - 11 = 2*17 - 3*11, so we have:
r = 2, s = 3, and in turn:
m = 8, k = 12. Hence:
11*12 + 2 = 17*8 - 2, evaluating either side gives us:
134, which is the desired power 87 (mod 187).
#### johng
##### Well-known member
MHB Math Helper
Hi,
I tend to agree with caffeinemachine that the easiest computation is via the Chinese remainder theorem.
Given m and n relatively prime, the unique x (mod mn) that satisfies x = a (mod m) and x = b (mod n) is of the form a + km for some k with 0 <= k < n.
(All of the given values are different for the range of k's and so one must be b (mod n)).
So I want -2 + 17k = 2 (mod 11)
6k = 4 (mod 11)
k = 8 (mod 11) -- the multiplicative inverse of 6 is 2 mod 11.
So x = -2 +8*17 = 136 - 2 = 134.
#### Deveno
##### Well-known member
MHB Math Scholar
Hi,
I tend to agree with caffeinemachine that the easiest computation is via the Chinese remainder theorem.
Given m and n relatively prime, the unique x (mod mn) that satisfies x = a (mod m) and x = b (mod n) is of the form a + km for some k with 0 <= k < n.
(All of the given values are different for the range of k's and so one must be b (mod n)).
So I want -2 + 17k = 2 (mod 11)
6k = 4 (mod 11)
k = 8 (mod 11) -- the multiplicative inverse of 6 is 2 mod 11.
So x = -2 +8*17 = 136 - 2 = 134.
Yes, but....
The ease of this is "hidden" in the step:
6k = 4 --> k = 8.
In this particular case, one can guess rather easily that [6-1]11 = 2 by trial-and-error. In general, finding a (multiplicative) inverse mod n (if it exists, which it assuredly will if n is prime, and the element inverted is not a multiple of n = p) involves exactly the same steps I outlined above. One often does not have a discrete log table lying around to read off which power of a generator a particular element is, and one HAS to resort to the division algorithm to find the inverse (finding primitive elements is not always a trivial task, although it IS easier for small numbers).
Don't misunderstand me, I find the application of the CRT a beautiful and elegant solution. At its heart, the CRT states that:
If (m,n) = 1, then $\Bbb Z_{mn} \cong \Bbb Z_m \times \Bbb Z_n$
However, while the ring-homomorphism one way is trivial to display, the inverse homomorphism is nontrivial (and amounts to actually solving the congruence pair). There is a slight difference between knowing a solution exists and exhibiting that solution.
Your calculations and mine are, of course, the same, as can be seen by taking the equation:
11*12 + 2 = 17*8 - 2 (mod 11).
I also do not know if the original poster has seen a PROOF of the CRT, with an understanding of what it means for "compound" moduli. In all fairness (both to you and caffeinemachine) I admit sometimes an understanding of more advanced methods makes problems like these easier to deal with. Do I run the risk of under-estimating the original poster's sophistication? Of course. I am the first to admit the laziest proof is not always the shortest.
If the exponent has been larger, my way would undeniably be more cumbersome. I certainly mean no disrespect to either of your fine answers. Knowing we can reduce an exponent b in ab (mod n), by taking b (mod φ(n)), is something well worth remembering, if one has been exposed to Euler's theorem (which we can assume the OP has, by the thread title), and I note in passing that Fermat's "Little Theorem" (which caffeinemachine DOES invoke) is just a special case of Euler's Theorem.
One hopes that the original poster gains more from our discussion of solution techniques here than just the answer to his problem.
#### johng
##### Well-known member
MHB Math Helper
Yes, but....
The ease of this is "hidden" in the step:
6k = 4 --> k = 8.
In this particular case, one can guess rather easily that [6-1]11 = 2 by trial-and-error. In general, finding a (multiplicative) inverse mod n (if it exists, which it assuredly will if n is prime, and the element inverted is not a multiple of n = p) involves exactly the same steps I outlined above. One often does not have a discrete log table lying around to read off which power of a generator a particular element is, and one HAS to resort to the division algorithm to find the inverse (finding primitive elements is not always a trivial task, although it IS easier for small numbers).
Given the parameters of the CRT, namely m, n are relatively prime, I'm trying to solve a + mk = b for k. So of course m does have an inverse mod n. Granted, the finding of such inverse requires a little work, but it's not exorbitant. A slight extension of the Euclidean algorithm for gcd's produces x and y with mx + ny = 1, and so x is the inverse of m. I find the extension a bit cumbersome to do by hand, but it's a Programming I exercise to encode (maybe toward the end of the course). Even for large integers, the gcd algorithm is "usually" quite fast; as you probably know the worst case is for consecutive Fibonacci numbers. But even here, it's logarithmic.
#### Deveno
##### Well-known member
MHB Math Scholar
Yes, that is what I was getting at. If the exponent is large-ish (say > 1000 for example), using a gcd algorithm is going to be MUCH faster than direct computation.
But with an exponent of only 7, direct computation can be done straight-forwardly, without involving more abstract results. Is this preferable? It depends on your point of view, I suppose. One thing that CAN be said about your solution/caffeinemachine's solution is that it is more widely applicable to a greater range of problems | 2021-11-29T00:04:56 | {
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https://math.stackexchange.com/questions/4459036/mathematical-inclusion-and-exclusion-of-elements-from-a-given-set-a | # Mathematical "inclusion" and "exclusion" of elements from a given set $A$?
If I have a set A, comprising of numbers from 1 to 10: $$A = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$$. Let's say I want to make another set by "including" all even numbers:
$$\{2, 4, 6, 8, 10\}$$
Or I wanted to make a different set by "excluding" all odd numbers:
$$\{2, 4, 6, 8, 10\}$$
These sets are of course the same. So is it correct to say that inclusion/exclusion are synonymous when it comes to set theory, as they're just different ways of building a set?
This might sound trivial, but I have a reason for asking: I want to understand if inclusion and exclusion are "commutative" properties, i.e. it doesn't matter in which order you apply them.
For example, let's say I make an operation to "filter" my set, by including all even numbers as we did before, producing set B
$$B = \{2, 4, 6, 8, 10\}$$
And then a separate operation to "exclude" any numbers less than 6 from set B, resulting in set C:
$$C = \{6, 8, 10\}$$
What if I started with A and applied the operations the other way around? First remove all numbers less than 6:
$$B = \{6, 7, 8, 9, 10\}$$
Then filter B to "include" only even numbers:
$$C = \{6, 8, 10\}$$
It seems intuitively to me that the result will always be the same no matter which order you apply the operation. Is this true for all cases no matter the set, however? Is there a way to prove that applying "filters" to a set (I'm not sure of the proper term) will always be commutative?
So in summary, there are two questions here:
1. Are the notions of "inclusion" and "exclusion" really synonymous from the point of view of applying an operation to a set to produce a subset?
2. Will applying these operations to produce a subset of a set always be commutative, i.e. produce the same result?
• It looks lile you are taking repeated intersections of sets. Intersection is both associative and commutative. May 26 at 10:25
• @CrackedBauxite I've had a think about this. Is it intersection, or is it a set difference / relative complement? If I've got the set $\{1, 2, 3, 4, 5\}$ and I exclude all odd numbers, I've got a set of odd numbers $\{1, 3, 5\}$. It's my original set $\{1, 2, 3, 4, 5\} - \{1, 3, 5\} = \{2, 4\}$
– Lou
May 26 at 16:45
• But if I were to do "include all even numbers" from the set $\{1, 2, 3, 4, 5\}$, then I have a set of even numbers $\{2, 4\}$. $\{1, 2, 3, 4, 5\} \cap \{2, 4\} = {2, 4}$ I think - but please correct me if my logic is wrong
– Lou
May 26 at 16:46
• So I think "include" as an operation represents an intersection, but "exclude" represents a set difference?
– Lou
May 26 at 16:48
• Set differences are also intersections. $A\setminus B = A\cap B^c$, where $B^c$ is the complement of $B$. May 27 at 8:38
@KevinS offers an excellent answer from a logic point of view. Here's another that relies on the idea of a filter. That's a concept useful in programming (particularly in lisp). You pass the items in your set through a filter that lets some through and blocks others.
In this sense "inclusion" and "exclusion" are really different ways to describe the same result. You can specify what you keep or what you reject. "Keep (just) the odds" is the same as "reject (only) the evens".
If you have two filters each of which is described independently of the other and refers only to properties of the things you are filtering then you can filter in either order.
The independence matters. If you think that the wine must match the main course then your options will depend on whether you see the wine list or the menu first.
• You've hit upon my actual use case - I'm doing filtering in Python and trying to understand if "inclusion" and "exclusion" for my use case are order-sensitive, or commutative operations. You've all helped me understand that they are commutative and therefore order should not matter.
– Lou
May 26 at 10:52
This is perhaps easier to see in the math logic. When you apply conditions to a set, you get subsets that satisfy those conditions (possibly the emptyset if the conditions aren't met). In your example, define: $$A := \{n\in\mathbb{N}\text{ }|\text{ }1\leq n\leq 10\}.$$ Then $$B$$ and $$C$$ are had by adding conditionals in the set definition: $$B := \{n\in A\text{ }|\text{ }\exists k\in\mathbb{N}: n=2*k\}$$ and $$C:= \{n\in B\text{ }|\text{ }n\geq 6\}.$$ $$\implies C = \bigg\{n\in \mathbb{N}\text{ }\bigg|\text{ }(1\leq n\leq 10)\wedge(\exists k\in\mathbb{N}: (n=2*k))\wedge(n\geq 6)\bigg\}.$$ From this viewpoint, the reductions made were simply conjuctions (a logical operation). Conjunction is certainly commutative.
Also, the term "exclusion" is synonymous with set difference: $$X-Y:= X\cap Y^c,$$ whereas "inclusion" usually refers to an injective map: $$\iota: S\hookrightarrow X$$ which can be used as an identifier of a subset. I wouldn't say the two notions are synonymous.
• I'm a beginner in set theory, would you mind clarifying what you mean about an "injective map"? I understand the term set difference.
– Lou
May 26 at 14:21
• It seems as though the terms used might have multiple meanings as they apply to math and programming. An injection is a 1-1 function. The math notion I gave generalizes to categories (using equivalence classes of monomorphisms to identify sub-objects). This is not inherently programmable and is best used for theory. May 26 at 21:34
To your first question, consider using the set-builder notation to build a subset. That is, define $$B\subseteq A$$ such that $$B = \{a\in A : \Phi(a)\}$$, where $$\Phi$$ is the logical formula which dictates what how we pick the values of $$A$$.
Now say that $$A$$ is some collection of numbers, and $$\Phi$$ is the rule "include all even numbers". Then some other rule $$\Psi$$ which says "do not include all not even numbers" is completely logically equivalent to $$\Phi$$.
And this generalises nicely by considering that, if we impose a condition like "even number", any element of a set will satisify that condition, or not satisfy that condition. We cannot have an element which does neither.
The second question follows as others have pointed out, by writing the conditions as operations which are commutative. | 2022-06-25T17:57:08 | {
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https://math.stackexchange.com/questions/1962884/chance-of-each-player-getting-an-ace-when-the-dealer-has-12-cards | # Chance of each player getting an ace when the dealer has 12 cards?
4 players are playing a card game wherein each of them gets 10 cards, and the dealer gets 12 cards. What is the probability that each player gets an ace?
I want the use the $p = \dfrac{n_A}{N}$ method, where $n_A$ equals the favourable outcomes and $N$ equals all possible outcomes.
Starting with $N$, I figured we could consider the dealer to be a fifth player, and considering we don't care about the order of the players we'd get:
$$N = \dfrac{52!}{(10!)^4\times12!} \times \dfrac{1}{5!}$$
Now for $n_A$, the aces can be divided among the players in $4!$ ways, and each of the players would still get 9 other cards from a total of 48, with the dealer getting the remaining twelve, thus giving us: $$n_A = 4! \times \dfrac{48!}{(9!)^412!} \times \dfrac{1}{5!}$$
But if we calculate $p$ this way we get a probability of $\approx 3\%$, which is just intuitively orders of magnitude too large to be correct, so I am sure I made a mistake somewhere. Can anyone help me spot it and then explain what I did wrong?
• Why the "$\pm$" in $\pm3\%$? – Barry Cipra Oct 10 '16 at 23:09
• @BarryCipra I meant "approximately equal to" but I couldn't find the sign for that on my keyboard, so I thought ± would be the next best thing – YakSal Tafri Oct 11 '16 at 7:01
• The way to get $\approx$ is to type $\approx$. Changed it. – BruceET Oct 11 '16 at 7:54
I wouldn't consider the dealer as a fifth player but instead, let me guide you through another way to get the answer using combinatorics.
We should start by counting $N$ as ${52 \choose 10}{42 \choose 10}{32 \choose 10}{22 \choose 10}$ for the number of ways the dealer can give ten cards from 52 to each of the four players.
$N = {52 \choose 10}{42 \choose 10}{32 \choose 10}{22 \choose 10} \approx 971089585681469963688868551062400$
Now for $n_A$ we will consider all four aces given in $4!$ and count for the number of ways nine cards from the remaining 48 can be given to each of the four players as ${48 \choose 9}{39 \choose 9}{30 \choose 9}{21 \choose 9}$.
$n_A = 4!{48 \choose 9}{39 \choose 9}{30 \choose 9}{21 \choose 9} \approx 35869963456698493441273194240000$
Hence
$p = {n_A \over N} = {4!{48 \choose 9}{39 \choose 9}{30 \choose 9}{21 \choose 9}\over {52 \choose 10}{42 \choose 10}{32 \choose 10}{22 \choose 10}} ={400 \over 10829} \approx 0.036$
Comment: Because 0.036 doesn't seem to match the answer you anticipated, and because there was at least one false start towards a combinatorial answer, I decided to simulate the 'deal' a million times in R statistical software, and see what proportion of deals gave one ace to each player. I got $0.037 \pm 0.0004.$ So I think @AlfredoLozano's method is correct (but note that $400/10829 = 0.03693785 \approx 0.037).$
My deck has 1's for Aces and 0's for all other cards for simplicity counting results, but the sample function treats each 'card' as distinct. The m-vector each.1 is 'logical' with elements TRUE and FALSE; its mean is the proportion of its TRUEs.
m = 10^6; each.1 = logical(m)
deck=c(1,1,1,1, rep(0,48))
for (i in 1:m) {
d = sample(deck, 40)
each.1[i] = (sum(d[1:10]==1)&sum(d[11:20]==1)&sum(d[21:30]==1)&sum(d[31:40]==1))
}
mean(each.1)
## 0.037124
Here is another way to get the answer. Imagine the dealer deals the first $10$ cards to player A, the next $10$ to player B, the next $10$ to player C, the next $10$ to player D, and keeps the rest for himself. Clearly the cards can be shuffled in any of $52!$ ways. The question becomes, in how many ways can the dealer "stack" the deck so that each player gets an Ace?
First, remove the Aces from the deck and shuffle the remaining $48$ cards, which can be done in $48!$ ways. Then shuffle the $4$ Aces, in $4!$ ways. Then insert the first Ace so it's one of the first $10$ cards in the deck, the second Ace so it's one of the next $10$ cards, and so forth. All this can be done in $48!\times4!\times10^4$ ways. So the the probability of each player getting an Ace is
$${48!\times4!\times10^4\over52!}={24\times10^4\over52\times51\times50\times49}=24\left(10\over52\right)\left(10\over51\right)\left(10\over50\right)\left(10\over49\right)\approx25\left(1\over5\right)^4=0.04\%$$
The exact answer, as a reduced fraction, is ${400\over10829}=0.0369378\ldots$ | 2019-08-23T11:23:46 | {
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http://mathhelpforum.com/discrete-math/133340-partial-orders-question-print.html | # Partial orders - Question
• Mar 11th 2010, 12:38 PM
gate13
Partial orders - Question
Good day to all,
We have just started relations and I came across the following question in my textbook:
Let X = {1, 2}. List all the partial orders that can be defined on X.
My solution:
I began by computing X x X (Cartesian product) which gave me:
X x X ={(1,1), (1,2), (2,1), (2,2)}
Since we are looking for partial orders, the relation has to be simultaneously reflexive, antisymmetric and transitive.
If the relation R on X is reflexive then it must contain (1,1) and (2,2)
The ordered pairs (1,2) and (2,1) cannot belong to R for if they did and R is also antisymmetric then that would imply 1=2, which is false.
Therefore I concluded that the list of partial orders is the set:
{(1,1), (2,2)}
I was wondering if my logic is flawed and if so what errors have I committed?
Finally, is it possible in this problem to determine the number of partial orders (cardinality)?
Any advice would be greatly appreciated.
Kindest regards
• Mar 11th 2010, 12:44 PM
Plato
Would $\{(1,1),(1,2),(2,2)\}$ also work?
Why or why not?
• Mar 11th 2010, 12:58 PM
gate13
Thank you Plato for your quick response.
I am not sure. The definition of antisymmetric states:
for every a,b in X ((a,b) belongs to R and (b,a) belongs to R implies a=b)
If (1,2) belongs to R and (2,1) does not belong to R (based on the set you listed) then the hypothesis of the implication is false which means that the implication is true. If this is a valid reasoning could we not say the same for the set: {(1,1), (2,1), (2,2)}
Slightly confused!
• Mar 11th 2010, 01:06 PM
Plato
But $(2,1)\notin\{(1,1),(1,2),(2,2)\}$. Is it?
Antisymmetric says that in both $(a,b)\in R~\&~(b,a)\in R$ then it must be true that $a=b$.
• Mar 11th 2010, 01:35 PM
gate13
There is something that I am obviously not understanding. (2,1) does not belong to the set you listed.
I believe the set http://www.mathhelpforum.com/math-he...26e4344e-1.gif is antisymmetric, since for it not to be antisymmetric we would have to have: http://www.mathhelpforum.com/math-he...0853db63-1.gif and a different than b. Therefore the set you provided is a partial order as well (reflexive, transitive and antisymmetric).
I apologize as I realize this may be obvious to many.
• Mar 11th 2010, 01:38 PM
Plato
Quote:
Originally Posted by gate13
I believe the set http://www.mathhelpforum.com/math-he...26e4344e-1.gif is antisymmetric, since for it not to be antisymmetric we would have to have: http://www.mathhelpforum.com/math-he...0853db63-1.gif and a different than b. Therefore the set you provided is a partial order as well (reflexive, transitive and antisymmetric).
That is correct. And there is one more p.o. on the set.
What is it?
• Mar 11th 2010, 01:43 PM
gate13
I believe the other partial order would be :
{(1,1), (2,1), (2,2)} for the same reasons listed previously.
• Mar 11th 2010, 02:01 PM
gate13
I just wanted to thank you Plato for your time and explanations. Many thanks. | 2016-09-28T15:51:27 | {
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https://mathhelpboards.com/threads/graduate-potw.3659/ | ### Welcome to our community
#### dwsmith
##### Well-known member
So on the recent graduate problem of the week, I saw that $\int_0^{\infty}\frac{\sin x}{x}dx = \frac{\pi}{2}$, but so does, $\int_0^{\infty}\frac{\sin^2 x}{x^2}dx = \frac{\pi}{2}$.
How can they both be the same?
#### Chris L T521
##### Well-known member
Staff member
So on the recent graduate problem of the week, I saw that $\int_0^{\infty}\frac{\sin x}{x}dx = \frac{\pi}{2}$, but so does, $\int_0^{\infty}\frac{\sin^2 x}{x^2}dx = \frac{\pi}{2}$.
How can they both be the same?
Let us use integration by parts to compute $\displaystyle\int_0^{\infty}\frac{\sin^2 x}{x^2}\,dx$. At the end, we will need to use the fact that $\displaystyle\int_0^{\infty}\frac{\sin x}{x}\,dx=\frac{\pi}{2}$
Let $u=\sin^2x$ and $\,dv=\dfrac{\,dx}{x^2}$. Then $\,du=2\sin x\cos x\,dx=\sin(2x)\,dx$ and $v=-\dfrac{1}{x}$. Therefore,
$\int_0^{\infty}\frac{\sin^2 x}{x^2}\,dx = \left[-\frac{\sin^2 x}{x}\right]_0^{\infty}+\int_0^{\infty}\frac{\sin(2x)}{x}\,dx=\int_0^{\infty}\frac{\sin(2x)}{x}\,dx.$
(We note that $|\sin x|\leq 1\implies |\sin^2 x|\leq 1$ and thus $\displaystyle\lim_{x\to\infty} \frac{\sin^2 x}{x}\sim \lim_{x\to\infty} \frac{1}{x}=0$; We also note that $\displaystyle\lim_{x\to 0}\frac{\sin^2 x}{x}=\lim_{x\to 0}\frac{\sin x}{x}\cdot\lim_{x\to 0}\sin x=0$. Hence, that's why the $\displaystyle\left[-\frac{\sin^2 x}{x}\right]_0^{\infty}$ term goes to zero.)
Now let $t=2x\implies\,dt=2\,dx$. Therefore,
$\int_0^{\infty}\frac{\sin(2x)}{x}\,dx\xrightarrow{t=2x}{} \int_0^{\infty}\frac{\sin t}{t/2}\frac{\,dt}{2}=\int_0^{\infty}\frac{\sin t}{t}=\frac{\pi}{2}.$
And thus, we also have that $\displaystyle\int_0^{\infty}\frac{\sin^2 x}{x^2}\,dx =\frac{\pi}{2}$.
I hope this makes sense!
#### ZaidAlyafey
##### Well-known member
MHB Math Helper
$$F(a)=\int^{\infty}_0\frac{\sin^2(ax)}{x^2}$$
Differentiate w.r.t a :
$$F'(a)=\int^{\infty}_0 \frac{\sin(2ax)}{x}$$
Let 2ax=t
$$F'(a)=\int^{\infty}_0 \frac{\sin(t)}{t}=\frac{\pi}{2}$$
$$F(a)=\frac{\pi}{2}a+C$$
Putting a =0 we get C = 0 hence
$$\int^{\infty}_0\frac{\sin^2(ax)}{x^2}=\frac{\pi \cdot a}{2}$$
So for a =1 we get our result :
$$\int^{\infty}_0\frac{\sin^2(x)}{x^2}=\frac{\pi}{2}$$
#### ZaidAlyafey
##### Well-known member
MHB Math Helper
If your question is why such thing happen , then I don't know , to me it is pretty strange !
If you see the graph of both functions , then you have no indications ...
#### dwsmith
##### Well-known member
I just thought it was strange. When I took Theory of Complex Variables, I had the $\int_0^{\infty}\frac{\sin^2x}{x^2}dx = \frac{\pi}{2}$ exercise so I was surprised to see that $\frac{\sin x}{x}$ lead to the same conclusion.
#### ZaidAlyafey
##### Well-known member
MHB Math Helper
In complex analysis $$\int^{\infty}_0 \dfrac{1-\cos(x)}{x^2}$$ and $$\int^{\infty}_0 \frac{\sin(x)}{x}$$ are conventional exercises to solve by contour integration .... | 2020-10-01T12:46:29 | {
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http://math.arizona.edu/~restrepo/475A/Notes/sourcea-/node21.html | Next: Contraction Mapping Up: SOLUTION OF NONLINEAR EQUATIONS Previous: Secant Method
## Fixed Point Iteration
def: Fixed Point is the value such that .
Fixed Point problems and root-finding problems are equivalent: let . Hence, if a function has a fixed point then has a root.
Three Problems:
1. What functions have a fixed point?
2. How do we determine the fixed point?
3. Is the fixed point unique?
Example has a fixed point at each ....just plot.
Example has 2 fixed points in ....plot, again.
Theorem (Existence and Uniqueness): If and then has a fixed point in .
Suppose, in addition, that exists on and that a positive constant exists with
then the fixed point in is unique.
Proof:
If or , then existence of fixed point is clear. Suppose not, then it must be true that and . Define . Then is continuous on and
The Intermediate Value Theorem implies that there exist for which . Thus, is fixed point of .
holds and that and are both fixed points in with . By the Mean Value Theorem a number exists between and such that
This contradiction comes from and the fixed point is unique.
Fixed-Point Iteration
Pick a and generate a sequence such that If the sequence converges to and is continuous then by the theorem above:
The algorithm is depicted in Figure 18
Fixed Point Algorithm
Input: TOL,
Output: or message of failure
Step 1: Set
Step 2: While do Steps 3 - 6
Step 3: Set % Compute
Step 4: if TOL then
output ; % found not
Stop
Step 5: Set
Step 6: Set Update .
Step 7: Output (Iterations exceeded. )
END
Theorem: (Fixed Point Iteration) Let and suppose . Suppose in addition that is continuous on with
if then for any in , the sequence
converges only linearly to the unique fixed point in
Proof: Fixed Point Theorem says . Since exists on apply mean value Theorem to to show that for any
since , and . Since is continuous on we have
thus
Remark: Can get higher-order convergence when
Theorem: Suppose solution of . and continuous and strictly bounded by on an open interval containing . Then such that . The sequence converges quadratically:
Fixed Point Iteration: let and suppose . Suppose in addition that exists on with
(6)
if is any number in , then
converges to unique fixed point in .
Proof: From fixed point theorem, a unique fixed point exists in . Since maps into itself, the sequence is defined and .
Using (6) and the Mean Value Theorem
By induction
(7)
Since
and
Corollary If satisfies hypothesis of Fixed Point Iteration theorem, a bounds for the error involved in using to approximate are given by
Proof: (a) follows from (7):
For
therefore for
Since then
Since then
Remark: Rate depends on . The smaller , the faster it converges.
Example: Consider for . This function is illustrated in Figure 19. Get matlab code used in the example.
First we wish to ensure that the function maps into itself.
Next we look at the derivative of
This fulfills the requirements for a unique fixed point to exist in . It also ensures that if we start with any non-negative value we will converge to the fixed point. The table below shows the first ten iterations for three different values of . Figure 20a and Figure 20b illustrate the iteration history and the logarithm of the error, for a case starting with .
Figure 21a and Figure 21b illustrate the iteration history and the logarithm of the error, for a case starting with .
Finally, Figure 22a and Figure 22b illustrate the iteration history and the logarithm of the error, for a case starting with .
The iterations for the three different starting points all appear to converge on . The log error plots are straight lines and they all have the same slope, indicating the same rate of convergence.
We can also prove analytically that is the fixed point of . A fixed point of satisfies
We can rearrange this to get which has one real root, .
Example: Consider the function for . We will start with the initial value and consider what happens for various values of . -Figure show the iterations for , respectively.
They also plot on the same graph as so we can see the fixed point, and finally plot the log of the error for each value of . We can see that for both and the iteration converges to a fixed point. The log error plots are straight lines with the slope showing the convergence rate (Question: Why does the log error plot flatten off for ?). In the case we can see that it converges faster than the case . For the fixed point does not converge but seems to bounce around different values in the interval . In fact for values of between and we get all sorts of interesting beheviour. For more information on this click here. But when we make less than 0.5 the iteration is able to escape from the interval and once it does this it increases rapidly with each iteration until it causes an overflow error in the computer.
(b) . (c) . (d) .">Get matlab code used in the example.
Now lets see whether we can understand what is happening. First let us look at the range of the function
This shows why the iterations blow up for less than 0.5. For the range is not within the domain of (i.e. ) and so points may 'escape'. However for any value of greater than a half the range is mapped to within the domain.
Next we need to look at the derivative of
The magnitude of the derivative is only less than one for all values of if . Thus for any value of greater than two the fixed point theorem holds and we have guaranteed convergence. We know, however, that we still get convergence to a fixed point for some values of less than two. What is happening in these cases?
If the magnitude of the derivative will be less than one for . As long as the fixed point lies within this interval the theorem tell us that there will be a region around the fixed point where iterations will converge to the fixed point. This is the case as long as . As it turns out, we may still start at any point within and we will eventually arrive at the fixed point although convergence takes longer and longer the closer is to the critical point.
For values of the fixed point still exists but it becomes unstable (i.e. If you start close to the fixed point and iterate you will move away from it rather than towards it).
If we plot and the line on the same graph we can see that there is only one fixed point within the interval for all values of . In fact we can calculate the value of the fixed point analytically by solving .
This is a simple quadratic equation with two solutions
For only the smaller of the two solutions lies within the interval and is the unique fixed point.
Subsections
Next: Contraction Mapping Up: SOLUTION OF NONLINEAR EQUATIONS Previous: Secant Method
Juan Restrepo 2003-04-12 | 2016-06-27T13:10:24 | {
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https://www.jiskha.com/questions/304464/what-substitution-could-i-use-to-integrate-a-a-2-x-2-3-2-dx | Math
What substitution could I use to integrate
a/(a^2 + x^2)^3/2 dx
1. 👍
2. 👎
3. 👁
1. Let u = x/(x^2 + a^2)^1/2
and you will find that
(1/a^2)* du
= integral of dx/(x^2+a^2)^3/2
which is the integral you want.
Therefore u/a^2
= (x/a^2)/(x^2 + a^2)^1/2
1. 👍
2. 👎
2. Computer program says the answer is
x/(a*(a^2 + x^2)^(1/2))
which is slightly different from your answer. Thanks so much for the help on this one. I was really stuck.
1. 👍
2. 👎
3. I did x = a tan u
dx = a (sec u)^2 du
int of a/(a^2 + x^2)^3/2 dx
= int of (a sec u)^2/(a^2 + (a tan u)^2)^3/2 du
= int of (a sec u)^2/(a sec u)^3 du
= int of (cos u)/a du
= (sin u)/a + K
since u = atan (x/a)
= x/(a*(a^2 + x^2)^(1/2)) + K
Thanks again...
1. 👍
2. 👎
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https://math.stackexchange.com/questions/2484250/all-invariant-subspaces-of-a-linear-operator-t | # All invariant subspaces of a linear operator $T$
Let $T \in \mathscr L(\mathbb F^n)$ such that $T(x_1,x_2,...,x_n)=(x_1,2x_2,...,nx_n)$. Then find all the invariant subspaces of $T$. Clearly, $Null$ $T$ and $Range$ $T$ are two invariant subspaces.
Also, all the subspaces spanned by the eigen vectors form $1$-dimensional invariant subspaces. In this case the eigen values of $T$ are $i$, where $i$$\in\{1,2,...,n\} and the corresponding eigen vector is of form (0,...,0,a,0,...,0) where a(\neq 0)\in \mathbb F is the i^{th}-component of the vector. But what about the invariant subspaces of other dimensions? For instance, if W is an invariant subspace of T of dimension k, then T|_W is a linear operator on W. So, for any w\in W, Tw\in W. If w\in span(e_1,e_2,...,e_k) then Tw\in span(e_1,2e_2,...,ke_k). Since, dim W= dim Tw=k, we can say that Tw=span(e_1,2e_2,...,ke_k)\in span(e_1,e_2,...,e_k)=W i.e Tw\in W. So, can we conclude from here that there will be invariant subspaces of all dimensions under T, given by span(e_1,e_2,...,e_k), where 1\leqslant k\leqslant n (which are precisely 2^n in number)? • What do you think the invariant subspaces are? – Demophilus Oct 22 '17 at 13:32 • @Demophilus If W is an invariant subspace of T of dimension k, then T|_W is a linear operator on W. So, for any w\in W, Tw\in W. If w\in span(e_1,e_2,...,e_k) then Tw\in span(e_1,2e_2,...,ke_k). Since, dim W= dim Tw=k, we can say that Tw=span(e_1,2e_2,...,ke_k)\in span(e_1,e_2,...,e_k)=W i.e Tw\in W. So, can we conclude from here that there will be invariant subspaces of all dimensions under T, given by span(e_1,e_2,...,e_k), where 1\leqslant k\leqslant n? – JackT Oct 22 '17 at 13:52 ## 2 Answers I'll work in characteristic zero to avoid quirks due to the field. Suppose that V is an invariant subspace, and let v\in V. Assume for a moment that all entries of V are nonzero. We have that v,Tv,T^2v,T^3v,\ldots,T^{n-1}v\in V. If \alpha_0T^0v+\cdots+\alpha_{n-1}T^{n-1}v=0, we get a linear system with matrix$$ A=v_0v_1v_2\cdots v_{n-1}\begin{bmatrix} 1&1&1&\cdots&1\\1&2&3&\cdots&n-1\\ 1&2^2&3^2&\cdots&(n-1)^2\\ \vdots&\vdots&\vdots&\cdots&\vdots\\ 1&2^{n-1}&3^{n-1}&\cdots&(n-1)^{n-1} \end{bmatrix} $$This is (one of) the well-known Vandermonde Matrix, which is invertible. So the only possible solution of the system is given by \alpha_0=\alpha_1=\cdots=\alpha_{n-1}=0, and thus v,Tv,\ldots,T^{n-1}v are linearly independent, which means that \dim V=n and so V=\mathbb F^n. So any invariant subspace will be made of vectors with at least one entry equal to zero. On the "nonzero part" of the subspace we can repeat the above reasoning, to conclude (as you suggested in your comment) that any invariant subspace of T is of the form$$ W=\text{span}\,\{e_j:\ j\in K\} $$for some K\subset\{1,\ldots,n\}. • :Thanks a lot... – JackT Oct 22 '17 at 14:07 Well, you can get 2^n (instead of just n) subspaces off the bat, by considering \operatorname{span}(B), where B ranges over all subsets of \lbrace e_1, \ldots, e_n \rbrace. Included in this are the whole space and the trivial space (spanned by the empty set), which are the range and nullspace respectively (unless the field has non-zero characteristic). The question is, are there any others? First, suppose \mathbb{F} = \mathbb{C}, and U is an invariant subspace, so that T|_U is an operator. It also will be diagonalisable. To see this, suppose \lambda is an eigenvalue for T|_U, and consider$$\operatorname{null}(T|_U - \lambda I|_U)^2 \subseteq \operatorname{null}(T - \lambda I)^2 = \operatorname{null}(T - \lambda I) = \operatorname{null}(T_U - \lambda I|_U),$$so the generalised eigenspace is no larger than the eigenspace. Hence a basis for$U$, consisting of eigenvectors for$T|_U$(and hence$T$) exists, so$U$must be of the above form. If$\mathbb{F} = \mathbb{R}$, the same argument mostly works. You just need to tiptoe around the complex case. If you extend the map to the complexification of the space, the argument works exactly as is. If$\mathbb{F}\$ is a different field, I'm not sure if I can help.
• You are absolutely correct. – JackT Oct 22 '17 at 14:11 | 2020-01-22T14:51:15 | {
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https://math.stackexchange.com/questions/2416560/is-the-set-2-3-4-open-in-some-metric-spaces-and-not-open-in-others | # Is the set $\{2, 3, 4\}$ open in some metric spaces and not open in others?
I just want to check my understanding. This is from Baby Rudin:
2.18 Definition Let $$X$$ be a metric space. All points and sets mentioned below are understood to be elements and subsets of $$X$$.
$$(a)$$ A neighborhood of $$p$$ is a set $$N_r(p)$$ consisting of all $$q$$ such that $$d(p, q) for some $$r>0$$. The number $$r$$ is called the radius of $$N_r( p)$$
$$(e)$$ A point $$p$$ is an interior point of $$E$$ if there is a neighborhood $$N$$ of $$p$$ such that $$N \subset E$$
$$(f)$$ $$E$$ is open if every point of $$E$$ is an interior point of $$E$$.
Suppose we have the metric space with set $$X=\{1, 2, 3, 4, 5\}$$ and distance function $$d(x, y)=|x-y|$$. Now $$2$$ is an interior point of $$\{2, 3, 4\}$$ because $$N_{0.5}(2)=\{2\} \subset \{2, 3, 4\}$$ (and a similar argument can be made for $$3$$ and $$4$$ as well.
But if our metric space is $$\mathbb{R}$$ with the same distance function, then $$\{2, 3, 4\}$$ is not open because no neighborhood of $$2$$ is a subset of $$\{2, 3, 4\}$$, so $$2$$ is not an interior point of $$\{2, 3, 4\}$$, right?
• Right. That's completely correct. Sep 4 '17 at 16:01
• Why "no neighborhood of $2$ is a subset of $\{2,3,4\}$"? You are right, but you need to add some explanation. Sep 4 '17 at 16:01
• @Krish When we are dealing with the set $\mathbb{R}$, every neighborhood will contain numbers which are not integers.
– Ovi
Sep 4 '17 at 16:03
• @Krish: Because any neighborhood of $2$ (by the definition given) has the form $(2-r,2+r)$ for some $r>0.$ This will readily contain some non-integer rational number. Sep 4 '17 at 16:03
• @CameronBuie sorry!!! But I was just checking whether OP understood the reason clearly or not. (+1) for the question. Sep 4 '17 at 16:07
## 1 Answer
You're absolutely right. Nicely reasoned!
Another thing to consider is that even when the underlying space is the same, using a different metric may yield different open sets. Letting our metric space be $\Bbb R,$ but with the distance function $$\delta(x,y):=\begin{cases}0 & x=y\\1 & x\neq y,\end{cases}$$ we can show that (for example) $\{2\}$ is a neighborhood of $2$ with radius $\frac12,$ and by similar reasoning conclude that $\{2,3,4\}$ is once again open. | 2021-09-26T08:51:51 | {
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https://thorstenharte.de/the-motion-of-a-conical-pendulum-is.html | the motion of a conical pendulum is. Here are a number of highest rated Pendulum Period Equation pictures upon internet. A) What is the magnitude of the torque (N m) and direction (from the listed answers) of the torque exerted on the ball about the center of the circle of motion?. Expression for Period of Conical Pendulum: Let us consider a conical pendulum consists of a bob of mass 'm' revolving in a horizontal circle with constant speed 'v' at the end of a string of length 'l'. Periodic events and periodic motion have long served as standards for units of time. Consider a mass m performing circular motion …. Torsion pendulum: the weights can be moved as shown to change the rotational inertia, and therefore the period. Conical pendulum is a small heavy mass attached with massless, flexible and inextensible string suspended from a rigid support and the mass is constraint to . According to Big Al, one of our first astronauts was successful because of his family background-his mother was spacey, and his father mooned people …. The contents of this file are shown in Table 1 below. If the object is held by a piece of string, the string marks out a conical …. Foucault's pendulum is another example of motion relative to the earth which exhibits the fact that the earth is not a Newtonian base. Define conical pendulum obtain an expression for the angle ma…. You are to investigate the motion …. The bob does not show any vertical motion (moves in horizontal circle only). Its construction is similar to an ordinary pendulum…. Derive the general differential equation of motion for the pendulum …. The resulting equation of motion …. This example shows how to simulate a three link conical compound pendulum made up of cylindrical bars connecting particles at the joints. The motion of a pendulum--be it the simple, spherical or the conical--is a concrete example wherein the three types of motions discussed above are realized physically. a conical pendulum – the dotted pendulum bob shows the equilibrium position and the grey arrows show its trajectory for fixed R and theta. The figure at the right shows an idealized pendulum, with a "massless" string or rod of length L and a bob of mass m. This article introduces a pendulum element to a 3-spring vibration isolator to achieve a high-static-low-dynamic (HSLD) …. The effective length of a second’s pendulum is 99. If the angle the string makes with the vertical axis is 45. The plane and the supporting string trace a conical pendulum…. Uniform Circular Motion - Conical Pendulum. Substituting into the equation for SHM, we get. The pendulum has also allowed astronomers and geologists to measure the motion…. 1: Comparison of the simple pendulum and a true pendulum for a 10 o amplitude. Theoretically, The time period of a conical pendulum …. 1 The Conical Pendulum A small ball of mass m is suspended from a string of length L. James Webb is a Conical Pendulum Orbiting the Sun : ja…. Consider a bob of mass 'm' tied to one end of a string of length 'l l ' and other end fixed to rigid support (S). Besides, a restoring force always acts on the bob of the pendulum …. This is similar to a conical pendulum. The string of a conical pendulum sweeps out a right-circular cone. However the most interesting variable is, the length of the swinging pendulum…. Add your answer and earn points. Suppose that an object of mass is attached to …. Swinging during a fixed period as a gravity pendulum of a certain weight, the swing of the pendulum decays faster if the damping is larger, and more slowly if the damping is smaller [1]. CONICAL PENDULUM A conical pendulum consists of a weight (or bob) fixed on the end of a string or rod suspended from a pivot. For the conical pendulum gravity mass and the angular velocity determine the angle between the string and the vertical axis of the pendulum . A conical pendulum is formed by attaching a mass m to a string of length L, then allowing the mass to swing in a horizontal circle. The motion is adjusted to be horizontal near the center of mass of the hanging portion. A mass oscillating on a spring is an example of an object moving with simple harmonic motion. The movement of a pendulum is called simple harmonic motion: when moved from a starting position, the pendulum feels a …. a weight connected by a rod with a fixed point; Horrocks devised a model of planetary motion …. It doesn't take much effort to keep the mass moving at a constant angular velocity at a constant radius. In our diagram the radius of the circle, r, is equal to L, the length of the pendulum…. Examples of Oscillations are all around us. On 1 December 1659 Christiaan Huygens outlined the steps by which he had come to the momentous discovery that motion along an inverted cycloid is …. The pendulum must swing freely so that it swings backwards and forwards in the same plane. The time period of a simple pendulum: It is defined as the time taken by the pendulum to finish one full oscillation and is denoted by “T”. For example, suspending a bar from a thin wire and winding it by an angle \theta, a torsional torque \tau = -\kappa\theta is produced, where \kappa is a characteristic property of the wire, known as the torsional constant. If a simple pendulum is fixed at one end and the bob is rotating in a horizontal circle, then it is called a conical pendulum. Example at 20G mass moves as a conical pendulum with 8x vibrating length and speed V if the radius of the circular motion is 5x to find: i) the string tension (I assume g = 10 ms-2, (2 A conical pendulum …. then acceleration of the body is proportional to displacement, but in the opposite direction of displacement. Let G be the centre of gravity of a compound pendulum of mass m that oscillates about a point O with OG = h If the pendulum …. 012 kg, the string has length L = 0. Relevant Equations: The Lagrangian is defined as the difference of the kinetic energy T and the potential energy U. (PDF) Turning points of the spherical pendulum and the gol…. This paper represents a continuation of the theoretical and computational work from an earlier publication, with the present calculations using exactly the same physical values for the lengths L (0. To set the pendulum in motion…. Velocity And Acceleration In Shm Get Our FREE Chrome Extension. If the… A bar magnet has a magnetic moment 2. However, this doesn't affect the period of pendulum. θ=suspension angle, r =radius of bob's circular motion, h =vertical height of suspension above the plane of the bob's motion…. The formula for the period T of a pendulum is T = 2π L g , where L is the length of the pendulum …. The precession of a Foucault pendulum viewed as a beat phenomenon of a conical pendulum …. diameter range Rasp 8" (10 Piece) 5 for Wood, 5 Steel 4 x Second Cut: 200 mm (8") Round, …. Translation in spherical coordinates: where q = longitude, j = colatitude and …. (Because the string sweeps out the surface of a cone, the system is known as a conical pendulum. Time period for conical pendulum= T = 2λ√r/gtan0. A conical pendulum is the simple pendulum whirled . March 2013 by Sam Categories: Mechanics | Tags: conical, pendulum | Comments Off on Horizontal Circular Motion (Conical Pendulum). Consider a conical pendulum with a mass m, attached to a string of length L. For large motions it is a chaotic system, but for small motions it is a simple linear system. Study on the Motion Characteristics of two. How forces affect motion, and the nature of . Healing, Let's GO!!! – Pendulum …. Files: 2 movies (384 X 384 pixels). A rigid body mounted on a fixed horizontal axis, about which it is free to rotate under the influence of gravity. So, the expression for the time period of small oscillations is equal to 2π× √(length of the pendulum…. Tie the thread to straw and tape the straw on the table in such a way that around half of an inch hangs over the edge. P is moving around the blue circle with angular velocity w. The simple harmonic motion is defined as a motion …. In Figure 1 we see that a simple pendulum has a small-diameter bob and a string that has a very small mass but is strong enough not to stretch …. The Amazing Pendulum 2014. 1 Uniform Circular Motion Uniform circular motion is the motion of an …. As the motion of the bob is a horizontal circular motion, the resultant force must be horizontal and directed towards the centre C of the circular motion…. A conical pendulum consists of a bob of mass m in motion in a circular path in a horizontal plane as shown in Figure. This paper explains why the paths in the rotating frame are circular, and the difference between the periods for the right- and left-handed rotations is . The object revolves with constant speed vin a hor-izontal circle of radius r,as shown in Figure 6. The SHO and Circular Motion • We can now see that the equation of motion of the simple pendulum at small angles—which is a simple harmonic oscillator is nothing but the x-component of the steady circularmotion of the conical pendulum • The simple pendulum is the. This report shows how to find an approximate of ‘g’ using the simple pendulum experiment. The spin or conical motion of the bob may cause a twist in the thread, thereby …. The course follows the typical progression of topics of a first-semester university physics course: Kinematics, Newton’s Laws, Energy, and Momentum. Progressive education for money. 79]: to analyze the motion of a pendulum moving in a horizontal circle (a conical pendulum). The motion is regular and repeating, an example of periodic motion. In order to prove this fact consider a simple pendulum having a bob of mass ' m ' and the length of pendulum is ' l '. Spherical pendulum and vertical pendulum are the special. Step 1: Derive the Equation of Motion. The other is a conical pendulum which involves a pendulum …. What is the formula for speed of pendulum at any point. First at the request of the Directors of the Indies Company I undertook for finding longitudes to construct clocks of which the sure and constant motion would be equal to a three-foot pendulum…. The motion of the bob of simple pendulum simple harmonic motion if it is given small displacement. Single and Double plane pendulum Gabriela Gonz´alez 1 Introduction We will write down equations of motion for a single and a double plane pendulum, following Newton’s equations, and using Lagrange’s equations. A mass on a spring has the vibrating spring mode resonantly coupled to the pendulum …. It also uses a simple pivot, perhaps a knife edge support, while a conical pendulum requires a more complex support with two-directional movement. The pendulum bob moves in a horizontal circle with constant speed v, θ = suspension angle, . Coasting Through a Vertical Loop (with Reaction Forces) | QT Embedded | Media | Old Embedded |. The horizontal component of tension balances this centripetal. The mass is set in motion in a horizontal circular path about the vertical axis. Model the ball in the pendulum is a particle. The swinging incense burner called a censer, also known as a thurible, is an example of a pendulum. Answer Text: The conical pendulum: -It consists of a small massive object tied to the end of a thin string tied to affixed rigid support. Answer: Let us consider a conical pendulum consists of a bob of mass 'm' revolving in a horizontal circle with constant speed 'v' at the end of a string of length 'l'. A conical pendulum consists of an object attached to a string and moving in a horizontal circle. The time taken by the pendulum to reach x = a/2 from the mean postion will be: Q7. This is different from Kepler's 3rd law where P ∝ r**(1/2), where r is the mean orbital radius. As shown in the figure above the driving force is F=-mgsintheta where the -ve sign implies that the. circular motion physics gcse amp a level revision and. Figure 6 illustrates this concept. the constant angular velocity of the bob. Circular motion with conical pendulum Number 135710-EN Topic Mechanics, two-dimensional motion Version 2017-02-17/HS Type Student exercise …. Bohr Magneton Spin Magnetic Moment Of Lectron Calculator. iii) The bob of a conical pendulum under goes (A) Rectilinear motion …. 85 m and the angle with the vertical is 37°. A conical pendulum is a weight (or bob) fixed on the end of a string (or rod) suspended from a pivot. 2 The Conical Pendulum A small object of mass mis suspended from a string of length L. 10 L m a) Draw a labeled free-body diagram for the pendulum …. We will derive the equation of motion for the pendulum using the rotational analog of Newton's second law for motion about a fixed axis, which is …. Seat pan and sprinkle some the list …. In simple languages definition of a conical pendulum can be summed up as a conical pendulum consists of a weight (or bob) fixed on the end of a string or rod suspended from a pivot. Hang a bowling ball from a ceiling hook for a large conical pendulum to rotate at a steady slow speed. This motion of the pendulum is called motion of the first type. This video is made to understand Circular Motion. Rectilinear motion in vertical circle. Based on the previous notation, the following formulas can be obtained: Definition of average velocity. Standard conical pendulum lengths that have been investigated experimentally (and subsequently published) usually. A conical pendulum is a simple pendulum with the bob describing a circle and the string a cone · The centripetal force in a conical pendulum of semi vertical . It spins around the vertical with angle and with angular …. In the overhead view of above figure, a long uniform rod of mass 0. If during a periodic motion, the particle moves to and fro on the same path, the motion is vibratory or oscillatory. An elliptical (nonplanar) motion may precess due to anharmonicity in the restoring force. Basically, this is a mass on a string attached to a rubber stopper. The equation of SHM for a simple pendulum …. A conical pendulum has length 50 cm. A conical pendulum is a mass attached to a nearly massless string that is held at the opposite end and swung in . April 25, 2018 Boris Sapozhnikov. The actual form of a pendulum …. By applying Newton's secont law for rotational systems, the equation of motion for the pendulum …. My Dashboard; Modules; WEEK 9 : CM2-CU8: Conical Pendulum; VIDEO : Conical Pendulum (Motion …. 2004): if the length of [a conical] pendulum be /, the semi-major axis of the ellipse described by the pendulum …. 12 S G Kamath The motion of a pendulum--be it the simple, spherical or the conical--is a concrete example wherein the three types of motions discussed above are realized physically. Because the bobs would lift in response to a faster speed (because they were basicly a conical pendulum…. What is a conical pendulum?. In the coordinate system that rotates with the wire, there will be fictitious Coriolis and centrifugal forces, in addition This Article is brought to you for free and …. This acceleration is called centripetal acceleration, and equals v 2 /r, where v is the speed of the object and r is the radius of. Suppose, further, that the object is given an initial horizontal velocity such that it executes a horizontal circular orbit of radius with angular velocity. The figure below shows a "conical pendulum", in which the bob (the small object at the lower end of the cord) moves in a horizontal circle at …. THE COMPOUND PENDULUM The term “compound” is used to distinguish the present rigid-body pendulum from the “simple” pendulum of Section 3. Motion Of An Object Attached To A Spring. The plane of its motion, with respect to the earth, rotated slowly clockwise. Best answer Expression for tension in the string of a conical pendulum: i. To describe the motion of a conical pendulum in terms of its tangential velocity. Motion of a horse pulling a cart on a straight road. 68 m and the angle between the string and vertical is 35°. Conical pendulum illustrates uniform circular motion, and the other cases are representative of a non-uniform circular motion. A conical pendulum consists of a bob of mass 'm' revolving in a horizontal circle with constant speed 'v' at the end of a string of length 'l'. A) What is the magnitude of the torque (N. A simple pendulum consists of a relatively massive object - known as the pendulum bob - hung by a string from a fixed support. This result is true for all horizontal conical pendulums for which the angle, θ, is measured from the pendulum's position of vertical equilibrium. Please help me!!! I cnt solve thisA bob of mass is suspended from a fixed point with a masslessstring oflength (i. Rotational Inertia Stick Demo. The sections on mechanics in this book are basically arranged in that order. The pendulum motion is induced by the weight of the hanging mass, which is moved initially. let 'h' be the depth of the bob below the support. The deeper it sits, the wider the arch of the revolving pendulum, the slower the clock will go. Conical Pendulum: Its time period, tension i…. Cm 4 the conical pendulum (shared). Conical pendulum is a small heavy mass attached with massless, flexible and inextensible string suspended from a rigid support and the mass is constraint to whirl in a horizontal circle with constant speed. Fresnel Reflectance Of S Polarized Light. Oscillation motion is a periodic motion in which the particles move to and fro on a particular predetermined path within equal time intervals. In the table below, let us look at the various differences between simple and compound pendulum. I can go to a very small angle and then it's kind of barely moving, really slow like that. dimensions of the conical pendulum, the magnitudes of the forces acting during the conical pendulum motion and a triangular construction involving the conical pendulum period. Huygens rotated a conical pendulum of length l (when the angle θ was small) he. So a pendulum can evidently be said to have a certain amount of oscillatory motion in 1 direction plus a variable amount of angular momentum which results in a periodic amount of lateral motion in the orthogonal direction. A simple pendulum is an idealized body consisting of a particle suspended by a light inextensible cord. When pulled to one side of its equilibrium position and released, the pendulum swings in a vertical plane under the influence of gravity. Thus together with the string the bob traces out a cone. Example: Motion of moon around earth; Motion of a piston in a cylinder; Motion of a simple pendulum etc. What Is The Difference Between A Simple Pendulum And A Co…. that a conical pendulum whose initial motion was elliptical, was compelled to process in the same direction as the oscillation of its mass (Olsson 1978, 1981; Gray et arl. ERIC is an online library of education research and information, sponsored by the Institute of Education Sciences (IES) of the U. OCR past papers can be found at: https://www. Correct answer - The bob of a conical pendulum undergoes a) rectilinear motion in horizontal plane b) uniform motion in a horizontal circle c) The bob of a conical pendulum undergoes a) rectilinear motion in horizontal plane …. Location: Cabinet: Mechanic (ME) Bay: Shelf: #1 Abstract: A wire is suspended from a ceiling mount and a bowling ball attached at the bottom. (7) determines whether the plates of the pendulum …. Sneak a peak! Assault a graveyard. And silence those who arrive in the tuna. Derivation of the equations of motion. A motorized, plastic plane* is suspended from a thin string and “flies” in a circular path with a constant speed. The string is whirled in a horizontal circle, then the arrangement is called a conical pendulum. We first determine the time period of one complete rotation in conical pendulum. The characteristic of this motion can be obtained in terms of the length of the string and the angle with respect to the vertical. A particle of mass m, just completes the. Added an answer on August 11, 2020 at 1:32 pm. N2 - Students often find mechanics a difficult area to grasp. In this paper, we present evidence to show that the dynamics of rigid solid bodies is not a closed discipline, particularly in the field of …. ' Centripetal force (Fc) is the result of gravity and tension. The point of intersection satisfies the system of two linear equations: 12 12 57 22solutions …. Description : This set of 2 movies illustrates the dynamics of the conical pendulum (object suspended at the end of a string, moving in uniform circular motion). 22 A torsional pendulum consists of a rigid body suspended by a string or …. The time required for the bob of the conical pendulum to travel one revolution. Introduction The spherical pendulum is a mechanical system of considerable pedagogical interest [1–5]. Obtain an expression for its time period. Ballistic Pendulum lab report; Angular Motion lab report; Lab Report 9; Other related documents. 1 Use a search engine such as Google to research the history and uses of one of the following materials: Tin Glass Cement Titanium Carbon fiber Present the …. in the variable θ about the point (a near-conical pendulum). Determine the period of the pendulum …. In a conical pendulum, the bob is rotated with different angular v…. If the Lagrangian is approximated by keeping terms up to cubic order, the system has three independent constants of motion…. Torque and angular momentum of a conical pendulum joe_coolish May 19, 2011 May 19, 2011 #1 joe_coolish 9 0 Homework Statement A ball (mass m = 250 g) on the end of an ideal string is moving in circular motion as a conical pendulum as in the figure. Circular Motion A conical pendulum is a mass attached to a nearly mass-less string that is held at the opposite end and swung in horizontal circles. A motion of a Conical Pendulum. So, that's what I wanna talk to you about in this video. A particle is acted upon by a force of constant magnitude which is always perpendicular to the velocity of the particle, the motion of the particle …. 3: A torsional pendulum consists of a rigid body suspended by …. Circular MotionUnit: Circular MotionSubject: Physics Grade XI. Caption: A schematic diagram of "a conical pendulum in motion. In the corresponding motions of the pendulum, close to conical motion…. A conical pendulum is formed by attaching a ball of mass m to a string of length L, then allowing the ball to move in a horizontal circle of radius r. Classify the following as motion along a straight line, circular or oscillatory motion: a. 75, and inverted pendulum, had a mass of 1000 kg and mechanical 1. [45] Researchers at the University of Chicago and Argonne National Laboratory have invented an innovative way for different types of quantum technology …. The conical pendulum was first studied by the English scientist Robert Hooke around 1660 as a model for the orbital motion of. The is an example of a 3D holonomic system. A special case of circular motion is the conical pendulum where the pendulum's "bob" is swung in a horizontal circle of radius Rwhile attached to a string of length L. If we take Tas time period of rotatory motion then velocity can also be expressed as: v= 2ˇr T: (2) Now replace the expression of vfrom Eq. A small ball of mass m is suspended from a string of length L. Draw a neat labelled diagram of conical pendulum. In this case, the moment of inertia has to be around the CM: I = 1 12 M L 2 leading to. Set the object traveling in a horizontal circular motion at a constant angle θ , as measured outward from when the object hangs straight down. A conical pendulum consists of an object moving in uniform circular motion at the end of a string of negligible mass (see Figure 1). Hang a bowling ball from a ceiling hook for a large conical pendulum …. Physics Lab 8: The Flying Pig – Centripetal Force Section: Name: Purpose To show the net force for a conical pendulum is mv 2 /r. uniform circular motion can be demonstrated with conical pendulum. Obtain the equations of motion of the system using Lagrange’s equations. I’m assuming that the pendulum bob is connected by an inelastic string of negligible mass, which allows me to fix the length of the string as. The correct option is : (a) Angular momentum of bob is …. You also should be able to find setup descriptions/figures in most textbook chapters on centripetal force. It is being regulated by a rotating pendulum that is called ‘conical’ after the shape of the silhouette of the motion. In this video, AAFREEN MAM has tried to explain the fundamentals necessary for solving physics questions. Of course, these different types of motion can be combined: for instance, the motion of a properly bowled bowling ball consists of a combination of trans-lational and rotational motion, whereas wave propagation is a combination of translational and oscillatory motion. The pendulum bob moves in a horizontal circle with a constant speed v. It is at ® rest under the action of …. Transcribed image text: PROJECTILE MOTION AND BALLISTIC PENDULUM Objective: The purpose of this experiment is to use the laws of conservation of energy and linear momentum to determine the velocity of a projectile, use this result to predict the projectile's range when fired in a uniform gravitational field and then compare this range to a measured value. 4b, which consisted of a particle at the end of a massless string. ANGULAR MOMENTUM OF CONICAL PENDULUM. Simple harmonic motion Solutions. Hello, The degree of freedom of a system = Number of directions in which it can move. When the string is horizontal there in …. In this Lesson, the sinusoidal nature of pendulum motion …. Introduction: A conical pendulum is a pendulum that is spun around in a circle instead of swinging backwards and forwards, such as like a traditional pendulum. A small mass is suspended by a cord and set into motion …. The pendulum does this because of inertia, which is the tendency of mass to stay in motion when a force acts upon it. For calculating the time period of a conical pendulum, we need to use the expression of Newton's second law of motion…. Answer (1 of 3): Consider a conical pendulum with a bob of mass m, length l, at an angle \theta with the vertical, going round with a uniform velocity v and …. Request PDF | The conical pendulum: The tethered aeroplane | The introductory physics lab curriculum usually has one experiment on uniform circular motion …. A simple pendulum is one which can be considered to be a point mass suspended from a string or rod of negligible mass. The Conical Pendulum Download PDF A great activity for physics classes investigating centripetal force and uniform circular motion. Calculate… A simple pendulum mounted on a car. In this Lesson, the sinusoidal nature of pendulum motion is. To determine the acceleration due to gravity by means of a conical pendulum. Coupled pendula: Three varieties are available. February 15, 2018 Boris Sapozhnikov. Laboratory 8: Conical Pendulum –Activity L θ r Figure 2: Conical pendulum The conical pendulum, shown in Figure 2, con-sists of a mass m at the end of a …. 2:50 Breaking the force of tension into its components. Conical pendulum – measuring g Number 13573 0-EN Topic Me ch anics , two -dimensional motion Version 201 7-02-17 / HS Type Student exercise Suggested for grade 11 -12 p. Circular Motion and the Conical Pendulum Introduction. 5 THE CONICAL PENDULUM: A STUDY OF. A conical pendulum, a thin uniform rod of length. Consider about the figure of a conical pendulum. Kinetic energy = (1/2) Mω2(A2 – x2) Potential energy = 1/2 Mω2x2. The modulationequationsare found to be soluble in terms of elementary …. There's a 1:1 bevel that drives the pendulum. simple pendulum moves in to and fro motion. 0 degrees, then the angular speed of the conical pendulum …. Each of five modules contains reading links to a free textbook, complete video lectures, conceptual quizzes, and a set of homework problems. It doesn't take much effort to keep the mass moving at a constant angular velocity in a circle of constant radius. Take a video of the motion as viewed from the side. Ordinary pendulum clocks had been invented by Christiaan Huygens in 1656, and by ordinary, we mean that the pendulum swings back and forth in a vertical plane. With an elliptical path it represents a motion …. A heavy spherical mass (approximately 1. Estimation of parameters of various damping models in pla…. The problem of the conical pendulum is to consider a mass attached to one end of a light inextensible string of length with the other end attached at the top of a vertical rod. The path of periodic motion may be rectilinear, open/closed curvilinear. The motion of a simple pendulum is like simple harmonic motion …. CiteSeerX - Document Details (Isaac Councill, Lee Giles, Pradeep Teregowda): The three-dimensional motion of the elastic pendulum or swinging spring is investigated in this study. pendulum, body suspended from a fixed point so that it can swing back and forth under the influence of gravity. The swing of the pendulum will have to be limited to the distance of the support rods. motion answer key, describing and measuring motion, measuring motion gizmo lesson info explorelearning, download guided and study acceleration motion answers pdf, describing and measuring motion worksheet answer key 1, chapter 1 matter in motion section 1 measuring motion, describing motion …. The string makes a constant angle with respect to the vertical; as a result, the mass moves with constant speed in a horizontal circle. Equipment and Supplies Flying Pig (or equivalent), stopwatch, meter stick Discussion When an object travels at constant speed along a circular path, we say it has uniform circular motion …. In a typical experiment, the conical motion is confined to the horizontal plane, however, it is possible to extend the mathematical analysis to three dimensions (Barenboim & Oteo, 2013). The bob of a conical pendulum is attached to a fixed point A,A by a string of length 50, c, m,50cm and is moving in a circular path, as shown in the diagram . Uniform motion in a vertical circle. Discuss the pendulum's motion in the effective potential diagram. You can modify simulation parameters directly in MATLAB workspace. Physics Lab report on circular motion. Take your string back about 40 - 50 cm. The initial amplitude of the pendulum must be small compared to pendulum …. When the ball at the end of the string swings to its lowest point, the string is cut by a sharp …. For this exercise, assume that angles not small, and wº =2. This result is true for all horizontal conical pendulums for which the angle, θ, is measured from the pendulum…. Name: Laboratory Section: The goal of this laboratory is to study uniform circular motion. The pendulum is perhaps the simplest experimental devices ever constructed, and yet for all its simplicity it has historically enabled scientists to both investigate and enumerate gravity; the fundamental force that shapes the very universe. The study of this lab revolves around the generation, propagation and reception of mechanical waves and vibrations. This can be done side-by-side, front and back, clockwise, counterclockwise, in an elliptical motion, or even in a bobbing movement up and down, which often indicates a strong affirmative action. A simple pendulum performs simple harmonic motion about x = 0 with an amplitude a and time period T. When a pendulum swings in one direction and then comes back to its original starting point, this is called a period. Sir Isaac Newton (1642-1727) established the scientific laws that govern 99% or more of our everyday experiences. A conical pendulum is constructed by attaching a mass to a string 2. The time period of a simple pendulum: It is defined as the time taken by the pendulum …. The constraint on a bead on a uniformly rotating wire in a force free space is. 290 7 Lagrangian and Hamiltonian Mechanics 7. And I love these pure math ones. This work consisted of studying the motion of a rigid conical pendulum …. There are several different ways to configure a Twin Elliptic Pendulum…. ii) For a particle having a uniform circular motion, which of the following is constant (A) Speed (B) Acceleration (C) Velocity (D) Displacement. You have constructed what is known as a physical pendulum. Non-uniform circular motion Up: Circular motion Previous: Centripetal acceleration The conical pendulum Suppose that an …. 1k points) motion in a plane; class-11; 0 votes. We identified it from honorable source. When a pendulum is set in motion, gravity causes a restoring force that will accelerate it toward the center point, Conical Pendulum. Measure the period using the stopwatch or period timer. Slider-crank mechanism, arrangement of mechanical parts designed to convert straight-line motion to rotary motion, as in a reciprocating piston engine, or to convert rotary motion to straight-line motion, as in a reciprocating piston pump. ) Find an expression for v in terms of the geometry in Figure 6. Expression for its time period: Consider the vertical section of a conical pendulum having bob (point mass) of mass m and string of length ‘L’. Place the pendulum assembly on to hand driven rotator such that the set screw is tightened onto the flat part of the hand driven rotator. This resultant force is the centripetal force. The bob of a pendulum of length l is pulled aside from its equilibrium position through …. It is then put into an elliptical orbit thus acting as a conical pendulum. 500-m string is moving in a uniform circular motion in a horizontal plane of radius …. In 1851, Jean-Bernard-Leon Foucault suspended a 67 metre, 28 kilogram pendulum from the dome of the Pantheon in Paris. This necessary centripetal acceleration required for the motion of the conical pendulum is provided by the x component of tension which is the …. The motion of a pendulum is a classic example of mechanical energy conservation. An object at the end of a string is used as a conical pendulum. Course of Theoretical Mechanics, Higher Education Press, Bei Jing, Chapter 5. and the relevant moment of inertia is that about the point of suspension. The stiff string (actually a piece of aluminum wire) is required for otherwise the pendulum will not remain in the same (rotating) plane and the motion becomes closer to a spinning spherical pendulum. "A long-period conical pendulum for vibration isolation," Physics Letters A 222, 141-147 (1996). State technological advantages of Nano Moment of inertia, area meter 4m Moment of inertia, mass kilogram-meter 2kg m Momentum, linear kilogram …. 18 Determine the equations of motion …. Friction: Pulling a Box on a Horizontal Surface. The figure shows a conical pendulum, in which the bob (the small object at the lower end of the cord) moves in a horizontal circle at a constant speed. Mass performing vertical circular motion under gravity. Do Problem 21 if the two masses are different. Science; Physics; Physics questions and answers; PROJECTILE MOTION AND BALLISTIC PENDULUM Objective: The purpose of this experiment is to use the laws of conservation of energy and linear momentum to determine the velocity of a projectile, use this result to predict the projectile's range when fired in a uniform gravitational field and then compare this range to a measured value. You must make a mark on the wall or your piece of paper to make sure that you let it go from the …. the motion of the mechanical system can be described by the generalized coordinates x ( t ) (translation M ), r ( t ) (elongation of the spring pendulum), and ˜ ( t ) (link rotation). The motion of conical pendulum is - 49146832 mukulsony1701 mukulsony1701 19. The string's motion follows a conical …. Vintage Jupiter Wall 31 Day Clock w/Pendulum and Key 26” Shop pendulum …. (You do NOT need to find the frequency of the oscillations. Simulation- Conical Pendulum: 3D. The is horizontal, so there is no J verLieal motion…. (hint: first calculate the tension using the net y- forces = 0 and use newton's second law of motion …. Pendulums are also seen at many gatherings in eastern Mexico where they mark the turning of the tides on the day which the tides are at their highest point. The conical pendulum lab allows students to investigate the physics and mathematics of uniform circular motion. Static and Kinetic Friction on an Inclined Plane. He also explained our relationship to the Universe through his Laws of Motion …. A simple pendulum is a special case of a conical pendulum in which angle made by the string with vertical is zero i. 130 m) for the conical pendulum…. 6-53 shows a conical pendulum, in which the bob (the small object at the lower end of the cord) moves in a horizontal circle at constant speed. The pendulum is a simple mechanical system that follows a differential equation. A physical pendulum consists of a uniform rod of length d and mass m pivoted at one end. Additionally, what are the 4 types of motion? These four are rotary, oscillating, linear and reciprocating. pendulum mass for a fixed length, and varied the pendulum length for a fixed mass. The simple pendulum is another mechanical system that moves in an oscillatory motion. g = acceleration due to gravity. This means that the forces in line with the arm of the pendulum must be equal and opposite since there is no motion in this direction and we see that T = m g cos. The goals of this lab are to verify that centripetal acceleration is given by a = v 2 /r and to show that the period of a conical pendulum is given by the theoretical equation:. There are two forces acting on the bob: its weight and the tension in the string (Figure 39). Determine the horizontal and vertical components of the force exerted by the wire on the pendulum…. oPhysics: Interactive Physics Simulations. An analysis of the motion presents that the equilibrium states of the pendulum are determined by the pendulum angular speed. Other rides, such as the rotor ride, Enterprise wheel, and Ferris wheel, spin the rider in circular motion either horizontally or vertically. The pendulum consists of a string and a bob (a weight, generally spherical) at the end. The reasons for this, however, and consequently the focus …. Robert Hooke's conical pendulum from the modern viewpoint of. A conical pendulum consists of a simple pendulum moving in a horizontal circle as shown in the figure. Let m be the lower mass and let M be the sum of the two masses. The motion of a conical pendulum in a rotating frame. What tangential speed, v, must the bob have so that it moves in a horizontal; Question: To describe the motion of a conical pendulum in terms of its tangential velocity. In both cases, an elliptical motion is induced. Find the maximum height above the ground that the ball reaches. Instead of swinging back and forth, the bob is to move in a horizontal circle at constant speed v, with the wire making a fixed angle β with the vertical direction (Fig. It is Robert Hooke who first studied the motion of a spherical pendulum in that a conical pendulum whose initial motion was elliptical, . Theoretically, The time period of a conical pendulum is directly proportional to the square root of its length. pendulum altogether in absolute measurements of g. The oscillatory motion of a simple pendulum: Oscillatory motion is defined as the to and fro motion of the pendulum in a periodic fashion and the centre point of oscillation known as equilibrium position. State the expression for its periodic time in terms of length. The mass is moving about the rod in uniform circular motion …. In your text, it is shown that the period of rotation T of a conical pendulum of length, L, is given by € T=2π Lcos(θ) g, where θ is the angle the pendulum …. A simple pendulum consists of a point mass suspended on a string or wire that has negligible mass. As the Bravais pendulum is a conical pendulum oscillating at Earth's surface, its motion is also discussed. Its construction is similar to an ordinary . 6-53 shows a conical pendulum, in which the bob (the small object at the lower end of the cord) moves in a horizontal …. The conical pendulum was first. ω: Angular velocity of rotational motion (rad/s) T: period of rotational motion (seconds) Find the period of a conical pendulum. The string makes an angle θ with the vertical. All of the energy in the pendulum is gravitational potential energy and there is no kinetic energy. The pendulum always moves in one angular direction. L2 : Motion of Pendulum - Motion and Time, Science, Class 7; Video | 04:40 min. Popular; Recent; Comments; Pendulum with spring animation MATLAB. The motion occurs in a vertical plane and is driven by a gravitational force. The moment of intertia I for a point mass rotating around a pivot (radius l) is ml², yielding our equation of motion: This is a second-order, non …. ࡱ > 3 5 2 #` p bjbjm m 5 6 > > > x , Q D D D D H S U U U E $h o ~ E> D H ; 7 7 7 @ 8D > H S 7 S 7 7 & > 7 H 8 @h ?Z 7 S ! 0Q 7 p 7 > 7 7 7 Q$ R f t AP Physics Lab Brockport High School NY USA Circular Motion: The Conical Pendulum Mr Keefer Objectives: Determine the acceleration of gravity g from the circular motion of a conical pendulum…. swing with horizontal circular motion such that r equals the length of the pendulum?. This pendulum is used as a model to analyze the motion of planets. Korean pendulum wall clocks. The reason that the pendulum oscillates about the vertical is that if the pendulum is displaced, the force of gravity pulls down on the pendulum. Expression for Period of Conical Pendulum: Let us consider a conical pendulum consists of a bob of mass 'm' revolving in a horizontal circle …. The time it takes the pendulum …. The period of the motion of a pendulum is virtually independent of its amplitude and depends primarily on the geometry of the pendulum …. Question 3 The figure shows a conical pendulum, in which th…. Consider a small body of mass m suspended from a rigid support with the help of a string of length l. Consider a conical pendulum with a line length ‘L’ and a rotation radius ‘r. Conical Pendulum Time Period: It consists of a string OA whose upper-end O is fixed and bob is tied at the other free end. The forces which are acting on the mass. The period of a simple pendulum …. The ball revolves with constant speed v in a horizontal circle of radius r as shown in the figure. In this experiment a mass is attached to a string and made to spin in a circle of fixed radius, the time period of the motion …. Correction of the excercise 4:00. Lab – Flying Pigs Purpose: To show that the net force for a conical pendulum is mv 2 /r Equipment/Supplies: flying pig apparatus stopwatch meterstick Discussion: Any object moving in uniform circular motion …. Equations for the following principal physical parameters. The restoring torque is supplied by the shearing of the string or wire. In other words, the vertical component of the tension force . The bob does not show any vertical motion …. The length L of the string is 1. A Bob of a conical pendulum undergoes option A rectilinear motion in a horizontal plane option B uniform motio… Get the answers you …. Differential equation of the motion (derived from Newton's second law):. This text is useful for students to follow many details on the motion of the Foucault pendulum in inertial and rotational frames and help in the study of the particle motion …. • As viewed from above, it moves in a circle, the centripetal force. The dimensionless factor of $$2\pi$$ can be derived using an in-sight from Huygens [15, p. doing this isaac physics question on circular motion, can anyone help The bob of a conical pendulum is attached to a fixed point A,A by a string of length. The pendulum is initially at rest in a vertical position. Imagine a conical pendulum in steady circular motion with small angle θ. The bob moves in a single plane, back and forth. This can be determined by measuring the time required for the pendulum to reoccupy a given position. (a) Write a MATLAB script, which solves (*) for. The equations of motion of the pendulum were derived using the Lagrangian method. The motion of the pendulum depends on its total energy E = T +V. 9 0 m and negligible mass and the bob follows a circular path of the circumference. eu Viaduktvej 35 · DK-6870 Ølgod Fax +45 7524 6282 www. The canonical example of simple harmonic motion is the motion …. This is a simulation of a double pendulum. Spherical pendulum and vertical pendulum are the special cases of conical pendulum. Geometrically, the arc length, s, is directly proportional to the magnitude of the central angle, θ, according to the formula s = rθ. October 28, 2018 Boris Sapozhnikov. As a result, the motion of the pendulum …. n Review Part A (Figure 1)A massless string of length L is fixed to a point in the ceiling and suspends a bob with mass m (i. h is the distance from the plane of the circular motion to where the string is attached. Is there any limitation on semivertical angle in a conical pendulum? Thank You. study of circular motion in conical pendulum purpose the purpose of this experiment is to study the effect of the. Class 7 Science Chapter 13 Motion and Time Textbook Exercise Questions and Answers. The 1 ∶ 1 ∶ 2 resonant elastic pendulum is a simple classical system that displays the phenomenon known as Hamiltonian monodromy. Let the bob be displaced from its mean position and whirled around a horizontal circle of radius 'r' with constant angular velocity 'ω'. If the weight is made to describe a horizontal circle so that the string describes a right circular cone, whose axis is a vertical through ( O ) , then system is called a conical pendulum. A new twist for the conical pendulum A new twist for the conical pendulum Moses, Thomas; Adolphi, Natalie L. Then the period of the simple pendulum is given by. We use a conical pendulum in this experiment. Jean Bernard Léon Foucault was the leading experimental physicist of his day. Imagine a basic conical pendulum. In this work, planar free vibrations of a single physical pendulum are investigated both experimentally and numerically. But, because for small θ, we can say that θ ≈ sinθ, we can re-write the DE as:. You can apply similar considerations to a simple pendulum, which is one on which all the mass is centered on the end of a string. The equations for a simple pendulum show how to find the frequency and period of the motion. This allowed the reduction of apparent gravity, thus increasing its natural period, a similar approach to that of Galileo more than two centuries earlier when he used an inclined plane to study the motion of falling objects. Equations for a Simple Pendulum. An inverted pendulum is simply a pendulum which has its fixed end located below the vibrating mass. Let us consider a conical pendulum …. The pendulum is in steady circular motion with constant angular velocity wk. the pendulum bob has an initial velocity. The Chaotic Motion of a Double Pendulum. In two or three attempts one could realize a satisfactory circular orbit. Other rides, such as the rotor ride, Enterprise wheel, and Ferris wheel, spin the rider in circular motion …. When this happens the pendulum is called a "conical pendulum" because the cord sweeps out (or creates in space) a conical surface with the apex at the …. Two pendula coupled by a spring (shown to the right) will show normal modes, and transfer of energy between the single pendula swinging modes. One end of a string ( AO ) is attached to a fixed point ( O ) and a weight ( W ) is tied to the other end ( A ). Uniform Accelerated Motion. Use the Tracker Software to analyze the motion of a conical pendulum, as viewed from the side. Let – ( N ) is the centre of horizontal circle of radius ( r ) …. Exploring Projectile Motion Concepts. Let the string subtends an angle θ with the vertical. We use centripetal acceleration here, because we have a circular motion. Conical pendulum clocks are part of the clock history but as they are not easy to implement you don’t see too many around, and as far as I know there are no wooden conical pendulum clocks. In this lab you will • test the theory of simple harmonic motion in the case of a simple pendulum and a Hooke’s law spring. The kinetic energy would be KE= ½mv2 ,where m is the mass of the pendulum, and v is the speed of the pendulum. Consider an object of mass m tied to a string of length l and whirled in a horizontal circle of. Because the bobs would lift in response to a faster speed (because they were basicly a conical pendulum) it could be. The upward force produced by the tension balances the downward force from the weight of the pendulum:. 2 is annotated with the mathematical symbols that will be used in the analysis of the period of the pendulum. The massless thread is only an idealization. A simple pendulum theoretically has the mass of the bob concentrated at one point, but this is impossible to achieve exactly in practice. 17 A simple pendulum of length l and mass m is pivoted to the block of mass M which slides on a smooth horizontal plane, Fig. Physics Lab 9: The Flying Pig – Centripetal Force Section: Name: Purpose To show the net force for a conical pendulum is mv 2 /r. Once the modules are completed, the course ends with. A ball is attached to a string and swung so that it travels in a horizontal circle. pendulum synonyms, Compensation pendulum; compound pendulum; Conical pendulum; and twisting his legs round it in sailor fashion, slipped down eight or ten feet, where his weight gave it a motion not un-like that of a pendulum…. tal plane with its string sweeping the. Conical Pendulum – If a simple pendulum is fixed at one end and the bob is rotating in a horizontal circle, then it is called a conical pendulum. The accel-eration is directed along the radius of the dashed circle and towards the shaft. He realized that the pendulum …. Periodic Motion Lab – The Conical Pendulum Purpose The goals of this lab are to verify that centripetal acceleration is given by a = v2/r and to show that the period of a conical pendulum is given by the theoretical equation: Procedure A small mass is suspended by a cord and set into motion …. Experiment with Conical Pendulum. In the figure below, a simple pendulum is represented at various positions of its motion. We have two contributions: kinetic energy of rotation around the CM and kinetic energy …. Unfortunately, Java cannot plot the motion of the pendulum just by using the angle q – it uses (x, y) coordinates to plot shapes. VIDEO : Conical Pendulum (Motion in a Horizontal Circle) Skip To Content. It is the analogue of the conical motion of a spherical pendulum with a fixed point of suspension. Smart pendulum leveling system One-button operation 530 Ft. motion of the conical pendulum is just the sum of the motion of two pendulums. In conical pendulum the bob does not oscillate back and forth but it moves in a circle. A conical pendulum is a mass attached to a nearly massless string that is held at the opposite end and swung in horizontal circles. Find Physics textbook solutions? 500 Selected Problems In Physics for JE… 877 solutions Selina - Concise Physics - Class 9 1224 solutions Lakhmir Singh, Manjit Kaur - Physics 10. | 2022-05-23T15:19:47 | {
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http://math.stackexchange.com/questions/771771/proving-units-in-a-ring | # Proving units in a ring
Suppose $R$ is a ring with no zero divisors and with identity $1_R$ not equal to $0_R$. Suppose that $a,b$ are in $R$ and that $ab$ is a unit. Prove that $b$ is a unit.
My thoughts: I know a unit is basically a unit that (for this example) would mean $abu = 1_R$ for some nonzero $u$ in $R$. I am really stuck after that. Not seeing a clear path to manipulate the variables to prove b is a unit by itself.
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Well, write it this way: $\;1=u(ab)=(ua)b\implies b\;$ is a unit...:) – DonAntonio Apr 27 at 19:00
@Don That yields only a left inverse $\,c = ua\,$ for $\,b.\,$ But conjugation shows it is a right inverse too - see my answer. – Bill Dubuque Apr 27 at 23:25
As the others have pointed out the calculation $$1=u(ab)=(ua)b$$ shows that $ua$ is a left inverse to $b$. Consider the product $b(ua)$. We have $$ua=1(ua)=((ua)b)(ua)=(ua)(b(ua)),$$ so $$(ua)(1-b(ua))=0.$$ As the ring has no zero divisors this implies that either $ua=0$ or $b(ua)=1$. But if $(ua)=0$, then $1=(ua)b=0$ which is a contradiction. The claim follows.
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+1 for actually using the fact the ring has no zero divisors... – Goos Apr 27 at 19:30
@Goos: Well, the claim is false without something extra, and that was the only piece available :-) All: Sorry about spoiling the problem. I really should have come up with an appropriate hint. – Jyrki Lahtonen Apr 27 at 19:45
yes, but there were two maybe three people claiming "solutions" without using that fact. So +1 for a legitimate proof that it is a unit and not just having a left inverse. – Goos Apr 27 at 20:10
@Jyrki Re: hints. See my answer for one way I often hint at it. It's not easy to give a hint for this without spilling the beans. – Bill Dubuque Apr 27 at 23:05
As discussed in the comments, since $ab$ is a unit then $uab = 1$ for some $u \in R$, so $ua$ is a left inverse for $b$. It remains to show that $ua$ is also a right inverse for $b$, i.e., $bua = 1$. Taking the equation $1 = uab$ and multiplying both sides by $ua$ on the right, we have $$ua = uabua \implies 0 = ua - uabua = ua(1 - bua) \, .$$ Since $R$ has no zero divisors, then either $ua = 0$ or $1 - bua = 0$. But again, $R$ has no zero divisors, so we must have $1 - bua = 0$, hence $1 = bua$. Thus $ua$ is a two-sided inverse for $b$.
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Hint $\$ Conjugate a one-sided inverse $\,bc=1\,$ to the other side via $\ (bc\!-\!1)b\, =\, b(cb\!-\!1)$
- | 2014-12-22T14:26:47 | {
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http://mathhelpforum.com/advanced-algebra/29103-centre-group-z-g.html | # Thread: centre of a group Z(G)
1. ## centre of a group Z(G)
Hey there, i have a question on the center of a group.
The centre Z(G) of a group G is defined by $\displaystyle Z(G) = g \epsilon G: \forall x \epsilon G, xg = gx$
(i) Show that Z(G) is normal subgroup of G
(ii) By considering the Class Equation of G acting on itself by conjugation show that if $\displaystyle |G| = p^n$ ( p prime) then $\displaystyle Z(G) \neq {1}$
(iii) If G is non abelian show that G/Z(G) is not cyclic.
(iv) Decude that any group of order $\displaystyle p^2$ is abelian.
(V) Deduce that a gorup of oder $\displaystyle p^2$ is isomorhpic either to $\displaystyle C_{p2}$ or to $\displaystyle C_p \times C_p$
any hints would be greatly appreicated and ill atempt the question once i have a clear idea of how to do them. consequently, i will post them online once i have worked them out. cheers guyz
2. Originally Posted by joanne_q
(i) Show that Z(G) is normal subgroup of G
Let $\displaystyle x\in \text{Z}(G)$ prove that $\displaystyle gxg^{-1} \in \text{Z}(G)$ for any $\displaystyle g\in G$.
(ii) By considering the Class Equation of G acting on itself by conjugation show that if $\displaystyle |G| = p^n$ ( p prime) then $\displaystyle Z(G) \neq {1}$
The conjugacy class equation says,
$\displaystyle |G| = |\text{Z}(G)| + \sum [G:\text{C}(x)]$.
Where the $\displaystyle x$'s are taken from distinct conjugacy classes of more than one element.
We know that the left hand side is divisible by $\displaystyle p$ so the right hand side is divisible by $\displaystyle p$. Now $\displaystyle \text{C}(x)$, the centralizer, is not whole $\displaystyle G$ because we are picking those $\displaystyle x$, this means $\displaystyle 1\leq |\text{C}(x)| \leq p^{n-1}$. This means $\displaystyle p$ divides the index of $\displaystyle \text{C}(x)$ under $\displaystyle G$. Thus, this proves that $\displaystyle \text{Z}(G)$ is divisible by $\displaystyle p$. Thus, the center is non-trivial.
3. Originally Posted by joanne_q
(iii) If G is non abelian show that G/Z(G) is not cyclic.
Contrapositive is easier. Let $\displaystyle H=\text{Z}(G)$. If $\displaystyle G/H$ is cyclic then there is $\displaystyle aH$ which generates the group $\displaystyle G/H$. Let $\displaystyle x,y\in G$. Note $\displaystyle xH,yH\in G/H$ thus $\displaystyle xH=a^nH$ and $\displaystyle yH=a^mH$. This means $\displaystyle x = a^n z_1$ and $\displaystyle y=a^mz_2$ where $\displaystyle z_1,z_2\in H$. But then $\displaystyle xy = a^n z_1 a^mz_2 = a^{n+m}z_1z_2$ and $\displaystyle yx = a^m z_2 a^n z_1 = a^{n+m}z_1z_2$ because $\displaystyle z_1,z_2$ commute with everything. Thus $\displaystyle G$ is abelian.
(iv) Decude that any group of order $\displaystyle p^2$ is abelian.
By Burnside's lemma (that is (ii)) we have that the center is non-trivial, forming the factor group we have a cyclic group. Thus the original needs to be abelian.
4. Originally Posted by joanne_q
(V) Deduce that a gorup of oder $\displaystyle p^2$ is isomorhpic either to $\displaystyle C_{p2}$ or to $\displaystyle C_p \times C_p$
Let $\displaystyle |G|=p^2$. Pick $\displaystyle a\not = 1$. Form the subgroup $\displaystyle H=\left< a \right>$ if $\displaystyle H = G$ then the group is cyclic and proof is complete. Otherwise choose $\displaystyle b\in G\setminus H$ and form $\displaystyle K=\left< b\right>$. Now $\displaystyle H\cap K = \{ 1\}$ this means $\displaystyle HK = G$*. Also $\displaystyle H,K\triangleleft G$ because the group is abelian. This means $\displaystyle G\simeq H\times K \simeq \mathbb{Z}_p \times \mathbb{Z}_p$.**
*)Because $\displaystyle |HK| = |G|$ by using the fact $\displaystyle |HK||H\cap K| = |H||K|$.
**)Theorem: If $\displaystyle H,K$ are normal subgroups with $\displaystyle H\cap K = \{ 1 \}$ and $\displaystyle HK = G$ then $\displaystyle G\simeq H\times K$.
5. Thanks a lot for the help really appreciate it. makes it easier to understand the subject aswell.
For part (ii), your solution was:
Originally Posted by ThePerfectHacker
The conjugacy class equation says,
$\displaystyle |G| = |\text{Z}(G)| + \sum [G:\text{C}(x)]$.
Where the $\displaystyle x$'s are taken from distinct conjugacy classes of more than one element.
We know that the left hand side is divisible by $\displaystyle p$ so the right hand side is divisible by $\displaystyle p$. Now $\displaystyle \text{C}(x)$, the centralizer, is not whole $\displaystyle G$ because we are picking those $\displaystyle x$, this means $\displaystyle 1\leq |\text{C}(x)| \leq p^{n-1}$. This means $\displaystyle p$ divides the index of $\displaystyle \text{C}(x)$ under $\displaystyle G$. Thus, this proves that $\displaystyle \text{Z}(G)$ is divisible by $\displaystyle p$. Thus, the center is non-trivial.
here is my solution, is it correct aswell?
$\displaystyle G \equiv |Z(G)| (mod p)$ since Z(G) is a fixed point set.
Now $\displaystyle |Z(G)| \equiv p^n(mod p)$, |Z(G)|=0.
So Z(G) has atleast p elements.
6. for part (iii) i believe you have proved the opposite of the question? i.e. G/Z(G) is cyclic and thus abelian...
to prove that it is non cyclic, do all the points you stated have to be contradicted..?
Originally Posted by ThePerfectHacker
Contrapositive is easier. Let $\displaystyle H=\text{Z}(G)$. If $\displaystyle G/H$ is cyclic then there is $\displaystyle aH$ which generates the group $\displaystyle G/H$. Let $\displaystyle x,y\in G$. Note $\displaystyle xH,yH\in G/H$ thus $\displaystyle xH=a^nH$ and $\displaystyle yH=a^mH$. This means $\displaystyle x = a^n z_1$ and $\displaystyle y=a^mz_2$ where $\displaystyle z_1,z_2\in H$. But then $\displaystyle xy = a^n z_1 a^mz_2 = a^{n+m}z_1z_2$ and $\displaystyle yx = a^m z_2 a^n z_1 = a^{n+m}z_1z_2$ because $\displaystyle z_1,z_2$ commute with everything. Thus $\displaystyle G$ is abelian.
7. Originally Posted by joanne_q
For part (ii), your solution was
Have you ever done the conjugacy class equation? There is a way around it if you never done it that way, I think I have an idea of what you might have done.
Let $\displaystyle G$ be a finite $\displaystyle p$-group. Let $\displaystyle G$ act on a non-empty finite set $\displaystyle X$. Then $\displaystyle |X|\equiv |X^G|(\bmod p)$ where $\displaystyle X^G$ is the invariant subset fixed by $\displaystyle G$.
here is my solution, is it correct aswell?
$\displaystyle G \equiv |Z(G)| (mod p)$ since Z(G) is a fixed point set.
Now $\displaystyle |Z(G)| \equiv p^n(mod p)$, |Z(G)|=0.
So Z(G) has atleast p elements.
Let $\displaystyle G$ act on itself by conjugation (i.e. $\displaystyle X=G$ and $\displaystyle g*x = gxg^{-1}$). Also $\displaystyle G$ is a finite $\displaystyle p$-group which fits the above result thus $\displaystyle |G| \equiv |G^G| (\bmod p)$ but $\displaystyle G^G = \text{Z}(G)$ because that is the subset left fixed under conjugation. Thus, $\displaystyle |G| \equiv |\text{Z}(G)|(\bmod p)$ which means the center needs to be divisible by $\displaystyle p$, i.e. it cannot be trivial.
8. Originally Posted by joanne_q
for part (iii) i believe you have proved the opposite of the question? i.e. G/Z(G) is cyclic and thus abelian...
to prove that it is non cyclic, do all the points you stated have to be contradicted..?
