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https://math.stackexchange.com/questions/3426080/a-set-of-elements-in-a-reduced-unity-ring | # A set of elements in a reduced unity ring
Let $$(A,+,\cdot)$$ be a unity ring with the property that if $$x \in A$$ and $$x^2=0$$ then $$x=0$$. Consider the set $$M=\{a\in A | a^3=a\}$$. Prove that:
a) $$2a\in Z(A)$$, $$\forall a\in M$$, where $$Z(A)$$ denotes the centre of the ring $$A$$;
b) $$ab=ba$$, $$\forall a,b\in M$$.
My attempts revolved around the fact that an idempotent element in a reduced ring is central.
So, since for $$a\in M$$ we have that $$(a^2)^2=a^2$$, it follows that $$a^2\in Z(A)$$, $$\forall a\in M$$.
The next thing I wanted to use in order to solve a) was that $$Z(A)$$ is a subring of $$A$$, so if I had proved that $$(a+1)^2 \in Z(A)$$, $$\forall a\in M$$, then we would have reached the desired conclusion. However, I couldn't prove this and I honestly doubt that it is true.
Another idea that I had was to prove that $$M$$ is a subring of $$A$$. Of course, this didn't work out because I cannot even prove that $$M$$ is closed under addition. Again, I don't know if this is true and it most likely isn't.
As for b), I think that a) should be of use, but I don't know how. It is a well-known problem that a ring with $$x^3=x$$ for any $$x$$ in that ring is commutative, but since $$(M,+,\cdot)$$ is almost definitely not a ring, this doesn't help.
EDIT: Is there any chance that this question is simply wrong? I tended to believe this before asking it here too, but since nobody has made any progress on it until now I am even more inclined to think so.
• If you honestly doubt that $(a+1)^2 \in Z(a)$ for all $a \in M$, then you honestly doubt that part (a) is true, since if part (a) is true (and since you've already noted $a^2 \in Z(a)$ for all $a \in M$), you get $(a+1)^2 = a^2+2a+1 \in Z(a)$ for all $a \in M$. – mathworker21 Nov 9 at 18:50
• @mathworker21 I agree with you. But since I have spent hours and hours trying to prove that $(a+1)^2 \in Z(A)$ for all $a\in M$ and nothing worked out, then I am inclined to believe that this statement may actually be wrong. Yet, since I cannot provide a counterexample, here I am, still hoping that someone better than me at rings will solve it.... – Math Guy Nov 9 at 21:34
• I don't know. I don't think the proof that any ring with $x^3 = x$ for all $x$ implies the ring is commutative is that easy... I could imagine spending hours failing to find that proof. By the way, where did you find this problem? – mathworker21 Nov 9 at 21:35
• @mathworker21 I know the "classical" one where $x^3=x$ for all $x$ holds in a ring, but since here we do not have a ring it doesn't really help I think. The problem is from a magazine from my country. – Math Guy Nov 9 at 21:37
• you missed my point. my point was that working on these kinds of problems for hours and failing doesn't mean they are false. (Of course that is always true, but I find the statement more meaningful here). The evidence/example I gave was the $x^3=x$ problem. For that problem, I could imagine working on it for several hours without finding the solution. And of course that problem is true. – mathworker21 Nov 9 at 21:54
## 1 Answer
We show $$M \subseteq Z(A)$$. This immediately gives (a) and (b).
Lemma 1: $$yx = 0 \implies xzy = 0$$ for any $$z$$.
Proof: $$(xzy)(xzy) = xz(yx)zy = 0$$.
Lemma 2: $$x^2 = x$$ implies $$x \in Z(A)$$.
Proof: For any $$y \in A$$, a short computation shows $$(xy-xyx)(xy-xyx) = 0 = (yx-xyx)(yx-xyx)$$.
Lemma 3: $$a \in M \implies a^2 \in Z(A)$$.
Proof: $$(a^2)^2 = a^4 = a^2$$, so use Lemma 2.
Claim: For any $$a \in M$$, $$a \in Z(A)$$.
Proof: Since $$(a-1)[a(a+1)]=0$$, Lemma 1 implies that for any $$b \in A$$, $$0 = a(a+1)b(a-1) = (a^2b+ab)(a-1).$$ Also, $$a(a+1)(a-1)=0$$ implies $$0 = ba(a+1)(a-1) = (ba^2+ba)(a-1) = (a^2b+ba)(a-1),$$ where the last equality used Lemma 3. Subtracting gives: (1) $$0 = (ba-ab)(a-1)$$. The exact same argument shows: (2) $$0 = (a-1)(ba-ab)$$. (1) immediately implies $$0 = (ba-ab)(a-1)b = (ba-ab)(ab-b)$$, and (2) with Lemma 1 implies $$0 = (ba-ab)b(a-1) = (ba-ab)(ba-b)$$. Subtracting the two results gives $$(ba-ab)^2 = 0$$.
• Thank you ! Could you tell me how you came up with this solution? – Math Guy Nov 10 at 11:23
• @MathGuy well, I played around with it for a while, so I exhausted many different approaches. Then I eventually stumbled upon something like Lemma $1$, first in the form of $0 = a(a+1)b(a-1)$ and realized it started giving equations I hadn't seen/derived before. So I knew I had something good. Then it was just a matter of finishing up, which was pretty easy (the proof of the claim is rather short). – mathworker21 Nov 10 at 13:02 | 2019-11-18T21:24:59 | {
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https://mathematica.stackexchange.com/questions/136962/numerical-approximation-of-pi | # Numerical approximation of $\pi$
Points are randomly scattered inside the unit square, some fall within the unit circle with probability $P=\pi/4$.
so $P$ is approximated by the fraction $$P\approx \frac{\text{Number of red points}}{\text{Number of all points}}$$ this leads $$\pi \approx 4\frac{\text{Number of red points}}{\text{Number of all points}}$$ (see following image)
There is a code for this:
tinyColor[color_, point_] := {PointSize[Small], color, Point[point]}
colorChoose[point_] :=
If[Norm[point] <= 1, tinyColor[Red, point], tinyColor[Blue, point]]
darts = RandomReal[{0, 1}, {40000, 2}];
coloredDarts =ParallelMap[colorChoose, darts];
insides = Map[Boole[Norm[#] <= 1] &, darts];
piapprox = Accumulate[insides]/Range[Length[darts]]
inner = Select[darts, Norm[#] <= 1 &];
outer = Select[darts, Norm[#] > 1 &];
Show[Plot[Sqrt[1 - x^2], {x, 0, 1}, Filling -> Axis, AspectRatio -> 1,
PlotLabel -> n == Length[darts] TildeTilde[π, 4.0*piapprox[[-1]]]],
ListPlot[{inner, outer},
PlotStyle -> {{PointSize[Tiny], Red}, {PointSize[Tiny], Blue}},
ImageSize -> {500, 500}]]
I tried to simplify this problem:
pts = RandomPoint[Rectangle[], 40000];
ListPlot[pts, AspectRatio -> 1, PlotStyle -> Blue]
The problem is following:
How can I split set of points pts into two parts, "inside the circle " and "outside the circle"?
• You should also investigate RegionMember, e.g., rf = RegionMember[Disk[]]; rf[darts] – Carl Woll Feb 10 '17 at 22:26
You can also use Select:
ptsin = Select[pts, Norm[#] < 1 &];
N[Length[ptsin]/Length[pts]]*4
(* 3.1496 *)
How can I split set of points pts into two parts, "inside the circle " and "outside the circle"?
{in, out} = SortBy[GatherBy[pts, Norm[#] < 1 &], Norm[#[[1, 1]]] &];
ListPlot[{in, out}, AspectRatio -> 1, PlotStyle -> {Red, Blue}]
• What's the purpose of SortBy? – anderstood Feb 4 '17 at 18:27
• The approximation of $\pi$ can be recovered with {in, out} = GatherBy[pts, Norm[#] < 1 &]; 4*Length[in]/(Length[out] + Length[in]) // N. – anderstood Feb 4 '17 at 18:30
• @anderstood, depending on the Norm of the first element in lst, the the first list in the output produced by GatherBy may be the "inside sublist" or the "outside sublist". Sorting the output of GatherBy makes the first list the "inside" one. – kglr Feb 4 '17 at 18:32
• Comment to my own comment: it should be {in, out} = SortBy[GatherBy[pts, Norm[#] < 1 &], Norm[#[[1, 1]]] &], cf kglr's comment above. – anderstood Feb 4 '17 at 18:55
Playing with Norm as shown in other answers:
pts = RandomReal[1, {40000, 2}];
4 True/(True + False) /. CountsBy[pts, Norm[#] < 1 &]
3.1474
That doesn't help with drawing the graphic however.
You can compute the norm of the point and verify if it is inside the circle.
At = 4;
d = 2;
totalpoints = 40000;
pts = RandomReal[{-1, 1}, {totalpoints, 2}];
pointsinsidecircle = Select[pts, Norm[#] < 1 &];
counter = Length[pointsinsidecircle];
approxpi = (4. At counter/totalpoints)/d^2;
Print["approx \[Pi] = ", approxpi]
ListPlot[{pts, pointsinsidecircle}, AspectRatio -> 1,
PlotStyle -> {Blue, Red}]
(*approx \[Pi] = 3.1328*)
• You could you Norm directly, and avoid AppendTo which is very slow. Using Select is probably a much faster option. – anderstood Feb 4 '17 at 18:24
• @anderstood i have changed the answer. Thank you. – Diogo Feb 4 '17 at 18:30
• Tip: Don't save graphics as JPEGs, they become fuzzy and lose color fidelity. Use PNGs. – Rahul Feb 10 '17 at 21:19
A slightly different approach:
inside = Pick[pts, Map[# ∈ Disk[] &, pts], True];
outside = Complement[pts, inside];
Also as pointed in a comment above:
inside = Pick[#, RegionMember[Disk[]][#], True] &@pts
outside = Complement[pts, inside];
` | 2019-12-07T21:34:15 | {
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https://math.stackexchange.com/questions/2247973/knowing-that-for-any-set-of-real-numbers-x-y-z-such-that-xyz-1-the-ineq | Knowing that for any set of real numbers $x,y,z$, such that $x+y+z = 1$ the inequality $x^2+y^2+z^2 \ge \frac{1}{3}$ holds.
Knowing that for any set of real numbers $x,y,z$, such that $x+y+z = 1$ the inequality $x^2+y^2+z^2 \ge \frac{1}{3}$ holds.
I spent a lot of time trying to solve this and, having consulted some books, I came to this:
$$2x^2+2y^2+2z^2 \ge 2xy + 2xz + 2yz$$ $$2xy+2yz+2xz = 1-(x^2+y^2+z^2)$$ $$2x^2+2y^2+2z^2 \ge 1 - x^2 -y^2 - z^2$$ $$x^2+y^2+z^2 \ge \frac{1}{3}$$ But this method is very unintuitive to me and I don't think this is the best way to solve this. Any remarks and hints will be most appreciated.
• This does seem like one of the best ways to tackle this problem. Why do you find this method counterintuitive? – vrugtehagel Apr 23 '17 at 12:56
• @vrugtehagel It entails using a property which is not directly derived from the problem. – ILoveChess Apr 23 '17 at 12:58
• Are you comfortable with calculus? Here's a completely different approach. $z = 1 - x - y$ so $x^2 + y^2 + z^2 = x^2 + y^2 +(1 - x - y)^2$. Now use some calculus to find when that has a maximum. It will be when $x = y = \frac{1}{3}$ hence also $z = \frac{1}{3}$. – badjohn Apr 23 '17 at 13:03
• – Martin Sleziak Apr 23 '17 at 21:29
Cauchy- Schwarz works: $$x^2+y^2+z^2=\frac{1}{3}(1^2+1^2+1^2)(x^2+y^2+z^2)\geq\frac{1}{3}(x+y+z)^2=\frac{1}{3}$$
$x^2+y^2+z^2$ only depends on the squared distance of $(x,y,z)$ from the origin and the constraint $x+y+z=1$ tells us that $(x,y,z)$ lies in a affine plane. The problem is solved by finding the distance between such plane and the origin: since the plane is orthogonal to the line $x=y=z$,
$$\min_{x+y+z=1}x^2+y^2+z^2 = \left(\frac{1}{3}\right)^2+\left(\frac{1}{3}\right)^2+\left(\frac{1}{3}\right)^2 = \frac{1}{3}$$ and we are done.
• clear: the sphere from the origin is tangent to the plane in x=y=z=1/3 – G Cab Apr 23 '17 at 14:49
This is not a proof in itself, but if you've studied statistics, then you've seen a proof that
$$0\le V(X)=E(X^2)-E(X)^2$$
If we now consider a random variable $X$ with three equally likely values, $X=x,y$, and $z$, then we have
$$E(X)={x+y+z\over3}\qquad\text{and}\qquad E(X^2)={x^2+y^2+z^2\over3}$$
If, in addition, we assume $x+y+z=1$, then we have $E(X)={1\over3}$, which implies $E(X^2)\ge\left(1\over3\right)^2={1\over9}$, or $x^2+y^2+z^2\ge{1\over3}$.
Excellent algebraic have been given (I voted for them). Intuition can here be obtained through visual proofs.
The equation $x+y+z=1$ defines a plane. $x^2+y^2+z^2=1/3$ defines a sphere. The following visualization depicts the plane in blue, the sphere in red.