No, there is no contradiction argument. I proved the contrapositive statement. | 2018-04-22T11:21:14 | {
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http://zveg.1upload.de/r-venn-diagram-6-sets.html | # R Venn Diagram 6 Sets
Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1. On a mission to transform learning through computational thinking, Shodor is dedicated to the reform and improvement of mathematics and science education through student enrichment, faculty enhancement, and interactive curriculum development at all levels. The circles or ovals represent events. Venn diagrams with three curves are used extensively in various medical and scientific disciplines to visualize relationships between data sets and facilitate data analysis. In the offset portions of the circles, the student lists those traits whch differ between the two items. Here is a Venn diagram of the sets S, T and V We can see, for example, that the element "casey" is in both set S and set T, but not in set V. To create a simple Venn diagram, you can just pass in the list with the specified set and overlap values into the venneuler() function. This Venn diagram shows all possible intersections of five sets. The Unattainable Triangle Good Good Fast Fast Cheap Cheap More expensive. Venn diagram, also known as Euler-Venn diagram is a simple representation of sets by diagrams. Entity Relation. We can represent this as, 01. Solution to Example 1. The remaining numbers in A are 7 and 9. The red part of each Venn diagram is the resulting set of a given. pseudovenn(dataset_dict, **kwargs) which plots a Venn-like intersection of six circles (not all intersections are present in such a plot, but many are). 5 circles representing 5 data sets). See [14] for a list of open problems related to Venn diagrams. In R, the VennDiagram package is the best option to build one. number and integers have no elements in common because the whole. Problem-solving using Venn diagram is a widely used approach in many areas such as statistics, data science, business, set theory, math, logic and etc. In the second example in the above Venn diagram, Set A is totally contained within set B How can we explain this situation? Suppose that sets A and B contain the following members: set A = {1,2} set B = {1,2,3,4,5,6,7,8} All members of set A are also members of set B. Let $$\text{C} =$$ student belongs to a club and $$\text{PT} =$$ student works part time. For example, the below six-way Venn diagram shows the distribution of shared gene families among six genomes, taken from D’Hont et al. A = { 2, 4, 6, 8 }. Repurposing to Use Segments. We will do this for the first column of the Venn diagram figure given previously. Venn Diagrams (H) - Version 2 January 2016. Venn Diagram Horz 03 03 - Venn Diagram is a high-resolution transparent PNG image. A Venn diagram, also called primary diagram, set diagram or logic diagram, is a diagram that shows all possible logical relations between a finite collection of different sets. , ISBN-10: 0321867327, ISBN-13: 978-0-32186-732-2, Publisher: Pearson. fill (Optional) Fill color of the sets. Some people want to create quantitative Venn diagrams, where the overlap (and perhaps the relative size of the circles) is data-driven. Venn diagrams are a great way to visualize the structure of set relationships. The Venn diagram above is an example. How many students are not involved in either band or sports? 2. 5 overlapping fragmented circle shapes. Venn diagram, graphical method of representing categorical propositions and testing the validity of categorical syllogisms, devised by the English logician and philosopher John Venn (1834–1923). 8-6 currently have issues preventing their full use in KNIME. (a) Label the Venn diagram to show the sets A and B where n(A B) = 18. TimeLine/Curve Chart. Venn Diagram Set Diagram. Thank you for this. 4 - Page 91 12 including work step by step written by community members like you. Typically overlapping shapes, usually circles, are used, and an area-proportional or. Manage My Favorites. The term Venn diagram is not foreign since we all have had Mathematics, especially Probability and Algebra. A Venn diagram represents a set as the interior of a. 3, with aesthetically pleasing results. First step: Install & load “VennDiagram” package. Linguistics. 9 it is also used to pass additional aesthetics parameters for the ggplot2 graphics. For example, the “days of the week” is a set and Monday is a member of this set. This is similar to the logical “and” Venn Diagrams Venn Diagrams use topological areas to stand for sets. venn: Draw a Venn Diagram with Five Sets in VennDiagram: Generate High-Resolution Venn and Euler Plots rdrr. Venn diagrams don’t usually explain anything, they just represent. Note that many edges overlap, so the diagram is infinitely intersecting. 4-set means I should be able to specify 4 sets. newpage() # make new page for plot. The Venn diagram above is an example. What goes where? What do the plants, flowers, and trees have in common? Students use Venn diagrams to find out!. Chapter 4 Probability and Venn diagrams 2 1 The Venn diagram shows the whole numbers from 1 to 12. For example, the following Venn diagrams is based on 4 sets (I, II, III and IV), but is not scaled: (it was generated using the R package VennDiagram with the code:. VennDiagram-internal. Start studying Sets and Venn Diagrams. Venn Diagrams. 43 ewlength\@[email protected] \@[email protected] The height of the entire Venn diagram. Create beautiful venn diagrams using this tool and download as image (png or jpg). A set is a collection of objects and its objects are called members. Draw the figure below. These Venn diagram word problems worksheet pdfs feature two sets representing the quantities of the data. All of the number Page 6/25. Althoughitcontainsallthedisjointintersections,thevisualizationof interactingcharacteristicsisabsent(13and4872sharethesamearea;thetotalsizeof“Poplar”isthelargest, however,thetotalareaofitisthethirdsmallest);in(d),diagramisdepictedbydifferentlyshapedtriangles. Using Venn Diagrams in Set Theory. I think there should be 2 sets of numbers. Nov 19, 2013 - Venn diagrams began as schematic devices used in logic theory to represent set collections & their respective relationships. Note that many edges overlap, so the diagram is infinitely intersecting. In a class there are:. From the above Venn diagram, what is the set S ∩ T? answer choices {casey, drew, jade, glen} {alex, casey, drew,hunter} {casey, drew} {drew, jade} Tags: Report Quiz. They have the incredible harmonies as if they too were born to sing together. Please refer to the Jupyter notebook for demos and a brief explanation of the interface; a more complete documentation. Venn diagrams show + - = underneath. Venn-Euler diagrams The combination of rectangles and circles are called Venn-Euler diagrams or simply Venn-diagrams. ER Diagram stands for Entity Relationship Diagram, also known as ERD is a diagram that displays the relationship of entity sets stored in a database. A complete Venn diagram represents the union of two sets. Currently they can only be accessed by passing the parameters s or likesquares to the low level creation function. For four sets, it is getting more difficult. If the assessment focus is to interpret a Venn diagram: Ask questions about the similarities and differences that the Venn diagram illustrates. They're also an example of a technique that works very well for a particular purpose, but that entirely fails outside its well-defined scope or when the number of sets gets too large. For the simplest Venn diagram you draw two intersecting circles. The union operations are reflected to the Venn diagram. We can represent this as, 01. Moreover, union operations between sets can be made. Inside the foldable I have students explain what element would be included given the Universal Set, Set A, and Set B. be the set of windy days, W R. ii) in set A but not in set B. a) Illustrate these sets using a Venn diagram. If we have two or more sets, we can use a Venn diagram to show the logical relationship among these sets as well as the cardinality of those sets. v) in set A and set B. Venn Diagrams are basically a representation of different elements in the form of circles. Probabilities from Venn diagrams: True or False & Matching Questions. Moreover, union operations between sets can be made. Perform the below mention boolean algebraic operation for the given set of elements. In this figure, the big rectangle shows the universal set $S$. We use circles to represent the sets, and enclose our diagram in a rectangle. One can solve counting problems involving unions and intersections of sets also with the help of Venn diagrams. The "Cylinder Venn Diagram" below gives a clear representation of different regions of the British Islands. You can also move and resize the circles by dragging and dropping. 96 like wine A, 93 like wine B, 96 like wine C, 92 like A and B, 91 like B and C, 93 like A and C, 90 like all three wines. While there are more than 30 symbols used in set theory , you don't need to memorize them all to get started. It is easy to make a Venn diagram for three sets. It consist of overlapping circle(s) in a box. These diagrams depict elements as points in the plane, and sets as regions inside closed curves. A Venn diagram (also sometimes also called primary diagram or set diagram) is a diagram that depicts all possible logical relations between a collection of sets. A number of variants on the squares type are implemented. These are placed inside A, but outside B. Venn diagrams are fairly intuitive and best learned through examples. Show sets F and G on a Venn diagram. This is an extremely important tool in logical analysis of business and scientific concepts. Draw a Venn diagram simply showing the sets of male and female dogs. I’ve done this one for you. Maui Math Circle Session 5 Venn Diagram Page 2 ! Problem!Set:!Draw!a!Venn!Diagram!for!each!problem. The Venn diagram shows information about the choices the guests made. This handy pack contains 3 different Venn diagram templates to help students explore probability, statistics and representation of data. Venn diagrams for Sets. The intersection of events A and B, written as P(A ∩ B) or P(A AND B) is the joint probability of at least two events, shown below in a Venn diagram. A Venn diagram uses overlapping circles or other shapes to illustrate the logical relationships between two or more sets of items. They may be used by those companies to build a. Venn diagrams are very useful in visualizing relation between sets. Contributed by: Marc Brodie (Wheeling Jesuit University) (March 2011). Free Venn diagram with 5 circles for PowerPoint. The VennDiagram package allows to build Venn Diagrams thanks to its venn. Via , users can specify additional parameters, mainly for the outer borders of the sets, as specified by par () , and since version 1. • Each area of a Venn diagram. Compute classification counts and draw a Venn diagram. Worksheet 2. Hint: You will likely need to use functions described in the setdiff help page to do this. The similar concept applies for venn diagrams with three sets. As we know set is the collection of unique things which are called the elements of the set. If they overlap it means that there are some items in both sets. This can be cumbersome I agree. What is a Venn Diagram? A Venn diagram is a diagram that shows the relationship between and among a finite collection of sets. A s imple symmetric Venn diagram consisting of five ellipses was given in [5]. Set diagram. On a mission to transform learning through computational thinking, Shodor is dedicated to the reform and improvement of mathematics and science education through student enrichment, faculty enhancement, and interactive curriculum development at all levels. A Venn diagram usually looks like the on underneath. In particular, Venn Diagrams are used to demonstrate De Morgan's Laws. CRAN - Package VennDiagram A set of functions to generate high-resolution Venn and Euler plots. A Venn diagram is a simple illustration that uses ovals to picture the universe of data that an analysis begins with and the subsetting, unions, and intersections If themes represent set-based operations, then a Venn diagram is a perfect way to document the themes that a subrelease will support. D/V Alpha/Beta Series 3Ø WIRING DIAGRAMS Diagram DD1. Shown below is a 6-Venn diagram formed entirely from curves drawn from axis-aligned edges. ∩: Intersection of two sets. Use it for drawing Venn and Euler diagrams. Venn Diagram Set Diagram. This is where entities that have all the qualities of the overlapping sets. Therefore, set A is a subset of Set B. A venn diagram makes a really good work to study the intersection between 2 or 3 sets. ID: 1213502 Language: English School subject: Math Grade/level: Grade 6 Age: 9-11 Main content: Sets Other contents: Venn Diagrams Add to my workbooks (12) Download file pdf Embed in my website or blog. There are two versions of the final poster, showing examples. This is an extremely important tool in logical analysis of business and scientific concepts. It has infinitely many members, including 2, 4, 6, 8, and so on. For example, the “days of the week” is a set and Monday is a member of this set. These Venn diagram word problems worksheet pdfs feature two sets representing the quantities of the data. It seems VennDiagram can not work on my data. L1S1 Name : Score : Venn Diagram 1) A B U 1 50 7 30 12 35 3 22 9 8 13 10 2 15 6 28 21 B' A U B A' B U A B U z f t a d q r u y x v k j o g i h p c e b m 3) A' A U B' A' B' U A B U June March October September January May July August 2) (A U B)' A A B U 4) A B U. Overlaps between these circles represent the intersection between the two sets. You Might Also Like The equilibrium membrane potentials to be expected across a membrane at 37 ∘ C, with a NaCl concentration of 0. However, making such a worksheet is a tedious task. However, the use of Venn diagrams in the field of statistics has been quite limited. Venn Diagram, also called Primary Diagram, Logic Diagram or Set Diagram, is widely used in mathematics, statistics, logic, computer science and business analysis for representing the logical relationships between two or more sets of data. And, venneuler may have generated a wrong Venn diagram to me. Overlapping areas indicate elements common to both sets. 4 Visualizing with Venn – A Solidify Understanding Task Creating Venn diagram’s using data while examining the addition rule for probability (S. He represented these relationships using diagrams, which are now known as Venn diagrams. To explain, we will start with an example where we use whole numbers from 1 to 10. If they overlap it means that there are some items in both sets. You are well known about the triangles that they are having three sides. The basic Venn diagram used in presentations shows two partially overlapping shapes, usually circles or ovals, and text to show what belongs to only one shape and what is common to both shapes. One good method to test quickly syllogisms is the Venn Diagram technique. In the UCAT, you will either have to interpret Venn diagrams or you will be given a piece of text and the answers will contain different Venn diagrams and you have to say which one represents the text. _____ _____ f. Show sets F and G on a Venn diagram. A while ago I wrote a small library for displaying Venn and Euler diagrams when trying to learn Javascript. Cards #25– 26 ask students to shade in the described region. Correct-Skipped. I wish there were some magic where you could get a sharp, durable cutter quickly. 🤔 Find out what you don't know with free Quizzes 🤔 Start Quiz Now!. 01 M on the left, given the following conditions. Your browser doesn't support canvas. We can sometimes use partial information about numbers in some of the regions to derive information about numbers in other regions or other sets. B’∩C DIAGRAM 2 2. 6 set Venn diagram with limited overlap. A 6-set venn diagram is never going to look very nice or be very easy to interpret, but here is an R package that can do up to 9 sets: https However depending on what you want/preferences you might want to choose a different package, such as Vennerable, VennDiagram , Venneuler, etc This link. Venn diagrams are also an effective tool for illustrating comparisons and commonalities. In this question , we have to find the region that is represented by the intersection of sets A and B. com/kaz_yos/venn library(VennDiagram) ## Warning: package 'VennDiagram' was built under R version 3. Venn diagrams show + - = underneath. Its very difficult to read and understand venn diagram. In this example I create a Venn diagram from product-ownership data from a survey. The following examples should help you understand the notation, terminology, and concepts relating Venn diagrams and set notation. The region inside the curve represents the elements that belong to the set, while the region outside the curve represents the elements that are excluded from the set. Follow us and share your feedback on Twitter, Reddit, Facebook and on our forums. For example, Figure 2 shows the primary diagram about sets A and B. Universal sets 6. using venn diagram , show that (A-B) , (A intersect B) and (B-A) are disjoint sets , taking A= {2,4,6,8,10,12} and B = {3,6,9,12,15} Priyanka Kedia, Meritnation Expert added an answer, on 24/8/14. This is a venn diagrams pratice exercise test- Click the START button to begin. 01 M on the left, given the following conditions. Venn diagrams are another way of presenting probability information. The universal setξ= A∪B ∪C. Venn diagrams are used to show logical relations between sets. com/techreports/2000/HPL-2000-73. Example: Constructing a Venn diagram for Three Sets continued Now determine the numbers that go in region VI. 140 like tea, 120 like coffee and 80 like both tea and coffee. This poster is an updated version of a previous Venn Diagram paper I presented in 2008, and the improvements are. 1 Draw a Venn diagram representing the relationship between isosceles triangles (I), right-angled triangles (R) and equilateral triangles (E). Observe the given Venn diagram and write the following sets. Venn Diagram. Counting & Venn Diagram Part: The counting principle, venn diagram, ven diagram formula, Sets, Venn Diagrams & Counting - Arizona State University, Venn diagram Unknown The counting Principle: If two jobs need to be completed and there are M ways to do the first job & N ways to do the second Job, then there are M*N ways to do one job followed. INTERSECTION OF SETS. The Venn diagram in kindergarten? Yes! Venn diagrams can be used effectively by our youngest students. Make Venn Diagram in R with Correctly Weighted Areas Venn diagrams are incredibly intuitive plots that visually display the overlap between groups. More Results. N, the set of natural numbers, and P, the set of positive integers D. Answer the word problems, once you have read and analyzed the three-set Venn diagrams displayed here. You Might Also Like The equilibrium membrane potentials to be expected across a membrane at 37 ∘ C, with a NaCl concentration of 0. The circle intersections illustrate qualities shared by the overlapping datasets. Members receive unlimited access to 49,000+ cross-curricular educational resources, including interactive activities, clipart, and abctools custom worksheet generators. Tip: Always start filling values in the Venn diagram from the innermost value. This kind of diagram serves as a graphical representation of how one item is particularly similar or linked to the next. Drawing Venn Diagrams. svg/434px-Venn_diagram_showing_Greek. It is given that n(P) = 165,n(Q)=105 and n (Q) = 95. This is a 5 Circle Venn Diagram template. In the Venn diagram, describe the overlapping area using a complete sentence. This class assumes you are already familiar with diagramming categorical propositions. Five-set Venn diagrams also require the use of ovals or you'll need to overlay a three-set Venn with a recursive curve. ca/~cos/venn/VennTriangleEJC. For ex-ample, Figure 4. Here the set is. Based on their sides and the angles, it is classified into different types. As noted by Henderson, symmetric Venn diagrams with n curves canno t exist for values of n that are composite. A Venn diagram, also called primary diagram, set diagram or logic diagram, is a diagram that shows all possible logical relations between a finite collection of different sets. In the SVG file, hover over a triangle to select it. Venn diagram symbols. A common use for Venn diagrams is the identification of shared and unique elements in different sets. Horizontal mode is identical to the vertical mode The Nested Venn diagram shows unique and shared relationships of eight sets by inlaying four unique-shared diagrams into the other four sets'. Other interactions are also available, such as color changing and export diagram in SVG and PNG format. A complete Venn diagram represents the union of two sets. Rahim described the set as follows: • M = {all of the foods he eats} • D = {his favourite desserts}. Often, they serve to graphically organize things, highlighting how the items are similar and different. The same procedures using set B completes region III. ii) in set A but not in set B. It then draws the result, showing each set as a circle. Try out the latest Rainbow Six updates on the Test Server and earn and exclusive charm through the BugHunterProgram. and 3; together have size 25, so the overlap between W and R is 10. I need to draw venndiagram for 6 sets by R. refer to the name plate data for correct connection For delta ( ) wired motors L1 L2. Calculate and draw custom Venn diagrams. They can also be useful for showing how subsets within a larger set are differentiated. Grouping and collection of things, in mathematical terms, is known as a set. The cardinality of the sets and intersection sets is represented by their corresponding circle (polygon) sizes. R # Imagine you have more than two sets and you would want to find the overlapping elements in different sets # and you would like to see the overlap using VennDiagram. 我想画一个维恩图，但是用R(维恩图库)它被限制在5个元素上。 Do you know how can draw a 6-sets venn diagram ?. 4 Classroom Task: 9. If anything it helps as a language to communicate this. Draw a Venn diagram showing the relationships. 1) A 2) B 3) A ∪ B 4) U 5) A' 6) B' 7) (A ∪ B)'. These diagrams depict elements as points in the plane, and sets as regions inside closed curves. On a mission to transform learning through computational thinking, Shodor is dedicated to the reform and improvement of mathematics and science education through student enrichment, faculty enhancement, and interactive curriculum development at all levels. For example, we can represent the event "roll an even number" by the set {2, 4, 6}. Our video lesson will also guide you with the steps to draw a Venn diagram. How to create a venn diagram with 6 sets of numbers whole,counting,integers,rational,irrational,real. practical usefulness of Venn diagrams diminishes but interesting mathematical questions arise. An ER diagram shows the relationship among entity sets. C D 4 10 8 6 1 2 7 3 5 9 A number is chosen at random from. 30 students are asked if they have a dog or cat. Create beautiful venn diagrams using this tool and download as image (png or jpg). See more ideas about venn diagram, diagram, venn diagram template. They are extensively used to teach Set Theory. Example 6: Shade the portion of the Venn diagram that represents the given set. A Venn diagram, named after John Venn, is a diagram representing all possible logical relations between a finite collection of different sets. A diagram that shows sets and which elements belong to which set by drawing regions around them. Venn Diagrams A Venn diagram is a drawing in which sets are represented by geometric figures such as circles and rectangles. A Venn diagram (also known as a set diagram or logic diagram) is a diagram that shows all possible logical relations between a finite collection of different sets. In business, Venn diagrams are used to compare products, processes, services, and pretty much anything that can be represented in sets. Venn diagrams can be used to help find probabilities when events are not mutually exclusive. Venn diagrams are used to determine conditional probabilities. vennCounts(x, include="both") vennDiagram(object, include="both", names=NULL Each column of x corresponds to a contrast or set, and the entries of x indicate membership of each row in each set or alternatively the significance. Sets and Venn Diagrams October 26, 2012 What is a Set? Mathematics, at its very core, can be described ENTIRELY in terms of sets. They show all of the possible mathematical or logical relationships between sets (groups of things). s} B = {s,t,r,s} C = {t,s,t,r} D = {s,r,s,t} a) A and B b) A and C c) B. 3, with aesthetically pleasing results. A typical venn diagram is shown in the figure below: In the figure, set A contains the multiples of 2 which are. The R visualization code provided in this Power BI desktop file will take a dynamic set of columns (based on the values you add in the fields pane), perform the overlap analysis, and display the diagram. Through the dark background, you may feel serious and formal. Venn diagrams are very useful constructs made of two or more circles that sometimes overlap. Introduced by Venn (1880), the Venn diagram has been popularized in texts on elementary logic and set theory (e. A Venn diagram is a graphical way of representing the relationships between sets. Nice Looking Five Sets Venn Diagrams Stack Overflow. It becomes very hard to read with more groups than that and thus must be avoided. The best way to explain how the Venn diagram works and what its formulas show is to give 2 or 3 circles Venn diagram examples and problems with solutions. In the diagram below, there are two sets, A = {1, 5, 6, 7, 8, 9, 10, 12} and B = {2, 3, 4, 6, 7, 9, 11, 12, 13}. gif 404!396 pixels. A Venn diagram shows all possible logical relations between data sets. If 30 play cricket and 20 both, the correct way of representing above by Venn diagram is. ” The first step is to list the twelve months of the year:. I wish there were some magic where you could get a sharp, durable cutter quickly. http://positivemaths. Create beautiful venn diagrams using this tool and download as image (png or jpg). The Venn diagram shows information about the choices the guests made. Other interactions are also available, such as color changing and export diagram in SVG and PNG format. Venn Diagram in case of three elements. ER Diagrams contain different symbols that use rectangles to represent entities, ovals to define attributes and diamond shapes to represent. Alternative Method (Using venn diagram) : Step 1 : Venn diagram related to the information given in the question Let M and S represent the set of students who like math and science respectively. Let the universal set U be all the elements in sets A, B, C and D. 4) Some numbers are Real numbers. OR FT from their Venn diagram If no marks in (b) award SCI for 5, 2 and 2 or identifying the correct regions by listing the correct numbers Penalise incorrect natation once and 3. I will quickly show you how to draw a Venn diagram composed of three distinct datasets in R. R = {x| x is a factor of 24} S = { } T = {7, 9, 11} Summary In this lesson, you learned about sets, subsets, the universal set, the null set and the cardinality of the set. They are extensively used to teach Set Theory. This is where entities that have all the qualities of the overlapping sets. Solved Examples. VENN DIAGRAM. All of the number Page 6/25. , the set of all elements being considered in a particular discussion. RE: In sets and Venn Diagrams, what do the symbols, Ø, Є and n mean? For example in the question: Q={2,4,6,8,10} and R={5,10,15,20}. To understand the right way to use Venn Diagrams, it helps to go back in time. The union operations are reflected to the Venn diagram. A Venn diagram (named after mathematician John Venn in 1880) is a method used to sort items into groups. Default is 0. You should habe 2 sets of numbers - say 5, 10, 15, 20, 25, 30 and another set that could be 20, 25, 30, 35, 40, 45. • Use black ink or ball-point pen. In these diagrams, the universal set is represented by a rectangle, and other sets of interest within the universal set are depicted by oval regions, or sometimes by circles or other shapes. While there are more than 30 symbols used in set theory, you don’t need to memorize them all to get started. mapping: the aesthetics mapping. The following examples should help you understand the notation, terminology, and concepts relating Venn diagrams and set notation. Watch our Venn diagram chapter video to get a detailed explanation of the significance of a Venn diagram in sets. These diagrams make use of circular shapes in various colors which appeal visually to any audience and can easily represent the size, value and relationship between different sets of information. The upper diagram to the. Venn and Euler diagrams in (c) and (d) both include six data sets: in (c), irregular polygons are drawntoillustratesixwoodyspecies. It is easy to understand the union and intersection of sets with the help of Venn diagrams. Use the Venn diagram below to fill in the missing statement. Venn diagrams use circles to represent sets and to illustrate the relationship between a finite collection of different sets. Answer the following using Venn Diagrams. Polish your personal project or design with these Venn Diagram transparent PNG images, make it even more personalized and more attractive. Play Venn diagram quizzes on Sporcle, the world's largest quiz community. Single speed motors. Aside from the innermost face and the outermost face, a rotationally symmetric $$n$$-Venn diagram can be partitioned into $$n$$ congruent clumps, each of size $$(2^n-2)/n$$; in this case, we call the clump a cluster—it is like a. Example 6 Provide a Venn diagram for the following syllogism: Some M are P All M are S. diagram function from VennDiagram package. …The first one, on the top left, is Hacking Skills. Coordinates of vertices are from http://sue. This is the currently selected item. Place Value Roald Dahl Themed Year 2 PowerPoint. As with many other diagrams in these pages, regions are coloured by weight. We can show the Universal Set in a Venn Diagram by putting a box around the whole thing: Now you can see ALL your ten best friends, neatly sorted into what sport they play (or not!). ” The type of three circle Venn Diagram we will need is the following: Image Source: Passy’s World of Mathematics. Sets within the universal set are usually represented by circles. Write down the numbers that are in set. Venn diagram, also known as Euler-Venn diagram is a simple representation of sets by diagrams. The trick is to make them user-friendly, hands-on, and developmentally appropriate as a tool even kindergarten students can use with ease. There is also some new language to be learnt: Symbols that represent AND (the intersection of sets), OR (the union of sets) etc. "A Venn diagram or set diagram is a diagram that shows all possible logical relations between a finite collection of sets. io diagram, select Venn in the left hand side of the template diagram. In this figure, the big rectangle shows the universal set $S$. You then have to use the given information to populate the diagram and figure out the remaining information. 9 it is also used to pass additional aesthetics parameters for the ggplot2 graphics. Also, it’s not possible to show more than 6 sets in a Venn diagram. Children need to think about how to sort something according to the two rules. From the adjoining Venn diagram, find the following sets. ER Diagrams contain different symbols that use rectangles to represent entities, ovals to define attributes and diamond shapes to represent. Venn Diagrams are very useful for visualizing the relationships between groups. Compare and contrast your data. In particular, Venn Diagrams are used to demonstrate De Morgan's Laws. These diagrams depict elements as points in the plane, and sets as regions inside closed curves. Multiplication rule. Venn diagrams are plots used to graphically display intersections between two or more groups. iv) in set B but not in set A. Venn diagrams are particular cases of Euler diagrams showing all possible combinations. A Venn diagram (also called primary diagram, set diagram or logic diagram) is a diagram that shows all possible logical relations between a finite collection of different sets. In this example I create a Venn diagram from product-ownership data from a survey. I have following data need to be visualized by a venn diagram Total counts: 1668 Counts for group A: 62 Counts for group B: 24 (Group B is a subgroup of group A, all counts in group B are included in group A) Counts for group C: 267 (including group A, but excluded group B) How to display the proportion(%) of overlap between each other by venn diagram? request the graph display 4 ways (total. newpage() # make new page for plot. An ER diagram shows the relationship among entity sets. These are standard Venn diagrams for comparing and contrasting two items. 1 By Juan Carlos Oliveros BioinfoGP, CNB-CSIC: 1. In this tutorial, I'll show how to plot a three set venn diagram using R and the ggplot2 package. Sal uses a Venn diagram to introduce two-way frequency tables. In this figure, the big rectangle shows the universal set $S$. You will need a two circle Venn Diagram for your answer. Results: We present a visualization approach for set relationships based on Venn diagrams. ca/~cos/venn/VennTriangleEJC. First download the Venn diagrams in excel zip file from here [xls version here]. Abstract: The Venn Diagram technique is shown for typical as well as unusual syllogisms. The section where the two sets overlap has the numbers contained in both Set A and B, referred to as the intersection of A and B. A Venn diagram, also called primary diagram, set diagram or logic diagram, is a diagram that shows all possible logical relations between a finite collection of different sets. Once we understand how to read the Venn Diagram we can use it in many applications. A subreddit for Venn diagram enthusiasts. Functions in venn. Find: a P(B) b P(A B) c P(A B) 2 The Venn diagram shows the whole numbers from 1 to 10. Venn Diagram Word Problems - Three Sets. 6 for all Venn diagrams with up to five sets, and it automatically decreases to 0. C B A Example The following Venn diagram shows the number of elements in each region for the. none of the above ____ 6. How many girls are on both the. Venn Diagrams (H) - Version 2 January 2016. 43 ewlength\@[email protected] \@[email protected] The height of the entire Venn diagram. Guide students toward an understanding of the Venn diagram by letting them physically manipulate hoola. Some of the worksheets for this concept are Grade 3 questions venn diagrams, Venn diagrams, Venn diagram l1s1, Ss, Venn diagram, Write details that tell how the subjects are different in, Chapter 3 1 venn diagrams, Math 211 sets practice work part 1 shade the region. Using Venn Diagrams in Set Theory. Venn diagrams show + - = underneath. Remember that, To maximize overlap, Union should be as small as possible; Calculate the surplus = n( A) +n(B) +n(C)-n(A or B or C). Follow us and share your feedback on Twitter, Reddit, Facebook and on our forums. We can represent this as, 01. vennCounts produces an object of class "VennCounts". 1 #13 To shade the set we need to compare the Venn diagram for A with the Venn diagram for B′, and bear in mind the meaning of union. They are much easier to take in at a glance than truth tables (for example) which are equivalent, but boring and difficult to interpret quickly. Any values that belong to more than one set will be placed in the sections where the circles overlap. I have following data need to be visualized by a venn diagram Total counts: 1668 Counts for group A: 62 Counts for group B: 24 (Group B is a subgroup of group A, all counts in group B are included in group A) Counts for group C: 267 (including group A, but excluded group B) How to display the proportion(%) of overlap between each other by venn diagram? request the graph display 4 ways (total. 01 M on the left, given the following conditions. Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1. See full list on mathsisfun. In this section we introduce the ideas of sets and Venn diagrams. 2 shows an example of a Venn diagram. The Venn diagram of A and B looks something like this: Here are some examples of set operations and their Venn Diagrams: A∪B A∩B A∪B (A∪B) A three-set Venn diagram looks. DIFFERENCE OF SETS. A = { set of even numbers between 0 and 10} 02. Select number of sets and update the venn diagram parameters Venn diagram maker tool is completely free to use. Note that the total of the entries is 52, and the complete set is denoted by ε. The above diagram is called Venn Diagram where there are two sets A and B and U is the universal set. Sets are shown as regions inside circles or other closed curves, and common elements of the sets are shown as intersections of these circles. number and integers have no elements in common because the whole. (ii) Sketch all possible Venn Diagram if csbsa. If a is an element of the set A then we write a 2 A,ifa is not an element of a set A,thenwe write a/2 A. Worked example J. Using Venn diagrams allows children to sort data A Venn diagram is when the two sorting circles overlap in the middle. I’m going to have to do another post on Venn Diagram activities, but it inspired me to put together some printable blank Venn Diagram templates that I know will get used!. After coming upon these answers: Create a Venn Diagram, How to plot Venn diagrams with Mathematica?, which I Stack Exchange Network Stack Exchange network consists of 176 Q&A communities including Stack Overflow , the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. It's just a rough guide from my limited knowledge on the hololive girls (would include guys too but there's limited space and I only know a few of them). 140 like tea, 120 like coffee and 80 like both tea and coffee. We use circles to represent the sets, and enclose our diagram in a rectangle. The section of the circles which overlap contain numbers that have properties pertaining to both of the circles. All other elements represent the consonants. Some, but not all, students in the school take swimming. Editable graphics with text placeholder. • 8 have a dog, but not a cat. Remark 1It is easy to manipulate Venn diagrams. n (M′ ∪ C′) = _____ b. A Venn diagram could be used, for example, to compare two companies. A Venn diagram is a mathematical illustration that shows all of the possible mathematical or logical relationships between sets. In a school of 320 students, 85 students are in the band, 200 students are on sports teams, and 60 students participate in both activities. Instructions. Venn Diagrams. Sort by one or two conditions. This means we start out with the second premise: All M are S. Media in category "6-set Venn diagrams" The following 7 files are in this category, out of 7 total. Venn Diagram and Shape Sorting Lesson Plan. Linguistics. A set of functions to generate high-resolution Venn and Euler plots. This particular geometric configuration for five category Venn diagrams was popularized by Adrian Dusa's venn package for R. Sets are represented by circles included in a rectangle that represents the universal set, i. Five-Set Diagrams. 5 Three-Set Venn Diagram. This can be cumbersome I agree. Stacked Venn diagram. Venn diagrams are listed in the probability section of the new Key Stage 3 Programme of Study and occur in GCSE reference P6 which involves representing sets and combinations of sets using Venn diagrams, as well as through the use of tables and grids. Venn's diagrams drawing tool for comparing up to four lists of elements. Note that although there are no elements shown inside. vi) in set A or set B. The chart is also known as 'Set diagram' or 'Logic diagram'. Now when you try to open the file, you must enable macros (in excel 2007, you may want to set the security to low and then reopen the file) 3. It generally consists of a box that represents the sample space S together with circles or ovals. It’s easy to create Venn and Euler diagrams in draw. Golf a Venn Diagram generator. The description of some sets is given and you are asked to draw a Venn diagram to illustrate the sets. Tried the R approach but then ran into this problem (fresh from the Console) WARN Table to R 0:57:35 R Version 3. Using Venn diagrams allows children to sort data A Venn diagram is when the two sorting circles overlap in the middle. Set theory makes wide use of Venn diagrams. I need to draw venndiagram for 6 sets by R. Solved Examples. The second set of branches represents the second draw. In a typical Venn diagram, 2 or more sets (each containing various elements) are represented as circles overlayed onto each other. Then each set in the problem is represented by a circle. In R, the VennDiagram package is the best option to build one. , 10 is a multiple of 3 and 5. VennTure can generate six-sets Venn diagrams with a graphic user interface (GUI), yet it consumes large amounts of memory and has low computational efficiency. This diagram on political parties process is a good example of a three-set Venn diagram, with text inside the circles to help further explain each topic, and the center being the "sweet spot" where. Probabilities from Venn diagrams: True or False & Matching Questions. A Venn diagram is a chart that compares two or more sets (collections of data) and illustrates the differences and commonalities between them with overlapping circles. A Venn diagram (also called primary diagram, set diagram or logic diagram) is a diagram that shows all possible logical relations between a finite collection of different sets. A 6-set venn diagram is never going to look very nice or be very easy to interpret, but here is an R package that can do up to 9 sets: https However depending on what you want/preferences you might want to choose a different package, such as Vennerable, VennDiagram , Venneuler, etc This link. These are diagrams that make use of geometric shapes to show relationships between sets. This means that as the number of contours increase, Euler diagrams are typically less visually complex than the equivalent Venn diagram, particularly if the number of non-empty intersections is small. You should habe 2 sets of numbers - say 5, 10, 15, 20, 25, 30 and another set that could be 20, 25, 30, 35, 40, 45. In a group of 40 students 6 are left-handed, 18 have size 8 feet and 2 are left-handed with size 8 feet. Venn diagrams are used to determine conditional probabilities. A Venn diagram could be used, for example, to compare two companies. (a) How many guests had custard? (b) How many guests had ice cream and custard? (c) How many guests went to the wedding? 5. We can now use the commands in this package for generating Venn diagrams. A s imple symmetric Venn diagram consisting of five ellipses was given in [5]. confusing as a presentation tool because of the number of. 3 shows the difference. This lesson covers how to use Venn diagrams to solve probability problems. A triple Venn diagram is a diagram that consists three intertwined circles that represents thoughts and the relation of each from one another. 9 Venn diagrams Venn Diagrams are the diagrams which represent the relationship between sets. First step: Install & load “VennDiagram” package. Diagram 15 is a Venn diagram that showsthe result of a survey carried out by a company on a group of 255 people about their favourite handphones. In Venn diagrams the curves are overlapped in every possible way, showing all possible relations between the sets. A Venn diagram consists. N, the set of natural numbers, and I, the set of integers B. Here is a famous example: a six-set venn diagram published in Nature that shows the relationship between the banana's genome. 96 like wine A, 93 like wine B, 96 like wine C, 92 like A and B, 91 like B and C, 93 like A and C, 90 like all three wines. A Venn diagram shows all possible logical relations between different sets or groups of data. A Venn diagram is an illustration that uses circles to show the relationships among things or finite groups of things. Find: (i) (ii) (iii) Answers: (i) (ii) (iii) Question 7: If , , and. Creates a Venn diagram with five sets. Since all numbers in set A have been placed, there are no numbers in region I. Some of the worksheets for this concept are Grade 3 questions venn diagrams, Venn diagrams, Venn diagram l1s1, Ss, Venn diagram, Write details that tell how the subjects are different in, Chapter 3 1 venn diagrams, Math 211 sets practice work part 1 shade the region. The shaded area of figure is 5. Figure 2: Venn’s Primary diagrams. A complete Venn diagram represents the union of two sets. Find how many people can speak both English and Hindi ? speak Does union of sets include intersection of sets ?. The closest I've seen is the Unicorn profile folks are posting about on Wood Central. In set theory, we commonly use Venn diagrams, developed by the logician John Venn ( ). Venn diagrams are useful in any situation requiring a direct comparison of two or more categories or concepts. Venn diagrams are being used to study and analyze the similarities and differences among languages. Venn Diagrams The Venn diagram, is a convenient way to illustrate definitions within the algebra of sets. Use it for drawing Venn and Euler diagrams. You also learned to use the Venn diagram to show relationships between sets. 6 set Venn diagram. Sets and Venn Diagrams October 26, 2012 What is a Set? Mathematics, at its very core, can be described ENTIRELY in terms of sets. A thorough introduction to shading regions of venn diagrams and using them to calculate probabilities. They can be interactive and fun, and here are a few ways to make a Venn diagram: Draw the circles on a blank piece of paper and fill in the information. Venn diagrams with complements, unions and intersections. This Site Might Help You. While there are more than 30 symbols used in set theory, you don’t need to memorize them all to get started. Characteristics of Venn Diagram. Creates a Venn diagram with five sets. colors: named list of colors for sets (one set=one color) na. possible interactions. Venn diagrams are a great way to visualize the structure of set relationships. VENN DIAGRAM. Revision notes on 'Set Notation & Venn Diagrams' for the Edexcel IGCSE Maths exam. That's ok though- sometimes a traditional Venn just won't work. Last modified September 6, 2017 Prism offers tools to draw circles and text, and you can use those to create informal Venn diagrams. Use MyDraw to create your own 5 set Venn diagram. Drawing these diagrams manually is difficult. It allows multiple sets (four sets for venn, 3 sets for Euler diagrams), customizable colours and fonts, simple syntax and and best of all the size of the circles is proportional to the size of the data sets (at least when comparing 2 data sets). In the center, the student lists the items shared in common. A Venn diagram is an illustration used to depict logical relationships between different groups or sets. Which Venn diagram correctly describes the relationship between the set of integers and the set of whole numbers A. Venn Diagrams Venn diagrams are very useful for sorting and processing data. Creates a Venn diagram with five sets. Student: The Venn Diagram below shows the number of girls on the soccer and track teams at a high school. In fact, the following three are the perfect foundation. 0 and Rserve <= 1. Venn diagram definition is - a graph that employs closed curves and especially circles to represent logical relations between and operations on sets and the terms of propositions by the inclusion, exclusion, or intersection of the curves. A Euler diagram resembles a Venn diagram, but does not neccessarily. If you wish to give an awesome presentation, using diagrams is great because they make your data look nicer and help your These cookies may be set through our site by our advertising partners. Use one of our Venn diagram templates to show all possible relations between your sets of data. Moreover, union operations between sets can be made. outwards_adjust: the multiplier defining the distance from the centre. Drawing Venn diagrams that is not specifically always two circles. 1) A 2) B 3) A ∪ B 4) U 5) A' 6) B' 7) (A ∪ B)'. OR FT from their Venn diagram. Venn diagrams are another way of presenting probability information. (a) Draw a Venn diagram for the two sets, P and R. The applicant was not hired. In a Venn diagram any set is depicted by a closed region. An entity set is a group of similar entities and these entities can have attributes. Jan 25, 2016 - Printable Venn diagram worksheets for primary grades 1 to 7 math students, based on the Singapore math curriculum. This Site Might Help You. Venn Diagrams (H) - Version 2 January 2016. Odd numbers and numbers greater than 10. Learn all about Venn diagrams and make your own with Canva. Textbook Authors: Blitzer, Robert F. venn (Required) A Venn object. So how did it start and The Venn diagram has emerged as a useful and versatile learning tool in education. There are 40 people preferred both brand P and brand Q, 35 people preferred both brand Q and brand R, 60 people preferred both brand P and brand R. A Venn diagram represents each set by a circle, usually drawn inside of a containing box representing the universal set. Venn diagrams, also called Set diagrams or Logic diagrams, are widely used in mathematics, statistics, logic,. First step: Install & load “VennDiagram” package. A triple Venn diagram is a diagram that consists three intertwined circles that represents thoughts and the relation of each from one another. c) t – x – z + r +b. 4-set means I should be able to specify 4 sets. Venn diagrams are helpful for. It's just a rough guide from my limited knowledge on the hololive girls (would include guys too but there's limited space and I only know a few of them). A venn diagram uses circles that. Members receive unlimited access to 49,000+ cross-curricular educational resources, including interactive activities, clipart, and abctools custom worksheet generators. The section where the two sets overlap has the numbers contained in both Set A and B, referred to as the intersection of A and B. Originally used as a way to show the differences and similarities. In order to properly celebrate John Venn's 180th birthday, today your task will be creating a program that outputs a Venn Diagram!. It seems VennDiagram can not work on my data. Use of Venn Diagrams: Venn diagrams or set diagrams are specially designed diagrams that show all possible logical relations between a finite collection of sets or aggregation of things. none of the above ____ 6. Coordinates of vertices are from http://sue. The universal setξ= A∪B ∪C. Venn diagrams also help us to convert common English words into mathematical terms that help add precision. Five percent of the students work part time and belong to a club. A B A B Ç Venn Diagrams Try this one!. venn: Draw a Venn Diagram with Five Sets in VennDiagram: Generate High-Resolution Venn and Euler Plots rdrr. Look at this Venn diagram. Repurposing to Use Segments. The following example (see example 1. Force Directed Tree. You can find 100s of templates and examples to be used freely in the diagram community of Creately. Venn diagrams were invented for use in a branch of mathematics called set theory. Set A = {2,4,6,8,10,12,14,16}. The term Venn diagram is not foreign since we all have had Mathematics, especially Probability and Algebra. Manage My Favorites. For instance: Out of forty students, 14 are taking English Composition and 29 are taking Chemistry. I am interested in learning how to do a 2-way venn diagram in R from a table generated with cuffdiff data. | 2021-04-15T04:33:06 | {
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https://mathhelpboards.com/threads/expected-number-of-light-bulbs-on.5655/ | # Expected number of light bulbs on
#### oyth94
##### Member
There are 50 light bulbs in a room each with its own switch. If a light bulb is on, Dick turns it off and if it is off , he turns it on. Initially all light bulbs are off . After 50 flips and assuming that Dick chooses switches to be flipped randomly, what is the expected number of light bulbs on in the room rounded to the nearest natural number?
To be honest, i tried many times but i can;t seem to arrive at an answer. i am stuck. any help would be appreciated!!!
so i have started with this:
Let Ei be the event that bulb i is on after 50 flips. This is the case if switch i is chosen an odd number of times. Compute Pr(Ei) and let Xi be the indicator random variable for Ei. Then E[Xi]=Pr(Ei). Now, let X be the total number of bulbs on after 50 flips. X=∑Xi, so E[X]=∑E[Xi]=n⋅Pr(Ei).
Last edited:
#### chisigma
##### Well-known member
There are 50 light bulbs in a room each with its own switch. If a light bulb is on, Dick turns it off and if it is off, he turns it on. Initially all light bulbs are off. After 50 flips and assuming that Dick chooses switches to be flipped randomly, what is the expected number of light bulbs on in the room rounded to the nearest natural number?
To be honest, i tried many times but i can;t seem to arrive at an answer. i am stuck. any help would be appreciated!!!
so i have started with this:
Let Ei be the event that bulb i is on after 50 flips. This is the case if switch i is chosen an odd number of times. Compute Pr(Ei) and let Xi be the indicator random variable for Ei. Then E[Xi]=Pr(Ei). Now, let X be the total number of bulbs on after 50 flips. X=∑Xi, so E[X]=∑E[Xi]=n⋅Pr(Ei).
The core of the problem is the meaning of 'Dick chooses switches to be flipped randomly'... if it means that any bulb has probability 1/2 to be switched by Dick at any iteration it is evident that at the end of each iteration the expected number of switched on bulbs is 25...
Kind regards
$\chi$ $\sigma$
#### Opalg
##### MHB Oldtimer
Staff member
There are 50 light bulbs in a room each with its own switch. If a light bulb is on, Dick turns it off and if it is off, he turns it on. Initially all light bulbs are off. After 50 flips and assuming that Dick chooses switches to be flipped randomly, what is the expected number of light bulbs on in the room rounded to the nearest natural number?
Fascinating problem! I have wasted half my Sunday morning on it.
Let's generalise it a bit, to the case where there are $k$ switches. Denote by $E(k,n)$ the expected number of lit bulbs after $n$ random flips, given that initially all light bulbs were off. I found $50$ much too large a number to work with, so I did some experiments with smaller numbers. For $k=4$, I found that the values of $E(4,n)$, for $n=1,2,3,4,5,6$ are $$1,\ \frac32,\ \frac74,\ \frac{15}8,\ \frac{31}{16}, \frac{63}{32},$$ which makes it seem likely that $E(4,n) = \dfrac{2^n-1}{2^{n-1}}$.
I then tried $k=6$ and found something even more interesting: for $n=1,2,3,4$, the values of $E(6,n)$ are $$1,\ \frac53,\ \frac{19}9,\ \frac{65}{27},$$ which very strongly suggests that $E(6,n) = \dfrac{3^n-2^n}{3^{n-1}}.$
If so, then it seems inevitable that the general formula must be $\boxed{E(2k,n) = \dfrac{k^n - (k-1)^n}{k^{n-1}}}.$
I'm not prepared to put money on it, but I'm 99% sure that $$E(50,50) = \dfrac{25^{50} - 24^{50}}{25^{49}} = 25 - 24\Bigl(\frac{24}{25}\Bigr)^{\!49} \approx 21.75.$$
Now it's time for Sunday lunch, so I'll leave it to others to see if those results can be proved.
#### oyth94
##### Member
Fascinating problem! I have wasted half my Sunday morning on it.
Let's generalise it a bit, to the case where there are $k$ switches. Denote by $E(k,n)$ the expected number of lit bulbs after $n$ random flips, given that initially all light bulbs were off. I found $50$ much too large a number to work with, so I did some experiments with smaller numbers. For $k=4$, I found that the values of $E(4,n)$, for $n=1,2,3,4,5,6$ are $$1,\ \frac32,\ \frac74,\ \frac{15}8,\ \frac{31}{16}, \frac{63}{32},$$ which makes it seem likely that $E(4,n) = \dfrac{2^n-1}{2^{n-1}}$.
I then tried $k=6$ and found something even more interesting: for $n=1,2,3,4$, the values of $E(6,n)$ are $$1,\ \frac53,\ \frac{19}9,\ \frac{65}{27},$$ which very strongly suggests that $E(6,n) = \dfrac{3^n-2^n}{3^{n-1}}.$
If so, then it seems inevitable that the general formula must be $\boxed{E(2k,n) = \dfrac{k^n - (k-1)^n}{k^{n-1}}}.$
I'm not prepared to put money on it, but I'm 99% sure that $$E(50,50) = \dfrac{25^{50} - 24^{50}}{25^{49}} = 25 - 24\Bigl(\frac{24}{25}\Bigr)^{\!49} \approx 21.75.$$
Now it's time for Sunday lunch, so I'll leave it to others to see if those results can be proved.
Thank you so much! I arrived with the same answer but I used either Poisson or Binomial. But my approach I didn't involve expected value...which I should...
#### zzephod
##### Well-known member
Fascinating problem! I have wasted half my Sunday morning on it.
Let's generalise it a bit, to the case where there are $k$ switches. Denote by $E(k,n)$ the expected number of lit bulbs after $n$ random flips, given that initially all light bulbs were off. I found $50$ much too large a number to work with, so I did some experiments with smaller numbers. For $k=4$, I found that the values of $E(4,n)$, for $n=1,2,3,4,5,6$ are $$1,\ \frac32,\ \frac74,\ \frac{15}8,\ \frac{31}{16}, \frac{63}{32},$$ which makes it seem likely that $E(4,n) = \dfrac{2^n-1}{2^{n-1}}$.
I then tried $k=6$ and found something even more interesting: for $n=1,2,3,4$, the values of $E(6,n)$ are $$1,\ \frac53,\ \frac{19}9,\ \frac{65}{27},$$ which very strongly suggests that $E(6,n) = \dfrac{3^n-2^n}{3^{n-1}}.$
If so, then it seems inevitable that the general formula must be $\boxed{E(2k,n) = \dfrac{k^n - (k-1)^n}{k^{n-1}}}.$
I'm not prepared to put money on it, but I'm 99% sure that $$E(50,50) = \dfrac{25^{50} - 24^{50}}{25^{49}} = 25 - 24\Bigl(\frac{24}{25}\Bigr)^{\!49} \approx 21.75.$$
Now it's time for Sunday lunch, so I'll leave it to others to see if those results can be proved.
I can report that simulation gives the mean number off after 50 flips of the 50 switchs is 21.74 with a standard error ~0.034
(another exercise on my Python learning curve)
.
Last edited:
#### Opalg
##### MHB Oldtimer
Staff member
I now see how to prove the formula for $E(2k,n)$, and it's really quite easy. When the $(n+1)$th switch is flipped, the probability that it is already on is $\dfrac{E(2k,n)}{2k}$, in which case the flip turns it off, and the number of "on" switches is reduced by $1$. Likewise, the probability that the $(n+1)$th switch is off is $\dfrac{2k-E(2k,n)}{2k}$, in which case the flip turns it on, and the number of "on" swiches is increased by $1$. Therefore $$E(2k,n+1) = \frac{E(2k,n)}{2k}(E(2k,n)-1) + \frac{2k-E(2k,n)}{2k}(E(2k,n)+1) = \frac{(k-1)E(2k,n) + k}k.$$ Now assume as an inductive hypothesis that $E(2k,n) = \dfrac{k^n-(k-1)^n}{k^{n-1}}$. Then $$E(2k,n+1) = \frac{(k-1)E(2k,n) + k}k = \frac{(k-1)k^n-(k-1)^{n+1} + k^n}{k^n} = \frac{k^{n+1} - (k-1)^{n+1}}{k^n},$$ which completes the inductive step. The base case $E(2k,1) = 1$ is easy to check, so the result is proved. | 2021-05-11T19:31:42 | {
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https://math.stackexchange.com/questions/2406516/frac12n-binomnn-frac12n1-binomn1n-frac122n | # $\frac{1}{2^n}\binom{n}{n}+\frac{1}{2^{n+1}}\binom{n+1}{n}+...+\frac{1}{2^{2n}}\binom{2n}{n}=1$: short proof?
The identity $\frac{1}{2^n}\binom{n}{n}+\frac{1}{2^{n+1}}\binom{n+1}{n}+...+\frac{1}{2^{2n}}\binom{2n}{n}=1$ arises from a question on probability in my textbook. A proof by induction on $n$, which exploits the fact that $\binom{a}{b}+\binom{a}{b+1}=\binom{a+1}{b+1}$, is straightforward but not enlightening.
Is it possible to find any very clever approaches? Via a combinatorial or probabilistic interpretation, for instance?
Suppose you are flipping a coin until you get $n+1$ heads or $n+1$ tails. For $k=0,\dots,n$, what is the probability that you are done after exactly $n+1+k$ flips?
If the last flip was a tails, this means that in the former $n+k$ flips there were exactly $n$ tails, and this happens with probability $\frac{1}{2}\cdot\frac{1}{2^{n+k}}\binom{n+k}{n}$ (the first $\frac{1}{2}$ is due to assuming the last flip is tails). Same for heads, so the probability of finishing after exactly $n+1+k$ flips is $\frac{1}{2^{n+k}}\binom{n+k}{n}$.
Now just note that the number of flips always ends up between $n+1$ and $n+1+n$ (by pigeonhole principle).
Updated Solution
Here's a neater solution!
\begin{align} \sum_{k=0}^n \frac 1{2^{n+k}}\binom {n+k}n &=\frac 1{2^{2n}}\sum_{k=0}^n \binom {k+n}k 2^{n-k}\\ &=\frac 1{2^{2n}}\sum_{k=0}^n \binom {k+n}k \sum_{j=0}^{n-k}\binom {n-k}j\\ &=\frac 1{2^{2n}}\sum_{l=0}^n \binom {2n-l}{n-l}\sum_{j=0}^l \binom lj &&(l=n-k)\\ &=\frac 1{2^{2n}}\sum_{j=0}^n \sum_{l=j}^n\binom {2n-l}n\binom lj\\ &=\frac 1{2^{2n}}\sum_{j=0}^n \binom {2n+1}{n+j+1} &&(*)\\ &=\frac 1{2^{2n}}\sum_{j=n+1}^{2n+1} \binom {2n+1}j\\ &=\frac 1{2^{2n}}\cdot \frac 12\sum_{j=0}^{2n+1}\binom {2n+1}j &&\text{(by symmetry)}\\ &=\frac 1{2^{2n+1}}\cdot 2^{2n+1}\\ &=1\;\;\color{red}{\blacksquare}\end{align}
$\qquad \qquad \quad ^*\displaystyle\scriptsize\text{using }\sum_{r} \binom {a-r}{c}\binom {b+r}{d}=\binom {a+b+1}{c+d+1}$
Solution posted earlier
Here's a direct algebraic proof without using induction.
\begin{align} \sum_{k=0}^n \binom {k+n}k x^k(x+y)^{n-k} &=\sum_{k=0}^n \binom {k+n}k x^k\sum_{j=0}^{n-k}\binom {n-k}Jy^jx^{n-k-j}\\ &=\sum_{k=0}^n \binom{k+n}k\sum_{j=0}^{n-k}\binom{n-k}jy^jx^{n-j}\\ &=\sum_{\ell=0}^n\binom{2n-\ell}{n-\ell}\sum_{j=0}^\ell\binom{\ell}jy^jx^{n-j} &&(\ell=n-k)\\ &=\sum_{j=0}^n\sum_{\ell=j}^n (-1)^{n-\ell}\binom{-n-1}{n-\ell}(-1)^{\ell-j}\binom{-j-1}{\ell-j}y^jx^{n-j}\\ &=\sum_{j=0}^n(-1)^{n-j}y^jx^{n-j}\sum_{\ell=j}^n\binom{-n-1}{n-\ell}\binom{-j-1}{\ell-j}\\ &=\sum_{j=0}^n(-1)^{n-j}y^jx^{n-j}\binom{-n-j-2}{n-j} &&\text{(Vandermonde)}\\ &=\sum_{j=0}^n(-1)^{n-j}y^jx^{n-j}\cdot (-1)^{n-j}\binom{2n+1}{n-j}\\ &=\sum_{j=0}^n \binom{2n+1}{n-j}y^jx^{n-j}\\ &=\sum_{i=0}^n \binom{2n+1}i x^i y^{n-i}\\ \text{Put }x=y=1:\hspace{4cm}\\ \sum_{k=0}^n\binom {k+n}k2^{n-k} &=\sum_{i=0}^m\binom{2n+1}i &&(i=n-j)\\ &=\frac 12\cdot 2^{2n+1} &&\text{(by symmetry)}\\ &=2^{2n}\\ \sum_{k=0}^n \binom {n+k}k 2^{-k}&=2^n\\ \sum_{k=0}^n\frac 1{2^{n+k}} \binom {n+k}n&=1\\ \end{align}
• @robjohn - Thank you - very kind of you! Sep 1, 2017 at 7:13
Here is an answer based upon generating functions. It is convenient to use the coefficient of operator $[z^n]$ to denote the coefficient of $z^n$ in a series. This way we can write e.g. \begin{align*} [z^k](1+z)^n=\binom{n}{k} \end{align*}
We obtain \begin{align*} \color{blue}{\sum_{k=0}^n\binom{n+k}{k}\frac{1}{2^{n+k}}} &=\sum_{k=0}^n\binom{2n-k}{n-k}\frac{1}{2^{2n-k}}\tag{1}\\ &=\sum_{k=0}^\infty[z^{n-k}](1+z)^{2n-k}\frac{1}{2^{2n-k}}\tag{2}\\ &=2^{-2n}[z^n](1+z)^{2n}\sum_{k=0}^\infty\left(\frac{2z}{1+z}\right)^{k}\tag{3}\\ &=2^{-2n}[z^n](1+z)^{2n}\cdot\frac{1}{1-\frac{2z}{1+z}}\tag{4}\\ &=2^{-2n}[z^n](1+z)^{2n+1}\cdot\frac{1}{1-z}\tag{5}\\ &=2^{-2n}\sum_{k=0}^n[z^k](1+z)^{2n+1}\tag{6}\\ &=2^{-2n}\sum_{k=0}^n\binom{2n+1}{k}\tag{7}\\ &=2^{-2n}\frac{1}{2}2^{2n+1}\tag{8}\\ &\color{blue}{=1} \end{align*}
Comment:
• In (1) we change the order of summation $k \rightarrow n-k$.
• In (2) we apply the coefficient of operator. We also set the limit to $\infty$ without changing anything since we are adding zeros only.
• In (3) we do a rearrangement and apply the formula $[z^{p-q}]A(z)=[z^p]z^qA(z)$.
• In (4) we apply the geometric series expansion.
• In (5) we do some simplifications.
• In (6) we do the Cauchy multiplication with the geometric series $\frac{1}{1-x}$ and restrict the upper limit of the sum with $n$ since other terms do not contribute to $[z^n]$.
• In (7) we select the coefficient of $z^k$.
• In (8) we apply the binomial theorem. | 2022-08-12T06:54:38 | {
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http://uqqp.erci.pw/hard-probability-questions.html | # Hard Probability Questions
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Brain teaser game: Hard Logic Probability Puzzle. population. The probability of an event occurring is the chance or likelihood of it occurring. Model Question Paper Mathematics Class XII Time Allowed : 3 hours Max: Marks: 100 General Instructions (i) The question paper consists of three parts A, B and C. | 2019-11-19T15:26:21 | {
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https://math.stackexchange.com/questions/176887/when-chessboards-meet-dominoes | # When chessboards meet dominoes
You probably have heard about the following brainteaser :
Consider a $$8\times 8$$ chessboard. Remove two extreme squares (top-left and bottom-right e.g.). Can you fill the remaining chessboard with $$1\times 2$$ dominoes ?
The answer lies in a coloring argument. The problem, however, does not use the fact that the chessboard is colored. I would like to know improved examples of coloring problems like this one.
For instance, could you extend the problem with Tetris-like L-shaped dominoes and solve it using more than two colors ?
• Well, with 2x3 "dominos" you notice that 6 doesn't divide 64... Jul 30 '12 at 16:44
• 2x3 means L shape. Jul 30 '12 at 16:45
• Could you clarify: When you talk about using Tetris pieces, you are talking about an ordinary chessboard, not one with two pieces removed? Jul 30 '12 at 16:55
• @SiliconCelery : as you can add two L-shaped dominos to get a 2x4 rectangle, they can fit a regular chessboard. Concerning my question I guess the problem should be different from the beginning, it should probably substract each of the extreme squares to get 60 = 15 x 4 squares Jul 30 '12 at 17:01
• The 35 hexominoes cannot be packed into a rectangle as you have two more squares of one color than the other. See en.wikipedia.org/wiki/Hexomino. I have seen many math puzzles that exploit coloring. For trominoes you often color in strips of three colors or diagonally like a checkerboard with three colors. Jul 30 '12 at 17:59
Solomon Golomb's book Polyominoes presents a number of arguments of this type.
One that I remember is: a square is deleted from an 8×8 checkerboard. Can the remaining 63 squares be covered by 21 1×3 rectangles? The answer involves coloring the checkerboard in three colors in alternating diagonal stripes:
This colors the 64 squares of the checkerboard with 21 green squares, 21 yellow squares, and 22 blue squares. Each 1×3 rectangle must cover exactly one square of each color. The deleted square therefore cannot be any of the green or yellow ones, nor any of the squares equivalent to one of these under a rotation or reflection of the checkerboard:
(This is four copies of the first diagram, superimposed, with suitable rotations.)
This eliminates all but 4 squares from consideration, namely the four bright blue ones in the previous diagram. So the only solutions involve deleting one of these four blue squares.
There are a number of analogous arguments about polyhexes that depend on a three-coloring of a hexagonal lattice:
For example, there are three different trihexes, which are made by joining three hexagons; two of these are guaranteed to cover exactly one cell of each color, no matter how they are placed.
I once wasted a lot of time trying to make myself a set of tetrominoes by marking up a 4x5 rectangle and cutting it apart, and I felt rather foolish when I realized that a straightforward checkerboard coloring shows that this is impossible. There are 5 tetrominoes, and four of them must cover two black and two white squares each. The 5th is T-shaped, and must cover three black squares and one white (or vice versa).
So they cannot possibly tile a 4×5 rectangle, which has equal numbers of black and white squares.
• Very nice example of coloring pattern. Jul 30 '12 at 19:38
A woodworm is sitting at the centre of a cube that's divided into $3^3$ identical cubelets. The woodworm can go from the centre of one cubelet to the centre of another in any edge-parallel direction. The woodworm would like to eat its way through the cube such that it visits the centre of each cubelet exactly once. Is this possible?
• Same principle, I see. Awesome ! Jul 30 '12 at 19:06
I searched for math.SE questions and answers that involve colouring of tilings and the like; here are two interesting ones that I found:
The Mathematics of Tetris
A tiling puzzle/question | 2021-09-21T03:11:12 | {
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https://discourse.julialang.org/t/solve-for-upper-lower-bound-of-a-numerical-definite-integral/58164 | # Solve for upper/lower bound of a numerical definite integral
Hi, I’m trying to solve for b, given f, a and k, please help, Is there a way to do it in julia?
quadgk(f,a,b)=k
Thanks
2 Likes
There’s always a Newton iteration:
b = some_initial_guess
tol = sqrt(eps())
while true
δb = (quadgk(f,a,b, rtol=tol*0.1)[1] - k) / f(b)
b -= δb
abs(δb) ≤ tol * abs(b) && break
end
Note that I’m using the fact that \frac{d}{db} \int_a^b f(x) dx = f(b).
10 Likes
@stevengj, this is excellent.
Below a summary of different solutions, including yours and your advice to run Optim properly (even if not recommended for this problem kept it for sake of completeness).
f(x) = exp(x)*sin(x)
# forward modelling
a = 0
# b = π : is UNKNOWN to be found given f, a and k
k = 0.5*(1.0 + exp(1)^π) # k = 12.0703463163896
# SOLUTION-1: Newton solution by @stevengj
b = 1.0 # initial guess
tol = sqrt(eps())
while true
δb = (quadgk(f,a,b, rtol=tol*0.1)[1] - k) / f(b)
b -= δb
abs(δb) ≤ tol * abs(b) && break
end
println("Solution = ", b) # 3.14159268 / solution π= 3.14159265
# SOLUTION-2: use Optim to minimizing g(x)^2
using Optim
h(x) = ((g.(x)).^2)[1]
dh(x) = (2*g.(x).*f.(x))[1] # derivative
# Checks:
g(pi) ≈ 0 # true
h(pi) ≈ 0 # true
dh(pi) ≈ 0 # true
b = [1.0] # initial value / solution π= 3.14159265
result = optimize(h, b, Newton(), Optim.Options(g_tol = 1e-16))
Optim.minimizer(result) # 3.14159265
result = optimize(h, dh, b, Newton(); inplace=false) # using derivative
Optim.minimizer(result) # 3.1415795 # does not improve accuracy
# SOLUTION-3: use NLsolve to find roots
using NLsolve
dg(x) = f.(x) # derivative
# Check:
g([pi]) ≈ 0 # true
b = [1.0] # initial value / solution π= 3.14159265
nlsolve(g, b; ftol=1e-16) # 3.14159264
nlsolve(g, dg, b; ftol=1e-16) # 3.14159263 # same accuracy...
# SOLUTION-4: use DifferentialEquations and interpolation
using DifferentialEquations, Plots, Dierckx
u0 = 0; tspan = (a, 4.) # initial conditions & timespan
du(u,p,t) = f(t) # define the system
prob = ODEProblem(du, u0, tspan) # define the problem
sol0 = solve(prob, dtmax=0.01) # solve it
i = argmin(abs.(sol0.u .- k)) # index of time-step the closest to solution
tsol = LinRange(sol0.t[i-1], sol0.t[i+1],20) # spline it around the solution
spl = Spline1D(tsol, sol0.(tsol) .- k)
b = mean(roots(spl)) # 3.1415926536
1 Like
Yes; it looks like the problem with your code is that Optim.optimize is expecting a vector of initial guesses, not a scalar. Try passing [b] instead of b.
However, this is a root-finding problem g(b) = 0. It is almost always a mistake to try to solve a root-finding problem by using an optimization algorithm to minimize |g|². You should use a root-finding algorithm like those in NLsolve.jl.
Moreover, in this case you know the derivative analytically, as I mentioned above, in which case you should certainly provide it to the root-finding algorithm.
However, for a 1d root-finding problem (b is a scalar) with an analytically known derivative, all of the generic root-finding packages will basically boil down to a Newton iteration like the one I wrote. The clever algorithms are mainly for the case where you don’t know the derivative, or have lots of derivatives (a big Jacobian) and can’t affort to compute them all or to invert the Jacobian.
Alternatively, in 1d, especially for smooth functions, there are often more clever algorithms available. For example, you could use ApproxFun.jl to construct a high-accuracy polynomial approximation of your integrand f, call cumsum to compute its integral, and use the ApproxFun.roots function to find all of the places where the integral equals k.
6 Likes
Rather than minimize |g|², one could apply g(b)=0 as an equality constraint to an optimization algorithm that supports such constraints, e.g. NLopt.jl or JuMP.jl. I often find it convenient to deal with roots the same way I deal with optimization. Sometimes you can leave out an objective function altogether, or just specify a constant like 0, to be “minimized” subject to the constraints (to some tolerance). However, I’ve never compared performance to a root finder.
2 Likes
Tried to follow it the best possible and have updated code above for 3 different methods: Newton by yourself, Optim and NLsolve.
Including the derivative did not seem to help much, but any blame should surely be on me. BR
*PS: code above now runs but better not shaking it too much…
1 Like
Yes, solving the optimization problem \min_x 1 subject to the constraint g(x)=0 is, of course, exactly equivalent to the root-finding problem g(x)=0, and a sufficiently clever NLP algorithm will perform steps essentially equivalent to Newton steps (perhaps limited by a trust region). They almost certainly aren’t going to be better than Newton steps, however, and in many cases will be worse (because they may spend a lot of effort doing computational steps that are only needed for more general NLPs).
In a single-variable problem where the derivative is known, if you aren’t going to do something more clever (e.g. Chebyshev fits, complex analysis, continued fractions, etc.) you might as well just use Newton. (Either implementing it yourself — it’s just a few lines — or calling a canned implementation like the one in Roots.jl).
6 Likes
There is another way. Define R as
$$\int_a^{R(x)}f=x$$
you get
$$\frac{d}{dx}R(x) = \frac{1}{f(R(x))},\quad R(a)=0$$
You can use an ODE solver to get b=R(k). This is the main trick of my paper on PDMP.
@rveltz, it is only me, or your Latex is not displaying properly?
3 Likes
It gives:
1 Like
@rveltz, mathematically this looks pretty much like @stevengj’s solution above. The difference is that the differential equations solver will not use Newton’s method?
Would like to try this when time allows, but would be surprised if it beats Newton in terms of accuracy and fast convergence.
I would say it is better in that you dont need to recompute \int_a^bf if you know that the solution b_{sol} is greater (smaller) than b. You “waste” less computation
1 Like
Yeah, it seems to me that there is a lot of redundancy in re-computing the full integral at each step. It should be enough to compute the integral only over the ‘correction interval’. (One should probably take some care with accumulated errors, however.)
2 Likes
@rveltz, in your nice solution how do you handle the cases with singularities where the function f() has zeros and 1/f() explodes to ∞?
Hum, I dont know! I guess the Newton solver will sufer as well. I’d say you have to use a stiff ODE solver that can handle finite time explosion.
1 Like
The topic “Solve equation” doesn’t do justice to the interesting things discussed here. @carloslesmes can I suggest editing the topic title to something more specific? For example, “Solve for upper/lower bound of a numerical definite integral” or some such.
1 Like
Am I looking at this crazy, or wouldn’t the correct boundary condition be R(0) = a instead of R(a) = 0?
1 Like
nope you are right! Thank you
Newton did not suffer for f(x)=exp(x)sin(x) in the example above with a zero at the solution pi. However, someone else is suffering to have your method working for that f(x) | 2022-05-24T08:34:57 | {
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https://nbviewer.ipython.org/url/ignite.byu.edu/che263/lectureNotes/interpolation.ipynb?flush_cache=true | # Interpolation¶
In [2]:
import numpy as np # numerics
import matplotlib.pyplot as plt # plotting
%matplotlib inline
## Key concept¶
• Given two arrays of $x_g$, $y_g$ data points (e.g., a plot of points).
• You want to estimate the $x_w$, $y_w$ value of an intermediate data point.
• You know the value of $x_w$ (cause it's where you want the point).
• You need to interpolate to find the corresponding $y_w$.
In [3]:
#### Just plotting some data to visualize
x_given = np.array([0,1,2,3,4,5,6,7,8,9,10])
y_given = np.cos(x_given**2.0/8.0) + 1
plt.rc('font', size=16)
plt.plot(x_given, y_given, 'o:')
plt.plot([2.5,2.5],[-0.1,1.6], '--', color='gray')
plt.plot([-0.1,2.5],[1.6,1.6], '--', color='gray')
plt.plot([2.5], [1.6], '*', markersize=20)
plt.xlim([-0.1,11]); plt.ylim([-0.1,2.2])
plt.xlabel('x'); plt.ylabel('y')
plt.text(2.6,0, r"$x_w$", fontsize=16); plt.text(0,1.7, r"$y_w$", fontsize=16);
### Question¶
• How to get intermediate values, that is, values between those that are part of the given data?
• Fit a curve to the data, then evaluate from that curve.
• Take the closest data point
• Linear interpolation
• Higher order interpolation
• (Others?)
### Question¶
• How does curve fitting differ from interpolation?
• Normally, interpolation is done locally, between a few (often two) bounding points.
• Curve fitting involves making a best fit curve everywhere
• Usually, a best-fit curve will not pass through all data points.
• Useful when there is noise and we want the underlying curve, or a model equation for the data.
• Interpolation normally considers the points themselves, not a single curve through everything.
• Even with noisy data, we might want to interpolate an intermediate point.
• If you want an intermediate point to some given data, use interpolation.
• If you want to fit a model though some scattered data, and then evaluate the model, use curve fitting.
### Examples?¶
• Think of some examples where you might use this.
## Linear Interpolation Exercise¶
• Given the xg and yg data below.
• Write a function called Linterp that takes the following arguments:
• xg an array of given x data
• yg an array of given y data corresponding to xg
• xw the value of x we want to interpolate at
• The function returns yw corresponding to xw
• Assume xg are uniformly spaced, and ascending.
Questions
• What is the linear interpolation formula at a given location?
• We write the equation for a line based on two points, then evaluate the line at the intermediate point:
• equate slopes, then solve for y_w: $$\frac{y_w-y_0}{x_w-x_0} = \frac{y_1-y_0}{x_1-x_0},$$ $$y_w = y_0 + (x_w-x_0)\frac{y_1-y_0}{x_1-x_0}.$$
In [4]:
#-------------------- Set some "given" x, y data, (normally given to us)
xg = np.array([0,1,2,3,4,5,6,7,8,9,10.]) # given x data
yg = np.cos(xg**2.0/8.0)+1 # given y data
print("xg = "+np.array2string(xg, formatter={'float_kind':lambda x: f"{x:4.2f}"}))
print("yg = "+np.array2string(yg, formatter={'float_kind':lambda x: f"{x:4.2f}"}))
#-------------------- interpolate to xw=2.5
xw = 2.5
xg = [0.00 1.00 2.00 3.00 4.00 5.00 6.00 7.00 8.00 9.00 10.00]
yg = [2.00 1.99 1.88 1.43 0.58 0.00 0.79 1.99 0.85 0.24 2.00]
### New library¶
In [4]:
from scipy.interpolate import interp1d
• interp1d takes the given x array and the given y array as arguments.
• Returns a function.
• Call that function wherever you want to interpolate to.
In [14]:
xg = np.array([0,1,2,3,4,5,6,7,8,9,10]) # given x data
yg = np.cos(xg**2.0/8.0)+1 # given y data
#---------------
f_interp = interp1d(xg, yg)
#---------------
xw = 2.5
yw = f_interp(xw)
print(yw)
1.6543795393445195
### Exercise¶
• Take the previous xg, yg data and plot the data as points.
• Plot 1000 points of the underlying function used to get xg and yg, called xx
• Also, plot values of a linear interpolant at the same 1000 points.
In [5]:
xg = np.array([0,1,2,3,4,5,6,7,8,9,10]) # given x data
yg = np.cos(xg**2.0/8.0)+1 # given y data
#### Try replacing¶
f_interp = interp1d(xg, yg)
#### with this¶
f_interp = interp1d(xg, yg, kind='cubic')
• This uses a cubic spline interpolant instead of a linear interpolant.
• That is, we use cubic functions between points that match up smoothly at interfaces.
Try this:
• Interpolate an x value that is outside of the bounds of the given xg data.
• What happens?
### Extrapolation¶
• As listed above, you will get an error if you try to call f_interp with an x value that is outside of the upper and lower bounds of the original xg array.
• This can be avoided using the fill_value='extrapolate' argument, like so:
• f_interp = interp1d(xg, yg, fill_value='extrapolate'
## Summary¶
1. Import library
2. You have some data from somewhere, xg, yg
3. Get ther interpolant function fi=interp1d(xg, yg)
4. Set desired xw intermediate points to interpolate at (can be an array).
5. Perform the interpolation to yw=fi(xw)
In [6]:
from scipy.interpolate import interp1d
xg = np.array([1,2,3,4,5]) # some data you have
yg = np.array([11, 2.2, 3.3, -88, 9])
fi = interp1d(xg,yg)
xw = 2.5
yw = fi(xw)
## Question¶
• What if you have given xg and yg arrays and you know the yw and want to interpolate to the corresponding xw?
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https://cs.stackexchange.com/questions/135784/big-oh-and-big-omega-when-n-and-log-n-terms-are-in-fn | # Big Oh and Big Omega when $n$ and $\log n$ terms are in $f(n)$
having problems with big oh and big omega functions when there is a $$\log n$$ added or subtracted. For example how do I deal with $$n+\log n$$ or $$n-\log n$$ when I have to determine whether the function is in $$\Omega(n)$$ or in $$\Omega(n^2)$$? For example, is $$n-\log n$$ in $$\Omega(n)$$ or in $$\Omega(n^2)$$?
I cannot ignore the log function and am not sure how to deal with it. Polynomials and logs when multiplied I find OK. But I have a mental block over this one so help would be appreciated
You are right that you cannot a priori ignore additional terms, although morally you can as the "smaller" terms do not contribute to the asymptotic growth.
As an example $$f(n) :=n+\log n$$ is in $$\Theta(n)$$. Why? Going back to the definition, we want to show that $$f(n)$$ is in $$O(n)$$ and in $$\Omega(n)$$. Showing $$f(n)\in \Omega(n)$$ is immediate, as $$f(n) \geq n$$ for all $$n>0$$.
To show that $$f(n)$$ is in $$O(n)$$, simply notice that for large enough $$n$$ (say $$n>N$$ for some constant $$N$$) we have $$\log n \leq n$$ and thus $$f(n)\leq 2n$$ for all $$n>N$$. By definition $$f(n)\in O(n)$$.
Actually, you CAN ignore the log function in a sum or substraction, because the log is always asymptotically negligible in front of $$n^x$$, for every $$x > 0$$.
If you consider the function $$f(n) = n^x + \alpha\log n$$ (where $$x > 0$$ and $$\alpha \in \mathbb{R}$$), then there exists $$A, B \in \mathbb{R}_+^*$$ such that $$An^x \leq f(n) \leq Bn^x$$. That means that $$f\in \Omega(n^x)$$ and $$f\in \mathcal{O}(n^x)$$.
• Agreed with the f(n) =n^2 or cubed -logn as the polynomial grows faster we can ignore the log n. f(n) grows faster than log n. Is it sufficient then to just prove that f(n) =n and is in Big omega and ignore the log term on the basis that as Tassle says logn<n . is this sufficient for a formal proof Feb 22 '21 at 15:40
• Sorry, I didn't really get your question… I'll try to answer what I understood. When you have a sum $f(n) = g(n) + h(n)$ with $h \in o(g)$ (see en.wikipedia.org/wiki/Big_O_notation#Little-o_notation), then $f\in \Omega(g)$ and $f\in \mathcal{O}(g)$. This is the case in your question, since $\log \in o(n^x)$ for $x > 0$, and yes it is sufficient for a proof. Feb 22 '21 at 15:48
In your example, $$n-\log(n)=\theta(n)$$, that is $$n-\log(n) = \Omega(n)$$ and also $$n-\log(n)=O(n)$$.
You can see this like that:
• $$n-\log(n)\le n = O(n)$$
• $$n-\log(n) \ge n - 0.5n = \Omega(n)$$
• as n-logn is always less than n the function can only be in Big Oh . It cannot be in Big omega as if I divide by n^2 to fiind c I get 1-logn > or equal to c. As n gets greater than 2 n-logn gets more negative and c is no longer a constant so it cannot be in Big Omega. Am I on the right track? Feb 24 '21 at 9:09
• From a certain point $log(n)\le0.5n$. This means that $n-\log(n)$ will actually not be negative, but actually if you substitute the inequality you get $n-\log(n)\ge 0.5n$, which is $\Omega(n)$. However, you are correct that it is not $\Omega(n^2)$. Feb 24 '21 at 10:20
• sorry. I meant n-logn is not in Big Omega (n) not Big Omega (n^2). My original question is to prove (n-logn) = Ω(n) true or false. I was trying to prove it false by contradiction. Hence my reasoning of c needing to be a positive constant for all n> no.I am not sure where you came up with the 0.5n either. I plotted the graphs and n-logn is always under f(n)= n when n>1 Feb 24 '21 at 10:41
• As my answer showed, it is indeed always lower than $n$. But, in the same time, it is bigger than $0.5n$. Try to draw this graph as well Feb 24 '21 at 11:32
• You are right and thank you for your help . In fact from what you have said I can prove that n-logn =big theta(n) . Feb 24 '21 at 12:07 | 2022-01-16T20:47:51 | {
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http://math.stackexchange.com/questions/67315/help-understanding-why-a-complete-totally-bounded-metric-space-implies-every-in/67355 | # Help understanding why a complete, totally bounded metric space implies every infinite subset has a limit point
I'm reading the following proof. Properties II and III are in my title, that a complete, totally bounded metric space implies every infinite subset has a limit point.
I have two questions near the end. Why does $d(x_n,x)\lt 2/n$? And secondly, why is $x$ a limit point of $A$? What other point in the neighborhood of $x$ is also in $A$? I don't get why they mention that $3/n\to 0$ as $n\to\infty$. How does that imply it's a limit point?
-
Please give a reference when you quote text from somewhere else. Otherwise, how can anyone else look it up? – Carl Mummert Sep 25 '11 at 1:53
There’s a slight error in the proof: the claim should be that $d(x_n,x) \le 2/n$. To see this, let $\epsilon$ be any positive real number; the sequence of $x_m$’s converges to $x$, so there is a positive integer $m > n$ such that $d(x_m,x) < \epsilon$. By the triangle inequality $$d(x_n,x) \le d(x_n,x_m)+d(x_m,x) < \frac2n + \epsilon.\tag{1}$$ Thus, $$d(x_n,x) < \frac2n + \epsilon$$ for every $\epsilon > 0$, and hence $d(x_n,x) \le \dfrac2n$.
This small error doesn’t affect the next step of the argument: if $y \in B(x_n,1/n)$, then $d(x_n,y) < 1/n$, so $$d(x,y) \le d(x,x_n)+d(x_n,y) \le \frac2n + d(x_n,y) < \frac2n + \frac1n = \frac3n,$$ $y \in B(x,3/n)$, and therefore $B(x_n,1/n) \subseteq B(x,3/n)$.
Now $B(x_n,1/n)\cap A$ is infinite for each $n$, and $B(x_n,1/n) \subseteq B(x,3/n)$, so $B(x,3/n)\cap A$ is infinite for each $n$. Since $3/n\to 0$ as $n\to\infty$, for any $\epsilon > 0$ there is an $n_\epsilon$ such that $3/n_\epsilon < \epsilon$. But then $B(x,\epsilon)\cap A \supseteq B(x,3/n_\epsilon)\cap A$, which is infinite. Thus, every nbhd of $x$ contains infinitely many points of $A$.
Added: To clarify, it is in fact true that $d(x_n,x)<2/n$ for each $n$; it just doesn’t follow directly from the fact that $d(x_m,x_n) < 2/m$ when $m<n$, as the weaker inequality does. If we want the strict inequality, we can modify the argument that I gave by choosing $m>n$ so that $1/m<\epsilon/2$ and $d(x_m,x)<\epsilon/2$ and then replacing $(1)$ by $$d(x_n,x)\le d(x_n,x_m)+d(x_m,x)<\frac1n+\frac1m+\frac{\epsilon}{2}<\frac1n+\epsilon.$$ Since $\epsilon$ can be chosen arbitrarily small, we conclude that $d(x_n,x)\le 1/n$ for each $n$.
I don't think this is an error. They had established $d(x_n,x_m) \leq 1/m+1/n \lt 2/n$ for $n \lt m$ already (with $n$ and $m$ interchanged), so if you fix $n$ and let $m \to \infty$ then you get strict inequality. – commenter Sep 25 '11 at 1:01 @commenter: Their statement isn’t false, but it doesn’t follow from the fact that $d(x_n,x_m)<2/m$ when $m I think it worth settling some terminology. This is from Topology: a first course by James R. Munkres. He identifies three versions of compactness for a topological space: (A) compactness, then (B) your property (which he calls "limit point compactness") and a milder condition (C) he calls "sequential compactness" which is that every infinite sequence of points has a convergent subsequence. He proves that (A) implies (B) implies (C). He also proves that the three conditions are equivalent for a metric space. Finally, a well-known theorem is that a metric space is compact if and only if it is complete and totally bounded. So the strongest condition is what is usually discussed in this setting. However, (A) then implies (B). Finally, and this is not obvious, the product of two compact topological spaces is compact, and the product of two sequentially compact spaces is sequentially compact. What is unexpected is that the product of two limit point compact spaces need not be limit point compact. However, if they are metric spaces, the result does hold then. - To address the first question (the other is addressed in the post by Brian). [Also, there's no error.] Consider$\epsilon=1/(2n)$. Then there is some$N>0$such that$d(x_m,x)<1/(2n)$for all$m \geq N$. Let$m=\max\{N,2n\}$. Then$d(x_n,x) \leq d(x_n,x_m)+d(x_m,x) < [(1/n)+(1/m)]+1/(2n) \leq (1/n) + (1/2n) + (1/2n) = 2/n$. - While the assertion in question is true, the presentation is in my view seriously flawed, in that it suggests a justification that doesn’t in fact work. – Brian M. Scott Sep 25 '11 at 1:36 I'm not sure this is quite what you're looking for, but I believe the author just jumped from$d(x_n,x_m) < 2/n$for$n | 2013-06-19T09:32:20 | {
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http://santotirso.pt/manhas/email-security-fpex/esq5j.php?1b5b77=equation-of-a-circle-given-two-points | Hint: Find the equation of the line passing through the two given points. This calculator can find the center and radius of a circle given its equation in standard or general form. What is the Equation of a Circle? Different forms equations of straight lines. Equation of circle when endpoints of the diameter are given : (x - x 1) (x - x 2) + (y - y 1) (y - y 2) = 0. Enter Circle Equation Find the equation of a circle that has a diameter with the endpoints given by the points A(1,1) and B(2,4) Step 1: Find the Midpoint ( h , k ) of AB : Write Standard Form When Given Two Points You. How To Calculate The Equation Of A Circle From 3 Points Tessshlo. Also, it can find equation of a circle given its center and radius. 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https://stats.stackexchange.com/questions/218019/how-to-find-a-standard-deviation-determined-by-a-normal-distribution-probability | # How to find a standard deviation determined by a Normal distribution probability?
The question is
A liquid drug is marketed in phials containing a nominal 1.5ml but the amounts can vary slightly. The volume in each phial may be modeled by a normal distribution with the mean 1.55ml and standard deviation $$\sigma$$ ml. The phials are sold in packs of 5 randomly chosen phials . It is required that in less than 0.5% of the packs will the total volume of the drug be less than 7.5ml. Find the greatest possible value of $$\sigma$$.
I need to find the greatest possible value of the standard deviation ($$\sigma$$). I worked out the following:
$$\mu= 1.55*5 = 7.75.$$
We are asked to find value of $$\sigma$$ such that probability of (total volume of $$5$$ packs $$\lt 7.5)\lt0.5\%$$
$$P(X\lt7.5)\lt0.005.$$
After standardizing, $$P(X\le\frac{7.5-7.75}{\sigma/5})<0.005$$ and I found $$\sigma=0.2170.$$ However, the answer provided is $$0.0434.$$
• Please add the self-study tag, read its tag-wiki, and indicate the specific help you need at the point you struck difficulty. Jun 9 '16 at 2:37
• what's CTL? ... ... Also please check the details of the question, it looks like you may have a mistake somewhere. Where did the 7.75 in your working come from? Please show more detail/explanation of what you're doing. (As far as possible your responses should result in edits to your question) Jun 9 '16 at 2:38
• How have you approached/engaged it so far? Any partly successful paths? Where else have you looked for answers? Jun 14 '16 at 2:06
• Interestingly, neither answer is correct.
– whuber
Jun 14 '16 at 13:59
Among the objectives of good introductory statistics courses is learning how to think about the Normal distribution. This question provides a nice example.
The key is to use units of measurement that are adapted to the distribution. That is, let the mean be the zero point and let the standard deviation be one unit. This is what a "Z score" measures.
In light of this, let's parse the question. To do so, I will use two fundamental facts: expectations add ("linearity of expectation") and variances of independent variables also add:
• The mean volume of one pack is 1.55 ml, whence the mean volume of five packs must be five times as large, or 7.75 ml: this is the zero point.
• Since the unknown variance of a single pack is $$\sigma^2,$$ the variance of the sum of five independent packs is $$5\sigma^2.$$ Therefore the standard deviation of the sum--the unit of measurement we must adopt--is $$\sqrt{5\sigma^2} = \sigma\sqrt{5}.$$
The question stipulates that in less than 0.5% of cases should the total be less than 7.5 ml. For the (standard) Normal distribution we remember (or can compute) that exactly 0.5% of cases are $$2.57\ldots$$ or more less than the mean. An example of this computation is
qnorm(0.5/100)
in R or
=NORMSINV(0.5/100)
in Excel, for instance.
One aim of the introductory course is to help you reach the point where such considerations are automatic: you can do them in your head correctly, apart (perhaps) from the arithmetical calculations.
This preliminary work enables us to rephrase the question like this:
What unit of measurement, given by $$\sigma\sqrt{5}$$ for a five-pack of drugs, will re-express an amount of $$7.5$$ ml as being $$2.57$$ less than $$7.75$$ ml?
The solution obviously is
$$\sigma\sqrt{5} = (7.75 - 7.5)/2.57\ldots = 0.097\ldots,$$
implying
$$\sigma = \frac{0.097\ldots}{\sqrt{5}} = 0.0433797\ldots$$
Comparing this result to the question shows that the work in the question was entirely correct up to the point where "$$\sigma/5$$" appeared: the square root was lost. This suggests remembering to think in terms of variances rather than standard deviations.
Comparing this result to the older answers that were posted also shows how they were basically moving in the correct direction but made mistakes along the way, too. Because arithmetical mistakes are easy to make, when one has the chance it's a good idea to check probabilistic calculations with simulations. For instance, the following R statement generates a large number of five-packs of drugs as described in the question (using the answer I obtained) and, to check my answer, computes the fraction with totals less than 7.5 ml:
mean(colSums(matrix(rnorm(5*1e6, 1.55, -0.25/qnorm(0.5/100) / sqrt(5)), nrow=5)) < 7.5)
(You can see all the data from the question embedded in this expression, along with the value 1e6 giving the number of five-packs to simulate.) When I run and re-run this code (which takes less than a second each time), I consistently obtain results between 0.0048 (0.48%) and 0.0052 (0.52%), in satisfactory agreement with the intended 0.5% target.
I think your understanding of the variance of the sum is mistaken. The variance of the 5-pack sum is 25 times the variance of the single pack.
• The only way you could justify the factor of $25$ is to suppose the five packs are perfectly correlated. Assuming, as is more likely the intent, that the five-pack sum can be modeled as the total of five independent Normal variables $X_1+\cdots+X_5$, its variance will be $$\operatorname{Var}(X_1+\cdots+X_5)=\sigma^2 +\cdots+\sigma^2=5\sigma^2.$$ Consequently its standard deviation will be $\sigma\sqrt{5}$, which (by following the path outlined in the question) leads directly to the correct answer.
– whuber
Jun 14 '16 at 14:01
\begin{align} 5\times 1.55 &= 7.75 \\ 5\times SD &= 5SD \end{align} Problem statement: $$P(X<7.5)<0.005$$ \begin{align} \frac{(7.5-7.75)}{(5SD^2)^{1/2}} &< 0.005 \\[8pt] \frac{-0.25}{2.576} &= (5SD^2)^{1/2} \\[5pt] &-0.0970 / 5^{(1/2)} \end{align} $$0.0434$$ as the standard deviation can never me negative, take the mod value.
• Welcome to Stats.SE. You may give hints but please do not give the full answer. Furthermore, can you please edit your post and explain the key steps in the solution and use MathJax in the formulas? Apr 26 '19 at 10:51 | 2021-12-08T01:12:44 | {
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http://mathhelpforum.com/math-topics/36935-problem-solving-sucessive-tetrahedral-numbers.html | # Math Help - Problem solving - sucessive tetrahedral numbers
1. ## Problem solving - sucessive tetrahedral numbers
would you guys be able to give me a hand with this problem solving question, i have no idea where to start...
Thanks
Zac
2. Hello, Zac!
12. The first four tetrahedral numbers are: . $1,\;4,\;10,\;20$
Find the pattern, then predict the next 3 terms of the sequence.
Take the difference of consecutive pairs of terms.
$\begin{array}{cccccccc}\text{Sequence:} & 1 && 4 && 10 && 20 \\
\text{Difference:} & & 3 & & 6 & & 10 \end{array}$
They are adding on consecutive "triangular" numbers.
. . $\begin{array}{ccc}3 & = & 1+2 \\ 6 &=&1+2+3\\ 10 &=&1+2+3+4 \end{array}$
We will add on the next three triangular numbers:
. . $\begin{array}{ccc}1+2+3+4+5 &=& 15 \\ 1+2+3+4+5+6 &=& 21 \\ 1+2+3+4+5+6+7 &=& 28\end{array}$
$\begin{array}{cccccccccccccc}\text{Sequence:} & 1 && 4 && 10 && 20 && {\color{blue}35} && {\color{blue}56} && {\color{blue}84}\\
\text{Difference:} & & 3 & & 6 & & 10 && +15 && +21 && +28\end{array}$
What is the $n^{th}$ term of the sequence?
I'll skip all the algebra . . .
$\text{The }n^{th}\text{ term is: }\;a_n \;=\;\frac{n(n+1)(n+2)}{6}$
3. Originally Posted by Soroban
I'll skip all the algebra . . .
$\text{The }n^{th}\text{ term is: }\;a_n \;=\;\frac{n(n+1)(n+2)}{6}$
as this is a problwm solving questions and according to my sheet i must show the algebra would you mind putting it up?
thanks so much!
Zac
4. Hello, Zac!
This will take a while . . . better sit down.
Take differences of consecutive terms,
. . then take differences of the differences, and so on.
$\begin{array}{cccccccccccccc}\text{Sequence} & 1 && 4 && 10 && 20 && 35 && 56 && 84 \\
\text{1st diff.} & & 3 && 6 && 10 && 15 && 21 && 28 \\
\text{2nd diff.} & & & 3 && 4 && 5 && 6 && 7 \\
\text{3rd diff.} & & & & 1 && 1 && 1 && 1 \end{array}$
The third differences are constant.
. . This tells us that the generating fuction is a cubic.
The general cubic function is: . $f(n) \;=\;an^3 + bn^2 + cn + d$
We will use the first four terms of the sequence . . .
$\begin{array}{ccccc}f(1)\:=\:1\!: & a + b + c + d & = & 1 & [1] \\
f(2) \:=\:4\!: & 8a + 4b + 2c + d &=& 4 & [2] \\
f(3) \:=\:10\!: & 27a + 9b + 3c + d &=& 10 & [3] \\
f(4) \:=\:20\!: & 64a + 16b + 4c + d &=& 20 & [4] \end{array}$
$\begin{array}{ccccc}\text{Subtract [2] - [1]:} & 7a + 3b + c &=& 3 & [5] \\
\text{Subtract [3] - [2]:} & 19a + 5b + c &=& 6 & [6] \\
\text{Subtract [4] - [3]:} & 37a + 7b + c &=& 10 & [7] \end{array}$
$\begin{array}{ccccc}\text{Subtract [6] - [5]:} & 12a + 2b &=& 3 & [8] \\
\text{Subtract [7] - [6]:} & 18a + 2b &=& 4 & [9] \end{array}$
$\begin{array}{ccccccc}\text{Subtract [9] - [8]:}& 6a \;=\; 1 & \Rightarrow & \boxed{a \;=\;\frac{1}{6}} \end{array}$
Substitute into [8]: . $12\left(\frac{1}{6}\right) + 2b \:=\:3\quad\Rightarrow\quad\boxed{ b \:=\:\frac{1}{2}}$
Substitute into [5]: . $7\left(\frac{1}{6}\right) + 3\left(\frac{1}{2}\right) + c \:=\:3 \quad\Rightarrow\quad \boxed{c\:=\:\frac{1}{3}}$
Substitute into [1]: . $\frac{1}{6} + \frac{1}{2} + \frac{1}{3} + d \:=\:1 \quad\Rightarrow\quad\boxed{ d\:=\:0}$
Hence, the function is: . $f(n) \;=\;\frac{1}{6}n^3 + \frac{1}{2}n^2 + \frac{1}{3}n \;=\;\frac{n}{6}(n^2+3n+ 2)$
. . Therefore: . $\boxed{f(n) \;=\;\frac{n(n+1)(n+2)}{6}}$ | 2015-05-26T14:19:26 | {
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http://math.stackexchange.com/questions/29023/value-of-sum-limits-n-xn | # Value of $\sum\limits_n x^n$
Why is $\displaystyle \sum\limits_{n=0}^{\infty} 0.7^n$ equal $1/(1-0.7) = 10/3$ ?
Can we generalize the above to
$\displaystyle \sum_{n=0}^{\infty} x^n = \frac{1}{1-x}$ ?
Are there some values of $x$ for which the above formula is invalid?
What about if we take only a finite number of terms? Is there a simpler formula?
$\displaystyle \sum_{n=0}^{N} x^n$
Is there a name for such a sequence?
This is being repurposed in an effort to cut down on duplicates, see here: Coping with abstract duplicate questions.
and here: List of abstract duplicates.
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This is a geometric series. See en.wikipedia.org/wiki/Geometric_series#Formula – lhf Mar 25 '11 at 15:47
Please replace the starting index $n=1$ by $n=0$ (otherwise the resulting sum is $0.7$ times what you wrote). – Did Mar 25 '11 at 17:59
@Jonas: Fair point: done. – Arturo Magidin Mar 25 '11 at 19:18
@Didier: That was my fault upon edit; fixed. – Arturo Magidin Mar 25 '11 at 19:19
@Arturo: Thanks. Sorry, I shouldn't have deleted my comment; I did so after I saw you had changed the title, and only then saw your comment. (To anyone who is confused: I had pointed out that including "geometric series" in the title gave additional context that made lhf's and yoyo's first sentences seem a little strange.) – Jonas Meyer Mar 25 '11 at 19:41
By definition, a "series" (an "infinite sum") $$\sum_{n=k}^{\infty} a_n$$ is defined to be a limit, namely $$\sum_{n=k}^{\infty} a_n= \lim_{N\to\infty} \sum_{n=k}^N a_n.$$ That is, the "infinite sum" is the limit of the "partial sums", if this limit exists. If the limit exists, equal to some number $S$, we say the series "converges" to the limit, and we write $$\sum_{n=k}^{\infty} a_n = S.$$ If the limit does not exist, we say the series diverges and is not equal to any number.
So writing that $$\sum_{n=0}^{\infty} 0.7^n = \frac{1}{1-0.7}$$ means that we are asserting that $$\lim_{N\to\infty} \sum_{n=0}^N0.7^n = \frac{1}{1-0.7}.$$
So what your question is really asking is: why is this limit equal to $\frac{1}{1-0.7}$? (Or rather, that is the only way to make sense of the question).
In order to figure out the limit, it is useful (but not strictly necessary) to have a formula for the partial sums, $$s_N = \sum_{n=0}^N 0.7^n.$$ This is where the formulas others have given come in. If you take the $N$th partial sum and multiply by $0.7$, you get $$\begin{array}{rcrcrcrcrcrcl} s_N &= 1 &+& (0.7) &+& (0.7)^2 &+& \cdots &+& (0.7)^N\\ (0.7)s_N &= &&(0.7) &+& (0.7)^2 &+&\cdots &+&(0.7)^N &+& (0.7)^{N+1} \end{array}$$ so that $$(1-0.7)s_N = s_N - (0.7)s_N = 1 - (0.7)^{N+1}.$$ Solving for $s_N$ gives $$s_N = \frac{1 - (0.7)^{N+1}}{1-0.7}.$$ What is the limit as $N\to\infty$? The only part of the expression that depends on $N$ is $(0.7)^{N+1}$. Since $|0.7|\lt 1$, then $\lim\limits_{N\to\infty}(0.7)^{N+1} = 0$. So, $$\lim_{N\to\infty}s_N = \lim_{N\to\infty}\left(\frac{1-(0.7)^{N+1}}{1-0.7}\right) = \frac{\lim\limits_{N\to\infty}1 - \lim\limits_{N\to\infty}(0.7)^{N+1}}{\lim\limits_{N\to\infty}1 - 0.7} = \frac{1 - 0}{1-0.7} = \frac{1}{1-0.7}.$$ Since the limit exists, then we write $$\sum_{n=0}^{\infty}(0.7)^n = \frac{1}{1-0.7}.$$
More generally, a sum of the form $$a + ar + ar^2 + ar^3 + \cdots + ar^k$$ with $a$ and $r$ constant is said to be a "geometric series" with initial term $a$ and common ratio $r$. If $a=0$, then the sum is equal to $0$. If $r=1$, then the sum is equal to $(k+1)a$. If $r\neq 1$, then we can proceed as above. Letting $$S = a +ar + \cdots + ar^k$$ we have that $$S - rS = (a+ar+\cdots+ar^k) - (ar+ar^2+\cdots+a^{k+1}) = a - ar^{k+1}$$ so that $$(1-r)S = a(1 - r^{k+1}).$$ Dividing through by $1-r$ (which is not zero since $r\neq 1$), we get $$S = \frac{a(1-r^{k+1})}{1-r}.$$
A series of the form $$\sum_{n=0}^{\infty}ar^{n}$$ with $a$ and $r$ constants is called an infinite geometric series. If $r=1$, then $$\lim_{N\to\infty}\sum_{n=0}^{N}ar^{n} = \lim_{N\to\infty}\sum_{n=0}^{N}a = \lim_{N\to\infty}(N+1)a = \infty,$$ so the series diverges. If $r\neq 1$, then using the formula above we have: $$\sum_{n=0}^{\infty}ar^n = \lim_{N\to\infty}\sum_{n=0}^{N}ar^{N} = \lim_{N\to\infty}\frac{a(1-r^{N+1})}{1-r}.$$ The limit exists if and only if $\lim\limits_{N\to\infty}r^{N+1}$ exists. Since $$\lim_{N\to\infty}r^{N+1} = \left\{\begin{array}{ll} 0 &\mbox{if |r|\lt 1;}\\ 1 & \mbox{if r=1;}\\ \text{does not exist} &\mbox{if r=-1 or |r|\gt 1} \end{array}\right.$$ it follows that: \begin{align*} \sum_{n=0}^{\infty}ar^{n} &=\left\{\begin{array}{ll} 0 &\mbox{if a=0;}\\ \text{diverges}&\mbox{if a\neq 0 and r=1;}\\ \lim\limits_{N\to\infty}\frac{a(1-r^{N+1})}{1-r} &\mbox{if r\neq 1;}\end{array}\right.\\ &= \left\{\begin{array}{ll} \text{diverges}&\mbox{if a\neq 0 and r=1;}\\ \text{diverges}&\mbox{if a\neq 0, and r=-1 or |r|\gt 1;}\\ \frac{a(1-0)}{1-r}&\mbox{if |r|\lt 1;} \end{array}\right.\\ &=\left\{\begin{array}{ll} \text{diverges}&\mbox{if a\neq 0 and |r|\geq 1;}\\ \frac{a}{1-r}&\mbox{if |r|\lt 1.} \end{array}\right. \end{align*}
Your particular example has $a=1$ and $r=0.7$.
Since this recently came up (09/29/2011), let's provide a formal proof that $$\lim_{N\to\infty}r^{N+1} = \left\{\begin{array}{ll} 0 &\mbox{if |r|\lt 1;}\\ 1 & \mbox{if r=1;}\\ \text{does not exist} &\mbox{if r=-1 or |r|\gt 1} \end{array}\right.$$
If $r\gt 1$, then write $r=1+k$, with $k\gt0$. By the binomial theorem, $r^n = (1+k)^n \gt 1+nk$, so it suffices to show that for every real number $M$ there exists $n\in\mathbb{N}$ such that $nk\gt M$. This is equivalent to asking for a natural number $n$ such that $n\gt \frac{M}{k}$, and this holds by the Archimedean property; hence if $r\gt 1$, then $\lim\limits_{n\to\infty}r^n$ does not exist. From this it follows that if $r\lt -1$ then the limit also does not exist: given any $M$, there exists $n$ such that $r^{2n}\gt M$ and $r^{2n+1}\lt M$, so $\lim\limits_{n\to\infty}r^n$ does not exist if $r\lt -1$.
If $r=-1$, then for every real number $L$ either $|L-1|\gt \frac{1}{2}$ or $|L+1|\gt \frac{1}{2}$. Thus, for every $L$ and for every $M$ there exists $n\gt M$ such that $|L-r^n|\gt \frac{1}{2}$ proving the limit cannot equal $L$; thus, the limit does not exist. If $r=1$, then $r^n=1$ for all $n$, so for every $\epsilon\gt 0$ we can take $N=1$, and for all $n\geq N$ we have $|r^n-1|\lt\epsilon$, hence $\lim\limits_{N\to\infty}1^n = 1$. Similarly, if $r=0$, then $\lim\limits_{n\to\infty}r^n = 0$ by taking $N=1$ for any $\epsilon\gt 0$.
Next, assume that $0\lt r\lt 1$. Then the sequence $\{r^n\}_{n=1}^{\infty}$ is strictly decreasing and bounded below by $0$: we have $0\lt r \lt 1$, so multiplying by $r\gt 0$ we get $0\lt r^2 \lt r$. Assuming $0\lt r^{k+1}\lt r^k$, multiplying through by $r$ we get $0\lt r^{k+2}\lt r^{k+1}$, so by induction we have that $0\lt r^{n+1}\lt r^n$ for every $n$.
Since the sequence is bounded below, let $\rho\geq 0$ be the infimum of $\{r^n\}_{n=1}^{\infty}$. Then $\lim\limits_{n\to\infty}r^n =\rho$: indeed, let $\epsilon\gt 0$. By the definition of infimum, there exists $N$ such that $\rho\leq r^N\lt \rho+\epsilon$; hence for all $n\geq N$, $$|\rho-r^n| = r^n-\rho \leq r^N-\rho \lt\epsilon.$$ Hence $\lim\limits_{n\to\infty}r^n = \rho$.
In particular, $\lim\limits_{n\to\infty}r^{2n} = \rho$, since $\{r^{2n}\}_{n=1}^{\infty}$ is a subsequence of the converging sequence $\{r^n\}_{n=1}^{\infty}$. On the other hand, I claim that $\lim\limits_{n\to\infty}r^{2n} = \rho^2$: indeed, let $\epsilon\gt 0$. Then there exists $N$ such that for all $n\geq N$, $r^n - \rho\lt\epsilon$. Moreover, we can assume that $\epsilon$ is small enough so that $\rho+\epsilon\lt 1$. Then $$|r^{2n}-\rho^2| = |r^n-\rho||r^n+\rho| = (r^n-\rho)(r^n+\rho)\lt (r^n-\rho)(\rho+\epsilon) \lt r^n-\rho\lt\epsilon.$$ Thus, $\lim\limits_{n\to\infty}r^{2n} = \rho^2$. Since a sequence can have only one limit, and the sequence of $r^{2n}$ converges to both $\rho$ and $\rho^2$, then $\rho=\rho^2$. Hence $\rho=0$ or $\rho=1$. But $\rho=\mathrm{inf}\{r^n\mid n\in\mathbb{N}\} \leq r \lt 1$. Hence $\rho=0$.
Thus, if $0\lt r\lt 1$, then $\lim\limits_{n\to\infty}r^n = 0$.
Finally, if $-1\lt r\lt 0$, then $0\lt |r|\lt 1$. Let $\epsilon\gt 0$. Then there exists $N$ such that for all $n\geq N$ we have $|r^n| = ||r|^n|\lt\epsilon$, since $\lim\limits_{n\to\infty}|r|^n = 0$. Thus, for all $\epsilon\gt 0$ there exists $N$ such that for all $n\geq N$, $| r^n-0|\lt\epsilon$. This proves that $\lim\limits_{n\to\infty}r^n = 0$, as desired.
In summary, $$\lim_{N\to\infty}r^{N+1} = \left\{\begin{array}{ll} 0 &\mbox{if |r|\lt 1;}\\ 1 & \mbox{if r=1;}\\ \text{does not exist} &\mbox{if r=-1 or |r|\gt 1} \end{array}\right.$$
The argument suggested by Srivatsan Narayanan in the comments to deal with the case $0\lt|r|\lt 1$ is less clumsy than mine above: there exists $a\gt 0$ such that $|r|=\frac{1}{1+a}$. Then we can use the binomial theorem as above to get that $$|r^n| = |r|^n = \frac{1}{(1+a)^n} \leq \frac{1}{1+na} \lt \frac{1}{na}.$$ By the Archimedean Property, for every $\epsilon\gt 0$ there exists $N\in\mathbb{N}$ such that $Na\gt \frac{1}{\epsilon}$, and hence for all $n\geq N$, $\frac{1}{na}\leq \frac{1}{Na} \lt\epsilon$. This proves that $\lim\limits_{n\to\infty}|r|^n = 0$ when $0\lt|r|\lt 1$, without having to invoke the infimum property explicitly.
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I think you mean "which is not zero since $r\neq 1$." – robjb Mar 26 '11 at 23:20
@Rob: Yes; thank you. – Arturo Magidin Mar 27 '11 at 0:22
+1 As its rare to see math researchers caring for even elementary questions. – Please Delete Account Mar 27 '11 at 2:18
Ignore my edit, I seem to have potatoes on my eyes. Sorry. – Lagerbaer Mar 31 '11 at 5:05
Arturo, I will mention one more way of proving that $r^n$ converges to $0$ as $n \to \infty$ if $|r| < 1$ that looks somewhat similar to your proof in the divergent case. Take $|r| = \frac{1}{1+a}$ for some $a > 0$. Then $$|r^n - 0| = |r|^n = \frac{1}{(1+a)^n} \leq \frac{1}{1+an} \leq \frac{1}{an} ,$$ which can be made sufficiently small by taking $n$ large enough. – Srivatsan Sep 30 '11 at 4:13
it's called a geometric series. let $-1<x<1$ and let $S_n=1+x+x^2+...+x^n$. then $xS_n=x+...+x^{n+1}=S^n-1+x^{n+1}$. move stuff around to get $$S_n=\frac{1-x^{n+1}}{1-x}$$ take the limit as $n\to\infty$ (noting that $x^n\to0$ if $|x|<1$)
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In yoyo's answer, you should also take into account that $x^{n+1}\to 0$ as $n\to\infty$ if $|x| < 1$. – knightofmathematics Mar 25 '11 at 16:48
...and $0^0=1$ (: – yoyo Mar 25 '11 at 19:03
If u expand your summation you get series $$1+0.7+0.7^2+\dots$$ as it is geometric series (you can see it here http://en.wikipedia.org/wiki/Geometric_series) $$\sum_{n=0}^{\infty}{0.7^n}=\frac{1}{1-0.7}$$
Or $S=1+x+x^2+\dots$
$xS_n=x+x^2+\dots=S-1$ now u take
$S-xS=1$
$S(1-x)=1 \implies S=\frac{1}{1-x}$
here your $x=0.7$
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The index in the sum should be $n$, not $i$. – Arturo Magidin Mar 25 '11 at 17:10
oops..i'll edit it – amul28 Mar 26 '11 at 3:54
I find the proof here lovely.
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The link here no longer seems to work. – Baby Dragon Oct 6 '13 at 1:19
@BabyDragon Here is the updated link. – David Mitra Oct 6 '13 at 9:12
## protected by AryabhataMar 31 '11 at 17:18
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https://mathoverflow.net/questions/278629/a-combinatorial-identity | # A combinatorial identity
I hope this is a suitable MO question. In a research project, my collaborator and I came across some combinatorial expressions. I used my computer to test a few numbers and the pattern was suggesting the following equation for fixed integers $K\geq n>0$.
$$\dfrac{K!}{n!K^{K-n}}\sum\limits_{ \begin{subarray}{c} k_1+\dotsb+k_{n}=K \\ k_i \geq 1 \end{subarray}} \prod\limits_{i=1}^n \dfrac{k_i^{k_i-2}}{(k_i-1)!}=\displaystyle {K-1\choose n-1}.$$
We tried to think of a proof but failed. One can probably move these $K!, n!$ to the right and rewrite the RHS, or maybe move $K!$ into the summation to form combinatorial numbers like $K\choose k_1,k_2,\dotsc,k_n$. We don't know which is better.
The questions are:
1. Anyone knows a proof for this identity?
2. In fact the expression that appears in our work is $\sum\limits_{ \begin{subarray}{c} k_1+\dotsb+k_{n}=K \\ k_i \geq 1 \end{subarray}} \sigma_p(k_1,\dotsc,k_n) \prod\limits_{i=1}^n \dfrac{k_i^{k_i-2}}{(k_i-1)!}$, where $p$ is a fixed integer and $\sigma_p(\dotsc)$ is the $p$-th elementary symmetric polynomial. The equation in the beginning simplifies this expression for $p=0,1$. Is there a similar identity for general $p$?