From that, you can imagine that the question could be rephrased as (in a mundane way): for any point in the plane, its distance from the $(0,0,0)$ origin is higher than $1/\sqrt{3}$? So the sphere should remain "below" the plane. Except when they meet. The symmetry of the problem tells you that the tangency point has equal coordinates $(1/\sqrt{3},1/\sqrt{3},1/\sqrt{3})$, which is one of the motivations behind the equation $(3x-1)^2+(3y-1)^2+(3z-1)^2$.
Would the sphere be bigger (a radius higher than $1/\sqrt{3}$), it would intersect the plane in more than a single tangency point.
All in all, this resorts to finding the distance of the plane to the origin, which is exactly where the sphere and the plane meet. So what you are looking at is the distance of the plane to the origin.
If the plane is given by $ax+by+cz+d$, the signed distance of a point $(x_0,y_0,z_0)$ to it is (Point-Plane Distance):
$$D = \frac{ax_0+by_0+cz_0+d}{\sqrt{a^2+b^2+c^2}}$$
which in your case gives exactly $1/\sqrt{3}$. Any point in the plane is farther to $(0,0,0)$ than $|D|$.
Hint:
Expand $(3x-1)^2+(3y-1)^2+(3z-1)^2\geqslant 0$ And simplify.
If you rewrite your proof as:
$3 ( x^2+y^2+z^2 ) \ge ( x^2+y^2+z^2 ) + ( 2xy+2yz+2zx ) = (x+y+z)^2 = 1$.
You would find that it is not so unintuitive after all.
Since you know that $x+y+z = 1$, a natural thing to do to the original equation is to homogenize it; namely make all terms have the same degree. This gives us "$3 ( x^2+y^2+z^2 ) \ge (x+y+z)^2$, and expanding out immediately tells us the solution.
In general, a technique that works for many cyclic polynomial inequalities is to try to 'smooth' the terms out. Terms that are just a power of one variable are the 'biggest', and using inequalities like "$x^2+y^2 \ge 2xy$" will 'mix' powers and thereby 'reduce' them.
It is enough to prove the result for $x,y,z$ positive since $$\frac{|x|+|y|+|z|}{3} \geq \frac{x+y+z}{3}$$
One can use the generalized AM-GM inequality: If $$M_p = \left(\frac{x^p+y^p+z^p}{3}\right)^{\frac{1}{p}}$$ then for $p < q$, $M_p \leq M_q$ with equality holding if and only if $x=y=z$. Here, using $M_2 \geq M_1$, we get $$\left(\frac{x^2+y^2+z^2}{3}\right)^{\frac{1}{2}} \geq \frac{|x|+|y|+|z|}{3} \geq \frac{x+y+z}{3} = \frac{1}{3}$$
We can minimize $x^2 + y^2 + z^2$ subject to the constraint $x+y+z = 1$ using Lagrange multipliers, we then find that $x = y = z = \frac{1}{3}$, therefore $x^2 + y^2 + z^2\geq\frac{1}{3}$.
• This only shows that this is a local minimum, right? – Carsten S Apr 24 '17 at 14:05
One more way to prove it is by substituting out for $z$:
$$x^2+y^2+z^2 = x^2+y^2+(1-x-y)^2=2x^2+2y^2+2xy-2x-2y+1$$ Now, substitute $x=\hat x+a$ and $y=\hat y+b$. We get $$2\hat x^2+2\hat y^2+2\hat x\hat y+(4a+2b-2)\hat x+(2a+4b-2)\hat y+(2(a^2+ab+b^2-a-b)+1)$$ We can eliminate the order 1 terms by letting $4a+2b-2=2a+4b-2=0$, which gives $a=b=\frac13$. With this substitution, we have $$2(\hat x^2+\hat x\hat y+\hat y^2)+\frac13$$ We can express the bracketed term as a sum of squares, giving us $$3(\hat x+\hat y)^2+(\hat x-\hat y)^2 + \frac13$$ and we can see that the smallest value this can take is $\frac13$. Indeed, it takes this value when $\hat x=\hat y=0$ - that is, when $x=y=\frac13$.
Ok, you want an intuitive proof, not a better proof. That's fine.
First note that for $x=y=z$, we have $x^2+y^2+z^2 = 3\left(\frac13\right)^2= \frac13$. Next let us show that $x^2+y+2+z^2$ is not minimal if $x\ne y$ or $y\ne z$. Indeed if $a\ne b$ then $$a^2 + b^2 - \left(\left(\frac{a+b}2\right)^2+\left(\frac{a+b}2\right)^2\right) = \frac12\left( a^2 + b^2 -2ab \right)=\frac12(a-b)^2>0,$$ so we can replace two of the numbers be their averages and lower the sum of squares.
Unfortunately we not yet quite done. We are done if we can show that there is a global minimum of $x^2+y^2+z^2$, given that $x+y+z=1$. Because then that global minimum must also be a local minimum, but for that we cannot have $x\ne y$ or $y\ne z$, so $x=y=z$ must be the global minimum, and its value is $\frac13$.
For that we can first argue that we can restrict ourselves to $x,y,z\ge0$ (e.g. if $x<0$ replace $(x,y,z)$ by $\frac1{-x+y+z}(-x,y,z)$, which yields a smaller value since $-x+y+z = 1 - 2x > 1$) and then appeal to the compactness of $\{(x,y,z)\colon \text{$x+y+z=1$,$x,y,z\ge0$}\}$.
I did not say that this would be pretty, but it is pretty intuitive to me :)
Here's another way to approach this. It's easy to see that the value of $\frac13$ is obtained when each of $x, y, z$ is $\frac13$. We want to show that as the variables deviate from this point (with their sum still being 1) the value cannot decrease.
So we look at the deviations from $\frac13$: $x=\frac13+\epsilon_1$, $y=\frac13+\epsilon_2$, $z=\frac13+\epsilon_3$ with $\epsilon_1+\epsilon_2+\epsilon_3=0$. you have
$x^2+y^2+z^2=\\ (\frac13+\epsilon_1)^2+(\frac13+\epsilon_2)^2+(\frac13+\epsilon_3)^2=\\\left(\frac19+\frac23\epsilon_1+\epsilon_1^2\right)+\left(\frac19+\frac23\epsilon_2+\epsilon_2^2\right)+\left(\frac19+\frac23\epsilon_3+\epsilon_3^2\right)=\\ \left(\frac19+\frac19+\frac19\right)+\frac23(\epsilon_1+\epsilon_2+\epsilon_3)+(\epsilon_1^2+\epsilon_2^2+\epsilon_3^2)=\\ \frac13+(\epsilon_1^2+\epsilon_2^2+\epsilon_3^2) \ge \frac13$
The desired inequality follows by noting that from the convexity of $f(t) = t^{2},$ we have $f\left(\dfrac{x+y+z}{3}\right) \leq \dfrac{f(x)+f(y)+f(z)}{3}$ | 2019-05-20T06:30:12 | {
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http://math.stackexchange.com/questions/293921/combinatorics-and-probability-problem | Combinatorics and Probability Problem
The problem I am working on is:
An ATM personal identification number (PIN) consists of four digits, each a 0, 1, 2, . . . 8, or 9, in succession.
a.How many different possible PINs are there if there are no restrictions on the choice of digits?
b.According to a representative at the author’s local branch of Chase Bank, there are in fact restrictions on the choice of digits. The following choices are prohibited: (i) all four digits identical (ii) sequences of consecutive ascending or descending digits, such as 6543 (iii) any sequence start-ing with 19 (birth years are too easy to guess). So if one of the PINs in (a) is randomly selected, what is the prob-ability that it will be a legitimate PIN (that is, not be one of the prohibited sequences)?
c. Someone has stolen an ATM card and knows that the first and last digits of the PIN are 8 and 1, respectively. He has three tries before the card is retained by the ATM (but does not realize that). So he randomly selects the $2nd$ and $3^{rd}$ digits for the first try, then randomly selects a different pair of digits for the second try, and yet another randomly selected pair of digits for the third try (the individual knows about the restrictions described in (b) so selects only from the legitimate possibilities). What is the probability that the individual gains access to the account?
d.Recalculate the probability in (c) if the first and last digits are 1 and 1, respectively.
---------------------------------------------
For part a): The total number of pins without restrictions is $10,000$
For part b): The number of pins in either ascending or descending order is $10 \cdot 1 \cdot 1 \cdot 1$, because once the first digit is known, then the three other spots containing digits are already spoken for. The number of pins where each slot contains the same digit is $10 \cdot 1 \cdot 1 \cdot 1$, because once the first digit is known there is only one option left to the rest of the slots. The number of pins that have their first and second slot occupied by 1 and 9, respectively, is $1 \cdot 1 \cdot 10 \cdot 10 \cdot$. So, if R is the set that contains these restricted pins, then $|R| = 130$; and if N is the set that contains the non-restricted ones, meaning R and N are complementary sets, then $|N| = 10,000 - 130$. Hence, the probability is then $P(N) = 9780/10000 = 0.9870.$ However, the answer is $0.9876$. What did I do wrong?
For part c): The sample space, containing all of the outcomes of the experiment that will take place, is $|N|=9870$. When it says that the thief won't use the same pair of digits in each try, does that not allow him trying the pin 8 5 2 1 in one try and the pin 8 2 5 1 in another try?
-
Ascending or descending is $14$, not $20$. That takes care of the disagreement. – André Nicolas Feb 3 '13 at 20:44
@AndréNicolas I didn't get 14 or 20, I got 10. Your saying the answer is 14, how did you get that? – Mack Feb 3 '13 at 20:56
@Eli: You got $2\cdot10=20$ for ascending and descending together, and André is saying that it should be $2\cdot7=14$ for both together. – joriki Feb 3 '13 at 21:04
Oh, I see. The 2 corresponds to the two choices (ascending or descending); and the 7 corresponds to the fact that you can't start the 4-digit pin with 7,8, or 9 when ascending, and you can't start a 4-digit pin with 0,1, or 2 when descending. – Mack Feb 3 '13 at 21:14
For b): Which is the descending sequence starting with $1$ that you counted?
For c): Good question; the problem is badly worded in that regard. Taking it literally, I'd tend to interpret it as referring to unordered pairs, but since it makes little sense to couple two different PINs in this manner, I suspect that they actually mean ordered pairs. However, note that the answer doesn't depend on this.
I understand neither why the question says that the thief knows the restrictions, nor why you say that the sample space has size $9870$. The thief knows that the first and last digits are $8$ and $1$, respectively; that's not compatible with any of the sequences excluded by the restrictions, and it doesn't allow for $9870$ possibilities.
-
I’d interpret is as referring to ordered pairs, partly from the language, and partly from the overall level of difficulty of the exercise. – Brian M. Scott Feb 3 '13 at 20:41
@joriki I don't really understand what you are asking when you say, "Which is the descending sequence starting with 1 that you counted?" – Mack Feb 3 '13 at 21:00
@Eli: Exactly. And $10-3=7$. – joriki Feb 3 '13 at 21:08
@Eli: I found another version of the book online with the answers included. It has $.0337$ for d., which is also wrong, but can be interpreted as a rounded version of the correct result $3/89$ (see Metin's answer). So it seems they just round results without indicating it (which is rather bad style). So it seems likely that $0.0333$ is a rounded version of $3/90$. That leaves the question how they arrive at $90$ options. Perhaps they meant to say $1$ and $8$ instead of $8$ and $1$; then the birth year rule would lead to a count of $90$. – joriki Feb 4 '13 at 8:35
@Eli: That would also explain why part c already mentions that the thief knows the restrictions, even though that's irrelevant for the question as posed. – joriki Feb 4 '13 at 8:37
For d): We have the case: $1$ * * $1$.
But the thief knows, by prohibition (i), it can not be $1111$. Thus he eleminates $1$ possibility.
Also he knows, by (iii), the second digit can not be $9$. There are exactly $10$ number of the form $19$ * $1$, namely, $1900$, $1910$, $1920$... So, at this stage he eleminates $10$ possibilities.
All in all, if he had no restrictions, there were $100$ choices for the form $1$ * * $1$. But he excluded $10 + 1 = 11$ of them and get 89 possible choices. Since he has 3 chances, the resulting possibility is $3/89$.
-
Just my two cents,
10^4 possibilities.
There are 14 ascending and descending groups of 4.
Keyspace 9,876
If the badguy knows two of the four spaces, he only has to guess through entropy 10^2. (None of the restrictions meet up with the range 8xx1)
3 Tries in 100
:)
-
(0123)(1234)(2345)(3456)(4567)(5678)(6789) And then the reverse of those... – Ben Sep 9 '13 at 5:33
Regarding part [c], I do not think that there is something wrong with the book. As @Mack said in one of his comments, thief has to guess only two second numbers. Since the given restrictions actually do not apply on this order of numbers, he has to guess among 100 (10*10) possible combinations of the numbers.
Since he has 3 tries, the probability that he will gain desired access is 3/100.
According to Wolfram Alpha, 3/100 is exactly 0.03, which is the same as written solution from the book.
Correct me if I am wrong.
- | 2015-09-04T06:16:35 | {
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https://engineering.stackexchange.com/questions/41139/condition-for-equilibrium-of-a-rigid-body-on-a-horizontal-surface | # Condition for equilibrium of a rigid body on a horizontal surface
The following question is from a past paper from Further Mechanics and it has been bothering me immensely I have spent hours cracking at a way to make sense of it, the question is in two parts
Part 1:
An object consists of a uniform solid circular cone, of vertical height 4r and radius 3r, and a uniform solid cylinder, of height 4r and radius 3r. The circular base of the cone and one of the circular faces of the cylinder are joined together so that they coincide. The cone and the cylinder are made of the same material.
Find the distance of the centre of mass of the object from the end of the cylinder that is not attached to the cone.
I first roughly sketched the object as such
Then finding the $$\bar{x}$$ of each part seperately:
For the cone I used the standard result of the centre of mass being $$\frac{1}{4}r$$ away from the base:
$$\bar{x}_{Cone}=\frac{1}{4}.4r + 4r=5r$$
For the cylinder I derived $$\bar{x}$$ via integration:
$$y=3r$$
$$\bar{x}_{cylinder}=\frac{\int_0^{4r} xdV}{\int_0^{4r}dV}$$
$$\because dV=y^2 \pi dx$$
$$\implies \bar{x}_{cylinder}=\frac{\int_0^{4r} 9r^2 \pi x dx}{\int_0^{4r}9r^2 \pi dx}$$
$$\implies \bar{x}_{cylinder}=\frac{9r^2 \pi \int_0^{4r} x dx}{9r^2 \pi\int_0^{4r} 1 dx}$$
$$\implies \bar{x}_{cylinder}=\frac{\frac{1}{2}\left[x^2 \right]^{4r}_0}{\left[x \right]^{4r}_0}$$
$$\therefore \bar{x}_{cylinder}=2r$$
Then by taking the weighted average of both objects I arrived at the correct value for the distance of the centre of gravity from the base of the cylinder which was
$$\bar{x}=\frac{11}{4}r$$
However the next part has had me at a complete loss for the better part of the day, it states that:
Show that the object can rest in equilibrium with the curved surface of the cone in contact with a horizontal surface.
I tried coming up with a rough sketch for this also
However I do not understand how to tackle this question at all, all I know is that for a body to be at equilibrium on such a surface, the centre of mass must pass through the point of suspension, however clearly this does not give an insight on how to answer this.
The condition for this question given in the marking scheme is that
Can someone explain what this means and why this is the case?
Given the geometry of the object, the center of gravity is very close to the surface between the cylinder and the cone. More specifically the distance between the center of gravity and the base of the cone is $$\frac{1}{4}r$$.
Basically what you need to prove is that the angle of the cone $$\phi$$ is such, so that when the object is tilted the weight crosses over the last $$\frac{1}{4}r$$.
So you need to prove that the angle $$\theta$$ is smaller that angle $$\phi$$.
Angle $$\theta$$ is calculated as: $$\tan\theta = \frac{\text{distance to be covered}}{\text{cylinder radius}}= \frac{\frac{11}{4}r}{3r}=\frac{11}{12} \Rightarrow \theta =42.5 [deg]$$
while Angle $$\phi$$ is calculated as: $$\tan\phi= \frac{\text{height of cone}}{\text{cone radius}}= \frac{4r}{3r}=\frac{4}{3} \Rightarrow \phi =53.1[deg]$$
Therefore, since $$\phi> \theta$$, the object becomes vertical enough so that the weight passes through the coned surface.
– NMech
Mar 23, 2021 at 9:52
• Reminds me of a crafty problem my undergrad QM prof posed. I'll adapt it to this particular object: using Heisenberg's Uncertainty Principle, how long will this object remain stable resting on the cone? It's not intended as a "real-world" problem but rather to see how you might use paired-parameters to guesstimate the stability. Mar 24, 2021 at 12:43
The CG is found by assuming the mass of the cone as 1/3m and the cylinder m, then
$$\bar{X}= \frac{2r*m+5r*m/3}{4/3m} =r*11/4$$
The cone side length (like the sharp tip of a pencil) is $$5r$$, the side of a 3,4,5 triangle, and the interior half tip angle is 36.87 degrees.
The CG is $$5.25r$$ from the tip of the cone and at rotation to rest on the side of the cone will be at $$5.25r*cos36.87= 4.19r<5r$$
I let you prove the condition.
It is well within the footprint of the pencil shape so it is stable. | 2023-01-27T01:50:05 | {
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https://math.stackexchange.com/questions/2773993/transport-equation-u-t-xu-x-u-0-with-ux-0-0-cosx-0 | # Transport equation $u_t + xu_x + u = 0$ with $u(x_0, 0) = \cos(x_0)$
I have been studying PDEs using Peter Olver's textbook. I have learnt how to solve equations such as $u_t + 2u_x = \sin(x)$ subject to an initial condition such as $u(0,x) = \sin x$. Letting $\epsilon = x - 2t$ and $u(t,x) = v(t,\epsilon)$, I then plug this into the transport equation.
However, I am not sure how to define a characteristic to solve the following equation
$$u_t + xu_x + u = 0, \qquad u(x_0, 0) = \cos(x_0)$$
because it has a variable 'speed' term $x$ and it is also not homogenous because of the term $u$.
A solution would be very helpful so I can see how to approach these problems.
• If you change coordinates to $x = e^y$ then $u_y = xu_x$ so your PDE in these new coordinates simplifies to $u_t + 2u_y + u = 0$. You can even simplify it further by taking $v = e^{t} u$ then $v_t + 2v_y = 0$ which is the equation you say you already know how to solve. – Winther May 9 '18 at 17:48
$$u_t + xu_x = -u$$ $\frac{dt}{1}=\frac{dx}{x}=\frac{du}{-u}$
First characteristics, from $\quad \frac{dt}{1}=\frac{dx}{x}$ :
$$x\,e^{-t}=c_1$$
Second characteristics, from $\quad\frac{dx}{x}=\frac{du}{-u}$ : $$x\,u=c_2$$ General solution of the PDE : $\quad x\,u=F(x\,e^{-t})$
$$u(x,t)=\frac{1}{x}F(x\,e^{-t})$$ $F$ is an arbitrary function, to be determined according to the boundary condition.
Condition : $\quad u(x_0, 0) = \cos(x_0)=\frac{1}{x_0}F(x_0\,e^{0})$
$F(x_0)=x_0\cos(x_0)$. Now the function $F$ is determined, i.e.: $F(X)=X\cos(X)$.
We put it into the above general solution , where $X=x\,e^{-t}$ , thus $F(x\,e^{-t})=(x\,e^{-t})\cos(x\,e^{-t})$ :
$$u(x,t)=\frac{1}{x}(x\,e^{-t})\cos(x\,e^{-t})=e^{-t}\cos(x\,e^{-t})$$
• Hi, thank you. Can you explain how you obtain your general solution of the PDE from the two characteristics? – PhysicsMathsLove May 11 '18 at 10:37
• The general solution can be expressed on various forms. For example on the form of implicit equation : $\Phi(c_1,c_2)=0$, or $c_1=f(c_2)$ or $c_2=F(c_1)$ or other forms. All functions introduced are arbitrary, but they are related one to the other. Doesn't matter the form chosen, the final result is the same after applying the boundary condition. – JJacquelin May 11 '18 at 10:51
• Ok I am still not sure how you obtain $xu = F(xe^{-t})$ from your characteristics, I.e. how it depends on $xe^{-t}$? – PhysicsMathsLove May 11 '18 at 10:57
• $c_2=F(c_1)$ with $c_2=xu$ and $c_1=xe^{-t}$ gives $xu=F(xe^{-t})$. – JJacquelin May 11 '18 at 11:02
• Why do you know one constant is a function of the other? – PhysicsMathsLove May 11 '18 at 11:12
Let us apply the method of characteristics. We get the characteristic equations $$\frac{\text{d} t}{\text{d} s} = 1 \, , \qquad \frac{\text{d} x}{\text{d} s} = x \, , \qquad \frac{\text{d} u}{\text{d} s} = -u \, .$$ Letting $t(0) = 0$, we know $t=s$. Letting $x(0) = x_0$, we get $x(t) = x_0\, e^t$. Since $u(0) = \cos(x_0)$, we have $u(t) = \cos(x_0)\, e^{-t}$. Finally, $$u(x,t) = \cos(x\, e^{-t})\, e^{-t} \, .$$
Note that \begin{align} \frac{\rm d}{{\rm d}t}u(e^t,t+t_0)&=\frac{\partial u}{\partial t}(e^t,t+t_0)+e^t\frac{\partial u}{\partial x}(e^t,t+t_0)\\ &=\left(\frac{\partial u}{\partial t}+x\frac{\partial u}{\partial x}\right)(e^t,t+t_0)\\ &=-u(e^t,t+t_0) \end{align} holds for all $t$ and $t_0$. Thus $$\frac{\rm d}{{\rm d}t}u(e^t,t+t_0)+u(e^t,t+t_0)=0,$$ or equivalently, $$\frac{\rm d}{{\rm d}t}\left(e^tu(e^t,t+t_0)\right)=0.$$ Therefore, $$e^tu(e^t,t+t_0)=e^{-t_0}u(e^{-t_0},-t_0+t_0)=e^{-t_0}u(e^{-t_0},0)=e^{-t_0}\cos e^{-t_0},$$ or equivalently, $$u(e^t,t+t_0)=e^{-t-t_0}\cos e^{-t_0}.$$ Finally, let $x=e^t>0$ and $\tau=t+t_0$. This gives $t=\log x$ and $t_0=\tau-\log x$. Hence $$u(x,\tau)=e^{-\log x-\tau+\log x}\cos e^{-\tau+\log x}=e^{-\tau}\cos\left(e^{-\tau}x\right).$$ | 2020-02-28T09:40:32 | {
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https://mathematica.stackexchange.com/questions/159043/count-the-pairs-in-a-set-of-data | # count the pairs in a set of Data
I have a list of Data, e.g.,
list1={0, 1, 1, 1, 1, 2, 2, 3, 3}
list2={0, 1, 1, 1, 1, 2, 2, 3, 3, 0}
and I want to get as an output True or False if I have only even number of Data of similar or not. in the above example I like to write a code that returns False for list1 and True for list2, since in list1 I have only one 0, but in the list2 there are even number of each number.
Thanks!
## 4 Answers
Using Counts
f = And @@ EvenQ[Values[Counts[#]]] &
{f[list1], f[list2]}
{False, True}
• +1. Or f = Counts /* AllTrue[EvenQ] – WReach Nov 1 '17 at 14:27
• Yours is much concise. I think you should post it as a separate answer. – Anjan Kumar Nov 1 '17 at 14:30
• I did, but when I saw yours, I deleted it :) I shall undelete. – WReach Nov 1 '17 at 14:36
Given:
f = Counts /* AllTrue[EvenQ];
Then:
f[list1]
(* False *)
f[list2]
(* True *)
You can use the Tally function:
Tally[list]
tallies the elements in list, listing all distinct elements together with their multiplicities.
So as
And@@EvenQ[Tally[list1][[;; , 2]]]
or
And@@EvenQ[Tally[list2][[;; , 2]]]
check[list_] := If[
Cases[EvenQ[Count[list, #] & /@ Union[list]], False] == {},
True, False]
check[list2]
True | 2019-09-20T17:32:13 | {
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https://stats.stackexchange.com/questions/48133/sum-standard-deviation-vs-standard-error | # Sum standard deviation vs standard error
I'm having difficulty in determining what exactly the difference is between the 2, especially when given an exercise and I have to choose which of the 2 to use. These is how my text book describes them:
Sum standard deviation
Given is a population with a normally distributed random variable $X$. When you have a sample $n$ from this population the population is:
$X_{sum} = X_1 + X_2 ... + X_n$ with
$\mu_{Xsum} = n \times \mu_x$ and $\sigma_{Xsum} = \sqrt{n} \times \sigma_x$.
Standard error
When you have a normally distributed random variable $X$ with mean $\mu_X$ and standard deviation $\sigma_X$ and sample length $n$, the sample mean $\bar{X}$ is normally distributed with $\mu_{\bar{x}} = \mu_X$ and $\sigma_{\bar{x}} = \dfrac{\sigma_X}{\sqrt{n}}$
These 2 are awefully similair to me to the point I can't at all decide which to use where. Here are the problems where I discovered I couldn't:
Problem 1
A filling machine fills bottles of lemonade. The amount is normally distributed with $\mu = 102 \space cl$.
$\sigma$ = $1.93\space cl$.
• Calculate the chance that out of 12 bottles the average volume is $100 \space cl$.
The problem itself is easy, however the troublesome part is what to choose for the standard deviation of the sample. Here they use $\dfrac{1.93}{\sqrt{12}}$ which I can live with, until I encountered the second problem.
Problem 2
A tea company puts 20 teabags in one package. The weight of a teabag is normally distributed with $\mu = 5.3 \space g$ and $\sigma = 0.5 \space g.$
• Calculate the chance that a package weighs less than 100 grams.
Here I thought they'd also use $\dfrac{0.5}{\sqrt{20}}$, but instead they use $\sqrt{20} \times 0.5$.
Can someone clear up the confusion?