----------Update----------
Question 2 is perhaps too vague, and I'd like to make it a bit more specific. Probably I should have written this down in the beginning, but I feared this is too long and unmotivated. But after seeing people's skills, I'm very tempted to leave it here in case somebody has remarks. In fact, question 2 partly comes from the effort to find a proof for the following (verified by computer).
$$\frac{1}{K!} \prod_{r=1}^{K} (r+1 -x)= \sum_{n=1}^K \frac{(-1)^n}{n!} \left[ \sum_{p=0}^n K^{n-p} \prod_{r=1}^p (x +r-4) \left( \sum\limits_{ \begin{subarray}{c} k_1+\dotsb+k_{n}=K \\ k_i \geq 1 \end{subarray}} \sigma_p(k_1,\dotsc,k_{n}) \prod\limits_{i=1}^n \dfrac{k_i^{k_i-2}}{(k_i-1)!} \right) \right],$$ Where $x$ is a fixed number (in our case, an integer).
• Combinatorial reformulation: consider all the trees $T$ on $\{0,1,\dots,K\}$ for which the vertices $1,2,\dots,p$ are in different components of $T\setminus \{0\}$. For each such a tree take a summand $(-K)^{\deg(0)-p}$. The sum equals $(K-2)!/(p-2)!$ for $p\geqslant 2$ and 0 for $p=0,1$. – Fedor Petrov Aug 15 '17 at 9:35
This is the answer to the first question, I wrote a long answer to Question 2 as a separate answer.
Note that $A:=\sum_{k_i>0,k_1+\dots+k_n=K}\frac{K!}{n!k_1!\dots k_n!} \prod k_i^{k_i-1}$ is a number of forests on the ground set $\{1,2,\dots,K\}$ having exactly $n$ connected components and with a marked vertex in each component ($k_i$ correspond to the sizes of components.) Add a vertex 0 and join it with the marked vertices. Then we have to count the number of trees on $\{0,1,\dots,K\}$ in which $0$ has degree $n$. Remember that the sum of $z_0^{d_1-1}\dots z_K^{d_K-1}$ over all trees on $\{0,\dots,K\}$, where $d_i$ is degree of $i$, equals $(z_0+\dots+z_K)^{K-1}$. Substitute $z_{1}=\dots=z_K=1$ and take a coefficient of $z_0^{n-1}$. It is $\binom{K-1}{n-1}K^{K-n}$.
• Thank you! This is a really cool argument, especially for me who doesn't have a lot of experiences in graph theory. I will mark this as the correct answer, but of course any further ideas and comments about question 2 are welcomed as well. – Honglu Aug 13 '17 at 18:41
• You need to talk to someone who's familiar with Volume 2 of Stanley. (This is more algebraic combinatorics than graph theory.) – Alexander Woo Aug 13 '17 at 23:10
• Sorry about my ignorance and thank you for pointing out the right words! I am thinking about asking around, and your comment definitely helps. – Honglu Aug 13 '17 at 23:58
Here's another proof. We first rewrite the identity (by setting $k_i=j_i+1$) as $$\sum_{j_1+\cdots +j_n=K-n}\prod_{i=1}^n \frac{(j_i+1)^{j_i-1}}{j_i!} = n\frac{K^{K-n-1}}{(K-n)!}. \tag{1}$$
Let $F(x)$ be the formal power series satisfying $F(x)= e^{xF(x)}$. It is well known (and easily proved, e.g., by Lagrange inversion) that $$F(x)^n = \sum_{j=0}^\infty n(j+n)^{j-1}\frac{x^j}{j!}.\tag{2}$$ In particular, $$F(x) = \sum_{j=0}^\infty (j+1)^{j-1}\frac{x^j}{j!}.$$ So the left side of $(1)$ is equal to the coefficient of $x^{K-n}$ in $F(x)^{n}$, which by $(2)$ is equal to the right side of $(1)$.
• Cool! I actually updated my question to include a more general identity that I want to prove. I kept thinking there is some generating function lurking behind, and I'm just looking at the coefficients. Any thoughts will be greatly appreciated. – Honglu Aug 14 '17 at 16:04
Here is a generating-function proof of your conjectured identity (and an answer to question 2).
The main ingredient is a formula for the appearing symmetric sums.
Let $T(z)$ (the tree function'') be the formal power series satisfying $T(z)=z\,e^{T(z)}$.
If $F$ is a formal power series the coefficients of $G(z):=F(T(z))$ are given by (Lagrange inversion) $$[z^0]G(z)=[z^0] F(z) \mbox{ , } [z^k]G(z)=\tfrac{1}{k} [y^{k-1}] F^\prime(y)\,e^{ky} =[y^k](1-y)F(y)\,e^{ky}\mbox{ for } k\geq 1\;.$$ In particular (as is well known) $$T(z)=\sum_{n\geq 1}\frac{n^{n-1}}{n!}z^n \;\mbox{ and }\; \frac{T(z)}{1-T(z)}=\sum_{n\geq 1}\frac{n^{n}}{n!}z^n$$ Thus $T(z)\left(1+\frac{t}{1-T(z)}\right)=\sum_{n\geq 1} \frac{(1+tn)n^{n-1}}{n!}$. Therefore \begin{align*}S_{p,n}(K):&=\sum_{{k_1+\ldots +k_n=K \atop k_i\geq 1}} \sigma_p(k_1,\ldots,k_n)\prod_{i=1}^n \frac{k_i^{k_i-1}}{k_i!}\\ &=[t^p]\sum_{k_1+\ldots +k_n=K \atop k_i\geq 1} \prod_{i=1}^n \frac{(1+tk_i) k_i^{k_i-1}}{k_i!}\\ &=[t^p z^K]\, T(z)^n \left(1+\frac{t}{1-T(z)}\right)^n \end{align*} and $$S_{p,n}(K)={n \choose p} [z^K]\frac {T(z)^n}{(1-T(z))^p}={n \choose p}[y^K]\,y^n\, \frac{(1-y)}{(1-y)^p}\,e^{Ky}\;\;\;\;\;(*)$$
Now consider the sum ($m:=x-4$) $$R(K,m):=\sum_{n=0}^K\frac{(-1)^n}{n!}\bigg[\sum_{p=0}^n K^{n-p}\,\left(\prod_{r=1}^p (m+r)\right) S_{n,p}(K)\bigg]$$ Since $K\geq 1$ the sum remains unchanged if we start the summation at $n=0$ (all $S_{0,p}(K)$ are $0$). Using that and $(*)$ gives \begin{align*} R(K,m):&=\sum_{n=0}^K\frac{(-1)^n}{n!}\bigg[\sum_{p=0}^n K^{n-p}\,\left(\prod_{r=1}^p (m+r)\right) S_{n,p}(K)\bigg]\\ &=[y^K]\,(1-y) \sum_{n=0}^K\frac{(-1)^n}{n!}\sum_{p=0}^n K^{n-p}\,p!{m+p \choose p}{n \choose p}\frac{y^n}{(1-y)^p}\,e^{Ky}\\ &=[y^K]\,(1-y) \sum_{p=0}^K\sum_{n=p}^K\frac{(-1)^n}{(n-p)!} K^{n-p}{m+p \choose p}\frac{y^n}{(1-y)^p}\,e^{Ky}\\ &=[y^K]\,(1-y) \sum_{p=0}^K {m+p \choose p}\frac{(-1)^p y^p}{(1-y)^p}\bigg[\sum_{n=p}^K\frac{(-1)^{n-p}}{(n-p)!} K^{n-p}y^{n-p}\,e^{Ky}\bigg]\\ &=[y^K]\,(1-y) \sum_{p=0}^K {m+p \choose p}\frac{(-1)^p y^p}{(1-y)^p}\bigg[1 +\mathcal{O}(y^{K-p+1})\bigg]\\ \end{align*} where $\mathcal{O}(y^{K-p+1})$ denotes a formal power series which is a multiple of $y^{K-p+1}$.
Taking into account the respective factors $y^p$ the terms in these series do not contribute to the coefficient $[y^K]$. Therefore
\begin{align*} R(K,m)&=[y^K]\,(1-y) \sum_{p=0}^K {m+p \choose p}\frac{(-1)^p y^p}{(1-y)^p}\\ &=[y^K]\,(1-y) \sum_{p\geq 0} {m+p \choose p}\frac{(-1)^p y^p}{(1-y)^p}\\ &=[y^K]\,(1-y) \left(\frac{1}{1+\tfrac{y}{1-y}}\right)^{m+1}\\ &=[y^K]\,(1-y)^{m+2}\\ &=(-1)^K\,{m+2 \choose K},\,\mbox{ as desired }. \end{align*}
• Nice! I really learned a lot from all the answers and appreciate everybody's effort. In particular, this generating function method seems to have some other applications in our project. I got another quick question. Let $H_n=\sum\limits_{k=1}^n 1/k$. Do you know whether we can similarly describe the formal series $\sum_{n\geq 1} \dfrac{H_nn^n}{n!}z^n$ using functional equations just like your $T(z)$ and others? – Honglu Aug 15 '17 at 22:33
• Thank you. (1) The formal series $\sum_{n\geq 1}\frac{n^n}{n!}p(n) z^n$ can be expressed as a rational function of $T(z)$ if $p$ is a polynomial in $n$ and $\tfrac{1}{n}$. My first guess is that it will not be easy to treat the case $p(n)=H_n$ via Lagrange inversion. (2) $T(z)$ is certainly not "my" function. $T(z)=-W(-z)$ where $W$ is "Lambert's W-function", and $T(z)=z e^{T(z)}$ goes back to Eisenstein. – esg Aug 16 '17 at 17:57
• Would you mind telling me your real name by email? Because if we decide to post anything about this work in the future, we will acknowledge you (also Fedor and other people). I just temporarily added my email in my MO profile. We are still working on an ultimate combinatorial expression that includes all my questions as special cases. It's too long to post in a comment, but if you are interested, I will be glad to send it to you by email as well. Of course interests from other people will also be welcomed, just let me know in the comment or send me an email. – Honglu Aug 24 '17 at 16:09
Here is the answer to Question 2. It may be probably simplified.
Denote $y=3-x$, then we rewrite your identity as $$\binom{y+K-2}K=\frac{(y-1)y(y+1)\dots (y+K-2)}{K!}=c_0\binom{y}0+c_1\binom{y}1+\dots+c_K\binom{y}K,$$ where $$c_p=p!\sum_{n=p}^K(-K)^{n-p}\frac1{n!}\sum\limits_{ \begin{subarray}{c} k_1+\dotsb+k_{n}=K \\ k_i \geq 1 \end{subarray}} \sigma_p(k_1,\dotsc,k_{n}) \prod\limits_{i=1}^n \dfrac{k_i^{k_i-2}}{(k_i-1)!}.$$ On the other hand, by Vandermonde--Chu identity we have $$\binom{y+K-2}K=\sum_{i=2}^{K}\binom{y}i\binom{K-2}{K-i},$$ so your identity is equivalent to the formula $$\sum_{n=p}^K(-K)^{n-p}\frac{K!}{n!}\sum\limits_{ \begin{subarray}{c} k_1+\dotsb+k_{n}=K \\ k_i \geq 1 \end{subarray}} \sigma_p(k_1,\dotsc,k_{n}) \prod\limits_{i=1}^n \dfrac{k_i^{k_i-2}}{(k_i-1)!}=\frac{K!}{p!}\binom{K-2}{K-p},$$ I multiplied both parts by $K!/p!$. Note that $$\frac{K!}{n!}\sum\limits_{ \begin{subarray}{c} k_1+\dotsb+k_{n}=K \\ k_i \geq 1 \end{subarray}} \sigma_p(k_1,\dotsc,k_{n}) \prod\limits_{i=1}^n \dfrac{k_i^{k_i-2}}{(k_i-1)!}$$ is a number of the trees $T$ on $\{0,1,\dots,K\}$ such that degree of 0 equals $n$ and $p$ vertices in different components of $T\setminus\{0\}$ are marked. Indeed, if these components $A_1,\dots,A_n$ are enumerated (this corresponds to the multiple $n!$) and $i$-th component $A_i$ has $k_i$ vertices, then we have $\frac{K!}{k_1!\dots k_n!}$ ways to choose $A_i$, $\sigma_p(k_1,\dotsc,k_{n})$ ways to mark $p$ vertices in different components, $k_i^{k_i-1}$ ways to make a tree on $A_i$ and choose a vertex in $A_i$ joined with 0.
Note that each (out of $\binom{K}p$ sets) set of $p$ marked vertices makes the same contribution to the sum. So, we may suppose that the marked set is $\{1,2,\dots,p\}$ and we have to prove that the sum of $(-K)^{n-p}$ over admissible trees (where the tree $T$ is admissible if $1,2,\dots,p$ are in different components of $T\setminus \{0\}$) equals $\frac1{\binom{K}p}\frac{K!}{p!}\binom{K-2}{K-p}=(p-1)p\dots (K-2)$.
We start to prove this from the case $p=0$, $p=1$, where the restriction that $1,2,\dots,p$ are in different components of $T\setminus \{0\}$ disappears. Then the sum $z_0^{n-1}z_1^{d_1-1}\dots z_K^{d_K-1}$, $d_i=\deg(i)$, over all trees on $\{0,\dots,K\}$ equals, as is well known and easy to prove, to $(z_0+\dots+z_K)^{K-1}$. Substituting $z_0=-K$, $z_1=\dots=z_K=1$ we get the result.
Now we deal with the more involved case $p\geqslant 2$. Denote $K=p+m$ and consider the variables $z_0,z_1,\dots,z_p,z_{p+1},\dots$ (infinitely many for simplicity of notations). Denote $s=z_0+z_1+\dots$, write $\sigma_i$ for the $i$-th elementary symmetric polynomial of $z_{p+1},z_{p+2},\dots$. Denote $\varphi_0=1$, $\varphi_m=s\varphi_{m-1}+(p-1)p\dots (p+m-2)\sigma_m$ for $m\geqslant 1$. I claim that the sum of $z_0^{n-p}z_1^{d_1-1}\dots z_{p+m}^{d_{p+m}-1}$ over all admissible trees equals $\varphi_m(z_0,z_1,\dots,z_{p+m},0,0,\dots)$.
Note that this implies our claim, as follows from the substitution $z_0=-K=-p-m,z_1=\dots=z_{p+m}=1$.
The proof is on induction in $m$. Base $m=0$ is clear. For the induction step, look at coefficients of any specific monomial $z_0^{n-p}z_1^{d_1-1}\dots z_{p+m}^{d_{p+m}-1}$. Consider two cases:
1) $d_i=1$ for a certain index $i\in \{p+1,\dots,p+m\}$, without loss of generality $i=p+m$. This corresponds to the case when $p+m$ has degree 1, such a vertex may be joined with any of other vertices, and removing corresponding edge we get a tree (it remains admissible) on $\{0,1,\dots,K-1\}$. This corresponds to the summand $s\varphi_{m-1}$: namely, $z_j\varphi_{m-1}$ corresponds to the edge between $p+m$ and $j$; $j=0,1,\dots,p+m-1$.
2) $d_{p+1},\dots,d_{p+m}$ are greater than 1. Then they are all equal to 2, since the degree of the whole monomial equals $m$. In this case there are $p(p+1)\dots (p+m-1)$ admissible trees (well, they are all admissible for such a choice of degrees and we may either apply the above formula for all trees, or prove it by induction, or as you wish). It remains to prove that the coefficient of $z_{p+1}\dots z_{p+m}$ in the function $\varphi_m$ equals $p(p+1)\dots (p+m-1)$. Since $\varphi_m=s\varphi_{m-1}+(p-1)p\dots (p+m-2)\sigma_m$, it is equivalent to proving that the coefficient of $z_{p+1}\dots z_{p+m}$ in $s\varphi_{m-1}$ equals $p(p+1)\dots (p+m-1)-(p-1)p\dots (p+m-2)=mp(p+1)\dots(p+m-2)$. We should take some $z_j$, $p+1\leqslant j\leqslant p+m$, from the multiple $s=\sum z_i$, and for each choice of $j$ we have a coefficient of $z_j^{-1}\cdot z_{p+1}\dots z_{p+m}$ in $\varphi_{m-1}$ equal to $p(p+1)\dots(p+m-2)$ - by induction (base $m-1=0$ is clear).
• Not sure whether you get notifications from comments in other answers or not. To be sure I just repeat some messages here. I really appreciate all your answers here. Ultimately we want to simplify a bigger expression that includes the RHS of my question 2 as a special case. It's pretty long and currently still over my head. I don't know if you have the time and the interest to take a look. But in case you do, you are welcomed to send me an email (in my profile). – Honglu Aug 24 '17 at 19:55 | 2019-09-21T20:09:25 | {
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https://au.mathworks.com/help/matlab/ref/power.html | # power, .^
Element-wise power
## Description
example
C = A.^B raises each element of A to the corresponding powers in B. The sizes of A and B must be the same or be compatible.
If the sizes of A and B are compatible, then the two arrays implicitly expand to match each other. For example, if one of A or B is a scalar, then the scalar is combined with each element of the other array. Also, vectors with different orientations (one row vector and one column vector) implicitly expand to form a matrix.
C = power(A,B) is an alternate way to execute A.^B, but is rarely used. It enables operator overloading for classes.
## Examples
collapse all
Create a vector, A, and square each element.
A = 1:5;
C = A.^2
C = 1×5
1 4 9 16 25
Create a matrix, A, and take the inverse of each element.
A = [1 2 3; 4 5 6; 7 8 9];
C = A.^-1
C = 3×3
1.0000 0.5000 0.3333
0.2500 0.2000 0.1667
0.1429 0.1250 0.1111
An inversion of the elements is not equal to the inverse of the matrix, which is instead written A^-1 or inv(A).
Create a 1-by-2 row vector and a 3-by-1 column vector and raise the row vector to the power of the column vector.
a = [2 3];
b = (1:3)';
a.^b
ans = 3×2
2 3
4 9
8 27
The result is a 3-by-2 matrix, where each (i,j) element in the matrix is equal to a(j) .^ b(i):
$\mathit{a}=\left[{\mathit{a}}_{1}\text{\hspace{0.17em}}{\mathit{a}}_{2}\right],\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\mathit{b}=\left[\begin{array}{c}{\mathit{b}}_{1}\\ {\mathit{b}}_{2}\\ {\mathit{b}}_{3}\end{array}\right],\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\text{\hspace{0.17em}}\mathit{a}\text{\hspace{0.17em}}.ˆ\text{\hspace{0.17em}}\mathit{b}=\left[\begin{array}{cc}{{\mathit{a}}_{1}}^{{\mathit{b}}_{1}}& {{\mathit{a}}_{2}}^{{\mathit{b}}_{1}}\\ {{\mathit{a}}_{1}}^{{\mathit{b}}_{2}}& {{\mathit{a}}_{2}}^{{\mathit{b}}_{2}}\\ {{\mathit{a}}_{1}}^{{\mathit{b}}_{3}}& {{\mathit{a}}_{2}}^{{\mathit{b}}_{3}}\end{array}\right].$
Calculate the roots of -1 to the 1/3 power.
A = -1;
B = 1/3;
C = A.^B
C = 0.5000 + 0.8660i
For negative base A and noninteger B, the power function returns complex results.
Use the nthroot function to obtain the real roots.
C = nthroot(A,3)
C = -1
## Input Arguments
collapse all
Operands, specified as scalars, vectors, matrices, or multidimensional arrays. A and B must either be the same size or have sizes that are compatible (for example, A is an M-by-N matrix and B is a scalar or 1-by-N row vector). For more information, see Compatible Array Sizes for Basic Operations.
• Operands with an integer data type cannot be complex.
Data Types: single | double | int8 | int16 | int32 | int64 | uint8 | uint16 | uint32 | uint64 | logical | char
Complex Number Support: Yes
collapse all
### IEEE Compliance
For real inputs, power has a few behaviors that differ from those recommended in the IEEE®-754 Standard.
MATLAB® IEEE
power(1,NaN)
NaN
1
power(NaN,0)
NaN
1
## Compatibility Considerations
expand all
Behavior changed in R2016b
## Extended Capabilities
### Topics
Introduced before R2006a | 2021-04-21T18:39:05 | {
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http://audio-upgrade.pl/sum-of-sequence-calculator.html | Sum Of Sequence Calculator
Use this calculator to determine the future value of your savings and lump sum. In geometric progressions where |r| < 1 (in other words where r is less than 1 and greater than –1), the sum of the sequence as n tends to infinity approaches a value. This calculator will find the infinite sum of arithmetic, geometric, power, and binomial series, as well as the partial sum, with steps shown (if possible). The Arithmetic series of finite number is the addition of numbers and the sequence that is generally followed include – (a, a + d, a + 2d, …. Approach : Read input number asking for length of the list using input() or raw_input(). Sum of 32-bit integer quantities is not computed by using 64-bit results, and overflow can occur for the LINQ to SQL translation of Sum. This free number sequence calculator can determine the terms (as well as the sum of all terms) of an arithmetic, geometric, or Fibonacci sequence. Sequences and summations CS 441 Discrete mathematics for CS M. Objects might be numbers or letters. Set Sum = 0. The limit of a sequence is the value the sequence approaches as the number of terms goes to infinity. Partial Sums of an Arithmetic Sequence. All term in the sequence meet a specific logical rule which needs to be recognised in order to find the missing terms. Because a series consists of evenly-spaced numbers, the median and mean (average) of the series will be the same. arithmetic series. This way, each term can be expressed by this equation: xₐ = xₐ₋₁ + xₐ₋₂. ): A series of numbers in which ratio of any two consecutive numbers is always a same number that is constant. Calculates the sum of the infinite geometric series. Arithmetic Sequences and Sums Sequence. The first of the examples provided above is the sum of seven whole numbers, while the latter is the sum of the first seven square numbers. We start with the formula for PV of a future value (FV) single lump sum at time n and interest rate i,. lets say you want to sum the numbers from 1 to 100 all you have to do is to calculate this (100)x(101) /2. What Is The Formula For Calculating Arithmetic Sequence? If the initial term of an arithmetic sequence is a 1 and the common difference of successive members is d , then the nth term of the sequence is given by:. The sequence begins 1, 1, 2, 3, 5, and each succeeding term is the sum of the previous two terms. If only a single number for value1 is supplied, SUM returns value1. Example 1: Find the sum of squares of the numbers from 0 to 5000. I want to calculate sum of the average daily value off each product. Work out: a. Sample Output: Input The Value For Nth Term: 5 1*1 = 1 2*2 = 4 3*3 = 9 4*4 = 16 5*5 = 25 The Sum Of The Above Series Is: 55 ASAP🙏 This problem has been solved!. skipna bool, default True. 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The arithmetic sequence calculator uses arithmetic sequence formula to find sequence of any property. The problem is that the number of cells to be summed will vary; for one run of the macro it could be 100 cells, while on the next it could be 300 and on the third only 25. std ([ddof]) Calculate window standard deviation. Sequence calculator allows to calculate online the terms of the sequence whose index is between two limits. Geometric Progression (G. Sequence: By a sequence, we mean an arrangement of number in definite order according to some rules. Build your own widget. Improve your math knowledge with free questions in "Find the sum of an arithmetic series" and thousands of other math skills. The sum of a geometric series is indeed an interesting place to start this discussion. You also generate a sum in each step, which you reuse in the next step. Home; Java Tutorial; Language; Data Type; Operators; Statement Control. Find the sum of first n terms of the series (i) 3 +33 +333 +. Then we update the value of variable sum as - sum = sum + c = 0 + 0 = 0. To concatenate a series of iterables, consider using itertools. help me how to calculate the sum of a series in Matlab. The Fibonacci sequence typically has first two terms equal to x₀ = 0 and x₁ = 1. In other words, how to take the value of a cell located in one worksheet and add it to the value of another cell located in another worksheet to come up with the total of the respective cells. The sequence begins 1, 1, 2, 3, 5, and each succeeding term is the sum of the previous two terms. Calculate the sum of series 1^2+2^2+3^2+4^2++n^2 (n>0) using the both for and while loop structure. Subtract the smaller fraction from the larger fraction to calculate the. Sum of 32-bit integer quantities is not computed by using 64-bit results, and overflow can occur for the LINQ to SQL translation of Sum. The partial sum approach of course involves a "trick" as well -- finding an expression for the dang partial sum. To find sum of your series, you need to choose the series variable, lower and upper bounds and also input the expression for n-th term of the series. Example 2: Find the sum of squares of the first 100 numbers of the form prime minus one. This calculator for to calculating the sum of a series is taken from Wolfram Alpha LLC. As n tends to infinity, S n tends to The sum to infinity for an arithmetic series is undefined. Leonhard Euler was able to calculate the exact sum of the p-series with p 2: 2-2 32 42 Use this fact to find the sum of each series: 2 32 so. Recently, mandatory vote-by-mail has received a great deal of attention as a means of administering elections in the United States. 01 then we use the formula for the sum of the infinite geometric series S oo = a 1 / (1 - r),. Using for loop. Program description:- Find sum of series 1² + 2² + 3² + 4² + 5² +…. com presented a large number of task in mathematics that you can solve online free of charge on a variety of topics: calculation of integrals and derivatives, finding the sum of the series, the solution of differential equations, etc. You may find functions such as sum, ones, or dot useful. In other words, how to take the value of a cell located in one worksheet and add it to the value of another cell located in another worksheet to come up with the total of the respective cells. sum (axis = None, skipna = None, level = None, numeric_only = None, min_count = 0, ** kwargs) [source] ¶ Return the sum of the values for the requested axis. If the value of the sum (in the limiting sense) exists, then they say that the series. Calculate the first term. Guidelines to use the calculator If you select a n, n is the nth term of the sequence If you select S n, n is the first n term of the sequence For more information on how to find the common difference or sum, see this lesson arithmetic sequence. Our sum of series calculator or arithmetic series calculator is an online tool which you can find on Google. is 1 and any term is equal to the sum of all the succeeding terms. It is used to achieve the gas–liquid or liquid–liquid two-phase separation. NPV = Sum CF* ((1+2%)/(1+D))^N When N is infinite, after simplification, NPV = CF * 1 / (1 - r) where r = (1+2%)/(1+D)--A+ V. In general, one does not expect to be able to calculate an infinite sum exactly. In this case, you have the sequence. But there are some series. By definition, the first two numbers in the Fibonacci sequence are either 1 and 1, or 0 and 1, depending on the chosen starting point of the sequence, and each subsequent number is the sum of the previous two. Java program to calculate the sum of N numbers using arrays, recursion, static method, using while loop. Each number in series is called as Fibonacci number. Leonhard Euler continued this study and in the process solved many. Finding a Rule for a Sequence [07/24/2003] What is the next number in this sequence? 1, 3, 11, 67, ? Finding Sum Formula using Sequences of Differences [06/28/1998] Finding a formula for the sum of the first n fourth powers using sequences of differences. 02 Line 10 Column 10 "The Sum of the above Series for term "s is : ". Find the sum of 35 terms of an arithmetic series of which the first term is "a" and the fifteenth term is "9a. It is said that when Gauss was ten. We start with the formula for PV of a future value (FV) single lump sum at time n and interest rate i,. I evaluated the partial sum through a calculator, my answer was -19/30, or -. Sample Output: Input The Value For Nth Term: 5 1*1 = 1 2*2 = 4 3*3 = 9 4*4 = 16 5*5 = 25 The Sum Of The Above Series Is: 55 ASAP🙏 This problem has been solved!. After discussing the above two examples one will wonder if any sequence has the same faith (meaning, it gets closer to a number). Sum of this series is calculated by the formula : sum=(n*(n+1))/2. It will also check whether the series converges. So, the given series is an arithmetic progression whose common difference is d = 2. An arithmetic sequence is one in which the difference between successive members is a constant. demonstration: form input vectors for 10, ten thousand, and 1*10^(many) 0 10 3 0 10 5 0 10 15 0 10000. We therefore derive the general formula for evaluating a finite arithmetic series. The limit of a sequence is the value the sequence approaches as the number of terms goes to infinity. Geometric Progression, Series & Sums Introduction. By applying this calculator for Arithmetic & Geometric Sequences, the n-th term and the sum of the first n terms in a sequence can be accurately obtained. Sequence calculator allows to calculate online the terms of the sequence whose index is between two limits. The T-junction is a novel type of separator used in the petroleum and gas industry. What two things do you need to know to find the sum of an infinite geometric series? Find the sum of the infinite geometric series. asked by ryan on March 15, 2017; Algebra. Explain why or why not. Please help. The two sequences are placed in two consecu- tive rows. The sums we have looked at so far are finite sums with finite upper and lower limits. I know the inequality of [integral from n+1 to infinity of An] < [Total sum - Partial sum] < [integral from n to infinity of An]. Thanks for your help. You need to know both of these numbers in order to calculate the sum of the arithmetic sequence. 1 - Enter the first term A1 in the sequence, the common ratio r and n n the number of terms in the sum then press enter. Hope this helps!. S = 10 1 2, a 1= 1 2 Write. +N^2 in C Programming Language. Such series appear in many areas of modern mathematics. This formula shows that a constant factor in the summands can be taken out of the sum. Free Summation Calculator. However, when the series has an infinite number of terms the summation is more complicated and the series may or may not have a finite. Let the variable equal the first term in the sequence, and equal the last term in the sequence. Observe that for the geometric series to converge, we need that $$|r| 1$$. Excel's Calculation Process. Any sequence that has a common second difference is a quadratic sequence. So the second term common to both sequences is 1+28 = 29. arithmetic series. asked by ryan on March 15, 2017; Algebra. Infinite Series calculator is a free online tool that gives the summation value of the given function for the given limits. In the opposite case, one should pay the attention to the «Series convergence test» pod. Sum of series in C language 1² + 2² + 3² + 4² + 5² +…. Linear sequences A number pattern which increases (or decreases) by the same amount each time is called a linear sequence. Sequences and summations CS 441 Discrete mathematics for CS M. s n = 2 − 3(0. Get comfortable with sequences in general, and learn what arithmetic sequences are. com Task : To find the sum of all the elements in a list. Calculates the 8-bit checksum for a sequence of hexadecimal bytes. On our site OnSolver. I understand the process in calculating a simple series like this one $$\\sum_{n=0}^4 2n$$ but I do not understand the steps to calculate a series like this one $$\\sum_{n=1}^x n^2$$ I have an aw. Note that a series is an indicated sum of the terms of a sequence!! In this section, we work only with finite series and the related sums. Learn Java by Examples: Write a program to calculate and print the sum of the following series: Sum(x) = 2 – 4 + 6 – 8 + 10 - 12 … - 100Learn Java by examples. See full list on mathsisfun. The free tool below will allow you to calculate the summation of an expression. S = 12, a 1= 2 8. This Arithmetic Sequence Calculator is used to calculate the nth term and the sum of the first n terms of an arithmetic sequence. First simplify the expression, X! receives great quick! x/x! = x/(x * (x-a million) * (x-2) *a million) hence x/x! = a million/(x-a million)! for x >= 2 and a million for x =a million Then note that that's the same to the enlargement of the Taylor series for e^x as a million + x + x^2/2! + x^3/3! +x^4/4! for x = a million. The sequence of partial sums of a series sometimes tends to a real limit. If you're behind a web filter, please make sure that the domains *. Free series convergence calculator - test infinite series for convergence step-by-step This website uses cookies to ensure you get the best experience. This is a free javascript calculator. Leonhard Euler continued this study and in the process solved many. \$1000000007\$ is a prime number, so division by \$2^L\$ modulo 1000000007 is a multiplication by a multiplicative inverse of 2 modulo 1000000007 (which is trivial to find) to the same power. 99805 Time Complexity: O(n). In general, one does not expect to be able to calculate an infinite sum exactly. As the top row increases, the bottom row decreases, so the sum stays the same. Infinite Geometric Series Calculator is a free online tool that displays the sum of the infinite geometric sequence. The T-junction is a novel type of separator used in the petroleum and gas industry. Sum definition is - an indefinite or specified amount of money. The first of the examples provided above is the sum of seven whole numbers, while the latter is the sum of the first seven square numbers. The first sum in the numerator is the sum of squared residuals for the first model (e. It is used to achieve the gas–liquid or liquid–liquid two-phase separation. On a higher level, if we assess a succession of numbers, x 1 , x 2 , x 3 ,. This document explains how to calculate the sum or total when working with cell data located in multiple worksheets. Explain why or why not. Finding a general expression for a partial sum by induction and then finding the limit of this partial sum is a perfectly valid technique. Question: Find the sum of (1/2) +(1/6) + (1/12) + + (1/9900) without using a calculator. Improve your math knowledge with free questions in "Find the sum of an arithmetic series" and thousands of other math skills. Get comfortable with sequences in general, and learn what arithmetic sequences are. var ([ddof]) Calculate unbiased window variance. 2, ALPHA ([I], 1, 1 0) EXE, and you should get an answer of 9. The z-transform of the unit pulse, = 1. Learn Java by Examples: Write a program to calculate and print the sum of the following series: Sum(x) = x/2 + x/5 + x/8 + … + x/100. BYJU'S online infinite geometric series calculator tool makes the calculation faster, and it displays the sum in a fraction of seconds. Sequences and summations CS 441 Discrete mathematics for CS M. Geometric Progression (G. Sum of Natural, Odd or Even Numbers Series Calculator getcalc. Java program to calculate the sum of GP series. Just enter the expression to the right of the summation symbol (capital sigma, Σ) and then the appropriate ranges above and below the symbol, like the example provided. This summation notation calculator can sum up many types of sequencies including the well known arithmetic and geometric sequencies, so it can help you to find the terms including the nth term as well as the sum of the first n terms of virtualy any series. Series (Find the sum) A finite Geometric Series (a limited number of terms, or Partial Sum) An infinite Geometric Series, if our infinite series is. Solving mathematical problems online for free. help me how to calculate the sum of a series in Matlab. The sums we have looked at so far are finite sums with finite upper and lower limits. To evaluate the Midpoint Sum, it gets messier. For some use cases, there are good alternatives to sum(). The sum of the members of a finite arithmetic progression is called an arithmetic series. Sn = (n/2) [a + l] Sn = (n/2) [2a + (n - 1)d] a = first term, n = number of terms of the series, d = common difference and l = last term. 0 / k for k in range(1, 10001)) What this code does: the innermost part is a generator expression, which computes the elements of a series 'on the fly'. Example 1: Find the sum of squares of the numbers from 0 to 5000. For example, in the sequence 10, 15, 20, 25, 30. Observe that for the geometric series to converge, we need that $$|r| 1$$. So again, a problem about earned interest might not be a perfect example, since you can withdraw your money at any instant and not only at whole number year values. Sample Output: Input The Value For Nth Term: 5 1*1 = 1 2*2 = 4 3*3 = 9 4*4 = 16 5*5 = 25 The Sum Of The Above Series Is: 55 ASAP🙏 This problem has been solved!. 1 26) a 12 = 28. Arithmetic Sequence. \) The series becomes $$\sum\limits_{n = 0}^\infty {\large\frac{{{u^n}}}{{n!}} ormalsize}. More Practice Problems with Arithmetic Sequence Formula Direction: Read each arithmetic sequence question carefully, then answer with supporting details. Given real (or complex!) numbers aand r, X1 n=0 arn= (a 1 r if jr <1 divergent otherwise The mnemonic for the sum of a geometric series is that it’s \the rst term divided by one minus the common ratio. Another approach could be to use a trigonometric identity. The sums we have looked at so far are finite sums with finite upper and lower limits. Guidelines to use the calculator If you select a n, n is the nth term of the sequence If you select S n, n is the first n term of the sequence For more information on how to find the common difference or sum, see this lesson arithmetic sequence. the starting principal you'll need to achieve the payouts desired:. org are unblocked. Math explained in easy language, plus puzzles, games, quizzes, worksheets and a forum. In column B, create the sequence that you are going to sum and name it seqa. A sequence is a list of numbers. After discussing the above two examples one will wonder if any sequence has the same faith (meaning, it gets closer to a number). Free Summation Calculator. [Note that the sequence of the sums S 1 (the sum of the first term), S 2 (the sum of the first two terms), S 3 (the sum of the first three terms),. The present value, PV, of a series of cash flows is the present value, at time 0, of the sum of the present values of all cash flows, CF. To determine if the series is convergent we first need to get our hands on a formula for the general term in the sequence of partial sums. Only this variable may occur in the sequence term. is 16 and sum of the next 3 terms is 128. The denominator doesn't change. This calculator for to calculating the sum of a series is taken from Wolfram Alpha LLC. How to use sum in a sentence. Because 1 is paired with 10 (our n), we can say that each column has (n+1). Alex's Arithmetic Sequence Sum Calculator is a very simple program, which allows you to get the sum of an Arithmetic Squence, it supports two types of sequences: ￭ type 1 - if you have the first. When the denominators are the same, add the numerators, or top numbers, to calculate the answer. arithmetic series. In column A, create the sequence 1, 2, 3,. ID 1001746465 1 1 1 1 23 4+5 1 There are several ways to implement this. Sum of Three Consecutive Integers calculator. I hope it is enough to get you going in the right direction. The sum of geometric series refers to the total of a given geometric sequence up to a specific point and you can calculate this using the geometric sequence solver or the geometric series calculator. Series calculator allows to calculate online the sum of the terms of the sequence whose index is between the lower and the upper bound. My dxp and csv file is uploaded in below. If you are struggling to understand what a geometric sequences is, don't fret! We will explain what this means in more simple terms later on and take a look at. Limit Comparison Test If lim (n-->) (a n / b n) = L, where a n, b n > 0 and L is finite and positive,. This value is equal to:. Solution: Given decimal we can write as the sum of 0. Leonhard Euler was able to calculate the exact sum of the p-series with p 2: 2-2 32 42 Use this fact to find the sum of each series: 2 32 so. Notice \t is used to provide 8 spaces (1 tab) between two values see in the output. It should be noted, that if the calculator finds sum of the series and this value is the finity number, than this series converged. hi i just want to confirm my method of using sum of pairs is correct. Because the Fibonacci value for 20000 has 4179 decimals and it needs quite an impressive amount of processing, the maximum allowed value is 20000. 0 / k for k in range(1, 10001)) What this code does: the innermost part is a generator expression, which computes the elements of a series 'on the fly'. As running variable, which is increased by 1 in each step, i is used. Excel does not calculate cells in a fixed order, or by Row or Column. We used a sum-over-states method to calculate the dynamic polarizabilities of 6S 1/2 ground state and highly-excited (nS 1/2 and nP 3/2 ) Rydberg state of cesium atoms, and identify corresponding magic detuning for optical wavelengths in the range of 850-1950 nm. Z Transforms of Common Sequences Ele 541 Electronic Testing Unit Pulse. We call an a term of the sequence. If you're seeing this message, it means we're having trouble loading external resources on our website. Representations of N as a sum of distinct elements from special sequences D. What Is Arithmetic Sequence? is a sequence of numbers such that the difference of any two successive members of the sequence is a constant. Infinite Geometric Series Calculator is a free online tool that displays the sum of the infinite geometric sequence. If the series has a finite number of terms, it is a simple matter to find the sum of the series by adding the terms. Sequence calculator allows to calculate online the terms of the sequence whose index is between two limits. this, use your calculator and examine high powers of numbers between 1 and -1. A geometric sequence is a sequence such that any element after the first is obtained by multiplying the preceding element by a constant called the common ratio which is denoted by r. The Corbettmaths video tutorial on finding the nth term for a fractional sequence. A series, which is not a list of terms like a sequence, is the sum of the terms in a sequence. A finite number of terms of an arithmetic sequence can be added to find their sum. Often the first numbers will be 1, but not always. If S n tends to a limit as n tends to infinity, the limit is called the sum to infinity of the series. But there are some series. I evaluated the partial sum through a calculator, my answer was -19/30, or -. Exclude NA/null. In mathematics, summation is the addition of a sequence of any kind of numbers, called addends or summands; the result is their sum or total. How to Find Sum of First n Terms When nth Term is Given ? A series whose terms are in Arithmetic progression is called Arithmetic series. Sum of this series is calculated by the formula : sum=(n*(n+1))/2. The Fibonacci sequence is named after Italian mathematician Leonardo of Pisa, known as Fibonacci. ID 1001746465 1 1 1 1 23 4+5 1 There are several ways to implement this. Work out: a. day, sum((random()*5)::integer) num FROM days -- left join other tables here to get counts, I'm using random group by days. The Sum (Summation) Calculator is used to calculate the total summation of any set of numbers. P Class 11 Engineering - Sum of n Terms of G. But there are some series. Calculate sum elements of sequence: sum. Doing so means that most of the depreciation associated with an asset is recognized in the first few years of its useful life. All term in the sequence meet a specific logical rule which needs to be recognised in order to find the missing terms. Thus the message becomes: Since we are using a 3 by 3 matrix, we break the enumerated message above into a sequence of 3 by 1. The first term of an infinite G. After all, yes 1/(1-x) has an honest-to-goodness explosion to infinity at x=1, but it makes perfectly good sense at x=-1, and tells us (what my calc students. Sum of elements in a list — programminginpython. Series (Find the sum) When you know the first and last term. [ Don't peek. Recently, mandatory vote-by-mail has received a great deal of attention as a means of administering elections in the United States. Write a program in C++ to calculate the sum of the series (1*1) + (2*2) + (3*3) + (4*4) + (5*5) + + (n*n). 02 Line 10 Column 42 Pic zz9 From N Highlight. I know the inequality of [integral from n+1 to infinity of An] < [Total sum - Partial sum] < [integral from n to infinity of An]. Number sequences questions usually consist of four to seven visible numbers along with a single missing number or, depending on the sequence's complexity level, 2 or 3 missing numbers. An infinite series has an infinite number of terms. Number sequences test practice for aptitude tests and psychometric IQ tests. The sum of a sequence is known as a series, and the harmonic series is an example of an infinite series that does not converge to any limit. ${s_n} = \sum\limits_{i = 1}^n i$. Sum of: from: to: Submit: Share a link to this widget: More. Solution: A series in which each number is sum of its previous two numbers is known as Fibonacci series. mean (*args, **kwargs) Calculate the window mean of the values. 23) a 21 = −1. The program solves Riemann sums using one of four methods and displays a graph when prompted. Fourth, we recall the sum of the X 2 and subtract 240. ID 1001746465 1 1 1 1 23 4+5 1 There are several ways to implement this. To be a good Java developer is to be fluent in Java 8. Sum of this series is calculated by the formula : sum=(n*(n+1))/2. Just enter the expression to the right of the summation symbol (capital sigma, Σ) and then the appropriate ranges above and below the symbol, like the example provided. The nineteenth term. 2, ALPHA ([I], 1, 1 0) EXE, and you should get an answer of 9. An arithmetic sequence is one in which the difference between successive members is a constant. To use this calculator, simply type in your list of inputs separated by commas (ie 2,5,8,10,12,18). Print the sum. Sum of sequence calculator. Sequence calculator allows to calculate online the terms of the sequence whose index is between two limits. Geometric Progression (G. asked by ryan on March 15, 2017; Algebra. For the finite sums series calculator computes the answer quite literally, so if there is a necessity to obtain a short expression we recommend computing a parameterized sum. This website uses cookies to ensure you get the best experience. Infinite Series calculator is a free online tool that gives the summation value of the given function for the given limits. I have made this small program to help me add numbers and find the sum, it runs but no output. The nth term of a sequence is given by U n = n 2 /(n + 1). 02 Line 10 Column 58 Pic zz. Everything you want to know about Java. I know the inequality of [integral from n+1 to infinity of An] < [Total sum - Partial sum] < [integral from n to infinity of An]. Guidelines to use the calculator If you select a n, n is the nth term of the sequence If you select S n, n is the first n term of the sequence For more information on how to find the common difference or sum, see this lesson arithmetic sequence. P Class 11 Engineering - Sum of n Terms of G. Write a program in C++ to calculate the sum of the series (1*1) + (2*2) + (3*3) + (4*4) + (5*5) + + (n*n). The sequence of these partial sums converges to also. 78 degrees Fahrenheit. The first of the examples provided above is the sum of seven whole numbers, while the latter is the sum of the first seven square numbers. com allows you to find the sum of a series online. For help creating a sequence, see Note 1I. If the range of a sum is finite, is typically assigned a sequence of values, with being evaluated for each one. Third, we square the sum of X (45 times itself = 2025) and divide it by N (number of scores). A finite number of terms of an arithmetic sequence can be added to find their sum. The 8-bit checksum is the 2's complement of the sum off all bytes. The first of the examples provided above is the sum of seven whole numbers, while the latter is the sum of the first seven square numbers. 01 then we use the formula for the sum of the infinite geometric series S oo = a 1 / (1 - r),. Such series appear in many areas of modern mathematics. Question: Find the sum of (1/2) +(1/6) + (1/12) + + (1/9900) without using a calculator. #sum_(k=1)^nk=(n(n+1))/2# and. The amount it increases or decreases by is known as the common difference. Add the numbers and divide by (n - 1) = 6 to get 95. "Mike" wrote: > I have a series of cashflows, forecast to grow at say 2% each year and > go on indefinitely. Objects might be numbers or letters. Using the nth term. Sum of a Series: If we are able to find the general term of the series successfully, then we can also. Work out: a. Which of the pairs of events below is dependent? Select the correct answer below: drawing a 7 and then drawing another 7 with replacement from a standard deck of cards rolling a 1 and then rolling a 6 with a standard die rolling a 3 and then rolling a 4 with. The sequence of these partial sums converges to also. Thus the message becomes: Since we are using a 3 by 3 matrix, we break the enumerated message above into a sequence of 3 by 1. An example of the sequence can be seen as follows:. Infinite Geometric Series Calculator is a free online tool that displays the sum of the infinite geometric sequence. BYJU’S online infinite series calculator tool makes the calculations faster and easier where it displays the value in a fraction of seconds. It will also check whether the series converges. The numerator above is a difference of two partial sums of the \2^n*a_{n}\ sequence. The two sequences are placed in two consecu- tive rows. This technique causes problems in several situations, however and cannot be universally relied upon. Using Wolfram Alpha 214 , put in 10 x x and you will get can be reduced to the much easier-to-calculate infinite sum 1 - x - x2 Series - A series is formed by the sum or addition of the terms in a two equations that allow easy calculation for an arithmetic series. The numerator above is a difference of two partial sums of the \2^n*a_{n}\ sequence. More Practice Problems with Arithmetic Sequence Formula Direction: Read each arithmetic sequence question carefully, then answer with supporting details. day, sum((random()*5)::integer) num FROM days -- left join other tables here to get counts, I'm using random group by days. For example, two-thirds plus four-thirds equals six-thirds. Average = Sum of terms / Number of terms. is 16 and sum of the next 3 terms is 128. Methods for Evaluating In nite Series Charles Martin March 23, 2010 Geometric Series The simplest in nite series is the geometric series. The sum of the first three terms of the sequence of terms common to the two given sequences is 3 times the second term of the sequence. The sum of geometric series refers to the total of a given geometric sequence up to a specific point and you can calculate this using the geometric sequence solver or the geometric series calculator. interval is [FIRST, LAST) NB. The sum of the members of a finite arithmetic progression is called an arithmetic series. Write a program in C++ to calculate the sum of the series (1*1) + (2*2) + (3*3) + (4*4) + (5*5) + + (n*n). Alternatively – put the first few terms of your sequence into WolframAlpha. Sample Output: Input The Value For Nth Term: 5 1*1 = 1 2*2 = 4 3*3 = 9 4*4 = 16 5*5 = 25 The Sum Of The Above Series Is: 55 ASAP🙏 This problem has been solved!. Next: Write a program in C++ to find the sum of series 1 - X^2/2! + X^4/4!- upto nth term. That's numberwang!. Our summation calculator can easily calculate the sum of any numbers you input. Number sequences test practice for aptitude tests and psychometric IQ tests. sin105 degrees B. the starting principal you'll need to achieve the payouts desired:. , x k , we can record the sum of these numbers in the following way:. Examples: SAS Statements. Lets say i have three sequences. Notice, that if x = 1 then, in the series, we are simply addi ng up an infinite number of 1's and of course the sum goes to infinity. Please help. An infinite series has an infinite number of terms. Beware: people often confuse the terms ‘sequence’ and ‘series’. An Efficient solution to solve the sum of geometric series where first term is a and common ration is r is by the formula :-sum of series = a(1 – r n)/(1 – r). Enter 2 numbers to add and press the = button to get the sum result. #a_n = (3/2)^n# Which means that #n#-th term is generates by raising #3/2# to the #n#-th power. Find the series. Sequences and summations CS 441 Discrete mathematics for CS M. [Note that the sequence of the sums S 1 (the sum of the first term), S 2 (the sum of the first two terms), S 3 (the sum of the first three terms),. That's numberwang!. This free number sequence calculator can determine the terms (as well as the sum of all terms) of an arithmetic, geometric, or Fibonacci sequence. +N^2 in C Programming Language. In other words, if you keep adding together the terms of the sequence forever, you will get a finite value. Products Classroom Activities Graphing Calculator Scientific Calculator Four Function Calculator Matrix Calculator Test Practice Geometry Tool. In other words, how to take the value of a cell located in one worksheet and add it to the value of another cell located in another worksheet to come up with the total of the respective cells. Series calculator allows to calculate online the sum of the terms of the sequence whose index is between the lower and the upper bound. sum_arithmetic_series is more general than required. what could be the problem? // Program that uses a switch statement. If this happens, we say that this limit is the sum of the series. P Class 11 Engineering - Sum of n Terms of G. All you have to do is write the first term number in the first box, the second term number in the second box, third term number in the third box and the write value of n in the fourth box after that you just have to click on the Calculate button, your result will be visible. Enter the sequence, the start value and end value from sigma notation and get a numerical sum. To add floating point values with extended precision, see math. What is the difficulty level of this exercise? Easy Medium Hard. About Sum (Summation) Calculator. Question: Find the sum of (1/2) +(1/6) + (1/12) + + (1/9900) without using a calculator. Calculate the sum of first n squares or the sum of consecutive square numbers from n 1 2 to n 2 2. A series is a special type of sequence: a. Notice, that if x = 1 then, in the series, we are simply addi ng up an infinite number of 1's and of course the sum goes to infinity. Calculate sum elements of sequence: sum. Thanks for your help. Math explained in easy language, plus puzzles, games, quizzes, worksheets and a forum. Bob has a need to use the SUM function in a macro in order to find the sum of all the values in a column. The sequence of partial sums of a series sometimes tends to a real limit. var ([ddof]) Calculate unbiased window variance. 23) a 21 = −1. The formula for the n-th term of a quadratic sequence is explained here. For example, some series don't sum monotonically to a limit -- the sum will bounce around it, and the summation's state can end up oscillatory forever. 02 Line 10 Column 42 Pic zz9 From N Highlight. skipna bool, default True. Third, we square the sum of X (45 times itself = 2025) and divide it by N (number of scores). Free Summation Calculator. +N^2 in C Programming Language. An example of the sequence can be seen as follows:. Instructions: Use this step-by-step Geometric Series Calculator, to compute the sum of an infinite geometric series by providing the initial term \(a$$ and the constant ratio $$r$$. Our sum of series calculator or arithmetic series calculator is an online tool which you can find on Google. Although SUM is specified as taking a maximum of 30 arguments, Google Sheets supports an arbitrary number of arguments for this function. About this calculator. (for example 2 series in the picture) 1 Comment. Free Arithmetic Sequences calculator - Find indices, sums and common difference step-by-step This website uses cookies to ensure you get the best experience. The sum function can be used as a series calculator, to calculate the sequence of partial sums of a series. Set Number = 0. The sums we have looked at so far are finite sums with finite upper and lower limits. The Arithmetic series of finite number is the addition of numbers and the sequence that is generally followed include – (a, a + d, a + 2d, …. day, sum((random()*5)::integer) num FROM days -- left join other tables here to get counts, I'm using random group by days. #sum_(k=1)^nb=nb# So. For example, the sum given by, means to sum an infinite number of terms as, The value of an infinite sum may be ∞ (in this case the sum is infinite). Definition: Arithmetic sequence is a list of numbers where each number is equal to the previous number, plus a constant. Using the while loop to calculate sum : While Loop « Statement Control « Java Tutorial. To add floating point values with extended precision, see math. Series calculator allows to calculate online the sum of the terms of the sequence whose index is between the lower and the upper bound. I can''t seem to find one of those. Build your own widget. A harmonic sequence is a sequence of numbers whose reciprocals form an arithmetic sequence. Here is the complete Java program with sample outputs. An Efficient solution to solve the sum of geometric series where first term is a and common ration is r is by the formula :-sum of series = a(1 – r n)/(1 – r). Let, t n be the n th term of AP, then (n+1) th term of can be calculated as (n+1) th = t n + D where D is the common difference (n+1) th - t n The formula to calculate N th term t n = a + (n – 1)d; where, a is first term of AP and d is the common difference. If Sum > Limit, terminate the repitition, otherwise. In the sequence a sub 1 (first term), a sub 2 (second term), a sub 3 (third term)…. Sum uses the standard Wolfram Language iteration specification. The last number n is input by the user. Σ is the symbol used to denote sum. Infinite Geometric Series Calculator is a free online tool that displays the sum of the infinite geometric sequence. Increment Number by one. Question: Find the sum of (1/2) +(1/6) + (1/12) + + (1/9900) without using a calculator. Sequence calculator allows to calculate online the terms of the sequence whose index is between two limits. 63333333 I'm having difficulty estimating how far this partial sum is away from the total sum. The blog talks about variety of topics on Embedded System, 8085 microprocessor, 8051 microcontroller, ARM Architecture, C2000 Architecture, C28x, AVR and many many more. Instructions: Use this step-by-step Geometric Series Calculator, to compute the sum of an infinite geometric series by providing the initial term $$a$$ and the constant ratio $$r$$. Arithmetic Sequence. After learning so much about development in Python, I thought this article would be interesting for readers and to myself… This is about 5 different ways of calculating Fibonacci numbers in Python [sourcecode language=”python”] ## Example 1: Using looping technique def fib(n): a,b = 1,1 for i in range(n-1): a,b = b,a+b return a print … Continue reading 5 Ways of Fibonacci in Python →. The 'nth' term is a formula with 'n' in it which enables you to find any term of a sequence without having to go up from one term to the next. 2, ALPHA ([I], 1, 1 0) EXE, and you should get an answer of 9. A sequence is a numbered list of values, produced by the calculation of a formula. C Program to calculate sum of 5 subjects and find percentage [crayon-5f51beda86fcb863628413/] Output : [crayon-5f51beda86fd2634443803/]. This is a free javascript calculator. The Fibonnacci numbers are also known as the Fibonacci series. Sequence Calculator, Sequence Examples. Improve your math knowledge with free questions in "Find the sum of an arithmetic series" and thousands of other math skills. Finding a general expression for a partial sum by induction and then finding the limit of this partial sum is a perfectly valid technique. How to find the sum of a finite Arithmetic Series! s n = n(t 1 + t n)/2 To find the sum of a finite arithmetic series, you need to know three things. Without using loops, calculate the sum of the following series for the first n terms. You may want to review the basics of geometric sequences or finding formulas. Notice, that if x = 1 then, in the series, we are simply addi ng up an infinite number of 1's and of course the sum goes to infinity. A good sequence to start with is the Fibonacci sequence. Repeat the following: a. About this calculator. sum_arithmetic_series FIRST LAST SKIP NB. #sum_(k=1)^nk=(n(n+1))/2# and. Definition: Geometric sequence is a list of numbers where each term is obtained by multiplying the previous term by a constant. How do we apply these useful rules to this question? First, calculate the average of the first and. This formula shows that a constant factor in the summands can be taken out of the sum. I have made this small program to help me add numbers and find the sum, it runs but no output. 2) ∧ 2 × 0. LED series current limiting resistor calculator - useful when designing circuits with a single LED or series/parallel LED arrays - for both the common small-current (20mA) LEDs and the more expensive, high power LEDs with currents up to a few Amperes. #sum_(k=1)^n(ak+b)# where a and b are constants. Free Summation Calculator. If S n tends to a limit as n tends to infinity, the limit is called the sum to infinity of the series. To add floating point values with extended precision, see math. 0 / k if k % 2 else -1. Fibonacci sequence formula; Golden ratio convergence; Fibonacci sequence table; Fibonacci sequence calculator; C++ code of Fibonacci function; Fibonacci sequence formula. 78 degrees Fahrenheit. Click the link for more information. The n-th partial sum of a series is the sum of the first n terms. asked by ryan on March 15, 2017; Algebra. White Rose Maths has prepared a series of Maths lessons online for Year 4, FREE videos and worksheets. This calculator computes n-th term and sum of geometric progression person_outline Timur schedule 2011-07-16 04:17:35 Geometric progression is a sequence of numbers where each term after the first is found by multiplying the previous one by a fixed non-zero number called the common ratio. We start with the general formula for an arithmetic sequence of $$n$$ terms and sum it from the first term ($$a$$) to the last term in the sequence ($$l$$):. Explain why or why not. This is a very versatile calculator that will output sequences and allow you to calculate the sum of a sequence between a starting item and an n-th term, as well as tell you the value of the n-th term of interest. Calculate the rolling quantile. Since an arithmetic sequence always has an unbounded long-term behavior, we are always restricted to adding a finite number of terms. For this reason, Sum evaluates to null instead of to zero for an empty sequence or for a sequence that contains only nulls. So 355 minus 289. If the above series converges, then the remainder R N = S - S N (where S is the exact sum of the infinite series and S N is the sum of the first N terms of the series) is bounded by 0< = R N <= (N. Limits capture the long-term behavior of a sequence and are thus very useful in bounding them. The sum to infinity of a geometric progression. This version provided Java with a functional aspect by introducing concepts such as functional interfaces, lambda expressions, streams, etc. Sum of 32-bit integer quantities is not computed by using 64-bit results, and overflow can occur for the LINQ to SQL translation of Sum. Besides finding the sum of a number sequence online, server finds the partial sum of a series online. NPV = Sum CF* ((1+2%)/(1+D))^N When N is infinite, after simplification, NPV = CF * 1 / (1 - r) where r = (1+2%)/(1+D)--A+ V. Solving mathematical problems online for free. Since there's two equally likely options, you'd expect a run to last for two flips. See full list on gigacalculator. Sum of sequence calculator. And, thanks to the Internet, it's easier than ever to follow in their footsteps (or just finish your homework or study for that next big. Therefore, the equation could be 4n. A geometric sequence refers to a sequence wherein each of the numbers is the previous number multiplied by a constant value or the common ratio. Calculate the sum of special series or number of sequences. Sum of Arithmetic Sequence Formula. com Task : To find the sum of all the elements in a list. Leonhard Euler continued this study and in the process solved many. Note that a series is an indicated sum of the terms of a sequence!! In this section, we work only with finite series and the related sums. Such series appear in many areas of modern mathematics. In the previous section we started looking at writing down a power series representation of a function. Geometric Sequence. The free tool below will allow you to calculate the summation of an expression. To determine if the series is convergent we first need to get our hands on a formula for the general term in the sequence of partial sums. Guidelines to use the calculator If you select a n, n is the nth term of the sequence If you select S n, n is the first n term of the sequence For more information on how to find the common difference or sum, see this lesson Geometric sequence. skipna bool, default True. This free number sequence calculator can determine the terms (as well as the sum of all terms) of an arithmetic, geometric, or Fibonacci sequence. Question: Find the sum of (1/2) +(1/6) + (1/12) + + (1/9900) without using a calculator. Store the result in an array. Perform addition to find the sum of two or more fractions. 8 Given two terms in an arithmetic sequence find the recursive formula. This calculator will find the infinite sum of arithmetic, geometric, power, and binomial series, as well as the partial sum, with steps shown (if possible). Hi everyone, the picture below is the sum of a series that have infinity limit, and i dont know the suitable function to deal with the problem. A sequence is a list of numbers, geometric shapes or other objects, that follow a specific pattern. For this reason, Sum evaluates to null instead of to zero for an empty sequence or for a sequence that contains only nulls. \$1000000007\$ is a prime number, so division by \$2^L\$ modulo 1000000007 is a multiplication by a multiplicative inverse of 2 modulo 1000000007 (which is trivial to find) to the same power. Calculate the sum of the series ∑ n = 1 ∞ a n whose partial sums are given. However, policy-makers disagree on the merits of this approach. Denote this partial sum by S n. An example of the sequence can be seen as follows:. Java program to calculate the sum of GP series. Fourth, we recall the sum of the X 2 and subtract 240. Work out: a. Online adding calculator. Sum of the series 1^1 + 2^2 + 3^3 + …. You may see the formula written as: Sum, S n, of n terms of an arithmetic series. The amount it increases or decreases by is known as the common difference. day ) select day, num, sum(num) over. +n and prints the result on the compiler screen. Two consecutive numbers in this series are in a ' Golden Ratio '. Flowchart. com allows you to find the sum of a series online. The Sum (Summation) Calculator is used to calculate the total summation of any set of numbers. 63333333 I'm having difficulty estimating how far this partial sum is away from the total sum. Such an argument was given by Nicolas Oresme (1323 - 1382 A. So the second term common to both sequences is 1+28 = 29. When the limit of partial sums exists, it is called the value (or sum) of the series. Dick and I both used tricks. It is capable of computing sums over finite, infinite and parameterized sequences. First simplify the expression, X! receives great quick! x/x! = x/(x * (x-a million) * (x-2) *a million) hence x/x! = a million/(x-a million)! for x >= 2 and a million for x =a million Then note that that's the same to the enlargement of the Taylor series for e^x as a million + x + x^2/2! + x^3/3! +x^4/4! for x = a million. You can also use this arithmetic sequence calculator as an arithmetic series calculator. If you are struggling to understand what a geometric sequences is, don't fret! We will explain what this means in more simple terms later on and take a look at. Here s how Wolfram Alpha solved it. In an Arithmetic Sequence the difference between one term and the next is a constant. Now, how do we determine the average of a sequence of integers? Rule #2: The average of a sequence of integers is the average of the first and last terms Applying the rules to find the sum of the sequence. #sum_(k=1)^n(ak+b)=sum_(k=1)^nak+sum_(k=1)^nb# We can factor the #a#. You need to know both of these numbers in order to calculate the sum of the arithmetic sequence. 2) ∧ 2 × 0. MATH 225N Week 4 Probability Questions and answers – Chamberlain College of Nursing Week 4 Homework Questions Probability 1. Only this variable may occur in the sequence term. Here s how Wolfram Alpha solved it. What Is Arithmetic Sequence? is a sequence of numbers such that the difference of any two successive members of the sequence is a constant. day ) select day, num, sum(num) over. Find the sum of first n terms of the series (i) 3 +33 +333 +. Free Arithmetic Sequences calculator - Find indices, sums and common difference step-by-step This website uses cookies to ensure you get the best experience. The sequence of these partial sums converges to also. For example, in the sequence 10, 15, 20, 25, 30. The denominator doesn't change. Calculate the rolling quantile. Thus the message becomes: Since we are using a 3 by 3 matrix, we break the enumerated message above into a sequence of 3 by 1. What two things do you need to know to find the sum of an infinite geometric series? Find the sum of the infinite geometric series. Bytes are provided as two-character strings. To find sum of your series, you need to choose the series variable, lower and upper bounds and also input the expression for n-th term of the series. Sum of Three Consecutive Integers calculator. Class 11 Commerce - Sum of terms of G. 2) ∧ 2 × 0. Let’s explore how we do that. Calculate the sum of series 1^2+2^2+3^2+4^2++n^2 (n>0) using the both for and while loop structure. Class 11 Commerce - Sum of terms of G. #sum_(k=1)^nk=(n(n+1))/2# and. On a higher level, if we assess a succession of numbers, x 1 , x 2 , x 3 ,. 63333333 I'm having difficulty estimating how far this partial sum is away from the total sum. If the value of the sum (in the limiting sense) exists, then they say that the series. After learning so much about development in Python, I thought this article would be interesting for readers and to myself… This is about 5 different ways of calculating Fibonacci numbers in Python [sourcecode language=”python”] ## Example 1: Using looping technique def fib(n): a,b = 1,1 for i in range(n-1): a,b = b,a+b return a print … Continue reading 5 Ways of Fibonacci in Python →. Calculators with newer operating systems actually have a summation function, which can by reached by pressing MATH 0. n must be a positive integer. In addition, when the calculator fails to find series sum is the strong indication that this series is divergent (the calculator prints the message like "sum diverges"), so our calculator also indirectly helps to get information about series convergence. In this case, a number of columns equal to the number of elements of the sequence x[n] should be skipped in order to accommo- date the results of the multiplication. Essentially the total variability in your dataset is the same as before, but now you have two different models to consider. Question: Find the sum of (1/2) +(1/6) + (1/12) + + (1/9900) without using a calculator. Free Summation Calculator. Fibonacci calculator The tool calculates F(n) - Fibonacci value for the given number, as well as the previous 4 values, using those to display a visual representation. The full sequence of keypresses to evaluate the Right-Hand Sum is: OPTN F4 [CALC] F6 F3 [Σ(] (1 + ALPHA ([I] × 0. This lesson assumes that you know about geometric sequences, how to find the common ratio and how to find an explicit formula. Some sources neglect the initial 0, and instead beginning the sequence with the first two ones. A geometric sequence refers to a sequence wherein each of the numbers is the previous number multiplied by a constant value or the common ratio. is 1 and any term is equal to the sum of all the succeeding terms. Unfortunately, the answer is NO. First simplify the expression, X! receives great quick! x/x! = x/(x * (x-a million) * (x-2) *a million) hence x/x! = a million/(x-a million)! for x >= 2 and a million for x =a million Then note that that's the same to the enlargement of the Taylor series for e^x as a million + x + x^2/2! + x^3/3! +x^4/4! for x = a million. This relationship of examining a series forward and backward to determine the value of a series works for any arithmetic series. Each number in series is called as Fibonacci number. | 2020-11-29T16:20:11 | {
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https://atekihcan.github.io/CLRS/02/P02-01/ | Insertion Sort on Small Arrays in Merge Sort
Although merge sort runs in $$\Theta(n \lg n)$$ worst-case time and insertion sort runs in $$\Theta(n^2)$$ worst-case time, the constant factors in insertion sort can make it faster in practice for small problem sizes on many machines. Thus, it makes sense to coarsen the leaves of the recursion by using insertion sort within merge sort when subproblems become sufficiently small. Consider a modification to merge sort in which $$n/k$$ sublists of length $$k$$ are sorted using insertion sort and then merged using the standard merging mechanism, where $$k$$ is a value to be determined.
1. Show that insertion sort can sort the $$n/k$$ sublists, each of length $$k$$, in $$\Theta(n \lg (n/k))$$ worst-case time.
2. Show how to merge the sublists in $$\Theta(n \lg(n/k))$$ worst-case time.
3. Given that the modified algorithm runs in $$\Theta(nk + n \lg(n/k))$$ worst-case time, what is the largest value of k as a function of n for which the modified algorithm has the same running time as standard merge sort, in terms of $$\Theta$$ notation?
4. How should we choose $$k$$ in practice?
#### Sorting Sublists
For input of size $$k$$, insertion sort runs on $$\Theta(k^2)$$ worst-case time. So, worst-case time to sort $$n/k$$ sublists, each of length $$k$$, will be $$n/k \cdot \Theta(k^2) = \Theta(nk)$$
#### Merging Sublists
We have $$n$$ elements divided into $$n/k$$ sorted sublists each of length $$k$$. To merge these $$n/k$$ sorted sublists to get a single sorted list of length $$n$$, we have to take 2 sublists at a time and continue to merge them.
This will result in $$\lg (n/k)$$ steps (refer to Figure 2.5 in page 38 of the chapter text). And in every step, we are essentially going to compare $$n$$ elements. So the whole process will run at $$\Theta(n \lg (n/k))$$.
#### Largest Value of k
For the modified algorithm to have the same asymptotic running time as standard merge sort, $$\Theta(nk + n \lg(n/k))$$ must be same as $$\Theta(n \lg n)$$.
To satisfy this condition, $$k$$ cannot grow faster than $$\lg n$$ asymptotically, if it does then because of the $$nk$$ term, the algorithm will run at worse asymptotic time than $$\Theta(n \lg n)$$).
So, let’s assume, $$k = \Theta(\lg n)$$ and see if we can meet the criteria …
\begin {aligned} \Theta(nk + n \lg(n/k)) & = \Theta(nk + n \lg n - n \lg k) \\ & = \Theta(n \lg n + n \lg n - n \lg (\lg n)) \\ & = \Theta(2n \lg n - n \lg (\lg n)) ^\dagger \\ & = \Theta(n \lg n) \end {aligned}
$$^\dagger\lg (\lg n)$$ is very small compared to $$\lg n$$ for sufficiently larger values of $$n$$.
#### Practical Value of k
To determine a practical value for $$k$$, it has to be the largest input size for which insertion sort runs faster than merge sort. To get exact value, we need to calculate the exact running time expressions with the constant factors and use the method described in Exercise 1.2.2. | 2023-03-29T01:15:31 | {
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https://math.stackexchange.com/questions/1646339/are-these-3-functions-linearly-independent-or-dependent | # Are these $3$ functions linearly independent or dependent?
Are the functions $f, g, h$ given below linearly independent? If they are not linearly independent, find a nontrivial solutions to the equations below:
$$f(x)=e^{2x}- \cos(9x), \quad g(x)=e^{2x}+ \cos(9x), \quad h(x)= \cos(9x)$$
My take so far is that, I know these functions are not linearly independent (meaning this is surely dependent) by Wronskian (as the Wronskian determinant isn't equal to zero), but I have no idea how to find nontrivial solutions to this question. I think I should have an answer as follows:
$$C_1(e^{2x}-\cos(9x)) + C_2(e^{2x}+\cos(9x)) + C_3\cos(9x) = 0 \qquad \text{ where } Cs \text{ are constant }$$
Could I get some help on finding those constants? I've tried $C_1$ as $-1$, $C_2$ as $1$ and $C_3$ as $0$ but that surely cannot be the answer.
• In Mathematica, Wronskian[{Exp[2 x] - Cos[9 x], Exp[2 x] + Cos[9 x], Cos[9 x]}, x] yields $0$, showing that the three functions are linearly dependent. – David G. Stork Feb 8 '16 at 18:19
Those functions are a red herring. ;-)
You basically have the vectors $$v-w,\quad v+w,\quad w$$ in some vector space. These vectors belong to the subspace spanned by $v$ and $w$, which has dimension at most $2$. So a set of three vectors is necessarily linearly dependent.
How to find a nonzero linear combination is easy: $$a(v-w)+b(v+w)+cw=0$$ gives $$(a+b)v+(-a+b+c)w=0$$ so we can choose $b=-a$ and so $c=2a$. If we take $a=1$, we obtain $b=-1$ and $c=2$.
You can take $C_1 = 1, C_2 = -1$ and $C_3 = -2$.
• I choose $C_1 = 1$ (without any reason). Then I wanted to cancel the two exponential so I had to choose $C_2 = -1$. Then I was left with $-2\cos(9x)$ therefore I had to pick $C_3=-2$. – Onil90 Feb 8 '16 at 18:10
$$\begin{bmatrix} f(x)\\ g(x)\\ h(x) \end{bmatrix}= \begin{bmatrix} 1 & -1\\ 1 & 1 \\ 0 & 1 \\ \end{bmatrix}\cdot \begin{bmatrix} -e^{2x}\\ \cos(9x) \end{bmatrix}$$ | 2020-06-04T18:32:01 | {
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https://cs.stackexchange.com/questions/80750/average-number-of-comparisons-in-sorted-insertion | # Average number of comparisons in sorted insertion
The questions is pretty simple: Given a sorted array of N elements, what is the average number of comparisons made in order to add a new element (let's call it x) in its correct position?
I am using linear search for that task.
So far, I've tried to solve it this way:
# of Comparisons Probability of A[i] > x (Not sure if correct)
1 1/n
2 1/n
3 1/n
4 1/n
... ...
n 1/n
Hence, the expected value would be:
$$E[x] = \sum_{x=1}^{n}x \cdot Pr(X=x)$$ $$E[x] = 1 \cdot Pr(X=1) + 2 \cdot Pr(X=2) + 3 \cdot Pr(X=3) + ... + n \cdot Pr(X=n)$$
Using $$Pr(X = n) = \frac{1}{n}$$ for all n,
$$E[x] = \frac{1}{n} \cdot \sum^{n}_{i=1}{i}$$
and finally $$E[x] = \frac{n+1}{2}$$
Still, I'm not sure if this is the correct way to solve it, since I'm not sure about my 1/n assumption.
Well, that would strongly depend on the algorithm that you use to find out where to insert the new element, and on the distribution of new elements.
Assuming that you do a linear search, and the new element could go to any position with the same probability, your result is close, but not quite exact. The new element can go into one of (n + 1) positions, not n. There may be up to n comparisons needed; n comparisons are needed both if the new element goes into the first array position, and if it goes into the second array position. So the result is
(1 + 2 + 3 + ... + (n-1) + n + n) / (n + 1) =
= (n (n+1) / 2 + n) / (n + 1)
= n (n + 3) / 2 / (n + 1)
= (n + 2 - 2 / (n + 1)) / 2 ≈ (n + 2) / 2,
which is just a tiny bit more than your answer (n + 1) / 2.
• Thanks a lot @gnasher729 ! I completely forgot to mention that I'd perform a linear search in order to find the position to insert the new number! – woz Sep 1 '17 at 23:29
• I don't understand what you mean by "n comparisons are needed both if the new element goes into the first array position, and if it goes into the second array position". Why do we need $n$ comparisons in order to insert $x$ into the first position? Assuming I compare by $x < A[i]$, a is single comparison $x < A[0]$ is enough to insert x into the first position. – fade2black Sep 1 '17 at 23:48
• @fade2black: Because you likely start comparing at the end of the array, which allows you to compare and move one element in the same iteration of the loop. If you start comparing at the beginning, then n comparisons are needed both if the new element is inserted just before, or just after the last element. – gnasher729 Sep 2 '17 at 22:38 | 2019-10-19T19:51:14 | {
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https://math.stackexchange.com/questions/2317625/show-which-of-6-2-sqrt3-and-3-sqrt2-2-is-greater-without-using-calculato | # Show which of $6-2\sqrt{3}$ and $3\sqrt{2}-2$ is greater without using calculator
How do you compare $6-2\sqrt{3}$ and $3\sqrt{2}-2$? (no calculator)
Look simple but I have tried many ways and fail miserably. Both are positive, so we cannot find which one is bigger than $0$ and the other smaller than $0$. Taking the first minus the second in order to see the result positive or negative get me no where (perhaps I am too dense to see through).
• sorry, multiply both sides by $6+2\sqrt{3}$ and use $(a-b)(a+b)=a^2-b^2$ – luka5z Jun 10 '17 at 19:00
• \begin{eqnarray*} \sqrt{a}+ \sqrt{b}= \sqrt{a+b+2 \sqrt{ab}} \end{eqnarray*} might be helpful. – Donald Splutterwit Jun 10 '17 at 19:01
• @Learner132 A lot of the solutions here add or subtract from both sides then square then compare to infer about the direction of the original identity. This is only guaranteed to work if both values are positive before squaring. E.g. consider the counterexample: $2.5>1 \rightarrow 2.5-2>1-2 \rightarrow 0.5 > -1 \rightarrow 0.25 > 1$. – Ian Miller Jun 11 '17 at 11:49
• I would just guesstimate. $\sqrt 3 \approx 1.75$, $\sqrt 2 \approx 1.4$, so $6 - 2 \sqrt 3 \approx 2.5$ and $-2 + 3 \sqrt 2 \approx 2.2$. – Robert Soupe Jun 15 '17 at 0:39
$6-2\sqrt 3 \gtrless 3\sqrt 2-2$
Rearrange: $8 \gtrless 3\sqrt 2 + 2\sqrt 3$
Square: $64 \gtrless 30+12\sqrt 6$
Rearrange: $34 \gtrless 12\sqrt 6$
Square: $1156 \gtrless 864$
• You might simplify by $2$ before squaring… – Bernard Jun 10 '17 at 19:11
• Thank you, good answer, but I doubt if math teacher is okay with this (> or <) symbol – Learner132 Jun 10 '17 at 19:29
• What's important about any symbol is its mathematical meaning. If you can the mathematical meaning of this symbol to your math teacher, I'll bet he/she is open to its usage. – Lee Mosher Jun 10 '17 at 19:57
• You should probably mention that the squaring steps are valid (both forward and backward) since the values on both sides are clearly positive. – Rory Daulton Jun 11 '17 at 0:33
• Yes, ordinarily you would start at the last step with a definite operator there (> in this case) and then work towards the inequality given to you. However, to find the necessary steps, it is often easier to work from what you’re given to something known, i.e., backwards. – BallpointBen Jun 11 '17 at 2:47
We have $\sqrt{3}\leq 1.8$ so $6-2\sqrt{3}\geq 2.4$, whereas $\sqrt{2}\leq 1.42$ so $3\sqrt{2}-2\leq 2.26$.
• I love your answer, I'll try to see if math teacher is okay with estimation – Learner132 Jun 10 '17 at 19:31
• @Learner132: Note that the estimates are easy to prove by squaring, if necessary. – user14972 Jun 10 '17 at 22:06
$$6-2√3 \sim 3√2-2\\ 8 \sim 3√2 +2√3 \\ 64 \sim 30+12√6\\ 34 \sim 12√6\\ 17 \sim 6√6\\ 289 \sim 36 \cdot 6\\ 289 > 216$$
• Thanks, quite understandable, but how I can explain ~ symbol to math teacher in this? – Learner132 Jun 10 '17 at 19:27
• This $\sim$ simbol is a placeholder for an undefined relationship, you can use whatever other symbol you prefer, while you dont multiply by -1 – Brethlosze Jun 10 '17 at 19:29
• I remember using that during school first courses without trouble – Brethlosze Jun 10 '17 at 19:29
• I understand you. Well, I hope math teacher is okay with it although it is not covered in the lecture. – Learner132 Jun 10 '17 at 19:40
• Teachers know how this work and what you thought... rewritting everything is not needed. – Brethlosze Jun 11 '17 at 6:17
Using simple continued fractions for $\sqrt {12}$ and $\sqrt {18}.$ Worth learning the general technique...Method described by Prof. Lubin at Continued fraction of $\sqrt{67} - 4$
$$\sqrt { 12} = 3 + \frac{ \sqrt {12} - 3 }{ 1 }$$ $$\frac{ 1 }{ \sqrt {12} - 3 } = \frac{ \sqrt {12} + 3 }{3 } = 2 + \frac{ \sqrt {12} - 3 }{3 }$$ $$\frac{ 3 }{ \sqrt {12} - 3 } = \frac{ \sqrt {12} + 3 }{1 } = 6 + \frac{ \sqrt {12} - 3 }{1 }$$
Simple continued fraction tableau:
$$\begin{array}{cccccccccc} & & 3 & & 2 & & 6 & \\ \\ \frac{ 0 }{ 1 } & \frac{ 1 }{ 0 } & & \frac{ 3 }{ 1 } & & \frac{ 7 }{ 2 } \\ \\ & 1 & & -3 & & 1 \end{array}$$
$$\begin{array}{cccc} \frac{ 1 }{ 0 } & 1^2 - 12 \cdot 0^2 = 1 & \mbox{digit} & 3 \\ \frac{ 3 }{ 1 } & 3^2 - 12 \cdot 1^2 = -3 & \mbox{digit} & 2 \\ \frac{ 7 }{ 2 } & 7^2 - 12 \cdot 2^2 = 1 & \mbox{digit} & 6 \\ \end{array}$$
Continued fraction convergents alternate above and below the irrational number, we get $$\frac{ 3 }{ 1 } < \sqrt {12} < \frac{ 7 }{ 2 }$$ Your first number was $6 - \sqrt {12},$ $$3 > 6 - \sqrt {12} > \frac{ 5 }{ 2 }$$ $$\frac{ 5 }{ 2 } < 6 - \sqrt {12} < 3$$
Next 18......................========================================
$$\sqrt { 18} = 4 + \frac{ \sqrt {18} - 4 }{ 1 }$$ $$\frac{ 1 }{ \sqrt {18} - 4 } = \frac{ \sqrt {18} + 4 }{2 } = 4 + \frac{ \sqrt {18} - 4 }{2 }$$ $$\frac{ 2 }{ \sqrt {18} - 4 } = \frac{ \sqrt {18} + 4 }{1 } = 8 + \frac{ \sqrt {18} - 4 }{1 }$$
Simple continued fraction tableau:
$$\begin{array}{cccccccccc} & & 4 & & 4 & & 8 & \\ \\ \frac{ 0 }{ 1 } & \frac{ 1 }{ 0 } & & \frac{ 4 }{ 1 } & & \frac{ 17 }{ 4 } \\ \\ & 1 & & -2 & & 1 \end{array}$$
$$\begin{array}{cccc} \frac{ 1 }{ 0 } & 1^2 - 18 \cdot 0^2 = 1 & \mbox{digit} & 4 \\ \frac{ 4 }{ 1 } & 4^2 - 18 \cdot 1^2 = -2 & \mbox{digit} & 4 \\ \frac{ 17 }{ 4 } & 17^2 - 18 \cdot 4^2 = 1 & \mbox{digit} & 8 \\ \end{array}$$
This time the number is $\sqrt {18} - 2.$
It is enough to use $$2 < \sqrt {18} - 2 < \frac{9}{4}$$
$$\color{red}{ 2 < \sqrt {18} - 2 < \frac{9}{4} < \frac{ 5 }{ 2 } < 6 - \sqrt {12} < 3 }$$
• Thank you, but I'll study it one day – Learner132 Jun 10 '17 at 19:51
Define
$a=6-2\sqrt 3>0$
$b=3\sqrt 2-2>0$
$a-b = 8 - (2\sqrt 3 + 3\sqrt 2)$
$(2\sqrt 3 + 3\sqrt 2)^2 = 30+12\sqrt 6 = 6×(5+2\sqrt 6) < 60 < 64$ because $6=2×3 < (5/2)^2$
$a-b > 8-8=0, a>b$
• Thank you, it took me a while to understand 2×3<(5/2)^2 but it is easier to explain to math teacher. – Learner132 Jun 10 '17 at 19:45
Here's yet another way, for those who aren't comfortable with the $\gtrless$ or $\sim$ notation.
We can use crude rational approximations to $\sqrt 2$ and $\sqrt 3$.
\begin{align} \left(\frac{3}{2}\right)^2 = \frac{9}{4} & \gt 2\\ \frac{3}{2} & \gt \sqrt 2\\ \frac{9}{2} & \gt 3\sqrt 2 \end{align}
And \begin{align} \left(\frac{7}{4}\right)^2 = \frac{49}{16} & \gt 3\\ \frac{7}{4} & \gt \sqrt 3\\ \frac{7}{2} & \gt 2\sqrt 3 \end{align}
Adding those two approximations, we get \begin{align} \frac{9}{2} + \frac{7}{2} = 8 & \gt 3\sqrt 2 + 2\sqrt 3\\ 6 + 2 & \gt 3\sqrt 2 + 2\sqrt 3\\ 6 - 2\sqrt 3 & \gt 3\sqrt 2 - 2 \end{align}
I'll use >=< to represent the unknown comparison.
$6-2\sqrt{3} >=< 3\sqrt{2}-2$
Lets start by adding two to both sides to reduce the number of numbers. This doesn't change the comparison result.
$8-2\sqrt{3} >=< 3\sqrt{2}$
Both sides are clearly positive ( $2\sqrt{3} < 6$ ) so we can square both sides without changing the comparison result.
In a more maginal case where we were unsure if the left hand side was positive we could have compared the two terms in the left hand side by squaring both of them and hence determined whether the left hand side was positive or negative.
$64 -32\sqrt{3} + 12 >=< 18$
Now lets collect terms.
$60 >=< 32\sqrt{3}$
Divide by four.
$15 >=< 8\sqrt{3}$
Square again.
$225 >=< 64 \times 3$
$225 > 192$
Therefore
$6-2\sqrt{3} > 3\sqrt{2}-2$
There is still some hope in taking the first minus the second in this case: $$6-2\sqrt{3} - (3\sqrt{2}-2) = 8 - (2\sqrt{3} + 3\sqrt{2})$$
So now the question boils down to if the expression with the square root exceeds $8$. We know that $8^{2} = 64$ and: $$(2\sqrt{3}+3\sqrt{2})^{2}=4*3+2(2\sqrt{3})(3\sqrt{2})+9*2 =30+12\sqrt{3}\sqrt{2}$$
Conversely: $$8^{2} = 64 = 28+36=28+6*6 = 28+6*\sqrt{6}\sqrt{6}$$
Subtracting them yields: $$28+6*\sqrt{6}\sqrt{6}-30+12\sqrt{3}\sqrt{2} = -2+6*\sqrt{6}\sqrt{6}-12\sqrt{3}\sqrt{2}$$
Thus this is only positive if the square root terms are positive and exceed 2. Comparing the square root terms: $$6*\sqrt{6}\sqrt{6}-12\sqrt{3}\sqrt{2} = 6*\sqrt{2}\sqrt{3}(\frac{\sqrt{2}}{\sqrt{2}})\sqrt{6}-12\sqrt{6}=12\frac{\sqrt{3}}{\sqrt{2}}\sqrt{6}-12\sqrt{6}=12\sqrt{6}(\frac{\sqrt{3}}{\sqrt{2}}-1)$$
We conclude that the square root term is positive since $\sqrt{3}>\sqrt{2}$ . But is it greater than $-2$? Or rather, how much do we need to multiply to the expression $\frac{\sqrt{3}}{\sqrt{2}}-1$ for it to be greater than $2$? $$n*(\frac{\sqrt{3}}{\sqrt{2}}-1)>2$$ $$n>\frac{2}{\frac{\sqrt{3}}{\sqrt{2}}-1}*\frac{\frac{\sqrt{3}}{\sqrt{2}}+1}{\frac{\sqrt{3}}{\sqrt{2}}+1}\approx\frac{2*(1.5+1)}{0.5} = 10$$ So the factor in front of the term $\frac{\sqrt{3}}{\sqrt{2}}-1$ should be at least 10. But since $12>10$ and $\sqrt{6}>1$, we conclude that $12\sqrt{6}>10$. Thus, substituting back to the original equation: $$-2+6*\sqrt{6}\sqrt{6}-12\sqrt{3}\sqrt{2} = -2+12\sqrt{6}(\frac{\sqrt{3}}{\sqrt{2}}-1) > 0$$ And so we conclude that: $$64>(2\sqrt{3}+3\sqrt{2})^{2}$$ and so, for positive root, $$8>(2\sqrt{3}+3\sqrt{2})$$ and $$8 - (2\sqrt{3} + 3\sqrt{2}) > 0$$
You may use other approaches. Notice that I split $8^{2} = 28+36$. Equally viable is to split $8^{2} = 30+34$, and you might have to use a similar (Squaring than rooting) trick to show that $34 > 12\sqrt{6}$, and then finally conclude that $8>(2\sqrt{3} + 3\sqrt{2})$. This alternative approach should be able to give you a smaller threshold compared to the approximation I made midway.
$6-2√3 \approx2(1.732) = 6 - 3.464 = 2.536$ $3√2-2 \approx 3(1.414) - 2 = 4.242 - 2 = 2.242$
This implies that $\dfrac 52$ is between the two quantities:
\begin{align} \dfrac{49}{4} &> 12 \\ \dfrac 72 &> 2\sqrt 3 \\ \dfrac{12}{2} &> 2\sqrt 3 + \dfrac 52 \\ 6 - 2\sqrt 3 &> \dfrac 52 \end{align}
and
\begin{align} \dfrac{81}{4} &> 18 \\ \dfrac 92 &> 3\sqrt 2 \\ \dfrac 52 &> 3\sqrt 2 - 2 \end{align}
It follows that $6-2√3 > 3\sqrt 2 - 2$.
• see this answer same concept but better (less digits, can be easily verified by hand, using lower and upper bounds instead of $\approx$)) and 12 hours earlier than your answer – miracle173 Jun 11 '17 at 14:16
• @miracle173 What you saw was only a part of what I intended to do. I needed some more time to think about the problem so I thought I had deleted everything. Obviously, I hadn't. What you see above is what I intended to do. – steven gregory Jun 11 '17 at 19:57 | 2021-06-20T06:22:49 | {
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https://math.stackexchange.com/questions/497944/find-the-average-velocity | # Find the average velocity
If a ball is thrown in the air with a velocity $34$ ft/s, its height in feet $t$ seconds later is given by
$$y = 34 t − 16 t^2 .$$
Find the average velocity for the time period beginning when $t = 2$ and lasting $0.5$ second, $0.1$ second, $0.05$ second, $0.01$ second and estimate the instantaneous velocity when $t = 2$.
I tried to solve by doing the following:
$y=34(2)-16(2)^2$
$y=68-64$
$y=a$
$4/0.5=8$ft/s
but I was told that answer is incorrect.
What did I do wrong?
• This is a limit approximation. If we say $y=f(x)$, then we can write that the first average is ${f(2.5)-f(2)\over 0.5}={-15-4\over 0.5}=-38$. – abiessu Sep 18 '13 at 21:50
• How did you get -15 for f(2.5) and 4 for f(2) ? – Grey Sep 18 '13 at 22:22
• I used $f(2.5) = 34(2.5)-16(2.5)^2 = 85-100=-15$ and $f(2)=34(2)-16(2)^2=68-64=4$. I saw the $62$ in the second term of your original post before it was edited and that $16\cdot 4$ is not $62$... – abiessu Sep 18 '13 at 22:26
• Thanks. And how to get get the instantaneous velocity when t = 2 ? – Grey Sep 18 '13 at 22:40
• That is the point of the exercise, first calculate $f(2.5)-f(2)\over 0.5$, then $f(2.1)-f(2)\over .1$, and so on, and use this series of values to estimate the instantaneous velocity at $t=2$. – abiessu Sep 18 '13 at 22:42
You are being asked for the average velocity over the time span $2$ to $2.5$ seconds, among others. To do that one, you need $y(2.5)$ and $y(2)$ Then the average velocity in that span is $\frac {y(2.5)-y(2)}{2.5-2}$
There is enough information here to complete the following process:
Given $y(t)=34t-16t^2$, we have:
$$y(2)=34\cdot 2-16\cdot 4=68-64=4$$
$$y(2.5)=34\cdot 2.5-16\cdot 2.5^2=85-100=-15$$
$$y(2.1)=34\cdot 2.1-16\cdot 2.1^2=71.4-70.56=0.84$$
$$y(2.05)=2.46$$ (using calculator)
$$y(2.01)=3.6984$$ (using calculator)
Now, the remaining task is to calculate $-19\over 0.5$, $-3.16\over 0.1$, $-1.54\over 0.05$, and $-0.3016\over 0.01$ and see if these numbers are approaching a particular number. I got each of these fractions by writing $y(2+t)-y(2)\over t$ for each of the values $t\in \{0.5,0.1,0.05,0.01\}$. | 2019-12-10T00:18:08 | {
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https://math.stackexchange.com/questions/2055398/mathematics-behind-this-card-trick | Mathematics behind this card trick
1. Suppose I have $21$ playing cards. I distribute them in $3$ columns and tell you to choose mentally a card. Then just indicate in which column the card is.
2. I pick up one of the columns which doesn't contain your card, then the column which contains your card then the remaining column.
3. Now I deal the cards in $3$ columns again, starting from the left to the right, and repeating the process until there are no cards left in my hand. I ask you to indicate to me in which column your card is.
4. I repeat step 2 then 3.
5. I repeat step 2.
Now by counting either way in the deck of $21$ cards, your card will be the 11th card.
I've tried using modulo to understand the problem but I'm stuck doing integer divisions. So does anyone have a simpler way to explain the trick.
Also could anyone explain why it works only for odd number of cards in each column and why it requires additional steps for more cards. e.g for $17 \cdot 3 = 51$ requires an additional step compared with $21$ cards?
EDIT: I forgot to add that the number of cards you have to count for the final step equals
$$1.5\text{ times the number of cards in each column} + 0.5$$
• If you do not have an odd number of cards in total, there is not a middle card Dec 12 '16 at 13:25
• If you have $n$ cards and $3$ columns, you need $\log_3(n)$ steps (rounded up) to distinguish them Dec 12 '16 at 13:26
• This related trick (and its explanation) might provide some insight Dec 12 '16 at 13:29
• I first learnt this card trick when I was 14. After learning some calculus, I was pleasant to realise that the fact that the card will eventually end up at the middle position is basically a consequence of the squeezing principle. Dec 13 '16 at 9:13
• @user1551 That seems like an overapplication of the squeezing principle.
– jwg
Dec 13 '16 at 10:30
Suppose that, when you first lay the cards on the table, the card I choose is at position $x$ in its column. You don't know $x$, but you know that $1\leq x\leq 7$.
Now, when you pick up the cards, my card will be at position $7+x$ in the full stack. The second time you lay the cards on the table, my card will appear at position $p_1=\lceil\frac{7+x}{3}\rceil$ in its column.
The second time you pick up the cards, my card will be at position $7+p_1$ in the full stack. The third time you lay the cards on the table, my card will appear at position $p_2=\lceil\frac{7+p_1}{3}\rceil$ in its column.
Finally, when you pick up the cards for the third time, my card will be at position $7+p_2$ in the full stack. Putting this all together, my card will be at position $$7+p_2=7+\left\lceil\frac{7+\lceil\frac{7+x}{3}\rceil}{3}\right\rceil$$ in the full stack. The trick is that this is equal to $11$ for all $x$ in the range $1\leq x\leq 7$.
For a proof of this last statement, as jpmc26 mentions, one can apply the identities $\lceil\frac{m+\lceil x\rceil}{n}\rceil=\lceil\frac{m+x}{n}\rceil$ and $\lceil n+x\rceil = n+\lceil x\rceil$ (for real $x$, integer $m$, and positive integer $n$) to show that $$7+\left\lceil\frac{7+\lceil\frac{7+x}{3}\rceil}{3}\right\rceil = 7+\left\lceil\frac{7+\frac{7+x}{3}}{3}\right\rceil = 7 + \left\lceil 3 + \frac{x+1}{9}\right\rceil = 10 + \left\lceil\frac{x+1}{9}\right\rceil \enspace,$$ which is clearly equal to $11$ for $1\leq x\leq 7$.
• You can use the fact that $\left \lceil \frac{7+\lceil y \rceil}{3} \right \rceil = \left \lceil \frac{7+ y }{3} \right \rceil$ (see here) to show that the last expression is equal to $7 + \left \lceil 3 + \frac{x + 1}{9} \right \rceil = 10 + \left \lceil \frac{x + 1}{9} \right \rceil$, which makes the fact it's always 11 much more obvious. Dec 12 '16 at 23:14
• Good point. I just checked that it holds for x=1 and for x=7, therefore it holds for x in between due to monotonicity. Should I edit the answer to include your remark? Dec 13 '16 at 8:17
• @EvangelosBampas: Yes! Dec 13 '16 at 12:47
• @EvangelosBampas Absolutely. (Thanks!) Comments are considered ephemeral/transient, and their primary purpose is for the improvement of a post. You can generally assume that useful content or clarification posted in them warrants an edit or an additional post. (In this case, I only provided some comparatively minor additions that just build on your answer and make the conclusion easier to see, so a separate post wouldn't be able to stand on its own.) Dec 13 '16 at 14:13
You can find a full explanation of this trick and related ones at
Gergonne’s Card Trick, Positional Notation, and Radix Sort
(Mathematics Magazine, February, 2010)
http://www.maa.org/sites/default/files/Bolker-MMz-201053228.pdf
The classic trick uses $27 = 3^3$ cards. Your $21$ card version is discussed on page 48.
• This answer comes dangerously close to "link only." Dec 12 '16 at 22:51
• I can see a total of 4 lines floating freely besides the link provided, which will definitly assure that the answer is too far from being called a link only if anything. Dec 13 '16 at 1:01
• @jpmc26 I don't like link only answers either, but the relevant content in the link is too long to post - I'd essentially be pasting in most of the paper. The link is to the journal, so stable. Dec 13 '16 at 2:06
After one deal-and-gather phase, your card is in the middle third of the deck, right?
When you deal out the cards again, where does that middle third end up? In the middle third (looking top-to-bottom) of the tableau.
To see this: on the first deal, pick a card, say, the 3 of hearts. Replace everything else in its column with red cards, and everything else in the other two columns with black cards. Gather and re-deal. You'll see a red "band" in the middle (top to bottom) of the new tableau.
Now you identify your card in that middle band. When you gather up the other cards, it'll once again be in the middle third of the deck, because there's one entire column-worth of cards in front of it, and one entire column-worth of cards behind it.
But you can say more than that: suppose your card was in the first column. Well, then, it was in the middle third of the first column, wasn't it? So if each column has $k$ cards, you've got $k$ cards in front of it (from one of the other piles), and another k/3 cards (from its own pile) in front of it, and the same for cards behind it. So now it's in the middle 1/9th of the pile.
And the next time it'll be in the middle 1/27 of the pile. And so on.
Now that analysis isn't QUITE right, because $k$ might not be divisible by 3. So instead of having k + k/3 cards in front of it after two deals, you have k + floor(k/3). For instance, if $k = 8$, then you'd have $8 + floor(8/3) = 8 + floor(2.66...) = 8 + 2 = 10$ cards in front of it. But aside from this minor glitch, what you get is the following:
after $p$ "passes" on a deck of $N$ cards, your card is in the middle (roughly) $(N/3^p)$ cards of the deck. When $3^p > N$, this means that your card is the middle card of the deck.
• +1, Definitely makes more sense to me than the algebra above. Dec 13 '16 at 1:00
Let's go step by step. After steps i) and ii) we already know that the chose card is in position $8,9,10,11,12,13,14$ in the pile. Then after dealing all the cards again in the given manner we know that the cards that were in the positions $8,9,10,11,12,13,14$ before dealing are now in places $3,4,5$ (actually it depends on the pile, but we can be sure that each of the mentioned cards is now at a position $3,4$ or $5$ in one of the three piles. So eventually after the second step we narrowed the choice to three cards.
After putting the cards in one pile again then we know that the chosen card will be in position $10,11,12$. And in the last deal obviously each of the cards will be in dealt in different pile, so we should be able to find out which one is the card for sure. But during the last deal the cards in positions $10,11,12$ are all in position $4$ in each of the new piles, hence after collecting all the cards, the chosen card will be in position $11$.
Eventually as you can see with each dealing we lower the number of possibilities from $n$ to $\lfloor\frac{n-1}{3}\rfloor + 1$. | 2022-01-20T09:58:50 | {
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# Inequalities trick
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Updated on: 26 Apr 2018, 03:34
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I learnt this trick while I was in school and yesterday while solving one question I recalled.
Its good if you guys use it 1-2 times to get used to it.
Suppose you have the inequality
$$f(x) = (x-a)(x-b)(x-c)(x-d) < 0$$
Just arrange them in order as shown in the picture and draw curve starting from + from right.
now if f(x) < 0 consider curve having "-" inside and if f(x) > 0 consider curve having "+" and combined solution will be the final solution. I m sure I have recalled it fully but if you guys find any issue on that do let me know, this is very helpful.
Don't forget to arrange then in ascending order from left to right. $$a<b<c<d$$
So for f(x) < 0 consider "-" curves and the ans is: $$(a < x < b)$$, $$(c < x < d)$$
and for f(x) > 0 consider "+" curves and the ans is: $$(x < a)$$, $$(b < x < c)$$, $$(d < x)$$
If f(x) has three factors then the graph will have - + - +
If f(x) has four factors then the graph will have + - + - +
If you can not figure out how and why, just remember it.
Try to analyze that the function will have number of roots = number of factors and every time the graph will touch the x axis.
For the highest factor d if x>d then the whole f(x) > 0 and after every interval of the roots the signs will change alternatively.
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Attachment:
Untitled.png [ 12.08 KiB | Viewed 4936 times ]
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Originally posted by gurpreetsingh on 16 Mar 2010, 10:11.
Last edited by Bunuel on 26 Apr 2018, 03:34, edited 1 time in total.
Edited.
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Re: Inequalities trick [#permalink]
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17 Oct 2010, 16:14
gurpreetsingh wrote:
I learnt this trick while I was in school and yesterday while solving one question I recalled.
Its good if you guys use it 1-2 times to get used to it.
So for f(x) < 0 consider "-" curves and the ans is : (a < x < b) , (c < x < d)
and for f(x) < 0 consider "+" curves and the ans is : (x < a), (b < x < c) , (d < x)
I don't understand this part alone. Can you please explain?
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Re: Inequalities trick [#permalink]
### Show Tags
17 Oct 2010, 17:19
Dreamy wrote:
gurpreetsingh wrote:
I learnt this trick while I was in school and yesterday while solving one question I recalled.
Its good if you guys use it 1-2 times to get used to it.
So for f(x) < 0 consider "-" curves and the ans is : (a < x < b) , (c < x < d)
and for f(x) < 0 consider "+" curves and the ans is : (x < a), (b < x < c) , (d < x)
I don't understand this part alone. Can you please explain?
Suppose you have the inequality
f(x) = (x-a)(x-b)(x-c)(x-d) < 0 you will consider the curve with -ve inside it.. check the attached image.
f(x) = (x-a)(x-b)(x-c)(x-d) > 0 consider the +ve of the curve
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Re: Inequalities trick [#permalink]
### Show Tags
22 Oct 2010, 06:33
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Yes, this is a neat little way to work with inequalities where factors are multiplied or divided. And, it has a solid reasoning behind it which I will just explain.
If (x-a)(x-b)(x-c)(x-d) < 0, we can draw the points a, b, c and d on the number line.
e.g. Given (x+2)(x-1)(x-7)(x-4) < 0, draw the points -2, 1, 7 and 4 on the number line as shown.
Attachment:
doc.jpg [ 7.9 KiB | Viewed 63852 times ]
This divides the number line into 5 regions. Values of x in right most region will always give you positive value of the expression. The reason for this is that if x > 7, all factors above will be positive.
When you jump to the next region between x = 4 and x = 7, value of x here give you negative value for the entire expression because now, (x - 7) will be negative since x < 7 in this region. All other factors are still positive.