• You should tag this as "homework" as well, since it seems to be a homework question. – Placidia Jan 20 '13 at 17:33
• @Placidia Are you kidding me?! This isn't homework, this is about understanding and differentiating 2 general concepts in statistics, which then could be implemented in homework questions.. like every other mathematical concept.. – JohnPhteven Jan 20 '13 at 17:36
• My textbook confused me; (Freedman, Pisani, Purves, Statistics, Fourth Edition). The chapter is titled "The Standard Error", uses $\sqrt{n} \times \sigma_x$ but the acronym is "SE", which must be a hint that my textbook is referring to "sum standard deviation"; indeed the textbook describes the experiment: "When drawing at random with replacement from a box of numbered tickets; the standard error for the sum of the draws is..." – The Red Pea Nov 7 '19 at 7:26
• ... more from the confusing part of Statistics textbook Freedman, Pisani, Purves, Fourth Edition: "In this book, we use SD for data and SE for chance quantities (random variables). This distinction is not standard and the term SD is often used in both situations" Indeed, as in this SE question, we refer to the "SD of X" ($\sigma_{X}$) and the "SD of sum of X" ($\sigma_{X_{sum}}$) – The Red Pea Nov 7 '19 at 7:37
The sum standard deviation is, as the name suggests, the standard deviation of the sum of $n$ random variables. The standard error you're talking about is just another name for the standard deviation of the mean of $n$ random variables. As you noted, the two formulas are closely related; since the sum of $n$ random variables is $n$ times the mean of $n$ random variables, the standard deviation of the sum is also $n$ times the standard deviation of the mean:
$\sigma_{X_{sum}} = \sqrt n\sigma_X = n \times \frac{\sigma_X}{\sqrt n} = n\times \sigma_\bar{X}$.
In the first problem you are dealing with a mean, the average of twelve bottles, so you use the standard deviation of the mean, which is called standard error. In the second problem you are dealing with a sum, the total weight of 20 packages, so you use the standard deviation of the sum.
Summary: use standard error when dealing with the mean (averages); use sum standard deviation when dealing with the sum (totals).
• But what I think is that theyask one about the sum of 12 bottles, and the mean of that sum? In other words, they're too similair to me.. – JohnPhteven Jan 20 '13 at 18:37
• There's no sum in question one. Each bottle is filled with an amount given by a normal distribution with mean 102, the question asks about the mean of twelve bottles. Where do you see a sum? – Jonathan Christensen Jan 20 '13 at 18:45
• Oh wait nevermind, I was being a little bit blind! In the first one they ask about the MEAN (i.e. average) out of a sample, in the second they transform a sample of 6 into '1' object (>namely, the box of teabags), with its own SD and M! – JohnPhteven Jan 20 '13 at 18:45
• I think I was writing my response the same time you were doing yours. Nice answer. – Placidia Jan 20 '13 at 18:46
• I meant 20, my brain is random, no idea how I got to 6 – JohnPhteven Jan 20 '13 at 18:48
The first standard deviation formula you gave is the SD for a sum. The standard error is the SD of the sample mean. Remember that: $\text{Var}(aX)=a^2 \text{Var}(X)$ and the variance of the sum is the sum of the variances (First formula). So
$\text{Var}(\bar{X})=\frac{n\sigma^2}{n^2}=\sigma^2/n$. Taking the square root gives the result.
Recall:
$\text{Var}(\sum X_i)=\sum (\text{Var}(X_i)=n \sigma^2.$ The Variance of the sums.
Problem 1 is looking for a statement about the sample mean; Problem 2 is about the sum, since the weight of the package is the sum of the weights of individual tea bags.
• Nice answer, +1, but I gave the other one a best answer since I read it first and it answered my question first. – JohnPhteven Jan 20 '13 at 18:48 | 2020-07-07T02:56:40 | {
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https://math.stackexchange.com/questions/558145/minimum-number-of-iterations-in-newtons-method-to-find-a-square-root | Minimum number of iterations in Newton's method to find a square root
I am writing an algorithm that evaluates the square root of a positive real number $y$. To do this I am using the Newton-Raphton method to approximate the roots to $f(x)=x^2-y$. The $n^{th}$ iteration gives $$x_n=\frac{x_{n-1}^2+y}{2x_{n-1}}$$ as an approximation to $\sqrt{y}$. I found that starting with an initial guess $x_0=1$ works pretty well generally, so an answer to the question below that assumes $x_0=1$ is fine.
My question: is there an exact expression for the minimum $N$ of iterations needed to attain a given precision $p$ in the approximate solution $x_N$? In other words I'm looking for the smallest integer $N$ such that $$\left|\frac{x_N-\sqrt y}{\sqrt y}\right|<p.$$
I've thought about this for a while and played around with the expression for the errors $\epsilon_n = x_n - \sqrt y$ which can be shown to satisfy $\epsilon_{n+1}=\epsilon_n^2\,/\,2x_n$, but I can't find an answer. I've looked around on Google but I couldn't find an answer either.
Any pointers to a solution online or help would be much appreciated. A follow-up would of course be: can $x_0$ be optimised (while being a simple enough expression in terms of $y$) in order to minimise $N$?
• I don't have an answer to your overall question, but here's a little bit of intuition: $x_n$ is the average of $x_{n-1}$ and $y/x_{n-1}$. One of these numbers must be less than $\sqrt{y}$, and one must be greater, so the average is a reasonable update. And of course, the closer $x_0$ is to $\sqrt{y}$, the better. Something like $x_0 = \lfloor \sqrt{y}\rfloor$ is a good guess (take the biggest perfect square less than $y$, and start with its square root). – BaronVT Nov 9 '13 at 16:58
• Try analyzing $x_n^2 - y$ rather than $x_n - \sqrt{y}$. I imagine you probably want to analyze the roundoff error too, rather than assuming addition, multiplication and division are exact, which makes the problem more complicated. – Hurkyl May 2 '14 at 8:21
There is such a formula: consider
$$\frac{x_n+\sqrt y}{x_n-\sqrt y}=\frac{\frac{x_{n-1}^2+y}{2x_{n-1}}+\sqrt y}{\frac{x_{n-1}^2+y}{2x_{n-1}}-\sqrt y}=\frac{(x_{n-1}+\sqrt y)^2}{(x_{n-1}-\sqrt y)^2}=\left(\frac{x_{n-1}+\sqrt y}{x_{n-1}-\sqrt y}\right)^2.$$
By recurrence,
$$\frac{x_n+\sqrt y}{x_n-\sqrt y}=\left(\frac{x_{0}+\sqrt y}{x_{0}-\sqrt y}\right)^{2^n}.$$
If you want to achieve $2^{-b}$ relative accuracy, $x_n=(1+2^{-b})\sqrt y$,
$$2^n=\frac{\log_2\frac{(1+2^{-b})\sqrt y+\sqrt y}{(1+2^{-b})\sqrt y-\sqrt y}}{\log_2\left|\frac{x_{0}+\sqrt y}{x_{0}-\sqrt y}\right|},$$
$$n=\log_2\left(\log_2\frac{2+2^{-b}}{2^{-b}}\right)-\log_2\left(\log_2\left|\frac{x_{0}+\sqrt y}{x_{0}-\sqrt y}\right|\right).$$
The first term relates to the desired accuracy. The second is a penalty you pay for providing an inaccurate initial estimate.
If the floating-point representation of $y$ is available, a very good starting approximation is obtained by setting the mantissa to $1$ and halving the exponent (with rounding). This results in an estimate which is at worse a factor $\sqrt 2$ away from the true square root.
$$n=\log_2\left(\log_2\left(2^{b+1}+1\right)-\log_2\left(\log_2\frac{\sqrt 2+1}{\sqrt 2-1}\right)\right) \approx\log_2(b+1)-1.35.$$ In the case of single precision (23 bits mantissa), 4 iterations are always enough. For double precision (52 bits), 5 iterations.
On the opposite, if $1$ is used as a start and $y$ is much larger, $\log_2\left|\frac{1+\sqrt y}{1-\sqrt y}\right|$ is close to $\frac{2}{\ln(2)\sqrt y}$ and the formula degenerates to $$n\approx\log_2(b+1)+\log_2(\sqrt y)-1.53.$$ Quadratic convergence is lost as the second term is linear in the exponent of $y$.
If your starting value of $y$ is correctly "scaled" between $1$ and $100$, then a good initial guess is
$X_0 = 1,1545 + 0,11545*y$ (linear approx) or if you prefer a second order approx. you can use:
$X_0 = 1,0 + y*( 0,18 - 0,0009*y)$
Using $X_0$, you need only 3 iterations (Newton / Heron) to get a Relative Error less than 0,0001 % !
Now, here is a fast method to get an excellent $X_0$ (with less than 1% error):
Store a pre-calculated constants table containing all the square roots with rounded 4 decimals in the range $[1 ; 100]$. So $table(i)=sqrt(i)$.
Use Simple Linear Interpolation to get a good $X_0$. Now, the Maximum Relative Error is still greater than 1% ! ( 1.5% near $y=1.4142$).
You can simply divide this Maximum Relative Error (MRE) by a factor of 2 by doing this trick:
Simply replace the 1st entry ($\sqrt(1)=1.0000$ by $\sqrt(1)=1.0075$ ) AND the second entry ( $\sqrt(2)=1.4142$ by $\sqrt(2)=1.4248$ ). That's all. You have a MRE of 0,75%! So, the WORST case is a $X_0$ with only 0.75% in the WHOLE RANGE $[1 ; 100]$.
Unscale your number, and with only ONE Newton / Heron iteration get a 0.0028% MRE in any case, and with two iteration you have a 0.00000004% MRE!!
That's Fast, Simple and very Accurate. Have fun. | 2019-07-21T09:39:28 | {
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https://cs.stackexchange.com/questions/105563/expected-value-of-the-distance-between-nodes-in-a-binary-tree | # Expected value of the distance between nodes in a binary tree
If there are 16 leaves in a full binary tree and two nodes $$a$$ and $$b$$ chosen at random, then what is the expected value of the distance between $$a$$ and $$b$$ in T?
My question here is, how do I correctly approach this question?
(Also tell me how to develop numerical skill in algorithm?)
Let's name the nodes $$\mathit{node0}$$ through $$\mathit{node15}$$. (with the implication that that's the order of the nodes)
I would approach this problem first by saying "if $$a$$ is $$\mathit{node0}$$, then what's the expected value of the distance between $$a$$ and $$b$$?" Then I'd ask the question "what if $$a$$ is $$\mathit{node1}$$?", etc. Presumably along the way there would be patterns and symmetries I could make use of so I wouldn't need to do a long calculation 16 times.
So let's take that first question: "if $$a$$ is $$\mathit{node0}$$, then what's the expected value of the distance between $$a$$ and $$b$$?"
First off, the distance between $$\mathit{node0}$$ and $$\mathit{node0}$$ is $$0$$. To $$\mathit{node1}$$, the distance is $$2$$. To nodes $$\mathit{node2}$$ or $$\mathit{node3}$$, the distance is $$4$$. To the four nodes $$\mathit{node4}$$ through $$\mathit{node7}$$, the distance is $$6$$. To the remaining eight nodes the distance is $$8$$.
So the expected distance when $$a$$ is $$\mathit{node0}$$ is $$(0 + 2 + 2*4 + 4*6 + 8*8)/16 = 6.125$$.
Now figure out the expected distance when $$a$$ is $$\mathit{node1}$$, and then other nodes. (there is an obvious pattern - prove it)
Then average over those results to find the answer.
Often, you will find a shorter or more elegant solution to a problem as you are working on it. This is normal, and should not be taken as a sign that your initial approach was wrong or that you should have seen the more elegant approach without trying your initial approach first, any more than a writer should expect to write an essay without first writing some notes or a rough draft.
Unfortunately, though writers are often trained in the idea of rough drafts and revisions, solutions to math or computer science problems are usually only presented in their final form without much indication of the bumbling around and blind corners that really went into discovering the elegant explanation.
• "To $\mathit{node1}$, the distance is $2$. To nodes $\mathit{node2}$ or $\mathit{node3}$, the distance is $4$." how?? Can u draw how node1 getting distance 2 and node2 and node3 both getting distance 4?? – Srestha Mar 14 at 16:23 | 2019-10-14T17:55:14 | {
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http://stats.stackexchange.com/questions/4959/how-do-you-calculate-the-expectation-of-left-sum-i-1n-x-i-right2 | # How do you calculate the expectation of $\left(\sum_{i=1}^n {X_i} \right)^2$?
If $X_i$ is exponentially distributed $(i=1,...,n)$ with parameter $\lambda$ and $X_i$'s are mutually independent, what is the expectation of
$$\left(\sum_{i=1}^n {X_i} \right)^2$$
in terms of $n$ and $\lambda$ and possibly other constants?
Note: This question has gotten a mathematical answer on http://math.stackexchange.com/q/12068/4051. The readers would take a look at it too.
-
The two copies of this question reference each other and, appropriately, the stats site (here) has a statistical answer and the math site has a mathematical answer. It seems like a good division: let it stand! – whuber Mar 4 '11 at 21:59
If $x_i \sim Exp(\lambda)$, then (under independence), $y = \sum x_i \sim Gamma(n, 1/\lambda)$, so $y$ is gamma distributed (see wikipedia). So, we just need $E[y^2]$. Since $Var[y] = E[y^2] - E[y]^2$, we know that $E[y^2] = Var[y] + E[y]^2$. Therefore, $E[y^2] = n/\lambda^2 + n^2/\lambda^2 = n(1+n)/\lambda^2$ (see wikipedia for the expectation and variance of the gamma distribution).
-
+1 Nice answer! – whuber Nov 27 '10 at 17:10
Thanks. A very neat way of answering the question (leading to the same answer) was also provided on math.stackexchange (link above in the question) a few minutes ago. – Wolfgang Nov 27 '10 at 17:24
The math answer computes the integrals using linearity of expectation. In some ways it's simpler. But I like your solution because it exploits statistical knowledge: because you know a sum of independent Exponential variables has a Gamma distribution, you're done. – whuber Nov 27 '10 at 21:19
I enjoyed it quite a bit and I am by no means a statistician or a mathematician. – Kortuk Nov 29 '10 at 18:04
very elegant answer. – Cyrus S Nov 30 '10 at 16:44
The answer above is very nice and completely answers the question but I will, instead, provide a general formula for the expected square of a sum and apply it to the specific example mentioned here.
For any set of constants $a_1, ..., a_n$ it is a fact that
$$\left( \sum_{i=1}^{n} a_i \right)^2 = \sum_{i=1}^{n} \sum_{j=1}^{n} a_{i} a_{j}$$
this is true by the Distributive property and becomes clear when you consider what you're doing when you calculate $(a_1 + ... + a_n) \cdot (a_1 + ... + a_n)$ by hand.
Therefore, for a sample of random variables $X_1, ..., X_n$, regardless of the distributions,
$$E \left( \left[ \sum_{i=1}^{n} X_i \right]^2 \right) = E \left( \sum_{i=1}^{n} \sum_{j=1}^{n} X_i X_j \right) = \sum_{i=1}^{n} \sum_{j=1}^{n} E(X_i X_j)$$
provided that these expectations exist.
In the example from the problem, $X_1, ..., X_n$ are iid ${\rm exponential}(\lambda)$ random variables, which tells us that $E(X_{i}) = 1/\lambda$ and ${\rm var}(X_i) = 1/\lambda^2$ for each $i$. By independence, for $i \neq j$, we have
$$E(X_i X_j) = E(X_i) \cdot E(X_j) = \frac{1}{\lambda^2}$$
There are $n^2 - n$ of these terms in the sum. When $i = j$, we have
$$E(X_i X_j) = E(X_{i}^{2}) = {\rm var}(X_{i}) + E(X_{i})^2 = \frac{2}{\lambda^2}$$
and there are $n$ of these term in the sum. Therefore, using the formula above,
$$E \left( \sum_{i=1}^{n} X_i \right)^2 = \sum_{i=1}^{n} \sum_{j=1}^{n} E(X_i X_j) = (n^2 - n)\cdot\frac{1}{\lambda^2} + n \cdot \frac{2}{\lambda^2} = \frac{n^2 + n}{\lambda^2}$$
-
This problem is just a special case of the much more general problem of 'moments of moments' which are usually defined in terms of power sum notation. In particular, in power sum notation:
$$s_1 = \sum_{i=1}^{n} X_i$$
Then, irrespective of the distribution, the original poster seeks $E[s_1^2]$ (provided the moments exist). Since the expectations operator is just the 1st Raw Moment, the solution is given in the mathStatica software by:
[ The '___ToRaw' means that we want the solution presented in terms of raw moments of the population (rather than say central moments or cumulants). ]
Finally, if $X$ ~ Exponential($\lambda$) with pdf $f(x)$:
f = Exp[-x/λ]/λ; domain[f] = {x, 0, ∞} && {λ > 0};
then we can replace the moments $\mu_i$ in the general solution sol with the actual values for an Exponential random variable, like so:
All done.
P.S. The reason the other solutions posted here yield an answer with $\lambda^2$ in the denominator rather than the numerator is, of course, because they are using a different parameterisation of the Exponential distribution. Since the OP didn't state which version he was using, I decided to use the standard distribution theory textbook definition Johnson Kotz et al … just to balance things out :)
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http://mathhelpforum.com/statistics/29269-help-probability-flipping-quarter.html | # Thread: Help with Probability flipping a quarter!!!!
1. ## Help with Probability flipping a quarter!!!!
So here is the problem. Maybe someone can help me. Thanks ahead of time anyone.
John and Judy play a game of flipping a quarter. If it comes up heads, John gets a point and if it comes up tails, Judy gets a point. They each bet $5 to play and the first person to get ten points wins the whole$10. At a point where John has 7 points and Judy has 5 points, the game is interrupted and they can't continue. John thinks he should win since he's ahead while Judy thinks she should win since she'd make a comeback. You are their friend, they agree to let you decide how to split up the money between them. Come up with a split that is most fair to both John and Judy.
2. Originally Posted by 2y4life
So here is the problem. Maybe someone can help me. Thanks ahead of time anyone.
John and Judy play a game of flipping a quarter. If it comes up heads, John gets a point and if it comes up tails, Judy gets a point. They each bet $5 to play and the first person to get ten points wins the whole$10. At a point where John has 7 points and Judy has 3 points, the game is interrupted and they can't continue. John thinks he should win since he's ahead while Judy thinks she should win since she'd make a comeback. You are their friend, they agree to let you decide how to split up the money between them. Come up with a split that is most fair to both John and Judy.
Calculate the probability $\displaystyle p_{John}$ that John would have won from this position.
Calculate the probability $\displaystyle p_{Judy}$ that Judy would have won from this position.
Then give John $\displaystyle \text{\$}10 \, p_{John}$and give Judy$\displaystyle \text{\$}10 \, p_{Judy}$.
3. Originally Posted by mr fantastic
Calculate the probability $\displaystyle p_{John}$ that John would have won from this position.
Calculate the probability $\displaystyle p_{Judy}$ that Judy would have won from this position.
Then give John $\displaystyle \text{\$}10 p_{John}$and give Judy$\displaystyle \text{\$}10 p_{Judy}$.
Yea...but is there a more simple way of finding that out? I tried and there are too many possibilities. John could flip another head and then flip 5 tails in a row and Judy would win or vice versa.
4. Originally Posted by 2y4life
So here is the problem. Maybe someone can help me. Thanks ahead of time anyone.
John and Judy play a game of flipping a quarter. If it comes up heads, John gets a point and if it comes up tails, Judy gets a point. They each bet $5 to play and the first person to get ten points wins the whole$10. At a point where John has 7 points and Judy has 5 points, the game is interrupted and they can't continue. John thinks he should win since he's ahead while Judy thinks she should win since she'd make a comeback. You are their friend, they agree to let you decide how to split up the money between them. Come up with a split that is most fair to both John and Judy.
This question is historically interesting ..... read this to see why.
5. Originally Posted by 2y4life
Yea...but is there a more simple way of finding that out? I tried and there are too many possibilities. John could flip another head and then flip 5 tails in a row and Judy would win or vice versa.
Let X be the random variable number of points John wins.
Let Y be the random variable number of points Judy wins.
Then:
Pr(John wins from this position) = Pr(X = 3) times Pr(Y < 5).
Pr(Judy wins from this position) = Pr(X < 3) times Pr(Y = 5).
X and Y both follow binomial distributions.
Edit: Actually it's slightly more complicated than this I just realised but I have to rush off now. I'll have more to say later unless someone else gets in first.
6. Originally Posted by mr fantastic
This question is historically interesting ..... read this to see why.
Yea, this was an extra credit problem for us to take home and figure out and my TA said that this problem came from two mathematicians named Blaise and Pierre.
I want to put this down as the answer but I'm not sure if this would constitute as a fair answer:
John- 7/12 x $10=$5.83
Judy- 5/12 x $10=$4.17
7. Originally Posted by mr fantastic
Let X be the random variable number of points John wins.
Let Y be the random variable number of points Judy wins.
Then:
Pr(John wins from this position) = Pr(X = 3) times Pr(Y < 5).
Pr(Judy wins from this position) = Pr(X < 3) times Pr(Y = 5).
X and Y both follow binomial distributions.
Edit: Actually it's slightly more complicated than this I just realised but I have to rush off now. I'll have more to say later unless someone else gets in first.
You only need to work out one of John winning or Judy winning (you know why, right).
Probability of John winning: You need to calculate Pr(X = 3) times Pr(Y < 5) for n = 3, 4, 5, 6 and 7.
Probability of Judy winning: You need to calculate Pr(X < 3) times Pr(Y = 5) for n = 5, 6 and 7. | 2018-06-25T10:56:08 | {
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https://math.stackexchange.com/questions/2915495/dimension-of-vector-space-countable-uncountable | # Dimension of vector space, countable, uncountable?
In set theory, when we talk about cardinality of a set we have notions like finite set, countably infinite and uncountably infinite sets.
Main Question
Let's talk about dimension of a vector space. In Linear Algebra I have always heard about two terms either a vector space is finite dimensional for example $\mathbb{R}^n$ or infinite dimensional for example $C[0,1]$.
Why don't we have notions like countably infinite dimensional vector space and uncountably infinite dimensional vector space.
May be, I am not able to see the bigger picture.
Extras
P.S. Long time ago, I was in a talk on Enumerative Algebraic geometry and the professor said, I always think of a positive natural number as a dimension of some vector space.
Don't we have then uncountable dimension vector space?/
• Some infinite-dimensional vector spaces have countable dimension, some have uncountable dimension. The dimension of a vector space is a well-defined cardinality. So, what's the question? – Lord Shark the Unknown Sep 13 '18 at 11:59
• I asked this as I have never heard of terms like countable dimension or uncountable dimension in books. @LordSharktheUnknown – StammeringMathematician Sep 13 '18 at 12:01
• An infinite countable dimmensionnal spaces cannot be complete but they exist. Consider the vector space of real sequences having finite support. – nicomezi Sep 13 '18 at 12:14
• "I have never heard of terms like countable dimension or uncountable dimension in books" --- This is because beginning and intermediate level linear algebra books rarely distinguish more precisely than "finite dimension" and "infinite dimension". It's usually only in graduate level algebra courses (e.g. probably all of the "third level" books in my answer to High-level linear algebra book) where you'll find the various notions of "algebraic dimension" defined as a cardinal number. – Dave L. Renfro Sep 13 '18 at 12:31
• Just because you don't see it in your text doesn't mean it doesn't exist. You're quite right that it makes sense! – Noah Schweber Sep 13 '18 at 12:31
We do have the notions of countable/uncountable dimensions. Just as a set can be finite or infinite (without specifying which infinite cardinality the set as) a vector space can be finite dimensional or infinite dimensional. We can then go one step more and ask, if the dimension is infinite, which infinite cardinal is it?
The definition of dimension of a vector space is the cardinality of a basis for that vector space (it does not matter which basis you take, because they all have the same cardinality). Then for any cardinal number $\gamma$, you can have a vector space with that dimension. For example, if $\Gamma$ is a set with cardinality $\gamma$, let $c_{00}(\Gamma)$ be the space of all $\mathbb{F}$-valued functions $f$ such that $$\text{supp}(f)=\{x\in \Gamma: f(x)\neq 0\}$$ is finite. Then let $\delta_x\in c_{00}(\Gamma)$ be the function such that $\delta_x(y)=0$ if $y\neq x$ and $\delta_x(y)=1$ if $y=x$. Then $(\delta_x)_{x\in \Gamma}$ is a basis for $c_{00}(\Gamma)$ with cardinality $\gamma$. If $\Gamma=\mathbb{N}$, we have a vector space with countably infinite dimension. If $\Gamma=\mathbb{R}$, we have a vector space with dimension equal to the cardinality of the continuum.
However, for infinite dimensional topological vector spaces (and for infinite dimensional Hilbert and Banach spaces in particular) the usual notion of a basis of limited use. This is because the coordinate functionals for an infinite basis do not interact very well with the topology (one can show that if $(e_i, e^*_i)_{i\in I}$ is a basis together with its coordinate functionals for an infinite dimensional Banach space, then only finitely many of the functionals $e^*_i$ can be continuous). Since the notion of a basis is not as useful in the infinite dimensional topological space case as it is in the finite dimensional case, you can see less emphasis on what the exact dimension is in this case.
However, in this situation you get into discussions of other types of coordinate systems (such as Schauder bases, FDDs, unconditional bases, etc.), which are different from the notion of an (algebraic) basis. You also can ask about density character instead of dimension, which is the smallest cardinality of a dense subset. This encodes topological information, while the purely algebraic notion of a basis does not. For example, infinite dimensional Hilbert space $\ell_2$ has no countable basis, it does have a countable, dense subset. So the dimension is that of the continuum, but the density character is $\aleph_0$.
The dimension of a vector space is the cardinality of a basis for that vector space. To say that a vector space has finite dimension therefore means that the cardinality of a basis for that vector space is finite. Since finite cardinalities are the same thing as natural numbers, we are safe in saying, for finite dimensional vector spaces, that the dimension is a natural number.
In general, some sets are countably infinite and some sets are uncountably infinite. So, applying this to those sets which happen to be bases of vector spaces, some vector spaces have countably infinite bases and therefore countably infinite dimension, and other vector space have uncountable infinite bases and therefore uncountably infinite dimension.
An example of a vector space over $\mathbb R$ of countably infinite dimension is $\mathbb R^{\infty}$ which is the space of infinite sequences of real numbers such that all but finitely terms in the sequence are equal to $0$. A countably infinite basis consists of $(1,0,0,0,...)$, $(0,1,0,0,...)$, $(0,0,1,0,...)$ and so on.
An example of a vector space over $\mathbb R$ of uncountably infinite dimension is the one you mention in your question, $C[0,1]$.
• That's okay. I do often write things wrong the first time. – Lee Mosher Sep 13 '18 at 12:18
• ... all (but finitely many) terms are zero. – Lee Mosher Sep 13 '18 at 12:19
• Had a hard time understanding your sentence, sorry for incovenience. – nicomezi Sep 13 '18 at 12:22
For the sake of a real world example: The electron in a hydrogen atom can take on countably many states. Each state is a basis vector for the span of possible electron states of hydrogen. There's are bound states and bound states are typically discrete. The relevant Schrodinger Equation also permits scattering states which have a continuous spectrum of possible energy states implying an uncountable vector space spanning the possible scattering states.