When you jump to the next region on the left between x = 1 and x = 4, expression will be positive again because now two factors (x - 7) and (x - 4) are negative, but negative x negative is positive... and so on till you reach the leftmost section.
Since we are looking for values of x where the expression is < 0, here the solution will be -2 < x < 1 or 4< x < 7
It should be obvious that it will also work in cases where factors are divided.
e.g. (x - a)(x - b)/(x - c)(x - d) < 0
(x + 2)(x - 1)/(x -4)(x - 7) < 0 will have exactly the same solution as above.
Note: If, rather than < or > sign, you have <= or >=, in division, the solution will differ slightly. I will leave it for you to figure out why and how. Feel free to get back to me if you want to confirm your conclusion.
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Save up to $1,000 on GMAT prep through 8/20! Learn more here > GMAT self-study has never been more personalized or more fun. Try ORION Free! Manager Joined: 29 Sep 2008 Posts: 105 Re: Inequalities trick [#permalink] ### Show Tags 22 Oct 2010, 11:45 19 20 if = sign is included with < then <= will be there in solution like for (x+2)(x-1)(x-7)(x-4) <=0 the solution will be -2 <= x <= 1 or 4<= x <= 7 in case when factors are divided then the numerator will contain = sign like for (x + 2)(x - 1)/(x -4)(x - 7) < =0 the solution will be -2 <= x <= 1 or 4< x < 7 we cant make 4<=x<=7 as it will make the solution infinite correct me if i am wrong Manager Joined: 10 Nov 2010 Posts: 142 Re: Inequalities trick [#permalink] ### Show Tags 11 Mar 2011, 06:29 VeritasPrepKarishma wrote: vjsharma25 wrote: How you have decided on the first sign of the graph?Why it is -ve if it has three factors and +ve when four factors? Check out my post above for explanation. I understand the concept but not the starting point of the graph.How you decide about the graph to be a sine or cosine waveform?Meaning graph starts from the +ve Y-axis for four values and starts from -ve Y-axis for three values. What if the equation you mentioned is (x+2)(x-1)(x-7)<0,will the last two ranges be excluded or the graph will also change? Retired Moderator Joined: 20 Dec 2010 Posts: 1877 Re: Inequalities trick [#permalink] ### Show Tags 11 Mar 2011, 06:49 24 15 vjsharma25 wrote: VeritasPrepKarishma wrote: vjsharma25 wrote: How you have decided on the first sign of the graph?Why it is -ve if it has three factors and +ve when four factors? Check out my post above for explanation. I understand the concept but not the starting point of the graph.How you decide about the graph to be a sine or cosine waveform?Meaning graph starts from the +ve Y-axis for four values and starts from -ve Y-axis for three values. What if the equation you mentioned is (x+2)(x-1)(x-7)<0,will the last two ranges be excluded or the graph will also change? I always struggle with this as well!!! There is a trick Bunuel suggested; (x+2)(x-1)(x-7) < 0 Here the roots are; -2,1,7 Arrange them in ascending order; -2,1,7; These are three points where the wave will alternate. The ranges are; x<-2 -2<x<1 1<x<7 x>7 Take a big value of x; say 1000; you see the inequality will be positive for that. (1000+2)(1000-1)(1000-7) is +ve. Thus the last range(x>7) is on the positive side. Graph is +ve after 7. Between 1 and 7-> -ve between -2 and 1-> +ve Before -2 -> -ve Since the inequality has the less than sign; consider only the -ve side of the graph; 1<x<7 or x<-2 is the complete range of x that satisfies the inequality. _________________ Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 8188 Location: Pune, India Re: Inequalities trick [#permalink] ### Show Tags 11 Mar 2011, 19:57 10 2 vjsharma25 wrote: I understand the concept but not the starting point of the graph.How you decide about the graph to be a sine or cosine waveform?Meaning graph starts from the +ve Y-axis for four values and starts from -ve Y-axis for three values. What if the equation you mentioned is (x+2)(x-1)(x-7)<0,will the last two ranges be excluded or the graph will also change? Ok, look at this expression inequality: (x+2)(x-1)(x-7) < 0 Can I say the left hand side expression will always be positive for values greater than 7? (x+2) will be positive, (x - 1) will be positive and (x-7) will also be positive... so in the rightmost regions i.e. x > 7, all three factors will be positive. The expression will be positive when x > 7, it will be negative when 1 < x < 7, positive when -2 , x < 1 and negative when x < -2. We need the region where the expression is less than 0 i.e. negative. So either 1 < x < 7 or x < -2. Now let me add another factor: (x+8)(x+2)(x-1)(x-7) Can I still say that the entire expression is positive in the rightmost region i.e. x>7 because each one of the four factors is positive? Yes. So basically, your rightmost region is always positive. You go from there and assign + and - signs to the regions. Your starting point is the rightmost region. Note: Make sure that the factors are of the form (ax - b), not (b - ax)... e.g. (x+2)(x-1)(7 - x)<0 Convert this to: (x+2)(x-1)(x-7)>0 (Multiply both sides by '-1') Now solve in the usual way. Assign '+' to the rightmost region and then alternate with '-' Since you are looking for positive value of the expression, every region where you put a '+' will be the region where the expression will be greater than 0. _________________ Karishma Veritas Prep GMAT Instructor Save up to$1,000 on GMAT prep through 8/20! Learn more here >
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Re: Inequalities trick [#permalink]
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07 Aug 2011, 07:24
gurpreetsingh wrote:
ulm wrote:
if we have smth like (x-a)^2(x-b)
we don't need to change a sign when jump over "a".
yes even powers wont contribute to the inequality sign. But be wary of the root value of x=a
This way of solving inequalities actually makes it soo much easier. Thanks gurpreetsingh and karishma
However, i am confused about how to solve inequalities such as: (x-a)^2(x-b) and also ones with root value.
could someone please explain.
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Re: Inequalities trick [#permalink]
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08 Aug 2011, 11:59
2
3
Asher wrote:
gurpreetsingh wrote:
ulm wrote:
if we have smth like (x-a)^2(x-b)
we don't need to change a sign when jump over "a".
yes even powers wont contribute to the inequality sign. But be wary of the root value of x=a
This way of solving inequalities actually makes it soo much easier. Thanks gurpreetsingh and karishma
However, i am confused about how to solve inequalities such as: (x-a)^2(x-b) and also ones with root value.
could someone please explain.
When you have (x-a)^2(x-b) < 0, the squared term is ignored because it is always positive and hence doesn't affect the sign of the entire left side. For the left hand side to be negative i.e. < 0, (x - b) should be negative i.e. x - b < 0 or x < b.
Similarly for (x-a)^2(x-b) > 0, x > b
As for roots, you have to keep in mind that given $$\sqrt{x}$$, x cannot be negative.
$$\sqrt{x}$$ < 10
implies 0 < $$\sqrt{x}$$ < 10
Squaring, 0 < x < 100
Root questions are specific. You have to be careful. If you have a particular question in mind, send it.
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Karishma
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Save up to $1,000 on GMAT prep through 8/20! Learn more here > GMAT self-study has never been more personalized or more fun. Try ORION Free! Manager Joined: 06 Apr 2011 Posts: 52 Location: India Re: Inequalities trick [#permalink] ### Show Tags 08 Aug 2011, 22:22 Quote: When you have (x-a)^2(x-b) < 0, the squared term is ignored because it is always positive and hence doesn't affect the sign of the entire left side. For the left hand side to be negative i.e. < 0, (x - b) should be negative i.e. x - b < 0 or x < b. Similarly for (x-a)^2(x-b) > 0, x > b Thanks Karishma for the explanation. Hope you wouldn't mind clarifying a few more doubts. Firstly, in the above case, since x>b could we say that everything with be positive. would the graph look something like this: positive.. b..postive.. a.. positive On the other hand if (x-a)^2(x-b) < 0, x < b, (x-a)^2 would be positive and for (x-b) if x<b the the left side would be negative. would the graph look something like this: negative.. b..postive.. a.. postive Am i right? If it is not too much of a trouble, could you please show the graphical representation. problems with \sqrt{x}.. this is all i could find (googled actually ): 1. √(-x+4) ≤ √(x) 2. x^\sqrt{x} =< (\sqrt{x})^x {P.S.: i tried to insert the graphical representation that i came up with, but i am a bit technically challenged in this area it seems} _________________ Regards, Asher Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 8188 Location: Pune, India Re: Inequalities trick [#permalink] ### Show Tags 09 Aug 2011, 03:28 1 2 Asher wrote: Firstly, in the above case, since x>b could we say that everything with be positive. would the graph look something like this: positive.. b..postive.. a.. positive On the other hand if (x-a)^2(x-b) < 0, x < b, (x-a)^2 would be positive and for (x-b) if x<b the the left side would be negative. would the graph look something like this: negative.. b..postive.. a.. postive So when you have $$(x - a)^2(x - b) < 0$$, you ignore x = a and just plot x = b. It is positive in the rightmost region and negative on the left. So the graph looks like this: negative ... b ... positive Am i right? If it is not too much of a trouble, could you please show the graphical representation. problems with \sqrt{x}.. this is all i could find (googled actually ): 1. √(-x+4) ≤ √(x) 2. x^\sqrt{x} =< (\sqrt{x})^x {P.S.: i tried to insert the graphical representation that i came up with, but i am a bit technically challenged in this area it seems} Squared terms are ignored. You do not put them in the graph. They are always positive so they do not change the sign of the expression. e.g. $$(x-4)^2(x - 9)(x+11) < 0$$ We do not plot x = 4 here, only x = -11 and x = 9. We start with the rightmost section as positive. So it looks something like this: positive... -11 ... negative ... 9 ... positive Since we need the region where x is negative, we get -11 < x < 9. Basically, the squared term is like a positive number in that it doesn't affect the sign of the expression. I would be happy to solve inequalities questions related to roots but please put them in a separate post and pm the link to me. That way, everybody can try them. _________________ Karishma Veritas Prep GMAT Instructor Save up to$1,000 on GMAT prep through 8/20! Learn more here >
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Re: Inequalities trick [#permalink]
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10 Aug 2011, 07:00
hey ,
can u please tel me the solution for this ques
a car dealership sells only sports cars and luxury cars and has atleast some of each type of car in stock at all times.if exactly 1/7 of sports car and 1/2 of luxury cars have sunroofs and there are exactly 42 cars on the lot.what is the smallest number of cars that could have roofs?
ans -11
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Re: Inequalities trick [#permalink]
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10 Aug 2011, 17:01
11
24
WoW - This is a cool thread with so many thing on inequalities....I have compiled it together with some of my own ideas...It should help.
1) CORE CONCEPT
@gurpreetsingh -
Suppose you have the inequality
f(x) = (x-a)(x-b)(x-c)(x-d) < 0
Arrange the NUMBERS in ascending order from left to right. a<b<c<d
Draw curve starting from + from right.
now if f(x) < 0 consider curve having "-" inside and if f(x) > 0 consider curve having "+" and combined solution will be the final solution. I m sure I have recalled it fully but if you guys find any issue on that do let me know, this is very helpful.
So for f(x) < 0 consider "-" curves and the ans is : (a < x < b) , (c < x < d)
and for f(x) > 0 consider "+" curves and the ans is : (x < a), (b < x < c) , (d < x)
If f(x) has three factors then the graph will have - + - +
If f(x) has four factors then the graph will have + - + - +
If you can not figure out how and why, just remember it.
Try to analyze that the function will have number of roots = number of factors and every time the graph will touch the x axis.
For the highest factor d if x>d then the whole f(x) > 0 and after every interval of the roots the signs will change alternatively.
Note: Make sure that the factors are of the form (ax - b), not (b - ax)...
example -
(x+2)(x-1)(7 - x)<0
Convert this to: (x+2)(x-1)(x-7)>0 (Multiply both sides by '-1')
Now solve in the usual way. Assign '+' to the rightmost region and then alternate with '-'
Since you are looking for positive value of the expression, every region where you put a '+' will be the region where the expression will be greater than 0.
2) Variation - ODD/EVEN POWER
@ulm/Karishma -
if we have even powers like (x-a)^2(x-b)
we don't need to change a sign when jump over "a".
This will be same as (x-b)
We can ignore squares BUT SHOULD consider ODD powers
example -
2.a
(x-a)^3(x-b)<0 is the same as (x-a)(x-b) <0
2.b
(x - a)(x - b)/(x - c)(x - d) < 0 ==> (x - a)(x - b)(x-c)^-1(x-d)^-1 <0
is the same as (x - a)(x - b)(x - c)(x - d) < 0
3) Variation <= in FRACTION
@mrinal2100 -
if = sign is included with < then <= will be there in solution
like for (x+2)(x-1)(x-7)(x-4) <=0 the solution will be -2 <= x <= 1 or 4<= x <= 7
BUT if it is a fraction the denominator in the solution will not have = SIGN
example -
3.a
(x + 2)(x - 1)/(x -4)(x - 7) < =0
the solution will be -2 <= x <= 1 or 4< x < 7
we cant make 4<=x<=7 as it will make the solution infinite
4) Variation - ROOTS
@Karishma -
As for roots, you have to keep in mind that given $$\sqrt{x}$$, x cannot be negative.
$$\sqrt{x}$$ < 10
implies 0 < $$\sqrt{x}$$ < 10
Squaring, 0 < x < 100
Root questions are specific. You have to be careful. If you have a particular question in mind, send it.
Refer - inequalities-and-roots-118619.html#p959939
Some more useful tips for ROOTS....I am too lazy to consolidate
<5> THESIS -
@gmat1220 -
Once algebra teacher told me - signs alternate between the roots. I said whatever and now I know why Watching this article is a stroll down the memory lane.
I will save this future references....
Please add anything that you feel will help.
Anyone wants to add ABSOLUTE VALUES....That will be a value add to this post
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Re: Inequalities trick [#permalink]
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11 Aug 2011, 22:57
1
3
sushantarora wrote:
hey ,
can u please tel me the solution for this ques
a car dealership sells only sports cars and luxury cars and has atleast some of each type of car in stock at all times.if exactly 1/7 of sports car and 1/2 of luxury cars have sunroofs and there are exactly 42 cars on the lot.what is the smallest number of cars that could have roofs?
ans -11
Please put questions in new posts. Put it in the same post only if it is totally related or a variation of the question that we are discussing.
Now for the solution:
There are 42 cars on the lot. 1/7 of sports cars and 1/2 of luxury cars have sunroofs. This means that 1/7 of number of sports cars and 1/2 of number of luxury cars should be integers (You cannot have 1.5 cars with sunroofs, right?)
We want to minimize the sunroofs. Since 1/2 of luxury cars have sunroofs and only 1/7 of sports cars have them, it will be good to have fewer luxury cars and more sports cars. Best would be to have all sports cars. But, the question says there are some of each kind at any time. So let's say there are 2 luxury cars (since 1/2 of them should be an integer value). But 1/7 of 40 (the rest of the cars are sports cars) is not an integer number. Let's instead look for the multiple of 7 that is less than 42. The multiple of 7 that is less than 42 is 35. So we could have 35 sports cars. But then, 1/2 of 7 (since 42 - 35 = 7 are luxury cars) is not an integer. The next smaller multiple of 7 is 28. This works. 1/2 of 14 (since 42 - 28 = 14 are luxury cars) is 7. So we can have 14 luxury cars and 28 sports cars. That is the maximum number of sports cars that we can have.
1/7 of 28 sports cars = 4 cars have sunroofs
1/2 of 14 luxury cars = 7 cars have sunroofs
So at least 11 cars will have sunroofs.
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Save up to $1,000 on GMAT prep through 8/20! Learn more here > GMAT self-study has never been more personalized or more fun. Try ORION Free! Manager Joined: 16 Feb 2012 Posts: 193 Concentration: Finance, Economics Re: Inequalities trick [#permalink] ### Show Tags 22 Jul 2012, 03:03 VeritasPrepKarishma wrote: mrinal2100: Kudos to you for excellent thinking! Correct me if I'm wrong. If the lower part of the equation$$\frac {(x+2)(x-1)}{(x-4)(x-7)}$$ were $$4\leq x \leq 7$$, than the lower part would be equal to zero,thus making it impossible to calculate the whole equation. _________________ Kudos if you like the post! Failing to plan is planning to fail. Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 8188 Location: Pune, India Re: Inequalities trick [#permalink] ### Show Tags 23 Jul 2012, 03:13 2 Stiv wrote: VeritasPrepKarishma wrote: mrinal2100: Kudos to you for excellent thinking! Correct me if I'm wrong. If the lower part of the equation$$\frac {(x+2)(x-1)}{(x-4)(x-7)}$$ were $$4\leq x \leq 7$$, than the lower part would be equal to zero,thus making it impossible to calculate the whole equation. x cannot be equal to 4 or 7 because if x = 4 or x = 7, the denominator will be 0 and the expression will not be defined. _________________ Karishma Veritas Prep GMAT Instructor Save up to$1,000 on GMAT prep through 8/20! Learn more here >
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Re: Inequalities trick [#permalink]
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25 Jul 2012, 22:21
pavanpuneet wrote:
Hi Karishma,
Just for my reference, say if the equation was (x+2)(x-1)/(x-4)(x-7) and the question was for what values of x is this expression >0, then the roots will be -2,1,4,7 and by placing on the number line and making the extreme right as positive...
----(-2)----(1)----(4)---(7)----then x>7, 1<x<4 and x<-2...Please confirm..
However, is say it was >=0 then x>7, 1<=x<4 and x<=-2; given that the denominator cannot be zero. Please confirm
Yes, you are right in both the cases.
Also, if you want to verify that the range you have got is correct, just plug in some values to see. Put x = 0, the expression is -ve. Put x = 2, the expression is positive.
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Re: Inequalities trick [#permalink]
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26 Jul 2012, 07:59
Thankyou Karishma.
Further, say if the same expression was (x+2)(1-x)/(x-4)(x-7) and still the question was for what values of x is the expression positive, then ... make it x-1 and with the same roots, have the rightmost as -ve. Then we look for the +ve intervals and check for those intervals if the expression is positive. for examples, in this case, -2<x<1 and 4<x<7 both depict positive interval but only first range satisfies the condition. Please confirm
However, if for the same equation as mentioned, say the expression was (x+2)(x-1)/(x-4)(x-7) >0 and then we were asked to give the range where this is valid, then we would also multiply the -ve sign and make is <0 and then make the range after extreme right root -ve and provide all the intervals where it is negative. Please confirm
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Re: Inequalities trick [#permalink]
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26 Jul 2012, 09:47
VeritasPrepKarishma wrote:
When you have (x-a)^2(x-b) < 0, the squared term is ignored because it is always positive and hence doesn't affect the sign of the entire left side. For the left hand side to be negative i.e. < 0, (x - b) should be negative i.e. x - b < 0 or x < b.
.
IMHO, it should be x<b and also x is not equal to a .
so, one shouldn't totally ignore the squared term. We can ignore it, if the expression is <=0
correct me, if I am wrong
my question is -
do we always have a sequence of + and - from rightmost to the left side. I mean is it possible to have + and then + again ?
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Re: Inequalities trick &nbs [#permalink] 26 Jul 2012, 09:47
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Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®. | 2018-08-18T10:58:47 | {
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https://www.physicsforums.com/threads/de-broglie-wavelength-of-30kev-electron.281225/ | # De Broglie wavelength of 30keV electron
1. Dec 23, 2008
### catkin
1. The problem statement, all variables and given/known data
The question is from Advanced Physics by Adams and Allday, section 8 Practice Exam Questions, question 30.
Estimate the de Broglie wavelength of an electron that has been emitted thermionically in a vacuum from a filament and then accelerated through a p.d. of 30.0 kV
2. Relevant equations
λde Broglie = h / p
E2 - p2c2 = m02c4
ETotal = m0c2 + K.E.
3. The attempt at a solution
I think the solution is valid; my concern is whether there is a better (= more elegant) way to do it.
The de Broglie wavelength is given by
λde Broglie = h / p
Where h is Planck's constant and p is momentum.
p could be found from p = γm0v but this would require finding v. More conveniently
E2 - p2c2 = m02c4
Where E is the total energy (E = m0c2 + K.E.)
Expanding E:
p2c2 = 2m0c2K.E. + K.E.2
Rearranging:
p = (1/c) √(2m0c2K.E. + K.E.2)
Substituting this p:
λde Broglie = hc / √(2m0c2K.E. + K.E.2)
Substituting values using SI units (including eV to J conversion factor 1.60E-19)
= 6.63E-34 * 3.00E+8 / Sqrt(( 2 * 9.11E-31 * 3.00E+8^2 * 30E+3 * 1.60E-19) + (( 30E+3 * 1.60E-19)^2 ))
= 7.0e-12 m ct2sf
2. Dec 23, 2008
### Redbelly98
Staff Emeritus
Looks good.
By the way, this is a little easier using eV energy units. That way you won't have all those 1.6e-19's to contend with:
hc = 1240 eV-nm (a good number to remember in the future)
moc2 = 511 keV or 511e3 eV
KE = 30.0e3 eV
so that
λdB = 1240 eV-nm / √[2 × 511e3 × 30.0e3 eV2 + (30.0e3)2eV2]
= 0.00698 nm
= 6.98 pm
3. Dec 23, 2008
### Andrew Mason
Since it is asking only for an estimate, I would first determine whether the electron is moving at relativistic speeds. If it is not, you can use classical mechanics to determine p:
A 30 kV potential difference gives the electron 30keV of kinetic energy. Since it has about 500 keV as rest mass, the relativisitic effect will be small (since you are only estimating) so I would use a non-relativistic approach.
As suggested by Redbelly, use the formula:
$$\lambda_{DeB} = \frac{hc}{pc}$$
For non-relativistic speeds,
$$pc = \sqrt{2 KE m_0c^2}$$ (this is just a fancy rearrangment of $v = \sqrt{2 KE/m}$).
where KE is in units of eV. Note $m_0c^2$ = 511 KeV. and hc = 1240 eV nm
Use that to work out the Debroglie wavelength, in nm:
$$\lambda_{DeB} = hc/pc = 1240/\sqrt{2 x 3x10^4 x 5.11 x 10^5} = 1240/1.75 x 10^5 = 7 x 10^{-3}nm$$
AM
4. Dec 23, 2008
### Redbelly98
Staff Emeritus
Excellent point.
5. Jan 18, 2009
### catkin
Thanks Redbelly and Andrew.
Sorry it has taken me so long to get back here.
Helpful points, both about choice of unit and working out early on that the extra complexity of a relativistic approach is unnecessary.
What is the rest mass to K.E. ratio heuristic that allows deciding a relativistic approach is unnecessary and how does it arise?
Best
Charles
6. Jan 19, 2009
### Redbelly98
Staff Emeritus
Loosely speaking: when the KE is small compared to the rest mass energy, non-relativistic approximations may be used. Or equivalently, when v is small compared to c.
I'm not aware of any universally-accepted cutoff point, such as "when KE is x% of the rest mass energy". But as you saw, with KE equal to 6% of mc2, the non-relativistic result was pretty close, within 2%, of the actual de Broglie wavelength.
7. Feb 3, 2009
### catkin
Thanks Redbelly :)
The relativistic/Newtonian choice is something of a judgement call on this one. In our studies so far we have chosen 1% discrepancy in the result as the (arbitrary) determinant. On that basis, we found the cut of point for electrons is ~25keV acceleration. This is consonant with 2% error for an electron accelerated through 30 keV.
The thing that troubled me was that the question is from an old exam paper and the time to answer (based on the marks available) don't allow for my solution except the examinee be very good!
Perhaps that introduces an new relativistic/Newtonian choice determinant -- the time available to answer. :)
Best
Charles | 2016-10-25T05:26:37 | {
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https://math.stackexchange.com/questions/2806398/probability-for-a-specific-value-for-sum-of-normally-distributed-random-variable | # Probability for a specific value for sum of normally distributed random variables
I am trying to solve an exercise using the probability density function for a sum of n random variables with the same mean and variance. I need to find the probability for the specific value of the sums of r.v.s:
Player B uses a fair coin to earn points, if it lands on heads he collects 1 point, if tails he collects 8 points. When he flips the coin 50 times, these points will add up cumulatively.
What is the probability that player B will have exactly 225 points?
I was thinking I could take the $PDF$ probability for $\sum^n_{i=1}(X_i)=225$ with a mean of $n*µ=225$ and standard deviation of $\sqrt{\sigma^2*n}=24.75$, which gives the value 0.016, as can be seen on the PDF plot:
However, my teacher solved this by looking at it logically, and counting any situation where Player B gets exactly 25 heads and 25 tails:
$P=($$50\atop25$$)*\frac{1}{2^{50}}=0.112$
I see that his answer makes sense and is correct, but I doubt I would be able to think logically in an exam situation.
Why is my answer not correct, is it because the r.v. is only approximately normally distributed?
Is there another general way of solving such an exercise if the amount of flips and the total sum of points had been different?
• Another user reminded me that i can use binomial PDF to solve this issue, which is a bit of a help. But I do still have the same questions as mentioned. – user102937 Jun 3 '18 at 11:28
• By binomial distribution there is no PDF (probability density function) but there is a PMF (probability mass function). Your teacher used this PMF and calculated $f(25)$ where $f$ denotes this PMF. It is defined by $f(x)=\binom{50}{x}p^x(1-p)^{50-x}$ where $p=0.5$ and $x$ takes values in $\{0,1,\dots,50\}$ – drhab Jun 3 '18 at 11:35
• Your method works fine, but you have to remember the continuity correction. You are approximating a discrete distribution with a continuous one. If you are working with the scores, then for $(24,25,26)$ tails you get $(218,225,232)$ as scores...so for your continuous variant you need to look between $225-3.5$ and $225+3.5$. If you do that you get $0.112462916$ as opposed to the exact value of $0.112275173$. – lulu Jun 3 '18 at 12:02
Your method works fine, but you have to remember the continuity correction. You are approximating a discrete distribution with a continuous one. Thus, for instance, your method gives a non-zero chance of getting $224$ points (say) though that is in fact impossible.
As you are working with the scores, then for $(24,25,26)$ tails you get $(218,225,232)$ as scores...so for your continuous variant you need to look between $225-3.5$ and $225+3.5$. If you do that you get $0.112462916$ as opposed to the exact value of $0.112275173$ ... not bad at all!
The first thing to notice is that $B$ throws 50 coins and makes 225 points. If $n_H$ and $n_T$ are the numbers of heads and tails, 50 throws make it so that $$n_H + n_T = 50.$$ Given that a heads gives 1 point and a tails 8 points, we can set up the equation $$1 \cdot n_H + 8 \cdot n_T = 225.$$
Solving the two equations together gives $n_t = 25$; so we are after the probability tha 25 tails pop up in 50 throws. You can either do it with combinations and permutations (as your teacher did), or plainly use the pmf for the binomial distribution that represents your situation solving for 25 successes in 50 trials.
If you want the graphics, below is the plot of the distribution $B(50,0.5)$ with a line through $0.11227...$ which is the required probability.
While it is true that the binomial distribution tends to a normal distribution, you cannot interpret the values of a probability density function as the probability of an event.
• I would not speak of the PDF of binomial distribution but of the PMF. It does concern a density, but this wrt the counting measure. Usually PDF's refer to densities wrt Lebesgue measure and PMF's to densities wrt counting measure. – drhab Jun 3 '18 at 11:40
• Agreed! It was a typo – Riccardo Sven Risuleo Jun 3 '18 at 21:23
The values $f(x)$ where $f$ denotes a PDF cannot be interpreted as probabilities (as you seem to think). Note for instance that they can take values that exceed $1$.
That is the mistake you made.
We have the equalities $H+T=50$ and $H+8T=225$ here leading to $H=25=T$, so the question can be rephrased as:
"By $50$ throws of a fair coin what is the probability on $25$ tails?"
This is the route taken by your teacher.
• Yes, okay i see. So i can use binomial distribution to check whether i get 25 successes. How about if i take the cdf for the value 225.5 and subtract the value for the cdf of 224.5? As far as i know, that cannot exceed 1. (Sorry that i keep sticking to the normal distribution assumption, i would just like to know) – user102937 Jun 3 '18 at 11:40
• You could calculate $P(\sum_{i=1}^{50}X_i=225)$ where $X_i$ takes values in $\{1,8\}$ by approaching it with normal distribution. If $\sum_{i=1}^{50}X_i\approx Y$ where $Y$ has normal distribution then an estimate of the probability would be $P(224.5<Y<225.5)$ as you suggest. – drhab Jun 3 '18 at 11:50 | 2020-01-21T23:45:08 | {
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https://mathematica.stackexchange.com/questions/58751/how-to-speed-up-my-project-euler-code?noredirect=1 | How to speed up my Project Euler code
The evaluation speed of Mathematica often depresses me. It did it again when I wrote a code to solve a Project Euler problem. https://projecteuler.net/problem=206
Here's my code:
ClearAll[a];
a =
Compile[{},
NestWhile[# + 1 &, 10^9, ! MatchQ[
IntegerDigits[#^2], {1, _, 2, _, 3, _, 4, _, 5, _, 6, _, 7, _,8, _, 9, _, 0}] &],
CompilationTarget -> "C"]
I think it is the simplest solution. I can't find other solutions which are simpler. But my Mathematica runs a long time before getting the answer. How can I speed this up? I've run into similar problems while I was solving other problems.
• Pattern matching is often slow, you might try something like IntegerDigits[#^2][[ ;; ;; 2]] == {1,2,3,4,5,6,7,8,9,0} instead. – C. E. Sep 3 '14 at 7:16
• #+1 can be replaced by #+10, because a square number ends with "0" must ends with "00" – wuyingddg Sep 3 '14 at 8:20
• If you realize that the maximum solution is 10 Floor[Ceiling[Sqrt[1929394959697989990]]/10] and step down by 10 from there, you'll get the solution almost immediately. – Mark McClure Sep 3 '14 at 12:38
• @MarkMcClure Yes but this is because you know the answer. It could also be very close to 102030405060708090. – anderstood Oct 21 '15 at 20:00
Compilation is a certainly a good idea if you're going the brute-force route. So let's first tackle that, which will get us the answer in roughly 30 seconds computation time. Afterwards we'll come up with better strategy, which can do it in 0.3 seconds.
Compilation
A performance pitfall when compiling functions is that sometimes the compiled function just calls the main evaluation loop, effectively gaining nothing. We can check this with CompilePrint:
Needs["CompiledFunctionTools"]
a // CompilePrint
(...)
Result = I3
1 I3 = MainEvaluate[(...)]
2 Return
The call to MainEvaluate is the culprit here.
So let's rewrite your function such that it does compile properly. Using Pickett's and wuyingddg suggestions, we end up with:
PerformSearch = Compile[
{
{startValue, _Integer},
{increment, _Integer}
},
NestWhile[
# + increment &,
startValue,
IntegerDigits[#^2][[-1 ;; 1 ;; -2]] =!= {9, 8, 7, 6, 5, 4, 3, 2, 1} &
]
]
Let's check if this did compile ok:
PerformSearch // CompilePrint
(...)
1 I4 = I0
2 I3 = I4
3 I7 = Square[ I3]
4 T(I1)2 = IntegerDigits[ I7, I5]]
5 T(I1)0 = Part[ T(I1)2, T(I1)1Span]
6 B0 = CompareTensor[ I6, R2, T(I1)0, T(I1)3]]
7 if[ !B0] goto 12
8 I3 = I4
9 I7 = I3 + I1
10 I4 = I7
11 goto 2
12 Return
Ok, that looks much better. We can now perform the search, but not after we've determined the range of possible solutions:
max = Floor @ Sqrt @ FromDigits @ Riffle[Range[9], 9] (* 138902662 *)
min = Ceiling @ Sqrt @ FromDigits @ Riffle[Range[9], 0] (* 101010102 *)
As an aside, note that max^2 is below $MaxMachineInteger on 64-bit systems. But on 32-bit it isn't, which causes PerformSearch to switch back to the uncompiled code. Keeping Mark McClure's comment in mind, we'll cheat a bit and start from the maximum to find immediately: result = PerformSearch[max, -1] 138901917 How much faster is this than starting from the minimum? (result - min)/(max - result) // N 50861.5 Roughly 50 thousand times faster, not bad! Starting from the minimum takes about 30 seconds on my machine. Last but not least, let's double-check if we've obtained the correct number: result^2 19293742546274889 If you multiply this with 100, you've got your desired integer. Non-brute-force The approach above scanned roughly 38 million (!) integers in the worst-case scenario (starting from the minimum). Other answers to the OP's question have shown that you can and should go one better than that. Here's my take on an efficient general solution, using an iterative step-by-step process: ClearAll[ FindIntegerRoots, CheckIfEmpty, SelectDigitSequences, InsertNewDigits, ConvertPattern ]; FindIntegerRoots[ pattern : { (0|1|2|3|4|5|6|7|8|9|Verbatim[_]).. }, power_Integer: 2 ] /; power > 1 := With[ { maxRoot = Floor[FromDigits[pattern /. Verbatim[_] -> 9]^(1/power)] }, FromDigits /@ SelectDigitSequences[ Fold[ CheckIfEmpty @ SelectDigitSequences[ InsertNewDigits @ #1, maxRoot, power, ConvertPattern[pattern, #2] ] &, {{}}, Range @ IntegerLength @ maxRoot ], maxRoot, power, ConvertPattern[pattern, Length @ pattern] ] ~Quiet~ {CompiledFunction::cfnlts} ~Catch~ "isEmpty" ]; CheckIfEmpty[{}] := Throw[{}, "isEmpty"]; CheckIfEmpty[nonEmpty_] := nonEmpty; SelectDigitSequences = Compile[ { {digitSequences, _Integer, 2}, {maxRoot, _Integer }, {power, _Integer }, {numDigits, _Integer }, {positions, _Integer, 1}, {compareDigits, _Integer, 1} }, Module[ { powersOfTen = 10^# & /@ Range[Length @ First @ digitSequences - 1, 0, -1] }, Select[ digitSequences, And[ # <= maxRoot, IntegerDigits[#^power, 10, numDigits][[positions]] == compareDigits ] & @ ( powersOfTen.# ) & ] ] ]; InsertNewDigits = With[ { range = List /@ Range[0, 9] }, Compile[ { {digitSequences, _Integer, 2} }, Outer[#1 ~Join~ #2 &, range, digitSequences, 1] ~Flatten~ 1 ] ]; ConvertPattern[pattern_, length_] := With[ { trimmed = pattern[[Length@pattern - length + 1 ;; -1]] }, Sequence @@ { length, Flatten @ Position[trimmed, _Integer], DeleteCases[trimmed, Verbatim[_]] } ]; While the code might be long, it is nevertheless fast: FindIntegerRoots @ Riffle[Range[9], _] // AbsoluteTiming {0.284622, {138901917}} A performance increase of 100 over the above brute-force method! But how efficient is it? Here's a nice graph of the amount of checks it performs in this case: That's about 296 thousand checks in total; certainly much less than 38 million! Some more concealed squares The nice thing about FindIntegerRoots is that it works on any pattern, not just the OP's case: FindIntegerRoots @ {_, 9} {3, 7} FindIntegerRoots @ { 1, _, 4} {12} FindIntegerRoots @ { 1, _, 6} {14} And if you're adventurous, you may even ask for roots of different powers: FindIntegerRoots[{_, _, _, 5}, 3] {5, 15} So let's have a bit of fun and search for more concealed squares. First, are there any partial concealed squares (i.e. of the form 1_2, 1_2_3, 1_2_3_4, etc)? FindIntegerRoots @ Riffle[Range @ #, _] & /@ Range @ 9 {{1}, {}, {}, {1312}, {}, {115256, 127334, 135254}, {}, {}, {138901917}} Flatten[%]^2 {1, 1721344, 13283945536, 16213947556, 18293644516, 19293742546274889} How about reverse concealed squares (i.e. 2_1, 3_2_1, 4_3_2_1, etc)? FindIntegerRoots @ Riffle[Reverse @ Range @ #, _] & /@ Range @ 9 {{1}, {}, {}, {}, {24171}, {}, {2657351, 2713399}, {}, {}} Flatten[%]^2 {1, 584237241, 7061514337201, 7362534133201} And finally, are there squares of the form 1_1_(...)_1_1? FindIntegerRoots @ Riffle[ConstantArray[1, #], _] & /@ Range[2, 9] {{11}, {119, 131}, {}, {}, {110369}, {}, {10065739}, {}} Flatten[%]^2 {121, 14161, 17161, 12181316161, 101319101616121} It seems there are many nice squares. If you happen to find a particularly nice one, let me know :) • Which version of Mathematica are you using? It's quite strange that your code generates the CompiledFunction::cfn warning in my v8 (Win 32bit & 64bit) and v9 (Win 32bit), but it does work without warning on Wolfram Cloud! – xzczd Sep 3 '14 at 14:15 • @xzczd I've checked this on v9 on a Mac. The issue here is 32 vs 64 bit; $MaxMachineInteger should namely be bigger than FromDigits@Riffle[Range[9], 9], which it isn't on 32 bit. – Teake Nutma Sep 3 '14 at 16:06
• That seems to be the reason, BTW the \$MaxMachineInteger in my v8 (Win 64 bit) is the same as that in 32 bit, not sure if it's just the nature of v8… – xzczd Sep 4 '14 at 1:56
• Is using Big O notation in "Roughly O(10^4) times faster" a correct statement? It is surely 10^4 times faster in this example, but what exactly does O(10^4) times faster mean? – user Sep 4 '14 at 15:29
• @bruce14 What I meant here is that it's roughly 4 orders of magnitude faster; faster than 10^3 and slower than 10^5. – Teake Nutma Sep 4 '14 at 17:15
This non-paralell strategy takes less than 4 seconds in my stone age laptop (and doesn't use the "reverse searching" trick)-
First we check for the possible 6 digits endings (after discounting the final "00" ending in the squared number):
set= 2 Position[IntegerDigits[Range[10^5 + 1, 10^6-1, 2]^2][[All, -5;; -1;; 2]], {7,8,9}]+ 10^5 - 1;
Then we complete the whole set of numbers to check by adding all the possible "heads" between the max and min:
fullSet = Total[Tuples[{Range[101, 138] 10^6, Flatten@set}], {2}];
And then a quick search gets our number:
Cases[IntegerDigits[fullSet^2], {1, _, 2, _, 3, _, 4, _, 5, _, 6, _, 7, _, 8, _, 9}]
(*{{1, 9, 2, 9, 3, 7, 4, 2, 5, 4, 6, 2, 7, 4, 8, 8, 9}}*)
(* 1929374254627488900 *)
Edit
We can still cut the time in half if we see that the number should end in 3 or 7 for its square to end in 9:
set = Join[10 (Position[IntegerDigits[Range[10^5 + #, 10^6, 10]^2][[All, -5 ;; -1 ;; 2]],
{7, 8, 9}] - 1) + # & /@ {3, 7}] + 10^5;
fullSet = Total[Tuples[{Range[101, 138] 10^6, Flatten@set}], {2}];
Cases[IntegerDigits[fullSet^2], {1, _, 2, _, 3, _, 4, _, 5, _, 6, _, 7, _, 8, _, 9}]
• +1 Very concise and fast (~1.3 times faster than mine!). – Teake Nutma Sep 4 '14 at 20:31
• @TeakeNutma Thanks! I halved the time in the edit.Less than 2 secs now on my poor man's machine :) – Dr. belisarius Sep 5 '14 at 2:41
• That's pretty fast. But alas, I've updated my answer and now it's 2x faster than yours :). – Teake Nutma Sep 5 '14 at 10:11
There is a method by using Catch and Throw, and it gives result in about 90s. I think it can be improve a lot by Paralilize or ParallelEvaluate, but failed to finish that.
i = 10^9 + {30, 70};
isMatchQ =
If[IntegerDigits[#^2][[1 ;; -1 ;; 2]] == {1, 2, 3, 4, 5, 6, 7, 8,
9, 0}, Throw[#]] &;
Catch[Do[isMatchQ /@ i; i += 100, {10^9}]]
Edit
I have found a way to speed up it by ParallelTable, this is the code:
len = ((Sqrt[1.93*10^16] - 1*10^8)/40 // IntegerPart)*10;
isMatchQ =
If[IntegerDigits[#^2][[1 ;; -1 ;; 2]] == {1, 2, 3, 4, 5, 6, 7, 8,
9}, Throw[#]] &;
ParallelTable[n = 10^8 + {3, 7} + len*i;
Catch[Do[isMatchQ /@ n; n += 10, {len / 10}]], {i, 0,
3}] // AbsoluteTiming
(*{42.523432, {Null, Null, Null, 138901917}}*)
I divided the whole range of n which makes 10^16 <= n^2 <= 1.93*10^16 into 4 parts, the length of each part is len. And I calculate these 4 parts in 4 kernels at the same time by ParallelTable, then get the result in a much shorter time, only 42.5s. (Sorry for my poor English....)
Here's a slight improvement of @wuyingddg's Catch&Throw method. The key point is to make Throw working inside ParallelDo using the method mentioned in this post:
SetSharedFunction[pThrow];
pThrow[expr_] := Throw[expr]
isMatchQ =
If[IntegerDigits[#^2][[1 ;; -1 ;; 2]] == {1, 2, 3, 4, 5, 6, 7, 8, 9, 0}, pThrow[#]] &;
Catch[ParallelDo[
isMatchQ /@ (i + {30, 70}), {i, 10^9, Sqrt[193 10^16], 100}]] // AbsoluteTiming
`
• It is much more brief than the ParallelTable one and spend nearly same time, pretty good! – wuyingddg Sep 3 '14 at 14:21 | 2021-01-23T02:24:08 | {
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https://la.mathworks.com/help/econ/dtmc.lazy.html | lazy
Description
example
lc = lazy(mc) transforms the discrete-time Markov chain mc into the lazy chain lc with an adjusted state inertia.
example
lc = lazy(mc,w) applies the inertial weights w for the transformation.
Examples
collapse all
Consider this three-state transition matrix.
$P=\left[\begin{array}{ccc}0& 1& 0\\ 0& 0& 1\\ 1& 0& 0\end{array}\right].$
Create the irreducible and periodic Markov chain that is characterized by the transition matrix P.
P = [0 1 0; 0 0 1; 1 0 0];
mc = dtmc(P);
At time t = 1,..., T, mc is forced to move to another state deterministically.
Determine the stationary distribution of the Markov chain and whether it is ergodic.
xFix = asymptotics(mc)
xFix = 1×3
0.3333 0.3333 0.3333
isergodic(mc)
ans = logical
0
mc is irreducible and not ergodic. As a result, mc has a stationary distribution, but it is not a limiting distribution for all initial distributions.
Show why xFix is not a limiting distribution for all initial distributions.
x0 = [1 0 0];
x1 = x0*P
x1 = 1×3
0 1 0
x2 = x1*P
x2 = 1×3
0 0 1
x3 = x2*P
x3 = 1×3
1 0 0
sum(x3 == x0) == mc.NumStates
ans = logical
1
The initial distribution is reached again after several steps, which implies that the subsequent state distributions cycle through the same sets of distributions indefinitely. Therefore, mc does not have a limiting distribution.
Create a lazy version of the Markov chain mc.
lc = lazy(mc)
lc =
dtmc with properties:
P: [3x3 double]
StateNames: ["1" "2" "3"]
NumStates: 3
lc.P
ans = 3×3
0.5000 0.5000 0
0 0.5000 0.5000
0.5000 0 0.5000
lc is a dtmc object. At time t = 1,..., T, lc "flips a fair coin". It remains in its current state if the "coin shows heads" and transitions to another state if the "coin shows tails".
Determine the stationary distribution of the lazy chain and whether it is ergodic.
lcxFix = asymptotics(lc)
lcxFix = 1×3
0.3333 0.3333 0.3333
isergodic(lc)
ans = logical
1
lc and mc have the same stationary distributions, but only lc is ergodic. Therefore, the limiting distribution of lc exists and is equal to its stationary distribution.
Consider this theoretical, right-stochastic transition matrix of a stochastic process.
$P=\left[\begin{array}{ccccccc}0& 0& 1/2& 1/4& 1/4& 0& 0\\ 0& 0& 1/3& 0& 2/3& 0& 0\\ 0& 0& 0& 0& 0& 1/3& 2/3\\ 0& 0& 0& 0& 0& 1/2& 1/2\\ 0& 0& 0& 0& 0& 3/4& 1/4\\ 1/2& 1/2& 0& 0& 0& 0& 0\\ 1/4& 3/4& 0& 0& 0& 0& 0\end{array}\right].$
Create the Markov chain that is characterized by the transition matrix P.
P = [ 0 0 1/2 1/4 1/4 0 0 ;
0 0 1/3 0 2/3 0 0 ;
0 0 0 0 0 1/3 2/3;
0 0 0 0 0 1/2 1/2;
0 0 0 0 0 3/4 1/4;
1/2 1/2 0 0 0 0 0 ;
1/4 3/4 0 0 0 0 0 ];
mc = dtmc(P);
Plot the eigenvalues of the transition matrix on the complex plane.
figure;
eigplot(mc);
title('Original Markov Chain')
Three eigenvalues have modulus one, which indicates that the period of mc is three.
Create lazy versions of the Markov chain mc using various inertial weights. Plot the eigenvalues of the lazy chains on separate complex planes.
w2 = 0.1; % More active Markov chain
w3 = 0.9; % Lazier Markov chain
w4 = [0.9 0.1 0.25 0.5 0.25 0.001 0.999]; % Laziness differs between states
lc1 = lazy(mc);
lc2 = lazy(mc,w2);
lc3 = lazy(mc,w3);
lc4 = lazy(mc,w4);
figure;
eigplot(lc1);
title('Default Laziness');
figure;
eigplot(lc2);
title('More Active Chain');
figure;
eigplot(lc3);
title('Lazier Chain');
figure;
eigplot(lc4);
title('Differing Laziness Levels');
All lazy chains have only one eigenvalue with modulus one. Therefore, they are aperiodic. The spectral gap (distance between inner and outer circle) determines the mixing time. Observe that all lazy chains take longer to mix than the original Markov chain. Chains with different inertial weights than the default take longer to mix than the default lazy chain.
Input Arguments
collapse all
Discrete-time Markov chain with NumStates states and transition matrix P, specified as a dtmc object. P must be fully specified (no NaN entries).
Inertial weights, specified as a numeric scalar or vector of length NumStates. Values must be between 0 and 1.
• If w is a scalar, lazy applies it to all states. That is, the transition matrix of the lazy chain (lc.P) is the result of the linear transformation
${P}_{\text{lazy}}=\left(1-w\right)P+wI.$
P is mc.P and I is the NumStates-by-NumStates identity matrix.
• If w is a vector, lazy applies the weights state by state (row by row).
Data Types: double
Output Arguments
collapse all
Discrete-time Markov chain, returned as a dtmc object. lc is the lazy version of mc.
collapse all
Lazy Chain
A lazy version of a Markov chain has, for each state, a probability of staying in the same state equal to at least 0.5.
In a directed graph of a Markov chain, the default lazy transformation ensures self-loops on all states, eliminating periodicity. If the Markov chain is irreducible, then its lazy version is ergodic. See graphplot.
References
[1] Gallager, R.G. Stochastic Processes: Theory for Applications. Cambridge, UK: Cambridge University Press, 2013. | 2020-08-15T08:40:51 | {
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https://math.stackexchange.com/questions/4258331/is-this-relation-symmetric-i-think-it-is-but-my-professor-claims-it-isnt/4258341 | # Is this relation symmetric? I think it is but my professor claims it isn't.
Is the relation $$R = \{(a,b),(a,c),(b,a),(b,c),(c,a),(c,b),(d,d)\}$$ symmetric?
My professor claims that if $$(d,d)$$ was not included, it would be symmetric, but the inclusion of $$(d,d)$$ ruins it because $$d$$ has to connect to another element of the relation.
• Two of the defining properties of equivalence relations are that they are both reflexive and symmetric, which would be useless if they were mutually exclusive :-P Sep 24 '21 at 4:12
• Hope your professor is willing to get corrected in this case! Sep 24 '21 at 8:39
• You could tell your professor that $(d,d)\in R$ "connects" to itself, ie, for the element $(\color{red}d,\color{green}d)$ in $R$, we have the element $(\color{green}d,\color{red}d)$ in $R$. There is no restriction that it must "connect" to a distinct element in $R$. As an example, the relation $R=\{(a,a)\}$ on the singleton set $\{a\}$ is an equivalence relation: reflexive, symmetric and transitive. Sep 24 '21 at 19:16
A very simple way to see if a relation $$R$$ is symmetric is to check its inverse relation $$R^{-1}$$, where $$(x,y)\in R\Leftrightarrow (y,x)\in R^{-1}$$
Since for the given relation, $$R=R^{-1}$$, it is symmetric.
$$\color{green}{\text{YOU ARE CORRECT.}}$$
Yes, that relation is symmetric. The definition of symmetry does not require each element to be connected to some other element; $$R$$ is symmetric iff for every $$x,y$$ such that $$(x,y)\in R$$ it is also the case that $$(y,x)\in R$$.
One way to see if a relation is symmetric is to draw the table what is in relation to what:
$$\begin{array}{r|c|c|c|c}R&a&b&c&d\\\hline a&F&T&T&F\\b&T&F&T&F\\c&T&T&F&F\\d&F&F&F&T\end{array}$$
and just see if the table is symmetrical across the main diagonal.
And - in this case it is. Thus, the relation is symmetric. Having $$(d,d)$$ in it has nothing to do with it, it would be symmetric either way (i.e. whether the entry for $$(d,d)$$ is "true" or "false"). | 2022-01-21T13:47:10 | {
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http://gpof.trattoriaborgo.it/midpoint-riemann-sum-sigma-notation.html | # Midpoint Riemann Sum Sigma Notation
We call Ln the left Riemann sum for the function f on the interval [a, b]. Similarly, the right Riemann sum is an overestimate. e−x2 dx by the Riemann sum with n = 4 subintervals and left end-points as sample points. Simmons computes R xdx. To express left, right, and midpoint Riemann sums in sigma notation, we must identify the points Ñk. One of the programs is opened so students can see that it is nothing more than the summation notation they had. First, determine the width of each rectangle. n n n b a x 4 0 4. 2 Area & Sigma Notation 2. Here ∆x = 3−1 10 = 0. How to Write Riemann Sums with Sigma Notation; How to Write Riemann Sums with Sigma Notation. 6255658911511259 1. Also, identify when an estimate is an overestimate or underestimate. If the sym- bol oo appears above the E, it indicates that the terms go on indefinitely. The same number of subintervals were used to produce each approximation. 603125 Extension – Area Programs Students will use the programs to compare area approximation methods. 2) Use the Midpoint Rule with n = 4 to approximate x 4) dx. (a) R 3 −1 xdx (b) R 4 2 x2dx. In each subinterval, choose a point c1, c2,cn and form the sum ; This is called a Riemann Sum ; NOTE LRAM, MRAM, and RRAM are all Riemann sums. rotation (2x2 matrices) row echelon form. 2 - Page 266 35 including work step by step written by community members like you. 6078493243021688 1. EXAMPLE 1: Find the area under the curve of the function f x ( ) =x +8 over the interval [0, 4] by using n rectangles. The same number of subintervals were used to produce each approximation. 35 4n+ 5k - 5 ОА. The Midpoint Rule summation is: $$\ds \sum_{i=1}^n f\left(\frac{x_i+x_{i+1}}{2}\right)\Delta x\text{. Then evaluate the Riemann sum he midpoint Riemann sum for f(x) = 3+ cOS TX on [0,5] with n= 50 dentify the midpoint Riemann sum. Write the sigma notation. Riemann Sums; The Definite Integral and FTC; Indefinite Integrals; 2 Techniques of Integration. Definite Integral from Summation Notation: To rewrite the given limit of summation notation, we'll compare the given expression with the formula of conversion and then substitute the values. 19) f(x) = x 2 - 2 , [0, 8], midpoint 19) 20) f(x) = cos x + 3 , [0, 2 ], left- hand endpoint 20) Express the sum in sigma notation. Wolfram|Alpha brings expert-level knowledge and capabilities to the broadest possible range of people—spanning all professions and education levels. Riemann Sums Name:_____ Date:_____ A Riemann sum Sn for the function is defined to be Sn = n k fck x 1 (). They saw how these come together when finding a Riemann Sum, as shown below. }$$ 6 Evaluating Riemann sums with data A car traveling along a straight road is braking and its velocity is measured at several different points in time, as given in the following table. 1 Riemann Sums and Area 3 2. Riemann Sums Using Sigma Notation With sigma notation, a Riemann sum has the convenient compact form f (il) Ax + f(Ñ2) Ax + + f (in) Ax Ef(Ñk) Ax. higher than the actual area B. f) General Riemann Sum. 2 Trapezoidal Rule Example Find the area under x3 using 4 subintervals using: left, right, midpoint and trapezoidal methods from [2, 3] Example – A free PowerPoint PPT presentation (displayed as a Flash slide show) on PowerShow. The notation is represented by the upper case version of the Greek letter sigma. Then study what happens to a nite sum approximation as the number of terms approaches in nity. Riemann sums in summation notation: challenge problem. The left and right Riemann sums of a function f on the interval [2, 6] are denoted by LEFT ( n ) and RIGHT( n ), respectively, when the interval is divided int… Enroll in one of our FREE online STEM summer camps. You can use sigma notation to write out the Riemann sum for a curve. One of the programs is opened so students can see that it is nothing more than the summation notation they had. 7) The table shows the velocity of a remote controlled race car moving along a dirt path for 8 seconds. Derivatives Derivative Applications Limits Integrals Integral Applications Riemann Sum Series ODE Multivariable Calculus Laplace Transform Taylor/Maclaurin Series Fourier Series Functions Line Equations Functions Arithmetic & Comp. 11) Use sigma notation to find the right Riemann Sum for f (x) = x 3 + 2 on [0, 3] with n = 30. The region bounded by y = x2, the x-axis, from x = 0 to x = 2. First, determine the width of each rectangle. Definite Integral from Summation Notation: To rewrite the given limit of summation notation, we'll compare the given expression with the formula of conversion and then substitute the values. Conic Sections. We call Ln the left Riemann sum for the function f on the interval [a, b]. Partition a, b by choosing These partition a, b into n parts of length ?x1, ?x2, ?xn. 1 Approximate the area under the curve f(x) = ln(x) between x= 1 and x= 5. The endpoints a and b are called the limits of integration. Is the midpoint Riemann sum an over or under approximation if the graph is: a. Also, identify when an estimate is an overestimate or underestimate. k be the midpoint of the kth subinterval (where all subintervals have equal width). The k" subinterval has width. 1 Riemann Sums and Definite Integrals. (k) (n-1) 4n 729 OC. In mathematics, the Riemann sum is defined as the approximation of an integral by a finite sum. By using this website, you agree to our Cookie Policy. Approximating the area under a curve using some rectangles. (Sigma notation for nite sums ) The symbol Xn k=1 a k denotes the sum a 1 + a 2 + + a n. The Midpoint Rule for definite integrals means to approximate the integral by using a midpoint Riemann Sum just as in 6. the left sum approximation to R 2π 0 sin(x)dx (a) with 8 equal subintervals (b) with n equal subintervals 2. where is the number of subintervals and is the function evaluated at the midpoint. The RiemannSum(f(x), x = a. Then evaluate the sum. So what happens if the "area" is below the x-axis as I mentioned before, area is inherently positive, but a Riemann Sum and therefore an integral can have negative values if the curve lies below the. Use the midpoint Riemann sum with n = 5 to nd an estimates on the area under the curve on the interval [0;10]. 2 Sigma Notation and Limits of Finite Sums In this section we introduce a convenient notations for sums with a large number of terms. Take a midpoint sum using only one sub-interval, so we only get one rectangle: The midpoint of our one sub-interval [0, 4] is 2. We have step-by-step solutions for your textbooks written by Bartleby experts!. By the way, you don’t need sigma notation for the math that follows. Thomas’ Calculus 13th Edition answers to Chapter 5: Integrals - Section 5. Partition a, b by choosing These partition a, b into n parts of length ?x1, ?x2, ?xn. Is your answer an over-estimate or an under-estimate? Split the interval [-4, 4] into two sub-intervals of length 4 and find the midpoint of each. Derivatives Derivative Applications Limits Integrals Integral Applications Riemann Sum Series ODE Multivariable Calculus Laplace Transform Taylor/Maclaurin Series Fourier Series Functions Line Equations Functions Arithmetic & Comp. Midpoint Rule: As noted above, the midpoint rule is a special case of Riemann sums where the interval integration [a, b] is divided n subintervals [x i-1, x i] each with length Dx = (b-a)/n. We are approximating an area from a to b with a=0 and b=5, n=5, right endpoints and f(x)=25-x^2 (For comparison, we'll do the same problem, but use left endpoints after we finish this. where is the number of subintervals and is the function evaluated at the midpoint. ” Example 1: Evaluate the Riemann sum for f ( x ) = x 2 on [1,3] using the four subintervals of equal length, where x i is the right endpoint in the i th subinterval (see Figure ). Do NOT evaluate your summation. Step 5 requires the formulas and properties of the sigma notation. 1 sigma notation and riemann sums 305 Area Under a Curve: Riemann Sums Suppose we want to calculate the area between the graph of a positive function f and the x-axis on the interval [a,b] (see below left). Is your answer an over-estimate or an under-estimate? Split the interval [-4, 4] into two sub-intervals of length 4 and find the midpoint of each. Choose the correct Riemann sum below. 1) as f x dx f m x 1 ( ) ( ) (a Riemann. ∫(1, 2) sin(1/x)dx. The sum on the right hand side is the expanded form. 6093739310551827. Arnold Schwarzenegger This Speech Broke The Internet AND Most Inspiring Speech- It Changed My Life. - Duration: 14:58. This is useful when you want to derive the formula for the approximate area under the curve. Write the midpoint Riemann sum in sigma nota-tion with n = 20. I expect you to show your reasoning clearly and in an organized fashion. (2k-1) (2n +1 -2K) k1 4n 729 OB. Math Help Boards: Sum Calculator. Similar formulas can be obtained if instead we choose c k to be the left-hand endpoint, or the midpoint, of each subinterval. 1 Approximate the area under the curve f(x) = ln(x) between x= 1 and x= 5. Sums can also be infinite, provided that the terms eventually get close enough to zero–this is an important topic in calculus. Most of the following problems are average. 6093739310551827. In the more compact sigma notation, we have Ln = Xn−1 i=0 f (xi)4x. Riemann Sums Using Sigma Notation With sigma notation, a Riemann sum has the convenient compact form f (il) Ax + f(Ñ2) Ax + + f (in) Ax Ef(Ñk) Ax. Partition a, b by choosing These partition a, b into n parts of length ?x1, ?x2, ?xn. It is applied in calculus to formalize the method of exhaustion, used to determine the area of a region. EXAMPLE 1Using Sigma Notationƒs1d + ƒs2d + ƒs3d +Á+ ƒs100d =a100i=1ƒsid. ) Let f (x) be defined on a, b. and we want to have 2 rectangles with the sample points being left endpoints, then we would assign some notation. rules of exponents. Left-Hand. Calculus Q&A Library se sigma notation to write the following Riemann sum. Let f(x) = 2/x a. 50 1 3+ cos 10 k = 1 2tk – T O A. Area under Curves 5. See math and science in a new way. Enjoy the videos and music you love, upload original content, and share it all with friends, family, and the world on YouTube. Here is how to set up the Riemann sum for the definite integral Z 3 1 x2 dx where n = 10: (1) Find ∆x = b−a n. Evaluate the sum using a calculator with n=20,50, and 100. Let's visualize rectangles in the left, right and midpoint Riemann sums for the function f = lambda x : 1/(1+x**2) a = 0; b = 5; N = 10 n = 10 # Use n*N+1 points to plot the function dx/2,N) x_right = np. How do you determine that it is defined on [0, 1] and what would the sigma notation look like for this? I understand how to calculate a Riemann sum, I am just not understanding how they get [0,1] from the given information :/. Other sums The choice of the $c_i$ will give different answers for the approximation, though for an integrable function these differences will vanish in the limit. Write the sum without sigma notation and evaluate it. }\) 6 Evaluating Riemann sums with data A car traveling along a straight road is braking and its velocity is measured at several different points in time, as given in the following table. scalar multiplication and matrices (Properties) scalar multiplication of vectors. 1 would become x1, and 2 would be x2. The LRAM uses the left endpoint, the RRAM uses the right endpoint and the MRAM uses the midpoint of intervals. We have step-by-step solutions for your textbooks written by Bartleby experts!. A 2 𝑛 2 + 4 𝑛 1 2 + 𝑖. Free math problem solver answers your algebra, geometry, trigonometry, calculus, and statistics homework questions with step-by-step explanations, just like a math tutor. The region bounded by y = x2, the x-axis, from x = 1 to x = 3. Approximating the area under a curve using some rectangles. 1 + √ x 1 − √ x dx 6. The application is intended to demonstrate the use of Maple to solve a particular problem. b, method = midpoint, opts) command calculates the midpoint Riemann sum of f(x) from a to b. Free midpoint calculator - calculate the midpoint between two points using the Midpoint Formula step-by-step This website uses cookies to ensure you get the best experience. For left Riemann sums, the left endpoints of the subintervals are 1) Ax, fork —. Riemann Sums; The Definite Integral and FTC; Indefinite Integrals; 2 Techniques of Integration. b, method = left, opts) command calculates the left Riemann sum of f(x) from a to b. (a) R 5 −1 xdx (b) R 2 1 x2dx 2. rotation (2x2 matrices) row echelon form. Similarly, the right Riemann sum is an overestimate. 6078493243021688 1. This behavior persists for more rectangles. So, for example, over here we could we could use the midpoint between x0 and x1 to find the height of the rectangle. Hence the Riemann sum associated to this partition is: Xn i=1 µ i n ¶2 1/n = 1 n3 Xn i=1 i2 = 1 n3 2n3 +3n2 +n 6 = 2+3/n+2/n2 6. Note particularly that since the index of summation begins at 0 and ends at n − 1, there are indeed n terms in this sum. 12+ 22+ 32+ 42+ 52+ 62+ 72+ 82+ 92+ 102+ 112=a11k=1k2,k 1aknThe index k ends at k n. Example: Estimate the area under the graph of f(x) = x2 + 1 over the interval [2;10] using 4 rectangles of equal width and midpoints. Riemann Sums. Constructing Accurate Graphs of Antiderivatives; The Second Fundamental Theorem of Calculus; Integration by Substitution; Integration by Parts; Other Options for Finding Algebraic Antiderivatives; Numerical Integration; 6 Using Definite Integrals. Therefore, the Riemann sum is: The upper-case Greek letter Sigma Σ is used to stand for sum. Know and understand the sum, di erence, constant multiple, and constant value rule for nite sums in Sigma notation. In each case where you used a Riemann sum to estimate an area, try to determine if you obtained an overestimate or an underestimate. The LRAM uses the left endpoint, the RRAM uses the right endpoint and the MRAM uses the midpoint of intervals. This is called a "Riemann sum". A summation is a sum of numbers that are typically defined by a function. and we want to have 2 rectangles with the sample points being left endpoints, then we would assign some notation. 15k 2 + 4 C. Calculus Q&A Library se sigma notation to write the following Riemann sum. A Riemann sum is an approximation of the area of a region that is found by dividing the region into rectangles or trapezoids. This is useful when you want to derive the formula for the approximate area under the curve. Use the programs on your calculator to find the value of the sum accurate to 3 decimal places. ) Find the limit of the Riemann Sum (found in part d) as the number of rectangles are increased to infinity. Series, Sigma Notation video (Leckie) Summation Notation (Paul) 1. The limit of Finite approximation to an Area. 2 - Sigma Notation and Limits of Finite Sums - Exercises 5. In mathematics, the Riemann sum is defined as the approximation of an integral by a finite sum. 3) Mid-point sums All value may be different but they represent a same quantity an approximated area under the curve. The LRAM uses the left endpoint, the RRAM uses the right endpoint and the MRAM uses the midpoint of intervals. EXAMPLE 1Using Sigma Notationƒs1d + ƒs2d + ƒs3d +Á+ ƒs100d =a100i=1ƒsid. Midpoint Riemann Sums: Suppose x i is the midpoint of the ith subinterval [x i 1;x i], that is x i = x i 1 + x i 2 for all i. For example, saying “the sum from 1 to 4 of n²” would mean 1²+2²+3²+4² = 1 + 4 + 9 + 16 = 30. Derivatives Derivative Applications Limits Integrals Integral Applications Riemann Sum Series ODE Multivariable Calculus Laplace Transform Taylor/Maclaurin Series Fourier Series Functions Line Equations Functions Arithmetic & Comp. Left, midpoint, and right Riemann sums were used to estimate the area between the graph of 𝑓(𝑥) and the x-axis on the interval [3, 7]. In Chapter 1, I introduce you to the Riemann sum formula for the definite integral. While summation notation has many uses throughout math (and specifically calculus), we want to focus on how we can use it to write Riemann sums. See full list on khanacademy. scalar product. The Midpoint Rule for definite integrals means to approximate the integral by using a midpoint Riemann Sum just as in 6. Use right Riemann sums to compute the following integrals. You always increase by one at each successive step. Take a photo of your homework question and get answers, math solvers, explanations, and videos. This is accomplished in a three-step procedure. Riemann sums in summation notation: challenge problem. 603125 Extension – Area Programs Students will use the programs to compare area approximation methods. you'll have to picture the above and below numbers because I can't show them on here. The a’s are. The Midpoint Rule for definite integrals means to approximate the integral by using a midpoint Riemann Sum (just as in 6. exactly equal to the actual area C. The first two arguments (function expression and range) can be replaced by a definite integral. Simplify the expression. A Task template for the Riemann Sum can be found in the following location: Tools>Tasks>Browse>Calculus>Integration>Riemann Sums and choose one of the options (Left, Right, or Midpoint). Write the midpoint Riemann sum in sigma notation for an arbitrary value of n. Use sigma notation to write a new sum $$R$$ that is the right Riemann sum for the same function, but that uses twice as many subintervals as $$S\text{. Riemann Sums: Sigma Notation Review This is a Riemann sum for f on the interval [a,b]. n n n b a x 4 0 4. Then evaluate the Riemann sum he midpoint Riemann sum for f(x) = 3+ cOS TX on [0,5] with n= 50 dentify the midpoint Riemann sum. Use a midpoint sum with 2 sub-intervals to estimate. a) Write a summation to approximate the area under the graph of 血岫捲岻噺捲戴髪の from x = 1 to x = 7 using the right endpoints of three subintervals of equal length. Substitution Rule; Powers of Trigonometric Functions; Trigonometric Substitutions; Integration by Parts; Partial Fraction Method for Rational Functions; Numerical Integration; Improper Integrals; Additional Exercises; 3 Applications of Integration. Step 2: Partition the interval into n subintervals. (n times) = cn, where c is a constant. Choose the correct answer below. 4 Financial applications of geometric sequences and series: compound interest, annual depreciation. n n n b a x 4 0 4. The endpoints are given by x 0 = a and x n = b. 2 - Sigma Notation and Limits of Finite Sums - Exercises 5. Sigma/summation notation. Definition S L: Left-endpoint Riemann Sum Choose the height of each rectangle on the interval [x i, x i +1]to be a b x y f H x L Area ≈ S R: Right-endpoint Riemann Sum Choose the height of each rectangle on the interval [x i, x i +1]to be a b x y f H x L Area ≈ S M: Midpoint Riemann Sum. One of the programs is opened so students can see that it is nothing more than the summation notation they had. Calculus Q&A Library se sigma notation to write the following Riemann sum. In sigma notation, we get that 1 2 3 99 = E k In This Module • We will introduce sigma notation — a compact way of writing large sums of like terms — and define the notion of a Riemann sum. 1 Riemann Sums and Area 3 2. Riemann Sums Name:_____ Date:_____ A Riemann sum Sn for the function is defined to be Sn = n k fck x 1 (). 318: 49-56 2. If the sym- bol oo appears above the E, it indicates that the terms go on indefinitely. The application is intended to demonstrate the use of Maple to solve a particular problem. Sigma Notation and Riemann Sums Sigma Notation: Notation and Interpretation of 12 3 14 1 n k nn k aaaaa a a (capital Greek sigma, corresponds to the letter S) indicates that we are to sum numbers of the form indicated by the general term. ) 2 −1 sin πx 4 dx 2. Use the programs on your calculator to find the value of the sum accurate to 3 decimal places. You always increase by one at each successive step. 0375 When n = 100, Left Riemann sum = 164. The remaining formulas are simple rules for working with sigma notation: ∗to be the midpoint of the Any Riemann sum is an approximation to an integral, but. So this is right over here. In mathematics, the Riemann sum is defined as the approximation of an integral by a finite sum. Option #1: If you noticed in step 2 above, we did not care if our subintervals were the same width. Σ 35 8n + 10k-5 n nV 2n n k=1 k=1 n 35 4n+ 5k OC. Sums and sigma notation. Similar formulas can be obtained if instead we choose c k to be the left-hand endpoint, or the midpoint, of each subinterval. Let's visualize rectangles in the left, right and midpoint Riemann sums for the function f = lambda x : 1/(1+x**2) a = 0; b = 5; N = 10 n = 10 # Use n*N+1 points to plot the function dx/2,N) x_right = np. root method. Riemann Sums: Sigma Notation Review This is a Riemann sum for f on the interval [a,b]. Evaluate the integral. By using this website, you agree to our Cookie Policy. Use a midpoint sum with 2 sub-intervals to estimate. Midpoint Riemann Sum. Riemann Sums, Sigma Notation and Writing Area as a Limit Lesson:Your AP Calculus students express the limit of a Riemann sum in integral notation and write integral notation as a limit of a Riemann sum. Option #1: If you noticed in step 2 above, we did not care if our subintervals were the same width. Choose The Correct Riemann Sum Below. Riemann Sums Using Sigma Notation With sigma notation, a Riemann sum has the convenient compact form f (il) Ax + f(Ñ2) Ax + + f (in) Ax Ef(Ñk) Ax. The subscript k is the summation index, and is a “dummy index”, in the sense that it can be replaced by any convenient letter. Find a closed form for a nite sum using the Gauss formula P n i=1 k= n(n+1) 2 and the formulas on pg. Left-Hand. 603125 Extension – Area Programs Students will use the programs to compare area approximation methods. Similarly, the right Riemann sum is an overestimate. How do you find Find the Riemann sum that approximates the integral #int_0^9sqrt(1+x^2)dx# using How do you Use a Riemann sum to approximate the area under the graph of the function #y=f(x)# on How do you use a Riemann sum to calculate a definite integral?. d) The Definite Integral as an Area and Average. right Riemann sum. Then evaluate the Riemann sum he midpoint Riemann sum for f(x) = 3+ cOS TX on [0,5] with n= 50 dentify the midpoint Riemann sum. [Analysis] Exercise 2. 1) as f x dx f m x 1 ( ) ( ) (a Riemann. Write the sigma notation. Series, Sigma Notation video (Leckie) Summation Notation (Paul) 1. Is your answer an over-estimate or an under-estimate? Split the interval [-4, 4] into two sub-intervals of length 4 and find the midpoint of each. Summation notation (or sigma notation) allows us to write a long sum in a single expression. EXAMPLE 1Using Sigma Notationƒs1d + ƒs2d + ƒs3d +Á+ ƒs100d =a100i=1ƒsid. The Midpoint Rule summation is: \(\ds \sum_{i=1}^n f\left(\frac{x_i+x_{i+1}}{2}\right)\Delta x\text{. As the partitions in [a,b] become finer and finer, our sum midpoint of. EXAMPLE 1: Find the area under the curve of the function f x ( ) =x +8 over the interval [0, 4] by using n rectangles. midpoint of each subinterval. Find the exact value of the definite integral. 10 50 1 3+ cos 10 k= 1 带) OB. With this notation, a Riemann sum can be written as \Sigma_{i=1}^n f(c_i)(x_i-x_{i-1}). [1,2] using left end, right end, and midpoint A right and left Riemann sum are used. (1) Z 6 −1 (2x −4) dx 9. Example: Estimate the area under the graph of f(x) = x2 + 1 over the interval [2;10] using 4 rectangles of equal width and midpoints. Computing Integrals using Riemann Sums and Sigma Notation Math 112, September 9th, 2009 Selin Kalaycioglu The problems below are fairly complicated with several steps. v2 (1 −v)6 dv 7. Derivatives Derivative Applications Limits Integrals Integral Applications Riemann Sum Series ODE Multivariable Calculus Laplace Transform Taylor/Maclaurin Series Fourier Series Functions Line Equations Functions Arithmetic & Comp. If you have a table of values, see midpoint rule calculator for a table. Riemann Sums; The Definite Integral; The Fundamental Theorem of Calculus; 5 Evaluating Integrals. Step 2: Partition the interval into n subintervals. We break the interval between 0 and 1 into n parts, each of width. - Duration: 14:58. By the way, you don't need sigma notation for the math that follows. + 15 •1 1 + 4 15 • 2 2 + 4 The value of the sum is. In the examples below, we'll calculate with. Works for Math, Science, History, English, and more. 35 4n+ 5k - 5 ОА. Stack Exchange network consists of 177 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Example 1: calculate Riemann sum for y = x^2 over the interval [0, 2] for 4 equal intervals. We call Ln the left Riemann sum for the function f on the interval [a, b]. The RiemannSum(f(x), x = a. 10 50 ( 2Tk – 1)). Examples Example 2. 1 Approximate the area under the curve f(x) = ln(x) between x= 1 and x= 5 using the left, right and midpoint rules with n= 4 intervals (rectangles). EXAMPLE 1Using Sigma Notationƒs1d + ƒs2d + ƒs3d +Á+ ƒs100d =a100i=1ƒsid. What is Meant by Riemann Sum? In mathematics, the. The Midpoint Rule for definite integrals means to approximate the integral by using a midpoint Riemann Sum (just as in 6. the left sum approximation to R 2π 0 sin(x)dx (a) with 8 equal subintervals (b) with n equal subintervals 2. 6078493243021688 1. Use sigma notation and the appropriate summation formulas to formulate an expression which represents the net signed area between the graph of f(x) = cosxand the x-axis on the interval [ ˇ;ˇ]. For a right Riemann sum, for , we determine the sample points as follows: Now, we can approximate the area with a right Riemann sum. The value of this left endpoint Riemann sum is _____, and it is an there is ambiguity the area of the region enclosed by y=f(x), the x-axis, and the vertical lines x = 4 and x = 8. Conic Sections. Watch the next lesson: https://www. Sigma Notation or Summation Notation. Similarly, the right Riemann sum is an overestimate. Adding up a bunch of terms can be cumbersome when there are a large number of terms. Your students will have guided notes, homework, and a content quiz on Riemann Sums and Sigma Nota. 603125 Extension – Area Programs Students will use the programs to compare area approximation methods. (All of them to start with. 2: Definition of a definite integral; Riemann sum; vocabulary (integrand, integral sign, differential, limits of integration) midpoint rule; trapezoidal rule (actually equivalent to the average of left and right rectangle rules). Sketch and nd the area of the region bounded by y = x3 and y = 4x2 4x. The Riemann sums are the called respectively the left, right, mid, upper and lower Riemann sum. Sums can also be infinite, provided that the terms eventually get close enough to zero–this is an important topic in calculus. It’s just a “convenience” — yeah, right. In sigma notation, we get that 1 2 3 99 = E k In This Module • We will introduce sigma notation — a compact way of writing large sums of like terms — and define the notion of a Riemann sum. Write the sigma notation. All other letters are constants with respect to the sum. 2 - Sigma Notation and Limits of Finite Sums - Exercises 5. The second part says that the definite integral of a continuous function from a to b can be found from any one of the function’s antiderivatives F as the number F(b)- F(a). 19) f(x) = x 2 - 2 , [0, 8], midpoint 19) 20) f(x) = cos x + 3 , [0, 2 ], left- hand endpoint 20) Express the sum in sigma notation. The subscript k is the summation index, and is a “dummy index”, in the sense that it can be replaced by any convenient letter. Compute sums expressed using sigma notation (Prob #5) Write sums using sigma notation (Prob #15) Compute sums using properties of sigma notation and formulas for common series (Prob #33) Section 5. simplified by using sigma notation and summation formulas to create a Riemann Sum. 10 50 ( 2Tk – 1)). Enjoy the videos and music you love, upload original content, and share it all with friends, family, and the world on YouTube. Take a photo of your homework question and get answers, math solvers, explanations, and videos. Use a midpoint sum with 2 sub-intervals to estimate. There are many ways to write a given sum in sigma notation. We're going to stick with RECTANGLES for the time being. 2 Riemann Sums and Integration Learning Goal 2. State the right Riemann Sum for the function on the given interval. 1) as f x dx f m x 1 ( ) ( ) (a Riemann. Midpoint Rule: As noted above, the midpoint rule is a special case of Riemann sums where the interval integration [a, b] is divided n subintervals [x i-1, x i] each with length Dx = (b-a)/n. se sigma notation to write the following Riemann sum. (2) Find the endpoints of. Sigma notation enables us to express a large sum in compact form: = an—I an The Greek capital letter (sigma) stands for "sum. d dx −2 x3 dv v2 4. Use right Riemann sums to compute the following integrals. 15k 2 + 4 C. }$$ Figure 1. 312 for the rst nsquares and the rst ncubes. Our courses show you that math, science, and computer science are – at their core – a way of thinking. and we want to have 2 rectangles with the sample points being left endpoints, then we would assign some notation. While the rectangles in this example do not approximate well the shaded area, they demonstrate that the subinterval widths may vary and the heights of the rectangles can be. There are a number of different types of Riemann sum that are important to master for the AP Calculus BC exam. Our courses show you that math, science, and computer science are – at their core – a way of thinking. Example: Estimate the area under the graph of f(x) = x2 + 1 over the interval [2;10] using 4 rectangles of equal width and midpoints. Then evaluate the Riemann sum he midpoint Riemann sum for f(x) = 3+ cOS TX on [0,5] with n= 50 dentify the midpoint Riemann sum. Find a closed form for a nite sum using the Gauss formula P n i=1 k= n(n+1) 2 and the formulas on pg. Following these steps gives you a Riemann Sum for f on the interval [a, b]. The first two arguments (function expression and range) can be replaced by a definite integral. When shown the Riemann Sum notation, each parameter was defined and discussed in detail, to include the Greek capital letter for sigma. Choose The Correct Riemann Sum Below. Can any one help how to find approximate area under the curve using Riemann Sums in R? It seems we do not have any package in R which could help. Deep bhayani on March 7, 2017 at 8:36 pm said: Riemann sum calculator There stand four temples in a row in a holy place. In sigma notation, we get that 1 2 3 99 = E k In This Module • We will introduce sigma notation — a compact way of writing large sums of like terms — and define the notion of a Riemann sum. See full list on khanacademy. Derivatives Derivative Applications Limits Integrals Integral Applications Riemann Sum Series ODE Multivariable Calculus Laplace Transform Taylor/Maclaurin Series Fourier Series Functions Line Equations Functions Arithmetic & Comp. On a definite integral, above and below the summation symbol are the boundaries of the interval, The numbers a and b are x -values and are called the limits of integration ; specifically, a is the lower limit and b is the upper limit. Download Free Mp4 LRAM, RRAM, and MRAM Tutorial TvShows4Mobile, Download Mp4 LRAM, RRAM, and MRAM Tutorial Wapbaze,Download LRAM, RRAM, and MRAM Tutorial Wapbase. Riemann Sums Using Sigma Notation With sigma notation, a Riemann sum has the convenient compact form f (il) Ax + f(Ñ2) Ax + + f (in) Ax Ef(Ñk) Ax. Each rectangle will have length ∆x =. 1: Areas and Distances Understand how rectangles are constructed to estimate the area underneath a curve using either the left hand endpoint. Stack Exchange network consists of 177 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Similar formulas can be obtained if instead we choose c k to be the left-hand endpoint, or the midpoint, of each subinterval. 2 - Sigma Notation and Limits of Finite Sums - Exercises 5. A 2 𝑛 2 + 4 𝑛 1 2 + 𝑖. For left Riemann sums, the left endpoints of the subintervals are 1) Ax, fork —. c) Convergence of Left and Right Sums for Monotonic functions and not Monotonic functions. 2a Sigma Notation and Area Approximation! Essential Learning Target Compute left, right and midpoint Riemann sums using either. The approximate value at each midpoint is below. Choose The Correct Riemann Sum Below. Thesymbol)Thus we can writeandThe sigma notation used on the right side of these equations is much more compact thanthe summation expressions on the left side. Right Riemann Sum. Is your answer an over-estimate or an under-estimate? Split the interval [-4, 4] into two sub-intervals of length 4 and find the midpoint of each. For example, saying “the sum from 1 to 4 of n²” would mean 1²+2²+3²+4² = 1 + 4 + 9 + 16 = 30. 2 Riemann Sums and Integration Learning Goal 2. 50 1 3+ cos 10 k = 1 2tk - T O A. (2k-1) (2n + 1 - 2) OF. Definition S L: Left-endpoint Riemann Sum Choose the height of each rectangle on the interval [x i, x i +1]to be a b x y f H x L Area ≈ S R: Right-endpoint Riemann Sum Choose the height of each rectangle on the interval [x i, x i +1]to be a b x y f H x L Area ≈ S M: Midpoint Riemann Sum. Compute the Riemann sum for R4 using 4 subintervals and right endpoints for the function on the interval [1,5]. Your students will have guided notes, homework, and a content quiz on Riemann Sums and Sigma Nota. Examples Example 2. ) We need Delta x=(b-a)/n Deltax is both the base of each rectangle and the distance between the endpoints. Use the Midpoint Rule with n = 3 to approximate Z 5 −1 (x2 −4) dx. In these sums, n is the number of subintervals into which the interval is divided by equally spaced partition points a = x0 < x1 < … < xn-1 < xn = b. 2: Definition of a definite integral; Riemann sum; vocabulary (integrand, integral sign, differential, limits of integration) midpoint rule; trapezoidal rule (actually equivalent to the average of left and right rectangle rules). Midpoint Riemann Sums: Suppose x i is the midpoint of the ith subinterval [x i 1;x i], that is x i = x i 1 + x i 2 for all i. The Midpoint Rule for definite integrals means to approximate the integral by using a midpoint Riemann Sum (just as in 6. Sigma Notation or Summation Notation. Then evaluate the Riemann sum he midpoint Riemann sum for f(x) = 3+ cOS TX on [0,5] with n= 50 dentify the midpoint Riemann sum. Midpoint Riemann Sums: Suppose x i is the midpoint of the ith subinterval [x i 1;x i], that is x i = x i 1 + x i 2 for all i. Now, find the endpoints. For example, say you’ve got f (x) = x2 + 1. 6255658911511259 1. The length of each subinterval is Δx = n b a. Integral Calculus Chapter 4: Definite integrals and the FTC Section 2: Riemann sums Page 8 Templated questions: 1. The approximate value at each midpoint is below. Write the correct sigma notation for any Riemann sum you encounter. Choose the correct answer below. As the partitions in [a,b] become finer and finer, our sum midpoint of. The values of the function are tabulated as follows; Left Riemann Sum LRS = sum_(r=1)^4 f(x)Deltax " " = Deltax { f(1) + f(2) + f(3) + f(4) } \\ \\ \\ (The LHS. All the four temples have 100 steps climb. your sketch the rectangles associated with the Riemann sum 4 k = 1 f(ck ) x k , using the indicated point in the kth subinterval for ck. c) Convergence of Left and Right Sums for Monotonic functions and not Monotonic functions. We draw rectangles using the values f(-2) = -4 and f(2) = -4, then add the values of the rectangles and get -4(4) + -4(4) = -32. Choose The Correct Riemann Sum Below. of de nite integrals using Riemann Sums in Sigma notation. lower than the midpoint area E. Is your answer an over-estimate or an under-estimate? Split the interval [-4, 4] into two sub-intervals of length 4 and find the midpoint of each. State the right Riemann Sum for the function on the given interval. where i is the index of summation, l is the lower limit, and n is the upper limit of summation. 3) Mid-point sums All value may be different but they represent a same quantity an approximated area under the curve. Use the Midpoint Rule with n = 3 to approximate Z 5 −1 (x2 −4) dx. Partition the interval into 4 subintervals of equal length. In previous entry, we talked about the Part 1 of Fundamental Theorem of Calculus. Use the programs on your calculator to find the value of the sum accurate to 3 decimal places. Right Riemann Sum. Riemann sums; Sigma notation Fundamental Theorem Riemann Sums Problem 2 Use a midpoint Riemann sum with 3 equal subintervals to approximate the area under y= 1 16 x. ) We need Delta x=(b-a)/n Deltax is both the base of each rectangle and the distance between the endpoints. Integral Calculus Chapter 4: Definite integrals and the FTC Section 2: Riemann sums Page 8 Templated questions: 1. The Midpoint Rule for definite integrals means to approximate the integral by using a midpoint Riemann Sum just as in 6. Now it is your turn to do the following problems. Riemann Sums. While the rectangles in this example do not approximate well the shaded area, they demonstrate that the subinterval widths may vary and the heights of the rectangles can be. EXAMPLE 1Using Sigma Notationƒs1d + ƒs2d + ƒs3d +Á+ ƒs100d =a100i=1ƒsid. So this is right over here. By comparing the sum we wrote for Forward Euler (equation (8) from the Forward Euler page) and the left Riemann sum \eqref{left_riemann}, we should be able to convince ourselves that they are the same when the initial condition is zero. Definite Integral from Summation Notation: To rewrite the given limit of summation notation, we'll compare the given expression with the formula of conversion and then substitute the values. For left Riemann sums, the left endpoints of the subintervals are 1) Ax, fork —. The sums we have been calculating - by adding together the areas of the rectangles drawn in each subinterval - are called Riemann sums, after the German mathematician Georg Friedrich Bernhard Riemann (1826-1866). scalar multiplication and matrices (Properties) scalar multiplication of vectors. The RiemannSum(f(x), x = a. If x k * is any point in the k th subinterval x k-1, x k, for k=1,2,…,n, then the. Choose the correct Riemann sum below. Right Riemann Sum. Left-Hand. Legal Notice: The copyright for this application is owned by Maplesoft. 15k 2 + 4 C. Conic Sections. The endpoints are given by x 0 = a and x n = b. higher than the actual area B. 12+ 22+ 32+ 42+ 52+ 62+ 72+ 82+ 92+ 102+ 112=a11k=1k2,k 1aknThe index k ends at k n. Typical choices are: left endpoints, right endpoints, midpoint, biggest value, smallest value. For example, saying “the sum from 1 to 4 of n²” would mean 1²+2²+3²+4² = 1 + 4 + 9 + 16 = 30. 0375 When n = 100, Left Riemann sum = 164. Step 2: Now click the button “Submit” to get the Riemann sum. Constructing Accurate Graphs of Antiderivatives; The Second Fundamental Theorem of Calculus; Integration by Substitution; Integration by Parts; Other Options for Finding Algebraic Antiderivatives; Numerical Integration; 6 Using Definite Integrals. 2 Riemann Sums and Area 1 5. v2 (1 −v)6 dv 7. (k)n-1) 729 OD (k-1) (n+1-K) n ket kot nº n n 729 OE. Every Riemann sum depends on the partition you choose (i. The same number of subintervals were used to produce each approximation. Option #1: If you noticed in step 2 above, we did not care if our subintervals were the same width. Evaluate the sum using a calculator with n=20,50, and 100. [Analysis] Exercise 2. In the more compact sigma notation, we have Ln = Xn−1 i=0 f (xi)4x. Write the midpoint Riemann sum in sigma notation for an arbitrary value of n. midpoint Riemann sum of f(x) over [a,b] using n intervals is larger than both the left and right Riemann sums of f(x) over [a,b] using n intervals. Thesymbol)Thus we can writeandThe sigma notation used on the right side of these equations is much more compact thanthe summation expressions on the left side. (1) Z 6 −1 (2x −4) dx 9. 2: Definition of a definite integral; Riemann sum; vocabulary (integrand, integral sign, differential, limits of integration) midpoint rule; trapezoidal rule (actually equivalent to the average of left and right rectangle rules). Here is the solution of a similar problem, which should give you an idea of how to write up your solution. (n times) = cn, where c is a constant. The LRAM uses the left endpoint, the RRAM uses the right endpoint and the MRAM uses the midpoint of intervals. Calculus Q&A Library se sigma notation to write the following Riemann sum. If your CAS can draw rectangles associated with Riemann sums, use it to draw rectangles associated with Riemann sums that converge to the integrals. Constructing Accurate Graphs of Antiderivatives; The Second Fundamental Theorem of Calculus; Integration by Substitution; Integration by Parts; Other Options for Finding Algebraic Antiderivatives; Numerical Integration; 6 Using Definite Integrals. The RiemannSum(f(x), x = a. Note particularly that since the index of summation begins at 0 and ends at n − 1, there are indeed n terms in this sum. Recall that a Riemann sum is an expression of the form where the x i * are sample points inside intervals of width. Is the midpoint Riemann sum an over or under approximation if the graph is: a. Then evaluate the sum. Midpoint Riemann Sum. How do you find Find the Riemann sum that approximates the integral #int_0^9sqrt(1+x^2)dx# using How do you Use a Riemann sum to approximate the area under the graph of the function #y=f(x)# on How do you use a Riemann sum to calculate a definite integral?. While summation notation has many uses throughout math (and specifically calculus), we want to focus on how we can use it to write Riemann sums. Riemann sums in summation notation calculator keyword after analyzing the system lists the list of keywords related and the list of websites with related content, in addition you can see which keywords most interested customers on the this website. Evaluate the following to practice the basic skills of Sigma Notation. simplified by using sigma notation and summation formulas to create a Riemann Sum. Thomas’ Calculus 13th Edition answers to Chapter 5: Integrals - Section 5. Right Riemann Sum. Partition a, b by choosing These partition a, b into n parts of length ?x1, ?x2, ?xn. There are many ways to write a given sum in sigma notation. Use These Values To Estimate The Value Of The Integral. You can use sigma notation to write out the Riemann sum for a curve. One method to approximate the area involves building several rect-angles with bases on the x-axis spanning the interval [a,b] and with. If we use the notation ‖‖𝑃 to denote the longest subinterval length we can force the longest subinterval length to 0 using a. Sigma Notation The sum of n terms a1,a2, midpoint, and right) is called a Riemann sum. 313-315) Practice problems: Text p. Midpoint Riemann Sum. summation symbol (an upper case sigma) Figure 5. 1) as f x dx f m x 1 ( ) ( ) (a Riemann. Use sigma notation and the appropriate summation formulas to formulate an expression which represents the net signed area between the graph of f(x) = cosxand the x-axis on the interval [ ˇ;ˇ]. Sums can also be infinite, provided that the terms eventually get close enough to zero–this is an important topic in calculus. How do you find Find the Riemann sum that approximates the integral #int_0^9sqrt(1+x^2)dx# using How do you Use a Riemann sum to approximate the area under the graph of the function #y=f(x)# on How do you use a Riemann sum to calculate a definite integral?. See full list on khanacademy. Definite Integral from Summation Notation: To rewrite the given limit of summation notation, we'll compare the given expression with the formula of conversion and then substitute the values. Livro de Calculo. Adding up a bunch of terms can be cumbersome when there are a large number of terms. asked by Bae on May 2, 2014; Calculus. Con-versely, given a Riemann sum in sigma notation, be able to iden-tify a function and an interval which give rise to that sum. Use n = 50 equal subdivisions. Simplify the Riemann Sum. (read “sigma a k as k runs from 1 to n” ). Evaluate the integral. exactly equal to the actual area C. of de nite integrals using Riemann Sums in Sigma notation. }\) Figure 1. row operation. Rolle's Theorem. 10 50 ( 2Tk – 1)). This is useful when you want to derive the formula for the approximate area under the curve. In order to check that the result does not depend on the sample points used, let’s redo the computation using now the left endpoint of each subinterval: Xn i=1 µ. In each case where you used a Riemann sum to estimate an area, try to determine if you obtained an overestimate or an underestimate. This formula includes a summation using sigma notation (Σ). " The index k tells us where to begin the sum (a the number below the E) and where to end (at the number above). Graph the function f(x) over the given interval. Is this Riemann sum an over-estimate or an under-estimate of the exact value? (b) Approximate the same integral by using Simpson’s Rule with n = 4 subintervals. See full list on khanacademy. ) In practice, evaluating a summation can be a little tricky. By the way, you don't need sigma notation for the math that follows. 6078493243021688 1. Sigma Notation or Summation Notation. Evaluate the integral. Figuring out the first (or last) element of the sum can rule out incorrect Riemann sums efficiently. The endpoints are given by x 0 = a and x n = b. It states in the book that it is recognized as a Riemann sum for a fn defined on [0,1]. f) General Riemann Sum. The subscript k is the summation index, and is a “dummy index”, in the sense that it can be replaced by any convenient letter. [Hint: Make the substitution u=(x-\mu) / \sigma, which will create two integrals. The sum would be the sum of the rectangles, so it would be the height times the width (change in x). The values of the function are tabulated as follows; Left Riemann Sum LRS = sum_(r=1)^4 f(x)Deltax " " = Deltax { f(1) + f(2) + f(3) + f(4) } \\ \\ \\ (The LHS. This gives the midpoint Riemann sum of f using n rectangles, which is denoted by M n. Use the programs on your calculator to find the value of the sum accurate to 3 decimal places. 9779070602600015 1. Therefore, always use a right-sum, with ci = a+i¢x. Use of the formulae for the nth term and the sum of the first n terms of the sequence. Legal Notice: The copyright for this application is owned by Maplesoft. The application is intended to demonstrate the use of Maple to solve a particular problem. scalar multiplication and matrices (Properties) scalar multiplication of vectors. root method. Here is how to set up the Riemann sum for the definite integral Z 3 1 x2 dx where n = 10: (1) Find ∆x = b−a n. First, determine the width of each rectangle. 2 Sigma Notation & Limits of Finite Sums NOTES SIGMA NOTATION a k =a 1 +a 2 +a 3 +!+a n−1 +a n k=1 n ∑ EX 1) Complete the table, given the following sums in sigma notation: The Sum in Sigma Notation The Sum Written Out, One Term for Each Value of k The Value of the Sum k k=1 5 ∑ (−1) k k k=1 3 ∑ k k=1 k+1 2 ∑ k2 k=4 k−1 5 ∑ EX. Stack Exchange network consists of 177 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. 22) _ 16 k = 2 6 22). Derivatives Derivative Applications Limits Integrals Integral Applications Riemann Sum Series ODE Multivariable Calculus Laplace Transform Taylor/Maclaurin Series Fourier Series Functions Line Equations Functions Arithmetic & Comp. In Chapter 1, I introduce you to the Riemann sum formula for the definite integral. XTRA Assignment 2 - Riemann sum tables w. The length of each subinterval is Δx = n b a. Simplify the expression. In previous entry, we talked about the Part 1 of Fundamental Theorem of Calculus. Example of finding a Riemann Sum for finite “n” using a table rather than sigma notation. The first two arguments (function expression and range) can be replaced by a definite integral. | 2020-10-27T06:47:08 | {
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http://nbkx.ihuq.pw/properties-of-matrix-multiplication-proof.html | Properties Of Matrix Multiplication Proof
The Transpose of the product of two matrices is equal to. There are four properties involving multiplication that will help make problems easier to solve. (b) Show that every eigenvector for Bis also an eigenvector for A. Algebraic Properties of Matrix Operations A. If A is an n×n symmetric orthogonal matrix, then A2 = I. Combined Calculus tutorial videos. addition and subtraction or multiplication and division. Note that the matrices in a matrix group must be square (to be invertible), and must all have the same size. Distributive Law of Set Theory Proof - Definition. The point of this note is to prove that det(AB) = det(A)det(B). Since our model will usually contain a constant term, one of the columns in the X matrix will contain only ones. Now, v^Tu is just the dot product of u and v, so it's a scalar. If A is an n×n symmetric matrix such that A2 = I, then A is orthogonal. c) If x 2 V then 0 ¢ x = 0. The transform of a sum is the sum of the transforms: DFT(x+y) = DFT(x) + DFT(y). However, matrix multiplication is not defined if the number of columns of the first factor differs from the number of rows of the second factor, and it is non-commutative, even when the product remains definite after changing the order of the factors. The examples below should help you see how division is not associative. Therefore there is no proof to why matrices are multiplied the way they do. multiplication, matrix-vector multiplication, and FFT. Matrix multiplication is not commutative. Let A, B, and C be three matrices. , x 6= 0) such that (A −λI)x =0. For example, 3 1 2 −1 0 = 3 6 −3 0. All for the high school levels of Grade 9, Grade 10, Grade 11, and Grade 12. The element, cij, of the product matrix is obtained by multiplying every element in the row i of matrix A by each element of column j of matrix B and then adding them. We do this first with simple numerical examples and then using geometric diagrams. Know the de nition and basic properties of a fundamental matrix for such a system. Let (A, B, C) be any choice of matrices. Commutative, Associative and Distributive Laws. It's the associative property of matrix multiplication. edu/˘schiu/ Matrix Multiplication: Warnings WARNINGS Properties above are analogous to properties of real numbers. In this page multiplication properties of matrices we are going to see some properties in multiplication. Matrix Multiplication Lesson. For example, the number of walks of length 2 is the number of vertices ksuch that there is an arc from ito kand an arc from k to j. The rst theorem stated that 0v = 0 for all vectors v. The cross product of two vectors a= and b= is given by Although this may seem like a strange definition, its useful properties will soon become evident. OLS in Matrix Form 1 The True Model † Let X be an n £ k matrix where we have observations on k independent variables for n observations. For example 4 + 2 = 2 + 4 Associative Property: When three or more numbers are added,. Pre‐requisite: Inverse of a matrix, addition, multiplication and transpose of a matrix. 7 Example (Matrix groups). This is the general linear group of 2 by 2 matrices over the reals R. Multiplication of Matrices. 1 (Properties of Matrix Addition and Scalar Multiplication) By (b), sum of multiple matrices are written as A + B + + M. jugate, there is some other matrix Q such that Since the associative law holds for matrix multiplication, the theorem is proved in the following way. Matrix multiplication is associative; for example, given 3 matrices A, B and C, the following identity is always true. Formula IA = A;BI = B whenever the products are de ned. Math Goals (Standards for. A product of permutation matrices is again a permutation matrix. We shall see the reason for this is a little while. Many of the basic properties of expected value of random variables have analogous results for expected value of random matrices, with matrix operation replacing the ordinary ones. This is a consequence of lemma 1. In this case, we have etA = PetDP 1 = P 2 4 e 1t e nt 3 5P 1: EXAMPLE 1. It is obvious that matrix multiplication is associative. then vTAv = vT(Av) = λvTv = λ Xn i=1. , , or is a unitary (orthogonal if real) matrix. , [A B]ij = [A]ij[B]ij; known asHadamard4 (or Schur) product,doesn’t workfor linear systems and linear transformations. By the definition and the lemma, all × determinant functions must return this value on this matrix. Multiplication of two sequences in time domain is called as Linear convolution. A square matrix A is said to be symmetric if AT = A and skew-symmetric if AT = A. b = a ∧b Properties of Rings. Matrix Multiplication: There are several rules for matrix multiplication. Symmetric matrices are also called selfadjoint. Let’s look at some properties of multiplication of matrices. Matrix Inverses and Solving Systems. Matrix Multiplication Suppose A and B are two matrices such that the number of columns of A is equal to number of rows of B. Write a proof for these Properties of Matrix Multiplication: Let A, B, C be matrices. Proofs of the properties of matrix multiplication. • Only the properties listed in Theorem 3. The continuous linear operators from into form a subspace of which is a Banach space with respect to. 2 The Matrix Trace; 3. Pre‐requisite: Inverse of a matrix, addition, multiplication and transpose of a matrix. n ×n zero matrix. 3 If A B and C are matrices of the correct size so that the required products are defined, and t e 2, then 3 = (AB)C (AB)T = NAT Proof 4. So, if A is invertible, your statement cannot be proved. The plural form of matrix is matrices. Let A be a squarematrix of ordern and let λ be a scalarquantity. Matrix multiplication is really useful, since you can pack a lot of computation into just one matrix multiplication operation. a * i = a, where i = 1 (identity element). If the zero matrix O is multiplied times any matrix A, or if A is multiplied times O, the result is O (see Exercise 16). , x 6= 0) such that (A −λI)x =0. Subsubsection Symmetry. Pre‐requisite: Inverse of a matrix, addition, multiplication and transpose of a matrix. Use the matrix multiplication calculator to multiply any types of matrix. edu/˘schiu/ Matrix Multiplication: Warnings WARNINGS Properties above are analogous to properties of real numbers. in this set is again in the set, so it is closed under multiplication. The properties of matrix addition and multiplication depend on the properties of the underlying number system. Say matrix A is an m×p matrix and matrix B is a p×n matrix. Game Theoretic Analysis of Multi-Processor Schedulers: Matrix Multiplication Example Oleksii Ignatenko 1 Institute of Software Systems NAS Ukraine o. 2 For any matrix A, we have (At)t = A. Characterizing Properties of an Orthogonal Matrix Theorem 6. Properties of the Matrix Inverse. If you believe your intellectual property has been infringed and would like to file a complaint, please see our Copyright/IP Policy. Sugilith-Kette(Rondell fac. Commutative property: When two numbers are added, the sum is the same regardless of the order of the addends. View History. Proof: All proofs can follow from basic arithmetic laws in R and previous definitions. ) Proof: No way. Identity Property Of Multiplication : The multiplication of any number and the identity value gives the same number as the result. juxtaposition of matrices will imply the "usual" matrix multiplication, and we will always use " " for the Hadamard product. Properties of Matrix Multiplication. In general, when the product AB and BA are possible. We look for an "inverse matrix" A 1 of the same size, such that A 1 times A equals I. Whatever A does, A 1 undoes. Math, Can you please tell me how to prove or derive the commutative property of addition (i. I would like to multiply a batched matrix X with dimension [batch_size, m, n] with a matrix Y of size[n,l], how should I do this?. A grade X student is aware of matrix but we will still start with the brief introduction of matrix and also discuss key rule for matrix multiplication. If A is an n n matrix and v and w are vectors in Rn, then the dot. Proposition (associative property) Multiplication of a matrix by a scalar is associative, that is, for any matrix and any scalars and. It is the purpose of this chapter to introduce the properties that characterise kernel functions. Let’s look at some properties of multiplication of matrices. Learn about the properties of matrix multiplication (like the distributive property) and how they relate to real number multiplication. In other words, if the order of A is m x n. By the definition and the lemma, all × determinant functions must return this value on this matrix. klmGCE Mathematics (6360) Further Pure 4 (MFP4) Textbook 10 1. Concept of elementary row and column operations. Which of the following statements illustrate the distributive, associate and the commutative property? Directions: Click on each answer button to see what property goes with the statement on the left. Here we sketch three properties of determinants that can be understood in this geometric context. MAT-0010: Addition and Scalar Multiplication of Matrices Introduction to Matrices. Matrices rarely commute even if AB and BA are both defined. Sorensen Math. What is the determinant of: (If you have forgotten about determinants, or wish you had, don't worry. Aside from this, though, matrix multiplication has all the properties you would like it to have. Perfect for acing essays, tests, and quizzes, as well as for writing lesson plans. Simple properties of addition, multiplication and scalar multiplication. 1 (Properties of Matrix Addition and Scalar Multiplication) By (b), sum of multiple matrices are written as A + B + + M. Basic properties of scalar multiplication are summarized at the end of this section. As such, it enjoys the properties enjoyed by triangular matrices, as well as other special properties. Matrix Calculations: Linear maps, bases, and matrix multiplication (where \" is the matrix multiplication of A and a vector v) Proof. For matrix multiplication to work, the columns of the second matrix have to have the same number of entries as do the rows of the first matrix. In case of convolution two signal sequences input signal x(n) and impulse response h(n) given by the same system, output y(n) is calculated. A symmetric matrix has real eigenvalues. Thus, if A = [aij]m×n and B = [bij]n×p are two matrices of order m × n and n × p respectively, then their product AB … [Read more] about How to Multiply Matrices. For example, addition and multiplication are binary operations of the set of all integers. I Eigenvectors corresponding to distinct eigenvalues are orthogonal. 25 Proof Using the remark 2. Multiplication of Matrices. The following properties are valid for the transpose: · The transpose of the transpose of a matrix is the matrix itself: (A T) T = A · Transpose of a scalar multiple: The transpose of a matrix times a scalar (k) is equal to the constant times the transpose of the matrix: (kA) T = kA T. position down into the subspace, and this projection matrix is always idempo-tent. Invertible matrices and proof of uniqueness of inverse, if it exists. Adjoint And Inverse Of A Matrix: In this article, you will know how to find the adjoint of a matrix and its inverse along with solved example questions. A product of permutation matrices is again a permutation matrix. Indeed, the matrix product A A T has entries that are the inner product of a row of A with a column of A T. To start practicing, just click on any link. The lesson may be something like this: Show this video to review how to multiply matrices. Interesting Properties of Matrix Norms and Singular Values $\DeclareMathOperator*{\argmax}{arg\,max}$ Matrix norms and singular values have special relationships. Invertible matrices and proof of the uniqueness of inverse, if it exists; (Here all matrices will have real entries). In this video, I wanna tell you about a few properties of matrix multiplication. 4 Row Operations and the LU Decomposition 102 Row Operations and Matrix. Matrix Multiplication. In this page multiplication properties of matrices we are going to see some properties in multiplication. The matrix B will be the inverse matrix of A. Matrix multiplication is one of the most fundamental problems in scientific computing and in parallel computing. at 24th St) New York, NY 10010 646-312-1000. Ad joint and inverse of a square matrix. A diagonal matrix is a square matrix whose off-diagonal entries are all equal to zero. Meaning of matrix multiplication In these examples we will explore the effect of matrix multiplication on the xy-plane. Matrix approach. Matrix multiplication Matrix multiplication is an operation between two matrices that creates a new matrix such that given two matrices A and B, each column of the product AB is formed by multiplying A by each column of B (Definition 1). error("Multiplying $r1 x$c1 and $r2 x$c2 matrix: dimensions do not match. Notice that the order of the matrices has been reversed on the right of the "=". General properties. Lemma 8 can be seen in the matrix equation R + ˇ 2 = R R 2; Lemma 9 in the. When A is multiplied by A-1 the result is the identity matrix I. If you installed a MATLAB, try to multiply two 1024*1024 matrix. We will apply most of the following properties to solve various Algebraic problems. However, virtually all of linear algebra deals with matrix multiplications of some kind, and it is worthwhile to spend some time trying to develop an intuitive understanding of the viewpoints presented here. These rules, forming matrix algebra, are naturally derivable from the properties of linear operators in the vector spaces which, as we have seen above, can be represented by matrices. In the next subsection, we will state and prove the relevant theorems. Sugilith-Kette(Rondell fac. If the zero matrix O is multiplied times any matrix A, or if A is multiplied times O, the result is O (see Exercise 16). But itdoes commute. Cayley’s defined matrix multiplication as, “the matrix of coefficients for the composite transformation T2T1 is the product of the matrix for T2times the matrix of T1”(Tucker, 1993). 12 Theorem 1. 3 Properties of matrix multiplication. So, matrix multiplication is just the image of composition of linear transformations under the identification of matrices with linear In particular, then, distributivity of matrix multiplication is really just distributivity of composition of linear transformations, which lends itself to a far more transparent. Exponential Matrix and Their Properties International Journal of Scientific and Innovative Mathematical Research (IJSIMR) Page 55 3. Institut de France, Binet also read a paper which contained a proof of the multiplication theorem but it was less satisfactory than that given by Cauchy. LECTURE 8: BASIC MATRIX ALGEBRA Prof. The cross symbol generally denotes the taking a cross product of two vectors, yielding a vector as the result, while the dot denotes taking the dot product of two vectors, resulting in a scalar. Properties of real symmetric matrices I Recall that a matrix A 2Rn n is symmetric if AT = A. Here, each element in the product matrix is simply the scalar multiplied by the element in the matrix. Matrix C and D below cannot be multiplied together because the number of columns in C does not equal the number of rows in D. Multiplication Properties of Exponents. The DCM matrix (also often called the rotation matrix) has a great importance in orientation kinematics since it defines the rotation of one frame relative to another. Properties Here is a list of all of the skills that cover properties! These skills are organized by grade, and you can move your mouse over any skill name to preview the skill. I can't understand / complete the proof that matrix multiplication is associative. References. A proof technique is introduced, which exploits the Grigoriev’s flow of the matrix multiplication function as well as some combinatorial properties of the Strassen computational directed acyclic graph (CDAG). The point of this note is to prove that det(AB) = det(A)det(B). If , the series does not converge (it is a divergent series). Our derivations use the fact that products of diagonal matrices are diagonal together with Bezout’s identity. Its subgroups are referred to as matrix groups or linear groups. When working with just real numbers or when working with scalars. Applying the associative and multiplicative identity properties, this is uv^T - u(v^Tu)v^T. We look for an “inverse matrix” A 1 of the same size, such that A 1 times A equals I. Matrix inverses Recall De nition A square matrix A is invertible (or nonsingular) if 9matrix B such that AB = I and BA = I. For example, addition and multiplication are binary operations of the set of all integers. The scalar is multiplied by each element of the matrix, giving us a new matrix of the same size. The definition of symmetric matrices and a property is given. by Marco Taboga, PhD. How to Multiply Matrices. 1 Matrix Operations Shang-Huan Chiu Department of Mathematics, University of Houston [email protected] Matrix multiplication is not commutative. Some simple properties of vector spaces Theorem Suppose that V is a vector space. Matrix worksheets include multiplication of square or non square matrices, scalar multiplication, associative and distributive properties and more. Aside from this, though, matrix multiplication has all the properties you would like it to have Theorem 2. Multiplies two matrices, if they are conformable. Matrix Multiplication Matrix multiplication: Why we do it the way we do it Because the most obvious way, i. By part (a), we have BAv = Av. Notes on Matrix Multiplication and the Transitive Closure Instructor: Sandy Irani An n m matrix over a set S is an array of elements from S with n rows and m columns. More generally, a nilpotent transformation is a linear transformation L of a vector space such that Lk = 0 for some positive integer k (and thus, Lj = 0 for all j ≥ k). , compressing one of the and will stretch the other and vice versa. Properties of matrix multiplication. In any column of an orthogonal matrix, at most one entry can be equal to 0. Boolean Logic Operations A Boolean function is an algebraic expression formed using binary constants, binary variables and Boolean logic operations symbols. Chapter 2: Determinants. 4 - Properties of Matrix-Matrix Multiplication Maggie Myers Robert A. Related Calculus and Beyond Homework Help News on Phys. To multiply a row by a column, multiply the first entry of the row by the first entry of the column. This feature is not available right now. It's the associative property of matrix multiplication. There are four properties involving multiplication that will help make problems easier to solve. I need help with a simple proof for the associative law of scalar multiplication of a vectors. We already know that = ad − bc; these properties will give us a c d formula for the determinant of square matrices of all sizes. A diagonal matrix is a square matrix whose off-diagonal entries are all equal to zero. Properties of Matrix Multiplication. Matrix multiplication. Exponential Matrix and Their Properties International Journal of Scientific and Innovative Mathematical Research (IJSIMR) Page 55 3. (b) Show that every eigenvector for Bis also an eigenvector for A. Each element in a matrix is called an entry. ImA = A = AIn (identity for matrix multiplication). There is a larger class of objects that behave like vectors in Rn. Matrix Formulation of the DFT. I Eigenvectors corresponding to distinct eigenvalues are orthogonal. any nice properties at all, nevertheless the single most important property of this multiplication is This proof gives no clue why the mulitplication should be associative, but it leaves no doubt that it is Theorem 2 Matrix multiplication is associative. We will discover that a given matrix may have more than one identity matrix. Adjoint And Inverse Of A Matrix: In this article, you will know how to find the adjoint of a matrix and its inverse along with solved example questions. This is the ultimate guide to Boolean logic operations & DeMorgan’s Theorems. ThematerialisstandardinthatthesubjectscoveredareGaussianreduction, vector spaces, linear maps, determinants, and eigenvalues and eigenvectors. A matrix is an example of what, in the general context of vector spaces, is called a linear operator. Then check if another mathematical object satis es the same properties. Matrices, vectors, vector spaces, transformations. Leila Wehbe Kernel Properties - Convexity. PROOF of [email protected] = of (Q -1 Q of = = z of Q ([email protected]) x of -1) U = [email protected] Matrix Notation for Geometric Transformations One important application of matrix algebra is in expressing the transfor-. Example: This matrix is 2×3 (2 rows by 3 columns): When we do multiplication To multiply an m×n matrix by an n×p matrix, the ns must be the same, and the result is an m×p matrix. proof of properties of trace of a matrix. Search this site. This is called the identity property of multiplication. • The set of all even integers forms a commutative ring under the usual addition and multiplication of integers. Except for the lack of commutativity, matrix multiplication is algebraically well-behaved. The first concerns the multiplication between a matrix and a scalar. We used the following 4 equations to estimate breeding values:. Similarly, if E0 is the matrix obtained by performing a column. Aside from this, though, matrix multiplication has all the properties you would like it to have. a * i = a, where i = 1 (identity element). When A is multiplied by A-1 the result is the identity matrix I. The area model for multiplication establishes the groundwork for helping visual learners in the conceptual understanding of the traditional algebraic skills of polynomial multiplication and factoring. GOLDVARD AND L. There are many types of matrices like the Identity matrix. when we multiply two matrices the number of columns of the first matrix should be equal to the number of rows of the second matrix. It is not for rectangular matrices of different sizes as it's not even defined for both "directions"! For square matrices, if it is not commutative for any pair of matrices, it is not commutative in general. I can't understand / complete the proof that matrix multiplication is associative. Matrix groups consist of matrices together with matrix multiplication. To start practicing, just click on any link. Proof of (d). Jabba's got it. weassociatewithˇthen n permutation matrix A Ai Properties areflectormatrixissymmetric Proof thesquareddistanceof. The rows of A form an orthonormal basis for Rn. elimination [5]. The product will be a 2×4 matrix, the outer dimensions. We shall see the reason for this is a little while. Groups, in general. Thus, the transpose of a row vector is a column vector and vice-versa. rules of matrix addition, multiplication of a matrix by a number, and matrix multiplication. You look at the proof and gure out the crucial step and properties which make the proof work. DMAM Distributivity across Matrix Addition, Matrices If α∈C and A,B∈Mmn,. In this lesson, we will learn about the identity matrix, which is a square matrix that has some unique properties. We use matrix techniques to give simple proofs of known divisibility properties of the Fibonacci, Lucas, generalized Lucas, and Gaussian Fibonacci numbers. Proof: Notice …{fill in your definition and try to decide if the answer fits the type of vector in V (ie. Rotation matrix, matrices multiplication, determinant, orthogonal matrices, equiangular rotations. Matrix is a way to organize data in the form of rows and columns. So, an extra property that relates ∥AB∥ to ∥A∥ and ∥B∥ is needed. proof of properties of trace of a matrix. Now we can explore some basics properties of the Hadamard Product. Nilpotent matrix. (This section can be omitted without affecting what follows. The performance will be dramatically improved if you use O(N^2. of its properties. In particular, ~a×~b 6= ~b×~a. The first concerns the multiplication between a matrix and a scalar. It is not for rectangular matrices of different sizes as it's not even defined for both "directions"! For square matrices, if it is not commutative for any pair of matrices, it is not commutative in general. The smallest such k is sometimes called the index of N. Thus, multiplication of matrix is not commutative. Multiplication of powers: Distribution of powers: If , then. Explicitly, suppose is a matrix and is a matrix, and denote by the product of the matrices. Matrix multiplication not commutative. That is the dot product, but stuffed into a $$1\times 1$$ matrix! Subsubsection Other Properties. Then certainly the product. Since the multiplication both ways generates the Identity matrix, then we are guaranteed that the inverse matrix obtained using the formula is the correct answer!. Then A−1 exists and we have. Matrix multiplication can be done only when the number of columns of is equal to the number of rows of. and of B and C have sizes for which the indicated sums and products are defined. Both the statement of this theorem and the method of its proof will be important for the study of differential equations in the next section. Proof: All proofs can follow from basic arithmetic laws in R and previous definitions. Observation: The following property is an obvious consequence of this definition. This number is then the scalar product of the two vectors. The properties are the commutative, associative, additive identity and distributive properties. Properties of Matrices. If I is the unit matrix, then AI = IA = A (I is called multiplicative identity). ,1966-67 American Hockey League Media Guide,2019 TOPPS UFC KNOCKOUT PURPLE SHIRT FIGHTER WORN RELIC #/25 JESSICA AGUILAR. The proposed approach is. For the A above, we have A 2 = 0 1 0 0 0 1 0 0 = 0 0 0 0. Multiplication of matrices generally falls into two categories, Scalar Matrix Multiplication and Vector Matrix Multiplication. Countable and uncountable sets. Preliminaries Consider the following situation: A is a matrix, possible augmented, and U is the reduced row echelon form of A. (This section can be omitted without affecting what follows. The determinant of a matrix inverse. One of the main contributions of [2] was to demonstrate that several diverse families of non-abelian groups support the reduction of n £ n matrix multiplication to group al-gebra multiplication. For matrix multiplication to work, the columns of the second matrix have to have the same number of entries as do the rows of the first matrix. If A is an n×n symmetric orthogonal matrix, then A2 = I. We used the following 4 equations to estimate breeding values:. If A is an nxm matrix and O the mxk zero-matrix, then AO = O. Properties of Matrix Multiplication Multiplication can only occur between matrices A and B if the number of columns in A match the number of rows in B. It is not for rectangular matrices of different sizes as it's not even defined for both "directions"! For square matrices, if it is not commutative for any pair of matrices, it is not commutative in general. Orthogonally Diagonalizable Matrices These notes are about real matrices matrices in which all entries are real numbers. Rank and nullity of a matrix We had seen in previous chapter that the number of non-zero rows in the rows in the row-echelon form of a matrix play an important role in finding solutions of linear equation. For two matrices A and B. Matrix multiplication: if A is a matrix of size m n and B is a matrix of. Notice that matrix multiplication is not generally commutative, i. Commutative Property of Multiplication(CPM)-It states that changing the order of the factors will not affect the product. In other words, regardless of the matrix A, the exponential matrix eA is always invertible, and has inverse e A. Properties of matrix multiplication. Since our model will usually contain a constant term, one of the columns in the X matrix will contain only ones. The set GL(n,F) of n×nmatrices with nonzero determi-nant is a group when equipped with matrix multiplication. Institut de France, Binet also read a paper which contained a proof of the multiplication theorem but it was less satisfactory than that given by Cauchy. 1 Vector Spaces Let F be a field (such as the real numbers, R, or complex numbers, C) with elements called scalars. Likewise, a scalar product can be taken outside the transform: DFT(c*x) = c*DFT(x). For the A above, we have A 2 = 0 1 0 0 0 1 0 0 = 0 0 0 0. In a 2 2 matrix multiplication, M= 7 in Strassen, m= n= p= 2. 5: A latin square of side nis an nby narray in which each cell contains a. Matrix Multiplication Suppose A and B are two matrices such that the number of columns of A is equal to number of rows of B.