The following theorem is an example where we need to distinguish between countably/uncountably infinite dimensional vector spaces. Some important theorems, such as Hilbert Nullstallensatz, can be deduced from it.
Let $$A$$ be an associative, not necessarily commutative, $$\mathbb{C}-$$algebra with unit. For $$a \in A$$ define $$\text{Spec } a = \{\lambda \in \mathbb{C} | a-\lambda \text{ is not invertible}\}$$ Assume that $$A$$ has no more than countable dimension over $$\mathbb{C}$$. Then
(a) If $$A$$ is a division algebra, then $$A=\mathbb{C}$$
(b) For all $$a \in A$$ we have $$\text{Spec } A \neq \emptyset$$; furthermore, $$a \in A$$ is nilpotent if and only if $$\text{Spec } A = \{0\}$$
(Adapted from Representation Theory and Complex Geometry by Chriss and Ginzburg, theorem 2.1.1.)
The proof uses the fact that for any $$a \in A$$, $$\{(a - \lambda)^{-1} | \lambda \in \mathbb{C}\}$$ is an uncountable family of elements of $$A$$. But $$A$$ has only countable dimension over $$\mathbb{C}$$, so they are not linearly indenpendent over $$\mathbb{C}$$.
A weak version of Hilbert Nullstellensatz:
Let $$A$$ be a finitely generated commutative algebra over $$\mathbb{C}$$. Then any maximal ideal of $$A$$ is the kernel of an algebra homomorphism $$A \to \mathbb{C}$$
This follows directly from the above theorem: $$A$$ finitely generated $$\Longrightarrow$$ $$A$$ has countable dimension over $$\mathbb{C}$$ $$\Longrightarrow$$ $$A/\mathfrak{m}$$ has countable dimension over $$\mathbb{C}$$ $$\Longrightarrow$$ $$A/\mathfrak{m}=\mathbb{C}$$
(We can also deduce the strong version of Nullstellensatz from it but need more argument.)
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http://openstudy.com/updates/55a2f823e4b05670bbb5452a | ## anonymous one year ago HELP ME PLEASE Which of the following exponential functions goes through the points (1, 6) and (2, 12)? f(x) = 3(2)x f(x) = 2(3)x f(x) = 3(2)−x f(x) = 2(3)−x
1. ybarrap
Plug in x=1 and see if it equals 6 Plug in x=2 and see if it equals 12 Make sense?
2. anonymous
not really? can you guide me through it? (xD I dont want to be an answer-hogger/wanter.. I dont want the answer, just explanations :) ) @freckles
3. anonymous
Do you understand that coordinates are in the form (x,y)?
4. anonymous
yes. @BMF96
5. anonymous
f(x) = y
6. anonymous
What don't you understand?
7. jim_thompson5910
If $\Large f(x) = 7(3)^x$ (for example), then what is the value of f(x) when x = 2? In other words, what is f(2) equal to?
8. anonymous
f(2)=441?? #CONFUSED
9. jim_thompson5910
Replace each x with 2 $\Large f(x) = 7(3)^x$ $\Large f(2) = 7(3)^2$ $\Large f(2) = 7(9)$ $\Large f(2) = 63$ Do you see how I got f(2) to be equal to 63?
10. jim_thompson5910
btw you square first and then you multiply
11. anonymous
ok, I forgot that rule :) (like DUHR, Bella, get a grip)
12. jim_thompson5910
Since $$\Large f({\color{red}{2}}) = {\color{blue}{63}}$$ from my example, this means the point $$\Large ({\color{red}{x}},{\color{blue}{y}})=({\color{red}{2}},{\color{blue}{63}})$$ lies on the function curve of f(x)
13. anonymous
okayyy....
14. anonymous
$$f(x)=y= \begin{cases} 3(2)^x\\ 2(3)^x\\ 3(2)^{-x}\\ 2(3)^{-x} \end{cases}\qquad \qquad \begin{array}{llll} x&y \\\hline\\ {\color{brown}{ 1}}&3(2)^{\color{brown}{ 1}}\\ &2(3)^{\color{brown}{ 1}}\\ &3(2)^{-{\color{brown}{ 1}}}\\ &2(3)^{-{\color{brown}{ 1}}}\\ {\color{brown}{ 2}}&3(2)^{\color{brown}{ 2}}\\ &2(3)^{\color{brown}{ 2}}\\ &3(2)^{-{\color{brown}{ 2}}}\\ &2(3)^{-{\color{brown}{ 2}}} \end{array}$$
15. jim_thompson5910
so what ybarrap said at the top, you plug in each x coordinate into each function and see if you get the correct corresponding y coordinates Let's say we pick choice B at random $\Large f(x) = 2(3)^x$ The first point is (1,6). To test if this point lies on the function f(x) curve, we plug in x = 1 and see if y = 6 pops out $\Large f(x) = 2(3)^x$ $\Large f(1) = 2(3)^1$ $\Large f(1) = 2(3)$ $\Large f(1) = 6$ It does, so (1,6) is definitely on this curve. How about (2,12)? Let's check $\Large f(x) = 2(3)^x$ $\Large f(2) = 2(3)^2$ $\Large f(2) = 2(9)$ $\Large f(2) = 18$ Nope. The point (2,18) actually lies on this function curve and NOT (2,12). So we can rule out choice B.
16. jim_thompson5910
So what just happened was that I've proven that the function $$\Large f(x) = 2(3)^x$$ does NOT go through both points (1,6) and (2,12).
17. anonymous
okay, so its A, C, or D lol.... so lets try to rule out A...
18. anonymous
@jim_thompson5910
19. jim_thompson5910
what did you get so far in checking choice A?
20. anonymous
that A is, in fact NOT the answer!?
21. jim_thompson5910
if x = 1, then what is f(1) ?
22. anonymous
f
23. anonymous
Wait, so it IS A!!!!!
24. jim_thompson5910
$\Large f(x) = 3(2)^x$ $\Large f(1) = 3(2)^1$ $\Large f(1) = \underline{ \ \ \ \ \ \ \ } \text{ (fill in the blank)}$
25. anonymous
6
26. jim_thompson5910
27. anonymous
12
28. jim_thompson5910
Good. Choice A is definitely the answer. As practice, why not go through C and D and eliminate them. With choice C, if x = 1, then what is f(x) equal to? | 2016-10-21T22:13:56 | {
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https://stats.stackexchange.com/questions/214127/simulating-random-variables-from-a-discrete-distribution | # Simulating random variables from a discrete distribution
I have the following discrete distribution where $p$ is a known constant:
$p(x,p)= \frac{(1-p)^3}{p(1+p)}x^2p^x , (0<p<1), x=0, 1, 2, \ldots$ .
How can I sample from this distribution?
– Tim
May 23, 2016 at 15:03
• Does this even sum to 1 for some value of $p$? (I doubt so) May 23, 2016 at 15:19
• @Elvis It does sum to 1. May 23, 2016 at 15:41
• @MatthewGunn My bad. May 24, 2016 at 6:43
This answer develops a simple procedure to generate values from this distribution. It illustrates the procedure, analyzes its scope of application (that is, for which $p$ it might be considered a practical method), and provides executable code.
### The Idea
Because
$$x^2 = 2\binom{x}{2} + \binom{x}{1},$$
consider the distributions $f_{p;m}$ given by
$$f_{p;m}(x) \propto \binom{x}{m-1}p^x$$
for $m=3$ and $m=2$.
A recent thread on inverse sampling demonstrates that these distributions count the number of observations of independent Bernoulli$(1-p)$ variables needed before first seeing $m$ successes, with $x+1$ equal to that number. It also shows that the normalizing constant is
$$C(p;m)=\sum_{x=m-1}^\infty \binom{x}{m-1}p^x = \frac{p^{m-1}}{(1-p)^m}.$$
Consider the probabilities in the question,
$$x^2 p^x = \left( 2\binom{x}{2} + \binom{x}{1} \right)p^x = 2 \binom{x}{2}p^x + \binom{x}{1} p^x =2 C(p;3) f_{p;3}(x) + C(p;2) f_{p;2}(x).$$
Consequently, the given distribution is a mixture of $f_{p;3}$ and $f_{p;2}$. The proportions are as $$2C(p;3):C(p;2) = 2p:(1-p).$$ It is simple to sample from a mixture: generate an independent uniform variate $u$ and draw $x$ from $f_{p;2}$ when $u \lt (1-p)/(2p+1-p)$; that is, when $u(1+p) \lt 1-p$, and otherwise draw $x$ from $f_{p;3}$.
(It is evident that this method generalizes: many probability distributions where the chance of $x$ is of the form $P(x)p^x$ for a polynomial $P$, such as $P(x)=x^2$ here, can be expressed as a mixture of these inverse-sampling distributions.)
### The Algorithm
These considerations lead to the following simple algorithm to generate one realization of the desired distribution:
Let U ~ Uniform(0,1+p)
If (U < 1-p) then m = 2 else m = 3
x = 0
While (m > 0) {
x = x + 1
Let Z ~ Bernoulli(1-p)
m = m - Z
}
Return x-1
These histograms show simulations (based on 100,000 iterations) and the true distribution for a range of values of $p$.
### Analysis
How efficient is this? The expectation of $x+1$ under the distribution $f_{p;m}$ is readily computed; it equals $m/(1-p)$. Therefore the expected number of trials (that is, values of Z to generate in the algorithm) is
$$\left((1-p) \frac{2}{1-p} + (2p) \frac{3}{1-p}\right) / (1-p+2p) = 2 \frac{1+2p}{1-p^2}.$$
Add one more for generating U. The total is close to $3$ for small values of $p$. As $p$ approaches $1$, this count asymptotically is
$$1 + 2\frac{1 + 2p}{(1-p)(1+p)} \approx \frac{3}{1-p}.$$
This shows us that the algorithm will, on the average, be reasonably quick for $p \lt 2/3$ (taking up to ten easy steps) and not too bad for $p \lt 0.97$ (taking under a hundred steps).
### Code
Here is the R code used to implement the algorithm and produce the figures. A $\chi^2$ test will show that the simulated results do not differ significantly from the expected frequencies.
sample <- function(p) {
m <- ifelse(runif(1, max=1+p) < 1-p, 2, 3)
x <- 0
while (m > 0) {
x <- x + 1
m <- m - (runif(1) > p)
}
return(x-1)
}
n <- 1e5
set.seed(17)
par(mfcol=c(2,3))
for (p in c(1/5, 1/2, 9/10)) {
# Simulate and summarize.
x <- replicate(n, sample(p))
y <- table(x)
# Compute the true distribution for comparison.
k <- as.numeric(names(y))
theta <- sapply(k, function(i) i^2 * p^i) * (1-p)^3 / (p^2 + p)
names(theta) <- names(y)
# Plot both.
barplot(y/n, main=paste("Simulation for", format(p, digits=2)),
border="#00000010")
barplot(theta, main=paste("Distribution for", format(p, digits=2)),
border="#00000010")
}
@dsaxton's approach is known as inverse transform sampling and is probably the way to go for a problem like this. To be a bit more explicit, the approach is:
1. Draw $u$ from uniform distribution on (0,1).
2. Compute $x = F^{-1}(u)$ where $F^{-1}$ is the inverse of the cumulative distribution function.
Computing $x = F^{-1}(u)$ is equivalent to finding the integer $x$ that is the solution to: $$\text{minimize} \quad x \quad \text{subject to} \sum_{j=0}^x \frac{(1 - p)^3}{p(1+p)} j^2p^j \geq u$$
Quick pseudo code to do this numerically:
1. Construct a vector $\boldsymbol{m}$ such that $m_j = \frac{(1 - p)^3}{p(1+p)} j^2p^j$.
2. Create a vector $\boldsymbol{c}$ such that $c_j = \sum_{k=0}^j m_j$.
3. Find the minimum index $x$ such that $c_x \geq u$.
Draw $u$ from a uniform$(0, 1)$ distribution and let $x$ be the smallest value of $k$ for which $\sum_{j=0}^{k} \frac{(1 - p)^3}{p (1 + p)} j^2 p^j > u$. Then $x$ will be a realization from the desired distribution.
• May be clearer to write $x = \min k \text{ subject to } \sum_j^k \ldots > u$. As written, it's not terribly clear what is being minimized and that $x$ equals the optimal $k$. May 23, 2016 at 16:10 | 2022-05-24T03:25:51 | {
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https://math.stackexchange.com/questions/2566206/notation-for-cartesian-product-except-one-set/2566431 | # Notation for cartesian product except one set?
Let's say I have a list of sets $S_i$, for $i=1,\ldots,n$. We often write the cartesian product of all these sets, with the exception of $S_k$ as:
$$S=S_1\times\cdots\times S_{k-1}\times S_{k+1}\times\cdots\times S_n$$
Is there a more succinct way to write it?
In Wikipedia (https://en.wikipedia.org/wiki/Cartesian_product), I found something, which might be what you are looking for: $\prod_{n=1}^k \Bbb{R} = \Bbb{R}\times \Bbb{R} \times\cdots\times \Bbb{R} = \Bbb{R}^k$. So maybe something like this one is also valid: $$\prod_{\scriptstyle i = 1\atop\scriptstyle i \ne k}^nS_i$$
where $S_i$ is the $i^\text{th}$ set of the list you mentioned.