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Why Joy Behar is Wrong to Slam “White Men” in Congress over Judge Kavanaugh Allegations September 20 2018 by Patrice Lee Onwuka
New York Times Attempts to Smear Amb. Nikki Haley Over $50k Curtains September 17 2018 by Patrice Lee Onwuka Coming This Holiday Season: New Perks for Retail Workers Courtesy of the Economy September 17 2018 by Patrice Lee Onwuka Why Serena Williams’ Claim of Sexism at the US Open Falls Flat September 10 2018 by Patrice Lee Onwuka ‘Cosby Show’ Star has a Message About Work Everyone Should Hear September 5 2018 by Patrice Lee Onwuka Why Did It Take Public Outcry for Twitter to Punish Threats Against Dana Loesch’s Kids? August 29 2018 by Patrice Lee Onwuka Polls: Americans Cool to Plastic Straws but Don’t Want Them Banned August 27 2018 by Patrice Lee Onwuka Will Congress Stop the FDA from Going After Our “Milk”? August 23 2018 by Patrice Lee Onwuka Another Reason to Keep ICE: The Arrest of Another Nazi War Criminal on US Soil August 22 2018 by Patrice Lee Onwuka Trump EPA Rolls Back Costly Obama-Era Energy Regulations August 21 2018 by Patrice Lee Onwuka Net Neutrality is Dead and the Internet is Getting Faster August 15 2018 by Patrice Lee Onwuka No Alexandria Ocasio-Cortez, An Invitation to Debate is Not Catcalling August 14 2018 by Patrice Lee Onwuka NYC’s Vote to Curb Uber, Lyft Growth is About Saving Taxis, Transit Not Helping Passengers August 9 2018 by Patrice Lee Onwuka Small Business Owners are Sitting Pretty in Today’s Economy August 8 2018 by Patrice Lee Onwuka 2.8 Million Fewer Americans on Food Stamps Since President Trump Took Office August 6 2018 by Patrice Lee Onwuka The Left Admits Americans Want Opportunities Not Handouts July 30 2018 by Patrice Lee Onwuka Millions of Americans are Unbanked Just Like This TV Talk Show Host July 26 2018 by Patrice Lee Onwuka Jeff Sessions Takes Colleges to Task for Coddling Students July 26 2018 by Patrice Lee Onwuka Look at What Tax Reform 2.0 Promises for Women and Families July 25 2018 by Patrice Lee Onwuka San Francisco Has a Crappy Problem on Their Hands, but Will the City Do Anything About It? July 20 2018 by Patrice Lee Onwuka Here’s Why Alexandria Ocasio-Cortez Is Wrong on Unemployment, Capitalism July 18 2018 by Patrice Lee Onwuka U.S. Should Make Unfair Drug Pricing, Foreign Practices a Priority in Trade Negotiations July 11 2018 by Patrice Lee Onwuka Trump Administration Slashes ACA Navigator Outreach Program -- And For Good Reasons July 11 2018 by Patrice Lee Onwuka The Left’s Hatred of Anything Trump Targets St. Jude Family July 9 2018 by Patrice Lee Onwuka Abolish ICE Protester Scales Lady Liberty Making for an Unhappy Fourth of July July 5 2018 by Patrice Lee Onwuka Suck This Up: Seattle Bans Plastic Straws July 2 2018 by Patrice Lee Onwuka Supreme Court Strikes Big Blow to Public Sector Unions, Upholds Free Speech of Public Workers June 28 2018 by Patrice Lee Onwuka Linda Sarsour Invokes Dr. Martin Luther King Jr. to Justify Harassment of Trump Officials June 27 2018 by Patrice Lee Onwuka No Maxine Waters, This is Not a Moment for Harassment or Incivility June 26 2018 by Patrice Lee Onwuka Making Kids Programming Better Through Common-Sense Broadcast TV Reforms June 22 2018 by Patrice Lee Onwuka A New Proposal to Replace the Broken Healthcare Law June 21 2018 by Patrice Lee Onwuka Trump Administration Expands Cheaper, Quality Healthcare Plans June 20 2018 by Patrice Lee Onwuka D.C. Will Kill Tips with New Minimum Wage June 19 2018 by Patrice Lee Onwuka S.E. Cupp Gets It Right: "Not all Women are Democrats" June 18 2018 by Patrice Lee Onwuka Today's Dads Are More Involved with Child Care Than Their Granddads June 15 2018 by Patrice Lee Onwuka Kathy Griffin Shaming Kevin Hart For Not Being Political is What's Wrong with Hollywood Elitism June 14 2018 by Patrice Lee Onwuka Optimism Among Small Business Owners Skyrockets Thanks to Tax Cuts June 13 2018 by Patrice Lee Onwuka Seattle Reverses Job-Killing Head Tax on Big Employers After Residents Speak Out June 12 2018 by Patrice Lee Onwuka Net Neutrality Ends Today. Here's Why It's Not the End of the World. June 11 2018 by Patrice Lee Onwuka It's Not Just Kate Spade and Anthony Bourdain, Suicide Rates Rise in Almost Every State. What's Going On? June 11 2018 by Patrice Lee Onwuka Is Anyone Buying Samantha Bee's Latest Apology for Her Vulgar Slur About Ivanka Trump? June 7 2018 by Patrice Lee Onwuka Women in Saudi Arabia are One Step Closer to Driving June 6 2018 by Patrice Lee Onwuka 5 Principles to Celebrate from the Supreme Court's Colorado Baker Ruling June 5 2018 by Patrice Lee Onwuka Women's March Organizers Tamika Mallory, Linda Sarsour Lash Out Against Israel June 4 2018 by Patrice Lee Onwuka NFL Finally Takes a Stand for the National Anthem May 25 2018 by Patrice Lee Onwuka Americans Report Doing Better Financially Under Year 1 of Trump May 24 2018 by Patrice Lee Onwuka Consumers, Small Banks to Gain Relief from Bi-Partisan Roll Back of Dodd-Frank Banking Regulations May 23 2018 by Patrice Lee Onwuka U.S. Baby Bust Leads to Lowest Birth Rates in 30 Years May 18 2018 by Patrice Lee Onwuka 3 Things to Know About the Senate Vote on Net Neutrality May 17 2018 by Patrice Lee Onwuka Hysteria Comes to Black Hair Care Products – Again May 16 2018 by Patrice Lee Onwuka Seattle's Business 'Head Tax' Has Even Starbucks, Amazon in an Uproar May 16 2018 by Patrice Lee Onwuka Oprah's Liberal Bias Clouded Good Advice from Her Commencement Address May 14 2018 by Patrice Lee Onwuka On Mother's Day, Advice from Some Moms May 12 2018 by Patrice Lee Onwuka Tennis Star Venus Williams Rejects "Feminist" Label May 10 2018 by Patrice Lee Onwuka Michelle Obama is Still Stumped by How Women Voted in 2016 May 7 2018 by Patrice Lee Onwuka Are You One of Third of Americans Who Say the Internet is Bad or Mixed for Society? May 1 2018 by Patrice Lee Onwuka In This Winning Economy, Cities Pay Young People to Move There May 1 2018 by Patrice Lee Onwuka Kim Kardashian and Kanye West Break the Internet for a Good Reason This Time - Free Thought April 26 2018 by Patrice Lee Onwuka Nancy Pelosi Chokes on "Crumbs" Comments April 25 2018 by Patrice Lee Onwuka 8 States Just Hit the Lowest Unemployment in Their History April 24 2018 by Patrice Lee Onwuka Kanye West Supports "Free-Thinking" Conservative Woman and Sets Twitter on Fire April 23 2018 by Patrice Lee Onwuka When Government Makes It Harder to Work, Fewer People Can Move Up the Ladder April 19 2018 by Patrice Lee Onwuka On Tax Day, Here's How the Federal Government Spends Every Dollar of Your Money April 17 2018 by Patrice Lee Onwuka Uber, Lyft Drivers Angry with Seattle City Council for Colluding with Taxis to Raise Rates April 17 2018 by Patrice Lee Onwuka They're Back! Share of Adult Children Living at Home Highest in 75 Years April 12 2018 by Patrice Lee Onwuka 3 Quotes from Mark Zuckerberg's Senate Grilling that Should Concern You April 11 2018 by Patrice Lee Onwuka 5 Drivers of the Pay Gap – They Are Not Gender Discrimination April 10 2018 by Patrice Lee Onwuka Diamond and Silk Labeled as "Unsafe" for Facebook, Maybe It's Time for A New Social Media Option April 10 2018 by Patrice Lee Onwuka Jimmy Kimmel Apologizes, But Not For His Melania Trump Joke April 9 2018 by Patrice Lee Onwuka How Many Americans Would Give Up Voting for a 10-Percent Pay Raise? It’s Surprising! April 6 2018 by Patrice Lee Onwuka Why Does Jimmy Kimmel Get Away with Mocking Melania Trump's Accent? April 5 2018 by Patrice Lee Onwuka There's No Room for Pregnancy Discrimination on the Job April 3 2018 by Patrice Lee Onwuka How the Blockbuster 'Roseanne' Reboot Gives a Voice to Middle America March 29 2018 by Patrice Lee Onwuka Tariffs: A New Bargaining Chip for Better Trade Deals? March 28 2018 by Patrice Lee Onwuka Joe Biden is Right to Attack Giving Every American a Universal Basic Income March 27 2018 by Patrice Lee Onwuka Ivanka Trump Has a Strategy to Get People into Good-Paying Jobs March 27 2018 by Patrice Lee Onwuka Should Responding to After-Hours Emails Be Banned? March 26 2018 by Patrice Lee Onwuka Cardi B Just Said What We're All Thinking About Taxes March 23 2018 by Patrice Lee Onwuka 3 Ways the Trump White House is Helping Millennials March 23 2018 by Patrice Lee Onwuka Google Shares How It Closed Unexplained Pay Gaps March 20 2018 by Patrice Lee Onwuka Jim Carrey Slams Sarah Sanders' Looks in a New Portrait March 19 2018 by Patrice Lee Onwuka VP Pence Celebrates Women as Everyday Heroes with IWF March 15 2018 by Patrice Lee Onwuka Joy Behar Finally Publicly Apologizes to VP Pence, Christians for Disparaging Jokes March 15 2018 by Patrice Lee Onwuka Vice President Pence Has an Answer for Joy Behar's Religious Intolerance March 13 2018 by Patrice Lee Onwuka Nancy Pelosi Back Peddles on Tax Bonuses Being "Crumbs" March 9 2018 by Patrice Lee Onwuka Why Tariffs on Steel, Aluminum are the Wrong Move March 9 2018 by Patrice Lee Onwuka 5 Things Women Cannot Do in Some Countries on International Women's Day March 8 2018 by Patrice Lee Onwuka Ride-sharing Replaces Buses in This City and Residents Love It March 7 2018 by Patrice Lee Onwuka How New Tariffs Could Raise the Prices of Your Everyday Goods March 6 2018 by Patrice Lee Onwuka Hollywood's Activism and Hypocrisy Were on Display at the Oscars One Pin at a Time March 5 2018 by Patrice Lee Onwuka 7 Facts About Women in the Workforce for Women's History Month March 1 2018 by Patrice Lee Onwuka Poll: Most Inappropriate Office Behavior is Not What You Think It is February 26 2018 by Patrice Lee Onwuka Why Amy Poehler Shouldn't Fight for a Hike in Tipped Minimum Wages February 26 2018 by Patrice Lee Onwuka Europe’s Gender Quotas for Corporate Boards Failed to Do Much for Women February 20 2018 by Patrice Lee Onwuka Should Healthy, Employable Americans Keep Medicaid Forever? February 16 2018 by Patrice Lee Onwuka 5 Companies Expanding Parental Leave Thanks to Tax Cuts February 15 2018 by Patrice Lee Onwuka The Downside of Another Fast-food Worker Protest for$15 Minimum Wages February 12 2018 by Patrice Lee Onwuka
Uber Has a Gender Earnings Gap, but There's a Good Explanation February 9 2018 by Patrice Lee Onwuka
Survey: #MeToo Backlash Against Women is Real February 7 2018 by Patrice Lee Onwuka
Ryan Seacrest Has a Can’t-Miss #MeToo Message About False Allegations February 6 2018 by Patrice Lee Onwuka
On the Anniversary of FMLA, Here’s Why There's Little to Celebrate February 5 2018 by Patrice Lee Onwuka
T-Mobile’s Super Bowl Ad Sounded Tone Deaf February 5 2018 by Patrice Lee Onwuka
Have You Checked Your Paycheck? You Might Be Surprised February 1 2018 by Patrice Lee Onwuka
6 Times Democrats Didn’t Cheer for Middle-Class Wins in Trump's State of the Union Address January 31 2018 by Patrice Lee Onwuka
6 Things to Watch for at President Trump’s State of the Union Address January 30 2018 by Patrice Lee Onwuka
Survey: New Tech Moms Have Great Experience Returning to Work Post Baby January 30 2018 by Patrice Lee Onwuka
Here are Six Political Moments from the 2018 Grammys January 29 2018 by Patrice Lee Onwuka
Debbie Wasserman Schultz, Nancy Pelosi Pooh-pooh $1,000 Bonuses from Tax Reform January 26 2018 by Patrice Lee Onwuka Walt Disney Heiress Thrashed Tax Reform, but Disney Company Gives Employees Tax Reform Bonuses, Investments January 25 2018 by Patrice Lee Onwuka Cashier-free Amazon Store Opens in Nation's First$15-Minimum-Wage City January 24 2018 by Patrice Lee Onwuka
Disney Credits Tax Reform for Its $50 Mil Education Investment,$1,000 Bonuses January 24 2018 by Patrice Lee Onwuka
Congress Made New Changes to 529 Plans that Parents Will Celebrate January 23 2018 by Patrice Lee Onwuka
Kristen Bell Opens Girl Power Sag Awards with Cheap Shot at Melania Trump January 22 2018 by Patrice Lee Onwuka
Feminist Slams ‘Fixer Upper’ Stars Chip & Joanna for Having 5th Child January 17 2018 by Patrice Lee Onwuka
Costco Shocks Customers by Exposing the High Price of Seattle’s Soda Tax January 16 2018 by Patrice Lee Onwuka
Condoleeza Rice Warns Against #MeToo Creating Snowflakes and Triggering Backlash January 15 2018 by Patrice Lee Onwuka
Walmart Announces Higher Starting Wages, Bonuses, and Benefits Thanks to Tax Reform January 11 2018 by Patrice Lee Onwuka
Utility Companies Cut Your Utility Bills Thanks to Tax Cuts January 11 2018 by Patrice Lee Onwuka
Say Goodbye to Busboys at Red Robin Thanks to Minimum Wage Hikes January 10 2018 by Patrice Lee Onwuka
USPS is Losing Money, But Blame It on Washington Not Amazon January 5 2018 by Patrice Lee Onwuka
No, Iceland Didn’t Just Make It Illegal to Pay Women Less Than Men January 5 2018 by Patrice Lee Onwuka
California Has Brewed Up a Common-Sense Solution to Drunk Driving January 4 2018 by Patrice Lee Onwuka
Top Hollywood Women Launch Time’s Up as the Answer to #MeToo, but Don’t Be Fooled January 3 2018 by Patrice Lee Onwuka
What You Need to Know About New York's Paid Leave Plan January 3 2018 by Patrice Lee Onwuka
18 States Hike Minimum Wage in 2018, Young Female Workers May Suffer January 2 2018 by Patrice Lee Onwuka
Ivanka Trump Visit Triggers Connecticut Parents to Pull Kids Out of School December 22 2017 by Patrice Lee Onwuka
Capitol Hill Just Passed a #GoodDealforWomen Through Tax Reform December 19 2017 by Patrice Lee Onwuka
California’s $15 Minimum Wage to Cost 400,000 Jobs – Many for Women December 18 2017 by Patrice Lee Onwuka Pew: Gender Discrimination, Sexual Harassment are Problems but Not Universal December 15 2017 by Patrice Lee Onwuka Alyssa Milano, John Oliver Aren’t for Saving the Internet but Internet Control December 14 2017 by Patrice Lee Onwuka Feminism is 2017’s Word of the Year But Women Still Don’t Know What It Means December 13 2017 by Patrice Lee Onwuka San Francisco Embraces Everyone but Delivery Robots December 12 2017 by Patrice Lee Onwuka Jobs Report: One Million More Women Are Working Since November 2016 December 8 2017 by Patrice Lee Onwuka What Paris Hilton Could Teach Chelsea Handler About Grace in Crisis December 8 2017 by Patrice Lee Onwuka Kellyanne Conway Had a #MeToo Moment Too, But No One Paid Attention Because of Politics December 7 2017 by Patrice Lee Onwuka Mark Zuckerberg’s Paternity Leave Tells Us 3 Things About Paid Leave, Millennials December 6 2017 by Patrice Lee Onwuka Sheryl Sandberg Calls Out the Chilling Effect Between Sexes Following Harassment Revelations December 5 2017 by Patrice Lee Onwuka Three Ways to Maximize Your Last Paychecks of 2017 December 1 2017 by Patrice Lee Onwuka Meghan Markle Giving Up Her Career for Prince Harry Doesn’t Make Her a Bad Woman November 29 2017 by Patrice Lee Onwuka Susan Sarandon is Still Anti-Hillary, The Left Blasts Her For It November 28 2017 by Patrice Lee Onwuka FCC’s Thanksgiving Blessing was to End Internet Neutrality Regulations November 27 2017 by Patrice Lee Onwuka Black Friday Protests Dwindle Exposing This Uncomfortable Truth November 27 2017 by Patrice Lee Onwuka Kathy Griffin’s Sadistic Trump Joke Left Her Jobless, She’s Not a Victim November 21 2017 by Patrice Lee Onwuka New York Wants to Dictate When You Work November 20 2017 by Patrice Lee Onwuka Bernie Sanders Blatantly Gives Bill Clinton, Al Franken a Pass on Sexual Harassment November 20 2017 by Patrice Lee Onwuka Do Winter Blizzards Call for Paid Leave Too? November 15 2017 by Patrice Lee Onwuka ‘The Simpsons’ Calling Kellyanne Conway a Nazi is a Sexist, Racist Cheap Shot November 14 2017 by Patrice Lee Onwuka How Zero Tolerance Polices Backfire in the Age of #MeToo November 13 2017 by Patrice Lee Onwuka Three G.I. Janes from American History that Make Women Proud on Veterans Day November 10 2017 by Patrice Lee Onwuka TSA Failing Undercover Weapons Tests – Again, and It’s Disturbing November 10 2017 by Patrice Lee Onwuka NAACP Exploits NFL Protests to Ban "Star-Spangled Banner" as National Anthem November 9 2017 by Patrice Lee Onwuka Women in Congress Have an Answer to Paid Leave November 8 2017 by Patrice Lee Onwuka Sexist LA Times Columnist Attacks Sarah Huckabee Sanders’ Looks November 6 2017 by Patrice Lee Onwuka Explosive: DNC, Hillary Clinton Campaign Rigged the Party Nomination for Her November 3 2017 by Patrice Lee Onwuka Eva Longoria is Desperately Wrong to Push the Latina Pay Gap November 2 2017 by Patrice Lee Onwuka Shame on Cosmo for Normalizing Incest – Again November 2 2017 by Patrice Lee Onwuka Poll: Most Women Love Tech but Only Half Think It’ll Improve Their Life November 1 2017 by Patrice Lee Onwuka 2018 Obamacare Premiums to Leave Even Wider Hole in Your Wallet October 30 2017 by Patrice Lee Onwuka Walmart Debuts Robots to Make (Holiday) Shopping Better October 30 2017 by Patrice Lee Onwuka NAACP Issues Inane Flying-While-Black Advisory, Please Keep It to Yourselves October 27 2017 by Patrice Lee Onwuka President Trump’s Opioid Attack Plan: Personal, Practical, and Policy October 26 2017 by Patrice Lee Onwuka UPenn Teacher Helps Black Women by Discriminating Against White Men October 24 2017 by Patrice Lee Onwuka Campus Sexual Assault: The Distressing Fight Mothers Face for Their Sons, Advice to Young Women October 23 2017 by Patrice Lee Onwuka IRS to Hold Tax Returns Hostage to Enforce Obamacare October 20 2017 by Patrice Lee Onwuka ABC’s Misleading Workplace Sexual Harassment Poll Pushes an Agenda October 19 2017 by Patrice Lee Onwuka ‘The Big Bang Theory’ Actress Counsels Actresses to be Modest, Triggers Feminist Backlash October 17 2017 by Patrice Lee Onwuka No, NFL Protests Aren't About the Gender Pay Gap October 16 2017 by Patrice Lee Onwuka Ivanka Trump to Boost Global Women Entrepreneurs Via World Bank Fund October 16 2017 by Patrice Lee Onwuka President Trump’s Anti-Regulation Crusade may be Working in Americans’ Eyes October 13 2017 by Patrice Lee Onwuka Lawmakers Try to Stop Secretary DeVos on Campus Sexual Assault Policy October 13 2017 by Patrice Lee Onwuka Black Lives Matter Group Opposes Free Speech for Every American October 12 2017 by Patrice Lee Onwuka Harvey Weinstein’s Open Secret Exposes Hypocrisy on the Left October 10 2017 by Patrice Lee Onwuka Marc Jacobs Doubles Down on Free Expression for Designers, Artists October 10 2017 by Patrice Lee Onwuka On Columbus Day, Give Him a Break October 9 2017 by Patrice Lee Onwuka Company Behind Fearless Girl Statute Admits Gender Pay Issues October 9 2017 by Patrice Lee Onwuka Three Bright Spots for Women in Today’s Jobs Report October 6 2017 by Patrice Lee Onwuka Half of Americans Can’t Afford More than$100 in Healthcare Premiums October 4 2017 by Patrice Lee Onwuka
One Tech Company Changed Approach to Promotions, Women Won Big October 3 2017 by Patrice Lee Onwuka
Librarian Rejects Dr. Seuss Books Because Melania Trump Sent Them September 29 2017 by Patrice Lee Onwuka
Amber Rose’s Slutwalk is a Stunt Not Empowerment September 29 2017 by Patrice Lee Onwuka
Do We Want a Feminist ‘Will & Grace’ Reboot? September 26 2017 by Patrice Lee Onwuka
Hillary Clinton Slams Women Who Support Trump as “Publicly Disrespecting Themselves” September 26 2017 by Patrice Lee Onwuka
Betsy DeVos Ends Obama-Era Policies on School Sexual Assault September 22 2017 by Patrice Lee Onwuka
Washington Wants Black Households to Trade Privacy for Cash September 22 2017 by Patrice Lee Onwuka
Half of College Women Get This Wrong about Hate Speech September 20 2017 by Patrice Lee Onwuka
Joe Biden Slams Socialist Universal Basic Income Idea September 19 2017 by Patrice Lee Onwuka
Millennials Would Trade Right to Vote for Student Loan Freedom September 18 2017 by Patrice Lee Onwuka
Feminists Attack Starbucks's Pumpkin Spice Lattes for Funding White Supremacy September 15 2017 by Patrice Lee Onwuka
DeVos Backlash Incites Dangerous Language September 14 2017 by Patrice Lee Onwuka
Over One in Three Americans Can’t Name Any 1st Amendment Rights September 12 2017 by Patrice Lee Onwuka
Miss America Pageant Gets Political and Blatantly Anti-Trump September 11 2017 by Patrice Lee Onwuka
Melania Trump Staffs Fewer People than Michelle Obama – A Savings for Taxpayers September 8 2017 by Patrice Lee Onwuka
Hillary Clinton Asks 'What Happened?' in 2016 September 7 2017 by Patrice Lee Onwuka
Kim Kardashian Still Eschews One Label: Feminist September 5 2017 by Patrice Lee Onwuka
Jerry Springer Rides the $15 Minimum Wage Wagon September 5 2017 by Patrice Lee Onwuka A$20 Question: Will Harriet Tubman be on the Bill? September 1 2017 by Patrice Lee Onwuka
Estee Lauder Faces Heat Over Paid Leave for Dads August 31 2017 by Patrice Lee Onwuka
No, Melania’s Shoes are Not the Story, Texas is August 30 2017 by Patrice Lee Onwuka
Illinois Governor Shuts Down $15 Minimum Wage August 28 2017 by Patrice Lee Onwuka Amy Schumer's Equal Pay Problem August 25 2017 by Patrice Lee Onwuka Poll: Nearly Three Out of Four Americans Would Die to Defend Freedom of Speech August 25 2017 by Patrice Lee Onwuka Here’s Another Way Women Suffer from Minimum Wage Increases August 24 2017 by Patrice Lee Onwuka Oops! Social Security Website Misdirects$11 Mil to Thieves August 23 2017 by Patrice Lee Onwuka
D.C. Raises Minimum Wage, Diners Foot the Bill August 21 2017 by Patrice Lee Onwuka
What Dr. Ben Carson Shows Us about Confronting Hateful Ideas August 17 2017 by Patrice Lee Onwuka
Hope Brings Another Woman to Trump Leadership Circle August 17 2017 by Patrice Lee Onwuka
Helen Mirren’s Feminism is about Anti-Trump and Potty-Mouths August 16 2017 by Patrice Lee Onwuka
CFPB Crackdown May Trap Borrowers in a Different Debt Trap August 14 2017 by Patrice Lee Onwuka
Sheryl Sandberg Leans In to One-Size-Fits-All Paid Leave Plan August 10 2017 by Patrice Lee Onwuka
Café Charges 18% Man Tax to Close the Gender Pay Gap August 9 2017 by Patrice Lee Onwuka
College STEM Women Claim to be Sexism Victims August 7 2017 by Patrice Lee Onwuka
A Sexism Rating Coming to a Movie Near You August 3 2017 by Patrice Lee Onwuka
IRS Rehires Hundreds of Fired Workers Despite Serious Transgressions August 3 2017 by Patrice Lee Onwuka
Study: MD County to Lose 47,000 jobs because of $15 Minimum Wage August 2 2017 by Patrice Lee Onwuka Adam Carolla Schools Congress on Campus Free Speech July 31 2017 by Patrice Lee Onwuka How One Tech Company is Taking on the Future of Work July 28 2017 by Patrice Lee Onwuka Governor Christie Shuts Down Costly Paid Leave Expansion Bill July 27 2017 by Patrice Lee Onwuka Teachers’ Union Boss Plays the Race Card against Secretary DeVos July 25 2017 by Patrice Lee Onwuka Cutting the Strings on Obamacare Enrollment Assistance July 24 2017 by Patrice Lee Onwuka Not Even Death Stops a Government Check July 21 2017 by Patrice Lee Onwuka Who Cares for the Caregivers? July 19 2017 by Patrice Lee Onwuka College Diversity Officials Rake in Major Dough July 18 2017 by Patrice Lee Onwuka DHS Hides Spending$3.6 Mil of Taxpayer Dollars on Pricey Conferences July 17 2017 by Patrice Lee Onwuka
Study: Marrying Before Having Kids is a Strong Path to Success for Millennials July 13 2017 by Patrice Lee Onwuka
$2 Mil NJ Welfare Fraud Exposes Holes in Our Social Safety Net July 7 2017 by Patrice Lee Onwuka “WTF” is Silicon Valley’s Next Answer to the Democratic Party July 6 2017 by Patrice Lee Onwuka Fraud Alert: Federal Free Phone, Internet Program Wastes$100 Mil Annually on Fake, Dead Enrollees July 3 2017 by Patrice Lee Onwuka
Cosmo Leaves Conservatives Off Their Short List of Female Presidential Candidates June 28 2017 by Patrice Lee Onwuka
Report: Seattle’s Minimum Wage Hike Gave Low-Wage Workers a $125 Monthly Pay Cut June 27 2017 by Patrice Lee Onwuka Obama Comes Out to Bash Senate Repeal Bill June 23 2017 by Patrice Lee Onwuka Maybe Profit Motivates Tech Companies More than Politics June 22 2017 by Patrice Lee Onwuka Ivanka Trump Hits Capitol Hill about Paid Leave June 20 2017 by Patrice Lee Onwuka Supreme Court Upholds Offensive Names as Free Speech June 20 2017 by Patrice Lee Onwuka Silicon Valley CEOs Ignore Calls to Boycott Trump Tech Summit June 19 2017 by Patrice Lee Onwuka What the Return of Pro-Net Neutrality Democrat Means for Internet Regs Rollback June 15 2017 by Patrice Lee Onwuka Is the Federal Consumer Cop Asleep on the Job? June 15 2017 by Patrice Lee Onwuka Obama Health Records Program Leaves Taxpayers on the Hook for Mispayments June 15 2017 by Patrice Lee Onwuka Just in Time for July 4th New York State Greenlights Uber, Lyft June 9 2017 by Patrice Lee Onwuka Filter, Don’t Pilfer June 8 2017 by Patrice Lee Onwuka What Michelle Obama Gets Wrong About Self-Interest June 8 2017 by Patrice Lee Onwuka Harvard Revokes Acceptances Over Online Comments June 6 2017 by Patrice Lee Onwuka Bill Maher’s N-Word Episode June 5 2017 by Patrice Lee Onwuka Hillary Clinton's Whine Tour Stops in Silicon Valley June 2 2017 by Patrice Lee Onwuka Maryland Governor Shuts Down Paid Sick-Leave Bill May 26 2017 by Patrice Lee Onwuka Chelsea Clinton Equates Child Marriage to Climate Change May 26 2017 by Patrice Lee Onwuka Obamacare’s Death Spiral Claims Another Healthcare Insurer May 25 2017 by Patrice Lee Onwuka Top 10 Unauthorized Federal Programs on the Chopping Block in Trump Budget May 25 2017 by Patrice Lee Onwuka Supreme Court Ends Judge Shopping for Patent Trolls May 23 2017 by Patrice Lee Onwuka State of Texas Trumps Austin over Uber, Lyft Regulations May 22 2017 by Patrice Lee Onwuka Undoing Net Neutrality Starts Today May 18 2017 by Patrice Lee Onwuka San Francisco May Stall Delivery Robots May 18 2017 by Patrice Lee Onwuka Miss USA Stands by Her Comments – Good for Her May 17 2017 by Patrice Lee Onwuka Miss USA Slammed for Saying Healthcare is a Privilege Rather than a Right May 15 2017 by Patrice Lee Onwuka Don’t Be Tone Deaf on Hearing Aids May 11 2017 by Patrice Lee Onwuka Grads Boo Betsy DeVos at Commencement May 11 2017 by Patrice Lee Onwuka TV Comedian John Oliver Piles on to Save Obama Internet Rules May 10 2017 by Patrice Lee Onwuka Automation Coming to a City near You May 9 2017 by Patrice Lee Onwuka Your Zip code Is a Good Predictor for How Long You Live May 9 2017 by Patrice Lee Onwuka Tech Bosses: Expect Significant Automation of Jobs by 2022 May 5 2017 by Patrice Lee Onwuka Can Another Tech Squad to Transform How Washington Works? May 5 2017 by Patrice Lee Onwuka Tech Founder to Ivanka Trump: Don’t Use My Story May 3 2017 by Patrice Lee Onwuka Are Saving and Debt Repayment the New Normal for Americans? May 2 2017 by Patrice Lee Onwuka This Immigrant is Showing Up to Work on May Day May 1 2017 by Patrice Lee Onwuka FCC Announces Net Neutrality Reversal Plan April 27 2017 by Patrice Lee Onwuka Copyright Office Reform Bill Passes House April 27 2017 by Patrice Lee Onwuka FCC Expected to Kick Net Neutrality to the Curb April 26 2017 by Patrice Lee Onwuka Delivery Robots Get Legislative Green Light in Two States April 25 2017 by Patrice Lee Onwuka Harvard Study: Higher Minimum Wages Trigger Restaurant Closures April 24 2017 by Patrice Lee Onwuka Not Done Rolling Back Washington’s Red Tape Yet April 19 2017 by Patrice Lee Onwuka Three Celebrities Feeling the Pain on Tax Day April 18 2017 by Patrice Lee Onwuka Female-Led FTC Takes Reducing Regulations Seriously April 18 2017 by Patrice Lee Onwuka New York Pilots Free College Tuition Plan April 13 2017 by Patrice Lee Onwuka Poll: Americans Comfortable with AI in Most--But Not All--Instances April 11 2017 by Patrice Lee Onwuka No Phone Calls on Airplanes After All April 11 2017 by Patrice Lee Onwuka Amazon to Hire 30,000 Part-time, Work-from-Home Jobs April 10 2017 by Patrice Lee Onwuka Judge Blocks Seattle’s Plan to Unionize Uber April 7 2017 by Patrice Lee Onwuka Stop the Misinformation on the Roll-Back of Privacy Rules April 5 2017 by Patrice Lee Onwuka The Challenge of Counterfeit Drugs April 4 2017 by Patrice Lee Onwuka Cali Lawmaker Wants to Push Gig-Economy Workers into Unions April 3 2017 by Patrice Lee Onwuka FCC Reverses Federal Power Grab Over LifeLine Program (AKA ObamaPhones for the Internet) March 30 2017 by Patrice Lee Onwuka Here’s How Uber is Boosting Diverse Applicants March 30 2017 by Patrice Lee Onwuka Congress Scraps Obama-Era Privacy Rules March 30 2017 by Patrice Lee Onwuka Should We Worry About Robots Replacing Workers? March 27 2017 by Patrice Lee Onwuka Black Leaders Tell Trump What They Have to Lose March 23 2017 by Patrice Lee Onwuka Congress to Free the Copyright Office from the Thumb of the Library of Congress March 23 2017 by Patrice Lee Onwuka Why Everything from Dresses to Avocados May Get More Expensive March 22 2017 by Patrice Lee Onwuka Trump, Clinton Campaigns Depended Upon Sharing Economy on the Campaign Trail March 21 2017 by Patrice Lee Onwuka Before You Blame the Wage Gap, Check Your Attitude March 20 2017 by Patrice Lee Onwuka Trump Hacks Away at the TSA March 20 2017 by Patrice Lee Onwuka Trump Pumps the Brakes on Obama-Era Fuel Efficiency Regs March 16 2017 by Patrice Lee Onwuka Students Push Back on Feminist Professor’s Indoctrination March 15 2017 by Patrice Lee Onwuka President Trump: May Be Time to Say "You're Fired!" to Duplicative Agency Employees March 14 2017 by Patrice Lee Onwuka Austin’s Struggle with Transportation After Uber, Lyft Ride Out of Town March 13 2017 by Patrice Lee Onwuka Steve Case: American Workers Feeling Left Behind, Really Have Been Left Behind March 13 2017 by Patrice Lee Onwuka Analogies Fly Over Net Neutrality March 9 2017 by Patrice Lee Onwuka These are the Women A Day Without Women Hurts March 8 2017 by Patrice Lee Onwuka Can Facebook’s New Tool Crack Down on “Fake News”? March 7 2017 by Patrice Lee Onwuka Is Regulating the Internet a Good Idea? March 6 2017 by Patrice Lee Onwuka Dr. Ben Carson Confirmed as HUD Head March 3 2017 by Patrice Lee Onwuka Robots Green-Lighted to Deliver Your Packages March 2 2017 by Patrice Lee Onwuka Three Key Messages to Black America from Trump’s Address March 1 2017 by Patrice Lee Onwuka Reports: GM Lobbies to Keep Competing Self-Driving Tech Off the Road February 28 2017 by Patrice Lee Onwuka New Self-driving Truck Company Speeds Up the Technology February 27 2017 by Patrice Lee Onwuka Small Internet Companies Gain Regulatory Relief from Washington February 24 2017 by Patrice Lee Onwuka Women Have a New Ally in Getting Them to Work February 24 2017 by Patrice Lee Onwuka Federal Watchdog: Obama’s Techie Squad Flouted the Rules February 22 2017 by Patrice Lee Onwuka Adulting School: Millennials Are Signing Up for Classes February 22 2017 by Patrice Lee Onwuka Former AG Eric Holder Leading Internal Investigation at Uber February 21 2017 by Patrice Lee Onwuka Whoopi Snubs the Mean Girls at New York Fashion Week in Defense of Tiffany Trump February 17 2017 by Patrice Lee Onwuka Humana Calls It Quits on ObamaCare, Leaving One State in Trouble February 16 2017 by Patrice Lee Onwuka Congress Opens the Door to Self-Driving Cars February 15 2017 by Patrice Lee Onwuka Linda McMahon Confirmed to Head the SBA February 14 2017 by Patrice Lee Onwuka Cabs Threaten to Halt Wheelchair Access Over Uber, Lyft February 14 2017 by Patrice Lee Onwuka Trump Models: The Next Target in the Fashion World Purge February 13 2017 by Patrice Lee Onwuka Senate Dems Come Out Swinging to Protect Washington’s Control of the Internet February 8 2017 by Patrice Lee Onwuka A Surprise Stand for Competition Over Cronyism in Battle for NYC Streets? February 8 2017 by Patrice Lee Onwuka Lifeline Internet Assistance Program (AKA Obamaphones for the Internet) Takes a Hit February 7 2017 by Patrice Lee Onwuka Big Mac ATM’s, the Answer to$15 Minimum Wages? February 2 2017 by Patrice Lee Onwuka
New FCC Chairman Gets to Work Cutting Red Tape February 2 2017 by Patrice Lee Onwuka
Trump Transportation Secretary Chao Confirmed, Despite Democratic Protest February 1 2017 by Patrice Lee Onwuka
Cutting the Red Tap: President Trump Orders 2-for 1 Rule on Regulations January 31 2017 by Patrice Lee Onwuka
Chelsea Handler's Garbage-Pail Mouth is Now Shaming Melania Trump's English January 27 2017 by Patrice Lee Onwuka
Break out the Dow 20K Hats? January 27 2017 by Patrice Lee Onwuka
Apple Hit with Another Lawsuit over Distracted Driving January 25 2017 by Patrice Lee Onwuka
Getting American Workers Ready for Automation’s Coming Shake-Up January 24 2017 by Patrice Lee Onwuka
Uber Ponies Up $20M to Feds Over Driver Pay January 23 2017 by Patrice Lee Onwuka Americans Aren’t Really Warming Up to Self-Driving Cars January 19 2017 by Patrice Lee Onwuka The Government’s “Slant” on Trademarks Gets Challenged January 19 2017 by Patrice Lee Onwuka Uber Sues Seattle to Block Drivers from Unionizing January 18 2017 by Patrice Lee Onwuka Inauguration Reminds Us Why Home Sharing is a Win for Travelers January 17 2017 by Patrice Lee Onwuka Jesse Jackson PUSHes Uber on Diversity Statistics January 13 2017 by Patrice Lee Onwuka 5 Times President Obama’s Rhetoric Didn’t Match His Kumbaya Message January 11 2017 by Patrice Lee Onwuka Jenna Bush Microaggressess at the Golden Globes? January 10 2017 by Patrice Lee Onwuka Rolling Back Obama’s Internet Privacy Rules January 9 2017 by Patrice Lee Onwuka Update: City of San Diego Says Restaurants Can't Publicly Blame Surcharge on Minimum Wage Hike January 6 2017 by Patrice Lee Onwuka San Diego Restaurants Slap Diners with Surcharges Thanks to Minimum Wage Increases January 6 2017 by Patrice Lee Onwuka The Uber Hook-Up Effect January 5 2017 by Patrice Lee Onwuka Christmas Eve Distracted Driving Fatality Could Trigger Government Action January 3 2017 by Patrice Lee Onwuka Cali Regulators Drive Uber’s Self-Driving Cars Away December 27 2016 by Patrice Lee Onwuka Will Seattle’s Driver Unionization be the End of Uber, Ridesharing? December 21 2016 by Patrice Lee Onwuka Not Fake News: Some UVA Students Sign Fake Anti-Christmas Petition December 20 2016 by Patrice Lee Onwuka Pinterest Misses Female Recruitment Quantity Quotas, But Hits on Quality December 19 2016 by Patrice Lee Onwuka No Michelle Obama, “Hope” Doesn’t Start and End with Your Husband December 19 2016 by Patrice Lee Onwuka Uber, San Francisco Regulators Crash Head-on over Self-Driving December 16 2016 by Patrice Lee Onwuka More on the Shaming of Wonder Woman December 15 2016 by Patrice Lee Onwuka Environmental Group Smears Black Beauty Products December 14 2016 by Patrice Lee Onwuka San Francisco Mayor Sides with Airbnb December 12 2016 by Patrice Lee Onwuka Airbnb Can’t Beat ‘Em, So Now They’re Lobbying ‘Em December 9 2016 by Patrice Lee Onwuka The Supremes Side with Samsung Against Apple in High-Profile Patent Case December 7 2016 by Patrice Lee Onwuka Cell Phones on the Road: A Serious Problem, but Do the Feds Need to Solve It? December 6 2016 by Patrice Lee Onwuka Airbnb Took on the Big Apple and Lost December 5 2016 by Patrice Lee Onwuka Obama’s Overtime Rule is Down, but Not Out December 2 2016 by Patrice Lee Onwuka Taxpayers Will Forgive Billions More in Student Loan Debt December 2 2016 by Patrice Lee Onwuka Uber Drivers Join Minimum Wage Protesters November 30 2016 by Patrice Lee Onwuka Negative Online Reviews Get New Federal Protection November 30 2016 by Patrice Lee Onwuka #GivingTuesday: the Hangover Cure for a Weekend of Spending November 29 2016 by Patrice Lee Onwuka Some Maine Restaurant Workers Worried about Effects of Minimum Wage Hike November 28 2016 by Patrice Lee Onwuka Judge Blocks Obama Workplace Overtime Rules November 23 2016 by Patrice Lee Onwuka Pew: One in Four Americans Makes Bank through the Gig Economy November 21 2016 by Patrice Lee Onwuka U.S. Small Businesses: Good News and Bad News November 21 2016 by Patrice Lee Onwuka Arizona’s New Minimum Wage Packs a One-Two Punch for Nonprofits November 18 2016 by Patrice Lee Onwuka Washington State Minimum Wage Passed and Childcare Costs Skyrocket November 17 2016 by Patrice Lee Onwuka Hi, I’m Ben, Your Senator and Uber Driver! November 15 2016 by Patrice Lee Onwuka Facebook Ends Multicultural Targeted Ads November 14 2016 by Patrice Lee Onwuka Oprah’s Guarded Hope for the New President Triggers the Left November 14 2016 by Patrice Lee Onwuka Star-Power Failed to Energize Millennials at the Polls November 10 2016 by Patrice Lee Onwuka Mom-preneur Faces Charges for Selling Ceviche? November 9 2016 by Patrice Lee Onwuka Uber is Back on the Road in Philly November 7 2016 by Patrice Lee Onwuka Airbnb’s Self-Policing Strategy against Discrimination November 4 2016 by Patrice Lee Onwuka Sanders and Warren Ignore that America is Better Off Now While Promoting Socialism • Evening Edit October 14 2019 by Patrice Lee Onwuka Biden campaign plans to lose Iowa, New Hampshire, and Nevada • Rising October 11 2019 by Patrice Lee Onwuka Is Warren trying to win black voters by floating Gillum as VP • Rising October 11 2019 by Patrice Lee Onwuka US Businesses Should Stand Behind Hong Kong • Making Money October 9 2019 by Patrice Lee Onwuka Left Leaning Outlets Turn on NYT Following Kavanaugh Hit Piece • Evening Edit September 23 2019 by Patrice Lee Onwuka Partisan Party Politics Turns American Voters Off • Lou Dobbs Tonight September 18 2019 by Patrice Lee Onwuka Bashing Billionaires Dissuades Everyone from Trying to Reach the Top • Making Money September 13 2019 by Patrice Lee Onwuka Dems Have a Hard Case to Make Heading Into 2020 • Your World with Neil Cavuto September 9 2019 by Patrice Lee Onwuka Dems Prioritize Gun Reform But Its All Talk • Evening Edit September 9 2019 by Patrice Lee Onwuka Talking Tech: Is Being a "Big Company" a Bad Thing? • Making Money with Charles Payne September 6 2019 by Patrice Lee Onwuka Media Needs to Find Balance Between Criticism and Support of President • Evening Edit September 3 2019 by Patrice Lee Onwuka Democrats Continue to Focus on Emotion Over Fact: Biden and AOC • Trish Regan Primetime August 30 2019 by Patrice Lee Onwuka Is the DNC rigging the debate rules? • Rising August 29 2019 by Patrice Lee Onwuka Is South Carolina Biden's Hail Mary? • Rising August 29 2019 by Patrice Lee Onwuka 4 in 10 People Won't Trust the 2020 Election Results if Their Team Loses • Lou Dobbs Tonight August 28 2019 by Patrice Lee Onwuka Taking Border Security Seriously with Approved Funding for The Wall • Evening Edit August 27 2019 by Patrice Lee Onwuka Women Are Winning in This Economy • Coast to Coast August 22 2019 by Patrice Lee Onwuka Language Police: In San Francisco and on Facebook • Lou Dobbs Tonight August 22 2019 by Patrice Lee Onwuka Harry Reid Denounces Trajectory of Dem Party • Making Money with Charles Payne August 21 2019 by Patrice Lee Onwuka Judge Trump’s actions, not words — and other commentary August 19 2019 by Patrice Lee Onwuka Conservative commentator rips Shapiro over criticism of people with multiple jobs August 16 2019 by Patrice Lee Onwuka How Will the Economy Effect the 2020 Conversation? • Coast to Coast August 15 2019 by Patrice Lee Onwuka Israel Should Say "Thanks, But No Thanks" to Reps. Omar and Tlaib • Lou Dobbs Tonight August 15 2019 by Patrice Lee Onwuka We Need to Act Now to Steer Away from Red Flags Ahead • Making Money August 14 2019 by Patrice Lee Onwuka "In God We Trust" Is Important to Display in School • Lou Dobbs Tonight August 14 2019 by Patrice Lee Onwuka Political Talking Points Stir Up Violent Threats Against SoulCycle Owner • Evening Edit August 9 2019 by Patrice Lee Onwuka Pitting Consumers Against Businesses in the Name of Politics • Coast to Coast August 8 2019 by Patrice Lee Onwuka Democratic Socialists Cannot Defeat Capitalism with Proper Pronouns and Complaints • Lou Dobbs August 6 2019 by Patrice Lee Onwuka Don't pander to women; offer real opportunity August 1 2019 by Patrice Lee Onwuka Bernie Ignores Legitimate Questions About his Health Care Plans • Evening Edit July 31 2019 by Patrice Lee Onwuka The Washington Post Plays Scapegoat for Those Still Hung Up on 2016 Election • Lou Dobbs Tonight July 30 2019 by Patrice Lee Onwuka Food Stamps Cut Because Booming Economy Means Less People Need Them • Fox & Friends First July 24 2019 by Patrice Lee Onwuka The Squad Tries Pushing the Democrats Further Left Over Israel • Evening Edit July 24 2019 by Patrice Lee Onwuka$15 Minimum Wage Robs Women of Jobs, Opportunity, Economic Mobility July 18 2019 by Patrice Lee Onwuka
"Squad" Pushes Democrat Party Closer to Socialism • Coast to Coast July 17 2019 by Patrice Lee Onwuka
"Woke Activism" is No Way to Get Things Done in Congress • Fox & Friends First July 15 2019 by Patrice Lee Onwuka
Cory Gardner, the women’s vote and religion: Western Conservative Summit takeaways July 14 2019 by Patrice Lee Onwuka
Democrats Have Placed All Their Eggs in the Identity Politics Basket • Evening Edit July 11 2019 by Patrice Lee Onwuka
Why Restaurant Workers are Rising Up to Save Tipped Wages July 9 2019 by Patrice Lee Onwuka
Women's World Cup Champions Demand Equal Pay, Do the Numbers Add Up? • After The Bell July 8 2019 by Patrice Lee Onwuka
The Folly of Demanding Amazon’s Tax Returns • Liberty Watch Radio July 7 2019 by Patrice Lee Onwuka
Democrats are Obsessed Over the Past as the Candidates Take Aim at Each Other • Making Money July 5 2019 by Patrice Lee Onwuka
Supreme Court Reaches Verdict on 2020 Census Question • Your World with Neil Cavuto June 27 2019 by Patrice Lee Onwuka
Democrats Ignore Booming Economy and Foreshadow Recession • Evening Edit June 25 2019 by Patrice Lee Onwuka
Economists Determine the Value of "Free" Social Media Apps • Making Money June 24 2019 by Patrice Lee Onwuka
Race is a Central Issue for 2020, Biden is Learning the Hard Way • Coast to Coast June 20 2019 by Patrice Lee Onwuka
Hearing on Reparations Shows Just How Complicated This Issue Is • Trish Regan Primetime June 19 2019 by Patrice Lee Onwuka
President Trump Prioritizes Legal Immigrants During Border Crisis • Lou Dobbs Tonight June 18 2019 by Patrice Lee Onwuka
Lets Take an Objective Look at the Russian Interference Allegations • PoliticsNation June 15 2019 by Patrice Lee Onwuka
Liberals Distract Voters From Booming Economy with Social Issues • Lou Dobbs Tonight June 14 2019 by Patrice Lee Onwuka
Why Kamala Harris’ ‘Equal Pay’ Proposal Will Hurt, Not Help Women June 13 2019 by Patrice Lee Onwuka
Republicans and Democrats at odds on overtime pay June 13 2019 by Patrice Lee Onwuka
Deep Fake Videos Can Have Dangerous Repercussions • Evening Edit June 12 2019 by Patrice Lee Onwuka
Bernie Sanders Bashes Walmart Execs Claiming Minimum Wage Isn't High Enough • Coast to Coast June 5 2019 by Patrice Lee Onwuka
Mandated Paid Time Off Is Not Only Bad for Most Businesses, It's Bad for Many Women June 5 2019 by Patrice Lee Onwuka
Democrats Throw Money at Homelessness But the Situation Still Gets Worse, Why? • Lou Dobbs Tonight June 5 2019 by Patrice Lee Onwuka
Stossel: The Paid Leave Fairy Tale • Reason TV June 4 2019 by Patrice Lee Onwuka
Crisis at the Border and Draining the Swamp, Trust Trump to Create Change • Making Money May 31 2019 by Patrice Lee Onwuka
Kamala Harris’ New Proposal to Punish Companies Renews Wage Gap Debate • Bold Politics May 24 2019 by Patrice Lee Onwuka
Conflict in D.C. Does Effect Markets, But Consumers Continue to Spend • Coast to Coast May 23 2019 by Patrice Lee Onwuka
The Gender Wage Gap and What You Can Do About It May 22 2019 by Patrice Lee Onwuka
Top Dems Blow Any Chance of Bipartisan Reform • Evening Edit May 22 2019 by Patrice Lee Onwuka
Women's Organization Announces The 2019 Empowered Woman of the Year May 21 2019 by Patrice Lee Onwuka
Kamala Harris Threatens to Fine Businesses that Contribute to the "Pay Gap" • Fox & Friends May 21 2019 by Patrice Lee Onwuka
Low Birth Rates Could Have Devastating Economic Impacts • Fox & Friends May 16 2019 by Patrice Lee Onwuka
Policy analyst on falling birth rates: Rhetoric from the left contributing to 'change in norms' May 16 2019 by Patrice Lee Onwuka
Sanctuary Cities Putting Illegal Immigrants Ahead of Citizens • Lou Dobbs Tonight May 8 2019 by Patrice Lee Onwuka
Trump Should Look to Private Sector for Federal Reserve Board Nominees • Coast to Coast May 2 2019 by Patrice Lee Onwuka
Every Aspect of the Economy is Thriving From Jobs, Wages, To Taxes• Fox & Friends First April 29 2019 by Patrice Lee Onwuka
Bernie Struggles to Win Over Black Women • Evening Edit April 25 2019 by Patrice Lee Onwuka
Patrice Onwuka on Plastic Bag Bans, Net Neutrality and Michelle Obama • Americhicks April 23 2019 by Patrice Lee Onwuka
Many Americans Don't Care About the Mueller Report • Cavuto Live April 20 2019 by Patrice Lee Onwuka
Obama Playbook Will Not Be Enough to Get Joe Biden in the White House • Fox & Friends April 20 2019 by Patrice Lee Onwuka
Mueller Report Finally Available: What did it say and what is next? • WAMU April 18 2019 by Patrice Lee Onwuka
Opportunity Zones Create Economic Prosperity • Making Money with Charles Payne April 17 2019 by Patrice Lee Onwuka
Left Tries Desperately To Keep Collusion Narrative Alive • Coast to Coast April 15 2019 by Patrice Lee Onwuka
As support for reparations grow, so does pushback from some black Americans April 12 2019 by Patrice Lee Onwuka
Democrats Are Selective When Deciding Who is Innocent or Guilty • Fox & Friends Saturday April 6 2019 by Patrice Lee Onwuka
Being Fiscally Conservative Includes Being Smart on All Aspects of the Economy • Coast to Coast April 3 2019 by Patrice Lee Onwuka
Don't Fall for the Equal Pay Day Myth • Larry O'Connor Show April 2 2019 by Patrice Lee Onwuka
Presidential Hopefuls Playing Politics with Joe Biden's Victims • Evening Edit April 1 2019 by Patrice Lee Onwuka
Joe Biden's 2020 Doom • Cheddar Politics April 1 2019 by Patrice Lee Onwuka
Marco Rubio's New Family Leave Bill is a Budget Neutral Conservative Plan • Coast to Coast March 27 2019 by Patrice Lee Onwuka
The gender pay gap still exists - but it's shrinking, according to a new study March 27 2019 by Patrice Lee Onwuka
Millions of American off food stamps under President Trump • OANN March 23 2019 by Patrice Lee Onwuka
The Electoral College Matters, and Here's Why • The Evening Edit March 20 2019 by Patrice Lee Onwuka
Threats to Drag Out Investigation Despite No Evidence Of Collusion • The Evening Edit March 13 2019 by Patrice Lee Onwuka
A Legislative Unicorn: Paid Family Leave Without New Taxes or Entitlements March 12 2019 by Patrice Lee Onwuka
Congress considers parental leave programs March 12 2019 by Patrice Lee Onwuka
The Ivanka Trump effect: Why Republicans are now championing paid family leave, and what their new proposal looks like March 12 2019 by Patrice Lee Onwuka
Joni Ernst, Mike Lee roll out conservative paid parental leave idea March 12 2019 by Patrice Lee Onwuka
Raising the Topic of Paid Parental Leave March 12 2019 by Patrice Lee Onwuka
Republicans Propose Plan to Let Parents Use Social Security for Paid Family Leave After Having a Child March 12 2019 by Patrice Lee Onwuka
GOP senators unveil paid parental leave proposal March 12 2019 by Patrice Lee Onwuka
Support Choices That Women Make When Looking at Gender Pay Gap • Making Money with Charles Payne March 8 2019 by Patrice Lee Onwuka
Trump Previews 2020 in CPAC Speech • Fox and Friends First March 4 2019 by Patrice Lee Onwuka
Consumer Confidence Bouncing Back as Trump Looks Ahead to 2020 • Coast to Coast March 4 2019 by Patrice Lee Onwuka
Pelosi Should Second Guess New Member's Calls for Trump Impeachment • The Evening Edit February 27 2019 by Patrice Lee Onwuka
AOC Lives in Luxury Housing in D.C. Despite Her Views on Affordable Housing • Fox and Friends First February 21 2019 by Patrice Lee Onwuka
Bernie Attacks Someone Who Achieved the American Dream • Making Money with Charles Payne February 20 2019 by Patrice Lee Onwuka
Bernie Sanders Continues to Misrepresent the Reality of Socialism • The Evening Edit February 19 2019 by Patrice Lee Onwuka
Meet the Women Who Protested the Women’s March January 21 2019 by Patrice Lee Onwuka
Trump Putting Border Security on the Table with DACA Extension • America's News Headquarters January 19 2019 by Patrice Lee Onwuka
Women's March is Not Inclusive of All Ideas • Fox & Friends January 19 2019 by Patrice Lee Onwuka
USPS Relies on Amazon: What this means for consumers • Your World with Neil Cavuto December 5 2018 by Patrice Lee Onwuka
How Will Markets React to a Potential Trade Deal with China? • Coast to Coast November 29 2018 by Patrice Lee Onwuka
Streisand accuses Trump of misogyny while liberals turn on their own November 28 2018 by Patrice Lee Onwuka
What Pete Davidson’s SNL Apology Could Teach Us About Civility • Bold Politics November 16 2018 by Patrice Lee Onwuka
Is Amazon Guilty of Taking Part In Corporate Welfare? • The Evening Edit November 14 2018 by Patrice Lee Onwuka
'Economic damage' predicted after city levies big tax on hi-tech firms November 13 2018 by Patrice Lee Onwuka
Markets React to Midterm Election Results • Coast to Coast November 12 2018 by Patrice Lee Onwuka
Volatile Markets Will Only Marginally Influence Midterms • The Evening Edit October 26 2018 by Patrice Lee Onwuka
PETA Claims Milk is Racist • Trish Regan Primetime October 23 2018 by Patrice Lee Onwuka
How Will Booming Economy Be Affected by The Fed? • Coast to Coast October 17 2018 by Patrice Lee Onwuka
Kanye Brings Attention to Real Economic Issues • Coast to Coast October 11 2018 by Patrice Lee Onwuka
Will Kavanaugh Confirmation Drive Women Away from the GOP? Not Likely October 8 2018 by Patrice Lee Onwuka
Anti-Kavanaugh Harassment Tactics Overlooked by the Left • Coast to Coast October 4 2018 by Patrice Lee Onwuka
Statement on #CancelKavanaugh Protest: Activists Use Dr. Ford to Exploit Emotion Over a Deeply Personal Issue October 4 2018 by Patrice Lee Onwuka
Republican Women Explain Why They Stand By Brett Kavanaugh More Than Ever October 3 2018 by Patrice Lee Onwuka
Women Want Kavanaugh Confirmed if FBI Finds No Corroborating Evidence • FBN am October 3 2018 by Patrice Lee Onwuka
A Tale of Two Cities: A Snapshot of Washington D.C. On the Historic Kavanaugh-Ford Hearing Day September 27 2018 by Patrice Lee Onwuka
FBI Investigation Was an Ongoing Distraction During Hearings • After the Bell September 27 2018 by Patrice Lee Onwuka
'There's No Presumption of Innocence': Women Rally Both for and Against Kavanaugh September 27 2018 by Patrice Lee Onwuka
Dems Would Love for Trump to Fire Rosenstein • Fox & Friends First September 25 2018 by Patrice Lee Onwuka
Republicans Have Gone Out of Their Way to Try to Hear Blasey • Coast to Coast September 21 2018 by Patrice Lee Onwuka
Jacques: Sexism is a losing strategy September 14 2018 by Patrice Lee Onwuka
Good Trade Relations with China is Good for the Economy • Coast to Coast September 12 2018 by Patrice Lee Onwuka
Will Reforming SNAP Benefit Americans? • Bold Politics September 7 2018 by Patrice Lee Onwuka
Motivations Behind Trying to Flip a Russian Oligarch • Fox & Friends First September 3 2018 by Patrice Lee Onwuka
More Than Ever, Millennials Are Taking Care of Their Aging Family Members • Bold Life August 24 2018 by Patrice Lee Onwuka
Trump's Commitment to Diversity in the White House Continues After Omarosa • Andrew Wilkow Show August 22 2018 by Patrice Lee Onwuka
Statement: IWF Applauds Roll Back of Clean Power Plan August 22 2018 by Patrice Lee Onwuka
Omarosa’s Antics Won't Undermine Trump’s Progress for Black Americans • Afternoons with Cliff Kelley August 21 2018 by Patrice Lee Onwuka
NYT Publishes Pure Speculation on Mueller Investigation • Fox & Friends First August 20 2018 by Patrice Lee Onwuka
America IS Great: a lesson from real Immigrants • Fox & Friends August 18 2018 by Patrice Lee Onwuka
Harley-Davidson Staying in U.S. is Huge Victory • Coast to Coast August 17 2018 by Patrice Lee Onwuka
Is there a racial gender pay gap? • Bold Politics August 10 2018 by Patrice Lee Onwuka
Ocasio-Cortez Forgets Socialism Didn't Work in South America • Your World with Neil Cavuto August 9 2018 by Patrice Lee Onwuka
Rubio-Wagner family leave bill would draw from Social Security August 7 2018 by Patrice Lee Onwuka
Explaining the Economic Security for New Parents Act • KDKA August 3 2018 by Patrice Lee Onwuka
Rubio's New Paid Leave Bill is Great for Women • Fox & Friends First August 2 2018 by Patrice Lee Onwuka
Bernie Sanders "Medicare for All" plan would be a disaster • Your World with Neil Cavuto July 31 2018 by Patrice Lee Onwuka
How will new GDP numbers influence the midterm elections • Coast to Coast July 30 2018 by Patrice Lee Onwuka
New Bill Aims to Extend Trump Tax Cuts Beyond 2025 • Fox & Friends First July 25 2018 by Patrice Lee Onwuka
Sessions prepares students for diverse opinions at college • Making Money with Charles Payne July 24 2018 by Patrice Lee Onwuka
Student loan debt saddling non-saving Millennials July 24 2018 by Patrice Lee Onwuka
Only 4.9 Percent of People Have Two Jobs • Fox & Friends First July 19 2018 by Patrice Lee Onwuka
Companies thinking about intellectual theft when dealing with China • Coast to Coast July 18 2018 by Patrice Lee Onwuka
Cyber Security Needs to be Taken Seriously During Trump-Putin Summit • FBN am July 16 2018 by Patrice Lee Onwuka
What Do You Think Of Seattle's Plastic Straw Ban? • Bold Life July 6 2018 by Patrice Lee Onwuka
Hopeful About North Korean Cooperation • FBN am July 6 2018 by Patrice Lee Onwuka
#133 Reforms needed for children's television July 2 2018 by Patrice Lee Onwuka
How will Supreme Court nominations effect the primaries? • Coast to Coast June 29 2018 by Patrice Lee Onwuka
Trump is no excuse for intimidation of women June 28 2018 by Patrice Lee Onwuka
Protesters Want Media Attention, Not Policy Change • Your World with Neil Cavuto June 22 2018 by Patrice Lee Onwuka
Immigration bills will be hard to pass, but we can't ignore this issue • FBN AM June 20 2018 by Patrice Lee Onwuka
We should be concerned that foreign agents had access to Hillary's emails • Fox & Friends First June 15 2018 by Patrice Lee Onwuka
Free Trade Without Free Riders • Coast to Coast June 7 2018 by Patrice Lee Onwuka
Democrats Playing Politics in Primaries • Making Money with Charles Payne June 4 2018 by Patrice Lee Onwuka
IWF Statement on New Steel and Aluminum Tariffs May 31 2018 by Patrice Lee Onwuka
Trump had good reason to pull out of meeting with North Korea • Making Money with Charles Payne May 24 2018 by Patrice Lee Onwuka
Should Medical Marijuana Be Allowed in Classrooms? • Fox & Friends First May 21 2018 by Patrice Lee Onwuka
Media Bias After They Refuse to Accept Trump's Corrected Statement • Your World with Neil Cavuto May 17 2018 by Patrice Lee Onwuka
Lower Birth Rates: Student Debt, Societal Change, Technological Advances • The Intelligence Report May 17 2018 by Patrice Lee Onwuka
China's Relationship with North Korea Could Make All The Difference • FBN am May 16 2018 by Patrice Lee Onwuka
#127 Questioning the Status Quo: Black voters don't owe their vote to either party May 14 2018 by Patrice Lee Onwuka
Were there problems with Cohen's payments? • Coast to Coast May 10 2018 by Patrice Lee Onwuka
Removing "In God We Trust" to be Politically Correct • Fox & Friends First May 7 2018 by Patrice Lee Onwuka
Proof Mueller is Fishing for Evidence • FBN am May 7 2018 by Patrice Lee Onwuka
We Want to Ensure a Legitimate and Lawful Immigration Process • FBN am April 26 2018 by Patrice Lee Onwuka
Nancy Pelosi Chokes on "Crumbs" Comment Because It Just Isn't True • Coast to Coast April 25 2018 by Patrice Lee Onwuka
#123 Education Savings Accounts for Military Families Can Make All the Difference April 23 2018 by Patrice Lee Onwuka
Why are so many young adults choosing to stay at home? • BoldTV April 20 2018 by Patrice Lee Onwuka
Gov. Brown is Choosing Sanctuary over Safety • Risk & Reward April 19 2018 by Patrice Lee Onwuka
Comey has made enemies on both sides of the aisle • Risk & Reward April 19 2018 by Patrice Lee Onwuka
Bias training won't do much at Starbucks • Intelligence Report with Trish Reagan April 19 2018 by Patrice Lee Onwuka
Tax Cuts Will Play Major Role in Midterms • Coast to Coast April 5 2018 by Patrice Lee Onwuka
New Survey: What would Americans do for a 10% raise? • Fox & Friends First April 5 2018 by Patrice Lee Onwuka
Trump to rescind spending with a maneuver Clinton used nearly 20 years ago • FBN:am April 5 2018 by Patrice Lee Onwuka
Citizenship Question MUST be Answered Truthfully • Cavuto Live April 2 2018 by Patrice Lee Onwuka
#118 New Amendment Provides Protection Against the Discrimination of Pregnant Women April 2 2018 by Patrice Lee Onwuka
On the New Spending Bill: Entitlements Drive the National Debt • Risk & Reward March 23 2018 by Patrice Lee Onwuka
Regulations will Hinder Innovation on Social Media • Risk & Reward March 23 2018 by Patrice Lee Onwuka
White House Millennials Conference to Address Economy, Free Speech, and Opioids March 18 2018 by Patrice Lee Onwuka
In the #MeToo era, young conservative women look for their spot March 16 2018 by Patrice Lee Onwuka
Shake Ups in White House Happening for a Reason • FBN AM March 15 2018 by Patrice Lee Onwuka
Dems Will Not Continue to Embrace Middle America • Coast to Coast March 14 2018 by Patrice Lee Onwuka
We Don't Want to Abandon Free Trade • Cavuto Live March 10 2018 by Patrice Lee Onwuka
GOP takes on paid family leave March 10 2018 by Patrice Lee Onwuka
Issues in California are a Classic Case of Federalism • Risk & Reward March 9 2018 by Patrice Lee Onwuka
A Trade War is Going to Undermine Current Economic Progress • Coast to Coast March 5 2018 by Patrice Lee Onwuka
#113 What You Don't Know About Immigrants: A real immigrant's story March 2 2018 by Patrice Lee Onwuka
A conservative solution to paid leave • The National Discourse Podcast February 27 2018 by Patrice Lee Onwuka
CNN Pushes Agenda at Townhall • Risk & Reward February 23 2018 by Patrice Lee Onwuka
Americans Recognize the 2nd Amendment is Important to Protect • Coast to Coast February 20 2018 by Patrice Lee Onwuka
Republicans Produced Plan, Waiting for Action Across the Aisle • FBN AM February 13 2018 by Patrice Lee Onwuka
#MeToo and Working Women • PBS To The Contrary February 10 2018 by Patrice Lee Onwuka
Fundamentals of Our Economy Are Strong • Coast to Coast February 8 2018 by Patrice Lee Onwuka
Republicans Tired of Partisan Bickering • Coast to Coast January 31 2018 by Patrice Lee Onwuka
Unions Losing Battle Against Inevitable Future • The Intelligence Report January 25 2018 by Patrice Lee Onwuka
Did Schumer Cave on Gov Shut Down? • FBN am January 23 2018 by Patrice Lee Onwuka
Dems ‘Big Risk’ Not Worth It • Your World January 22 2018 by Patrice Lee Onwuka
Media Attacks Trump's Doctor • Risk & Reward January 19 2018 by Patrice Lee Onwuka
The Left Criticizing Tax Plan and Immigration • Risk & Reward January 12 2018 by Patrice Lee Onwuka
Concerns on Trade with China • Coast to Coast January 10 2018 by Patrice Lee Onwuka
Weighing in on Trump vs. Bannon • FBN AM January 5 2018 by Patrice Lee Onwuka
U.N. Funding Cut Good for Women • Forbes on Fox December 29 2017 by Patrice Lee Onwuka
Next Moves for the GOP • Coast to Coast December 26 2017 by Patrice Lee Onwuka
Saving Money After the Holidays • Fox & Friends First December 26 2017 by Patrice Lee Onwuka
Senate Passes Tax Bill • FBN AM December 20 2017 by Patrice Lee Onwuka
Are Black Women Leaving The Democratic Party to Become Independents? December 19 2017 by Patrice Lee Onwuka
Final GOP Tax Bill • Risk & Reward December 15 2017 by Patrice Lee Onwuka
Statement: Unnecessary internet regulation no more, FCC right to overturn net neutrality rules December 14 2017 by Patrice Lee Onwuka
Calls for Conyers to Resign • Risk & Reward Pt. 2 December 1 2017 by Patrice Lee Onwuka
Steinle's Killer Acquitted • Risk & Reward Pt. 1 December 1 2017 by Patrice Lee Onwuka
FCC Net Neutrality Vote Looms: How could things change? • The Big Picture December 1 2017 by Patrice Lee Onwuka
Time for Transparency: Taxpayer dollars used to protect Congressional misbehavior • Coast to Coast November 30 2017 by Patrice Lee Onwuka
More Bang for Your Buck • Fox & Friends First November 30 2017 by Patrice Lee Onwuka
Are some Dems willing to sacrifice protections for women to hold onto power? • Intelligence Report November 28 2017 by Patrice Lee Onwuka
Federal Judge blocks Trump's EO on Sanctuary Cities: Is Trump at a disadvantage? • Bulls & Bears November 25 2017 by Patrice Lee Onwuka
Has NFL commissioner lost control of the hemorrhaging league? • Bulls & Bears November 25 2017 by Patrice Lee Onwuka
Black Friday protests dwindle exposing this uncomfortable truth • Bulls & Bears November 25 2017 by Patrice Lee Onwuka
Sexual harassment is wrong no matter when it occurred and by whom • Varney & Co. November 24 2017 by Patrice Lee Onwuka
Bernie Sanders Blatantly Gives Bill Clinton, Al Franken a Pass on Sexual Harassment November 20 2017 by Patrice Lee Onwuka
There's Nothing Glamorous about “Auntie" Maxine Waters & Her “Impeach Him!” Chant November 17 2017 by Patrice Lee Onwuka
Twitter cracking down on Verified accounts • Fox & Friends First November 17 2017 by Patrice Lee Onwuka
'Saturday Night Live' says DNC is old and out of touch • Risk & Reward November 13 2017 by Patrice Lee Onwuka
Tax reform push: Will there be compromise? • FBN:AM November 13 2017 by Patrice Lee Onwuka
Are tech companies truly neutral platforms or do they reflect bias? • Fox & Friends First November 3 2017 by Patrice Lee Onwuka
Statement: GOP Tax Bill is a Win for Women November 2 2017 by Patrice Lee Onwuka
CNBC Slams DNC Chair Tom Perez, 'Get Off The Talking Points' On GOP Tax Plan November 2 2017 by Patrice Lee Onwuka
We cannot choose our history but we can choose not to repeat it • Risk & Reward October 31 2017 by Patrice Lee Onwuka
We're seeing the waning days of the NFL protests • Risk & Reward October 31 2017 by Patrice Lee Onwuka
This tax reform plan will help all workers • FBN:AM October 30 2017 by Patrice Lee Onwuka
West Wing Reads for 10/30/17 October 30 2017 by Patrice Lee Onwuka
Will any democrats get on board with tax reform? • FBN AM October 26 2017 by Patrice Lee Onwuka
What role is Bannon playing outside of the White House? • MSNBC Live October 23 2017 by Patrice Lee Onwuka
Podcast #98 Why sexual assault statistics on college campuses are misleading October 19 2017 by Patrice Lee Onwuka
Will new Trump/McConnell bromance last to get agenda through? • FBN AM October 17 2017 by Patrice Lee Onwuka
Hillary caring about "women's issues" a sham? • The Intelligence Report October 16 2017 by Patrice Lee Onwuka
More sexual harrassment allegations come out in Hollywood • Risk & Reward October 13 2017 by Patrice Lee Onwuka
Wait–Are Black Women Leaving The Democratic Party? October 11 2017 by Patrice Lee Onwuka
Eminem bashes Trump in new music video, but does it matter? • Coast To Coast October 11 2017 by Patrice Lee Onwuka
Statement: Abandoning The Clean Power Plan: A Destructive Regulatory Package October 10 2017 by Patrice Lee Onwuka
These four GOP Senators won't vote for Graham-Cassidy • FBN AM September 26 2017 by Patrice Lee Onwuka
These celebrities are joining in on the NFL anthem protests... • Risk & Reward September 26 2017 by Patrice Lee Onwuka
This sport is standing up for the National Anthem... • Risk & Reward September 25 2017 by Patrice Lee Onwuka
Does Graham-Cassidy healthcare bill have a chance? • FBN AM September 21 2017 by Patrice Lee Onwuka
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Policy Focus: Your Ideas, Your Rights - Intellectual Property in the 21st Century June 25 2017 by Patrice Lee Onwuka | 2019-12-13T23:04:26 | {
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https://questioncove.com/updates/4fc1326be4b0964abc832226 | Mathematics
OpenStudy (anonymous):
Find the angle between the vectors u and v if u = (1, 2) and v = (- 4, - 2)
jimthompson5910 (jim_thompson5910):
Use the formula $\Large \theta = \arccos\left(\frac{u\cdot v}{|u||v|}\right)$ where u and v are vectors.
OpenStudy (anonymous):
which number in my points do i use for each... for example which number in (1,2) would I use for which u
jimthompson5910 (jim_thompson5910):
In the numerator of that fraction, you're computing the dot product between u and v
jimthompson5910 (jim_thompson5910):
So what is the dot product between u = (1, 2) and v = (- 4, - 2)?
OpenStudy (anonymous):
-8
jimthompson5910 (jim_thompson5910):
good
jimthompson5910 (jim_thompson5910):
now you need to find |u| and |v|
OpenStudy (anonymous):
how do I do that?
jimthompson5910 (jim_thompson5910):
use the formula |x| = sqrt(a^2+b^2) where the vector x is x = (a,b)
OpenStudy (anonymous):
so it is sqrt(1^2+2^2) for u's and then sqrt(-4^2+-2^2) for v's?
jimthompson5910 (jim_thompson5910):
yes, so what do you get?
OpenStudy (anonymous):
i think u's is sqrt(5) and v's is 2sqrt(5)
jimthompson5910 (jim_thompson5910):
You got it, nice work So what is |u||v|?
OpenStudy (anonymous):
10... so then its arccos(-8/10)?
jimthompson5910 (jim_thompson5910):
yes
OpenStudy (anonymous):
which turns out to be 143.13degrees
jimthompson5910 (jim_thompson5910):
you nailed it
OpenStudy (anonymous):
You keep helping me sooo much. Thanks, I am taking a Pre-Calculus course by myself and sometimes it can be difficult to figure out all of the questions by myself :).
jimthompson5910 (jim_thompson5910):
you're very welcome, that must be pretty crazy lol
OpenStudy (anonymous):
yep... I'm almost done though (thankfully haha)
jimthompson5910 (jim_thompson5910):
well that's good
OpenStudy (anonymous):
is there a difference between a u surrounded by two of the straight lines as opposed to the just one in the question you just helped me with?
jimthompson5910 (jim_thompson5910):
Well technically it should have been $\Large \|u\|$ since the two vertical bars denote the norm or magnitude of a vector. But the absolute value bars say the same thing (they're just used in a more general sense). So either work in my opinion.
OpenStudy (anonymous):
oh, ok... thanks for clarifying. I am working on another type of question like that and wasn's sure if there was a difference.
jimthompson5910 (jim_thompson5910):
there's not much of a difference really | 2020-05-28T11:02:19 | {
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http://math.stackexchange.com/questions/237073/playing-roulette | # Playing roulette
Suppose we have a roulette wheel with $38$ total slots ($18$ black/red, $2$ neither). The ball lands in one slot selected uniformly at random, independent of all previous spins. An $\$x$bet on "black" pays$\$2x$ if the ball lands on black, and $\$0$otherwise. If you bet$1 on black for 100 consecutive spins, how much money will you end up with in expectation?
You walk up to the Roulette table with $\$15$and intend to walk away with$\$16$ using the following betting strategy. You will first bet $\$1$on black. If you win you have$\$16$ and walk away, otherwise you have $\$14$and bet$\$2$ on black on the next spin. If you win that bet you walk away, otherwise you bet $\$4$on the next spin. If you win that bet you walk away, otherwise you bet$\$8$ on the next spin and walk away, either with $\$16$if you win or$\$0$ if you lose.
What's the probability you will walk away with $\$16$? How much money are you expected to walk away with? My biggest problem with these kinds of questions is figuring out how to choose the random variable to calculate the expected values. I understand the formula for calculating the expected value, but translating a problem into those terms has been giving me a hard time. - ## 2 Answers Let's first look at the first question: If you bet$1 on black for 100 consecutive spins, how much money will you end up with in expectation?
So you want to know what your final return will be, at the end of 100 spins. Call this $R$. That is just giving it a name, but what is your final return? You can see that it is the sum of the returns from each bet. So let the return on the $i$th bet be $R_i$, then note that $R = R_1 + R_2 + \dots + R_{100}$. So the expected return is $E[R] = E[R_1] + E[R_2] + \dots + E[R_{100}]$ by linearity of expectation.
So to calculate $E[R]$, we'll be done if we calculate each $E[R_i]$. Let's try to calculate a particular $E[R_i]$. You bet $1$ dollar, and you get back $2$ if the ball lands on black, and $0$ if it doesn't. In other words, you gain $1$ dollar if it lands on black, and lose $1$ dollar if it doesn't. The probability of the former is $18/38$, and that of the latter is $20/38$. In other words, $R_i$ is $1$ with probability $18/38$, and $-1$ with probability $20/38$, so the expected value of $R_i$ is $E[R_i] = \frac{18}{38}(-1) + \frac{20}{38}(1) = \frac{-2}{38}$. Now, as this is the same for each $R_i$, we have $E[R] = E[R_1] + E[R_2] + \dots + E[R_{100}] = \left(\frac{-2}{38}\right)100 \approx -5.26$.
For the second question, let the amount you walk away with be $W$. Let $p = 18/38$, the probability that your bet on black succeeds. There are $5$ possible outcomes:
• you win your first bet: probability $p$
• you lose your first bet, and win your second: probability $(1-p)p$
• you lose your first two bets, and win the third: probability $(1-p)^2p$
• you lose your first three bets, and win the fourth: probability $(1-p)^3p$
• you lose all four bets: probability $(1-p)^4$
In the first four outcomes, you walk away with $16$ dollars, so the probability of that happening (let's call it $q$) is $q = p + (1-p)p + (1-p)^2p + (1-p)^3p = 1 - (1-p)^4 = 1 - (20/38)^4 \approx 0.92$.
[More simply, you could think of it as just two outcomes: (a) that you win some bet, which has probability $q = 1 - (1-p)^4$, and (b) that you win no bet (lose all bets), which has probability $(1-p)^4$.]
In other words, $W$ is $16$ with probability $q$, and $0$ with probability $1-q$. So the expected amount of money you walk away with is therefore $E[W] = q(16) + (1-q)0 = (1-(1-p)^4)16 \approx 14.77$.
[Aside: Note that this is less than the $15$ you came in with. This shows that you can't win in expectation even with your clever betting strategy; a consequence of the optional stopping theorem.]
-
Probability of getting a black on any particular spin is 18 in 38, ie $p = \frac{18}{38} = \frac{9}{19} = 0.474$. Let $X$ be the random variable 'number of blacks in 100 consecutive spins'. Then the expected value of X is $E(X) = 100p = \frac{900}{19} = 47.37$, that is, you expect to win 47.37 times in 100 spins. Since you get $\$$2 for each \$$1 bet, you will therefore expect to walk away with$\$2\times 47.37 = \$94.74$. For the second part of your question, the only way you can lose is if for four consecutive spins, you get non-blacks each time. So what is the probability of losing 4 times in a row? Well, for 1 spin, the probability of not getting black is$1-p = \frac{20}{38} = \frac{10}{19} = 0.526$so for 4 consecutive spins, the probability of not getting any black is$(\frac{10}{19})^{4} = \frac{10,000}{130,321} = 0.0767$. That is, you have about a 7.7% chance of losing with this betting strategy. That is, you have 92.3% chance of walking away with$\16.
-
Can you elaborate on why the expected value is just $100p$? Particularly, how does the linearity of expectations apply here? If we let $X_i = 1$ if we land on black and $X_i = 0$ if we don't land on black, how can we calculate the expected value then? – user1038665 Nov 14 '12 at 10:53
Also, for the second part, how would we calculate the "amount of money you are expected to walk away with"? – user1038665 Nov 14 '12 at 10:55
Expected value of a bet = wager x probability. – theo Nov 14 '12 at 11:02
If you bet 1 dollar 100 times on black you will win about 47.37 times and thus leave the casino with 94.74 of your original 100 dollars. – Hagen von Eitzen Nov 14 '12 at 11:03
The answer given is -\$(2/38)*(100), so I believe this is incorrect. – user1038665 Nov 14 '12 at 11:20 | 2014-11-26T02:11:33 | {
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https://stats.stackexchange.com/questions/507040/how-to-create-a-continuous-distribution-on-a-b-with-textmean-textmo?noredirect=1 | # How to create a continuous distribution on $[a, b]$ with $\text{mean} = \text{mode} = c$?
I would like to sample from a distribution on a specific domain $$[a, b]$$ with mean $$c$$ where $$a < c < b$$. Ideally, I would also like the mode (i.e. the "peak") of the distribution to be $$c$$ as well.
How to derive this distribution?
The best I can do is a triangular distribution with my desired mean, but the peak is in the wrong place. I imagine the distribution I want is smooth and skewed.
EDIT: If $$c$$ were at the midpoint of $$a$$ and $$b$$ then I think a truncated normal distribution is roughly what I'm looking for. So something similar where the tails go smoothly to zero at the endpoints of my interval when $$c$$ is not in the middle would be nice.
EDIT 2: I'm looking to perturb some parameters in a simulation, so really anything reasonable should be fine.
• There are infinitely many possible solutions, which enables you to apply stricter criteria to narrow it down. To that end, could you explain what use you intend to make of this family of distributions? BTW, "skewed" strongly suggests the peak should not coincide with the mean.
– whuber
Jan 28 at 20:11
• Good point. I did my best to clarify. Jan 28 at 20:37
• There are still infinitely many solutions, but another way would be to generate a beta RV with parameters that give it the desired skew and unimodality, and then shift and scale the whole distribution so it lies in [a,b].
– Noah
Jan 28 at 20:50
• Thanks! That looks like it will do just fine. Jan 28 at 20:53
• @Noah That looks a lot like an acceptable answer!
– Dave
Jan 28 at 20:53
I will describe every possible solution. This gives you maximal freedom to craft distributions that meet your needs.
Basically, sketch any curve you like for the density function $$f$$ that meets your requirements. Separately scale the heights of the left and right halves of it (on either side of $$c$$) to make their masses balance, then scale the heights overall to make it a probability density.
Here are some details.
Because the distribution is continuous, it has a density function $$f$$ with finite integral. By splitting $$f$$ at $$c$$ we can construct all such functions out of two separately chosen non-negative non-decreasing, not identically zero, integrable functions $$f_1$$ and $$f_2$$ defined on $$[0,1].$$ Here, for example, are two such functions:
Set $$f$$ to agree affinely with $$f_1$$ on $$[a,c]$$ and with the reversal of $$f_2$$ on $$[c,b].$$ This means there are two positive numbers $$\pi_i$$ for which
$$f\mid_{[a,c]}(x) = \pi_1 f_1\left(\frac{x-a}{c-a}\right);\quad f\mid_{[c,b]}(x) = \pi_2 f_2\left(\frac{b-x}{b-c}\right).$$
This construction guarantees $$c$$ is the unique mode. Moreover, if you want the tails to taper to zero, just choose $$f_i$$ that approach $$0$$ continuously at the origin.