• Possibly \prod_{i = 1\atop i \ne k}^n would be better? (with \atop separating lines instead of a comma) – CiaPan Dec 14 '17 at 11:15
• Yes, you are right. Thank you for the formatting :) – ArsenBerk Dec 14 '17 at 11:16
• I changed $\Bbb{R}$ x $\Bbb{R}$ x ... x $\Bbb{R}$ to $\Bbb{R} \times \Bbb{R} \times \cdots \times \Bbb{R}.$ That is proper MathJax usage. $\qquad$ – Michael Hardy Dec 14 '17 at 13:13
• In addition to the atop business, another way to notate the indices would be with $\in$. I could see that being especially convenient if these indices are already in a set or are repeated throughout the work. If $E=\{1,\cdots,n\}\setminus\{k\}$ then you can have $$\prod_{i\in E}S_i$$ – gen-ℤ ready to perish Dec 14 '17 at 13:42
• Another formatting comment: if you look carefully, your subscripts on the product symbol are in a smaller font than the superscript. IMHO it's better to avoid this by using \prod_{\scriptstyle i = 1\atop\scriptstyle i \ne k}^n: compare$$\prod_{i = 1\atop i \ne k}^nS_i\quad\hbox{and}\quad \prod_{\scriptstyle i = 1\atop\scriptstyle i \ne k}^nS_i$$ – David Dec 15 '17 at 1:36
I have seen a notation for this kind of construction during some of my math lectures (but can't find a reference right now). This was mostly in the context of differential forms (e.g. interior product with vector), but can be applied to your case: $$S_1\times \dotsm \times \widehat{S_k} \times\dotsm \times S_n := S_1\times \dotsm \times S_{k-1}\times S_{k+1} \times \dotsm S_n$$ The hat denotes the factor to be omitted. Note that this is not a universally standard notation, so even the professors that used it defined it at some point early in the lecture.
• This is what I would use, along with a parenthetical remark along the lines of "where the hat denotes the factor to be omitted." I very seldom see anything as formal as in the other answers. – Matthew Leingang Dec 14 '17 at 15:00
• Hatcher uses a notation similar to this in his text Algebraic Topology (see page 105 for an example). – Xander Henderson Dec 17 '17 at 2:52
In general, we can write
$$S_1 \times \dots \times S_n := \prod_{i=1}^n S_i$$
and then we can apply all conventions we are used to.
As for your question, this can be written as:
$$\prod_{i = 1 \atop i \neq k}^n S_i$$
Although it might not be common in set theory, it is common for game theorists to write $S_{-i}$ for $S_1 \times \cdots \times S_{i-1} \times S_{i+1} \times \cdots \times S_n$. See page 15 of chapter one of Osborne and Rubinstein's text on game theory, for example.
That notation is useful in game theory because, if $S_j$ represents the set of strategies available to player $j$, then one often needs to describe how all players except player $i$ have acted. Such a description will be a member of $S_i$. In particular, the notation becomes useful in defining a Nash equilibrium. See https://en.wikipedia.org/wiki/Nash_equilibrium. | 2020-10-01T13:42:49 | {
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http://radiojuventudnerja.es/f3y8d/euclidean-distance-between-two-vectors-32de3b | 2017) and the quantum hierarchical clustering algorithm based on quantum Euclidean estimator (Kong, Lai, and Xiong 2017) has been implemented. Applying the formula given above we get that: \begin{align} d(\vec{u}, \vec{v}) = \| \vec{u} - \vec{v} \| \\ d(\vec{u}, \vec{v}) = \| \vec{u} - \vec{w} +\vec{w} - \vec{v} \| \\ d(\vec{u}, \vec{v}) = \| (\vec{u} - \vec{w}) + (\vec{w} - \vec{v}) \| \\ d(\vec{u}, \vec{v}) \leq || (\vec{u} - \vec{w}) || + || (\vec{w} - \vec{v}) \| \\ d(\vec{u}, \vec{v}) \leq d(\vec{u}, \vec{w}) + d(\vec{w}, \vec{v}) \quad \blacksquare \end{align}, \begin{align} d(\vec{u}, \vec{v}) = \| \vec{u} - \vec{v} \| = \sqrt{(2-1)^2 + (3+2)^2 + (4-1)^2 + (2-3)^2} \\ d(\vec{u}, \vec{v}) = \| \vec{u} - \vec{v} \| = \sqrt{1 + 25 + 9 + 1} \\ d(\vec{u}, \vec{v}) = \| \vec{u} - \vec{v} \| = \sqrt{36} \\ d(\vec{u}, \vec{v}) = \| \vec{u} - \vec{v} \| = 6 \end{align}, Unless otherwise stated, the content of this page is licensed under. . The reason for this is because whatever the values of the variables for each individual, the standardized values are always equal to 0.707106781 ! The Euclidean distance between two random points [ x 1 , x 2 , . Determine the Euclidean distance between. The Pythagorean Theorem can be used to calculate the distance between two points, as shown in the figure below. The distance between two vectors v and w is the length of the difference vector v - w. There are many different distance functions that you will encounter in the world. In mathematics, the Euclidean distance between two points in Euclidean space is the length of a line segment between the two points. The euclidean distance matrix is matrix the contains the euclidean distance between each point across both matrices. A generalized term for the Euclidean norm is the L2 norm or L2 distance. Append content without editing the whole page source. u = < -2 , 3> . Solution. Wikidot.com Terms of Service - what you can, what you should not etc. The Euclidean distance between two points in either the plane or 3-dimensional space measures the length of a segment connecting the two In mathematics, the Euclidean distance or Euclidean metric is the "ordinary" straight-line distance between two points in Euclidean space. Otherwise, columns that have large values will dominate the distance measure. pdist2 is an alias for distmat, while pdist(X) is … You are most likely to use Euclidean distance when calculating the distance between two rows of data that have numerical values, such a floating point or integer values. If you want to discuss contents of this page - this is the easiest way to do it. If columns have values with differing scales, it is common to normalize or standardize the numerical values across all columns prior to calculating the Euclidean distance. To calculate the Euclidean distance between two vectors in Python, we can use the numpy.linalg.norm function: Applying the formula given above we get that: (2) \begin {align} d (\vec {u}, \vec {v}) = \| \vec {u} - \vec {v} \| = \sqrt { (2-1)^2 + (3+2)^2 + (4-1)^2 + (2-3)^2} \\ d (\vec {u}, \vec {v}) = \| \vec {u} - \vec {v} \| = \sqrt {1 + 25 + 9 + 1} \\ d (\vec {u}, \vec {v}) = \| \vec {u} - \vec {v} \| = \sqrt {36} \\ d (\vec {u}, \vec {v}) = \| \vec {u} - \vec {v} \| = 6 … Find out what you can do. 1 Suppose that d is very large. ml-distance-euclidean. Check out how this page has evolved in the past. By using this metric, you can get a sense of how similar two documents or words are. Computes the Euclidean distance between a pair of numeric vectors. View wiki source for this page without editing. A generalized term for the Euclidean norm is the L2 norm or L2 distance. Recall that the squared Euclidean distance between any two vectors a and b is simply the sum of the square component-wise differences. View and manage file attachments for this page. With this distance, Euclidean space becomes a metric space. Older literature refers to the metric as the Pythagorean metric. $d(\vec{u}, \vec{v}) = \| \vec{u} - \vec{v} \| = \sqrt{(u_1 - v_1)^2 + (u_2 - v_2)^2 ... (u_n - v_n)^2}$, $d(\vec{u}, \vec{v}) = d(\vec{v}, \vec{u})$, $d(\vec{u}, \vec{v}) = || \vec{u} - \vec{v} || = \sqrt{(u_1 - v_1)^2 + (u_2 - v_2)^2 ... (u_n - v_n)^2}$, $d(\vec{v}, \vec{u}) = || \vec{v} - \vec{u} || = \sqrt{(v_1 - u_1)^2 + (v_2 - u_2)^2 ... (v_n - u_n)^2}$, $(u_i - v_i)^2 = u_i^2 - 2u_iv_i + v_i^2 = v_i^2 - 2u_iv_i + 2u_i^2 = (v_i - u_i)^2$, $\vec{u}, \vec{v}, \vec{w} \in \mathbb{R}^n$, $d(\vec{u}, \vec{v}) \leq d(\vec{u}, \vec{w}) + d(\vec{w}, \vec{v})$, Creative Commons Attribution-ShareAlike 3.0 License. The primary takeaways here are that the Euclidean distance is basically the length of the straight line that's connects two vectors. 3.8 Digression on Length and Distance in Vector Spaces. Using our above cluster example, we’re going to calculate the adjusted distance between a … Euclidean metric is the “ordinary” straight-line distance between two points. Euclidean distance between two vectors, or between column vectors of two matrices. , y d ] is radicaltp radicalvertex radicalvertex radicalbt d summationdisplay i =1 ( x i − y i ) 2 Here, each x i and y i is a random variable chosen uniformly in the range 0 to 1. This victory. The corresponding loss function is the squared error loss (SEL), and places progressively greater weight on larger errors. The shortest path distance is a straight line. And now we can take the norm. Computes the Euclidean distance between a pair of numeric vectors. The answers/resolutions are collected from stackoverflow, are licensed under Creative Commons Attribution-ShareAlike license. We will now look at some properties of the distance between points in $\mathbb{R}^n$. It corresponds to the L2-norm of the difference between the two vectors. The Euclidean distance between 1-D arrays u and v, is defined as maximum: Maximum distance between two components of x and y (supremum norm) manhattan: Absolute distance between the two vectors (1 … I have the two image values G= [1x72] and G1 = [1x72]. And these is the square root off 14. A little confusing if you're new to this idea, but it … D = √ [ ( X2-X1)^2 + (Y2-Y1)^2) Where D is the distance. We determine the distance between the two vectors. if p = (p1, p2) and q = (q1, q2) then the distance is given by. u of the two vectors. Notify administrators if there is objectionable content in this page. Solution to example 1: v . is: Deriving the Euclidean distance between two data points involves computing the square root of the sum of the squares of the differences between corresponding values. ... Percentile. u, is v . Let’s assume OA, OB and OC are three vectors as illustrated in the figure 1. It can be computed as: A vector space where Euclidean distances can be measured, such as , , , is called a Euclidean vector space. General Wikidot.com documentation and help section. The squared Euclidean distance is therefore d(x SquaredEuclideanDistance is equivalent to the squared Norm of a difference: The square root of SquaredEuclideanDistance is EuclideanDistance : Variance as a SquaredEuclideanDistance from the Mean : Euclidean distance, Euclidean distance. . So there is a bias towards the integer element. The following formula is used to calculate the euclidean distance between points. View/set parent page (used for creating breadcrumbs and structured layout). Two squared, lost three square until as one. $\vec {u} = (2, 3, 4, 2)$. I need to calculate the two image distance value. The distance between two points is the length of the path connecting them. Source: R/L2_Distance.R Quickly calculates and returns the Euclidean distances between m vectors in one set and n vectors in another. {\displaystyle \left\|\mathbf {a} \right\|= {\sqrt {a_ {1}^ {2}+a_ {2}^ {2}+a_ {3}^ {2}}}} which is a consequence of the Pythagorean theorem since the basis vectors e1, e2, e3 are orthogonal unit vectors. Most vector spaces in machine learning belong to this category. Accepted Answer: Jan Euclidean distance of two vector. By using this formula as distance, Euclidean space becomes a metric space. $\vec {v} = (1, -2, 1, 3)$. Each set of vectors is given as the columns of a matrix. Installation $npm install ml-distance-euclidean. Euclidean distance.$\begingroup$Even in infinitely many dimensions, any two vectors determine a subspace of dimension at most$2$: therefore the (Euclidean) relationships that hold in two dimensions among pairs of vectors hold entirely without any change at all in any number of higher dimensions, too. Squared Euclidean Distance, Let x,yâRn. This system utilizes Locality sensitive hashing (LSH) [50] for efficient visual feature matching.$\endgroup$– whuber ♦ Oct 2 '13 at 15:23 The formula for this distance between a point X ( X 1 , X 2 , etc.) In simple terms, Euclidean distance is the shortest between the 2 points irrespective of the dimensions. Y = cdist(XA, XB, 'sqeuclidean') The Euclidean distance between two vectors, A and B, is calculated as: Euclidean distance = √ Σ(A i-B i) 2. You want to find the Euclidean distance between two vectors. We can then use this function to find the Euclidean distance between any two vectors: #define two vectors a <- c(2, 6, 7, 7, 5, 13, 14, 17, 11, 8) b <- c(3, 5, 5, 3, 7, 12, 13, 19, 22, 7) #calculate Euclidean distance between vectors euclidean(a, b) [1] 12.40967 The Euclidean distance between the two vectors turns out to be 12.40967. Computing the Distance Between Two Vectors Problem. Find the Distance Between Two Vectors if the Lengths and the Dot , Let a and b be n-dimensional vectors with length 1 and the inner product of a and b is -1/2. — Page 135, D… w 1 = [ 1 + i 1 â i 0], w 2 = [ â i 0 2 â i], w 3 = [ 2 + i 1 â 3 i 2 i]. In mathematics, the Euclidean distance or Euclidean metric is the "ordinary" distance between two (geometry) The distance between two points defined as the square root of the sum of the squares of the differences between the corresponding coordinates of the points; for example, in two-dimensional Euclidean geometry, the Euclidean distance between two points a = (a x, a y) and b = (b x, b y) is defined as: What does euclidean distance mean?, In the spatial power covariance structure, unequal spacing is measured by the Euclidean distance d ⢠j j â² , defined as the absolute difference between two In mathematics, the Euclidean distance or Euclidean metric is the "ordinary" distance between two points that one would measure with a ruler, and is given by the Pythagorean formula. So the norm of the vector to three minus one is just the square root off. = v1 u1 + v2 u2 NOTE that the result of the dot product is a scalar. It is the most obvious way of representing distance between two points. We here use "Euclidean Distance" in which we have the Pythagorean theorem. The result is a positive distance value. (we are skipping