For $$f$$ to be a probability density it must integrate to unity:
$$1 = \int_a^b f(x)\,\mathrm{d}x =\pi_1(c-a) \mu_1^{(0)} + \pi_2(b-c) \mu_2^{(0)}\tag{*}$$
where
$$\mu_i^{(k)} = \int_0^1 x^k f_i(x)\,\mathrm{d}x.$$
Moreover, $$c$$ is intended to be the mean, whence
\begin{aligned} c &= \int_a^b x f(x)\,\mathrm{d}x \\ &= \pi_1(c-a)((c-a)\mu_1^{(1)} + a\mu_1^{(0)}) + \pi_2(b-c)(-(b-c)\mu_2^{(1)} + b\mu_2^{(0)}). \end{aligned}\tag{**}
$$(*)$$ and $$(**)$$ establish a system of two linear equations in the $$\pi_i.$$ You can check it has nonzero determinant and a unit positive solution $$(\pi_1,\pi_2).$$ (This checking comes down to the fact that after normalizing the $$f_i$$ to be probability densities, their means $$\mu_i^{(1)}$$ must be in the interval $$[0,1].$$)
This figure plots the solution $$f$$ for the previous two functions where $$[a,b]=[-1,3]$$ and $$c=1/2:$$
By design, $$f$$ looks like $$f_1$$ to the left of $$c$$ and like the reversal of $$f_2$$ to the right of $$c.$$
The spike at the mode might bother you, but notice that it was never assumed or required that $$f$$ must be continuous. Most examples will be discontinuous. However, every distribution $$f$$ meeting your specifications can be constructed this way (by reversing the process: split $$f$$ into two halves, which obviously determine the $$f_i$$).
To demonstrate how general the $$f_i$$ can be, here is the same construction where the $$f_i$$ are (inverse) Cantor functions (as implemented in binary.to.ternary at https://stats.stackexchange.com/a/229561/919). These are not continuous anywhere.
NB: $$f_1$$ is binary.to.ternary; $$f_2(x) = f_1(x^{2.28370312}).$$ This illustrates one way to eliminate the discontinuity at $$c:$$ $$f_2$$ was selected from the one-parameter family of functions $$f_1(x^p),$$ $$p \gt 0,$$ and a value of $$p$$ was identified to make the left and right limits of $$f$$ at $$c$$ equal. This family "pushes" the mass of $$f_1$$ left and right by a controllable amount, thereby modifying the amount by which the right tail is scaled (vertically) in constructing $$f.$$
For those who would like to experiment, here is an R function to create $$f$$ out of the $$f_i$$ and code to create the figures. Three commands at the end check whether (1) $$f$$ is a pdf, (2) its mean is $$c,$$ and (3) its mode is $$c.$$
#
# Given numbers a. < b. < c. and non-negative, non-decreasing, not identically
# zero functions f.1 and f.2 defined on [0,1], construct a density function f
# on the interval [a., b.] with mean c. and unique mode c. that behaves like f.1
# to the left of c. and like the reversal of f.2 to the right of c.
#
# ... are optional arguments passed along to integrate.
#
create <- function(a., b., c., f.1, f.2, ...) {
cat("Create\n")
p.1 <- integrate(f.1, 0, 1, ...)$$value p.2 <- integrate(f.2, 0, 1, ...)$$value
mu.1 <- integrate(function(u) u * f.1(u), 0, 1, ...)$$value mu.2 <- integrate(function(u) u * f.2(u), 0, 1, ...)$$value
A <- matrix(c(p.1 * (c.-a.), p.2 * (b.-c.),
(c.-a.) * ((c.-a.) * mu.1 + a.*p.1),
(b.-c.) * (-(b.-c.) * mu.2 + b.*p.2)),
2)
pi. <- solve(t(A), c(1, c.))
function(x) {
ifelse(a. <= x & x <= c., 1, 0) * pi.[1] * f.1((x-a.)/(c.-a.)) +
ifelse(c. < x & x <= b., 1, 0) * pi.[2] * f.2((b.-x)/(b.-c.))
}
}
#
# Example.
#
a. <- -1
b. <- 3
c. <- 1/2
f.2 <- function(x) x^(2/3)
f.1 <- function(x) exp(2 * x) - 1
f <- create(a., b., c., f.1, f.2)
#
# Display f.1 and f.2.
#
par(mfrow=c(1,2))
curve(f.1(x), 0, 1, lwd=2, ylab="", main=expression(paste(f[1], ": ", x^{2/3})))
curve(f.2(x), 0, 1, lwd=2, ylab="", main=expression(paste(f[2], ": ", e^{2*x}-1)))
#
# Display f.
#
x <- c(seq(a., c., length.out=601), seq(c.+1e-12*(b.-a.), b., length.out=601))
y <- f(x)
par(mfrow=c(1,1))
plot(c(x, b., a.), c(y, 0, 0), type="n", xlab="x", ylab="Density", main=expression(f))
polygon(c(x, b., a.), c(y, 0, 0), col="#f0f0f0", border=NA)
curve(f(x), b., a., lwd=2, n=1201, add=TRUE)
#
# Checks.
#
integrate(f, a., b.) # Should be 1
integrate(function(x) x * f(x), a., b.) # Should be c.
x[which.max(y)] # Should be close to c.
• Lovely! But a quibble: I do not think this captures distributions in which the 'left' or 'right' halves themselves contain nondifferentiable or discontinuous points—I am especially thinking of densely nondifferentiable or densely discontinuous curves—which are essentially 'undrawable' except by approximation. I am not actually convinced this is a defensible quibble, though. :) More just musing… Jan 29 at 6:28
• @Alexis I cannot see any exceptions: I made no conditions on the $f_i$ concerning differentiability or even continuity. For instance, both $f_i$ could be Cantor functions. I will add an illustration.
– whuber
Jan 29 at 15:13
• I was thinking about Cantor functions! Quibble withdrawn (and thank you for entertaining my musing :). Jan 29 at 18:32
All you need to do is come up with a two-parameter family of distributions, express the mean and mode in terms of those parameters, and then solve for them both being $$c$$. Working off your triangle idea, we can have a pentagon instead (straight lines from $$a$$ to $$c$$ and from $$c$$ to $$b$$), giving us three parameters ($$y_1$$ at $$a$$, $$y_2$$ at $$b$$, and $$y_3$$ at $$c$$), but we don't really have three degrees of freedom. We have that the total probability mass from $$a$$ to $$c$$ is $$\frac 1 2 (y_1+y_3)(c-a)$$, and the mass from $$c$$ to $$b$$ is $$\frac 1 2 (y_2+y_3)(c-b)$$, so we need to have $$\frac 1 2 (y_1+y_3)(c-a)+\frac 1 2 (y_2+y_3)(c-b)=1$$. Then you just need to make sure that the mean is $$c$$ as well.
• The problem with this approach is that it can fail even in the nicest circumstances (as I pointed out in a comment to the question). For instance, it does not work with the Beta family of distributions: there are no solutions unless the Beta parameters are equal.
– whuber
Jan 29 at 15:18 | 2021-11-27T11:32:06 | {
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https://math.stackexchange.com/questions/1585105/power-series-at-x-0 | # Power series at $x=0$
$$\sum_{n=0}^∞n!x^{2n}$$
This power series has radius of convergence $R=0$ since the ratio test shows it diverges for all $x≠0$.
But what if $x=0$? Then we have $\sum_{n=0}^\infty n! \cdot 0^{2n}$
Surely this series is undefined since the first partial sum is $0!\cdot 0^{2\cdot0}$ and $0^0$ is undefined, right?
Well, my lecturer claims this series is convergent and is equal to 1.
But $\sum_{n=1}^\infty n!\cdot 0^{2n}$ clearly converges to 0, so he is claiming that $0!\cdot 0^0=1$?
So is he wrong? Or am I missing something here?
• $0^0 = 1$ by convention. – user258700 Dec 22 '15 at 2:01
• Really? I've never heard that. I've always thought it was just undefined. Why isn't $\frac00$ similarly defined by convention? – Refnom95 Dec 22 '15 at 2:08
• It's only when dealing with limits that $0^0$ is indeterminant. – Michael Burr Dec 22 '15 at 2:09
• It's done mostly in the context of power series. It's much neater to write $\sum_i \text{bla}_ix^i$ than $\text{bla}_0+\sum_{i>0}...$. – YoTengoUnLCD Dec 22 '15 at 2:20
There are a lot of situations where it is convenient to define $$0^0=1$$ and of course $0!=1$, so $0!\cdot0^0=1$.
In some situations, it's most appropriate to leave $0^0$ undefined. In particular, the function $x^y$ converges to different limits as $x\to0$ and $y\to0$ depending on which path is taken, so analysis does not yield a natural value for $0^0$ and it may not make sense to define $0^0=1$.
However, when dealing with discrete exponents (natural numbers), it's usually better to define $0^0=1$, because this is consistent with combinatorial or set-theoretic interpretations of exponentiation. For instance, the set of functions from the empty set to the empty set has cardinality $1$, so $0^0=1$ by the set-theoretic definition. Also, using the repeated multiplication definition of exponentiation, an empty product is always $1$, even when the product contains only zeroes. (For more information, see this Wikipedia article.)
It is important to consider where the $0^0$ in the expression comes from. In your case, it's the first term of a power series. A power series' notion of exponentiation is closest to the repeated multiplication interpretation, so taking $0^0$ makes sense. Consider that a power series is just a generalization of a polynomial. How would you evaluate $f(x)=x^2+x+1$ at $x=0$? What if this is written $f(x)=x^2+x^1+x^0$? | 2019-05-20T06:20:11 | {
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http://mathhelpforum.com/number-theory/124509-help-proof.html | # Math Help - Help with a proof.
1. ## Help with a proof.
Hi, I'm stuck at proving the following question...
Prove that for all n>0,
1/2 + 2/2^2 + 3/2^3 + ... + n/2^n = 2 - (n+2)/2^n
I've tried all sorts of different ways of solving this, but to no avail.
Any help is appreciated
2. Originally Posted by seven.j
Hi, I'm stuck at proving the following question...
Prove that for all n>0,
1/2 + 2/2^2 + 3/2^3 + ... + n/2^n = 2 - (n+2)/2^n
I've tried all sorts of different ways of solving this, but to no avail.
Any help is appreciated
$\sum_{k=1}^{n}\frac{k}{2^k}$. Note that $\sum_{k=1}
^n x^k=\frac{x^{n+1}-x}{x-1}$
. Differentiating both sides and multiplying by $x$ gives $\sum_{k=1}^{n}k\cdot x^k=...$ figure the right side out.
3. $\sum\limits_{j=1}^{n}{\frac{j}{2^{j}}}=\sum\limits _{j=1}^{n}{\sum\limits_{k=1}^{j}{\frac{1}{2^{j}}}} =\sum\limits_{k=1}^{n}{\frac{1}{2^{k}}\left( \sum\limits_{j=0}^{n-k}{2^{-j}} \right)}=\frac{1}{2^{n}}\sum\limits_{k=1}^{n}{\fra c{1}{2^{k}}\left( 2^{n+1}-2^{k} \right)},$ you can do the rest, those are finite geometric sums.
4. Hello, seven!
Here's one way . . .
Prove that for all $n>0\!:$
. . $\frac{1}{2}+ \frac{2}{2^2} + \frac{3}{2^3} +\:\hdots\:+ \frac{n}{2^n} \; =\; 2 - \frac{n+2}{2^n}$
$\begin{array}{ccccc}
\text{We have:} &S &=& \dfrac{1}{2} + \dfrac{2}{2^2} + \dfrac{3}{2^3} + \dfrac{4}{2^4} + \:\hdots\:+\dfrac{n}{2^n}\qquad\qquad \\ \\[-3mm]
\text{Multiply by }\dfrac{1}{2}\!: & \dfrac{1}{2}S &=& \quad\;\; \dfrac{1}{2^2} + \dfrac{2}{2^3} + \dfrac{3}{2^4} + \hdots + \dfrac{n-1}{2^n} + \dfrac{n}{2^{n+1}}\end{array}$
. . . $\text{Subtract: }\quad \frac{1}{2}S \;=\;\underbrace{\frac{1}{2} + \frac{1}{2^2} + \frac{1}{2^3} + \frac{1}{2^4} + \hdots + \frac{1}{2^n}}_{\text{geometric series}} - \frac{n}{2^{n+2}}$ .[1]
The geometric series has the sum: . $\frac{1}{2}\cdot\frac{1 - \left(\frac{1}{2}\right)^n}{1-\frac{1}{2}} \;=\;1 - \frac{1}{2^n}$
Then [1] becomes: . $\frac{1}{2}S \;=\;\left(1 - \frac{1}{2^n}\right) - \frac{n}{2^{n+1}} \;=\;1 - \frac{n+2}{2^{n+1}}$
Multiply by 2: . $S \;=\;2 - \frac{n+2}{2^n}$
5. Originally Posted by seven.j
Hi, I'm stuck at proving the following question...
Prove that for all n>0,
1/2 + 2/2^2 + 3/2^3 + ... + n/2^n = 2 - (n+2)/2^n
I've tried all sorts of different ways of solving this, but to no avail.
Any help is appreciated
This problem also can be done by induction pretty easily. if you're interested, you can try it that way. to me it was the most knee-jerk approach to try, and it worked out great. but you have lots of nice approaches here to choose from | 2015-05-07T08:51:11 | {
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https://themathhelp.com/threads/probability-of-heads-up.2947/ | #### puremath
##### Active Member
If Miranda flips a coin twice, what is the probability that both coins will land heads up?
Solution (not my work):
Probability of coin 1 landing heads up is 1/2. Probability of second coin landing heads up is 1/2.
P(both heads up) = (1/2)(1/2) = 1/4.
Question:
What indicates in the problem that 1/2 is multiplied by 1/2? Taking a test, a student might guess that he or she must add 1/2 to itself.
#### MarkFL
##### La Villa Strangiato
Staff member
Moderator
Math Helper
We are asked to find the probability that the coin is heads on the first flip AND the coin is heads on the second flip. The "AND" tells us to multiply.
puremath
#### puremath
##### Active Member
We are asked to find the probability that the coin is heads on the first flip AND the coin is heads on the second flip. The "AND" tells us to multiply.
Staff member
Moderator
Math Helper
Yes.
#### puremath
##### Active Member
We are asked to find the probability that the coin is heads on the first flip AND the coin is heads on the second flip. The "AND" tells us to multiply.
The word AND is not in the problem. This is the way I would type the question:
If Miranda flips a coin twice, what is the probability that the coin lands heads up on the first flip and heads up on the second flip? Now the word problem makes sense to me.
#### MarkFL
##### La Villa Strangiato
Staff member
Moderator
Math Helper
When it says both that means both the first flip AND the second flip are heads.
#### MarkFL
##### La Villa Strangiato
Staff member
Moderator
Math Helper
Actually, It could be worded better, as it first implies she is flipping the same coin twice, then it speaks of both coins. If I were to author such a problem, I might state:
If Miranda flips a coin twice, what is the probability that it lands heads up both times?
puremath
#### puremath
##### Active Member
When it says both that means both the first flip AND the second flip are heads.
When reading this the first time, it is not so clear.
#### puremath
##### Active Member
Actually, It could be worded better, as it first implies she is flipping the same coin twice, then it speaks of both coins. If I were to author such a problem, I might state:
If Miranda flips a coin twice, what is the probability that it lands heads up both times?
It is ok but my wording makes it clear to the reader that this problem involves AND and thus multiplication is the operation. Back to Cohen on Monday.
#### MarkFL
##### La Villa Strangiato
Staff member
Moderator
Math Helper
What is the probability that of the two flips, at least one is heads?
#### puremath
##### Active Member
What is the probability that of the two flips, at least one is heads?
P(at least one heads up) = 1 - P(not one is heads up). At the job now. I have a probability book that I also want to learn with you later on in my self-study trek.
#### MarkFL
##### La Villa Strangiato
Staff member
Moderator
Math Helper
You're on the right track. I will give you time to complete the question.
#### puremath
##### Active Member
You're on the right track. I will give you time to complete the question.
For two flips, there are 4 possibilities, only one of which (getting tails twice) does not have any head. So, the other 3 out of 4 cases would have one or two heads. I say 75% probability. Yes?
MarkFL
Staff member | 2020-04-10T06:57:07 | {
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https://math.stackexchange.com/questions/2476453/how-should-one-proceed-in-this-trigonometric-simplification-involving-non-intege | # How should one proceed in this trigonometric simplification involving non integer angles?
The problem is as follows:
Find the value of this function $$A=\left(\cos\frac{\omega}{2} +\cos\frac{\phi}{2}\right )^{2} +\left(\sin\frac{\omega}{2} -\sin\frac{\phi}{2}\right )^{2}$$ when $$\omega=33^{\circ}{20}'$$ and $$\phi=56^{\circ}{40}'$$.
Thus,
$$A=\left(\cos\frac{\omega}{2} +\cos\frac{\phi}{2}\right)^{2} +\left(\sin\frac{\omega}{2} -\sin\frac{\phi}{2}\right)^{2}$$
By solving the power I obtained the following:
$$A= \cos^{2}\frac{\omega}{2} +\cos^{2}\frac{\phi}{2} +2\cos\frac{\omega}{2}\cos\frac{\phi}{2} +\sin^{2}\frac{\omega}{2} +\sin^{2}\frac{\phi}{2} +2\sin\frac{\omega}{2}\cos\frac{\phi}{2}$$
I noticed some familiar terms and using pitagoric identities then I rearranged the equation as follows:
\begin{align} A &= \cos^{2}\frac{\omega}{2} +\sin^{2}\frac{\omega}{2} +\cos^{2}\frac{\phi}{2} +\sin^{2}\frac{\phi}{2} +2\cos\frac{\omega}{2}\cos\frac{\phi}{2} +2\sin\frac{\omega}{2}\cos\frac{\phi}{2} \\ &= 1+1+2\cos\frac{\omega}{2}cos\frac{\phi}{2} +2\sin\frac{\omega}{2}\cos\frac{\phi}{2} \end{align}
Since the latter terms are another way to write prosthapharesis formulas I did the following:
\begin{align} \cos\alpha +\cos\beta &= 2\cos\frac{\alpha+\beta}{2}\cos\frac{\alpha-\beta}{2} \\ \cos\alpha -\cos\beta &= -2\cos\frac{\alpha+\beta}{2}\cos\frac{\alpha-\beta}{2} \\ \\ \alpha+\beta &= \omega \\ \alpha-\beta &= \phi \end{align}
By solving the system I found: $$\alpha=\frac{\omega+\phi}{2}$$ and $$\beta=\frac{\omega-\phi}{2}$$.
Therefore by inserting these into the problem:
\begin{align} A &= 1+1 +2\cos\frac{\omega}{2}\cos\frac{\phi}{2} +2\sin\frac{\omega}{2}\cos\frac{\phi}{2} \\ &= 2 +2\cos\frac{\omega}{2}\cos\frac{\phi}{2} -\left(-2\sin\frac{\omega}{2}\cos\frac{\phi}{2}\right) \\ &= 2 +\cos\frac{\omega+\phi}{2} +\cos\frac{\omega-\phi}{2} -\left(\cos\frac{\omega+\phi}{2} -\cos\frac{\omega-\phi}{2}\right) \end{align}
By cancelling elements,
$$A=2 +2\cos\frac{\omega-\phi}{2}$$
However I'm stuck at trying to evaluate these values:
\begin{align} \omega &= 33^{\circ}{20}'\; \phi=56^{\circ}{40}' \\ \omega-\phi &= \left(33+\frac{20}{60}\right) -\left(56+\frac{40}{60}\right) = -23-\frac{20}{60} \end{align}
Therefore,
$$A=2+2\cos\left(\frac{-23-\frac{20}{60}}{2}\right).$$
However the latter answer does not appear in the alternative neither seems to be right. Is there something wrong on what I did?
• Just a remark: Use \cos and \sin instead of cos and sin. – MrYouMath Oct 17 '17 at 8:52
• @MrYouMath Thanks for the advise. I'll take note on that next time I post a similar question. I'm new on LaTex. – Chris Steinbeck Bell Oct 18 '17 at 2:31
There is a mistake in your expansion of the second bracket.
We should have
$A=\cos^2\frac{\omega}{2}+2\cos\frac{\omega}{2}\cos\frac{\phi}{2}+\cos^2 \frac{\phi}{2}+\sin^2\frac{\omega}{2}-2\sin\frac{\omega}{2}\sin\frac{\phi}{2}+\sin^2\frac{\phi}{2}$
$=2+2(\cos\frac{\omega}{2} \cos\frac{\phi}{2} -\sin\frac{\omega}{2}\sin\frac{\phi}{2})$ .
In the bracket we have the expansion for $\cos(\frac{\omega}{2}+\frac{\phi}{2})$ and fortunately $\omega+\phi=90^\circ$.
Thus $A=2+2\cos(45^\circ)$
$=2+\sqrt{2}$
Hint: You did a miscalculation:
$$A=\left (\cos\frac{\omega}{2}+\cos\frac{\phi}{2} \right )^{2}+\left (\sin\frac{\omega}{2}-\sin\frac{\phi}{2} \right )^{2}$$ $$=\cos ^2\frac{\omega}{2}+2\cos\frac{\omega}{2}\cos\frac{\phi}{2}+\cos^2 \frac{\phi}{2}+\sin ^2\frac{\omega}{2}-2\sin\frac{\omega}{2}\sin\frac{\phi}{2}+\sin^2 \frac{\phi}{2}$$ $$=2+2\cos\frac{\omega}{2}\cos\frac{\phi}{2}-2\sin\frac{\omega}{2}\sin\frac{\phi}{2}$$ $$=2+2\left(\cos\frac{\omega}{2}\cos\frac{\phi}{2}-\sin\frac{\omega}{2}\sin\frac{\phi}{2}\right)$$ $$=2+2\cos\left(\frac{\omega}{2}+\frac{\phi}{2}\right)$$
In the last step I used $\cos(x+y)=\cos x\cos y-\sin x \sin y$.
• Mind including the "last step" in your answer?. I mean the one involving inserting the angle values in the last formula. Usually this seems the most logical and obvious thing to do. But what to do when the angles are not "integers". Was the procedure I used correct?. I mean to decompose them into fractions. The latter part I'm not sure. – Chris Steinbeck Bell Oct 18 '17 at 2:34 | 2019-08-21T13:39:17 | {
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http://happyworldtrip.com/mru0g30/ea5dc6-sum-of-1-to-10 | Start with the binomial expansion of (k − 1) 2: (k-1)^2: (k − 1) 2: (k – roganjosh May 10 '17 at 19:49 (, Techniques for Adding the Numbers 1 to 100, Quick Insight: Intuitive Meaning of Division, Quick Insight: Subtracting Negative Numbers, Surprising Patterns in the Square Numbers (1, 4, 9, 16…), Learning How to Count (Avoiding The Fencepost Problem), Different Interpretations for the Number Zero. Now for the explanation: How many beans do we have total? (i) 2, 7, 12 , ., to 10 terms. We prove the formula 1+ 2+ ... + n = n(n+1) / 2, for n a natural number. Beginners Java program to find sum of odd numbers between 1 -100 ½n(n + 1),. Yep, you get the same formula, but for different reasons. Our math solver supports basic math, pre-algebra, algebra, trigonometry, calculus and more. Python Program to Calculate Sum of Odd Numbers from 1 to N using For Loop. my calculator seems to say 55 though, i don't get how the sum of n terms formula has failed as it's for an AP. Find all the evens, So, the sum from 50 + 52 + … 100 = (50 * 51) – (24 * 25) = 1950. There is a simple applet showing the essence of the inductive proof of this result. I like Steve's thinking, but sometimes I like to take a hammer to a thumbtack. Vandermonde convolution X k r k! We get the next biggest even number (n + 1) and take off the extra (n + 1)/2 “-1″ items: To add 1 + 3 + 5 + … 13, get the next biggest even (n + 1 = 14) and do, Let’s say you want the evens from 50 + 52 + 54 + 56 + … 100. 17 Answers 26 The main idea is that if you write all the numbers from 0 to 999999 down as six digit numbers (possibly prepending zeros) then all digits appear the same number of times. Well, we have 2 equal rows, we must have n/2 pairs. Math. The above method works, but you handle odd and even numbers differently. What is the sum of 1+1/10+1/100+1/1000... etc, where each time the denominator increases by an factor of 10 Through how many radians does it turn in one second? Each nth term of the series is the sum of the n terms i.e n*(n+1)/2. Then push the [Next] button to step through the stages of the proof. Calculate the sum … Q:-A wheel makes 360 revolutions in one minute. for(i=1;i<=n;i++) sum=sum+i; Sum of Natural Numbers using Formula. The loop structure should look like for(i=2; i<=N; i+=2). Yes. Submitted by IncludeHelp, on September 04, 2018 Given the value of N and we have to find sum of all numbers from 0 to N in C language. In case you’re wondering whether it “really” lines up, it does. Instead of writing out numbers, pretend we have beans. Output 2: Enter the value of n: 0 Enter a whole positive number! Below is the complete algorithm. What is the formula for the sum of odd numbers? How to calculate this $\sum\limits_{n=0}^{\infty}\frac{n}{2^n}e^{jwn}$ 0. 2. Python Program to Calculate Sum of Even Numbers from 1 to N using For Loop. This program to find the sum of n numbers allows the user to enter any integer value. sum = n(n+1)/2 In the above program, unlike a for loop, we have to increment the value of i inside the body of the loop. For instance the sum of the numbers from 1 to 10 is 55 whereas the sum of the digits is 46. Let’s say we mirror our pyramid (I’ll use “o” for the mirrored beans), and then topple it over: Cool, huh? home Front End HTML CSS JavaScript HTML5 Schema.org php.js Twitter Bootstrap Responsive Web Design tutorial Zurb Foundation 3 tutorials Pure CSS HTML5 Canvas JavaScript Course Icon Angular React Vue Jest Mocha NPM Yarn Back End PHP Python Java … Home Science Math History Literature Technology Health Law Business All Topics Random. The sum of all the odd integers from 1 to 50 is 625. To find sum of even numbers we need to iterate through even numbers from 1 to n. Initialize a loop from 2 to N and increment 2 on each iteration. 1. In this program, we first check number is odd or not. To add evens from 2 to 50, find 1 + 2 + 3 + 4 … + 25 and double it: So, to get the evens from 2 to 50 you’d do 25 * (25 + 1) = 650. The above formula is one core step of the idea. The partial sums of the series 1 + 2 + 3 + 4 + 5 + 6 + ⋯ are 1, 3, 6, 10, 15, etc.The nth partial sum is given by a simple formula: ∑ = = (+). = r +s n! Python Program to Sum the list with start 10; Program to Sum the list of float; Python Program to Sum the list of float with start 10.1; Program to calculate the sum of elements in a tuple. Let’s look at a small set: The average is 2. Python Program to Sum the list with start 10; Program to Sum the list of float; Python Program to Sum the list of float with start 10.1; Program to calculate the sum of elements in a tuple. We know that the even numbers are the numbers, which are completely divisible by 2. This Python program allows the user to enter the maximum limit value. Popular Questions of Class 11th mathematics. The so-called educator wanted to keep the kids busy so he could take a nap; he asked the class to add the numbers 1 to 100. You need to initialize the counter as 0 and while the loop executes you need to collect / sum them to the counter and finally outside the loop print/ echo the counter. This is demonstrated by the following code snippet. So, we can say the average of the entire set is actually just the average of 1 and n: (1 + n)/2. DESCRIPTION. In summation notation, this may be expressed as + ... Achilles could run at 10 m/s, while the tortoise only 5. The formula to find the sum of first n natural numbers is as follows. Geometric sum nX−1 k=0 ark = a 1− rn 1− r r 6= 1 Geometric series X∞ k=0 ark = a 1− r |r| < 1 3. Feb 21, 2015 . Feb 21, 2015 . Since 50 is an even number, and we want the sum of all the odd integers from 1 to 50, the number 50 won't be included in the sum. First Name. In a similar vein to the previous exercise, here is another way of deriving the formula for the sum of the first n n n positive integers. 3) Most importantly, this example shows there are many ways to understand a formula. We want to add 1 bean to 2 beans to 3 beans… all the way up to 5 beans. Thus, the sum of the integers from 1 to 100, which are divisible by 2 or 5, is 3050. 2, 7, 12, .,to 10 term We know that Sum of AP = /2 (2a + (n 1) d) Here n = 10, a = 2, & d = 7 2 = 2 Putting these in formula , Sum = /2 (2a + (n 1) d) = 10/2 (2 2 + (10 1) So we divide the formula above by 2 and get: Now this is cool (as cool as rows of numbers can be). for(i=1;i<=n;i++) sum=sum+i; Sum of Natural Numbers using Formula. 1) Adding up numbers quickly can be useful for estimation. If you can’t see any matches tap the remaining deck to turn over a new card. If we have 100 numbers (1…100), then we clearly have 100 items. 2) Compute some of digits in numbers from 1 … Output 3: Enter the value of n: -10 Enter a whole positive number! Here, we are implementing a C program that will be used to find the sum of all numbers from 0 to N without using loop. Learn more - Program to check odd numbers. Otherwise, we used the mathematical formula of Sum of Series 1 + 2+ 3+ … + N = N * (N + 1) / 2; TIP: If the function is not returning any value then used Void. C Program to Print Sum of all Even Numbers from 1 to n. This program allows the user to enter the maximum limit value. #SumOfNumbers #1to100 How can we calculate the sum of natural numbers? Well, the sum is clearly 1 + 2 + 3 + 4 + 5. 4. In mathematics, summation is the addition of a sequence of any kind of numbers, called addends or summands; the result is their sum or total. But of course, we don’t want the total area (the number of x’s and o’s), we just want the number of x’s. Interview Answers. Let’s add the numbers 1 to 9, but instead of starting from 1, let’s count from 0 instead: By counting from 0, we get an “extra item” (10 in total) so we can have an even number of rows. \sum_{k=1}^n (2k-1) = 2\sum_{k=1}^n k - \sum_{k=1}^n 1 = 2\frac{n(n+1)}2 - n = n^2.\ _\square k = 1 ∑ n (2 k − 1) = 2 k = 1 ∑ n k − k = 1 ∑ n 1 = 2 2 n (n + 1) − n = n 2. To see that, look at this oblong number, in which the base is one more than the height: case 1. please enter the maximum value: 10 The sum of Even numbers 1 to 10 = 30 The sum of odd numbers 1 to 10 = 25. case 2. please enter the maximum value: 100 The sum of Even numbers 1 to 100 = 2550 The sum of odd numbers 1 to 100 = 2500 . clear, insightful math lessons. Just make a for loop from starting 1 to 10, like given below. sum = n(n+1)/2 The program calculates the sum of numbers till the given input. By the way, there are more details about the history of this story and the technique Gauss may have used. Notice in both cases, 1 is on one side of the average and N is equally far away on the other. integer m 6= −1 5. How about even numbers, like 2 + 4 + 6 + 8 + … + n? 2. C Program to print sum of the natural numbers from 1 to 10 # include # include main( ) { int n,sum=0,i; clrscr( ); for (i=1; i<=10; i++) sum=sum+i; printf(“sum of natural numbers from 1 to 10 is %d\\n”,sum); getch( ); } The goal is to clear the tableau by matching pairs of cards that add up to ten. 0. how to find a sum of numbers in a sequence when some intermediate terms are not taken in to consideration? Ex 5.3 ,1 Find the sum of the following APs. However, our formula will look a bit different. Answer Add Tags. Though both programs are technically correct, it is better to use for loop in this case. For these examples we’ll add 1 to 10, and then see how it applies for 1 to 100 (or 1 to any number). We want to get rid of every number from 1 up to a – 1. input this case is 6. Since thousand squared = 1 million, we get million / 2 + 1000/2 = 500,500. sum of n terms= n/2 {2a(n-1)d} where n is no of terms, a is the first term and d is difference, subbed in you get 5{2x9}=90. There is a more simple way to find the sum of odd numbers from 1 to N. The first odd number is 1, then the next odd number is 3, i.e (1+2). Enter n value: 20 Sum of odd numbers from 1 to 20 is: 100. Example 1: Adding all items in a tuple, and returning the result ; Example 1: Starting with the 10, and adding all items in a tuple to this number: Example: Compute the sum and product of the numbers from 1 to 10. C programming, exercises, solution : Write a program in C to calculate the sum of numbers from 1 to n using recursion. The above formula is one core step of the idea. How about odd numbers, like 1 + 3 + 5 + 7 + … + n? I won’t. The difference between consecutive triangles increases by 1.. A formula for the triangular numbers. Telescoping sum X a≤k
sum of 1 to 10 2020 | 2022-01-25T08:22:50 | {
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http://math.stackexchange.com/questions/115558/nth-roots-of-unity | # $n$th Roots of Unity
I need to show that the product of all the $n$th roots of unity is $(−1)^{n+1}$.
Is there a way to do this by induction? If there is, I can't seem to figure it out. Are there other, perhaps more efficient, methods of proof?
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The problem with attempting to do it "by induction" is that the $n$th roots of unity are not related to the $(n+1)$th roots of unity (except for the single one that they have in common, $1$. – Arturo Magidin Mar 2 '12 at 4:07
If you're going to do an induction argument, it should be an induction on the number of (possibly repeated) prime divisors of $n$. I think this should actually be possible, with Wilson's Theorem stepping in as a base case. But, honestly, just about anything else seems easier....like all the answers given below. – Pete L. Clark Mar 2 '12 at 4:28
The $n^{th}$ roots of unity are precisely the roots of the polynomial $$x^n - 1$$ Now use the fact that the product of the roots of any polynomial is $(-1)^n$ times the constant coefficient.
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Is this fact that the product of the roots of any polynomial is $(−1)^n$ times the constant coefficient true for any polynomial? Can you elaborate on that? – Dominick Gerard Mar 2 '12 at 4:42
@DominickGerard: Consider writing the polynomial as a product of linear factors, then multiplying it out again. What will the constant term be? Example: $(z-z_1)(z-z_2) = z^2 - (z_1 + z_2) z + z_1 z_2$. – Antonio Vargas Mar 2 '12 at 4:58
@DominickGerard Yes, it holds for any polynomial with coefficients in an integral domain (including $\mathbb{Z}$ or any field): en.wikipedia.org/wiki/Vieta%27s_formulas – dls Mar 2 '12 at 5:01
Note that if $\zeta$ is an $n$th root of unity, then so is $\frac{1}{\zeta}$. Moreover, the only roots of unity for which $\zeta=\frac{1}{\zeta}$ are $\zeta=1$ and $\zeta=-1$ (since $\zeta=\frac{1}{\zeta}$ implies $\zeta^2=1$, hence $\zeta=1$ or $\zeta=-1$).
So, if we take the $n$th roots of unity, then we can pair off all of them except for $1$ and perhaps $-1$ (if $n$ is even), into pairs of the form $\{\zeta,\frac{1}{\zeta}\}$. The product of these two equals $1$; so when you do the product of all roots, you end up with a bunch of pairs that multiply to $1$, and you have $1$; and if $n$ is even, then you also have $-1$ leftover. Which means the product of all of them is...
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Well, you said "efficient" and "elementary number theory", and that triggered this in my mind: the result is a special case of
(Wilson's Theorem in a Finite Abelian Group): Let $(G,\cdot)$ be a finite abelian group, and let $P = \prod_{x \in G} x$. Then $P = 1$ (the identity element) unless $G$ has exactly one element $t$ of order $2$, in which case $P = t$.
This is something that others (in particular, Bill Dubuque) have talked about here and elsewhere before. I include a statement and proof of this in my notes/prebook on elementary(ish...) number theory: it is at the end of Appendix B.
Added: Not only is this a special case of a "generalized Wilson theorem", it's a special case that includes the classical Wilson Theorem! Since the unit group of $(\mathbb{Z}/p\mathbb{Z})^{\times}$ is cyclic of order $p-1$, computing $(p-1)! \pmod{p}$ is essentially the same as multiplying together all the $(p-1)$st roots of unity.
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+1 See these posts for more on the group-theoretic Wilson's theorem and involutions. – Bill Dubuque Mar 2 '12 at 4:13
You can prove it with de Moivre, on the basis that the roots are all of the form $e^{2\pi i k/n}$, so their product is $e^{2\pi i/n\sum{k}} = e^{\pi i(n-1)} = (-1)^{n-1} = (-1)^{n+1}$.
Another proof idea would be to notice that the roots of unity come in complex conjugate pairs, which is evident in the symmetry of the spacing of the roots in the complex plain. So the product of each pair is of course $\alpha \bar\alpha = |\alpha| = 1$ since all the roots fall on the unit circle. The only exceptions are $1,-1$. $1$ is always a root and won't change the product, and $-1$ is only a root when $n$ is even, so we conclude that the product is exactly $(-1)^{n+1}$. | 2014-03-12T07:14:37 | {
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https://forum.azimuthproject.org/discussion/2459/lecture-7-universal-properties-products-coproducts-and-exponentials-david-spivak | #### Howdy, Stranger!
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# Lecture 7 - Universal properties: products, coproducts and exponentials - David Spivak
edited January 21
## Comments
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1.
It was interesting to learn about the relationship between currying, uncurrying, and exponentials. I'd heard people talking about currying before and handwaving, but this lecture made it very concrete. The motivation of set exponentiation (the set of maps) {1, 2, 3}^{1,2} as "3"^"2" ~ 9 maps in total was nice as well.
I also liked the motivation of Either as a pushout from two functions. I recently did a few exercises using Either in scala and ended up getting very confused! So this sort of knowledge would probably have made that experience a bit less confusing.
I noticed that it was mentioned in passing that products, co-products and exponentials are all examples of adjunctions. And I just listened to the first 10 seconds of lecture 8, and I gather that they are discussed therein, so I look forward to reviewing that lecture too in time.
Comment Source:It was interesting to learn about the relationship between currying, uncurrying, and exponentials. I'd heard people talking about currying before and handwaving, but this lecture made it very concrete. The motivation of set exponentiation (the set of maps) {1, 2, 3}^{1,2} as "3"^"2" ~ 9 maps in total was nice as well. I also liked the motivation of Either as a pushout from two functions. I recently did a few exercises using Either in scala and ended up getting very confused! So this sort of knowledge would probably have made that experience a bit less confusing. I noticed that it was mentioned in passing that products, co-products and exponentials are all examples of [adjunctions](https://en.wikipedia.org/wiki/Adjoint_functors). And I just listened to the first 10 seconds of lecture 8, and I gather that they are discussed therein, so I look forward to reviewing that lecture too in time.
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2.
edited March 2020
Yes, it's a good one!
Ideas from the first five minutes (using David Spivak's terminology):
• A universal property is a "one-stop shop" for maps
• The product of sets $$A$$ and $$B$$ is the one-stop shop from maps into $$A$$ and $$B$$.
Meaning: suppose you have a set $$I$$, and maps $$f: I \rightarrow A$$ and $$g: I \rightarrow B$$, with $$f$$ in your "left-hand", and $$g$$ in your right.
These two functions are equivalent to a single function $$f \times g: I \rightarrow A \times B$$.
The set $$I$$ can be thought of as an "index" set which is used to pick out elements via the functions out of $$I$$.
When $$I$$ is a singleton, the "one-stop shop" gives you a single element in the product of $$A$$ and $$B$$, in lieu of two separate elements one from $$A$$ and the other from $$B$$.
Comment Source:Yes, it's a good one! Ideas from the first five minutes (using David Spivak's terminology): * A universal property is a "one-stop shop" for maps * The product of sets \$$A\$$ and \$$B\$$ is the one-stop shop from maps into \$$A\$$ and \$$B\$$. Meaning: suppose you have a set \$$I\$$, and maps \$$f: I \rightarrow A\$$ and \$$g: I \rightarrow B\$$, with \$$f\$$ in your "left-hand", and \$$g\$$ in your right. These two functions are equivalent to a single function \$$f \times g: I \rightarrow A \times B\$$. The set \$$I\$$ can be thought of as an "index" set which is used to pick out elements via the functions out of \$$I\$$. When \$$I\$$ is a singleton, the "one-stop shop" gives you a single element in the product of \$$A\$$ and \$$B\$$, in lieu of two separate elements one from \$$A\$$ and the other from \$$B\$$.
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3.
• The coproduct of $$A$$ and $$B$$ is the "one-stop shop" for maps into $$A$$ and $$B$$.
Comment Source:* The coproduct of \$$A\$$ and \$$B\$$ is the "one-stop shop" for maps into \$$A\$$ and \$$B\$$.
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4.
edited March 2020
Thinking out loud here...
The idea of the "one-stop shop for maps" generalizes nicely from the product to the limit.
Looking at the general case of the limit, which includes as special cases terminal objects, products, equalizers and pullbacks, brings clarity to the whole picture. And in the category Set, there is an explicit construction for all limits.
To review: suppose we are given a diagram D in the category C. Note: for this definition, D is not required to be a commuting diagram -- it's just a collection of objects and arrows.
The limit of D, if it exists, will be another object Lim(D) in C, along with "projection maps" from Lim(D) to each of the objects in D, which satisfies a couple of conditions. First, for every arrow $$f: X_1 \rightarrow X_2$$ in D, the triangle diagram with objects Lim(D), $$X_1$$, and $$X_2$$ and arrows $$f$$ and the two projections from Lim(D) to $$X_1$$ and $$X_2$$ -- this triangle diagram commutes.
Lim(D) along with these projections is called a cone over D. The cones over D form a category, with arrows going between the apexes of the cones, such that these arrows satisfy the following commutativity condition with respect to the cones. Suppose that $$X$$ and $$Y$$ are each the apex of a cone, and $$f: X \rightarrow Y$$. Then $$f$$ is defined to be a morphism between the two cones if for each object $$d$$ in D, the triangle formed by $$d$$, $$X$$ and $$Y$$, with arrows $$f$$ and the two projections to $$d$$ -- one projection within each cone -- commutes.
The second requirement on Lim(D) is that it be a terminal object in this category of cones.
Comment Source:Thinking out loud here... The idea of the "one-stop shop for maps" generalizes nicely from the product to the limit. Looking at the general case of the limit, which includes as special cases terminal objects, products, equalizers and pullbacks, brings clarity to the whole picture. And in the category Set, there is an explicit construction for all limits. To review: suppose we are given a diagram D in the category C. Note: for this definition, D is not required to be a _commuting_ diagram -- it's just a collection of objects and arrows. The limit of D, if it exists, will be another object Lim(D) in C, along with "projection maps" from Lim(D) to each of the objects in D, which satisfies a couple of conditions. First, for every arrow \$$f: X_1 \rightarrow X_2\$$ in D, the triangle diagram with objects Lim(D), \$$X_1\$$, and \$$X_2\$$ and arrows \$$f\$$ and the two projections from Lim(D) to \$$X_1\$$ and \$$X_2\$$ -- this triangle diagram commutes. Lim(D) along with these projections is called a cone over D. The cones over D form a category, with arrows going between the apexes of the cones, such that these arrows satisfy the following commutativity condition with respect to the cones. Suppose that \$$X\$$ and \$$Y\$$ are each the apex of a cone, and \$$f: X \rightarrow Y\$$. Then \$$f\$$ is defined to be a morphism between the two cones if for each object \$$d\$$ in D, the triangle formed by \$$d\$$, \$$X\$$ and \$$Y\$$, with arrows \$$f\$$ and the two projections to \$$d\$$ -- one projection within each cone -- commutes. The second requirement on Lim(D) is that it be a terminal object in this category of cones.
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edited March 2020
Now that's somewhat of an abstract story, which doesn't give you a feeling for how to actually find what this special cone Lim(D) actually is. And in general such an object may not exist at all.
But in the category Set it always exists, and can be constructed explicitly, as follows.
Comment Source:Now that's somewhat of an abstract story, which doesn't give you a feeling for how to actually find what this special cone Lim(D) actually is. And in general such an object may not exist at all. But in the category Set it always exists, and can be constructed explicitly, as follows.
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edited March 2020
Suppose we have a diagram $$D$$ in Set. Let the objects in $$D$$ be named $$S_1, \ldots S_n$$, where each $$S_i$$ is a set. Let the arrows in $$D$$ be $$f_1, \ldots f_k$$ -- these are functions.
Then Lim(D) consists of a subset of the Cartesian product of the set $$S_1, \ldots S_n$$ -- so it is a relation R between these sets. It consists of all tuples $$(x_1, \ldots, x_n)$$ that satisfy the all of the constraints given by the functions $$f_i$$. Suppose $$f_i: S_j \rightarrow S_k$$. Then the constraint given by $$f_i$$ stipulates that for a tuple $$(x_1, \ldots, x_n)$$, that $$f_i(x_j) = x_k$$.
The cone has relation R at its apex, and the ordinary projection functions from R to each of the sets $$S_1, \ldots, S_n$$ as the arrows from the apex R to the objects in the diagram $$D$$.
It is but an exercise to show that this cone with R at its apex is terminal within the category of cones over $$D$$.
Comment Source:Suppose we have a diagram \$$D\$$ in Set. Let the objects in \$$D\$$ be named \$$S_1, \ldots S_n\$$, where each \$$S_i\$$ is a set. Let the arrows in \$$D\$$ be \$$f_1, \ldots f_k\$$ -- these are functions. Then Lim(D) consists of a subset of the Cartesian product of the set \$$S_1, \ldots S_n\$$ -- so it is a relation R between these sets. It consists of all tuples \$$(x_1, \ldots, x_n)\$$ that satisfy the all of the constraints given by the functions \$$f_i\$$. Suppose \$$f_i: S_j \rightarrow S_k\$$. Then the constraint given by \$$f_i\$$ stipulates that for a tuple \$$(x_1, \ldots, x_n)\$$, that \$$f_i(x_j) = x_k\$$. The cone has relation R at its apex, and the ordinary projection functions from R to each of the sets \$$S_1, \ldots, S_n\$$ as the arrows from the apex R to the objects in the diagram \$$D\$$. It is but an exercise to show that this cone with R at its apex is terminal within the category of cones over \$$D\$$.
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7.
The relation R = Lim(D) is a one-stop shop for cones over D, in the following sense. Suppose that we have a cone C with apex A over D. Then C consists of a bunch of separate maps from A into the objects of D. The universal property for R = Lim(D) says that there a unique map m(C) from A into R, which gives a morphism from C into the universal cone with apex Lim(D).
m(C) is a single map that contains all of the information in C. That's because the projections from C into the objects of D can be reconstructed by composing m(C) with the standard projections from R into the objects of D.
So the multiple separate maps comprising the cone C can, without loss of information, be replaced by the single map m(C).
Comment Source:The relation R = Lim(D) is a one-stop shop for cones over D, in the following sense. Suppose that we have a cone C with apex A over D. Then C consists of a bunch of separate maps from A into the objects of D. The universal property for R = Lim(D) says that there a unique map m(C) from A into R, which gives a morphism from C into the universal cone with apex Lim(D). m(C) is a single map that contains all of the information in C. That's because the projections from C into the objects of D can be reconstructed by composing m(C) with the standard projections from R into the objects of D. So the multiple separate maps comprising the cone C can, without loss of information, be replaced by the single map m(C).
• Options
8.
edited March 2020
For an illustration of this construction, let's apply it to the equalizer.
Here, the diagram D consists of two objects $$X_1, X_2$$, and two maps $$f,g: X_1 \rightarrow X_2$$.
Note: here the fact that D is not required to be a commutative diagram is significant. If it were so required, then that would imply that $$f$$ and $$g$$ are equal. But we don't require that, as we are constructing the equalizer of distinct maps $$f$$ and $$g$$.
Then, as per the general construction given above, the relation R = Lim(D) will consist of all tuples $$(x,y)$$, such that $$f(x) = y$$, and $$g(x) = y$$. Putting these together, we get that:
$equalizer(f,g) = Lim(D) = \lbrace (x,y)\ |\ y = f(x) = g(x) \rbrace = \lbrace (x,y)\ |\ x \in dom(f) \cap dom(g),\ f(x) = g(x)\rbrace = f \cap g$
Which is to say: the equalizer of functions $$f$$ and $$g$$ is the relation (the graph representation) for the partial function that is obtained by restricting both functions to the set of points where they are both defined and both produce the same value.
Comment Source:For an illustration of this construction, let's apply it to the equalizer. Here, the diagram D consists of two objects \$$X_1, X_2\$$, and two maps \$$f,g: X_1 \rightarrow X_2\$$. Note: here the fact that D is not required to be a _commutative_ diagram is significant. If it were so required, then that would imply that \$$f\$$ and \$$g\$$ are equal. But we don't require that, as we are constructing the equalizer of distinct maps \$$f\$$ and \$$g\$$. Then, as per the general construction given above, the relation R = Lim(D) will consist of all tuples \$$(x,y)\$$, such that \$$f(x) = y\$$, and \$$g(x) = y\$$. Putting these together, we get that: \$equalizer(f,g) = Lim(D) = \lbrace (x,y)\ |\ y = f(x) = g(x) \rbrace = \lbrace (x,y)\ |\ x \in dom(f) \cap dom(g),\ f(x) = g(x)\rbrace = f \cap g\$ Which is to say: the equalizer of functions \$$f\$$ and \$$g\$$ is the relation (the graph representation) for the partial function that is obtained by restricting both functions to the set of points where they are both defined and both produce the same value.
• Options
9.
edited March 2020
We get the product as a special case of a limit, where the diagram $$D$$ consists only of objects, and no arrows.
In Set, this will be a collection of sets $$S_1, \ldots, S_n$$ -- and no arrows to impose constraints on the tuples in R = Lim(D).
So Lim(D) is the unconstrained set of tuples -- i.e., the Cartesian product $$S_1 \times \cdots \times S_n$$.
Comment Source:We get the _product_ as a special case of a limit, where the diagram \$$D\$$ consists only of objects, and no arrows. In Set, this will be a collection of sets \$$S_1, \ldots, S_n\$$ -- and no arrows to impose constraints on the tuples in R = Lim(D). So Lim(D) is the unconstrained set of tuples -- i.e., the Cartesian product \$$S_1 \times \cdots \times S_n\$$.
• Options
10.
What's the limit of an empty diagram $$D$$?
It's simply a terminal object in the category, for the following reason.
A cone over the empty diagram consists just of an apex object A, along with zero "leg" morphisms.
So the category of cones over the empty diagram coincides exactly with the ambient category C.
The limit is defined to be a terminal cone, which amounts to the same thing as a terminal object in C.
Comment Source:What's the limit of an empty diagram \$$D\$$? It's simply a terminal object in the category, for the following reason. A cone over the empty diagram consists just of an apex object A, along with zero "leg" morphisms. So the category of cones over the empty diagram coincides exactly with the ambient category C. The limit is defined to be a terminal cone, which amounts to the same thing as a terminal object in C.
• Options
11.
edited March 2020
Exercise 1. Use this construction to concretely explain the meaning of pullbacks in Set.
Recall that a pullback is defined as the limit of a diagram $$D$$, consisting of three objects $$A,B,C$$, and morphims $$f: A \rightarrow C$$ and $$g: B \rightarrow C$$.
Comment Source:Exercise 1. Use this construction to concretely explain the meaning of pullbacks in Set. Recall that a pullback is defined as the limit of a diagram \$$D\$$, consisting of three objects \$$A,B,C\$$, and morphims \$$f: A \rightarrow C\$$ and \$$g: B \rightarrow C\$$.
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12.
edited March 2020
\begin{CD} A \cap B @>\subseteq>>B \\@V{\subseteq}VV {}@VV{\subseteq}V \\ A @>>\subseteq> C \end{CD}
Here, $$C$$ is taken to be any set which contains both $$A$$ and $$B$$, and $$\subseteq$$ means the inclusion function.
Diagram from Example 1.8.3 of:
• Benjamin C. Pierce, Basic Category Theory for Computer Scientists, MIT Press, 1991.
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13.
edited March 2020
\begin{CD} f^{-1}(A) @>\subseteq>>B \\@V{f|_{f^{-1}(A)}}VV {}@VV{f}V \\ A @>>\subseteq> C \end{CD}
Here, $$A$$ is a subset of the codomain of $$f$$. The claim made by this diagram is that inverse image of $$A$$ under $$f$$ is the pullback of $$f$$ along the inclusion function for $$A$$.
Diagram from Example 1.8.2 of:
• Benjamin C. Pierce, Basic Category Theory for Computer Scientists, MIT Press, 1991.
• Options
14.
edited March 2020
@ChrisGoddard wrote:
I noticed that it was mentioned in passing that products, co-products and exponentials are all examples of adjunctions.
Interesting! Keep us posted if you dig further into this.
Comment Source:@ChrisGoddard wrote: > I noticed that it was mentioned in passing that products, co-products and exponentials are all examples of [adjunctions](https://en.wikipedia.org/wiki/Adjoint_functors). Interesting! Keep us posted if you dig further into this.
• Options
15.
edited March 2020
Problem 1: spell out the explicit construction for colimits in the category Set.
The path of analysis is similar to that used for the explicit construction for limits. But the result is not "symmetrical." In one of the lectures Bartosz commented on the asymmetry between product and coproduct in Set. He traced it back to some asymmetries in the structure of the morphisms in Set, which as we know are total functions. The asymmetry is that whereas each point in the domain must map -- be related to -- to just one point in the codomain, each point in the codomain can be related to zero points in the domain (where the function is not onto), or multiple points in the domain (where the function is not one-to-one).
Comment Source:Problem 1: spell out the explicit construction for colimits in the category Set. The path of analysis is similar to that used for the explicit construction for limits. But the result is not "symmetrical." In one of the lectures Bartosz commented on the asymmetry between product and coproduct in Set. He traced it back to some asymmetries in the structure of the morphisms in Set, which as we know are total functions. The asymmetry is that whereas each point in the domain must map -- be related to -- to just one point in the codomain, each point in the codomain can be related to zero points in the domain (where the function is not onto), or multiple points in the domain (where the function is not one-to-one).
• Options
16.
Problem 2: use this construction to give a concrete explanation for coequalizers, coproducts, pushouts and initial objects in Set.
Comment Source:Problem 2: use this construction to give a concrete explanation for coequalizers, coproducts, pushouts and initial objects in Set.
Sign In or Register to comment. | 2021-07-24T06:40:12 | {
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http://mathhelpforum.com/calculus/159010-finding-riemann-sums.html | # Math Help - Finding Riemann sums
1. ## Finding Riemann sums
The question:
An electrical signal S(t) has its amplitude |S(t)| tested (sampled) every 1/10 of a second. It is desired to estimate the energy over a period of half a second, given exactly by:
$(\int_0^\frac{1}{2} \! |S(t)|^2 \, \mathrm{d}t)^{\frac{1}{2}}$
The result of the measurement are shown in the following table:
$\begin{center}
\begin{tabular}{| l | c | c| c| c| c|}
\hline
t & .1 & .2 & .3 & .4 & .5\\ \hline
\| S(t) \| & 60 & 50 & 50 & 45 & 55 \\ \hline
e(t) & 5 & 3 & 7 & 4 & 10 \\
\hline
\end{tabular}
\end{center}$
a) Using the above data for S(t), set up the appropriate Riemann sum and compute an appropriate value for the energy.
My attempt:
The set P (set of partitions) is clearly equal to {0, 1/10, 1/5, 3/10, 4/10, 1/2} so dt is 1/10 i.e. the width of each rectangle is 1/10.
My problem is with producing lower and higher Riemann sums. I'm used to having a function defined as something like $x^2$ and producing a general sum given dx (in this case, dt). In this question, we don't know the function, we're just given values.
Does this mean I take the max/min of each interval (e.g. [0, 1/10], [1/10, 1/5] etc.) and work out the partitions by hand? I'm a bit confused.
Any help would be greatly appreciated!
2. Originally Posted by Glitch
The question:
An electrical signal S(t) has its amplitude |S(t)| tested (sampled) every 1/10 of a second. It is desired to estimate the energy over a period of half a second, given exactly by:
$(\int_0^\frac{1}{2} \! |S(t)|^2 \, \mathrm{d}t)^{\frac{1}{2}}$
The result of the measurement are shown in the following table:
$\begin{center}
\begin{tabular}{| l | c | c| c| c| c|}
\hline
t & .1 & .2 & .3 & .4 & .5\\ \hline
\| S(t) \| & 60 & 50 & 50 & 45 & 55 \\ \hline
e(t) & 5 & 3 & 7 & 4 & 10 \\
\hline
\end{tabular}
\end{center}$
a) Using the above data for S(t), set up the appropriate Riemann sum and compute an appropriate value for the energy.
My attempt:
The set P (set of partitions) is clearly equal to {0, 1/10, 1/5, 3/10, 4/10, 1/2} so dt is 1/10 i.e. the width of each rectangle is 1/10.
My problem is with producing lower and higher Riemann sums. I'm used to having a function defined as something like $x^2$ and producing a general sum given dx (in this case, dt). In this question, we don't know the function, we're just given values.
Does this mean I take the max/min of each interval (e.g. [0, 1/10], [1/10, 1/5] etc.) and work out the partitions by hand? I'm a bit confused.
Any help would be greatly appreciated!
left sum ... L = 0.1[S(0) + S(.1) + S(.2) + S(.3) + S(.4)]
right sum ... R = 0.1[S(.1) + S(.2) + S(.3) + S(.4) + S(.5)]
the table does not give a value for S(0), so it looks as though you can only compute the right sum.
3. Inside each interval you are given only two values- the values at the two endpoints. For the "lower sum" use the smaller of those two, for the "higher sum" use the larger.
In the first interval, from 0.1 to 0.2, you are given exactly two values- 60 and 50. The "lower sum" uses the smaller of those, 50, and the "higher sum" uses the higher, 60. Between .2 and .3 your two values are 50 and 50. Since those are the same, use 50 for both sums. Between .3 and .4, the two values are 50 and 45. Use 45 for the "lower sum" and 50 for the "higher sum". Finally, between .4 and .5, the two values are 45 and 55. Use 45 for the lower sum and 55 for the higher sum.
Summarizing, for the "lower sum" use 50, 50, 45, and 45. For the "higher sum" use 60, 50, 50, and 55.
4. For the lower sum I used: 0, 50, 50, 45, 45
I squared each (as per the function), and multiplied them by 1/10 to produce the area of each partition. My result was 905.
For the higher sum I used: 60, 60, 50, 50, 55
Again, I squared each, and multiplied by 1/10. My result was 1522.5
I took the average of the higher and lower sums, to get 1213.75. As per the question, the integral is to the power of 1/2, so I took the square root and got 34.84. However, the solution in my text is 36.95 (or square root of 1365).
I'm not sure where I went wrong. My answer is close, but incorrect. :/
Edit: I just realised that using '0' for the lower sum isn't correct. I used 60 instead, but I get 1393.75 which is slightly off the given solution. | 2014-04-20T00:43:13 | {
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https://math.stackexchange.com/questions/2714910/find-all-pairs-of-x-y-z-such-that-x-y-sqrtz2-2018 | # Find all pairs of $(x, y ,z)$ such that $x + y =\sqrt{z^{2} + 2018}, …$
Find all pairs of $(x,y,z)$, of real numbers, such that
$$x + y = \sqrt{z^{2} + 2018}$$ $$x + z = \sqrt{y^{2} + 2018}$$ $$y + z = \sqrt{x^{2} + 2018}$$
An attempt : Squaring we get $$(x + y)^{2} = z^{2} + 2018 \implies (x + y)^{2} - z^{2} = 2018$$ $$(x + z)^{2} = y^{2} + 2018 \implies (x + z)^{2} - y^{2} = 2018$$ $$(y + z)^{2} = x^{2} + 2018 \implies (z + y)^{2} - x^{2} = 2018$$ which also means $$(x + y - z)(x + y + z) = 2018$$ $$(x + z - y)(x + y + z) = 2018$$ $$(z + y - x)(x + y + z) = 2018$$ so $$\frac{2018}{x+y-z} = \frac{2018}{x+z-y} = \frac{2018}{y+z-x}$$
$$(x+y-z) = (x+z-y) = (y+z-x)$$ $$y-z = z-y \implies z = y$$ $$x-y= y-x \implies x=y$$ so my answer is $$(x, y, z), \:\:\: x=y=z$$ but with the 3 initial equations, we must also have $$(x+y) = \sqrt{z^{2} + 2018} \implies 4x^{2} = x^{2} + 2018$$ or $$x^{2} = 2018/3$$ so the solution is $$(x, y, z), \:\:\: x=y=z = \sqrt{2018/3}$$
Is this sufficient already? are there better techniques?
• Doesn't $x=y=z$ Imply $2x = \sqrt{x^{2} + 2018}$ from which you can solve for $x$? – Jim Haddocc Mar 30 '18 at 16:16
• The equation can be rewritten as $x + y + z = f(x) = f(y) = f(z)$ where $f(t) = t + \sqrt{t^2+2018}$. The function $f$ is strictly increasing, so $f(x) = f(y) = f(z) \implies \cdots$ – achille hui Mar 30 '18 at 16:22
• Indeed, we get $$x=y=z=\sqrt{\frac{2018}{3}}$$ – Dr. Sonnhard Graubner Mar 30 '18 at 16:25
• @PrathyushPoduval yes, thanks for that, i was meant that way. I have edited the post – Arief Anbiya Mar 30 '18 at 16:26
• Just the condition that $z^2+m$ is a square implies that$z^2+m \ge (z+1)^2$ or $z \le (m-1)/2$. – marty cohen Mar 30 '18 at 16:29
At the stage you have... $$(x + y - z)(x + y + z) = 2018$$ $$(x + z - y)(x + y + z) = 2018$$ $$(z + y - x)(x + y + z) = 2018$$ This implies... $$x+y-z=x+z-y=z+y-x=\dfrac{2018}{x+y+z}$$ ...unless $x+y+z=0$.
Notice that $x+y+z=0$ implies $x+y=-z$ and so $-z=\sqrt{z^2+2018}$ which means $z^2=z^2+2018$ and so $z=0$. Likewise, $x=y=0$ (contradiction since $(0,0,0)$ isn't a solution).
Thus $x+y-z=x+z-y=z+y-x$ and so you got that $x=y=z$. But then $x+y=\sqrt{z^2+2018}$ implies that $2z=\sqrt{z^2+2018}$ and so $4z^2=z^2+2018$ and so $x=y=z = \pm \sqrt{\dfrac{2018}{3}}$. But $x+y$ etc. are square roots (thus non negative). So there is only one solution.
• Thanks for the answer, but i think negative value of $x=y=z < 0$ is not allowed. This is because the 3 initial equations hold. – Arief Anbiya Mar 30 '18 at 16:42
• Silly me. You are correct. – Bill Cook Mar 30 '18 at 17:15 | 2020-10-30T17:29:41 | {
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https://math.stackexchange.com/questions/896186/how-can-i-prove-that-the-span-of-a-subspace-and-its-orthogonal-complement-is-the | # How can I prove that the span of a subspace and its orthogonal complement is the whole vector space?
The book Linear and Geometric Algebra explains the following theorem in a way that I haven't been able to figure out:
If $\mathbf{A}$ and $\mathbf{B}$ are subspaces of a vector space $\mathbf{B}$ then the set of all combinations $\mathbf{a} + \mathbf{b}$ such that, $\mathbf{a} \in \mathbf{A}$ and $\mathbf{b} \in \mathbf{B}$ is called the span of $\mathbf{A}$ and $\mathbf{B}$, written as $\text{span}(\mathbf{A}, \mathbf{B})$.
Furthermore let $\mathbf{U}^{\bot}$ be the subspace consisting of all vectors orthogonal to a subspace $\mathbf{U}$, in the sense that $\mathbf{u} \in \mathbf{U}^{\top}$ if and only if, $\mathbf{u} \perp \mathbf{v}$ for all vectors $\mathbf{v} \in \mathbf{U}$.
I have of course been able to prove that $\mathbf{U}^{\bot}$ is indeed a subspace for all subspaces $\mathbf{U}$, if this turns out to be useful.
The theorem I want to prove is: if $\mathbf{U}$ is a subspace of $\mathbf{V}$ then $\text{span}(\mathbf{U}, \mathbf{U}^{\perp}) = \mathbf{V}$.
The book mentioned above proves it as follows: If $\text{span}(\mathbf{U}, \mathbf{U}^{\perp}) \neq \mathbf{V}$ then there is a nonzero $\mathbf{u} \perp \text{span}(\mathbf{U}, \mathbf{U}^{\perp})$ because any orthonormal basis for a subspace of an inner product space can be extended into an orthonormal basis for the entire inner product space. In particular $\mathbf{u} \perp \mathbf{U}$, i.e. $\mathbf{u} \in \mathbf{U}^{\perp}$, a contradiction.
I understand how an orthonormal basis for a subspace of an inner product space can be extended into an orthonormal basis for the whole inner product space essentially using Gram-Schmidt orthogonalisation. I don't understand how this process allows you to go from, $\text{span}(\mathbf{U}, \mathbf{U}^{\perp}) \neq \mathbf{V}$ to $\exists \mathbf{u} \in \mathbf{U} : \mathbf{u} \perp \text{span}(\mathbf{U}, \mathbf{U}^{\perp})$. So my question would be how does this implication work?
• Possible duplicate of math.stackexchange.com/questions/878438/…
– Surb
Aug 13, 2014 at 13:34
• It seems to me that such statement holds only if $U$ is closed -- this is satisfied if everything is finite dimensional, for example.. Aug 13, 2014 at 13:35
• Here's a counter-example for infinite dimensional spaces: math.stackexchange.com/questions/636517/…
– Surb
Aug 13, 2014 at 13:37
• Sure, the book I'm using is only on finite dimensional vector spaces. Aug 13, 2014 at 13:37
• I don't think this question is a duplicate of that one because the proof in question is quite different, that one is constructive, this one is a proof by contradiction. Aug 13, 2014 at 13:41
You pick an orthonormal basis for $\text{span}(\mathbf{U}, \mathbf{U}^{\perp}) \neq \mathbf{V}$, say $e_j$, $1\leq j\leq n$. Extend this to an orthonormal basis for $\mathbf{V}$, $e_j$, $1\leq j\leq m$. Since $\text{span}(\mathbf{U}, \mathbf{U}^{\perp}) \neq \mathbf{V}$, $n<m$. But then $u=e_{n+1}\perp\text{span}(\mathbf{U}, \mathbf{U}^{\perp})$, by construction, and thus $u\perp \mathbf{U}$, i.e. $u\in \mathbf{U}^\perp$, and thus $u\perp u$, i.e. $\|u\|=0$. This is a contradiction, since $\|u\|=1$.
Note that $$Null(U)=U^\perp$$ now by the rank-nullity theorem we have $$Null(U)+Rank(U)=n$$, where n is the dimension of $$\mathbb{R}^n$$. Now we can use basis extension to combine the basis for $$U$$ and $$U^\perp$$. Since $$U+U^\perp$$ is a subspace of $$\mathbb{R}^n$$, and the basis for $$U+U^\perp$$ has n vectors (by rank nullity thorem before), the same amount of vectors needed to span $$\mathbb{R}^n$$, by the $$Dimension\text{ }Theorem$$ the basis for $$U\cup U^\perp$$ is also a basis to $$\mathbb{R}^n$$ (It takes any set of n linearly independent vectors for them to span a subspace of dimension n). Therefore $$U\cup U^\perp$$=$$\mathbb{R}^n$$ I'm not sure how rigorous this proof is, I like the one above better as well. I just forgot this and looked up a proof, but also wanted to find my own after reading Jonas'. This idea definitely works and I thought I'd share it as an alternative. | 2022-05-25T13:57:00 | {
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https://www.cs.utexas.edu/users/flame/laff/alaff/chapter10-launch.html | ## Unit10.1.1Subspace iteration with a Hermitian matrix
The idea behind subspace iteration is to perform the Power Method with more than one vector in order to converge to (a subspace spanned by) the eigenvectors associated with a set of eigenvalues.
We continue our discussion by restricting ourselves to the case where $A \in \Cmxm$ is Hermitian. Why? Because the eigenvectors associated with distinct eigenvalues of a Hermitian matrix are mutually orthogonal (and can be chosen to be orthonormal), which will simplify our discussion. Here we repeat the Power Method:
\begin{equation*} \begin{array}{lll} v_0 := \mbox{ random vector} \\ v_0^{(0)} := v_0 / \| v_0 \|_2 \amp \amp \mbox{ normalize to have length one } \\ {\bf for~} k:=0, \ldots \\ ~~~ v_0 := A v_0^{(k)} \\ ~~~ v_0^{(k+1)} := v_0 / \| v_0 \|_2 \amp \amp \mbox{ normalize to have length one } \\ {\bf endfor} \end{array} \end{equation*}
In previous discussion, we used $v^{(k)}$ for the current approximation to the eigenvector. We now add the subscript to it, $v_0^{(k)} \text{,}$ because we will shortly start iterating with multiple vectors.
###### Homework10.1.1.1.
You may want to start by executing git pull to update your directory Assignments.
Examine Assignments/Week10/matlab/PowerMethod.m which implements
[ lambda_0, v0 ] = PowerMethod( A, x, maxiters, illustrate, delay )
This routine implements the Power Method, starting with a vector x for a maximum number of iterations maxiters or until convergence, whichever comes first. To test it, execute the script in Assignments/Week10/matlab/test_SubspaceIteration.m which uses the Power Method to compute the largest eigenvalue (in magnitude) and corresponding eigenvector for an $m \times m$ Hermitian matrix $A$ with eigenvalues $1, \ldots, m \text{.}$
Be sure to click on "Figure 1" to see the graph that is created.
Solution
Watch the video regarding this problem on YouTube: https://youtu.be/8Bgf1tJeMmg. (embedding a video in a solution seems to cause PreTeXt trouble...)
Recall that when we analyzed the convergence of the Power Method, we commented on the fact that the method converges to an eigenvector associated with the largest eigenvalue (in magnitude) if two conditions are met:
• $\vert \lambda_0 \vert \gt \vert \lambda_1 \vert \text{.}$
• $v_0^{(0)}$ has a component in the direction of the eigenvector, $x_0 \text{,}$ associated with $\lambda_0 \text{.}$
A second initial vector, $v_1^{(0)} \text{,}$ does not have a component in the direction of $x_0$ if it is orthogonal to $x_0 \text{.}$ So, if we know $x_0 \text{,}$ then we can pick a random vector, subtract out the component in the direction of $x_0 \text{,}$ and make this our vector $v_1^{(0)}$ with which we should be able to execute the Power Method to find an eigenvector, $x_1 \text{,}$ associated with the eigenvalue that has the second largest magnitude, $\lambda_1$ . If we then start the Power Method with this new vector (and don't introduce roundoff error in a way that introduces a component in the direction of $x_0$), then the iteration will home in on a vector associated with $\lambda_1$ (provided $A$ is Hermitian, $\vert \lambda_1 \vert \gt \vert \lambda_2 \vert \text{,}$ and $v_1^{(0)}$ has a component in the direction of $x_1 \text{.}$) This iteration would look like
\begin{equation*} \begin{array}{lll} x_0 := x_0 / \| x_0 \|_2 \amp \mbox{ } \amp \mbox{ normalize known eigenvector } x_0 \mbox{ to have length one } \\ v_1 := \mbox{ random vector} \\ v_1 := v_1 - x_0^Hv_1 x_0 \amp \amp \mbox{ make sure the vector is orthogonal to } x_0 \\ v_1^{(0)} := v_1 / \| v_1 \|_2 \amp \amp \mbox{ normalize to have length one } \\ {\bf for~} k:=0, \ldots \\ ~~~ v_1 := A v_1^{(k)} \\ ~~~ v_1^{(k+1)} := v_1 / \| v_1 \|_2 \amp \amp \mbox{ normalize to have length one } \\ {\bf endfor} \end{array} \end{equation*}
###### Homework10.1.1.2.
Copy Assignments/Week10/matlab/PowerMethod.m into PowerMethodLambda1.m. Modify it by adding an input parameter x0, which is an eigenvector associated with $\lambda_0$ (the eigenvalue with largest magnitude).
[ lambda_1, v1 ] = PowerMethodLambda1( A, x, x0, maxiters, illustrate, delay )
The new function should subtract this vector from the initial random vector as in the above algorithm.
Modify the appropriate line in Assignments/Week10/matlab/test_SubspaceIteration.m, changing (0) to (1), and use it to examine the convergence of the method.
What do you observe?
Solution
Watch the video regarding this problem on YouTube: https://youtu.be/48HnBJmQhX8. (embedding a video in a solution seems to cause PreTeXt trouble...)
Because we should be concerned about the introduction of a component in the direction of $x_0$ due to roundoff error, we may want to reorthogonalize with respect to $x_0$ in each iteration:
\begin{equation*} \begin{array}{lll} x_0 := x_0 / \| x_0 \|_2 \amp \mbox{ } \amp \mbox{ normalize known eigenvector } x_0 \mbox{ to have length one } \\ v_1 := \mbox{ random vector} \\ v_1 := v_1 - x_0^Hv_1 x_0 \amp \amp \mbox{ make sure the vector is orthogonal to } x_0 \\ v_1^{(0)} := v_1 / \| v_1 \|_2 \amp \amp \mbox{ normalize to have length one } \\ {\bf for~} k:=0, \ldots \\ ~~~ v_1 := A v_1^{(k)} \\ ~~~ v_1 := v_1 - x_0^Hv_1 x_0 \amp \amp \mbox{ make sure the vector is orthogonal to } x_0 \\ ~~~ v_1^{(k+1)} := v_1 / \| v_1 \|_2 \amp \amp \mbox{ normalize to have length one } \\ {\bf endfor} \end{array} \end{equation*}
###### Homework10.1.1.3.
Copy PowerMethodLambda1.m into PowerMethodLambda1Reorth.m and modify it to reorthogonalize with respect to x0:
[ lambda_1, v1 ] = PowerMethodLambda1Reorth( A, x, v0, maxiters, illustrate, delay );
Modify the appropriate line in Assignments/Week10/matlab/test_SubspaceIteration.m, changing (0) to (1), and use it to examine the convergence of the method.
What do you observe?
Solution
Watch the video regarding this problem on YouTube: https://youtu.be/YmZc2oq02kA. (embedding a video in a solution seems to cause PreTeXt trouble...)
We now observe that the steps that normalize $x_0$ to have unit length and then subtract out the component of $v_1$ in the direction of $x_0 \text{,}$ normalizing the result, are exactly those performed by the Gram-Schmidt process. More generally, it is is equivalent to computing the QR factorization of the matrix $\left( \begin{array}{c|c} x_0 \amp v_1 \end{array} \right).$ This suggests the algorithm
\begin{equation*} \begin{array}{l} v_1 := \mbox{ random vector} \\ ( \left( \begin{array}{c | c} x_0 \amp v_1^{(0)} \end{array} ), R \right) := {\rm QR}( \left( \begin{array}{c | c} x_0 \amp v_1 \end{array} \right) ) \\ {\bf for~} k:=0, \ldots \\ ~~~( \left( \begin{array}{c | c} x_0 \amp v_1^{(k+1)} \end{array} \right), R ) := {\rm QR}( \left( \begin{array}{c | c} x_0 \amp A v_1^{(k)} \end{array} \right) ) \\ {\bf endfor} \end{array} \end{equation*}
Obviously, this redundantly normalizes $x_0 \text{.}$ It puts us on the path of a practical algorithm for computing the eigenvectors associated with $\lambda_0$ and $\lambda_1 \text{.}$
The problem is that we typically don't know $x_0$ up front. Rather than first using the power method to compute it, we can instead iterate with two random vectors, where the first converges to a vector associated with $\lambda_0$ and the second to one associated with $\lambda_1 \text{:}$
\begin{equation*} \begin{array}{l} v_0 := \mbox{ random vector} \\ v_1 := \mbox{ random vector} \\ ( \left( \begin{array}{c | c} v_0^{(0)}\amp v_1^{(0)} \end{array} ), R \right) := {\rm QR}( \left( \begin{array}{c | c} v_0 \amp v_1 \end{array} \right) ) \\ {\bf for~} k:=0, \ldots \\ ~~~( \left( \begin{array}{c | c} v_0^{(k+1)} \amp v_1^{(k+1)} \end{array} \right), R ) := {\rm QR}( A \left( \begin{array}{c | c} v_0^{(k)} \amp v_1^{(k)} \end{array} \right) ) \\ {\bf endfor} \end{array} \end{equation*}
We observe:
• If $\vert \lambda_0 \vert \gt \vert \lambda_1 \vert \text{,}$ the vectors $v_0^{(k)}$ will converge linearly to a vector in the direction of $x_0$ at a rate dictated by the ratio $\vert \lambda_1 \vert / \vert \lambda_0 \vert \text{.}$
• If $\vert \lambda_0 \vert \gt \vert \lambda_1 \vert \gt \vert \lambda_2 \vert, \text{,}$ the vectors $v_1^{(k)}$ will converge linearly to a vector in the direction of $x_1$ at a rate dictated by the ratio $\vert \lambda_2 \vert / \vert \lambda_1 \vert \text{.}$
• If $\vert \lambda_0 \vert \geq \vert \lambda_1 \vert \gt \vert \lambda_2 \vert$ then $\Span( \{ v_0^{(k)}, v_1^{(k)} \} )$ will eventually start approximating the subspace $\Span( \{ x_0, x_1 \} ) \text{.}$
What we have described is a special case of subspace iteration. The associated eigenvalue can be approximated via the Rayleigh quotient:
\begin{equation*} \lambda_0 \approx \lambda_0^{(k)} = {v_0^{(k)}}^H A v_0^{(k)} \mbox{ and } \lambda_1 \approx \lambda_1^{(k)} = {v_1^{(k)}}^H A v_1^{(k)} \end{equation*}
Alternatively,
\begin{equation*} A^{(k)} = \left( \begin{array}{c| c} v_0^{(k)} \amp v_1^{(k)} \end{array} \right)^H A \left( \begin{array}{c| c} v_0^{(k)} \amp v_1^{(k)} \end{array} \right) \mbox{ converges to } \left( \begin{array}{c| c} \lambda_0 \amp 0 \\ \hline 0 \amp \lambda_1 \end{array} \right) \end{equation*}
if $A$ is Hermitian, $\vert \lambda_1 \vert \gt \vert \lambda_2 \vert \text{,}$ and $v^{(0)}$ and $v^{(1)}$ have components in the directions of $x_0$ and $x_1 \text{,}$ respectively.
The natural extention of these observations is to iterate with $n$ vectors:
\begin{equation*} \begin{array}{l} \widehat V := \mbox{ random } m \times n \mbox{ matrix} \\ ( \widehat V^{(0)}, R ) := {\rm QR}( \widehat V ) \\ A^{(0)} = { V^{(0)}}^H A V^{(0)} \\ {\bf for~} k:=0, \ldots \\ ~~~( \widehat V^{(k+1)}, R ) := {\rm QR}( A \widehat V^{(k)} ) \\ ~~~ A^{(k+1)} = {\widehat V^{(k+1)}~}^H A \widehat V^{(k+1)} \\ {\bf endfor} \end{array} \end{equation*}
By extending the reasoning given so far in this unit, if
• $A$ is Hermitian,
• $\vert \lambda_0 \vert \gt \vert \lambda_1 \vert \gt \cdots \gt \vert \lambda_{n-1} \vert \gt \vert \lambda_{n} \vert \text{,}$ and
• each $v_j$ has a component in the direction of $x_j \text{,}$ an eigenvector associated with $\lambda_j \text{,}$
then each $v_j^{(j)}$ will converge to a vector in the direction $x_j \text{.}$ The rate with which the component in the direction of $x_p \text{,}$ $0 \leq p \lt n \text{,}$ is removed from $v_j^{(k)} \text{,}$ $n \leq j \lt m \text{,}$ is dictated by the ratio $\vert \lambda_p \vert / \vert \lambda_j \vert \text{.}$
If some of the eigenvalues have equal magnitude, then the corresponding columns of $\widehat V^{(k)}$ will eventually form a basis for the subspace spanned by the eigenvectors associated with those eigenvalues.
###### Homework10.1.1.4.
Copy PowerMethodLambda1Reorth.m into SubspaceIteration.m and modify it to work with an $m \times n$ matrix $V \text{:}$
[ Lambda, V ] = SubspaceIteration( A, V, maxiters, illustrate, delay );
Modify the appropriate line in Assignments/Week10/matlab/test_SubspaceIteration.m, changing (0) to (1), and use it to examine the convergence of the method.
What do you observe?
Solution
Watch the video regarding this problem on YouTube: https://youtu.be/Er7jGYs0HbE. (embedding a video in a solution seems to cause PreTeXt trouble...) | 2021-05-09T05:31:00 | {
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http://mathematica.stackexchange.com/questions/33652/uniformly-distributed-n-dimensional-probability-vectors-over-a-simplex | # Uniformly distributed n-dimensional probability vectors over a simplex
What's the right way to generate a random probability vector $p={p_1,\ldots,p_n} \in {(0,1)}^n$ where $\sum_i p_i=1$, uniformly distributed over the $(n-1)$-dimensional simplex?
What I have is
Intervals = Table[{0, 1}, {i, n}]
RandomPoint := Block[{a},
a = RandomVariate[UniformDistribution[Intervals]];
a/Total[a]];
But I am unsure that this is correct. In particular, I'm unsure that it's any different from:
RandomPoint := Block[{a},
a = Table[Random[], {i, n}];
a/Total[a]];
And the latter clearly will not distribute vectors uniformly. Is the first code the right one?
-
This question may be relevant. – Sjoerd C. de Vries Oct 8 '13 at 11:21
Thanks, @SjoerdC.deVries. That question seems to suggest that my first code is also incorrect? I'm assuming that that bunch of smart guys would have stumbled upon it. – Schiphol Oct 8 '13 at 11:42
Perhaps DirichletDistribution might help? – chuy Oct 8 '13 at 14:03
That question involved points on a sphere. Your constraint of $\sum{p_i}=1$ is different. – Sjoerd C. de Vries Oct 8 '13 at 14:39
I agree with chuy. From wikipedia: Dirichlet Distribution, we have: "When alpha = 1, the symmetric Dirichlet distribution is equivalent to a uniform distribution over the open standard K-1-simplex, i.e. it is uniform over all points in its support". Using DirichletDistribution is most likely then also the best implementation for most purposes. – Jacob Akkerboom Oct 8 '13 at 15:26
#/Total[#,{2}]&@Log@RandomReal[{0,1},{m,n}] will give you a sample of m points from a uniform distribution over an n-1-dimensional regular simplex. (An equilateral triangle is a 2-dimensional regular simplex.) Here's what m = 2000, n = 3 should look like, where {x,y} = {p[[2]]-p[[1]], Sqrt@3*p[[3]]} are the barycentric coordinates of the 3-element probability vector p:
Here's what you get if you omit the Log@ and normalize Uniform(0,1) variables, which is what both of the OP's examples do:
See for yourself. Try it with n = 2 and make a histogram of p[[1]]. Or use n = 3 and ListPlot the barycentric coordinates: {x,y} = {p[[2]]-p[[1]],Sqrt@3*p[[3]]}. – Ray Koopman Oct 9 '13 at 18:41
You can also use Mathematica's built-in DirichletDistribution: points = RandomVariate[DirichletDistribution[{1, 1, 1}], 2000] /. v_?VectorQ :> {v[[2]] - v[[1]], Sqrt[3] (1 - Total[v])}; and then ListPlot[points]. – chuy Oct 11 '13 at 18:28 | 2015-09-02T02:42:17 | {
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https://mathzsolution.com/use-of-without-loss-of-generality/ | # Use of “without loss of generality”
Why do we use “without loss of generality” when writing proofs?
Is it necessary or convention? What “synonym” can be used?
I think this is great question, as the mathematical use of “without loss of generality” often varies from its literal meaning. The literal meaning is when you rephrase a general statement
$$P(x)P(x)$$ is true for all $$x∈Sx \in S$$,
using another set (which is easier to work with)
$$P(z)P(z)$$ is true for all $$z∈Tz \in T$$,
where $$PP$$ is some property of elements in $$SS$$ and $$TT$$, and it can be shown (or is known) that $$S=TS=T$$.
For example:
• We want to show that $$P(x)P(x)$$ is true for all $$x∈Zx \in \mathbb{Z}$$. Without loss of generality, we can assume that $$x=z+1x=z+1$$ for some $$z∈Zz \in \mathbb{Z}$$. [In this case, $$S=ZS=\mathbb{Z}$$ and $$T={z+1:z∈Z}T=\{z+1:z \in \mathbb{Z}\}$$.]
• We want to show that $$P(x)P(x)$$ is true for all $$x∈Zx \in \mathbb{Z}$$. Without loss of generality, we can assume that $$x=5q+rx=5q+r$$ where $$q,r∈Zq,r \in \mathbb{Z}$$ and $$0≤r. [In this case, $$S=ZS=\mathbb{Z}$$ and $$T={5q+r:q∈Z and r∈Z and 0≤r.]
In the above instances, indeed no generality has been lost, since in each case we can prove $$S=TS=T$$ (or, more likely, it would be assumed that the reader can deduce that $$S=TS=T$$). I.e., proving that $$P(z)P(z)$$ holds for $$z∈Tz \in T$$ is the same as proving that $$P(x)P(x)$$ holds for $$x∈Sx \in S$$.
The above cases are examples of clear-cut legitimate usage of "without loss of generality", but there is a widespread second use. Wikipedia writes:
The term is used before an assumption in a proof which narrows the premise to some special case; it is implied that the proof for that case can be easily applied to all others (or that all other cases are equivalent). Thus, given a proof of the conclusion in the special case, it is trivial to adapt it to prove the conclusion in all other cases.
[emphasis mine.]
So, paradoxically, "without loss of generality" is often used to highlight when the author has deliberately lost generality in order to simplify the proof. Thus, we are rephrasing a general statement:
$$P(x)P(x)$$ is true for all $$x∈Sx \in S$$,
as
$$P(z)P(z)$$ is true for all $$z∈Tz \in T$$, and
if $$x∈Sx \in S$$, then there exists $$z∈Tz \in T$$ for which $$P(x)P(x)$$ is true if $$P(z)P(z)$$ is true.
For example:
• Let $$SS$$ be a set of groups of order $$nn$$. We want to show $$P(G)P(G)$$ is true for all $$G∈SG \in S$$. Without loss of generality, assume the underlying set of $$GG$$ is $${0,1,…n−1}\{0,1,\ldots n-1\}$$ for some $$n≥1n \geq 1$$. [Here, $$TT$$ is a set of groups with underlying set $${0,1,…n−1}\{0,1,\ldots n-1\}$$ that are isomorphic to groups in $$SS$$, and the reader is assumed to be able to deduce that property $$PP$$ is preserved by isomorphism.]
My personal preference is to replace the second case with:
"It is sufficient to prove $$P(z)P(z)$$ for $$z∈Tz \in T$$, since [[for some reason]] it follows that $$P(x)P(x)$$ is true for all $$x∈Sx \in S$$." | 2022-10-04T09:07:04 | {
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https://www.physicsforums.com/threads/integration-problems-intriguing-integrals.226048/ | # Integration problems: Intriguing Integrals
1. Apr 2, 2008
### BioCore
1. The problem statement, all variables and given/known data
$$\int$$x^3$$\sqrt{}a^2-x^2$$dx
2. Relevant equations
3. The attempt at a solution
I came this far with my solution:
a^5$$\sqrt{}sin^3\Theta cos^2\Theta$$d$$\Theta$$
a^5$$\sqrt{}sin^3\Theta-sin^5\Theta$$d$$\Theta$$
The answer given to me is:
-1/15(a^2 - x^2)^3/2 (2a^2 + 3x^2) + C
It seems that they used sine theta, and not cos theta. I have currently tried to do that but am not sure how I would get rid of the extra sin$$\Theta$$.
Thanks for the help.
1. The problem statement, all variables and given/known data
2. Relevant equations
3. The attempt at a solution
Last edited by a moderator: Apr 2, 2008
2. Apr 2, 2008
### rock.freak667
$$\int x^3 \sqrt{a^2-x^2}dx$$
let $x=acos\theta \Rightarrow dx=-asin\theta$
$$\int x^3 \sqrt{a^2-x^2}dx \equiv \int (acos\theta)^3\sqrt{a^2-a^2cos^2\theta} \times -asin\theta d\theta$$
$$\int - a^4cos^3\theta \sqrt{a^2(1-cos^2\theta)} \times -asin\theta d\theta$$
$$-a^5 \int cos^3\theta sin^2\theta d\theta$$
recall that $sin^2\theta +cos^2\theta=1$
and substitute for $sin^2\theta$
3. Apr 2, 2008
### BioCore
I would assume that using your solution I would have the right answer. The only problem is that in the solutions my Professor gave us, he has an intermediate step before the final solution such as this:
$$a^5 \int sin^3\theta cos^2\theta d\theta$$
Also in explaining to us Trig Substitution he placed $$\sqrt{a^2 - x^2}$$ as the bottom portion of the right triangle. Thus he stated that
x = asin$$\Theta$$
dx = acos$$\Theta$$
$$\sqrt{a^2 - x^2}$$ = acos\theta
Last edited by a moderator: Apr 2, 2008
4. Apr 2, 2008
### rock.freak667
Your teacher used x=sin$\theta$, I used x=cos$\theta$.
So in his triangle, $sin\theta=\frac{x}{a}$ which would make the adjacent side,$\sqrt{a^2-x^2}$. At the end of that integral you would get terms involving $cos\theta$
In my substitution, the opposite side would be $\sqrt{a^2-x^2}$ and at the end of my integral I would have terms involving $sin\theta$
So in theory it should all work out the same.
5. Apr 2, 2008
### BioCore
$$-a^5 \int \sqrt{1 - sin^2\theta} (1 - sin^2\theta) (sin^2\theta)d\theta$$
= $$-a^5 \int (1 - sin^2\theta)^3/2 (sin^2\theta)d\theta$$
= $$-a^5 \int sin^3\theta - sin^6\theta d\theta$$
= $$-a^5 [1/4sin^4\theta + 1/7sin^7\theta ]$$ +C
Now I know that I should substitute for the sine function, but is this so far correct? | 2018-03-21T06:01:56 | {
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http://math.stackexchange.com/questions/251370/proving-an-algorithms-correctness-in-determining-the-number-of-1-bits-in-a-bit | # Proving an algorithm's correctness in determining the number of 1 bits in a bit string
procedure bit count(S: bit string)
count := 0
while S != 0
count := count + 1
S := S ∧ (S − 1)
return count {count is the number of 1s in S}
Here S-1 is the bit string obtained by changing the rightmost 1 bit of S to a 0 and all the 0 bits to right of this to 1s.
So I understand why this is correct, and I have write a rough explanation;
After every iteration, the rightmost 1 bit in S, as well as all the bits to the right of it, is set equal to 0. Thus, after each iteration, the next right-most 1 is accounted for and set to 0, until the entire string is 0's and the loop breaks with the count equal to the number of 1's.
I know this kind of answer won't pass in any mathematics community, so I was hoping to write a formal proof, but I don't know how to go about doing that. My proof skills are particularly shoddy, so an explanation of the techniques involved would be greatly appreciated.
-
Is ^ and or xor? – copper.hat Dec 5 '12 at 7:04
@copper.hat: It has to be $\land$ for the algorithm to work. – Brian M. Scott Dec 5 '12 at 7:05
Here’s one way to make your informal explanation a bit more formal.
Let $\sigma=b_1b_2\dots b_n$ be a bit string of length $n$. For convenience let $[n]=\{1,\dots,n\}$. If $\sigma$ is not the $0$ string, let $k=\max\{i\in[n]:b_i=1\}$, so that $b_k=1$ and $b_i=0$ for $i>k$. Then $\sigma-1$ is the bit string $a_1a_2\dots a_n$ such that for each $i\in[n]$
$$a_i=\begin{cases} b_i,&\text{if }i<k\\ 0,&\text{if }i=k\\ 1,&\text{if }i>k\;. \end{cases}$$
Then $$\sigma\land(\sigma-1)=(b_1\land a_1)(b_2\land a_2)\ldots(b_n\land a_n)\;,$$ and for $i\in[n]$ we have
$$b_i\land a_i=\begin{cases} b_i,&\text{if }i<k\\ 0,&\text{if }i\ge k\;: \end{cases}$$
$b_k\land a_k=1\land 0=0$, and $b_i\land a_i=0\land 1=0$ for each $i>k$. In other words, if $$\sigma\land(\sigma-1)=c_1c_2\dots c_n\;,$$ then
$$c_i=\begin{cases} b_i,&\text{if }i<k\\ 0,&\text{if }i\ge k\;. \end{cases}$$
Recalling that $b_k=1$ and $b_i=0$ for $i>k$, we see that $\sigma$ and $\sigma\land(\sigma-1)$ differ only in the $k$-th bit, where $\sigma$ has a $1$ and $\sigma\land(\sigma-1)$ has a $0$.
If $|\sigma|_1$ is the number of $1$’s in $\sigma$, we’ve just shown that
$$|\sigma\land(\sigma-1)|_1=|\sigma|_1-1\;.$$
Thus, each pass through the loop adds $1$ to the count and reduces the number of $1$’s in the string by $1$. (In other words, $\text{count}+|\sigma|_1$ is a loop invariant for this loop.) Since $\text{count}$ starts at $0$ and $|\sigma|_1$ at the number of $1$’s in the input bit string, it’s now clear that after $|\sigma|_1$ passes through the loop, the input string will be the zero string, and $\text{count}$ will be the number of $1$’s in the input string. And since the input is now the zero string, we exit the loop with no further change in $\text{count}$.
-
What you observed is already quite good. If you would like to argue more formal I would suggest to prove first a lemma that says.
Lemma: If $S>0$ then the binary representation of $S$ has one digit 1 more than the binary representation of $S \land (S-1)$.
$Proof.$ Let the binary representation of $S\neq 0$ be the string $s=\text{bin}(S)$. We assume that $s$ ends on $v=10^k$ and that $s=uv$. Subtracting 1 from $S$ gives in the binary representation a string $u01^k$. A bit-wise AND of $S$ and $S-1$ is therefore the string $u0^{k+1}$, which proves the lemma.
Use this lemma and a loop invariant to prove the correctness of your algorithm.
-
Actually, I think you have a very good start.
What I think is needed are these ideas:
(1) Explicit specifying of all the locations in S where the one bits are.
(2) An analysis of what the critical step S := S ∧ (S − 1) does.
Here is my take on (2).
Suppose S, at this point, has k zero bits on the right with the k+1st bit a one. Then $S = a 2^k$, with a an odd integer (if $a$ were even, the j+1st bit would be zero).
Then $S-1 = a 2^k-1 = (a-1)2^k + 2^k-1$ so S ^ (S-1) = $(a-1)2^k$ since (1) the k lower bits of S are zero, and (2) since a is odd, a-1 is even, so the k+1st bit of S-1 is zero. Furthermore, since a is odd, all the bits of a-1 are the same as the bits of a except for the low order bit.
Thus the bits of S ^ (S-1) are the same as the bits of S except that the lowest one bit has been set to zero.
This lemma (which is what it really is) allows you to set up an inductive hypothesis that each time through the loop the current low order one bit has been set to zero and count is incremented by one.
At the end, count equals the number of one bits in the original number.
I have great admiration for whoever first came up with this, and I thank John Taylor for providing me with the opportunity of doing this analysis.
I had seen this algorithm before, but had never seen a proof, and I enjoyed working this out.
Aha! I see that Brian M. Scott has answered as I was entering my answer, and that our two analyses are essentially the same with quite different notations.
- | 2016-06-29T23:34:13 | {
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https://math.stackexchange.com/questions/823816/is-sum-limits-n-1-infty-frac-sin-nnn-convergent | # Is $\sum\limits_{n=1}^\infty \frac{|\sin n|^n}n$ convergent？
Is the series
$$\sum_{n=1}^\infty \frac{|\sin n|^n}n\tag{1}$$
convergent？
If one want to use Abel's test, is
$$\sum_{n=1}^\infty |\sin n|^n\tag{2}$$
convergent？
Thank you very much
The second series is divergent because the irrational number $\frac{\pi}{2}$ has an infinite number of convergents $\frac{p_n}{q_n}$ with an odd denominator, hence $\left|\frac{p_n}{q_n}-\frac{\pi}{2}\right|\leq\frac{1}{q_n^2}$ gives: $$\left|\sin(p_n)\right| = \left|\sin\left(\frac{\pi}{2}q_n+\frac{\theta}{q_n}\right)\right|=\left|\cos\frac{\theta}{q_n}\right|,\quad |\theta|\leq 1,$$ $$\left|\sin(p_n)\right|\geq 1-\frac{1}{q_n^2}$$ hence $\left|\sin n\right|^n$ is bigger that $\left(1-\frac{1}{n^2}\right)^n$ infinitely often and the partial sums of $\sum_{n\in\mathbb{N}}\left|\sin n\right|^n$ cannot be bounded. However, they cannot grow too fast: numerical experiments give
$$\sum_{n=1}^{N}\left|\sin n\right|^n=O\left(N^{1/2}\right)\tag{1}$$
(consistent with the Weyl bounds for the partial sums of $\sum_{n\in\mathbb{N}}e^{in^2}$) that is enough to state that $$\sum_{n=1}^{+\infty}\frac{\left|\sin n\right|^n}{n}$$ converges by partial summation. This is also consistent with the model for which $\left|\sin n\right|$ acts like $\sin X$ where $X$ is a random variable, uniformly distributed over $[0,\pi]$: $$E\left[\sum_{n=1}^{+\infty}\frac{(\sin X)^n}{n}\right]=\frac{1}{\pi}\int_{0}^{\pi}-\log(1-\sin\theta)\,d\theta=\frac{4K}{\pi}+\log 2<+\infty.$$ A possible strategy in order to prove $(1)$ is to prove first, through Weyl's equidistribution theorem, that, provided that $M$ is big enough:
$$\sum_{n=N+1}^{N+M}\left|\sin n\right|^r\approx \frac{M}{\pi}\int_{0}^{\pi}\sin^r\theta\,d\theta,\tag{2}$$
then exploit the fact that:
$$\int_{0}^{\pi}\sin^r \theta\,d\theta = O\left(\frac{1}{\sqrt{r}}\right).\tag{3}$$
By putting together $(2)$ and $(3)$ we get:
$$\sum_{i=1}^{M^2}\left|\sin n\right|^n\leq\sum_{j=0}^{M-1}\sum_{n=Mj+1}^{M(j+1)}\left|\sin n\right|^{Mj+1}\ll M\cdot\sum_{j=0}^{M-1}\frac{1}{\sqrt{Mj+1}}\ll M.\tag{4}$$
Moreover, since the supremum of the derivative of $\sin^r\theta$ over $[0,\pi]$ behaves like $\sqrt{\frac{r}{e}}$, keeping track of all the constants we get: $$\sum_{n=1}^{+\infty}\frac{\left|\sin n\right|^n}{n}\leq\frac{2}{\sqrt{\pi}}\left(1+\frac{1}{\sqrt{e}}\right)\zeta(3/2) = 4.73565\ldots$$ that is not too much far from the truth (given by partial summation and explicit numerical computations till $n=10^8$, with an error term $\frac{C}{10^4}$):
$$\sum_{n=1}^{+\infty}\frac{\left|\sin n\right|^n}{n}\leq \color{red}{2.151}.\tag{5}$$
Post-mortem addendum: it is interesting to study tailor-made numerical methods for the computation of such very slowly-converging series.
• Do you have an idea on how to prove the estimate with $O(N^{1/2+\epsilon})$? – Gabriel Romon Jun 16 '14 at 13:02
• @G.T.R: I extended my answer including my thoughts about proving the crucial estimate. – Jack D'Aurizio Jun 16 '14 at 16:50
• @Jack D'Aurizio: Writing in MMA Sum[Abs[Sin[n]]^n/n, {n, 1, 10^8}] I do not get an answer. How did you get to 2.151? Thank you! – TeM Jul 14 '17 at 20:10
• @TeM: just follow the last link. – Jack D'Aurizio Jul 14 '17 at 20:11
• It's not clear to me that this constitutes a rigorous proof; how is Weyl's equidistribution theorem sufficient here? – user41281 Nov 21 '17 at 1:36
Note: Another great answer to this problem is given by Terry Tao in this MO post.
Disclaimer:Not my solution. One elegant answer was posted by Robert Isreal on here . Apparently a holistic answer is not known. | 2019-11-17T23:45:25 | {
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# DIGITAL SUM of the number
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06 Sep 2010, 18:32
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DIGITAL SUM of the number
Given a number $$N_z$$ , all the digits of N are added to obtain a number $$N_y$$. All the digits of N are added to obtain a number $$N_x$$, and so on, till we obtain a single digit number $$N$$. This single digit number $$N$$ is called the digital sum of the original number $$N_z$$.
What is the digital sum of 84001 ?
Answer: 8+4+0+0+1 = 13 = 1+3 = 4 Hence, the digit sum of the number is 4.
Note: In finding the digital sum of a number we can ignore the digit 9 or the digits that add up to 9.
For example, in finding the digital sum of the number 246819, we can ignore the digits 2, 6, 1, and 9. Hence,
the digital sum of 246819 is = 4 + 8 = 12 = 1 + 2 = 3.
Digital Sum Rule of Multiplication: The digital sum of the product of two numbers is equal to the digital sum of the product of the digital sums of the two numbers.
Example: The product of 129 and 35 is 4515.
Digital sum of 129 = 3 and digital sum 35 = 8. Product of the digital sums = 3 × 8 = 24 Digital sum = 6.
Digital sum of 4515 is = 4 + 5 + 1 + 5 = 15 = 1 + 5 = 6. Digital sum of the product of the digital sums =
digital sum of 24 = 6 Digital sum of the product (4515) = Digital sum of the product of the digital sums
(24) = 6
Applications of Digital Sum :
1.Rapid checking of calculations while multiplying numbers :
Suppose a student is trying to find the product 316 × 234 × 356, and he obtains the number 26525064. A
quick check will show that the digital sum of the product is 3. The digital sums of the individual numbers
(316, 234 and 356) are 1, 9, and 5. The digital sum of the product of the digital sum is 1 × 9 × 5 = 45 = 4 + 5 = 9. ⇒ the digital sum of the product of the digital sums (9) ≠ digital sum of the 26525064 (3). Hence,
the answer obtained by multiplication is not correct.
Note: Although the answer of multiplication will not be correct if the digital sum of the product of the digital
sums is not equal to digital sum of the product, but the reverse is not true i.e. the answer of multiplication
may or may not be correct if the digital sum of the product of the digital sums is equal to digital sum of
the product.
Suppose you have a question : ( Very absurd eg by me but learn the concept )
What is value is A+B+C+D+E?
$$878373 * 7738838339 * 827287 = E6D35C50B5547A87623729$$
HINT: Find the digital sum on LHS and equate it with that of RHS.
2.Determining if a number is a perfect square or not :
The digital sum of the numbers which are perfect squares will
always be 1, 4, 7, or 9.
Note: A number will NOT be a perfect square if its digital sum is NOT 1, 4, 7, or 9, but it may or may not
be a perfect square if its digital sum is 1, 4, 7, or 9.
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08 Sep 2010, 22:18
dang.... what are some real life applications of this digital sum?
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09 Sep 2010, 01:18
Thats an interesting piece of information gurpreet !
Thanks
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17 Sep 2010, 02:41
Interesting piece of info Gurpreet.
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https://nbviewer.jupyter.org/url/www.maths.usyd.edu.au/u/olver/teaching/MATH3976/notes/27.ipynb | # Lecture 27: Error of quadrature¶
Last lecture we observed the following:
1. The right-hand rule had error $O(n^{-1})$ for $n$ sample point
2. The trapezium rule had error $O(n^{-2})$ for $n$ sample point
3. The trapezium rule had error $O(n^{-\alpha})$ for all choices of $\alpha \geq 0$ for smooth, periodic functions
This lecture we set out to prove these results.
## Right-hand rule convergence¶
Recall that the right-hand rule is defined by, for $x_k \triangleq kh$ and $h \triangleq 1/n$
$$\int_0^1 f(x) dx \approx h \sum_{k=1}^n f(x_k) = \sum_{k=1}^n \int_{x_{k-1}}^{x_k} (f(x) - f(x_k) ) dx$$
About the only tool available to analyse integrals is integration by parts:
\begin{align*} \int_a^b u(x) v'(x) dx &= [u(x) v(x)]_a^b - \int_a^b u'(x) v(x) \cr & = u(b) v(b) - u(a)v(a) - \int_a^b u'(x) v(x) \end{align*}
Consider the error in the first panel. Taking $u(x) = f(x)$ and $v(x) = x$ in the integration by parts formula yields:
\begin{align*} \int_0^h (f(x) - f(h)) dx = [(f(x) - f(h)) x]_0^h - \int_0^h f'(x) x dx \cr = - \int_0^h f'(x) x dx \end{align*}
Assume that $f'(x)$ is bounded in $[0,1]$, and define
$$M= \sup_{x \in [0,1]} |f'(x)|$$
We then get a bound on the error:
\begin{align*} \left|\int_0^h (f(x) - f(h)) dx\right| &= \left|\int_0^h f'(x) x dx\right| \leq \int_0^h |f'(x)| x dx \leq M \int_0^h x dx \cr &\leq {M h^2 \over 2} \end{align*}
This formula caries over to every other panel (by the same argument):
$$\left|\int_{x_{k-1}}^{x_k} (f(x) - f(x_k)) dx \right| \leq {M h^2 \over 2}$$
Thus we have
\begin{align*} \left\|\int_0^1 f(x) dx - h \sum_{k=1}^n f(x_k)\right\| &= \left\|\sum_{k=1}^n \int_{x_{k-1}}^{x_k} (f(x) - f(x_k) ) dx\right\| \cr &\leq \sum_{k=1}^n \left\|\int_{x_{k-1}}^{x_k} (f(x) - f(x_k) ) \right\| dx &\leq \sum_{k=1}^n {M h^2 \over 2} = {M h^2 n \over 2} = {M \over 2 n} = O(n^{-1}) \end{align*}
Thus we have proven that the right-hand rule converges like $O(n^{-1})$.
## Trapezium rule observed convergence, revisited¶
Recall the trapezium rule, here implemented for general intervals $[a,b]$:
In [2]:
using PyPlot
function trap(f,a,b,n)
h=(b-a)/n
x=linspace(a,b,n+1)
v=f(x)
h/2*v[1]+sum(v[2:end-1]*h)+h/2*v[end]
end
trap(f,n)=trap(f,0.,1.,n);
Consider integration of the following simple function:
In [3]:
f=x->exp(x).*cos(x)
g=linspace(0.,1.,1000)
plot(g,f(g));
As in last lecture, we can compare the error of trap to the exact integral, approximated by quadgk.
In [5]:
ex=quadgk(f,0.,1.)[1]
ns=round(Int,logspace(1,5,100)) # integers spaced logarithmically apart
errT=zeros(length(ns))
errπ=zeros(length(ns))
for k=1:length(ns)
n=ns[k]
errT[k]=abs(trap(f,n-1)-ex ) # error in non-periodic
errπ[k]=abs(trap(θ->cos(cos(θ-0.1)),0.,2π,n-1)-exπ ) # error in periodic
end
We see that for non-periodic functions we observe $O(n^{-2})$ convergence but for periodic functions we converge much faster:
In [6]:
loglog(ns,errT) # blue curve
loglog(ns,errπ) # green curve
loglog(ns,(1.0ns).^(-2)) # red curve, with same slope as blue curve
Out[6]:
1-element Array{Any,1}:
PyObject <matplotlib.lines.Line2D object at 0x304c5ed90>
The defining property of smooth, periodic functions is that the function and all its derivatives match at 0 and 2π. Now consider a function where only some of he derivatives match:
In [8]:
h=x->10f(x).*x.^2.*(1-x).^2
plot(g,h(g));
We see here that the convergence rate has converged to $O(n^{-4})$:
In [11]:
exf=quadgk(f,0.,1.)[1]
ns=round(Int,logspace(1,4,100))
errf=zeros(length(ns))
errh=zeros(length(ns))
for k=1:length(ns)
n=ns[k]
errf[k]=abs(trap(f,n-1)-exf )
errh[k]=abs(trap(h,n-1)-exh )
end
loglog(ns,errf) # blue curve
loglog(ns,1./ns.^2) # green curve
loglog(ns,errh) # red curve
loglog(ns,1./ns.^4) # aqua curve;
The conclusion is that derivatives at the endpoints dictate the convergence rate of the Trapezium rule.
# Bernoulli Polynomial¶
To prove this, we need to be clever about our choice of functions when we integrate by parts. Define the first three Bernoulli Polynomials by
$$B_0(x) = 1, B_1(x) = x-{1 \over 2}, B_2(x) = x^2 -x + {1 \over 6}$$
as depticted here:
In [12]:
g=linspace(0.,1.,1000)
B0=x->ones(x)
B1=x->x-1/2
B2=x->x.^2-x+1/6
plot(g,B0(g))
plot(g,B1(g))
plot(g,B2(g));
These polynomials satisfy two important properties:
1. $B_k'(x) = k B_{k-1}(x)$, just like monomials $x^k$
2. $B_2(-1) = B_2(1) = {1 \over 6} We Thus consider integration by parts with the Trapezium rule. Recall the Trapezium rule: $$\int_0^1 f(x) dx \approx h\left({f(x_0) \over 2} + \sum_{k=1}^{n-1} f(x_k) + {f(x_n) \over 2} \right) = \sum_{k=1}^n \int_0^1 \left(f(x_{k-1}) + (x-x_{k-1}) {f(x_k) - f(x_{k-1}) \over h}\right) dx$$ As in the right-hand rule, we begin with the error in the first panel: $$\int_0^h \left[f(x) - \left(f(0) + x {f(h) - f(0) \over h}\right) \right] dx.$$ Taking $$u(x) = f(x) - \left(f(0) + x {f(h) - f(0) \over h}\right)$$ and$v(x) = h B_1(x/h)$(so that$v'(x) = 1$) in the integration by parts formula: $$\int_0^h \left[f(x) - \left(f(0) + x {f(h) - f(0) \over h}\right) \right] dx = [u(x) v(x)]_0^h - \int_0^h u(x) v(x) dx = - h \int_0^h \left[f'(x) - {f(h) - f(0) \over h}\right] B_1(x/h) dx.$$ Here we used that$u(0) = u(h) = 0$to kill of the first terms. Now take$u(x) = f'(x) - {f(h) - f(0) \over h}$and$v(x) = h {B_2(x/h) \over 2}$(so that$v'(x) = B_1(x/h)): \begin{align*} - h \int_0^h \left[f(x) - {f(h) - f(0) \over h}\right] B_1(x/h) dx &= -{h^2 \over 2}\left[\left(f'(h) - {f(h) - f(0) \over h}\right) B_1(1) - \left(f'(0) - {f(h) - f(0) \over h}\right) B_1(-1)\right] \cr & \qquad +{h^2 \over 2} \int_0^h f''(x) B_2(x/h) dx \cr & = - {f'(h) - f'(0) \over 12} h^2 +{h^2 \over 2} \int_0^h f''(x) B_2(x/h) dx \end{align*} By the same logic, we have in each panel $$\int_{x_{k-1}}^{x_k} \left[f(x) - \left(f(x_{k-1}) + x {f(x_k) - f(x_{k-1}) \over h}\right) \right] dx = - {f'(x_k) - f'(x_{k-1}) \over 12} h^2 +{h^2 \over 2} \int_{x_{k-1}}^{x_k} f''(x) B_2((x-x_{k-1})/h) dx$$ Summing over every panel, and using the telescoping sum, gives us \begin{align*} \int_0^1 f(x) dx - h\left({f(x_0) \over 2} + \sum_{k=1}^{n-1} f(x_k) + {f(x_n) \over 2} \right) = -{h^2 \over 12} \sum_{k=1}^n \left(f'(x_k) - f'(x_{k-1}) - \int_{x_{k-1}}^{x_k} f''(x) B_2((x-x_{k-1})/h) dx\right) \cr -{h^2 \over 12} (f'(1) - f'(0)) + {h^2 \over 12} \sum_{k=1}^n \int_{x_{k-1}}^{x_k} f''(x) B_2((x-x_{k-1})/h) dx \end{align*} But on each panel we have $$\left|\int_{x_{k-1}}^{x_k} f''(x) B_2((x-x_{k-1})/h) dx\right| \leq \int_{x_{k-1}}^{x_k} |f''(x)| |B_2((x-x_{k-1})/h)| dx \leq M_2 {h \over 6}$$ whereM_2 = \sup_{x \in [0,1]} |f''(x)|$and we used the fact that$|B_2(x)| \leq {1 \over 6}\$ (Exercise). Thus we have
$$\left|\sum_{k=1}^n \int_{x_{k-1}}^{x_k} f''(x) B_2((x-x_{k-1})/h) dx\right| \leq {M_2 \over 6}$$
and hence we have
$$\int_0^1 f(x) dx - h\left({f(x_0) \over 2} + \sum_{k=1}^{n-1} f(x_k) + {f(x_n) \over 2} \right) = -{1 \over 12 n^2} (f'(1) - f'(0)) + O(n^{-2}) = O(n^{-2})$$
which proves our second observation. | 2020-06-02T08:32:52 | {
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https://eventthyme.net/how-to-find-average-velocity-calculus/ | in
# How To Find Average Velocity Calculus
Given, s = 3t2 − 6t. Average velocity is a vector quantity.
Force and Motion Worksheet Motion graphs, Worksheets and
### Use this information to guess the instantaneous velocity of the rock at time 1.
How to find average velocity calculus. So,displacement in between 2s and 5s is s = 3[t2]5 2 − 6[t]5 2 = 3(25 −4) − 6(5 − 2) = 45m. Yes, you're supposed to find the average velocity over each interval. The velocity at t = 10 is 10 m/s and the velocity at t = 11 is 15 m/s.
To find the instantaneous velocity at any position, we let t1=t t 1 = t and t2=t+δt t 2 = t + δ t. [0,3] b.[0,2] c.[0,1] d.[0,h] where h>0 is a real number Nevertheless, this is exactly how a gps determines velocity from position!
Velocity calculations used in calculator: For example, if you drive a car for a distance of 70 miles in one hour, your average velocity equals 70 mph. Solving for the different variables we can use the.
The average velocity formula describes the relationship between the length of your route and the time it takes to travel. To avoid these tedious calculations, we would really like to have a formula. Average velocity is defined as total displacement/ total time taken for that.
It is determined that its height (in feet) above ground t seconds later (for is given by find the average velocity of the rock over each of the given time intervals. Her average velocity is 8m west / 4s = 2 m/s west. = 12 miles per hour.
For example, if an object is tossed into the air we might find the following data for the height in feet, y, of the object as a function of the time in seconds, t, where t = 0 is when the object is released upward. How to find the change in position over the change in time for various intervals, and finding the slope of the secant line to find average velocity. If you recall from earlier mathematics studies, average velocity is just net distance traveled divided by time.
However, this technically only gives the object's average velocity over its path. Average velocity is defined in terms of the relationship between the distance traveled and the time that it takes to travel that distance. Using calculus, it's possible to calculate an object's velocity.
Although speed and velocity are often words used interchangeably, in physics, they are distinct concepts. So,average velocity = 45 5 −2 = 15ms−1. In other words, velocity is equal to rate of change of position vector with respect to time.
The average acceleration would be: I was showed the correct answer but i really need to learn what the process is to achieve the answer. The sum of the initial and final velocity is divided by 2 to find the average.
For example, let’s calculate a using the example for constant a above. If you are in need of technical support, have a question about advertising opportunities, or have a general question, please contact us by phone or submit a message through the form below. Bart walks west at 5 m/s for 3 seconds, then turns around and walks east at 7 m/s for 1 second.
Average velocity = 8 m west / 4s = 2 m/s west. The average velocity calculator uses the formula that shows the average velocity (v) equals the sum of the final velocity (v) and the initial velocity (u), divided by 2. The average velocity of an object is its change in position divided by the total amount of time taken.
Purchase calculus etf 6e hide menu show menu. In many common situations, to find velocity, we use the equation v = s/t, where v equals velocity, s equals the total displacement from the object's starting position, and t equals the time elapsed. (i) 0.1 seconds ft/s (ii) 0.05 seconds ft/s (iii) 0.01 seconds ft/s (b) estimate the instantaneous velocity (in ft/s) of the pebble after 2 seconds.
Using calculus to find acceleration. The average velocity formula and velocity units. An automobile travels 540 kilometers in 4 hours and 30 minutes.
Let's take a look at average velocity. Computing average velocities for smaller, and smaller, values of as we did above is tedious. Find the average velocity of the object over the following intervals.
Jane · 3 · mar 17 2018. The formula for finding average velocity is: Math video on how to compute the average velocity of an object moving in one dimension and how to represent average velocity on a graph.
The instantaneous velocity at an instant t or simply ‘velocity’ at an instant t is defined as limiting value of the average velocity as δt → 0, evaluated at time t. The average velocity of a body in a certain time interval is given as the displacement of the body in that time interval divided by time. Find the average velocity (in ft/s) of the pebble for the time period beginning when t = 2 and lasting the following amount of time.
The formula for calculating average velocity is therefore: Acceleration is measured as the change in velocity over change in time (δv/δt), where δ is shorthand for “change in”. In the previous section, we have introduced the basic velocity equation, but as you probably have already realized.
When calculating the average velocity, only the times and positions at the starting and ending points are taken into account. Speed (or rate, r) is a scalar quantity that measures the distance traveled (d) over the change in time (δt), represented by. If a ball is thrown into the.
Average velocity is the result of dividing the distance an object travels by the time it takes to travel that far. Example 2 finding average velocity a rock is dropped from a height of 64 ft. The average velocity of an object is its total displacement divided by the total time taken.
In other words, it is the rate at which an object changes its position from one place to another. In this case, we have $$s(t)=\frac{13}{t+2}$$ (if you mean something else, please say so). Average velocity = change in speed / change in time.
This is a question from my calculus book. The average velocity is different from the instantaneous velocity, and the two are many times not the same number. So if a particle covers a certain displacement $$\overrightarrow{ab}$$ in a time $$t_1$$ to $$t_2$$, then the average velocity of the particle is:
Velocity (v) is a vector quantity that measures displacement (or change in position, δs) over the change in time (δt), represented by the equation v = δs/δt.
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Calculating Average Speed WS Elementary school science | 2021-11-27T19:40:30 | {
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"url": "https://eventthyme.net/how-to-find-average-velocity-calculus/",
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https://math.stackexchange.com/questions/2992385/non-isomorphic-connected-unicyclic-graphs | # Non-isomorphic connected, unicyclic graphs
The question is: Find the number of non-isomorphic connected, unicyclic graphs(graphs with exactly one cycle) on 6 vertices.
According to this question does it mean that there is a general formula that enable us to find the number of non-isomorphic connected, unicyclic graphs on n vertices so that for vertices 6 may be particular?
This task doesn't suggest the existence of such a formula.
However, you can find a general formula for the number of non-isomorphic connected unicyclic graphs on $$n$$ vertices and specify it for $$n=6$$ to get the result.
But you can as well just count such graphs for $$n=6$$ directly.
• As I think hexagon is the only non-isomorphic connected unicyclic graphs on 6 vertices. Is that not? – 2468 Nov 10 '18 at 13:58
• No, you can have a smaller cycle as well. – Jakob B. Nov 10 '18 at 14:14
Using the notation from Analytic Combinatorics we have for the combinatorial class $$\mathcal{U}$$ of unicyclic non-isomorphic graphs the equation (we are attaching trees to the nodes of the single cycle where the root of the tree is merged into the cycle)
$$\def\textsc#1{\dosc#1\csod} \def\dosc#1#2\csod{{\rm #1{\small #2}}} \mathcal{U} = \textsc{DHD}_{\ge 3}(\mathcal{T})$$
where $$\mathcal{T}$$ is the class of unlabeled rooted trees:
$$\mathcal{T} = \mathcal{Z} \times \textsc{MSET}(\mathcal{T})$$
Here we have used the dihedral operator for the single cycle (which is a bracelet i.e. a necklace that can be turned over) and the unlabeled multiset operator. The class equation for trees immediately yields a functional equation via the exponential formula, and which is
$$T(z) = z \exp\left(\sum_{\ell\ge 1} \frac{T(z^\ell)}{\ell}\right).$$
It was proved at the following MSE link using this functional equation that these have recurrence
$$t_{n+1} = \frac{1}{n} \sum_{q=1}^{n} t_{n+1-q} \left(\sum_{\ell|q} \ell t_{\ell}\right).$$
Using the cycle index $$Z(D_q)$$ of the dihedral group we have
$$U(z) = \sum_{q\ge 3} Z(D_q; T(z)).$$
Therefore the number of non-isomorphic connected unicyclic graphs is
$$U_n = [z^n] \sum_{q=3}^n Z\left(D_q; \sum_{p=1}^n t_p z^p\right).$$
This yields the sequence
$$0, 0, 1, 2, 5, 13, 33, 89, 240, 657, 1806, 5026, 13999, \\ 39260, 110381, 311465, 880840, 2497405, 7093751, 20187313, \ldots$$
with two leading zeros because the smallest cycle is on three nodes. The data point to OEIS A001429 where the procedure is confirmed. In particular we get for six nodes
$$\bbox[5px,border:2px solid #00A000]{ U_6 = 13.}$$
Remark. The cycle index of the cyclic group is given by
$$Z(C_q) = \frac{1}{q} \sum_{k|q} \phi(k) a_k^{q/k}$$
and of the dihedral group
$$Z(D_q) = \frac{1}{2} Z(C_q) + \begin{cases} \frac{1}{2} a_1 a_2^{(q-1)/2} & q \quad \text{odd} \\ \frac{1}{4} \left( a_1^2 a_2^{(q-2)/2} + a_2^{q/2} \right) & q \quad\text{even.} \end{cases}.$$
This computation was done with the following Maple code.
with(numtheory);
pet_varinto_cind :=
proc(poly, ind)
local subs1, subs2, polyvars, indvars, v, pot, res;
res := ind;
polyvars := indets(poly);
indvars := indets(ind);
for v in indvars do
pot := op(1, v);
subs1 :=
[seq(polyvars[k]=polyvars[k]^pot,
k=1..nops(polyvars))];
subs2 := [v=subs(subs1, poly)];
res := subs(subs2, res);
od;
res;
end;
pet_cycleind_cyclic :=
proc(n)
local s, d;
s := 0;
for d in divisors(n) do
s := s + phi(d)*a[d]^(n/d);
od;
s/n;
end;
pet_cycleind_dihedral :=
proc(n)
local s;
s := 1/2*pet_cycleind_cyclic(n);
if type(n, odd) then
s := s + 1/2*a[1]*a[2]^((n-1)/2);
else
s := s + 1/4*(a[1]^2*a[2]^((n-2)/2) + a[2]^(n/2));
fi;
s;
end;
t :=
proc(n)
option remember;
if n=1 then return 1 fi;
q=1..n-1);
end;
U :=
proc(n)
option remember;
local res, m, tgf, dhdgf;
res := 0;
for m from 3 to n do
dhdgf :=
pet_varinto_cind(tgf, pet_cycleind_dihedral(m));
res := res +
coeff(expand(dhdgf), z, n);
od;
res;
end;
Addendum. We can also answer the question for the labeled case. The combinatorial classes are the same, only now we get the classic tree function $$T(z)$$ using
$$\mathcal{T} = \mathcal{Z} \times \textsc{SET}(\mathcal{T})$$
and functional equation
$$T(z) = z \times \exp T(z)$$
and the dihedral operator becomes
$$\sum_{\ell\ge 3} \frac{z^\ell}{|D_\ell|} = \sum_{\ell\ge 3} \frac{z^{\ell}}{2\ell} = -\frac{1}{2} z - \frac{1}{4} z^2 + \frac{1}{2} \log \frac{1}{1-z}.$$
We are thus interested in extracting the coefficient as follows,
$$n! [z^n] \left(-\frac{1}{2} T(z) - \frac{1}{4} T(z)^2 + \frac{1}{2} \log \frac{1}{1-T(z)}\right).$$
This has three components, the first is by Cayley
$$- n! [z^n] \frac{1}{2} T(z) = -\frac{1}{2} n^{n-1}.$$
The second is
$$- n! [z^n] \frac{1}{4} T(z)^2 = - (n-1)! [z^{n-1}] \frac{1}{2} T(z) T'(z) \\ = -\frac{(n-1)!}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^n} \frac{1}{2} T(z) T'(z) \; dz.$$
Letting $$w=T(z)$$ so that $$z= w \exp(-w)$$ and $$dw = T'(z) \; dz$$ this becomes for $$n\ge 2$$
$$-\frac{(n-1)!}{2\pi i} \int_{|w|=\gamma} \frac{\exp(nw)}{w^n} \frac{1}{2} w \; dw = - \frac{(n-1)!}{2} \frac{n^{n-2}}{(n-2)!} \\ = - \frac{1}{2} (n-1) n^{n-2}.$$
Finally for the third one we get
$$n! [z^n] \frac{1}{2} \log\frac{1}{1-T(z)} = (n-1)! [z^{n-1}] \frac{1}{2} \frac{1}{1-T(z)} T'(z) \\ = \frac{(n-1)!}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^n} \frac{1}{2} \frac{1}{1-T(z)} T'(z) \; dz.$$
With the same substitution as before we find
$$\frac{(n-1)!}{2\pi i} \int_{|w|=\gamma} \frac{\exp(nw)}{w^n} \frac{1}{2} \frac{1}{1-w} \; dw \\ = \frac{1}{2} (n-1)! \sum_{q=0}^{n-1} \frac{n^q}{q!}.$$
Collecting everything we obtain
$$\frac{1}{2} (n-1)! \sum_{q=0}^{n-1} \frac{n^q}{q!} - n^{n-1} + \frac{1}{2} n^{n-2}$$
or alternatively
$$\bbox[5px,border:2px solid #00A000]{ \frac{1}{2} (n-1)! \sum_{q=0}^{n-3} \frac{n^q}{q!}.}$$
This sequence is OEIS A057500:
$$0, 0, 1, 15, 222, 3660, 68295, 1436568, 33779340, 880107840, \\ 25201854045, 787368574080, 26667815195274, 973672928417280, \ldots$$ | 2020-02-23T18:25:02 | {
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https://www.gaussianwaves.com/2014/07/generating-basic-signals-gaussian-pulse-and-power-spectral-density-using-fft/ | # Generating Basic Signals – Gaussian Pulse and Power Spectral Density using FFT
(5 votes, average: 4.80 out of 5)
For more such examples check this ebook : Digital Modulations using Matlab: build simulation models from scratch – by Mathuranathan Viswanathan
## Introduction
Numerous texts are available to explain the basics of Discrete Fourier Transform and its very efficient implementation – Fast Fourier Transform (FFT). Often we are confronted with the need to generate simple, standard signals (sine, cosine, Gaussian pulse, squarewave, isolated rectangular pulse, exponential decay, chirp signal) for simulation purpose. I intend to show (in a series of articles) how these basic signals can be generated in Matlab and how to represent them in frequency domain using FFT.
## Gaussian Pulse : Mathematical description:
In digital communications, Gaussian Filters are employed in Gaussian Minimum Shift Keying – GMSK (used in GSM technology) and Gaussian Frequency Shift Keying (GFSK). Two dimensional Gaussian Filters are used in Image processing to produce Gaussian blurs. The impulse response of a Gaussian Filter is Gaussian. Gaussian Filters give no overshoot with minimal rise and fall time when excited with a step function. Gaussian Filter has minimum group delay. The impulse response of a Gaussian Filter is written as a Gaussian Function as follows
$$g(t) = \frac{1}{\sqrt{2 \pi } \sigma} e^{- \frac{t^2}{2 \sigma^2}}$$
The Fourier Transform of a Gaussian pulse preserves its shape.
\begin{align} G(f) & =F[g(t)]\ & = \int_{-\infty }^{\infty } g(t)e^{-j2\pi ft}\, dt\ & = \frac{1}{\sigma \sqrt{2 \pi } } \int_{-\infty }^{\infty } e^{- \frac{t^2}{2 \sigma^2}}e^{-j2\pi ft}\, dt\ &=\frac{1}{\sigma \sqrt{2 \pi } } \int_{-\infty }^{\infty } e^{- \frac{1}{2 \sigma^2}\left[t^2+j4 \pi \sigma^2 ft \right]}\, dt\ &=\frac{1}{\sigma \sqrt{2 \pi } } \int_{-\infty }^{\infty } e^{- \frac{1}{2 \sigma^2}\left[t^2+j4 \pi \sigma^2 ft + (j 2 \pi \sigma^2 f)^2 – (j 2 \pi \sigma^2 f)^2\right]}\, dt\ &=e^{ \frac{1}{2 \sigma^2}(j 2 \pi \sigma^2 f)^2}\frac{1}{\sigma \sqrt{2 \pi } } \int_{-\infty }^{\infty } e^{- \frac{1}{2 \sigma^2}\left[t+j 2 \pi \sigma^2 f \right]^2}\, dt\ &=e^{ \frac{1}{2 \sigma^2}(j 2 \pi \sigma^2 f)^2}=e^{ – \frac{1}{2}( 2 \pi \sigma f)^2} \end{align }
The above derivation makes use of the following result from complex analysis theory and the property of Gaussian function – total area under Gaussian function integrates to 1. By change of variable, let (u=t+j 2 \pi \sigma^2 f ). Then,
$$\frac{1}{\sigma \sqrt{2 \pi } }\int_{-\infty }^{\infty }e^{- \frac{1}{2 \sigma^2}\left[t+j 2 \pi \sigma^2 f \right]^2}\, dt =\frac{1}{\sigma \sqrt{2 \pi } }\int_{-\infty }^{\infty }e^{- \frac{1}{2 \sigma^2} u^2}\, du =1$$
Thus, the Fourier Transform of a Gaussian pulse is a Gaussian Pulse.
$$\frac{1}{\sqrt{2 \pi } \sigma} e^{- \frac{t^2}{2 \sigma^2}} \rightleftharpoons e^{ – \frac{1}{2}( 2 \pi \sigma f)^2}$$
## Gaussian Pulse and its Fourier Transform using FFT:
The following code generates a Gaussian Pulse with ( \sigma=0.1s ). The Discrete Fourier Transform of this digitized version of Gaussian Pulse is plotted with the help of (FFT) function in Matlab.
## Double Sided and Single Power Spectral Density using FFT:
Next, the Power Spectral Density (PSD) of the Gaussian pulse is constructed using the FFT. PSD describes the power contained at each frequency component of the given signal. Double Sided power spectral density is plotted first, followed by single sided power spectral density plot (retaining only the positive frequency side of the spectrum).
For more such examples check this ebook : Digital Modulations using Matlab: build simulation models from scratch – by Mathuranathan Viswanathan
## Recommended Signal Processing Books
• Vittorio Todisco
Hi, thanks again for your works here.
I have a problem, I’m plotting a 3D Gaussian pulse.
I have been able to create it and to make the fft of it, but the phase it’s not null and quite incomprehensible.
This is a quick example of what I get:
close all;
clear;
clc;
N = 64; % Size of a sunction
D = N/2; % to indicate origin at the center of the function
a = 8; % radius for cylindrical function
w = 0.4; % decay rate for exponential function
y = repmat(1:N,N,1);
x = y’;
r = sqrt((D-x).^2+(D-y).^2); % definition of radius
sig = 5; % Variance for gaussian function
g = (2*pi*sig^2)*exp(-((r.^2))./(2*sig^2));
figure;mesh(g); title(‘2D Gaussian Function’);
G = fft2(g);
G = fftshift(G);
figure; mesh(abs(G)); title(‘Fourier Transform of Gaussian function’);
fzf=atan2(imag(G),real(G));
figure;
mesh(fzf);
title(‘Fourier Phase of Gaussian function’);
• pankaj
Dear sir, this code is great for generating the gaussian pulse without using matlab toolboxes.
I want to ask that from the same code, if i want to generate the train of gaussian pulses, how can i generate (without using the matlab function pulstran)? kindly provide the baisc code for it. I have searched over the internet but not found. suppose I want to generate 12 gaussian pulses wtith sum time duration between them. plzz.
Hello this is Lingaraj, doing m-tech in belgaum. I have a problem in finding power spectral density using time domain approach,i have a written code found somewhere but i am not getting it. Please leave a reply as soon as possible…
• KP
Hi Viswanathan,
This really awesome article on Gaussian wave generation and analysis. I am trying to generate a Gaussian pulse of 20ns and plot the frequency response of it. When I try to simulate it in matlab, it gives me error saying out of memory. I think I am doing something wrong with the code:
function [ output_args ] = freqencySpectrum( ~ )
fs=0.00000001; // sampling at twice the highest frequency (20ns =50MHz, so sampling at 100MHz)
t=-30:1/fs:30;
sigma = 20E-9;
A=1;
p1 = A/(sqrt(2*pi*sigma^2));
x=p1*(exp(-(t.^2)/(2*sigma^2)));
figure(1);
subplot(4,1,1)
plot(t,x,’b’);
title([‘Gaussian Pulse sigma=’, num2str(sigma),’s’]);
xlabel(‘Time(s)’);
ylabel(‘Amplitude’);
L=length(x);
NFFT = 1024;
X = fftshift(fft(x,NFFT));
Pxx=X.*conj(X)/(NFFT*NFFT); %computing power with proper scaling
f = fs*(-NFFT/2:NFFT/2-1)/NFFT; %Frequency Vector
subplot(4,1,2)
plot(f,abs(X)/(L),’r’);
title(‘Magnitude of FFT’);
xlabel(‘Frequency (Hz)’)
ylabel(‘Magnitude |X(f)|’);
xlim([-10 10])
Pxx=X.*conj(X)/(L*L); %computing power with proper scaling
subplot(4,1,3)
plot(f,10*log10(Pxx),’r’);
title(‘Double Sided – Power Spectral Density’);
xlabel(‘Frequency (Hz)’)
ylabel(‘Power Spectral Density- P_{xx} dB/Hz’);
X = fft(x,NFFT);
X = X(1:NFFT/2+1);%Throw the samples after NFFT/2 for single sided plot
Pxx=X.*conj(X)/(L*L);
f = fs*(0:NFFT/2)/NFFT; %Frequency Vector
subplot(4,1,4)
plot(f,10*log10(Pxx),’r’);
title(‘Single Sided – Power Spectral Density’);
xlabel(‘Frequency (Hz)’)
ylabel(‘Power Spectral Density- P_{xx} dB/Hz’);
end
Can you help me with this code?
Thanks,
KP
• KP,
Sigma time is not equal to pulse width. Pulse width should be more than the sigma variation… In the following code, the sigma variation is assumed to be 20ns. The pulse width is 10*20ns. I have fixed the code for you.. Here it is
sigma = 20e-9%sigma variation of the pulse 20ns
pulseWidth = 10*sigma %double sided pulse width
fs = 40*1/pulseWidth %Very very high oversampling factor for smooth curve (40 points), FFT with 1024 points is sufficient to cover this
t=-(pulseWidth/2):1/fs:(pulseWidth/2)
A=1;
p1 = A/(sqrt(2*pi*sigma^2));
x=p1*(exp(-(t.^2)/(2*sigma^2)));
figure(1);
subplot(4,1,1)
plot(t,x,’b’);
title([‘Gaussian Pulse sigma=’, num2str(sigma),’s’]);
xlabel(‘Time(s)’);
ylabel(‘Amplitude’);
L=length(x);
NFFT = 1024;
X = fftshift(fft(x,NFFT));
Pxx=X.*conj(X)/(NFFT*NFFT); %computing power with proper scaling
f = fs*(-NFFT/2:NFFT/2-1)/NFFT; %Frequency Vector
subplot(4,1,2)
plot(f,abs(X)/(L),’r’);
title(‘Magnitude of FFT’);
xlabel(‘Frequency (Hz)’)
ylabel(‘Magnitude |X(f)|’);
Pxx=X.*conj(X)/(L*L); %computing power with proper scaling
subplot(4,1,3)
plot(f,10*log10(Pxx),’r’);
title(‘Double Sided – Power Spectral Density’);
xlabel(‘Frequency (Hz)’)
ylabel(‘Power Spectral Density- P_{xx} dB/Hz’);
X = fft(x,NFFT);
X = X(1:NFFT/2+1);%Throw the samples after NFFT/2 for single sided plot
Pxx=X.*conj(X)/(L*L);
f = fs*(0:NFFT/2)/NFFT; %Frequency Vector
subplot(4,1,4)
plot(f,10*log10(Pxx),’r’);
title(‘Single Sided – Power Spectral Density’);
xlabel(‘Frequency (Hz)’)
ylabel(‘Power Spectral Density- P_{xx} dB/Hz’);
• KP
Hi Mathuranathan,
Thank you very much for the reply. I am little confused about something in the code.
1. Why did you set the pulse width as “pulseWidth = 10*sigma %double sided pulse width”?
Why 10?
2. I have attached the output of the simulation. I see that the pulse width is like 60ns, how can I make it 20ns? Should I even lower the sigma value as the pulse width should be greater than the sigma value?
And in that case, should I change the sampling frequency (fs = 40*1/pulseWidth %) value as well? or 40 will be fine?
3. Finally, how should I interpret the output of single sided PSD, in the attached figure, I believe that pulse width is 60ns, so the frequency should like 16.66MHz, but in the plot how can I relate it to 16MHz?
Thanks
KP
• 1) Subtle difference exists between the terms sigma and pulse width.
Sigma is the standard deviation of the pulse. The term pulse width depends on where the measurement is taken. Usually, the width of the pulse at half-the maximum value is called Full-Width at Half maximum pulse duration (FWHM).
What I was intending is not the FWHM width. It is the full duration of the pulse considered for simulation. Note that the Gaussian pulse gradually tapers nears zero on either ends. As an approximation, we must stop at some point. I chose 10 times the value of sigma for this.
2) To have a full pulse width of 20ns, the sigma has to be lower.
3) What we plot is the frequency response of a single pulse. It breaks down the pulse in frequency domain and shows the different frequency components that make-up the pulse in time domain.
The connotation of frequency (that measures periodicity 60MHz => 16.66ns) is applicable only when the pulse is repeated periodically. You might need to repeat the pulse at a desired regular interval and then plot the frequency response to check for any spike in the frequency domain that is indicative of the frequency of the repeated pattern.
• KP
Thank you very much Mathuranathan, this will help.
• Golam Kibria Chowdhury
Hello,
Your example is absolutely fabulous. This is the first time I have seen theory and practical coding to understand.
1.) I would like to generate a pulse train of Gaussian pulse in time domain with a certain width (let’s say 20 ns in the above example) with a repetition interval of (let’s say 100 ns). In paper, I have to convolve with a Dirac comb. Again let’s say we limit this Dirac comb to a certain number of cycles (let us say 50 cycles).
How do I do it in Matlab?
2.) A further addition to this problem, (I cannot get the first one though) is: let us say we make another pulse offset with the pulse from above. Let us say we keep the pulsewidth the same (20 ns), limit the number of cyles to 50 (like above), but change the repetition interval from every other pulse. For the sake of clearly enquiring it, let me draw a diagram here
*****——————–*****——————–*****——————–**** (P1)
*****————————-*****—————*****————————-**** (P2)
Here P1 = pulse one
P2 pulse 2
both pulses are in phase at one time (let us say t = t1), then it shifted in time (let us say 4 ns) with the next pulse, then again it is in phase.
How could I do it matlab?
Thank you so much for clear explanation. I deeply appreciate it.
Kind regards
Chowdhury | 2017-10-22T08:02:01 | {
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http://math.stackexchange.com/questions/18532/finding-the-sum-fx-sum-n-2-infty-fracxnnn-1 | # Finding the sum $f(x)=\sum_{n=2}^{\infty} \frac{x^n}{n(n-1)}$
I'm trying to find $$f(x)=\sum_{n=2}^{\infty} \frac{x^n}{n(n-1)}$$
I found the radius of convergence of the above series which is $R=1$. Checking $x=\pm 1$ also yields a convergent series. Therefore the above series converges for all $x\in [-1, 1]$.
Using differentiation of the series term by term we get: $$f'(x)=\sum_{n=2}^{\infty} \frac{x^{n-1}}{n-1}=\sum_{n=1}^{\infty} \frac{x^{n}}{n}=-\log(1-x)$$ which also has $R=1$, and then, by integrating term by term we get: $$f(x)=\int_{0}^{x} f'(t)dt=-\int_{0}^{x} \log(1-t)dt=x-(x-1)\ln(1-x)$$
if I understand the theorems in my textbook correctly, the above formulas are true only for $x \in (-1, 1)$. It seems the above is correct since this is also what WolframAlpha says.
I'm abit confused though. At first, it seemed the above series converges for all $x\in [-1, 1]$ but in the end I only got $f(x)$ for all $|x|\lt 1$, something seems to be missing. What can I say about $f(-1)$ and $f(1)$?
-
Try using Abel's theorem.
-
I thought of Abel's theorem but I'm not sure on its usage, can I say that: $$f(1)=\lim_{x\to 1^{-}}f(x)=\lim_{x\to 1^{-}}(x-(x-1)\ln(1-x))=1$$ and $$f(-1)=\lim_{x\to -1^{+}}f(x)=\lim_{x\to 1^{+}}(x-(x-1)\ln(1-x))=\ln4-1$$ – dawson Jan 22 '11 at 17:55
If so, why do I need Tauber's theorem (it's not in my textbook)? – dawson Jan 22 '11 at 17:56
@dawson: No you don't need it. I am just having a braindead moment. – Aryabhata Jan 22 '11 at 18:00
@dawson: Since you noted the series is convergent, Abel's theorem tells you that the lim of f is same as the limit of the series. So you seem to have it right :-) – Aryabhata Jan 22 '11 at 18:07
On second thought, $\frac{|x|^n}{n(n-1)}\leq\frac{1}{n(n-1)}$ and $\sum \frac{1}{n(n-1)}$ converges, then by Weierstrass test the above series converges uniformly in $[-1, 1]$, does that make sense? – dawson Jan 22 '11 at 18:22
show 3 more comments
If you rewrite $\frac{1}{n(n-1)}$ in the form $\frac{1}{n-1}-\frac{1}{n}$, then you can rewrite the series in both cases $x = \pm 1$ and compute their values directly. You can then confirm that in both cases the value you compute coincides with the value $f(\pm 1)$. (In other words, rather than appealing to Abel's theorem, as Moron suggests, in this particular case you can verify it.)
[Caveat: In the case $x = -1$, you will need to use the familiar series for $\log 2$, and maybe the easiest way to prove this is by appealing to Abel's theorem (applied to the series for $\log (1 + x)$). So my approach probably doesn't really avoid Abel's theorem, at least for $x = -1$.]
- | 2014-03-16T04:45:14 | {
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https://math.stackexchange.com/questions/950949/a-simple-system-of-equations/950953 | # A simple system of equations
I'm trying to refresh my school math knowlegde and have trouble solving a simple system of equations:
$\begin{cases} x + xy + y = -3,\\ x - xy + y = 1. \end{cases}$
I derive $y$ from the second:
$y - xy = 1 - x$
$y(1-x)=(1-x)$
Hence,
$y = \frac {(1-x)}{(1-x)} = 1$, provided that x ≠ 1
Next, I substitute $y=1$ in the first equasion,
$x + x + 1 = -3$; $2x = -4$, $x=-2$.
The answer seems to be $(-2; 1)$.
The problem book, however, also lists a second answer, $(1; -2)$.
I feel that I've done something wrong. Give me a hint, please.
• You excluded the case $x=1$ in your workings. So you have to go back and check that too. – Mark Bennet Sep 29 '14 at 11:58
• Thank you, @MarkBennet! So, when I get an $x≠a$ condition, where $a$ is some number, in one of the equasions of the system, I should plug it into all other equasions, calculate the second variable, then check on all the equations if this $(x, y)$ combination solves them all? – CopperKettle Sep 29 '14 at 12:47
By saying "provided that $x \neq 1$", after getting a solution you have to go back to try out $x=1$.
Note that $$y(1-x)=(1-x)\iff (1-x)(y-1)=0\iff x=1\ \text{or}\ y=1.$$ Since you've already considered the case when $y=1$, you need to consider the case when $x=1$.
Add the two together to get $~2x+2y=-2\iff x+y=-1$. Now replace this value in either one of the initial two equations to get $xy$. And when you know both the Sum and the Product of two numbers, you can determine their values by solving the quadratic $~u^2-su+p=0$, where $s=x+y$ and $p=xy$, since this is what you get when expanding $~(u-x)(u-y)=0.~$ So all that's left to do now is applying the well-known quadratic formula. :-)
• Great, I see you're using Vieta's formulas in an imaginative way! (0: I will try to follow this route. As I understand, $u_1, u_2$ in this quadratic equation will equal $x, y$. I'm too dense to understand the line "since this is what you get when expanding $(u-x)(u-y)=0.$" though. (0: – CopperKettle Sep 29 '14 at 14:02
• I solved the quadratic, getting the result $(1; -2)$. How will I get (or logically derive) the second result, $(-2; 1)$, I wonder. – CopperKettle Sep 29 '14 at 14:43
• @CopperKettle: Your two equations are both symmetrical in x and y, so if $(a,b)$ is a solution, then so is $(b,a)$. As for your other remark, just open up the parentheses, and see what you get. – Lucian Sep 29 '14 at 15:20 | 2020-08-14T02:47:21 | {
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https://mathhelpboards.com/threads/linear-transformation.4753/ | Linear transformation
Kaspelek
New member
Where T(p(x)) = (x+1)p'(x) - p(x) and p'(x) is derivative of p(x).
a) Find the matrix of T with respect to the standard basis B={1,x,x^2} for P2.
T(1) = (x+1) * 0 - 1 = -1 = -1 + 0x + 0x^2
T(x) = (x+1) * 1 - x = 1 = 1 + 0x + 0x^2
T(x^2) = (x+1) * 2x - x^2 = 2x + x^2 = 0 + 2x + x^2
So, the matrix for T with respect to B equals
[-1 1 0]
[0 0 2]
[0 0 1].
b) Find a basis for kerT and hence write down dim(kerT).
c) Find a basis for ImT and hence write down dim(ImT).
d) Does the transformation have an inverse?
I've done part a, so any guidance on the rest would be greatly appreciated.
Active member
Where T(p(x)) = (x+1)p'(x) - p(x) and p'(x) is derivative of p(x).
a) Find the matrix of T with respect to the standard basis B={1,x,x^2} for P2.
T(1) = (x+1) * 0 - 1 = -1 = -1 + 0x + 0x^2
T(x) = (x+1) * 1 - x = 1 = 1 + 0x + 0x^2
T(x^2) = (x+1) * 2x - x^2 = 2x + x^2 = 0 + 2x + x^2
So, the matrix for T with respect to B equals
[-1 1 0]
[0 0 2]
[0 0 1].
b) Find a basis for kerT and hence write down dim(kerT).
As you've correctly stated, we may express the transformation in the form
$$\displaystyle A_T=\left[ \begin{array}{ccc} -1 & 1 & 0 \\ 0 & 0 & 2 \\ 0 & 0 & 1 \end{array} \right]$$
Now as for $$\displaystyle ker(T)$$, we proceed to find the kernel of this transformation in the same way as we would for any other transformation. That is, we would like to solve $$\displaystyle A_T x=\left[ \begin{array}{ccc}0 & 0 & 0\end{array} \right]^T$$, which requires we put the augmented matrix
$$\displaystyle \left( A_T|0 \right) = \left[ \begin{array}{ccc} -1 & 1 & 0 & | & 0 \\ 0 & 0 & 2 & | & 0 \\ 0 & 0 & 1 & | & 0 \end{array} \right]$$
in its reduced row echelon form (of course, this problem could be solved with a little intuition instead of row reduction, but the upshot of this method is that it works where your intuition might fail). You should end up with
$$\displaystyle \left[ \begin{array}{ccc} 1 & -1 & 0 & | & 0 \\ 0 & 0 & 1 & | & 0 \\ 0 & 0 & 0 & | & 0 \end{array} \right]$$
Which tells you that the kernel of the transformation is the set of all polynomials $$\displaystyle a+bx+cx^2$$ such that $$\displaystyle a-b=0$$ and $$\displaystyle c=0$$. Thus, we may state that the kernel of $$\displaystyle T$$ is spanned by the vector
$$\displaystyle \left[ \begin{array}{ccc}1 & 1 & 0\end{array} \right]^T$$, which is thus the basis of the kernel of T. Since there is one vector in the basis, the dimension of the kernel is 1.
Any questions about the process so far?
Last edited:
Active member
c) Find a basis for ImT and hence write down dim(ImT).
d) Does the transformation have an inverse?
For c), what we need to do is find the largest possible set of linearly independent column-vectors. If you wanted to use the rref (reduced row echelon form) that we computed previously, you simply choose the column vectors corresponding to the 1's (i.e. the "pivots") of the reduced matrix. That is, choosing the first and third columns, we find that $$\displaystyle \left[ \begin{array}{ccc}-1 & 0 & 0\end{array} \right]^T$$ and $$\displaystyle \left[ \begin{array}{ccc}0 & 2 & 1\end{array} \right]^T$$ form a basis of the image. It follows that $$\displaystyle dim(Im(T))=2$$.
For d), we simply note that the kernel of this transformation is not of dimension 0. This is enough to state that the transformation does not have an inverse.
Last edited:
Fernando Revilla
Well-known member
MHB Math Helper
Something better to express the solutions as vectors of $P_2$ instead of coordinates. That is, $B_{\ker T}=\{1+x\}$ and $B_{\mbox{Im }T}=\{-1,2x+x^2\}$.
Kaspelek
New member
That helped a lot guys, thanks a lot. | 2021-02-27T03:17:05 | {
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http://mathhelpforum.com/algebra/42773-solved-finding-sum-two-unknown-variables.html | # Thread: [SOLVED] Finding the sum of two unknown variables
1. ## [SOLVED] Finding the sum of two unknown variables
Hello everyone,
Given that $43p + 23q = 4323$, find $p + q$?
The question is to find the sum of the unknown variables. Now, the variables have to be integers, and as you can see from the easy example I made, the sum could be $43(100) + 23(1)=4323$ $\Rightarrow p+q = 101$
Of course, this is not the only solution. Try the following:
$43(54) + 23(87) = 4323$
$\Longrightarrow p + q = 141$
I suspect that there are even more integer solutions to this equation. My method for finding the values is the old simple bruteforce "technique", where I divide 4323 by highest coefficient and that's where I get a headstart. You can of course find out the pattern, but these "cooked" equations are no good for learning.
My question is: is there a logical process of solving for $p + q$? Other than bruteforce, of course. I asked my teacher today, and he just said "that's number theory." I looked up number theory, but I was kind of lost.
These type of questions are pretty common on the SAT I tests, and I love them. My brother and I used to exchange multivariable equations like that and try to find the sum all the time.
2. Hello,
What comes in my mind is that we have to solve for p and q before finding p+q... maybe there is another method...
This comes with the Euclidian algorithm, which will yield to Bézout's theorem (google for them).
43 & 23 are coprime, that is to say they have no common divider.
The theorem states that if 43 and 23 are coprime, then we can write it :
$43u+23v=1$, where u & v are integers (positive or negative).
-------------------------
Now, the thing is that you have to find a particular solution to the equation $43u+23v=1$, thanks to the Euclidian algorithm.
$43=23*1+20 \implies 20=43-23$
$23=20*1+3 \implies 3=23-20=23-(43-23)=-43+2*23$
$20=3*6+2 \implies 2=20-3*6=(43-23)-6*(-43+2*23)=7*43-13*23$
$3=2+1 \implies 1=3-2=(-43+2*23)-(7*43-13*23)=-8*43+15*23$
Thus $\boxed{u=-8 \text{ and } v=15}$
--------------------------
What's good with it is that we can multiply the equation $43u+23v=1$ by 4323 in order to get a particular solution for p and q in $43p+23q=4323$
--> $\boxed{p_1=-8*4323=-34584 \text{ and } q_1=15*4323=64845}$
---------------------------
MOO
So we have :
\begin{aligned} 4323&={\color{red}43}*p_1+{\color{red}23}*q_1 \\
&={\color{red}43}*p+{\color{red}23}*q \end{aligned}
$43p_1+23q_1=43p+23q$
$43(p-p_1)=23(q_1-q)$
Because 43 and 23 are coprime, we know by the Gauss theorem, that $q_1-q=43k$ and $p-p_1=23k$, $\forall \text{ integer (positive or negative) } k.$
Therefore the general solution is :
$p=p_1+23k$
$q=q_1-43k$
$\boxed{p+q=p_1+q_1-20k}$
Remember, $p_1=-34584$ and $q_1=64845$
------------------------------
$p_1$ and $q_1$ are ugly because 4323 is a large number, and I showed a general method.
If you have got a particular solution (100 and 1), set up $p_1=100$ and $q_1=1$ and take it from the red MOO.
I hope this is clear enough... If you have any question...
3. Hello, Chop Suey!
Here's a rather primitive solution . . .
Given that: . $43p + 23q =\: \:4323$, find $p + q.$
We have: . $43p + 23q \:=\:43(100) + 23 \quad\Rightarrow\quad 23q - 23 \:=\:43(100) - 43p$
Factor: . $23(q-1) \:=\:43(100-p) \quad\Rightarrow\quad q \;=\;\frac{43(100-p)}{23} + 1$
Since $q$ is an integer, $(100-p)$ must be divisible by 23.
This happens for: . $p \;=\;100,\:77,\:54,\:31\:\hdots \:100-23t$
. . .and we have: . $q \;=\;1,\:44,\:87,\:130,\:\hdots\:1 + 43t$
Therefore: . $p+q \;=\;(100-23t) + (1 + 43t) \;=\;101 + 20t\:\text{ for any integer }t$
4. I want to thank both of you, Moo and Soroban, for the astoundingly fast reply.
Moo: I'm going to look up those theorems you mentioned and discuss this further should I have any questions.
Soroban: It may be primitive, but you made me wish if I only approached this method before.
Thanks again
5. Here are the wikipedia links (so that you won't be mistaken ) :
Bézout's identity/lemma (not really theorem actually) : Bézout's identity - Wikipedia, the free encyclopedia
The way I used the Euclidian algorithm : Extended Euclidean algorithm - Wikipedia, the free encyclopedia
Ok...Gauss theorem = Euclid's lemma (they're the same, but I don't know the circumstances that made 2 different names lol)
If a positive integer divides the product of two other positive integers, and the first and second integers are coprime, then the first integer divides the third integer.
This can be written in notation:
If n|ab and gcd(n,a) = 1 then n|b.
Euclid's lemma - Wikipedia, the free encyclopedia | 2013-06-19T11:25:37 | {
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https://math.stackexchange.com/questions/898149/is-there-a-name-for-the-function-maxx-0/898812 | # Is there a name for the function $\max(x, 0)$?
Is there a name for the function $\max(x, 0)$?
For comparison, the function $\max(x, -x)$ is known as the absolute value (or modulus) of x, and has its own notation $|x|$
• should there be a name for this function? is it important enough to get its own name? is it used often enough to warrant an abbreviation? does it capture something so profound to be worth naming? If you answer any of these question positively, then you can improve your question. – Ittay Weiss Aug 15 '14 at 10:20
• In finance max(x-s,0) is the payoff of a call option where s is the strike price and x is the price of the underlying stock on the expiry date. – Colonel Panic Aug 15 '14 at 11:06
• Another note, I have seen the notation $[f]^+$ and $[f]^-$ for positive part and negative part respectively. – Brad Aug 15 '14 at 14:14
• math.stackexchange.com/q/25349 – Jonas Meyer Aug 15 '14 at 18:39
• The question asks for the name of the operation, not an example of something that is computed using it. The operation $\max(x,0)$ isn't called "payoff of a call option", just as the operation of multiplication isn't called "force", despite Newton's second law. – David Richerby Aug 16 '14 at 20:37
This is called the positive part of the real number $x$, and often denoted by $x^+$.
Likewise, the negative part of $x$ is $x^-=\max\{-x,0\}$ and the pair of nonnegative real numbers $(x^+,x^-)$ is fully characterized by the pair of identities $$x=x^+-x^-,\qquad\lvert x\rvert=x^++x^-.$$
• It should be noted that the negative part is, in fact, a positive number. For example the negative part of -3 is 3. – Michael Lugo Aug 15 '14 at 13:19
• Indeed the positive part and the negative part are both nonnegative numbers. – Did Aug 15 '14 at 15:51
• Similarly, the imaginary part of a complex number is a real number. – user133281 Aug 15 '14 at 22:40
• I would rather call this the "positive part of the identity function", because the "positive part" definition as you cite it refers to functions and not numbers. – example Aug 16 '14 at 13:34
• @example Thanks for your input. It just happens that this (i.e., the positive part of the real number $x$) is what mathematicians call it (i.e., $\max(x,0)$). – Did Aug 16 '14 at 13:42
Wikipedia calls this the ramp function and notes that it can be written using Macaulay brackets.
$\{x\} = \begin{cases} 0, & x < 0 \\ x, & x \ge 0. \end{cases}$
Since this is a math site, not a programming site, my answer may or may not be regarded as trivia. Anyway...
In computer graphics this function is called clamping. The general form is $\mathrm{clamp(x, lowerBound, upperBound)}$ and is defined as
function clamp(x, lowerBound, upperBound):
if(x < lowerBound)
return lowerBound
else if(x > upperBound)
return upperBound
else
return x
or $\mathrm{min( max(x, lowerBound), upperBound)}$.
$\max(x,0)$ is the special case $\mathrm{clamp}(x, 0, +\infty)$.
The clamping function is ubiquitous in computer graphics: You often need to confine a calculated value (e.g. a color intensity) into a range of valid values (e.g. $[0,1]$ or $[0,255]$).
• It looks like Java, only function is not Java. Replacing function with something like public static int makes it Java, a method to be inserted in some class. – MickG Aug 16 '14 at 9:40
• It's pseudocode. – Sebastian Negraszus Aug 16 '14 at 10:14
• It's pseudocode and can trivially easily be translated into Python, JS or anything else. @OutlawLemur It's not literal Python because of a few minor differences: a) function clamp(...) instead of def clamp(...): b) No colons at the end of the if/elif/else lines c) Conditional expressions enclosed by parentheses if(x < lowerBound) instead of if x < lowerBound: – smci Aug 17 '14 at 0:31
• @MickG Even if you replaced function with public static int, it would still be a long way away from being Java. – JLRishe Aug 17 '14 at 11:53
• @MickG: It is much closer to Python than Java. – Reid Aug 17 '14 at 19:55
You can check that:
$$\color{blue}{\max(x,0) = x \, H(x)}$$
where $H(x)$ is the Heaviside or unit step function. A name for this? Not a clue, but hope it helps.
• This is actually a pretty nice implementation. For example, here's a quick and dirty computation of the derivative for $x \neq 0$: $$\frac{d}{dx} \max(x,0) = \frac{d}{dx} H(x)\cdot x = H(x) \frac{d}{dx} x = H(x) \cdot 1 = H(x).$$ It works because $H(x)$ is basically a constant. More precisely, the function $$\begin{cases} 1 & x>0 \\ 0 & x < 0\end{cases}$$ is a locally constant openly-supported partial functions $\mathbb{R} \rightarrow \mathbb{R}$, which is exactly what you need in order to treat it as a constant insofar as differentiation is concerned. – goblin GONE Aug 5 '17 at 4:12
I have heard this function called the rectifier. This is a pretty exclusive field name though, and I wouldn't expect to see it anywhere outside of neural networks.
• I think the use in neural networks derives from the use of the term in electrical engineering: en.wikipedia.org/wiki/Rectifier – A. Donda Aug 16 '14 at 2:27
• @A.Donda Thanks for the information. – Did Aug 16 '14 at 13:05
You can use:
$$f(x) = \frac{x + |x|}{2}$$
• I suppose that's interesting, but it's not at all what the question is asking. – Ray Aug 15 '14 at 18:38
• The point is that the function doesn't need a name, it can be expressed in terms of $|x|$ and $x$. – Darth Geek Aug 15 '14 at 18:42
• So squaring doesn't need a name because it can be expressed in terms of $x$ and $2$? And factorial doesn't need a name because it can be expressed in terms of product? And... – David Richerby Aug 16 '14 at 20:40
• @DavidRicherby It's not about that. We call $x^2$ squaring but we don't have a name for $x^2+3$. We have a name for $n!$ but not one for $(-1)^nn!$. So perhaps a name for the function described is not really needed. It turns out it does have a name, but if it didn't and you needed to use it in a paper, for instance you can just give it a special letter and define it at the beggining of the paper and it would be ok. – Darth Geek Aug 17 '14 at 8:19
• How is (x+|x|)/2 better then max(x, 0) anyway? – Cthulhu Aug 17 '14 at 9:26 | 2020-12-05T17:21:24 | {
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https://www.physicsforums.com/threads/lowest-surface-to-volume-ratio-for-an-uncovered-vessel.617947/ | # Lowest surface to volume ratio for an uncovered vessel
1. Jul 2, 2012
### eee000
Hi,
It is well known that a sphere has the lowest surface to volume ratio. However, a related question is: What is the shape that gives the lowest surface to volume ratio if you do not include the "top" in the surface. That is, what is the maximal volume of an uncovered vessel of a fixed surface area?
For example, a cylinder whose height is equal to its radius, has a volume Pi R^2 h = Pi R^3 and a surface (without cover) of Pi R^2 + 2*Pi *R*h = 3 Pi R^2. If we fix the volume to unity, the surface in this case is 3.Pi^(1/3).
In comparison, a half sphere of unity volume has a smaller surface -- (18 Pi)^(1/3).
However, it's clear this is not the best shape - a sphere cut a bit above half its volume beats the half sphere.
So, what is the shape that gives the lowest surface to volume ratio if you do not include the "top" in the surface ?
Thanks!
2. Jul 2, 2012
### chiro
Hey eee000 and welcome to the forums.
I'm a little confused about what you mean by "top".
With a sphere, there is no discontinuous boundary like you have with say a cube so in these kinds of situations (where the surface is completely smooth), it's really hard to define a "top".
Based on this, do we need the shape to have something corresponding to a "top" where the top is an area that is "smooth" respect to the rest of the body?
3. Jul 2, 2012
### HallsofIvy
That depends strongly on how you define "top" for a general surface.
4. Jul 2, 2012
### eee000
Thank you,
The top I have in mind is defined by gravity. I want to have a vessel full of water that do not spill out. That is, if we define the z axis to be the direction of gravity, I think of a surface that may have holes in it, but the volume that we count is only integrated up to lowest z coordinate in the holes.
5. Jul 2, 2012
### Curious3141
There are well-known formulae for the volume and surface of a spherical cap: http://en.wikipedia.org/wiki/Spherical_cap
(removed wrong stuff).
Last edited: Jul 2, 2012
6. Jul 2, 2012
### eee000
Thank you for the spherical cap formulae, but I think your algebra is wrong. Fixing the sphere radius R, and parameterizing the cap by h, I find the volume to be V=Pi h^2 (R - h/3) and the surface S=2 Pi R h. The lowest ratio is obtained when h=3R/2, and then V=9Pi/8 R^3, and S=3Pi R^2. Setting the volume to unity (R=(8/9Pi)^(1/3)), I find S=(64 Pi/3)^(1/3), better than the half sphere with (18 Pi)^(1/3) and better than the full sphere with (36 Pi)^(1/3).
7. Jul 2, 2012
### Curious3141
You're right, I made a silly mistake.
What I should've done is simply expressed V/A as a function of h, and that's $\frac{V}{A} = \frac{1}{6}(3h - \frac{h^2}{r})$. Taking the derivative and setting it to zero yields your result of $h = \frac{3r}{2}$.
Intuitively, shouldn't this (partial sphere to a height of 1.5r) be the open vessel with the lowest A/V ratio?
8. Jul 2, 2012
### eee000
I'm sorry,I made a mistake too...
Since the surface to volume ratio scales as 1/V^(1/3) we need to do the optimization for fixed volume. This yields that the optimal shape is exactly the half sphere.
Obviously, my statement "(64 Pi/3)^(1/3), better than the half sphere with (18 Pi)^(1/3)" is wrong, as 64/3 > 18 ...
9. Jul 2, 2012
### Curious3141
I didn't see that, but I wasn't following that anyway, because I think your method of visualising the problem is unnecessarily confusing. Why not just let r = 1 and figure out the value of h that minimises the A/V ratio (or maximises the V/A ratio, as I did) for the open partial sphere? Now I'm sure that h = 1.5 (for r = 1) is the right answer.
10. Jul 2, 2012
### eee000
sure it is, but I think the correct formulation of the problem is that of a fixed volume, especially if one is interested in he general question - what is the best shape of all. Since the A/V ratio generally scales with 1/V^(1/3), comparison must be made for equal volumes.
11. Jul 2, 2012
### Curious3141
OK, I see where you're coming from. To state the problem rigorously, "For an open partial sphere of unit volume, what value of h/r will yield minimal surface area?"
Can we agree on that statement?
The way to go about it then is to put h = kr (where 0 < k <= 2), set V(h,r) = 1, then work out r in terms of k alone.
Then plug that expression for r into A(h,r) to get A in terms of k alone.
Finally, minimise A(k) over the domain (0,2].
I have to turn in now - I'm already up too late with a bad cold and I have to work tomorrow. So please go ahead (and if you've done it, please post the basic steps, if you can). If the problem remains unsolved tomorrow, and I'm free, I'll work on it.
12. Jul 2, 2012
### eee000
For the partial sphere, the solution is easy - the half sphere is optimal.
The open problem (to the extent of my minimal knowledge) is:
"For an open manifold of unit volume(*) , what shape will yield minimal surface(*) area?"
where Volume is the volume contained up to the first hole, and surface includes the whole manifold but the holes
13. Jul 2, 2012
### Curious3141
But have you proved it? Just asking.
EDIT: Proved it myself.
$A(k) = (2){(9\pi)}^{\frac{1}{3}}.k^{-\frac{1}{3}}.{(3-k)}^{-\frac{2}{3}}$
and this has a minimum at k = 1 (when h = r), and the minimum is ${(18\pi)}^{\frac{1}{3}}$, yielding a half-sphere of unit volume.
I cannot provide any insight on the more general problem.
Last edited: Jul 2, 2012
14. Jul 3, 2012
### haruspex
Suppose you have an optimal vessel and it holds more than a half-sphere of the same surface area of vessel. Necessarily the upper rim will lie in one horizontal plane. Create a cap for it by reflection in that plane. You now have a closed vessel of twice the volume and twice the area, and it would seem that it holds more fluid than a closed sphere of the same surface area.
15. Jul 3, 2012
### eee000
Thank you haruspex! That's what I was looking for.
I was thinking in more local terms, along the lines of Steiner's arguments, but your one-liner is much better. | 2018-05-24T20:13:45 | {
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https://socratic.org/questions/an-urn-contains-100-marbles-20-white-30-red-50-green-calculate-the-probability-o | # An urn contains 100 marbles: 20 white, 30 red, 50 green. Calculate the probability of selecting White, Red and Green marbles respectively. What is the probability of pulling a white, green, white and red marbles consecutively?
Apr 18, 2016
$\frac{950}{156849}$ or approximately 0.6%
#### Explanation:
Assuming the marbles are not replaced in the urn:
• The probability of the first marble being white is $\frac{20}{100}$
• The probability of the next marble being green is then $\frac{50}{99}$
• The probability of the next marble being white is $\frac{19}{98}$
• The probability of the next marble being red is $\frac{30}{97}$
So the probability of the sequence white, green, white, red is:
$\frac{20}{100} \cdot \frac{50}{99} \cdot \frac{19}{98} \cdot \frac{30}{97}$
$= \frac{10}{\textcolor{red}{\cancel{\textcolor{b l a c k}{50}}}} \cdot \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{50}}}}{99} \cdot \frac{19}{98} \cdot \frac{30}{97}$
$= \frac{10 \cdot 19 \cdot 30}{99 \cdot 98 \cdot 97}$
$= \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{3}}} \cdot 1900}{\textcolor{red}{\cancel{\textcolor{b l a c k}{3}}} \cdot 33 \cdot 98 \cdot 97}$
$= \frac{1900}{33 \cdot 98 \cdot 97}$
$= \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}} \cdot 950}{33 \cdot \textcolor{red}{\cancel{\textcolor{b l a c k}{2}}} \cdot 49 \cdot 97}$
$= \frac{950}{33 \cdot 49 \cdot 97}$
$= \frac{950}{156849} \approx 0.006$
That is approximately 0.6%
Aug 24, 2016
In support of Georg's solution
#### Explanation:
For probability questions of this type, if you are ever in doubt, draw a probability tree
$\textcolor{red}{\text{Assumption: this is selection without replacement}}$
From the diagram observe that the initial selection of
White ->20/100->20%
Red" "-> 30/100->30%
Green->50/100->50%
From the probability tree the overall sequenced sampling probability of white: green: white: red is:
$\frac{20}{100} \times \frac{50}{99} \times \frac{19}{98} \times \frac{30}{97}$
For what follows refer to George's solution | 2020-09-19T05:45:55 | {
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https://math.stackexchange.com/questions/2800632/why-is-finding-factors-until-half-of-the-number-enough | # Why is finding factors until half of the number enough?
If I want to find factors of a number except itself, at first I think that I divide it the number in turn $1, 2, 3, ... , n - 1$. However, after a while, noticed that division to get the factors is sufficient up to $n/2$(inclusive). The rest part is not needed. What is its reason? Why is it enough to find them? How can it be explained?
@Edit, to find prime number until $\sqrt n$ is sufficient, but what about perfect numbers? I think $n/2$ is good way to go perfect numbers?
For 6,
1 2 3 4 5
^
^
^|
Up to here, it's ok.
• By "perfect number", do you mean a number which is the sum of all its proper divisors, $n = \sum_{d|n,d\lt n} d$? Of course six is one such number, $6=1+2+3$. – hardmath May 29 '18 at 14:29
• @hardmath < Yes, exactly. – itsnotmyrealname May 29 '18 at 14:30
• Note that for finding perfect numbers, no one has ever found an odd perfect number. The question if one exists is "open". The exact form of even perfect numbers is known, and they have many factors of two. In that sense it is a good idea to test for an even perfect number by dividing out as many factors of two as possible. See this previous Question How to check for perfect numbers? and its answers for more details, involving Mersenne primes. – hardmath May 29 '18 at 14:46
• Note: to find all of the prime factors, it is sufficient to go up to SQRT(n). To find all factors, it is faster to first find all of the prime factors and then use them to generate all of the factors. – RBarryYoung May 29 '18 at 20:54
There can't possibly be any factors between $\frac n2$ and $n$. Suppose $\frac n2 < a < n$ and $a \cdot b = n$. What could $b$ be? If $b=1$, then $a \cdot b = a < n$. If $b \geq 2$, then $a \cdot b > \frac n2 \cdot 2 = n$. So $b$ can't be any positive integer; thus $a$ isn't a factor of $n$.
In fact, as long as you list both factors when you do the division, you can stop testing at $\sqrt n$. That's because it's impossible for both factors to be greater than $\sqrt n$: if $a,b > \sqrt n$, then $a \cdot b > \sqrt n \cdot \sqrt n = n$. So factors always come in pairs $a \cdot b = n$ with $a < \sqrt n$ and $b > \sqrt n$ (with the exception of $\sqrt n$ itself, if it's an integer). Here's how this method works to find the factors of $10$:
• $3 < \sqrt 10 < 4$, so we can stop testing at $3$
• Test $1$: $\frac{10}1 = 10$, so $10 = 1 \cdot 10$, giving us the two factors $1$ and $10$
• Test $2$: $\frac{10}2 = 5$, so $10 = 2 \cdot 5$, giving us the two factors $2$ and $5$
• Test $3$: $\frac{10}3$ is not an integer so we don't get any factors
Thus the full list of factors is $1,10,2,5$ (or, reordered, $1,2,5,10$)
This method works to find all the factors, not just the prime factors, so it's a perfect method for testing whether a number is perfect (in our example, $1+2+5=8\neq10$, so $10$ is not perfect).
• Thank you. But, what about $6$? $\sqrt 6 = 2.44$ Should we ceil the number to get half border which is $3$? – itsnotmyrealname May 29 '18 at 14:38
• @itsnotmyrealname The argument I gave shows that you never have to check anything greater than $\sqrt n$ (since one of the factors must be less or equal to than $\sqrt n$), so we don't have to check $3$ in this case: just $1$ and $2$. Checking $1$ gives us $1$ and $6$; checking $2$ gives us $2$ and $3$. – BallBoy May 29 '18 at 14:40
• @itsnotmyrealname Another comment: the savings are much larger for larger $n$. The next known perfect number is $n=28$. By the $\sqrt n$ method, we have to check only up to (and including) $5$, which is much less than the $14$ we'd need to check by the $\frac n2$ method! Try it and see if you can show that $28$ is perfect. – BallBoy May 29 '18 at 14:41
• Gotcha what you mean. You can glance at it, ideone.com/VYKxup – itsnotmyrealname May 29 '18 at 15:00
Actually to find the (prime) factors of a whole number $n$, it is enough to check for divisors up to $\sqrt n$. Going up to $n/2$ would be overkill.
One way to think about this is to ask, if $n$ is not a prime but instead composite, how big can the smallest divisor of $n$ be? If we have the factorization $n = a\cdot b$, we cannot have both $a$ and $b$ larger than $\sqrt n$.
To make a complete list of the factors of a whole number $n\gt 1$, you can do it by checking all $1\lt a \lt \sqrt n$ and putting both $a$ and $b$ in your list when the division $n/a = b$ is exact. If $\sqrt n$ happens to be an integer (i.e. $n$ turns out to be a perfect square when we extract $\sqrt n$), then put one copy of that integer in the list of divisors. The list (of proper divisors of $n$) will be complete when you finish by including $1$, which is always a divisor you want in your list.
• Yes, I realized it early. Thank you for your algorithm. There are a host of code snippets etc. but I wanted to get what there exist in underneath. – itsnotmyrealname May 29 '18 at 15:10
Suppose we can write $n = a*b$, so both $a$ and $b$ are factors. Notice that it must be the case that one of them is less than or equal to $n/2$. To see this, suppose both $a,b > n/2$. Then $a*b > n$, which is absurd. This means for any pair of factors, one of them must be less than $n/2$, hence you need only check up to and including $n/2$ to find them all.
$\frac{n}{x}<2$ if $x>\frac{n}{2}$. $2$ is the smallest prime. If $x$ was a factor then $x\times \text{something}=n$ where $\text{something}\ge 2$. | 2019-09-16T14:08:49 | {
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https://mathhelpboards.com/threads/program-for-approximating-nth-root-of-a-number.5323/ | # Program for Approximating nth root of a Number
#### Yuuki
##### Member
I have a question for a programming exercise I'm working on for C.
The problem is to "Write a program that uses Newton's method to approximate the nth root of a number to six decimal places." The problem also said to terminate after 100 trials if it failed to converge.
Q1. What does "converge" mean?
Does it mean the difference between two approximation can be made as small as I like?
Q2. On what condition should the program terminate?
There are two such conditions: 1) if the loop has been executed 100 times, 2) the difference between the "true" answer and the approximation is less than 0.000001.
I know how to set 1), but how should I express 2)?
Right now, I am setting the condition as
|approximation - root| < 0.00001,
but I feel it's kind of cheating, because I'm not supposed to know the real answer if I'm making approximations.
Are there any other any ways to express the condition, especially one involving the function f(x) = x^n - c (x is the nth root of c)?
#### Klaas van Aarsen
##### MHB Seeker
Staff member
Re: question on program for approximating nth root of a number
Hi Yuuki!
I have a question for a programming exercise I'm working on for C.
The problem is to "Write a program that uses Newton's method to approximate the nth root of a number to six decimal places." The problem also said to terminate after 100 trials if it failed to converge.
Q1. What does "converge" mean?
Does it mean the difference between two approximation can be made as small as I like
Yes. Basically.
More specifically, that you can get as close to the nth root as you want by just taking enough trials.
Q2. On what condition should the program terminate?
There are two such conditions: 1) if the loop has been executed 100 times, 2) the difference between the "true" answer and the approximation is less than 0.000001.
I know how to set 1), but how should I express 2)?
Right now, I am setting the condition as
|approximation - root| < 0.00001,
but I feel it's kind of cheating, because I'm not supposed to know the real answer if I'm making approximations.
Are there any other any ways to express the condition, especially one involving the function f(x) = x^n - c (x is the nth root of c)?
Exactly. You're not supposed to use the real root.
But what you can do is setting the condition as for instance
$$|\text{approximation} - \text{previous approximation}| < 0.0000001$$
If you achieve that, it is unlikely that the first 6 decimal digits will change in more iterations.
#### Yuuki
##### Member
Re: question on program for approximating nth root of a number
Thanks.
I set a new variable root0 to store the previous approximation, and set the condition as you said.
It worked beautifully.
#### zzephod
##### Well-known member
It is possible to estimate the error in an iterate directly (assuming it small anyway).
Let $$\displaystyle x$$ be the 6-th root of $$\displaystyle k$$ and $$\displaystyle x_n$$ an estimate of $$\displaystyle x$$ with error $$\displaystyle \varepsilon_n$$ such that:
$$\displaystyle x=x_n+\varepsilon_n$$
Then raising this to the 6-th power gives:
$$\displaystyle x^6=x_n^6 + 6 \varepsilon_n x_n^5 + O(\varepsilon_n^2)$$
Now ignoring second and higher order terms in $$\displaystyle \varepsilon$$ and rearranging we get:
$$\displaystyle \varepsilon_n=\frac{x_n^6-x^6}{6x_n^5}=\frac{x_n^6-k}{6x_n^5}$$
OK lets look at an example: Take $$\displaystyle k=66$$, and $$\displaystyle x_n=2$$, so $$\displaystyle x_n^6=64$$, then
$$\displaystyle \varepsilon_n=\frac{66-64}{6\times32}\approx 0.01042$$
which compares nicely with $$\displaystyle 66^{1/6}=2.01028...$$
The above is very similar (for similar read identical) to computing the next iterate and taking the difference of the iterates as an estimate of the error in the first.
Last edited:
#### Klaas van Aarsen
##### MHB Seeker
Staff member
It is possible to estimate the error in an iterate directly (assuming it small anyway).
It is also possible to specify an upper boundary for the remaining error in a specific iteration (in this specific case).
First off, after the first (positive) iteration, all iterations are guaranteed to be above the root.
The remaining error in those iterations is guaranteed to be less than the change in the approximation. | 2021-05-07T07:41:20 | {
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https://www.physicsforums.com/threads/diagonals-of-a-quadrilateral.789945/ | # Homework Help: Diagonals of a Quadrilateral
Tags:
1. Dec 31, 2014
### Born
1. The problem statement, all variables and given/known data
Problem 55 from Kiselevś Geometry - Book I. Planimetry: "Prove that each diagonal of a quadrilateral either lies entirely in its interior, or entirely in its exterior. Give an example of a pentagon for which this is false."
2. Relevant equations
3. The attempt at a solution
The pentagon part is pretty easy. I'm having trouble with the proof. A proof by contradiction seems to be the easiest way to solve this problem but I'd prefer a proof that also explains why this should be true.
I've tried using straight line properties (i.e. a straight line can be formed though any two points and it is unique) but I haven't gottten anywhere.
Thanks in advanced for any help!
Last edited: Dec 31, 2014
2. Dec 31, 2014
### Simon Bridge
Can you see why the pentagon can break the rule?
If a diagonal breaks the rule, then it must cross one of the sides - that help?
You can also look at the classes of quadrilateral and see how diagonals are formed in each case.
3. Dec 31, 2014
### MidgetDwarf
Maybe supplementary angles of a transversal are....
4. Jan 1, 2015
### lurflurf
the diagonal divides the plane in two and contains exactly two of the four points there are two cases both points lie on the same side or each lies on one side
5. Jan 2, 2015
### Born
Simon, MidgetDwarf, and lurflurf, I think you'll like what I've come up with. I'm sorry to not be able to show some pictures but I believe the written proof will suffice. Hope it's clear enough. Thank you for your help.
The three properties of straight lines in the proof are the following: (1) A straight line can be created from any two points, (2) this line is unique, and (3) if two straight lines coincide at least at two points, all their points coincide (making them the same line).
$\mathrm{Proof:}$
A quadrilateral has four vertices, each vertex point must connect to two others in order to form the sides of the quadrilateral. Labeling these four points A, B, C, and D and forming the following sides AB, BC, CD, and DA we create the quadrilateral ABCD. The diagonals of said quadrilateral will consequently be AC and BD.
If a diagonal were to not lie completely inside or outside the quadrilateral then it (the diagonal) must cross one of the sides of the quadrilateral (either to enter or to exit the figure).
The diagonal AC cannot cross the side AB, DA, BC, or CD because this would imply that the diagonal AD equals the respective side it crosses by property (3) (since AC would coincide with the point of the side it crosses and the point A or C). The same applies to BD and the side AB, DA, BC, or CD.
This implies that the diagonals of a quadrilateral cannot cross its sides.
Therefore the diagonals of a quadrilateral must either lie entirely inside or entirely outside. $\mathrm{QED}$
Last edited: Jan 2, 2015
6. Jan 2, 2015
### Born
$\mathrm{Follow\ up:}$
Concerning the pentagon: labeling the five vertex points A, B, C, D, E; and forming the sides AB, BC, CD, DE, and EA. A diagonal is made from point A to point D crossing the side BC. This is possible since the diagonal would only share one point with the side BC- (This is unlike the quadrilateral in which every diagonal would share two points of a side if said diagonal crossed said side)
Therefore a diagonal which lies partially outside and partially inside the figure is possible. $\mathrm{QED}$
Last edited: Jan 2, 2015 | 2018-07-16T00:10:17 | {
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https://math.stackexchange.com/questions/2188573/probability-of-a-team-winning-a-tournament | # Probability of a team winning a tournament
Problem
In the World Series of baseball, two teams (call them A and B) play a sequence of games against each other, and the first team to win four games wins the series. Let p be the probability that A wins an individual game, and assume that the games are independent. What is the probability that team A wins the series?
There will be at maximum seven games played to decide a clear winner.
This is a practice problem from a course on Probability. Its solution provided in the course gives two approaches.
First approach:
The Game will stop as any one of team wins.
$$P(A) = P(\text{A winning in 4 games}) + P(\text{A winning in 5 games}) + P(\text{A winning in 6 games}) + P(\text{A winning in 7 games})$$
For A to win, the last game must be won by team A.
$$\Rightarrow P(A) = p^4 + { 4 \choose 3}p^4q + { 5 \choose 3}p^4q^2 + { 6 \choose 3}p^4q^3$$
Second approach
Imagine telling the players to continue playing the games even after the match has been decided, then the outcome of the match won’t be affected by this, and this also means that the probability that A wins the match won’t be affected by assuming that the teams always play 7 games.
$$P(A) = P(\text{A winning 4 times in 7 games}) + P(\text{A winning 5 times in 7 games}) + P(\text{A winning 6 times in 7 games}) + P(\text{A winning 7 times in 7 games})$$
I am not able to follow the second approach. Why is second approach correct?
Update:
I am looking for an intuitive explanation for the second approach because in the second approach it appears that probabilities for winning 5, 6 and 7 matches are used which were not required for team A to win.
• Why is the second approach correct? In the same way that if I flip a coin one time the probability it is heads is the same as the probability the first coin flipped is heads when flipping a coin seven times in succession. If I ask what is the probability I win the very next game against my friend where I play only one game, the probability is the same as if i ask what is the probability I win the very next game against my friend when I play fifty games after the first only important game. – JMoravitz Mar 15 '17 at 22:40
• Because a team winning in the second version will also win in the first, and vice versa – Henry Mar 15 '17 at 22:40
• @JMoravitz Thanks. The second approach uses winning 5, 6, as well as 7 games. If I am correct then in the coin example, only the probability of getting heads the first time is considered even if we flipped it 7 times. – ovais Mar 15 '17 at 22:50
• Both ways to compute $P(A)$ produce a polynomial in $p$. Since they are both correct, the the two methods should produce the same polynomial; and they do: $-20p^7 + 70p^6 -84p^5 + 35p^4$. – Fabio Somenzi Mar 16 '17 at 0:51
Team $A$ wins the game if they win four matches times before $B$ wins four matches. Thus the first approach measures the probability of team A doing so when team $B$ wins zero, one, two, or three matches before $A$'s fourth victory.
$$P(A) = \left(\binom 30 p^3q^0+\binom 41 p^3q^1+\binom 52 p^3q^2+\binom 63 p^3q^3\right)p$$
Now imagine the teams kept playing for a full seven matches even after one of them wins the game. Team $A$ wins the game if they win at least four matches among those seven, since if they do so then team $B$ can win at most three matches before $A$ wins their fourth.
\begin{align}P(A) & = \binom 77 p^7q^0+\binom 76 p^6q^1+\binom 75 p^5q^2+\binom 74p^4q^3\end{align}
We can see these are equal by taking the terms of the first equation, and including the imagined games after victory. Then if you use the binomial theorem to expand...
\begin{align}P(A) &= \binom 30 p^4q^0(p+q)^3+\binom 41 p^4q^1(p+q)^2+\binom 52 p^4q^2(p+q)+\binom 63 p^4q^3 \\[1ex] &= \binom 30 p^4q^0(p+q)^3+\binom 41 p^4q^1(p+q)^2+\binom 52 p^5q^2+\left(\binom 52+\binom 63\right) p^4q^3\\[1ex] &\vdots \\[1ex] &= \binom 77 p^7q^0+\binom 76 p^6q^1+\binom 75 p^5q^2+\left(\binom 30+\binom 41+\binom 52+\binom 63\right) p^4q^3 \\[1ex]& = \binom 77 p^7q^0+\binom 76 p^6q^1+\binom 75 p^5q^2+\binom 74p^4q^3\end{align}
• Thanks, Graham, I am looking for an intuition that why the second approach is using the probability of winning 5,6 or 7 matches in computing the result. It appears that it may increase $P(A)$ but it does not as both answers are same. – ovais Mar 16 '17 at 10:59
As an addendum to Graham's answer, after suitable massaging of the formulae, we get, for a best-of-$n$ tournament with odd $n$:
$$\sum_{0 \leq k \leq \frac{n-1}{2}} \binom{n}{\frac{n+1}{2}+k} \sum_{0 \leq j \leq k} (-1)^j \binom{\frac{n+1}{2}+k}{j} p^{\frac{n+1}{2}+k} \enspace.$$
We can then generate plots like this for $P(A)$:
Our intuition that a larger $n$ favors the stronger team is confirmed.
My question was to understand the intuition behind why the second approach is correct, especially why the second approach is apparently using probability for winning in 5, 6 or 7 matches while computing $P(A)$.
To figure out I tried to understand the cases which first event - P(A wins 4 times in 7 games) - of the second approach missed because that can give some clue over why we were adding the probabilities for winning 5,6 or 7 matches.
In this approach, we imagined the game is continued for 7 matches. For simplicity let's assume that $p = q = \frac 1 2$.
In the first approach, we added probabilities of four disjoint events - P(A winning in 4 games), P(A winning in 5), .. , P(A winning in 7 games). The denominators of each of these cases - representing the total number of possible cases - are different - $2^4, 2^5, 2^6, 2^7$.
In the second approach, denominators in the four disjoint events - P(A winning 4 times in 7 games), P(A winning 5 times in 7 games), ... P(A winning 7 times in 7 games) - are same and equal to $2^7$ representing the total number of possible outcomes.
Since the winner is decided as soon as one of the team wins 4 games, in the second approach, we have over counted the number of possible outcomes.
To reach the correct answer, either we decrease or adjust the number of possible outcomes from over counted outcomes or adjust it by increasing the number of favourable outcomes.
Adjusting the denominator will transform the second approach into the first approach. Thus let's try adjusting the favourable outcomes.
To put it accurately, we are not adjusting the favourable outcomes but after adding total number of ways A can win 4 matches, all the other outcomes of the games irrespective of success or failure of A will be considered as favourable.
This is equivalent to: P(A wins 4 matches and loses 3) + P(A wins 5 matches and loses 2) + P(A wins 6 matches and loses 1) + P(A wins 7 matches and loses 0).
Thus we are adding probabilities of winning 5, 6, and 7 matches because now the numerator contains all possible outcomes for losing 0, 1,2 or 3 matches once team A has won and the denominator already counts all the possible outcomes in the 7 games. | 2019-05-24T05:04:43 | {
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https://math.stackexchange.com/questions/1725821/structure-of-gal-mathbbq-zeta-15-mathbbq/1725844 | Structure of $Gal(\mathbb{Q}(\zeta_{15})/\mathbb{Q})$?
$\zeta_{15}$ is $15$th primitive $n$th root of unity.
Question: Find the structure of the group $Gal(\mathbb{Q}(\zeta_{15})/\mathbb{Q})$
I know that if $p$ is prime then $G=Gal(\mathbb{Q}(\zeta_{p})/\mathbb{Q})=\mathbb{Z_{p}}^*$ but when $p$ is not prime, I am not show how to solve this if not prime.
Does my required group have any relation to some cyclic group $\mathbb{Z_n}$, i.e. abelian
Would really appreciate your guidance as I have not got my head around these concepts of cyclotomic fields and Galois structure. Thanks
• It's the same thing as for primes. The Galois group is $\Bbb{Z}_{15}^*$ – Crostul Apr 3 '16 at 11:11
• Why is this the case? I would like to be able to construct some kind of proof – thinker Apr 3 '16 at 11:11
• en.wikipedia.org/wiki/Cyclotomic_field – Crostul Apr 3 '16 at 11:12
• I have seen it, and it does not give an explanation in sufficient detail – thinker Apr 3 '16 at 11:13
• Google "cyclotomic fields". You will find a lot of proofs for this fact, and other things – Crostul Apr 3 '16 at 11:15
Let $G$ be the Galois group of $\Bbb Q(\zeta_n)$ over $\Bbb Q$. An element $f \in G$ is entirely determined by its image on $\zeta_n$.
Since $f(\zeta_n)^k=f(\zeta_n^k)=1 \iff \zeta_n^k=1$ (recall that $f$ is a field automorphism), you know that $f(\zeta_n)$ is a primitive $n$-th root of unity: $$f(\zeta_n)=\zeta_n^{k_f}$$ for some integer $1≤k_f≤n$ coprime with $n$.
Therefore, you have a well-defined map $$\alpha : G \to (\Bbb Z/n\Bbb Z)^* \qquad \alpha(f)=[k_f]_n$$ You can check that this is a group isomorphism. It is clearly injective. Since $|G|=[\Bbb Q(\zeta_n):\Bbb Q]=\text{deg}(\Phi_n)=\phi(n) = |(\Bbb Z/n\Bbb Z)^*|$, $\alpha$ is bijective.
Finally, $(f \circ g)(\zeta_n)=f(g(\zeta_n)) = f(\zeta_n^{k_g})=\zeta_n^{k_g \, k_f}$ shows that $$\alpha(f \circ g) = [k_{f \circ g}]_n = [k_f]_n[k_g]_n=\alpha(f)\alpha(g).$$
More generally, let $K$ be a field of characteristic coprime with $n$ (e.g. if $\text{car}(K)=0$), and suppose that $R_n = \{x \in \overline K \mid x^n=1\}$ denotes the set of $n$-th roots of unity in an algebraic closure $\overline K$ of $K$ (notice that $R_n$ is always a cyclic group for the multiplication, since it is a finite subgroup of $(K^*,\cdot)$).
Then the extension $K(R_n)$ over $K$ is Galois (it is separable because $n$ is coprime with the characteristic of $K$) and you can show similarly that the Galois group of $K(R_n)$ over $K$ embeds in $(\Bbb Z/n\Bbb Z)^*$.
• so in this case $\zeta_{n}$ acts as a cyclic generator? – thinker Apr 3 '16 at 14:10
• Yes, absolutely: the $n$-th roots of $1$ (i.e. $x$ such that $x^n=1$) are of the form $x=\zeta_n^j$, i.e. the set $R_n$ of the $n$-th roots of $1$ is actually the subgroup generated by $\zeta_n$. Indeed, $R_n$ is a finite subgroup (for the multiplication) of $\overline{ \Bbb Q}^*$ (where $\overline{\Bbb Q}$ is an algebraic closure of $\Bbb Q$. You may know that every finite subgroup of $(K^*,\cdot)$ (where $K$ is any field) is actually cyclic. – Watson Apr 3 '16 at 14:13
• Then, $\Bbb Q(\zeta_n)$ over $\Bbb Q$ is Galois because it is separable (any extension of a field of characteristic $0$, as $\Bbb Q$, is separable) and normal (it is the splitting field of $X^n-1$ : any root of $X^n-1$ belongs to $R_n = \langle \zeta_n \rangle$, and therefore to $\Bbb Q(\zeta_n)$). – Watson Apr 3 '16 at 14:16 | 2019-11-12T16:57:43 | {
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http://mathhelpforum.com/algebra/148292-sum-2-squares.html | Math Help - Sum of 2 squares
1. Sum of 2 squares
Sorry before I tell you the question here was the question that led to this question.
Show that (m^2+1)(n^2+1)=(m+n)^2+(mn-1)^2
Done that successfully.
Using this results, write 500050 as the sum of 2 square numbers.
I have no idea how to make a knowledgeable guess at the two squares so I've tried using random (well, not exactly, generic random numbers) to get an answer but to no avail
2. Originally Posted by Mukilab
Sorry before I tell you the question here was the question that led to this question.
Show that (m^2+1)(n^2+1)=(m+n)^2+(mn-1)^2
Done that successfully.
Using this results, write 500050 as the sum of 2 square numbers.
I have no idea how to make a knowledgeable guess at the two squares so I've tried using random (well, not exactly, generic random numbers) to get an answer but to no avail
Try m=7.
50 is very close to 49 and stands out that way..
3. Hello, Mukilab!
Show that: . $(m^2+1)(n^2+1)\:=\m+n)^2+(mn-1)^2" alt=" (m^2+1)(n^2+1)\:=\m+n)^2+(mn-1)^2" />
Done that successfully. . Good!
Using this results, write 500,050 as the sum of 2 square numbers.
Note that: . $500,050 \;=\;(50)(10,\!001) \;=\;(7^2+1)(100^2+1)$
$\text{Let }m = 7,\;n = 100 \text{ in the formula:}$
. . $(7^2 + 1)(100^2 + 1) \;=\;(7 + 100)^2 + (7\cdot100 - 1)^2 \;=\;107^2 + 699^2$
4. m=7 is a really nice guess
OK. $500050=10001*50$
and $10001=10000+1$
and $50=49+1$
Apply in your equation: $(m^2+1)(n^2+1)=(m+n)^2+(mn-1)^2$
Then....
5. That would make n a decial (10sqrt10)
6. Originally Posted by Soroban
Hello, Mukilab!
Note that: . $500,050 \;=\;(50)(10,\!001) \;=\;(7^2+1)(100^2+1)$
$\text{Let }m = 7,\;n = 100 \text{ in the formula:}$
. . $(7^2 + 1)(100^2 + 1) \;=\;(7 + 100)^2 + (7\cdot100 - 1)^2 \;=\;107^2 + 699^2$
Thanks Soroban!
Sorry, only saw the first post at the start
Any tips for future questions such as this?? Take away the integers so I'm left with the algebraic numbers (as you did with 49 and 100)?
Thanks again
7. Originally Posted by Mukilab
That would make n a decial (10sqrt10)
Maybe you entered 50050 / 50 instead of 500050 / 50? m=7 gives n=100 as shown above.
8. Originally Posted by Mukilab
Thanks Soroban!
Sorry, only saw the first post at the start
Any tips for future questions such as this?? Take away the integers so I'm left with the algebraic numbers (as you did with 49 and 100)?
Thanks again
Another way to go about expressing integers as sums of two squares is knowing that: the set of integers expressible as the sum of two squares is closed under multiplication, and an odd prime is expressible as the sum of two squares if and only if it is congruent to 1 (mod 4). See here and here. | 2016-04-30T04:09:47 | {
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https://www.physicsforums.com/threads/showing-that-v-is-a-direct-sum-of-two-subspaces.747625/ | # Showing that V is a direct sum of two subspaces
Hi guys, I have this general question.
If we are asked to show that the direct sum of ##U+W=V##where ##U## and ##W## are subspaces of ##V=\mathbb{R}^{n}##, would it be possible for us to do so by showing that the generators of the ##U## and ##W## span ##V##? Afterwards we show that their intersection is a zero-vector. For example:
##U## is a subspace generated by ##(0,1)## and ##W## is a subspace generated by ##(2,2)##. Clearly those generators span two dimensional ##V##, and their intersection is ##(0,0)##. Therefore the conclusion can be made that their direct sum is ##V##. Is this kind of reasoning okay?
## Answers and Replies
jbunniii
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Yes, this reasoning is valid. In general, ##V## is the direct sum of ##U## and ##W## if and only if ##V = U+W## and ##U \cap W = \{0\}##. However, this is not the definition of direct sum, it's a (simple) theorem which you should try to prove.
However, be aware that this only works for two subspaces. If you have three or more subspaces, say ##\{U_i\}_{i=1}^{N}##, then it's possible to have ##V = U_1 + \ldots U_N## and ##U_i \cap U_j = \{0\}## for all ##i \neq j##, but the sum is not direct. It's a good exercise to construct an example where this occurs.
Yes, this reasoning is valid. In general, ##V## is the direct sum of ##U## and ##W## if and only if ##V = U+W## and ##U \cap W = \{0\}##. However, this is not the definition of direct sum, it's a (simple) theorem which you should try to prove.
However, be aware that this only works for two subspaces. If you have three or more subspaces, say ##\{U_i\}_{i=1}^{N}##, then it's possible to have ##V = U_1 + \ldots U_N## and ##U_i \cap U_j = \{0\}## for all ##i \neq j##, but the sum is not direct. It's a good exercise to construct an example where this occurs.
Ah ok thanks for your explanation jbuniii. If it's not too much, I would also like to ask why this is true? ##V_1 - V_2 = V_1 + V_2## if we define subtraction to be ##\{ x-y | x \in V_1, y \in V_2 \}.## This doesn't feel intuitive at all, I thought the subtraction of two subspaces will somehow reduce the elements of the new subspace, why it could be equal to the sum of subspaces?
jbunniii
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Ah ok thanks for your explanation jbuniii. If it's not too much, I would also like to ask why this is true? ##V_1 - V_2 = V_1 + V_2## if we define subtraction to be ##\{ x-y | x \in V_1, y \in V_2 \}.## This doesn't feel intuitive at all, I thought the subtraction of two subspaces will somehow reduce the elements of the new subspace, why it could be equal to the sum of subspaces?
Since ##y \in V_2## if and only if ##y \in -V_2##, it doesn't matter whether you add or subtract elements of ##V_2##, you get the same result in both cases: ##V_1 - V_2 = V_1 + V_2##.
Admittedly, the notation ##V_1 - V_2## can seem a bit misleading. Fortunately, since it is the same as ##V_1 + V_2##, there's no reason to use ##V_1 - V_2##.
Note that you could even define ##aV_1 + bV_2 = \{a x + b y | x \in V_1, y \in V_2\}##, where ##a## and ##b## are nonzero scalars. Then ##V_1 - V_2## is just a special case with ##a=1## and ##b=-1##.
Once again we have ##aV_1 + bV_2 = V_1 + V_2## for any nonzero ##a,b##, because ##x \in aV_1## if and only if ##x \in V_1## and similarly, ##y \in V_2## if and only if ##y \in bV_2##. So we can just stick with the notation ##V_1 + V_2## since all the other "linear combinations" of ##V_1## and ##V_2## give the same result.
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jbunniii
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By the way, here's a bit of extra info in case it is helpful. Suppose we start with a collection of subspaces ##\{U_i\}_{i=1}^{N}## of ##V##. We may be interested in combining the ##U_i##'s to form a larger subspace. The naive thing to do would be to form the set theoretic union ##\cup_{i=1}^{N}U_i##. But this won't generally be a subspace.
So what we really want is the smallest subspace containing all of the ##U_i## (or equivalently, the smallest subspace containing ##\cup_{i=1}^{N}U_i##). This turns out to be exactly ##U_1 + \ldots + U_N##. Proof: certainly ##U_1 + \ldots + U_N## is a subspace containing each ##U_i##. If ##S## is another such subspace then it must contain all elements of the form ##u_1+\ldots+u_N## with ##u_i \in U_i##. Thus ##U_1 + \ldots + U_N \subset S## so ##U_1 + \ldots + U_N## is the smallest such subspace.
Now in general, the ##U_i##'s may not be "linearly independent" of each other: maybe there is some nonzero element of ##U_1## which can be expressed as a linear combination of elements of the other ##U_i##'s. But if they ARE linearly independent, then we say that the sum is a direct sum, and we write it as ##U_1 \oplus \ldots \oplus U_N## instead of ##U_1 + \ldots + U_N##. We can obtain a basis for a direct sum ##U_1 \oplus \ldots \oplus U_N## by simply selecting a basis ##B_i## for each ##U_i## and taking the union: ##B = \cup_{i=1}^{N} B_i##. Indeed, this union is disjoint because there are no common elements among the ##B_i##'s. In the finite-dimensional case, this immediately tells us that
$$\text{dim}(U_1 \oplus \ldots \oplus U_N) = \sum_{i=1}^{N}\text{dim}(U_i)$$
This is not true if the sum is not direct. In that case, all we can say is that
$$\text{dim}(U_1 + \ldots + U_N) \leq \sum_{i=1}^{N}\text{dim}(U_i)$$
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Wow thanks for your detailed explanation. I just want to make sure for this last time that I understand this correctly.
Since ##y \in V_2## if and only if ##y \in -V_2##, it doesn't matter whether you add or subtract elements of ##V_2##, you get the same result in both cases: ##V_1 - V_2 = V_1 + V_2##.
Can I also intuitively think that ##V_2## = ##-V_2## if ##V## is a subspace, because if you multiply every elements in the subspace with -1, then you will basically get the same subspace because every element has an additive inverse? (In other words the negatives will become the positives and vice-versa)
jbunniii
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Can I also intuitively think that ##V_2## = ##-V_2## if ##V## is a subspace, because if you multiply every elements in the subspace with -1, then you will basically get the same subspace because every element has an additive inverse? (In other words the negatives will become the positives and vice-versa)
Sure, the fact that ##V_2## is a subspace and therefore contains its additive inverses is exactly the reason why ##y \in V_2## if and only if ##y \in -V_2##, and this latter statement is the definition of set equality: we conclude that ##V_2 = -V_2##. | 2021-06-14T17:17:30 | {
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http://thatcomputerguy.us/bernie-movie-rijetop/33d176-use-set-notation-and-list-all-the-elements | Write in roster notation all the integer numbers that are greater than 0 and less than 10, non-inclusive: Write in roster notation all integer numbers that are less than two, inclusive. {x | x is an odd whole number less than 10} Express the set using set notation and the listing method. Purplemath. the set of all counting numbers less than or equal to 7. list all the elements of the set and use set notation? $\{x|x \in \mathbb Z, x>0, x \ is \ divisible \ by \ 5\}$ OR would it be written like this? if the pattern is obvious (a nasty word in mathematics). I have a question that asks to use set builder notation for positive integers divisible by 5. They are surrounded by braces and separated by commas. If the order of the elements is changed or any element of a set is repeated, it does not make any changes in the set. 8 years ago. (a) The set A of counting numbers between ten and twenty. They wrote about it on the chalkboard using set notation: P = ... Let P be the set of all members in the math club. The expressions "A includes x" and "A contains x" are also used to mean set membership, although some authors use them to mean instead "x is a subset of A". Favorite Answer . Recently Asked Questions Could you help me with this math question? For example, if the universal set is $$\mathbb{R}$$, we cannot list all the elements of the truth set of “$$x^2 < 4$$.” In this case, it is sometimes convenient to use the so-called set builder notation in which the set is defined by stating a rule that all elements of the set must satisfy. In set theory and its applications to logic, mathematics, and computer science, set-builder notation is a mathematical notation for describing a set by enumerating its elements, or stating the properties that its members must satisfy.. Example. A set can be written explicitly by listing its elements using set bracket. A set is created by using the set() function or placing all the elements within a pair of curly braces. It is used with common types of numbers, such as integers, real numbers, and natural numbers. This notation can also be used to express sets with an interval or an equation. Use set notation and the listing method to describe the set. The blade length is the radius of the circle formed by the area swept by the blades. ... {x | x is an even multiple of 5 that is less than 10} Example. More in general, given a tuple of indices, how would you use this tuple to extract the corresponding elements from a list, even with duplication (e.g. {9,10,11,12} List all the elements of the following set. {50, 47, 44, ...., 29} {50,47,44,41,38,35,32,29} List all the elements of the following set. For example, the number 5 is an integer, and so it is appropriate to write $$5 \in \mathbb{Z}$$. tuple (1,1,2,1,5) produces [11,11,12,11,15]). Sets in Python A set in Python is a collection of objects. It is a way to describe the set of all things that satisfy some condition (the condition is the logical statement after the “$$\st$$” symbol). List any subsets, and show the relationships among the sets and subsets in a Venn diagram. The sets in python are typically used for mathematical operations like union, intersection, difference and complement etc. Defining a set using this sort of notation is very useful, although it takes some practice to read them correctly. - the answers to estudyassistant.com . ) Use set notation, and list all the elements of the set. Relevance. List or Roster method, Set builder Notation, The empty set or null set is the set that has no elements. Summary: Set-builder notation is a shorthand used to write sets, often for sets with an infinite number of elements. List all the elements of the following set. The symbols shown in this lesson are very appropriate in the realm of mathematics and in mathematical logic. The set of all the fibers over the elements of Y is a family of sets indexed by Y. That is, x is an element of the intersection A ∩ B, if and only if x is both an element of A and an element of B. For example, for the function f(x) = … Let us now explain what a set notation is. Set notation uses curly brackets { } which are sometimes referred to as braces. Answer Save. (a) The set A of counting numbers between ten and twenty. $\{x|x \in \mathbb Z, x>0, x \ = \ x/5 \}$ discrete-mathematics. SET BUILDER Notation: Set-builder notation does not list all the elements of a set the way roster notation does. Set notation or ways to define a set. We can list each element (or "member") of a set inside curly brackets like this: Common Symbols Used in Set … List all of the elements of each set using the listing method. 2 Answers. An object that belongs to a set is called an element (or a member) of that set. They are different from lists or tuples in that they are modeled after sets in mathematics. 128k 24 24 gold badges 165 165 silver badges 216 216 bronze badges. Set Symbols. Roster Notation. If a and be are meant to be arbitrary elements of the set, not a and b specifically, then there are 5C2 ways of selecting a set, 5x4 / 2 = 10 ways. The inverse image of a singleton, denoted by f −1 [{y}] or by f −1 [y], is also called the fiber over y or the level set of y. The set of all even integers, expressed in set-builder notation. Use roster and rule notation to describe the set F, which consists of all the positive multiples of 5 that are less than 50. Two sets are equivalent if they contain the same number of elements. In set-builder notation, the set is specified as a selection from a larger set, determined by a condition involving the elements. 2. StartSet x | x is a natural number greater than 9 and less than 17 EndSet {x | x is a natural number greater than 9 and less than 17} Get Answer. There is more than one format for writing this sequence in Set Builder Notation. Answer to List all the elements of the following set. asked Apr 12 '10 at 11:35. In this notation, the vertical bar ("|") means "such that", and the description can be interpreted as "F is the set of all numbers n, such that n is an integer in the range from 0 to 19 inclusive". Instead, set-builder notation describes the properties of the elements of the set. For example, one can say “let $$A$$ be the set of all odd integers”. list all the elements of each set? {−29, −24 , -19 } Answers: 2 Get Other questions on the subject: Mathematics. Answer: 3 question List all the elements of the following set. share | cite | improve this question | follow | asked Jul 28 '17 at 8:58. Use set notation to list all the elements in the following set. Use set notation to list all the elements of the set. For example: The intersection of the sets {1, 2, 3} and {2, 3, 4} is {2, 3}. Venn diagrams can be used to express the logical (in the mathematical sense) relationships between various sets. The intersection of two sets A and B, denoted by A ∩ B, is the set of all objects that are members of both the sets A and B.In symbols, ∩ = {: ∈ ∈}. I think you got the point. use set notation and the listhing method to describe the set? The symbol 2 is used to describe a relationship between an element of the universal set and a subset of the universal set, and the symbol $$\subseteq$$ is used to describe a relationship between two subsets of the universal set. A rule works well when you find lots and lots of elements in the set. Objects placed within the brackets are called the elements of a set, and do not have to be in any specific order. Set example To create a set, we use the set() function. Three Sets. enclosing the list of members within curly brackets. (6,7,8...,14) 3. We use special notation to indicate whether or not an element belongs to a set, as shown below. a. to the nearest square foot, what is the area of this circle for model 3? Set notation is used to help define the elements of a set. Lv 7. Some Example of Sets . The cardinality or cardinal number of a set is the number of elements in a set. Would this be the correct way writing it? Other notations include f −1 (B) and f − (B). Denote each set by set-builder notation, using x as the variable. {2,4,8, ..., 256} The complete list is { } The objects are called members or elements of the set. share | improve this question | follow | edited Aug 10 '19 at 11:16. jpp. Use set notation and the listing method to describe the set. question: A pro golfer is ranked 48th out of 230 players that played in a PGA event in 2016. . The natural number less than 37 that are divisible by 5. Use set notation and the listing method to describe the set. The set of all whole numbers greater than 8 and less than 13. Here are some more examples: Example 0.3.1. A set is a collection of things, usually numbers. If that's what you want, you need to state that more clearly, but then all you have to do is to list the 10 possible pairs of two elements … Let us say the third set is "Volleyball", which drew, glen and jade play: Volleyball = {drew, glen, jade} But let's be more "mathematical" and use a Capital Letter for each set: S means the set of Soccer players; T means the set of Tennis players; V means the set of Volleyball players If … You can also use Venn Diagrams for 3 sets. listing method is {1,2,3,4,5,6,7} set notation is {x | 1 ≤ x ≤ 7 }-----I must point out that some textbooks include zero as a counting number. Creating a set. Notation and terminology. anonymous. A roster is a list of the elements in a set. Mathematics, 21.06.2019 19:00, bentonknalige. Describing sets One can describe a set by specifying a rule or a verbal description. Then $$A$$ is a set and its elements are all the odd integers. After school they signed up and became members. and listing method to describe the set. Here is a commonly used format: Set A = {x | x ∈ ℤ, 0 ≤ x 12} Two sets are equal if they contain the exact same elements although their order can be different. For example, a set F can be specified as follows: = {∣ ≤ ≤}. Show Video Lesson. python list. 1. the set of all counting numbers less than or equal to 6? A set whose elements all belong to another set; for example, the set of odd digits, O 5 {1, 3, 5, 7, 9}, is ... Indicate the multiples of 4 and 12, from 1 to 240 inclusive, using set notation. Related course Python Programming Bootcamp: Go from zero to hero. If the set contains a lot of elements, you can use an ellipsis ( . Sets are available in Python 2.4 and newer versions. Use set notation and the listing method to describe the set. We can create a set, access it’s elements and carry out these mathematical operations as shown below. Interval and Graphical Notation. You could define a set with a verbal description: All sets above are described verbally when we say, " The set of all bla bla bla "b. 2. Sample questions . - 17271819 Logician George Boolos strongly urged that "contains" be used for membership only, and "includes" for the subset relation only. Whether or not an element belongs to a set notation and the listing method to describe the.. From a larger set, and do not have to be in any specific order are equal if contain......., 29 } { 50,47,44,41,38,35,32,29 } list all the elements of the elements a! No elements help define the elements of the set ( ) function or placing all the elements each... Brackets are called the elements within a pair of curly braces '19 11:16.! Called an element ( or a member ) of that set 128k 24 24 badges. 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Follow | Asked Jul 28 '17 at 8:58 is more than one format for writing this sequence set...: set-builder notation describes the properties of the set ( ) function on the subject: mathematics the answers estudyassistant.com! Within a pair of curly braces } express the logical ( in the realm of mathematics and in logic... Such as integers, expressed in set-builder notation is a set, access ’. | 2021-09-22T11:59:07 | {
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https://math.stackexchange.com/questions/1364821/verify-that-binomn14-frac-left-substack-binomn2-displaystyle | # Verify that $\binom{n+1}{4} = \frac{\left(\substack{\binom{n}{2}\\{\displaystyle2}}\right)}{3}$ for $n \geq 4$
Verify that for $n \geq 4$ $$\dbinom{n+1}{4} = \frac{\left(\substack{\binom{n}{2}\\{\displaystyle2}}\right)}{3}$$
Now present a combinatoric argument for the above.
First, by verify does it mean check for some n > 3? if so, then n = 4 gives both sides value 5. I have tried expanding both sides and It just gets messy, but i'm sure that would be attempting to prove it?
I cant quite move ahead with this one, I have said that $\left(\substack{\binom{n}{2}\\{\displaystyle2}}\right)$ is the number of ways of choosing 2 pairs of objects from n objects. my reasoning for this is that $\dbinom{n}{2}$ is the number of ways of choosing 2 objects from n and hence $\left(\substack{\binom{n}{2}\\{\displaystyle2}}\right)$ is the number of ways of choosing 2 of these ways... What i am struggling to do is understand how the three comes into it.
• I think, in this case, "verify" means to prove algebraically, by manipulating the right hand side until it looks like the left. – Theo Bendit Jul 17 '15 at 18:08
• Please don't use display math in titles; it takes too much vertical space on the front page. – Rahul Jul 17 '15 at 18:09
• Hints: (1) How are two pairs of two things like one quadruple of things? (2) What if both pairs contain a common element? These are two independent hints. – Rahul Jul 17 '15 at 18:12
• before reading the answer below fully and do the bad thing, i would like to take your comment into consideration and work it out for myself. (1) A pair of things is like one quadruple of things when each pair are disjoint, two pairs of things are not like a quadruple of things when they have a common member. so to expand, if i have four objects a,b,c,d. there are three ways i can choose distinct pairs from these four objects. [ab][cd], [bc][ad], [ac][bd]. Now concerning (2), i am struggling to use this hint at the moment. – user197848 Jul 17 '15 at 18:47
"Verify" would mean verify algebraically (using formulas such as $\binom{n+1}{4}=\frac{(n+1)n(n-1)(n-2)}{24}$. It's not that hard if you just consider the factorizations of each side. However, there is a nice combinatorial argument. $\left(\substack{{\binom{n}{2}}\\{\displaystyle 2}}\right)$ is the number of ways to pick two pairs of elements from a set of $n$; we can combine these pairs, and if an element is common to both pairs, then treat one of the instances of the element as a new $(n+1)$-st element. This counts the number of ways to pick 4 elements from a set of $n+1$. However, note that all ways to pick 4 are counted exactly 3 times, so you must divide by 3 to get the equality that is desired.
To verify:
$${n\choose2} = \frac{n(n-1)}{2}$$
$${\frac{n(n-1)}{2}\choose2} = \frac{\frac{n(n-1)}{2}.(\frac{n(n-1)}{2}-1)}{2}$$
$${\frac{n(n-1)}{2}\choose2} = \frac{n(n-1).(n^2-n-2)}{8} = \frac{(n+1)n(n-1)(n-2)}{8}\tag1 - RHS$$
$${(n+1)\choose4} = \frac{(n+1)n(n-1)(n-2)}{4!} = \frac{RHS}{3}$$
Hence proved.
• AH i see where it went all wrong for me. thanks :) – user197848 Jul 17 '15 at 18:20
• You are welcome. See the other responder's answer for a combintorial argument. It makes sense. Good luck – Satish Ramanathan Jul 17 '15 at 18:22
Using \begin{align} \binom{n}{2} = \frac{n(n-1)}{2} \end{align} then \begin{align} \frac{1}{3} \, \binom{\binom{n}{2}}{2} &= \frac{1}{2} \binom{n}{2} \, \left(\frac{n}{2} - 1\right) \\ &= \frac{n(n-1)}{4!} \left( n(n-1)-2 \right) = \frac{(n+1)(n)(n-1)(n-2)}{4!} \\ &= \binom{n+1}{4}. \end{align}
As to the modified component of the question: The comments seem to be helpful
Algebraic Manipulation is actually not that bad. $\binom{n}{2}=n(n-1)/2$, so $$\binom{\binom{n}{2}}{2}=\binom{n(n-1)/2}{2}=(n(n-1)/2)(n(n-1)/2-1)/6=n(n-1)(n^2-n-2)/24=n(n-1)(n+1)(n-2)/24=(n+1)!/4!(n-3)!=\binom{n+1}{4}$$ | 2019-05-22T09:07:57 | {
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https://math.stackexchange.com/questions/1347223/help-finding-the-limit-of-this-series-frac14-frac18-frac116 | # Help finding the limit of this series $\frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \frac{1}{32} + \cdots$
How can I go about finding the limit of $$\frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \frac{1}{32} + \cdots = \sum_{k = 1}^{\infty} \frac{1}{2^{k+1}}?$$ Could I use the absolute value theorem? I have a feeling it converges to $0$ but I am not sure.
• Of the series or the sequence? – Sloan Jul 2 '15 at 17:26
• sorry for the confusion. The third one and the limit of the series. – Lil Jul 2 '15 at 17:30
• Divide every term by $2$ and see what happens. – Yves Daoust Jul 3 '15 at 12:28
• The sequence converges to zero, but the sum of the terms converges to a positive value. Be sure you make a distinction between the values of the individual terms and the sum of the terms. – John Molokach Jul 3 '15 at 16:12
You know the limit of $1+\frac12 +\frac14+\frac18+\ldots$, don't you?
Your sequence is just like that, but without the $1$ and the $\frac12$.
Using the formula for the geometric series $$\sum\limits_{k=0}^\infty q^k=\frac{1}{1-q}$$ with $|q|<1$ we have: $$\sum\limits_{k=0}^\infty \frac{1}{2^{n+1}}=\sum\limits_{k=0}^\infty \left(\frac{1}{2}\right)^{n+1}=\sum\limits_{k=0}^\infty \frac{1}{2}\cdot \left(\frac{1}{2}\right)^n = \frac {1}{2}\cdot\sum\limits_{k=0}^\infty \left(\frac{1}{2}\right)^n = \frac {1}{2}\cdot \frac{1}{1-\frac{1}{2}}=\frac{1}{2}\cdot 2=1.$$
• The problem was modified after you answered it. – N. F. Taussig Jul 3 '15 at 10:03
• Well, if we start with $k=1$ we can easily obtain $$\sum\limits_{k=1}^\infty \frac {1}{2^{n+1}}=\sum\limits_{k=0}^\infty \frac{1}{2^{n+1}}-\frac{1}{2^{0+1}}=1-\frac{1}{2}=\frac {1}{2}.$$ – Hirshy Jul 3 '15 at 10:17
If your series begins at $\frac{1}{4}$, then it should be:
The sequence $a_n = \dfrac{1}{2^{n+2}}$ converges to $0$ as $n \to \infty$.
The series $$\sum_{n=0}^{\infty} \frac{1}{2^{n+2}}=\frac{1}{4} \cdot \sum_{n=0}^{\infty} \frac{1}{2^{n}} = \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \ldots$$ is the sum of a geometric sequence with common ratio $\frac{1}{2}$ and first term $\frac{1}{4}$. The series is convergent since the absolute value of the common ratio, $|r| = \frac{1}{2} < 1$. The sum evaluates to $$\frac{\frac{1}{4}}{1- \frac{1}{2}} = \frac{1}{2}$$ by using the general formula for the sum of a geometric series with common ratio $r$ and first term $a$, namely:$$\sum_{n=0}^{\infty}ar^n = \frac{a}{1-r}$$ as long as $|r| < 1$.
• maybe I have the wrong formula.. my series is 1/4+1/8+1/16+1/32... – Lil Jul 2 '15 at 17:54
• It begins at 1/4 not 1/2 as everyone is commenting. – Lil Jul 2 '15 at 17:54
• Then your series is $$\sum_{n=0}^{\infty} \frac{1}{2^{n+2}}$$ since the sum of a geometric series starts from $n=0$ by convention. :-) – Zain Patel Jul 2 '15 at 17:57
• I don't understand why it would be 1/2^n+2 because when n=1 it would become 1/2^3 which is 1/8 but my first term is 1/4...? – Lil Jul 2 '15 at 17:58
• doesn't the sum of a geometric series begin at 1? or no? – Lil Jul 2 '15 at 17:58
Use $$\sum_{n=0}^\infty a r^n=\frac{a}{1-r},\quad |r|<1.$$ Take $r=\frac12$ and $a=\frac14$ since the first term of your series is $\frac14$ and the common ratio of your Geometric progression is $\frac12.$
• The problem was modified after you answered it. – N. F. Taussig Jul 3 '15 at 10:03
I will add an indefinite version of the sum. In difference calculus we have that the function $2^n$ is the analogous of $e^x$ on ordinary differential calculus.
We have that $\sum 2^n \delta n=2^n$ (you can see it by yourself if you do the difference and see that it doesnt change, i.e. $\Delta 2^n=2^{n+1}-2^n=2^n$).
In your sum, doesnt taking limits, we have that
$$\sum 2^{-n-1}\delta n=-2^{-n}$$
If we take limits then
$$\sum_{n=1}^{\infty} 2^{-n-1}=-2^{-n}\bigg\lvert_{1}^{\infty}=-2^{-\infty}+2^{-1}=\frac{1}{2}$$
Take care with notation because
$$\sum_{k\ge 0}2^k\ne\sum_{k\ge 0}2^n$$ | 2020-10-27T14:55:05 | {
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https://math.stackexchange.com/questions/1366021/every-normal-subgroup-is-the-kernel-of-some-homomorphism | # Every normal subgroup is the kernel of some homomorphism
Clearly the kernel of a group homomorphism is normal, but I often hear my professor mention that any normal subgroup is the kernel of some homomorphism.
This feels correct but isn't entirely obvious to me.
One thought I had is that for any normal subgroup $N$ of $G$, we could define the quotient homomorphism $\pi:G\to G/N$ since $G/N$ is a group.
I was imagining that we could consider $\pi^{-1}:G/N\to G$, whose kernel would then be $N$. However, $\pi^{-1}$ doesn't exist since $\pi$ is not a bijection in general.
So my question is this: is there an obvious way to define a homomorphism whose kernel is an arbitrary normal subgroup of $G$? Or does it depend on the particular group whether you can define such a homomorphism?
• No, you more or less have it, the kernel of $\pi\colon G\to G/N$ is $N$. No need to worry about $\pi^{-1}$, or if it exists. Jul 18, 2015 at 23:17
You only have to consider that $\pi:G\to G/N$, defined by $\pi(x)=xN$, is a homomorphism and the $\ker \pi$ is precisely $N$.
• overcomplicating as usual. thanks. Jul 18, 2015 at 23:18
• that's why we are meant to be here Jul 18, 2015 at 23:19
Anyway, if the kernel is non-trivial, the homomorphism is not injective!
You're confusing the inverse of an isomorphism, and the inverse image of a subset by a map. So, yes, if $$N$$ is a normal subgroup, it is the kernel of the canonical homomorphism : \begin{align*}\pi\colon G&\longrightarrow G/N,\\ g&\longmapsto gN. \end{align*} Indeed, $$\pi^{-1}(\bar 1)=\pi^{-1}(N)=N$$.
Your intuition is correct, and unfortunately, yes, $\pi^{-1}$ isn't a function. Could we somehow "make" it one?
Well, one way this is often done with "ordinary" functions, is to treat $f^{-1}$, for a given function $f: A \to B$, as not something like this:
$f^{-1}: B \to A$
but instead considering $f^{-1}$ as a function between power sets:
$f^{-1}: \mathcal{P}(B) \to \mathcal{P}(A)$
where for a subset $Y \subseteq B$, we define $f^{-1}(Y) = X = \{x \in A: f(x) \in Y\}$.
Applying this definition to $\pi^{-1}$, we get:
$\pi^{-1}(e_{G/N}) = \pi^{-1}(N) = \{g \in G: \pi(g) \in N\}$
$= \{g \in G: gN = N\} = \{g \in G: ge^{-1} \in N\} = \{g \in G: g \in N\} = N$ | 2022-05-24T16:35:15 | {
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https://en.khanacademy.org/math/ap-statistics/density-curves-normal-distribution-ap/stats-normal-distributions/a/basic-normal-calculations | If you're seeing this message, it means we're having trouble loading external resources on our website.
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## AP®︎/College Statistics
### Course: AP®︎/College Statistics>Unit 4
Lesson 4: Normal distributions and the empirical rule
# Basic normal calculations
AP.STATS:
VAR‑2 (EU)
,
VAR‑2.A (LO)
,
VAR‑2.A.3 (EK)
CCSS.Math:
Many measurements fit a special distribution called the normal distribution. In a normal distribution,
• approximately equals, 68, percent of the data falls within 1 standard deviation of the mean
• approximately equals, 95, percent of the data falls within 2 standard deviations of the mean
• approximately equals, 99, point, 7, percent of the data falls within 3 standard deviations of the mean
Problem 1
A large sample of females had their systolic blood pressure measured. The mean blood pressure was 125 millimeters of mercury and the standard deviation was 10 millimeters of mercury.
Which normal distribution below best summarizes the data?
Problem 2
What percent of females had blood pressures between 105 and 135 millimeters of mercury?
percent
Problem 3
The sample had a total of 400 females that participated.
About how many females in the sample had blood pressures higher than 145 millimeters of mercury?
approximately equals
females
## Want to join the conversation?
• What is the Gauss curve have to do with this?
• For the empirical rule to be valid the data must be normally distributed, so the rules for percentages in the problems above would not hold true if the data didn't follow a gaussian or normal distribution.
• I answered the last question in the following way. I figured out the z score which was 0.9772 for value that is two standard deviations above normal. Therefore one can deduce that the sample size that lies to the right of this should be 0.0228 (1-0.9772). That multiplied by 400 gives me 9.1 approximated to 9. What is wrong with this method?
• I agree that using the z-Score is the most accurate answer. The answer given is using approximations of the normal and are not as accurate.
• where does the 13.5% come from in question 3?
• Empirical rule!
Ok, so 95% of the observations are within 2 standard deviations of the mean, and 68% of the observations are within 1 standard deviation of the mean, right?
Let's find the difference between 2 s.d. and 1 s.d. It will be 95%-68%=27%. But you have to divide this 27% by 2 because you have to find the percentage between 105 and 115 millimeters. There comes 13.5%!
• why do i have E.D.?
• what if they ask you to solve for standard deviation with only knowing the range
• This question is not really meaningful for a normal distribution, since all normal distributions have infinite range.
For general data sets, knowing the range of a data set is not sufficient for finding its standard deviation. For example, the data sets 1,5,5,9 and 1,2,8,9 both have range 9-1=8, but 1,2,8,9 has the larger standard deviation because the values are spread out farther from the mean (5).
Have a blessed, wonderful day!
• Are these realistic statistics? They better not be, my math teacher commonly uses extremely unrealistic stats, it's funny.
• For the titles of Articles, Videos, and Skills; should the wording follow the same grammar rules as the title of a book does?
e.g.: Should the titles be capitalized with the exception of conjunctions like and/or/but/ect?
I'm slightly confused, because KA only capitalizes the first word, but I can't find anywhere on the web where it tells you how to properly right the title.
Any help?
I'm slightly confused...
• Khan Academy is an exception to the usually title rules. Sometimes websites might only capitalize the first word or not follow the exact rules as in a book for some videos and articles. You are correct though, that rule is used commonly in books, movies, etc.
(1 vote)
• Is there a table or cheat sheet I can use to help with the subject?
(1 vote)
• Here is a website that has z-tables for positive and negative z-scores as well as other things related to this subject: https://z-scoretable.com/.
There are plenty of z-tables on Google Images as well.
Hope this helps!😄 | 2023-03-23T02:23:21 | {
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http://summerofgodel.blogspot.my/ | ## Wednesday, December 31, 2014
### Almost all integers contain the digit 3
This is my last chance to blog this year. However, I did not get back into the proof of Gödel's incompleteness theorem yet. So instead, this post is motivated by this Numberphile video, which explains why almost all integers contain the digit 3. More precisely, the percentage of integers that contain at least one occurrence of the digit '3' in the set $$\mathbb{N}_u = \{0,1,2,3,\cdots,u-1\}$$ goes to 100% as the upper bound $$u$$ goes to infinity.
We are going to attack this problem with two different approaches. First, we'll use a recurrence relation approach. Second, we'll go over the combinatorial proof discussed in the video. Finally, we'll check that the two approaches do give the same answer.
In both approaches, we will use values of $$u$$ that are powers of 10. So if $$u = 10^n$$, for $$n \in \mathbb{N}^+$$, then the set under consideration $$\mathbb{N}_{10^n}$$, which we denote $$A_n$$, is the set of all of the (non-negative) integers containing at most $$n$$ digits.
Recurrence relation approach
In this approach, we use $$S_n$$, for $$n \in \mathbb{N}^+$$, to denote the subset of $$A_n$$ of all of the integers that contain at least one occurrence of the digit '3' and we use $$T_n$$ to denote $$|S_n|$$, the cardinality of $$S_n$$.
• Since $$A_1=\{0,1,2,3,4,5,6,7,8,9\}$$, $$S_1=\{3\}$$ and $$T_1=1$$.
• Since $$A_2=\{0,1,2,3,4,\cdots,98,99\}$$, $$S_2=\{ 3,13,23,30,31,32,33,34,\cdots,39,43,53,63,73,83,93\}$$$$=\{ 3,13,23,43,53,63,73,83,93\}$$$$\cup \{30,31,32,33,34,\cdots,39\}$$. In other words, we can split the set $$S_2$$ into two subsets, one whose elements all start with the digit '3' (there are exactly 10 such integers), and the other one whose elements do not start with the digit '3' (there are exactly $$9\cdot T_1$$ of those, since if the first digit is not '3' then the least significant digit must belong to $$S_1$$). Thus, $$T_2 = 10 + 9\cdot T_1 = 10 + 9 = 19$$.
• Similarly, since $$A_3=\{0,1,2,3,4,\cdots,998,999\}$$, $$S_3$$ is the union of two sets, one whose elements all start with the digit '3' (there are exactly $$10^2$$ such integers), and the other one whose elements do not start with the digit '3' (there are exactly $$9\cdot T_2$$ of those, since if the most significant digit is not '3' then the number formed by the other digits must belong to $$S_2$$). Thus, $$T_3 = 10^2 + 9\cdot T_2= 100 + 9\cdot 19 = 271$$.
• Since this reasoning clearly generalizes to all values of $$n \in \mathbb{N}^+$$, we obtain the following recurrence relation:
$\left\{ \begin{array}{ll} T_1 = 1 & \\ T_n = 10^{n-1} + 9T_{n-1} & \text{if}\ n>1 \end{array} \right.$ Now, let's solve this recurrence relation by iteration to get a closed-form formula for $$T_n$$:
$$\begin{array}{l@{0pt}l} T_1 & = 1\\ T_2 = 10^1 + 9T_1 & = 10+9 \\ T_3 = 10^2 + 9T_2 & = 10^2+9(10+9) \\ & = 10^2 + 9\cdot 10 + 9^2 \\ T_4 = 10^3 + 9T_3 & = 10^3+9(10^2 + 9\cdot 10 + 9^2) \\ & = 10^3+9\cdot 10^2 + 9^2 \cdot 10 + 9^3\\ T_5 = 10^4 + 9T_4 & = 10^4+9(10^3+9\cdot 10^2 + 9^2 \cdot 10 + 9^3)\\ & = 10^4+9\cdot 10^3+9^2\cdot 10^2 + 9^3 \cdot 10 + 9^4\\ \cdots &\\ T_n = 10^{n-1} + 9T_{n-1} & = 10^{n-1}+9\cdot 10^{n-2}+9^2\cdot 10^{n-3} + \cdots + 9^{n-2}\cdot 10^1 + 9^{n-1}\\ \end{array}$$
In conclusion, $$\displaystyle T_n = \sum_{i=0}^{n-1} \left( 9^{n-i-1}\cdot 10^i\right)$$ for $$n \in \mathbb{N}^+$$
Since this formula is rather ugly, let's turn to the combinatorial approach discussed in the Numberphile video.
Combinatorial approach
It is easy to determine the total number of integers with exactly $$n$$ digits without having to enumerate them, namely $$10 \times 10 \times \cdots \times 10 = 10^n$$, since there are exactly 10 digits to choose from for each position in the $$n$$-digit number. Note that this count actually includes all of the numbers with leading zeros, such as 01 and 00023, which are identical to 1 and 23, respectively. In other words, what we really have is $$\left|A_n\right|= 10^n$$.
Similarly, we can compute the total number of integers with at most $$n$$ digits that do not contain the digit '3', namely $$9 \times 9 \times \cdots \times 9 = 9^n$$, since there are now only 9 digits to choose from at each position in the integer.
In conclusion, $$T_n = 10^n -9^n$$.
This is both a much nicer and easier-to-derive closed-form formula than the one we obtained with the recurrence relation approach. But are the two formulas equal?
Let's check our work
Let $$P(n)$$, for $$n \in \mathbb{N}^+$$, denote: $$\displaystyle \sum_{i=0}^{n-1} \left( 9^{n-i-1}\cdot 10^i\right) = 10^n - 9^n$$
We now prove by mathematical induction that $$P(n)$$ holds for $$n \in \mathbb{N}^+.$$
Basis:
• $$\displaystyle \sum_{i=0}^{1-1} \left( 9^{1-i-1}\cdot 10^i\right) = \sum_{i=0}^{0} \left( 9^{1-i-1}\cdot 10^i\right) = 9^{1-0-1}\cdot 10^0 = 9^0 \cdot 10^0 =1$$
• $$10^1 - 9^1 = 10 - 9 = 1$$
• Therefore $$P(1)$$ holds
Inductive step:
• Assume that $$P(n)$$ holds for any $$n \in \mathbb{N}^+$$, that is: $\sum_{i=0}^{n-1} \left( 9^{n-i-1}\cdot 10^i\right) = 10^n - 9^n \quad \text{[inductive hypothesis]}$
• Let's compute the left-hand side of $$P(n+1)$$:
$\begin{array}{l@{0pt}l} \sum_{i=0}^{n} \left( 9^{n-i}\cdot 10^i\right) & = 9\left(\frac{1}{9}\sum_{i=0}^{n} \left( 9^{n-i}\cdot 10^i\right)\right)\\ & = 9\left(\sum_{i=0}^{n} \left( 9^{n-i-1}\cdot 10^i\right)\right)\\ & = 9\left(\sum_{i=0}^{n-1} \left( 9^{n-i-1}\cdot 10^i\right) + \sum_{i=n}^{n} \left( 9^{n-i-1}\cdot 10^i\right) \right)\\ & = 9\left(\sum_{i=0}^{n-1} \left( 9^{n-i-1}\cdot 10^i\right) + \left( 9^{n-n-1}\cdot 10^n\right) \right)\\ & = 9\left(\sum_{i=0}^{n-1} \left( 9^{n-i-1}\cdot 10^i\right) + \frac{10^n}{9} \right)\\ & = 9\left( \left(10^n - 9^n\right) + \frac{10^n}{9} \right) \quad \text{by the inductive hypothesis} \\ & = 9\cdot 10^n - 9\cdot 9^n + 10^n\\ & = 10\cdot 10^n - 9\cdot 9^n\\ & = 10^{n+1} - 9^{n+1}\\ \end{array}$
• Therefore $$P(n+1)$$ holds.
Main result
We are now in a position to prove our main result, namely that the percentage of integers that contain the digit '3' in the set $$A_n$$ goes to 100% as $$n$$ goes to infinity. This percentage is equal to
$$100\times\frac{T_n}{\left|A_n\right|}=100\times\left(\frac{10^n-9^n}{10^n}\right)=100\times\left(1-\frac{9^n}{10^n}\right)= 100 - 100\left(\frac{9}{10}\right)^n$$.
And this percentage goes to 100% as $$n$$ goes to infinity, since $$\displaystyle \lim_{n \to \infty}\left( \frac{9}{10}\right)^n = 0$$.
Discussion
Of course, all of the proofs above still hold if we had picked the digit '8' (say) instead of the digit '3'. In other words, it is also true that almost all integers contain the digit '8'. So, if we use $$E_n$$ to denote the cardinality of the subset of $$A_n$$ of all of the integers that contain at least one occurrence of the digit '8', $$E_n = 10^n-9^n$$. Similarly for the digit '5' (say): $$F_n = 10^n-9^n$$.
But, if both $$100\times\frac{T_n}{\left|A_n\right|}$$ and $$100\times\frac{E_n}{\left|A_n\right|}$$ go to 100% as $$n$$ goes to infinity, the sum of these two percentages will eventually exceed 100%! That is true. However, this sum double counts all of the integers that contain both the digit '3' and the digit '8'. The correct percentage of such integers is $$100\times\frac{10^n-8^n}{10^n}$$, which also goes to 100% as $$n$$ goes to infinity.
So there is no paradox here. When $$n$$ gets large enough, almost all of the integers will contain all of the digits 0 through 9. Note that all of these results are asymptotic. Indeed, it takes relatively large values of $$n$$ for these percentages to get anywhere close to 100%. For example, for the percentage of integers that contain the digit '3' to reach 90%, 95%, 99% and 99.99%, the number $$n$$ of digits in the integer must be larger than 21, 28, 43 and 87, respectively.
## Tuesday, July 9, 2013
### Robinson Arithmetic is Σ1-complete
Recall that Q (i.e., Robinson Arithmetic) is an axiomatized formal theory (AFT) of arithmetic couched in the interpreted formal language LA. Let L be a subset of LA and let T be some AFT of arithmetic.
We say that T is L-sound iff, for any sentence φ in L, if T ⊢ φ, then φ is true.
We say that T is L-complete iff, for any sentence φ in L, if φ is true, then T ⊢ φ.
In chapter 9, Peter Smith defines a subset of LA, called Σ1. Then, using the fact that Q is order-adequate, he proves that Q is Σ1-complete. This is important because the well-formed formulas (wff's) of Σcan express the decidable numerical properties and relations, and therefore Q will be sufficiently strong. Now to the details...
First, let's define a few interesting subsets of LA:
## Sunday, July 7, 2013
In chapter 9, Peter Smith defines the following concept: A theory T that captures the relation ≤ is order-adequate if it satisfies the following nine properties:
• O1: T ⊢ ∀x (0 ≤ x)
• O2: For any n, T ⊢ ∀x ((x = 0 ∨ x = 1 ∨ x = 2 ∨ ... ∨ x = n) → x ≤ n)
• O3: For any n, T ⊢ ∀x (x ≤ n → (x = 0 ∨ x = 1 ∨ x = 2 ∨ ... ∨ x = n))
• O4: For any n, if T ⊢ φ(0) and T ⊢ φ(1) and ... and T ⊢ φ(n) then T ⊢ (∀x ≤ n)φ(x)
• O5: For any n, if T ⊢ φ(0) or T ⊢ φ(1) or ... or T ⊢ φ(n) then T ⊢ (∃x ≤ n)φ(x)
• O6: For any n, T ⊢ ∀x (x ≤ n → x ≤ Sn)
• O7: For any n, T ⊢ ∀x (n ≤ x → (n = x ∨ Sn ≤ x))
• O8: For any n, T ⊢ ∀x (x ≤ n ∨ n ≤ x)
• O9: For any n>0, T ⊢ (∀x ≤ n-1)φ(x) → (∀x ≤ n)(x ≠ n → φ(x))
Then, we have the following theorem:
## Wednesday, July 3, 2013
### Robinson Arithmetic captures the "less-than-or-equal-to" relation
In this post, we'll start discussing the material in Chapter 9 of Peter Smith's book, namely up to section 9.3.
Before proceeding, review the definition of Robinson Arithmetic (denoted Q) as well as what it means for a theory to capture a numerical relation.
Now, we'll show that Robinson Arithmetic captures the ≤ numerical relation with the open well-formed formula: ∃x (x + m = n).
In other words, we are going to prove the following two-part theorem:
Theorem: If m and n are natural numbers, then:
1. If m ≤ n, then Q ⊢ ∃x (x + m = n)
2. If m > n, then Q ⊢ ¬ ∃x (x + m = n)
Recall that if m and n are natural numbers, then m and n are the numerical terms representing m and n, respectively, in the formal theory.
Proof sketch of part 1:
## Sunday, June 30, 2013
### Q - Robinson Arithmetic
Now that we are familiar with Baby Arithmetic (BA), we can make its language more expressive by allowing variables and quantifiers back into its logical vocabulary. When we do this, we simply obtain the interpreted language LA, that was described earlier.
Since we now have variables and quantifiers, we can replace the schemata of BA with regular axioms (see below). The resulting formal system of arithmetic is called Robinson Arithmetic and is often denoted by the letter Q, as described in Chapter 8 of Peter Smith's book. Here is his definition of Q:
## Friday, June 28, 2013
### Baby arithmetic
Since chapter 7 in Peter Smith's book is so short, I'll summarize it quickly and then start discussing chapter 8.
Chapter 7 starts by comparing the two incompleteness theorems discussed in previous chapters (let's call these "Smith's theorems") to Gödel's first incompleteness theorem as follows:
## Thursday, June 27, 2013
### Consistent, sufficiently strong formal systems of arithmetic are incomplete
In Chapter 6, Peter Smith proves another incompleteness theorem. To place this theorem in context, let's review some definitions about axiomatized formal theories (or AFTs) from earlier posts.
Let T be some AFT.
• T is consistent iff (if and only if) there is no sentence φ such that T proves both φ and ¬ φ.
• T is sound iff every theorem that T proves is true according to the interpretation that is built into T's language.
• T is (negation-)complete iff for every sentence φ in T's language, T proves either φ or ¬ φ.
• T is decidable iff there is an algorithm that, given any sentence φ in T's language, determines whether or not T proves φ.
• If needed, go here to review what it means for T's language to express a numerical property P or a numerical relation R.
• If needed, go here to review what it means for T to capture a numerical property P or a numerical relation R.
• If needed, go here to review what it means for T's language to be sufficiently expressive.
We'll use the following abbreviations:
## Monday, June 24, 2013
### Sound formal systems with sufficiently expressive languages are incomplete
In chapter 5, Peter Smith uses a counting argument to prove that sound axiomatized formal theories (AFTs) that can express a good amount of arithmetic are incomplete. In short: since their set of theorems is effectively enumerable but their set of true sentences is not, the two sets must be different. In a sound theory, the theorems are all true. Therefore, some truths must remained unproved in such theories.
Let's start with a definition. The language of an AFT of arithmetic is sufficiently expressive if and only if:
1. It can express quantification over numbers, and
2. It can express every effectively decidable two-place numerical relation.
Now, to the pivotal theorem of this chapter:
## Sunday, June 23, 2013
### An enumerable but not effectively enumerable set
Let's focus on the first half of chapter 5 in Peter Smith's book. First, Smith describes a new characterization of effectively enumerable sets. Second, Smith gives an example of a set of integers that is not effectively enumerable.
Let's define the numerical domain of an algorithm as the set of natural numbers with the following property: when the algorithm takes one of these numbers as input, it terminates and returns a result. Every algorithm has a numerical domain. If the input to some algorithm is not a single natural number, then the numerical domain of this algorithm is the empty set. Otherwise, the numerical domain is the set of those natural numbers on which the algorithm does not crash and does not go into an infinite loop.
It turns out that each numerical domain is effectively enumerable. In fact, the converse is also true, according to the following theorem:
## Saturday, June 22, 2013
### Capturing numerical properties in a formal language of arithmetic
Peter Smith starts Chapter 4 by describing LA, a formal language that is at the core of several AFTs (axiomatized formal theories) of arithmetic. Then Smith explains what it means for a formal language of arithmetic to "express" a numerical property and the stronger notion of "capturing" a numerical property.
Here is the definition of LA:
## Friday, June 21, 2013
### Formal systems or axiomatized formal theories
In chapter 3, Peter Smith defines formal systems or, as he calls them, axiomatized formal theories (I will use AFT as an abbreviation for this phrase).
A theory T is an AFT if...
## Wednesday, June 19, 2013
### Effective computability, decidability and enumerability
In Chapter 2, Smith covers familiar ground. So this post will be short. Here is a quick summary, section by section.
Section 1 reviews terminology pertaining to functions, namely the notions of domain and range, as well as special cases of functions, namely injective (or one-to-one) functions, surjective (or onto) functions and bijective functions (or one-to-one correspondences).
## Tuesday, June 18, 2013
### Peter Smith's overview of Gödel's incompleteness theorems
Finally! As was my initial goal, I now turn my attention to Peter Smith's book on the proof of Gödel's incompleteness theorems (by the way, I own the first edition of his book; so that is the one I will be referring to). I did not expect to spend that much time on Nagel and Newman's book. But it was worthwhile in getting the big picture.
Chapter 1 is an introductory and user-friendly discussion of topics that were mentioned in earlier posts, such as basic arithmetic, formal systems, (in)completeness, consistency, the statement of Gödel's incompleteness theorems and some of their implications. Therefore, in this post, I will just highlight the points where Smith provides new information or has a different perspective.
## Monday, June 17, 2013
### Proof sketch of Gödel's incompleteness theorems
In this post, I will follow the outline of the proof given in Section VII of Nagel and Newman's book. Recall that:
• a formal system F is consistent if there is no well-formed formula (wff) w such that F proves both w and ¬ w, and
• a formal system is (semantically) complete if it can prove all of the true wff's that it can express.
Now, Gödel's first incompleteness theorem can be paraphrased as:
Any consistent formal system that is expressive enough to model arithmetic is both incomplete and "incompletable."
and Gödel's second incompleteness theorem can be paraphrased as:
Any consistent formal system that is expressive enough to model arithmetic cannot prove its own consistency.
According to Nagel and Newman, a proof of the first theorem can be sketched in 4 steps:
## Friday, June 7, 2013
### Mappings and the arithmetization of meta-mathematics
In an earlier post, we discussed mappings in general and then described Gödel numbering, which is a mapping between the set of well-formed formulas (wff's) in a formal system and the set of positive integers.
In part B of section VII, Nagel and Newman describe a more interesting mapping used in Gödel's proof. In this mapping, the domain is the set of meta-mathematical statements about structural properties of wff's, while the co-domain is the set of wff's. This post describes how this mapping enabled Gödel to arithmetize meta-mathematics.
## Thursday, June 6, 2013
### Mappings and Gödel numbering
Section VI of Nagel and Newman's book describes mappings. A mapping is an operation that applies to two sets called the domain and the co-domain. More specifically, a mapping associates to each element of the domain one or more elements of the co-domain.
## Tuesday, May 28, 2013
### Absolute proof of consistency of FSN
In section IV of their great little book, Nagel and Newman discuss efforts by Gottlob Frege and then Bertrand Russell to reduce arithmetic to logic. This is clearly another attempt at a relative proof of consistency: if this reduction were successful, then arithmetic would be consistent provided logic is consistent.
Whether this latter statement is true or not, Whitehead and Russell's Principia Mathematica was a landmark achievement: it almost completed the first step in an absolute proof of consistency of arithmetic, since it led to the formalization of an axiomatic system for arithmetic.
In section V, Nagel and Newman describe a formalization of propositional (or sentential) logic, that is, a subset of the logic system in Principia Mathematica (but not a large enough subset to represent arithmetic). The bulk of this section first describes the formalization process, which yields the standard syntax and inference rules of propositional logic (including modus ponens) and then outlines an absolute proof of consistency of this formalized axiomatic system.
This absolute proof of consistency is a proof by contrapositive, which relies on the following true conditional statement:
## Sunday, May 26, 2013
### Absolute proofs of consistency and meta-mathematics
In earlier posts, we explained why consistency is an important property of axiomatic systems and discussed relative proofs of consistency, in which the proof of consistency of a system is based on the assumption that another axiomatic system is consistent. In this post, we introduce absolute proofs of consistency that do not make any assumptions about any other axiomatic system. Apparently, David Hilbert was the first to study and propose such proofs, according to Nagel and Newman's book (Section III, page 26).
Recall that an axiomatic system is consistent if it cannot derive both a theorem and its negation. What do we mean by the negation of a theorem? Let's take, as a simple example, the theorem: "6 is divisible by 3." Its negation is simply the following statement: "6 is not divisible by 3" or equivalently "It is not the case that 6 is divisible by 3." This second formulation of the negation, although less elegant in English, is preferable because the negation is added to the front of the original theorem. In a formal system, negation is handled by simply adding a symbol for the phrase "it is not the case that." Several symbols have been used for negation, such as ~ and ¬ . We'll use the latter here. So, if T is any theorem in some formal system, then the formula ¬T is the negation of T.
Remember that our formal system FS did not have a symbol for negation. So we will extend FS into a new formal system called FSN (for FS with Negation), whose alphabet is { E, 0, 1, ¬ }. FSN has exactly one axiom, namely the same as A1 in FSFSN also has the same inference rules as FS, namely IR1 and IR2. But it has one additional inference rule that uses negation. Here is the full description of FSN:
## Wednesday, May 22, 2013
### Relative proofs of consistency
In the last post, we defined and explained the importance of the property of consistency for an axiomatic system. The second half of Section II in Nagel and Newman's book describes ways of proving that an axiomatic system is consistent.
But first, note that the question of consistency of Euclidean geometry did not arise. Its axioms were supposed to describe the real world; and something that actually exists cannot be self-contradictory. In other words, existence (or truth) implies internal consistency.
The need for consistency proofs arose much later, with non-Euclidean geometries, which do not obviously model space as we experience it. Non-existence does not imply inconsistency. But any interesting abstract construct had better be internally consistent.
Second, checking the internal consistency of all of the theorems produced so far is typically not a valid proof of consistency, because (interesting) axiomatic systems generate an infinite number of theorems. The proof of consistency must guarantee that not a single theorem, including some that we have not yet produced and that we might never produce, contradicts any other theorem in the system.
One possible way to prove the consistency of an axiomatic system is model-based, where a model is a kind of interpretation.
## Monday, May 20, 2013
### Consistency of axiomatic systems
The "problem of consistency" is the topic of Section II of Nagel and Newman's book. This section defines "consistency" and explains when and why it became an important property of axiomatic systems.
The oldest and most famous axiomatic system is that of Euclid, in which he systematized all of the knowledge of geometry (and more) available to him over two thousand years ago. Based on five axioms, Euclid was able to rigorously prove a very large number of known and new theorems (called "propositions" in his Elements). His axioms were supposed to be intuitively true. The first four axioms dealt with line segments, lines, circles and angles (see this Wikipedia entry) and have been viewed as self-evident. In contrast, the fifth axiom, which was equivalent to the following statement: "Through a point outside a given line, only one parallel to the line can be drawn" (page 9), was not intuitively true (apparently because the two lines involved extend to infinity in two directions, similarly to asymptotes).
Since this proposed axiom was not obviously true, many mathematicians tried to prove that it logically follows from the first four axioms. Only in the nineteenth century was it demonstrated that it is NOT possible to prove the parallel axiom from the first four axioms.
This proof that it is impossible to prove a given statement is a great precursor of Gödel's incompleteness theorems.
## Friday, May 17, 2013
### Formal systems, axioms, inference rules, formal proofs
I'll start this post by describing a simple formal system that I made up.
formal system is comprised of axioms and inference rules. Each axiom and inference rule is defined syntactically, that is, by ordered sequences of symbols that follow strict syntactic rules but do not (necessarily) have any meaning. The set of all symbols allowed in a formal system are explicitly listed in its alphabet, simply, a finite, non-empty set of symbols.
My formal system, let's call it FS, uses the alphabet {E,0,1} and has only one axiom and two inference rules. Here is the axiom (in the box below):
## Wednesday, May 15, 2013
### Let's get started with Nagel and Newman (Section I)
Let's ease our way into this. I haven't read the Nagel and Newman book in a few years. But I remember it as a short and highly readable overview of the proof. I have a 1986 softcover edition by NYP Press. Today, I want to write about Section I - Introduction. It's really short: under 5 pages.
First, a short quote that makes me feel better about my past failures at REALLY understanding the proof of GIT:
## Tuesday, May 14, 2013
### I just want to understand the proof of Gödel's incompleteness theorems
So this is it! Summer 2013... This is when I get to master the proof of Gödel's incompleteness theorems (thereafter: GIT).
I discovered GIT as an undergraduate student in one of my Artificial Intelligence courses. My instructor was reading Gödel, Escher, Bach and was raving about the book. He even took some of our exam problems from it. But what I remember most is the outline of the proof of GIT, more precisely the first theorem. To be honest, the only part of it that stuck with me was the idea of numbering the formulas. Of course, at the time, I bought a copy of the book. I liked many parts of it but I never finished it.
Years later, while I was teaching computability theory and other computer science topics, I encountered GIT several times. At some point, I decided I wanted to understand the proof of GIT.
First, I read the classic "Gödel's Proof" by Nagel and Newman. That is a concise and user-friendly overview of the proof. But it was only a proof sketch. So I looked at some textbooks on formal logic, which were neither concise nor user-friendly. I still did not understand the proof.
Finally, a couple of years ago, I came across a promising volume: "Introduction to Gödel’s Theorems" by Peter Smith. I read the first few chapters but then ran out of time (and motivation) to finish it. The following year, I made another attempt: I had to start from scratch and did not get any farther into it the second time around... | 2018-03-22T10:03:31 | {
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https://joshhug.gitbooks.io/hug61b/content/chap9/chap94.html | ## Weighted Quick Union (WQU)
Improving on Quick Union relies on a key insight: whenever we call find, we have to climb to the root of a tree. Thus, the shorter the tree the faster it takes!
New rule: whenever we call connect, we always link the root of the smaller tree to the larger tree.
Following this rule will give your trees a maximum height of $\log N$, where N is the number of elements in our Disjoint Sets. How does this affect the runtime of connect and isConnected?
Let's illustrate the benefit of this with an example. Consider connecting the two sets T1 and T2 below:
We have two options for connecting them:
The first option we link T1 to T2. In the second, we link T2 to T1.
The second option is preferable as it only has a height of 2, rather than 3. By our new rule, we would choose the second option as well because T2 is smaller than T1 (size of 3 compared to 6).
We determine smaller / larger by the number of items in a tree. Thus, when connecting two trees we need to know their size (or weight). We can store this information in the root of the tree by replacing the -1's with -(size of tree). You will implement this in Lab 6.
#### Maximum height: Log N
Following the above rule ensures that the maximum height of any tree is Θ(log N). N is the number of elements in our Disjoint Sets. By extension, the runtimes of connect and isConnected are bounded by O(log N).
Why log N? The video above presents a more visual explanation. Here's an optional mathematical explanation why the maximum height is $\log_2 N$. Imagine any element $x$ in tree $T1$. The depth of $x$ increases by $1$ only when $T1$ is placed below another tree $T2$. When that happens, the size of the resulting tree will be at least double the size of $T1$ because $size(T2) \ge size(T1)$. The tree with $x$ can double at most $\log_2 N$ times until we've reached a total of N items ($2^{\log_2 N} = N$). So we can double up to $\log_2 N$ times and each time, our tree adds a level $\rightarrow$ maximum $\log_2 N$ levels.
You may be wondering why we don't link trees based off of height instead of weight. It turns out this is more complicated to implement and gives us the same Θ(log N) height limit.
### Summary and Code
Implementation Constructor connect isConnected
QuickUnion Θ(N) O(N) O(N)
QuickFind Θ(N) Θ(N) Θ(1)
QuickUnion Θ(N) O(N) O(N)
Weighted Quick Union Θ(N) O(log N) O(log N)
N = number of elements in our DisjointSets data structure | 2021-05-15T19:49:35 | {
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https://math.stackexchange.com/questions/2282473/euler-equations-for-the-incompressible-fluids | # Euler equations for the incompressible fluids
Incompressible fluid with constant density ρ fills the three-dimensional domain below the free surface $z = η(r)$ in cylindrical polar coordinates. The flow is axisymmetric and steady, and the only non-zero velocity component is $u_θ$. Gravity acts upon the fluid. The fluid in $r\lt a$ rotates rigidly about the z-axis with angular velocity $\Omega$ and the fluid $r \ge a$ is irrotational.
Use the radial and vertical components of the Euler equations to show that the pressure $p$ in the region $z < η$, $r < a$ satisfies
$$\frac{p}{\rho} = \frac{1}{2}{Ω^2}{r^2} −gz + \text{constant}$$ and find the constant.
$$\frac{∂u_r}{∂t}+\left(u_r\frac{∂}{∂r}+u_z\frac{∂}{∂z}\right)u_r-\frac{(u_θ)^2}{r}+\frac{1}{\rho}\frac{∂p}{∂r}=0$$
$$\frac{∂u_θ}{∂t}+\left(u_r\frac{∂}{∂r}+u_z\frac{∂}{∂z}\right)u_θ+\frac{u_θu_r} {r}=0$$
$$\frac{∂u_z}{∂t}+\left(u_r\frac{∂}{∂r}+u_z\frac{∂}{∂z}\right)u_z+\frac{1}{\rho}\frac{∂p}{∂z}=-g$$
I have the Euler equation in cylindrical coordinate. Then how should I figure out the $\Omega$. The next question is to show that the free surface position in $r<a$ is $$η=\frac{\Omega^2a^2}{g}(\frac{r^2}{2a^2}-1)$$ if it is helpful. Thank you so much!
• Are you sure the radial equations aren't available to you? That's a fair amount of effort to derive those. May 15 '17 at 19:14
Sincerely, if you derived the Euler equations for cylindrical coordinates, this is a very simple exercise for you.
With the hypothesis, many terms disappear, those involving $u_r=0,\;u_z=0$ and the partials wrt time (steady flow).
Further, the motion is rigid, meaning that all fluid parts have zero relative motion. If they are moving in circles, the angular velocity for all point in the fuid $\Omega$ is the same! You don't need to figure anything, it must be known. With all this, $u_\theta=\Omega r$
$\begin{cases} -\dfrac{(u_θ)^2}{r}+\dfrac{1}{\rho}\dfrac{\partial p}{\partial r}=0\\ \dfrac{1}{\rho}\dfrac{\partial p}{\partial z}=-g \end{cases}$
Integrating the second,
$\dfrac{p}{\rho}=-gz+f(r)\tag 1$
So is $\dfrac{1}{\rho}\dfrac{\partial p}{\partial r}=f'(r)$ or $\dfrac{p}{\rho}+C=f(r)$. With the first equation,
$-\dfrac{(u_θ)^2}{r}+f'(r)=0$ or $-\dfrac{(\Omega r)^2}{r}+f'(r)=0$ or
$-\Omega^2r+f'(r)=0$. Integrating, $f(r) = \dfrac{1}{2}{Ω^2}{r^2}+h(z)$
$\dfrac{p}{\rho}= \dfrac{1}{2}{Ω^2}{r^2}+h(z)-C\tag 2$
Comparing $(1)$ and $(2)$ we have that $h(z)=-gz+C$
$\dfrac{p}{\rho} = \dfrac{1}{2}{Ω^2}{r^2} −gz +C$
I could not recover the given solution for the free surface, although I've found some relation that produces the well known paraboloid for the free surfce of the Newton's bucket. To find $C$, we can consider the heigh $z$ for the free surface ($p=0$) at some $r$. We can set $z=0$ for $r =a$.
$0=\dfrac{1}{2}{Ω^2}{a^2}+C$ or $C=-\dfrac{1}{2}{Ω^2}{a^2}$
$\dfrac{p}{\rho} = \dfrac{1}{2}{Ω^2}{r^2} −gz-\dfrac{1}{2}{Ω^2}{a^2}=\dfrac{1}{2}Ω^2(r^2-a^2)-gz$
For the free surface $z=\eta(r)$, $p=0$
$\eta=\dfrac{1}{2g}Ω^2(r^2-a^2),\;r\lt a$
From other considerations, $\eta$ for $r\ge a$ is $\eta=-\dfrac{\Omega^2a^4}{2gr^2}$. As $\eta$ has to be continuous $\eta(a)=-\dfrac{\Omega^2a^2}{2g}$, for the formula we've found for $r\lt a$
$0=\dfrac{Ω^2a^2}{2}+\dfrac{\Omega^2a^2}{2}+C$ leading to $C=\Omega^2a^2$ and to $\eta=\dfrac{\Omega^2a^2}{g}(\dfrac{r^2}{2a^2}-1)$ for $r\lt a$
$$\eta= \begin{cases} \dfrac{\Omega^2a^2}{g}(\dfrac{r^2}{2a^2}-1)& x\lt a\\ -\dfrac{\Omega^2a^4}{2gr^2}&a\le x \end{cases}$$
I've sketched the function for the free surface (setting $\dfrac{\Omega^2a^2}{2g}=1$ and $a=1$)
• It is actually the next two question of this one. math.stackexchange.com/questions/2280167/… I try to get the constant and the free surface and the number matches when I set r=a, p=0 and use the free surface when r>=a. I get C=-1/2(\Omega) ^2a^2-(\Omega)^2a^2/2 and free surface as above. Can I just plug in the limit May 16 '17 at 7:40
• And if I want to draw a sketch for the free surface displacement then it is gonna be a paraboloid but when happen at r=a then? May 16 '17 at 7:49
• @stedmoaoa, the resultig function is smooth. May 16 '17 at 13:47
• Thank you so much it really makes sense. So the graph is like x^2-1 and -1/x^2 Then, how should I describe at r=a? May 16 '17 at 14:08
• Continuous and differentiable for all quantites. May 16 '17 at 17:37 | 2021-09-21T03:01:58 | {
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https://math.stackexchange.com/questions/1866962/a-statement-about-divisibility-of-relatively-prime-integers | # A statement about divisibility of relatively prime integers
I'm solving a problem, and the solution makes the following statement: "The common difference of the arithmetic sequence 106, 116, 126, ..., 996 is relatively prime to 3. Therefore, given any three consecutive terms, exactly one of them is divisible by 3." Why is this statement true? Where does it come from? Is it generalizable to other numbers?
• And here I thought it was something about "relatively prime statements" – Yuriy S Jul 21 '16 at 21:56
There is a simple, well-known theorem about that:
If $a$, $b$ and $m$ are integers, $m\ge 2$ and $\gcd(b,m)=1$ then the set $\{a,a+b,a+2b,\ldots,a+(m-1)b\}$ is a complete residue system.
This implies, for example, that the set contains exactly one multiple of $m$.
• Also, most of the other answers hint at a proof of that theorem. – Jeppe Stig Nielsen Jul 22 '16 at 8:04
Take $a\in\mathbb{N}$ and suppose that $a\equiv 1\pmod 3$. Then \begin{align*} a&\equiv 1\pmod 3\\ a+10\equiv 11&\equiv 2\pmod 3\\ a+20\equiv 21&\equiv 0\pmod 3 \end{align*}
If we instead have $b\in\mathbb{N}$ such that $b\equiv 2\pmod 3$, then adding $10$ to $b$ is sufficient. The case in which we have $\equiv 0\pmod 3$ is obviously trivial.
If $\,b\,$ is coprime to $\,m\,$ then $\,a+ib,\, i = 1,\ldots,m\,$ is a complete residue system mod $\,m\,$ since
$$\ a+ib \equiv a+j b \iff (i\!-\!j)b\equiv 0\overset{(b,m)=1}\iff i\!-\!j\equiv 0\iff i = j\,\ {\rm by}\,\ 1\le i,j\le m$$
Hence, being complete, it contains an element $\equiv 0,\,$ i.e. a multiple of $\,m.$
Remark $\$ When $\,b = 1\,$ we obtain the well-known special case that any sequence of $\,m\,$ consecutive integers contains a multiple of $\,m.$
If $d$ is relatively prime to $b$, then $a + d j$ is divisible by $b$ if and only if $j \equiv -a d^{-1} \mod b$. Thus exactly one of any $b$ consecutive terms of the arithmetic sequence $a, a+d, a+2d, \ldots$ is divisible by $b$.
In general let $d$ be the common difference of an arithmetic sequence, where $3\nmid d$, and let $a$, $a+d$, and $a+2d$ be three consecutive terms in that sequence. Now WLOG if we assume $3\nmid a$ then we can write $a$ and $d$ as: $$a=3k+1\quad\text{ or }\quad a'=3k+2\quad\text{ and }\quad d=3j+1\quad\text{ or }\quad d'=3j+2$$ for some $j$, $k\in\mathbb{Z}$, where primes have been added to distinguish the two possibilities for $a$ and $d$. Now look at the four possibilities we have for $a+d$, and $a+2d$: $$a+d=3(k+j)+2,\quad \quad a+2d=3(k+2j)+3\qquad (A)$$ $$a+d'=3(k+j)+3,\quad \quad a+2d'=3(k+2j)+5\qquad (B)$$
$$a'+d=3(k+j)+3,\quad \quad a'+2d=3(k+2j)+5\qquad (C)$$ $$a'+d'=3(k+j)+4,\quad \quad a'+2d'=3(k+2j)+6\qquad (D)$$
As can be seen, since we assumed $3\nmid a$, only one member of the three consecutive terms is divisible by $3$ on listing all possible combinations (A), (B), (C), and (D). It would make no difference if we assumed $3\nmid a+d$ or $3\nmid a+2d$ as our starting point as the problem is cyclic modulo $3$. By this look at the congruences of $a$, $a+d$, $a+2d\pmod{3}$ for each of (A), (B), (C), and (D) and you see we get the full set of residue classes $[0]$, $[1]$, $[2]$ modulo $3$ in some order, which is the reason we get one and only one number divisible by $3$.
Further for some $m$ for which $\gcd(m,d)=1$ look at $m$ consecutive terms of an arithmetic sequence module $m$: $$a,\ a+d,\ a+2d,\dotsc,\ a+(m-1)d\pmod{m}\qquad (E)$$ then $a+kd\equiv a+jd\pmod{m}$ iff $(k-j)d\equiv 0\pmod{m}$ for $0\le k,j\le m-1$, which since $[0]$, $[1]$, $[2],\dotsc,\ [m-1]$ form a complete residue system modulo $m$ we must have $[k]=[j]$ modulo $m$, and so in fact $k=j$ with respect to our $m$ consecutive terms with any distinct pair of terms incongruent modulo $m$, and only one unique number in the set being divisible by $m$. To see this imagine (E) rewritten with respect to residue classes modulo $m$ reordered so: $$a+[0],\ a+[1],\ a+[2],\dotsc,\ a+[m-1]\pmod{m}\qquad (E')$$ If $m\mid a$ then $m\equiv a+[0]\pmod{m}$, and we also see $m$ cannot divide any other number in the set; if $m\nmid a$ then we can write $a=mj+k$ where $j$, $k\in\mathbb{Z}$, and $\gcd(k,m)=1$, which is in the class $[mj+k]=[k]=[k']$ modulo $m$, and $[k']+[k'']=[0]$ modulo $m$, $0<k',k''\le m-1$, where $[k']$ and $[k'']$ are additive inverses in the complete residue system (note that $[0]$ is self inverse wrt addition). | 2019-04-24T11:48:59 | {
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http://mathhelpforum.com/geometry/167839-counting-sides.html | # Math Help - Counting sides
1. ## Counting sides
A rectangle that measures 4cm by 6cm is divided into 24 squares with sides 1cm in length. What is the total number of 1cm long sides in those 24 squares? (If 2 squares share a side, the side should be counted only once).
Now I know the answer is 58 and I drew a sketch and counted the sides but I was wondering if there was any shortcut to this problem that I was missing..? Setting up any kind of expression or equation or maybe a combinations problem of some sort?
2. Let use a vertical layout like this (for explanation):
****
****
****
****
****
****
First count all the right and bottom sides of each square: 24 * 2 = 48. There is no overlap in this counting.
However, we missed some sides with this count. Which ones? The entire top row and left column.
Well, the number of edges we did not count is:
# on top + # on left = 4 + 6 = 10
Grand total is: 48 + 10 = 58
So from this the general solution for number of sides in a n by m block is:
2nm + n + m = n(1 + m) + m(1 + n)
This suggest thats an alternative way is to count:
#horizontal edges + #vertical edges = n(1 + m) + m(1 + n)
3. Hi snow--your solution definitely seems to work but I think there is a column of 6 missing Thanks!
4. Hello, sarahh!
A rectangle that measures 4cm by 6cm is divided into 24 squares with sides 1cm in length.
What is the total number of 1cm long sides in those 24 squares?
(If 2 squares share a side, the side should be counted only once).
You made a sketch but you didn't see any pattern, did you?
Code:
* - * - * - * - * - * - *
| | | | | | |
* - * - * - * - * - * - *
| | | | | | |
* - * - * - * - * - * - *
| | | | | | |
* - * - * - * - * - * - *
| | | | | | |
* - * - * - * - * - * - *
The unit squares are in 4 rows and 6 columns.
Imagine these 24 squares formed with matches.
How many matches are used?
We see that there are 6 matches in each row.
And there are 4 + 1 rows.
Hence, there are: .6(4+1) horizontal matches.
We see that there are 4 matches in each column.
And there are 6 + 1 columns.
Hence, there are: .4(6 + 1) vertical matches.
We can generalize this problem.
Suppose there are R rows and C columns.
Then there are: .C(R+1) horizontal matches
. . . . . . . .and: .R(C+1) vertical matches.
Total: . $C(R+1) + R(C+1) \;=\; R + C + 2RC$ matches.
5. Originally Posted by sarahh
Hi snow--your solution definitely seems to work but I think there is a column of 6 missing Thanks!
There is a row of 6 and a column of 4 missing. Which is why a 10 (= 6 + 4) is added at the end.
6. Just for fun, here is another approach-- first do it the wrong way, then patch it up.
There are 24 squares in all, each with 4 sides, and each side is shared between two squares, so there are
$\frac{4 \times 24}{2} = 48$ sides.
But oops, the sides along the perimeter of the rectangle (2 * (4 + 6) = 20) are not shared! So we need to add those back in.
$48 + \frac{20}{2} = 58$ | 2014-08-28T04:24:57 | {
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https://math.stackexchange.com/questions/113536/fibonaccis-final-digits-cycle-every-60-numbers | Fibonacci's final digits cycle every 60 numbers
How would you go about to prove that the final digits of the Fibonacci numbers recur after a cycle of 60?
References:
The sequence of final digits in Fibonacci numbers repeats in cycles of 60. The last two digits repeat in 300, the last three in 1500, the last four in , etc. The number of Fibonacci numbers between and is either 1 or 2 (Wells 1986, p. 65).
http://mathworld.wolfram.com/FibonacciNumber.html
• Why do you conjecture that? Do you have a reference that asserts the statement? – Giovanni De Gaetano Feb 26 '12 at 9:52
• I added some references. – crazyGuy Feb 26 '12 at 9:57
• Neat! Have a look at this link. – Juan S Feb 26 '12 at 10:00
• Here is a link: oeis.org/A096363 – Stefan Gruenwald Dec 14 '18 at 20:59
Notice that:
$$F_{n+15} \equiv 7F_n \pmod{10}$$ for $$n\geq 1$$.
Also the order of $$7$$ mod $$10$$ is $$4$$ so the repetition in the digits of the Fibonacci numbers begins after place $$15\times 4 = 60$$.
• I got that recursion by just writing out the first $16$ Fibonacci numbers mod $10$ and noticing that you get $1,1,...,7,7,...$, this dictates that that next $13$ numbers must be $7$ times their corresponding numbers (mod $10$) in the first lot of $15$ numbers. – fretty Feb 26 '12 at 10:03
• How do we know that F_n+15 = 7F_n mod 10 ? and How do you get that 7 mod 10 is 4 ? thanks. – crazyGuy Feb 26 '12 at 10:07
• Write out the first 16 Fibonacci numbers mod 10: 1,1,...,7,7,... Now by the recursion of the fibonacci numbers the fact that this "multiple-repeat" has happened now tells us that the next 13 numbers must be the same as the 3rd-15th Fibonacci numbers but multiplied by $7$ mod 10. So an actual repeat will happen as soon as we have multipled by enough $7$'s to get $1$ mod 10. This is the "order" of $7$ mod $10$. It is easy to work out here, $7^2 \equiv 4$ mod $10$ and $7^4 \equiv 1$ mod $10$ so the order is $4$. – fretty Feb 26 '12 at 10:13
• @sic2 He's not saying that $7\mod10$ is $4$, but rather that $7^4\equiv 1\mod 10$ and that this is not true for any smaller power of $7$, which is a consequence of Euler's theorem. – Alex Becker Feb 26 '12 at 10:15
• Also this method generalises to the Fibonacci numbers mod $n$. Find the first $1,1,...,a,a,...$ then work out the order of $a$ mod $n$ and you will have your first place where it recurs. – fretty Feb 26 '12 at 10:29
Since each term in the Fibonacci sequence is dependent on the previous two, each time a $0\pmod{m}$ appears in the sequence, what follows must be a multiple of the sequence starting at $F_0,F_1,\dots=0,1,\dots$ That is, a subsequence starting with $0,a,\dots$ is $a$ times the sequence starting with $0,1,\dots$
Consider the Fibonacci sequence $\text{mod }2$: $$\color{red}{0,1,1,}\color{green}{0,1,1,\dots}$$ Thus, the Fibonacci sequence repeats $\text{mod }2$ with a period of $3$.
Consider the Fibonacci sequence $\text{mod }5$: $$\color{red}{0,1,1,2,3,}\color{green}{0,3,3,\dots}$$ Thus, the Fibonacci sequence is multiplied by $3\pmod{5}$ each "period" of $5$. Since $3^4=1\pmod{5}$, the Fibonacci sequence repeats $\text{mod }5$ with a period of $20=4\cdot5$.
Thus, the Fibonacci sequence repeats $\text{mod }10$ with a period of $60=\operatorname{LCM}(3,20)$.
Note that
$$\begin{pmatrix} F_{n+1}\\ F_{n+2} \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ 1 & 1 \end{pmatrix} \begin{pmatrix} F_n \\ F_{n+1} \end{pmatrix}$$
and
$$\begin{pmatrix} 0 & 1 \\ 1 & 1 \end{pmatrix}^{60} \equiv \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \mod 10.$$
One can verify that $60$ is the smallest power for which this holds, so it is the order of the matrix mod 10. In particular
$$\begin{pmatrix} F_{n+60}\\ F_{n+61} \end{pmatrix} \equiv \begin{pmatrix} F_n \\ F_{n+1} \end{pmatrix} \mod 10$$
so the final digits repeat from that point onwards.
• The same matrix applies to the Lucas Numbers, however, $\text{ mod }10$, they have a period of $12$. The fact that $\begin{pmatrix} 0&1\\1&1\end{pmatrix}^{60}=\begin{pmatrix}1&0\\0&1\end{pmatrix}\text{ mod }10$, says that the period of a sequence satisfying $a_n=a_{n-1}+a_{n-2}$ divides $60$. – robjohn Feb 26 '12 at 19:21
$F_{0}=F_1=F_{60}=F_{61}= 1 \mod10$
By inspection, these are the first two pairs of consecutive Fibonaccis for which this is true. Since the recurrence relation only takes into account the previous two terms and last digits only depend on previous last digits, this suffices to prove the claim.
I have found that there is an 11 number grouping. You must reduce numbers to a single digit (13=4 or 55=10=1). After 24 numbers into the sequence it repeats. To go a little deeper, a 9 is in the twelfth and twenty- fourth sequence and the real number is also divisible by 12 in these positions (which coincidentally is every twelfth number in the sequence).
Now if you do the multiplication table of numbers 1 through 8 and reduce all numbers to a single digit, you will find that 1 and 8 correspond in reverse with each other. The same for 2 and 7, also 4 and 5. 3 and 6 are different. You will notice that for every number sequence it will repeat after a 9 appears. So, Fibonacci sequence would be this:
1,1,2,3,5,8,4,3,7,1,8,9,8,8,7,6,4,1,5,6,2,8,1,9,1,1,2,3,5,8,4,3,7,1,8,9,8,8,7,6,4,1,5,6,2,8,1,9......
Remember how I said 1 and 8, 2 and 7, 4 and 5 correspond with each other by reducing the multiplication table to single digits and all numbers repeat a sequence after a 9?
If you move the eleven number sequence between the nines and set 1,1,2,3... above the sequence 8,8,7,6... You will find that they all correspond with the reducing method afore mentioned.
It's very interesting.
• What do you mean $11$-number grouping? Why are you reducing mod $9$ instead of mod $10$? How does this prove that it repeats with cycle $60$? – 6005 Oct 24 '16 at 2:58
• I apologize. This was not to prove or disprove a 60 cycle repeat. Maybe I posted in the wrong area? I just found this sequence repeat and I thought I would give it to bigger brains and see what they can make of it. – Michael Bunton Oct 24 '16 at 3:12
• 11 number grouping means the first 11 numbers in the sequence. The twelfth number is 144 and that reduces to 9 i.e.(1+4+4=9). Then a second set of eleven numbers followed by a reduced 9. – Michael Bunton Oct 24 '16 at 3:20 | 2019-10-14T01:11:53 | {
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https://math.stackexchange.com/questions/1396839/find-limit-a-n-frac1n1-frac1n2-frac1nn/1396843 | # find Limit $a_n= \frac{1}{n+1}+\frac{1}{n+2}+…+\frac{1}{n+n}$ [duplicate]
show sequence $a_n= \frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{n+n}$ converges to log2
my attempt:
1. sequence $a_n$ is monotonic increasing
2. 0<$a_n$<1/2
how to find limit?
• Oh come on, this must be the duplicate of three duplicates. – Vincenzo Oliva Aug 14 '15 at 9:39
$$\lim_{n\rightarrow \infty}\sum_{i=1}^{n}\frac 1 {n+i}=\lim_{n\rightarrow \infty}\sum_{i=1}^{n}\frac 1 n\frac 1 {1+\dfrac in}=\int_0^1\frac 1 {1+x}dx=\ln(1+1)-\ln(1+0)=\ln(2)$$ (Riemann sum)
• +1, but there should be a limit as $n \to \infty$ instead of the sum going to infinity. The sum goes fom $i=1$ to $i=n$ – 6005 Aug 14 '15 at 9:07
• @6005 Fixed, thank you! – GeorgSaliba Aug 14 '15 at 9:09
• ya i was hoping there would be some other way. riemann sum cannot be used to compute limit of sequence$a_n=1+1/2+1/3...+1/n−log(n+1)$ – ketan Aug 14 '15 at 9:19
• @ketan Then why not use the harmonic number $H_n$ that was already mentioned by @MichaelGaluza? But I suppose this becomes a slightly different question than the one you asked. – GeorgSaliba Aug 14 '15 at 9:21
• @ketan $a_n=1+1/2+1/3+...+1/n - \log(n+1)=H_n -\log(n+1)=\log(n)+\gamma + \epsilon_n-\log(n+1)=\log(n/(n+1))+\gamma +\epsilon _n$ – GeorgSaliba Aug 14 '15 at 9:23
\begin{align} \log(2) &=\sum_{k=1}^\infty\frac{(-1)^{k-1}}k\\ &=\lim_{n\to\infty}\sum_{k=1}^{2n}\frac{(-1)^{n-1}}k\\ &=\lim_{n\to\infty}\left(\vphantom{\sum_{k=1}^n}\right.\overbrace{\sum_{k=1}^{2n}\frac1k}^{\substack{\text{sum of all}\\\text{terms}}}-2\overbrace{\sum_{k=1}^n\frac1{2k}}^{\substack{\text{sum of even}\\\text{terms}}}\left.\vphantom{\sum_{k=1}^n}\right)\\[6pt] &=\lim_{n\to\infty}\left(\sum_{k=1}^{2n}\frac1k-\sum_{k=1}^n\frac1k\right)\\[6pt] &=\lim_{n\to\infty}\sum_{k=n+1}^{2n}\frac1k\\ \end{align}
• Nice overbrace. Is there a corresponding underbrace command or somethings similar? – mathreadler Aug 14 '15 at 9:23
• Yep. It is \underbrace :-) Then you use subscripts instead of superscripts. – robjohn Aug 14 '15 at 9:23
Here is classical proof of it. Let $H_n=\sum_{i=1}^n 1/i$. It's well-known that $$H_n = \ln n + \gamma + \epsilon_n,$$ where $\gamma$ is Euler–Mascheroni constant, $\epsilon_n\to 0$. So, $$a_n = H_{2n}-H_{n} = \ln 2 + \epsilon_{2n} - \epsilon_n \to \ln2.$$
• what is $\epsilon_n$? is it always zero? – ketan Aug 14 '15 at 9:28
• @ketan, no. It's sequence which tends to $0$. – Michael Galuza Aug 14 '15 at 11:39
Notice, $$a_n=\frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{n+3}+\ldots +\frac{1}{n+n}$$ $$=\sum_{r=1}^{\infty}\frac{1}{n+r}=\sum_{r=1}^{\infty}\frac{1}{n\left(1+\frac{r}{n}\right)}$$$$=\sum_{i=1}^{\infty}\frac{\frac{1}{n}}{1+\frac{r}{n}}$$ Now, let $\frac{r}{n}=x\implies \frac{1}{n}=dx\to 0$ $$\text{upper limit of x}=\lim_{n\to \infty}\frac{n}{n}=1$$ $$\text{lower limit of x}=\lim_{n\to \infty}\frac{1}{n}=0$$ Hence, using integration we get $$a_n=\int_{0}^{1}\frac{dx}{1+x}$$ $$=\left[\ln (1+x)\right]_{0}^{1}$$ $$=\left[\ln 2-\ln 1\right]=\ln 2$$ | 2020-10-30T07:23:17 | {
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http://mathhelpforum.com/number-theory/151472-proving-binomial-coeficient-always-natural-number.html | # Math Help - Proving the binomial coeficient is always a natural number?
1. ## Proving the binomial coeficient is always a natural number?
How would one go about proving that:
${ n \choose k }$
is always a natural number? That is, assuming $0 \leq k \leq n$ and that $n = 1, 2, 3, 4...$ and that $k = 1, 2, 3, 4...$; what I mean by that is that $n$ and $k$ are both positive integers. Its a question in my textbook that I bought early to get ahead on fall classes, so I don't have any teachers to ask about this question; not till fall atleast. I'm not asking for a straight out proof, but preferably some direction as to where I would go, which direction I should go to prove this? A litttle guidence is all I need, please don't simply give me the full answer, if you don't mind. Thanks in advance
2. Originally Posted by mfetch22
How would one go about proving that:
${ n \choose k }$
is always a natural number? That is, assuming $0 \leq k \leq n$ and that $n = 1, 2, 3, 4...$ and that $k = 1, 2, 3, 4...$; what I mean by that is that $n$ and $k$ are both positive integers. Its a question in my textbook that I bought early to get ahead on fall classes, so I don't have any teachers to ask about this question; not till fall atleast. I'm not asking for a straight out proof, but preferably some direction as to where I would go, which direction I should go to prove this? A litttle guidence is all I need, please don't simply give me the full answer, if you don't mind. Thanks in advance
Use the fact that $\displaystyle{\binom{n}{k}=\binom{n-1}{k}+\binom{n-1}{k-1}$ for $\displaystyle{k=1\ldots n-1}$ and $\displaystyle{\binom{n}{0}=\binom{n}{n}=1}$.
(This will be a proof by induction)
3. Originally Posted by mfetch22
How would one go about proving that:
${ n \choose k }$
is always a natural number? That is, assuming $0 \leq k \leq n$ and that $n = 1, 2, 3, 4...$ and that $k = 1, 2, 3, 4...$; what I mean by that is that $n$ and $k$ are both positive integers. Its a question in my textbook that I bought early to get ahead on fall classes, so I don't have any teachers to ask about this question; not till fall atleast. I'm not asking for a straight out proof, but preferably some direction as to where I would go, which direction I should go to prove this? A litttle guidence is all I need, please don't simply give me the full answer, if you don't mind. Thanks in advance
There's also the combinatorial proof. (The binomial coefficient is given as an "archetypal example" not far from the top of the page.) The formula(s) given for $\displaystyle { n \choose k }$ can be shown to equal either of the two equivalent quantities
1) the coefficient of the $\displaystyle x^k$ term in the polynomial expansion of the binomial power $\displaystyle (1 + x)^n$
2) the number of ways to choose k elements from a set of n elements. (The number of distinct k-subsets of {1, 2, ... , n}.)
Obviously both of these quantities are integers.
4. Another way is to use Legendre's formula, which states that ( $\mathbb{P}$ is the set of prime numbers)
$
n! = \prod_{p\in \mathbb{P}} p^{\sum_{i=1}^\infty \Big\lfloor\frac{n}{p^i}\Big\rfloor}
$
Thus for any prime $p$ we have:
$
\binom{n}{k} = \frac{n!}{k!(n-k)!} = \prod_{p\in \mathbb{P}} p^{\sum_{i=1}^\infty \Big\lfloor\frac{n}{p^i}\Big\rfloor-\Big\lfloor\frac{k}{p^i}\Big\rfloor-\Big\lfloor\frac{n-k}{p^i}\Big\rfloor}}
$
So all what's left is to show that $\lfloor{x+y}\rfloor \ge \lfloor{x}\rfloor + \lfloor{y}\rfloor$
so it follows that for all $i\in\mathbb{N}$:
$
\Big\lfloor\frac{n}{p^i}\Big\rfloor-\Big\lfloor\frac{k}{p^i}\Big\rfloor-\Big\lfloor\frac{n-k}{p^i}\Big\rfloor} \ge 0
$
and therefore $\binom{n}{k} \in \mathbb{N}$
5. Originally Posted by Unbeatable0
Another way is to use Leibnitz' formula, which states that ( $\mathbb{P}$ is the set of prime numbers)
$
n! = \prod_{p\in \mathbb{P}} p^{\sum_{i=1}^\infty \Big\lfloor\frac{n}{p^i}\Big\rfloor}
$
Thus for any prime $p$ we have:
$
\binom{n}{k} = \frac{n!}{k!(n-k)!} = \prod_{p\in \mathbb{P}} p^{\sum_{i=1}^\infty \Big\lfloor\frac{n}{p^i}\Big\rfloor-\Big\lfloor\frac{k}{p^i}\Big\rfloor-\Big\lfloor\frac{n-k}{p^i}\Big\rfloor}}
$
So all what's left is to show that $\lfloor{x+y}\rfloor \ge \lfloor{x}\rfloor + \lfloor{y}\rfloor$
so it follows that for all $i\in\mathbb{N}$:
$
\Big\lfloor\frac{n}{p^i}\Big\rfloor-\Big\lfloor\frac{k}{p^i}\Big\rfloor-\Big\lfloor\frac{n-k}{p^i}\Big\rfloor} \ge 0
$
and therefore $\binom{n}{k} \in \mathbb{N}$
Perfect , in fact $\lfloor{ x+y \rfloor} = \lfloor{ x \rfloor} + \lfloor{ y \rfloor} + \lfloor{ \{ x \} + \{ y \} \rfloor} \geq \lfloor{ x \rfloor} + \lfloor{ y \rfloor}$ .
6. Originally Posted by mfetch22
How would one go about proving that:
${ n \choose k }$
is always a natural number? That is, assuming $0 \leq k \leq n$ and that $n = 1, 2, 3, 4...$ and that $k = 1, 2, 3, 4...$; what I mean by that is that $n$ and $k$ are both positive integers. Its a question in my textbook that I bought early to get ahead on fall classes, so I don't have any teachers to ask about this question; not till fall atleast. I'm not asking for a straight out proof, but preferably some direction as to where I would go, which direction I should go to prove this? A litttle guidence is all I need, please don't simply give me the full answer, if you don't mind. Thanks in advance
If you are asking from the perspective of the Binomial Expansion,
then you can observe a pattern if you refrain from expressing the terms with indices...
$(a+b)^2=(a+b)(a+b)=aa+ab+ba+bb$
"a" times "b" is expressed in both possible ways.
$(a+b)^3=(a+b)(a+b)^2=(a+b)(aa+ab+ba+bb)$
$=aaa+aab+aba+abb+baa+bab+bba+bbb$
$=aaa+aab+aba+baa+abb+bab+bba+bbb$
If the "a" values were labelled $a_1,\ a_2,\ a_3$ we could arrange them,
hence we have all possible selections of 2 "a"'s with a "b", 2 "b"'s with an "a",
the only possible selection of 3 "a"s, the only selection of 3 "b"'s.
For higher powers, the same pattern exists.
Since this pattern is the same as "choose k from n available",
the binomial coefficient may be calculated using
$\binom{n}{k}$
There are k! arrangements of a selection.
As shown above, the number of selections is a natural number,
which can be calculated by "unarranging" the arrangements.
The number of arrangements of k from n is n(n-1)(n-2)...(n-[k-1])
which is $\frac{n!}{k!}$ as k! has been "cut off" the tail.
The number of arrangements is k! times the number of selections. | 2014-07-22T18:41:05 | {
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https://math.stackexchange.com/questions/2338799/c1-function-on-compact-set-is-lipschitz/2338828 | # $C^1$ function on compact set is Lipschitz
Let $\emptyset\ne A\subset\mathbb{R}^n$ be open and let $f \in C^1 (A, \mathbb{R})$ be a function. Let $\emptyset\ne K\subset A$ be compact and convex. I want to prove that $f$ is Lipschitz on $K$; that is, prove there exists a constant $c > 0$ such that $| f ( x ) - f( y ) | \leq c \, \| x - y \|, \forall x , y \in K$.
My approach:
Let $x,y\in K$ be two arbitrary points. Then, since $K$ is convex, the line segment between $x$ and $y$, i.e. $Co(x,y)$, is in $K$. Thus, by MVT, there exists a vector $b\in Co(x,y)$ such that
$$\text{(*) } |f(x)-f(y)|=|\langle x-y, (\nabla f)(b)\rangle|\le \|x-y\|\|(\nabla f)(b)\|$$
Now, since $K$ is compact, $f$ takes a maximum and a minimum values on $K$, so that $\exists b'\in K$ such that $\|(\nabla f)(b') \|\ge \|(\nabla f)(b) \|$, for all $b\in K$. Let $c\in \mathbb{R}$, $c:=(\nabla f)(b')$, then
$$|f(x)-f(y)|\le\|x-y\|\|(\nabla f)(b)\|\le\|x-y\|\|(\nabla f)(b')\|=c\|x-y\|$$
This implies that $f$ is Lipschitz on $K$ for all $x,y\in K$.
Please let me know if you think my proof is correct or not very much? I'm somewhat concerned about the part with the gradient - how exactly is the maximality of the norm of the gradient related to the EVT, that is to $f$ taking maximum and minimum values? As far as I can tell, the maximum norm of the gradient exists because that is the direction to the maximum (or minimum) point of $f$.
• What is $a$? It seems undefined to me . . . – Robert Lewis Jun 28 '17 at 0:24
• Sorry, it was supposed to be $y$. Fixed. – sequence Jun 28 '17 at 0:26
• I guess is because $||f'||$ is a real function on a compact set? – Li Chun Min Jun 28 '17 at 0:41
• @LiChunMin Can you please clarify your question? – sequence Jun 28 '17 at 0:47
• Looks good to me. You do not need the EVT. Since $\nabla (f)$ is continuous, the real-valued function $\|\nabla (f)\|$ is continuous .So $T=\{\| \nabla f(x)\|: x\in K\}$ is compact (because $K$ is compact) , so $T$ is a bounded real subset....So $\|f(x)-f(y)\|\leq \|x-y\|\cdot \sup T$ for all $x,y\in K$. – DanielWainfleet Jun 28 '17 at 2:48
Here's how I would do it:
for $x, y \in K$, let $\gamma(t):[0, 1] \to K$ be the line segment
$\gamma(t) = x + t(y - x); \tag{1}$
then
$\gamma(0) = x, \tag{2}$
$\gamma(1) = x +(y - x) = y, \tag{3}$
and
$\gamma'(t) = y - x; \tag{4}$
furthermore, since $K$ is convex, $\gamma(t)$ lies entirely within $K$, hence also in $A$.
Now, for $x, y \in K$, we have:
$\Vert f(y) - f(x) \Vert = \Vert \displaystyle \int_0^1 \dfrac{d(f(\gamma(t))}{dt}dt \Vert = \Vert \displaystyle \int_0^1 \nabla f(\gamma(t)) \cdot \gamma'(t) dt \Vert$ $\le \displaystyle \int_0^1 \Vert \nabla f(\gamma(t)) \cdot \gamma'(t) \Vert dt \le \displaystyle \int_0^1 \Vert \nabla f(\gamma(t)) \Vert \Vert \gamma'(t) \Vert dt.\tag{5}$
Since $K$ is compact and $\nabla f \in C^0(A, \Bbb R)$, so hence $\nabla f \in C^0(K, \Bbb R)$, $\Vert \nabla f \Vert$ is bounded by some $B$ on $K$, hence
$\displaystyle \int_0^1 \Vert \nabla f(\gamma(t)) \Vert \Vert \gamma'(t) \Vert dt \le \displaystyle \int_0^1 B \Vert \gamma'(t) \Vert dt; \tag{6}$
using (4),
$\displaystyle \int_0^1 B \Vert \gamma'(t) \Vert dt = \displaystyle \int_0^1 B \Vert y - x \Vert dt = B \Vert y - x \Vert; \tag{7}$
bringing together (5), (6), and (7) yields
$\Vert f(y) - f(x) \Vert \le B \Vert y - x \Vert, \tag{8}$
that is, $f(x)$ is Lipschitz continuous on $K$.
• Thank you, this is very elegant. Do you think my proof could possibly be correct? – sequence Jun 28 '17 at 1:32
• Yes, your method seems fine. Note it agrees with mine. – Robert Lewis Jun 28 '17 at 1:39
• "Since $K$ is compact and $\nabla f \in C^0(A, \Bbb R)$, so hence $\nabla f \in C^0(K, \Bbb R)$, $\Vert \nabla f \Vert$ is bounded by some $B$ on $K$.." Is there a proof for that or is it trivial? – abuchay May 14 '18 at 14:25
• @abuchay: Well, $\nabla f \in C^0(K)$ since $\nabla f \in C^0(A)$ and $K \subset A$. $\Vert \nabla f \Vert \in C^0(K)$ since $\Vert \cdot \Vert$ is itself continuous in its argument. Then the boundedness follows since continuous functions on compacta are bounded, which is standard result in elementary topology. – Robert Lewis May 14 '18 at 15:40
The convexity of $K$ is not needed. Suppose the conclusion fails. Then for each $m\in \mathbb N,$ there exist $y_m,x_m \in K$ such that
$$\tag 1 |f(y_m)- f(x_m)| > m|y_m-x_m|.$$
Because $K$ is compact, we can find a subsequence $m_k$ such that the sequences $y_{m_k},x_{m_k}$ converge to points $y,x\in K$ respectively.
Suppose $y\ne x.$ Then $|y-x| > 0.$ For large $k$ we then have
$$|f(y_{m_k})- f(x_{m_k})| > m_k|y_{m_k}-x_{m_k}|> m_k(|y-x|/2)\to \infty$$
This implies $f$ is not bounded on $K.$ But $f$ is continuous on $A,$ hence is continuous on $K,$ hence $f$ is bounded on $K$ by compactness. This contradiction shows $y=x.$
Because $A$ is open we can choose $r>0$ such that $B(x,r)\subset A.$ For large $k$ we then have $y_{m_k},x_{m_k}$ in the compact convex set $\overline {B(x,r/2}) \subset A.$ By $(1),$ your highlighted mean value inequality then fails in this last set, contradiction. Therefore $(1)$ cannot hold, proving the result.
• I dont understand why $m_k(|y-x|/2)\to \infty$. Why does it tend to infinity? If $m_k$ is a subsequence of $y_{mk}$ then $m_k$ also converges to $y$. Then, how can $m_k$ be a subsequence of $x_{mk}$ too? – John Keeper Oct 24 '18 at 9:46
• @JohnKeeper $m_k$ is a subsequence of $1,2,\dots$ hence $\to \infty$ – zhw. Oct 24 '18 at 21:32 | 2019-11-11T22:27:37 | {
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https://brilliant.org/discussions/thread/puzzle-paradox-in-probability/ | Here's a cute puzzle.
Player A tosses a coin indefinitely and stops when he encounters a consecutive sequence of HT. Player B plays a similar game, except he stops when he encounters HH consecutively. Is the expected number of tosses equal for both players? If not, who has more, and can you give an intuitive explanation why?
Note by C Lim
4 years, 8 months ago
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I think that player A is expected to be finished earlier. If he encounters an H, then he would have to keep tossing H's in order to continue. The chance of throwing $$n$$ consecutive H's decreases rapidly as $$n$$ increases.
If player B encounters an H, then he will have to toss a T to continue, but after tossing a T, he will continue tossing, because no matter what he tosses next, he doesn't encounter HH.
So, player A is doomed so to say whenever he tosses an H, because he'll have to keep tossing H's in order to continue. That's why he is expected to stop earlier than player B. I hope that makes sense.
- 4 years, 8 months ago
Yup that's correct! It would also be interesting to explicitly calculate the expected number of throws for both A and B.
- 4 years, 8 months ago
Player A is expected to throw his first H after $$2$$ tosses, because
$1 \cdot \frac{1}{2} + 2 \cdot \frac{1}{4} + 3 \cdot \frac{1}{8} + 4 \cdot \frac{1}{16} + 5 \cdot \frac{1}{32} + \dots = 2.$
Similarly, after having tossed his first H, player A is expected to throw his first T (and thus stopping) after another 2 tosses. So (hopefully) I can conclude that player H is expected to toss 4 times.
Player B is expected to roll his first H in two tosses. Then, there is a chance of 50% that he stops after the third toss, i.e. after tossing H again. However, if he tosses a T, then again, he is expected to toss his second H in two tosses, after which there is a 50% chance that he stops after the toss after that. It goes on and on. So, the amount of tosses player B is expected to do is
$3 \cdot \frac{1}{2} + 6 \cdot \frac{1}{4} + 9 \cdot \frac{1}{8} + 12 \cdot \frac{1}{16} + 15 \cdot \frac{1}{32} + \dots = 3 \cdot 2 = 6.$
And $$6 > 4$$.
I hope that what I did is allowed, I'm not completely convinced of that just yet. Could anyone confirm that what I did is correct (or show me what I've done wrong)?
- 4 years, 8 months ago
I'm not sure what you did is ok either, but your answer is correct! [ This suggests there's probably a way to rigourously justify your working. ]
Here's how I'd do it. To compute the expected number of tosses for HH, we define two values:
• let a = expected number of additional tosses, assuming previous was 'H';
• let b = expected number of additional tosses, assuming previous was 'T'.
Then $$a = \frac{b+1} 2 + \frac 1 2$$ since if the previous toss was 'H', there's a 50% chance the next is 'T' (in which case the expected number of tosses is b+1), and a 50% chance the next is 'H' (in which case the number of tosses is 1 since it ends).
Likewise, $$b = \frac{a+1} 2 + \frac{b+1}2$$ since if the previous toss was 'T', there's a 50% chance the next is 'H' (in which case expected number of tosses is a+1), and a 50% chance the next is 'T' (and the expected number of tosses is b+1).
Solving then gives us a=4, b=6. The expected number of tosses is then b=6, since we can assume the 0-th toss was 'T' with no loss of generality.
The equations for HT are similar:
$a = \frac {a+1} 2 + \frac 1 2, \quad b = \frac{a+1}2 + \frac{b+1} 2$
which gives us a=2, b=4.
[ P.S. To be pedantic, my working isn't 100% rigourous either, since it already assumes a, b are finite. ]
- 4 years, 8 months ago
That's really great. Two completely different ways of solving the problem :) I wouldn't know how to prove that $$a,b$$ are finite though.
- 4 years, 8 months ago
If you want a finiteness proof, use first principles. Let $$N_A$$ be the number of tosses until $$A$$ stops. The probability $$P[N_A>n]$$ is the probability that $$A$$ has not thrown HT in the first $$n$$ tosses. This can only happen if $$A$$ has thrown $$j$$ tails followed by $$n-j$$ heads, for some $$0 \le j \le n$$. Thus there are $$n+1$$ possible sequences, all equally likely, and so $P[N_A > n] = (n+1)2^{-n}$ Thus $P[N_A \ge n] \;=\; P[N_A > n-1] \; = \; n2^{1-n}$ Normal probability and series summation tricks give us $E[N_A] \;=\; \sum_{n=1}^\infty P[N_A \ge n] \; = \; \sum_{n=1}^\infty n2^{1-n} \; = \; 4$ The result for $$B$$ is a little more challenging. Let $$N_B$$ be the number of tosses until $$B$$ stops. The probability that $$N_B > n$$ is the probability that $$B$$ does not throw HH in the first $$n$$ tosses. This is the probability that the first $$n$$ tosses are made up of "HT"s and "T"s stuck together, plus the probability that the first $$n-1$$ tosses are made up of "HT"s and "T"s stuck together, followed by a single "H".
It is well-known that the number of ways of tiling a $$n\times1$$ chess-board using just $$2\times1$$ dominoes and $$1\times1$$ squares is the Fibonacci number $$F_{n+1}$$. (Just to be definite, $$F_0=0$$, $$F_1=1$$, etc.) Thus the probability that $$N_B > n$$ is the sum of the probabilities $$F_{n+1}2^{-n}$$ and $$F_n2^{-n}$$, and hence $P[N_B > n] \; = \; \frac{F_{n+2}}{2^n}$ and hence $P[N_B \ge n] \; = \; P[N_B > n-1] \; = \; \frac{F_{n+1}}{2^{n-1}} \qquad n \ge 1.$ Thus $E[N_B] \; = \; \sum_{n=1}^\infty P[N_B \ge n] \; = \; \sum_{n=1}^\infty \tfrac{F_{n+1}}{2^{n-1}} \; = \; 4G(0.5) - 2$ where $G(x) \;=\; \sum_{n=0}^\infty F_nx^n$ is the generating function of the Fibonacci numbers. It is well-known that $G(x) \; = \; \frac{x}{1-x-x^2}$ so that $$G(0.5) = 2$$. Thus $$E[N_B] = 6$$.
- 4 years, 8 months ago
Of course, once you showed that the number of ways is $$F_{n+2}$$, you can use the ratio test to conclude that the sum is finite, which is all we needed.
Nice way of calculating the sum through generating functions.
Staff - 4 years, 8 months ago
In a similar vein, check out High school math #8.
Staff - 4 years, 8 months ago
The absorbing Markov chain also gets the same answer.
This is the markov chain for ending the game with HH, for player B. I is the initial state, and the probabilities of moving to different states are shown:
$$\left(\begin{array}{cccccccc} & I & H & T & HT & TH & TT & HH \\ I & 0 & \frac{1}{2} & \frac{1}{2} & 0 & 0 & 0 & 0 \\ H & 0 & 0 & 0 & \frac{1}{2} & 0 & 0 & \frac{1}{2} \\ T & 0 & 0 & 0 & 0 & \frac{1}{2} & \frac{1}{2} & 0 \\ HT & 0 & 0 & 0 & 0 & \frac{1}{2} & \frac{1}{2} & 0 \\ TH & 0 & 0 & 0 & \frac{1}{2} & 0 & 0 & \frac{1}{2} \\ TT & 0 & 0 & 0 & 0 & \frac{1}{2} & \frac{1}{2} & 0 \\ HH & 0 & 0 & 0 & 0 & 0 & 0 & 1 \end{array}\right)$$
Matrix Q is:
$$\left(\begin{array}{rrrrrr} 0 & \frac{1}{2} & \frac{1}{2} & 0 & 0 & 0 \\ 0 & 0 & 0 & \frac{1}{2} & 0 & 0 \\ 0 & 0 & 0 & 0 & \frac{1}{2} & \frac{1}{2} \\ 0 & 0 & 0 & 0 & \frac{1}{2} & \frac{1}{2} \\ 0 & 0 & 0 & \frac{1}{2} & 0 & 0 \\ 0 & 0 & 0 & 0 & \frac{1}{2} & \frac{1}{2} \end{array}\right)$$
Fundamental matrix $$(I-Q)^{-1}$$ is:
$$\left(\begin{array}{rrrrrr} 1 & \frac{1}{2} & \frac{1}{2} & 1 & \frac{3}{2} & \frac{3}{2} \\ 0 & 1 & 0 & 1 & 1 & 1 \\ 0 & 0 & 1 & 1 & 2 & 2 \\ 0 & 0 & 0 & 2 & 2 & 2 \\ 0 & 0 & 0 & 1 & 2 & 1 \\ 0 & 0 & 0 & 1 & 2 & 3 \end{array}\right)$$
The required expectation is the sum of the first row, which is 6.
Similarly doing for the 'HT' case yields an expected value of 4.
- 4 years, 8 months ago | 2018-04-21T06:13:59 | {
"domain": "brilliant.org",
"url": "https://brilliant.org/discussions/thread/puzzle-paradox-in-probability/",
"openwebmath_score": 0.9759904146194458,
"openwebmath_perplexity": 660.6993215232189,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.978712648206549,
"lm_q2_score": 0.8757869916479466,
"lm_q1q2_score": 0.8571438058606087
} |
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