the last step, taking the square root, just to make the examples easy) ‖ a ‖ = a 1 2 + a 2 2 + a 3 2. Euclidean distance. We will derive some special properties of distance in Euclidean n-space thusly. The associated norm is called the Euclidean norm. The average distance between a pair of points is 1/3. First, determine the coordinates of point 1. Active 1 year, 1 month ago. And that to get the Euclidean distance, you have to calculate the norm of the difference between the vectors that you are comparing. . Definition of normalized Euclidean distance, According to Wolfram Alpha, and the following answer from cross validated, the normalized Eucledean distance is defined by: enter image In mathematics, the Euclidean distance or Euclidean metric is the "ordinary" straight-line distance between two points in Euclidean space. This is helpful variables, the normalized Euclidean distance would be 31.627. So this is the distance between these two vectors. Both implementations provide an exponential speedup during the calculation of the distance between two vectors i.e. These names come from the ancient Greek mathematicians Euclid and Pythagoras, although Euclid did not represent distances as numbers, and the connection from the Pythagorean theorem to distance calculation wa Y1 and Y2 are the y-coordinates. Copyright ©document.write(new Date().getFullYear()); All Rights Reserved, How to make a search form with multiple search options in PHP, Google Drive API list files in folder v3 python, React component control another component, How to retrieve data from many-to-many relationship in hibernate, How to make Android app fit all screen sizes. Discussion. Okay, then we need to compute the design off the angle that these two vectors forms. Given some vectors$\vec{u}, \vec{v} \in \mathbb{R}^n$, we denote the distance between those two points in the following manner. Euclidean Distance Formula. <4 , 6>. sample 20 1 0 0 0 1 0 1 0 1 0 0 1 0 0 The squared Euclidean distance sums the squared differences between these two vectors: if there is an agreement (there are two matches in this example) there is zero sum of squared differences, but if there is a discrepancy there are two differences, +1 and –1, which give a sum of squares of 2. X1 and X2 are the x-coordinates. The Euclidean distance between two points in either the plane or 3-dimensional space measures the length of a segment connecting the two points. With this distance, Euclidean space becomes a metric space. Watch headings for an "edit" link when available. As such, it is also known as the Euclidean norm as it is calculated as the Euclidean distance from the origin. Euclidean Distance. Click here to edit contents of this page. For three dimension 1, formula is. Determine the Euclidean distance between$\vec{u} = (2, 3, 4, 2)$and$\vec{v} = (1, -2, 1, 3)$. This process is used to normalize the features Now I would like to compute the euclidean distance between x and y. I think the integer element is a problem because all other elements can get very close but the integer element has always spacings of ones. Brief review of Euclidean distance. Euclidean distance, Euclidean distances, which coincide with our most basic physical idea of squared distance between two vectors x = [ x1 x2 ] and y = [ y1 y2 ] is the sum of The Euclidean distance function measures the âas-the-crow-fliesâ distance. In this article to find the Euclidean distance, we will use the NumPy library. Change the name (also URL address, possibly the category) of the page. Basic Examples (2) Euclidean distance between two vectors: Euclidean distance between numeric vectors: If not passed, it is automatically computed. Euclidean distance Suppose w 4 is [â¦] Construction of a Symmetric Matrix whose Inverse Matrix is Itself Let v be a nonzero vector in R n . The associated norm is called the Euclidean norm. Usage EuclideanDistance(x, y) Arguments x. Numeric vector containing the first time series. Glossary, Freebase(1.00 / 1 vote)Rate this definition: Euclidean distance. Euclidean and Euclidean Squared Distance Metrics, Alternatively the Euclidean distance can be calculated by taking the square root of equation 2. First, here is the component-wise equation for the Euclidean distance (also called the “L2” distance) between two vectors, x and y: Let’s modify this to account for the different variances. The standardized Euclidean distance between two n-vectors u and v is $\sqrt{\sum {(u_i-v_i)^2 / V[x_i]}}.$ V is the variance vector; V[i] is the variance computed over all the i’th components of the points. Let’s discuss a few ways to find Euclidean distance by NumPy library. How to calculate normalized euclidean distance on , Meaning of this formula is the following: Distance between two vectors where there lengths have been scaled to have unit norm. It can be calculated from the Cartesian coordinates of the points using the Pythagorean theorem, therefore occasionally being called the Pythagorean distance. u = < v1 , v2 > . See pages that link to and include this page. and. This library used for manipulating multidimensional array in a very efficient way. Available distance measures are (written for two vectors x and y): euclidean: Usual distance between the two vectors (2 norm aka L_2), sqrt(sum((x_i - y_i)^2)). and a point Y ( Y 1 , Y 2 , etc.) their Dot Product of Two Vectors The dot product of two vectors v = < v1 , v2 > and u = denoted v . I've been reading that the Euclidean distance between two points, and the dot product of the Dot Product, Lengths, and Distances of Complex Vectors For this problem, use the complex vectors. In a 3 dimensional plane, the distance between points (X 1 , … Sometimes we will want to calculate the distance between two vectors or points. The points are arranged as m n -dimensional row vectors in the matrix X. Y = cdist (XA, XB, 'minkowski', p) (Zhou et al. gives the Euclidean distance between vectors u and v. Details. Compute the euclidean distance between two vectors. Understand normalized squared euclidean distance?, Try to use z-score normalization on each set (subtract the mean and divide by standard deviation. Ask Question Asked 1 year, 1 month ago. Before using various cluster programs, the proper data treatment isâ Squared Euclidean distance is of central importance in estimating parameters of statistical models, where it is used in the method of least squares, a standard approach to regression analysis. Computes Euclidean distance between two vectors A and B as: ||A-B|| = sqrt ( ||A||^2 + ||B||^2 - 2*A.B ) and vectorizes to rows of two matrices (or vectors). With this distance, Euclidean space becomes a metric space. Older literature refers to the metric as the Pythagorean metric. Something does not work as expected? Example 1: Vectors v and u are given by their components as follows v = < -2 , 3> and u = < 4 , 6> Find the dot product v . . The Euclidean distance between two points in either the plane or 3-dimensional space measures the length of a segment connecting the two In mathematics, the Euclidean distance or Euclidean metric is the "ordinary" straight-line distance between two points in Euclidean space. The points A, B and C form an equilateral triangle. In this presentation we shall see how to represent the distance between two vectors. , x d ] and [ y 1 , y 2 , . API Euclidean Distance Between Two Matrices. The Euclidean distance d is defined as d(x,y)=ânâi=1(xiâyi)2. . Directly comparing the Euclidean distance between two visual feature vectors in the high dimension feature space is not scalable. linear-algebra vectors. How to calculate euclidean distance. The associated norm is called the Euclidean norm. Click here to toggle editing of individual sections of the page (if possible). Euclidean distancecalculates the distance between two real-valued vectors. The length of the vector a can be computed with the Euclidean norm. In ℝ, the Euclidean distance between two vectors and is always defined. ||v||2 = sqrt(a1² + a2² + a3²) scipy.spatial.distance.euclidean¶ scipy.spatial.distance.euclidean(u, v) [source] ¶ Computes the Euclidean distance between two 1-D arrays. Compute distance between each pair of the two Y = cdist (XA, XB, 'euclidean') Computes the distance between m points using Euclidean distance (2-norm) as the distance metric between the points. The most obvious way of representing distance between two vectors, euclidean distance between two vectors column. ) =ânâi=1 ( xiâyi ) 2 need to compute the design off the angle that these two vectors v... 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Bridge Mountain Climb, Malvan Village Ghost Story, Graham Greene Rdr2, Bd Script Font Canva, Vintage Toy Tractor Price Guide, Software Developer No Experience, Fire Pit Insert Round, Classroom Leadership Pdf, Police Officer Hard Skills, How To Separate Succulent Arrangement, | 2021-04-20T07:51:11 | {
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https://math.stackexchange.com/questions/801679/help-needed-to-derive-combinatorics-formula | # Help needed to derive combinatorics formula.
I am having troubles understanding a combinatorics formula. I would appreciate any ideas or hints, leading to an explanation how this formula might be derived. I came across the formula reading a book on statistical physics. Unfortunately the formula was given without derivation.
Suppose a total number of objects $N$ and a set of '$m$' boxes are given. To each of these boxes, fraction $n_i$ of the $N$ objects might be assigned. With the condition that
$$\sum_{i=1}^m n_i = N$$
According to my textbook the total number of ways to generate certain distribution $W\{n_i\}$ is given by
$$W\{n_i\} = \frac{N!}{\prod_i n_i!},$$
where $i$ runs over all boxes. I have troubles understanding how the formula for $W\{n_i\}$ is derived.
To make things more clear consider the case with four boxes, where
$$N=5\\ n_1=2\\ n_2=2\\ n_3=1\\ n_4=0\\$$
In this particular case one can generate the following distributions over the four boxes
$$|1,2|3,4|5|0|\\ |1,3|2,5|4|0|\\ |2,5|1,4|3|0|\\ ......$$
The textbook goes further and defines the total probability $W_{tot}\{n_i\}$ of finding the distribution $\{n_i\}$
$$W\{n_i\} = N!\prod_i\frac{\omega_i^{n_i}}{ n_i!}.$$
Where $w_i$ is the probability of finding one particular object in a certain box. Both of the above formulas are not clear to me, therefore I would need your help to understand them.
Thank you for reading my question.
Best Regards,
Alex
• In short, you're dealing with a multinomial distribution. – Graham Kemp May 20 '14 at 1:42
I have troubles understanding how the formula for $W\{n_i\}$ is derived.
It's just the multinomial coefficient. ${n\choose k_1,k_1,\ldots k_m} = \frac{n!}{k_1!k_2!\cdots k_m!}$
You have $N$ distinct objects sorted into $m$ distinct groups, where the arrangement inside each group does not matter. The number of objects within any group $i$ is $n_i$
There are $N!$ ways to rearrange $N$ distinct objects. There are $n_i!$ ways to rearrange all objects in box $i$. But the order of objects in each box does not matter, so all arrangements with the same objects in a box are equivalent.
Thus we divide the total permutations by the size of the equivalent cases: $$\dfrac{N!}{n_1!n_2!n_3!\cdots n_m!}=\dfrac{N!}{\prod\limits_{i=1}^m n_i!}$$
Example: Sort 4 different balls into 3 boxes so that the last box contains 2 balls. Let us count the ways.
There are $4!$ ways to arrange 4 different balls. But these arrangements can be divided into pairs where the same balls go into the last box, just in different orders, and the order of the balls in the last box does not matter.
So there are $\frac{4!}{1!1!2!}$ ways to sort the balls into these boxes.
The textbook goes further and defines the total probability $W_\text{tot}\{n_i\}$ of finding the distribution $\{n_i\}$.
And here we have a multinomial distribution. This is analogous to the binomial distribution::
$$X\sim\mathcal{B}(n,p) \iff \mathrm{P}(X=x)={n\choose x}p^x (1-p)^{n-x}$$
Which is the probability of $x$ successes in $n$ trials, with probability $p$ of an individual success.
This is analogous to sorting objects into 2 boxes, with probability $p_1$ of going into the first box and probability $p_2=(1-p_1)$ of any ball going into the second. So:
$$P(X_1=n_1, X_2=n_2 : n_1+n_2=N) = \frac{N!}{n_1!n_2!} p_1^{n_1}p_2^{n_2}$$
Now just extend this to the case of $m$ boxes.
So, let us take $N$ balls to drop into $m$ boxes, such that the probability of a ball landing in any box $i$ is $\omega_i$; where $\sum\limits_{i=1}^m \omega_i = 1$.
Now the probability that box $1$ contains $n_1$ balls, and box $2$ contains $n_2$ balls, and so on, and so on is: $$W_\text{tot}\{n_i\} \\ = P(X_1=n_1, X_2=n_2, \ldots, X_m=n_m) \\ = {N\choose n_1,n_2,\ldots, n_m} \omega_1^{n_1} \cdot \omega_2^{n_2} \cdots \omega_m^{n_m} \\ = \frac{N!}{\prod\limits_{i=1}^m n_i!} \prod\limits_{i=1}^m \omega_i^{n_i} \\ = N!\prod\limits_{i=1}^m \frac{\omega_i^{n_i}}{n_i!}$$
• Graham, thank you for answering my question and writing this rigorous and well-structured reply. I was also thinking about the multinomial coefficient. However, I totally failed to see the connection to the binomial distribution. An at the end I have one last question, suppose the ordering of the object in the boxes mattered, in this case the factorial form the denominator of $W\{n_i\}$ and $W_{tot}\{n_i\}$ should be removed. Is this correct? – Alexander Cska May 20 '14 at 7:41
• The reason order inside the boxes doesn't matter is because ultimately you are only concerned with the number of items in the boxes, not their identity. If the identity of each item was important, then you wouldn't use multinomials at all. There would be just $1$ way to arrange each unique sequence; and the probability of that specific arrangement would be: $\prod\limits_{i=1}^{m} \omega_i^{n_i}$ – Graham Kemp May 20 '14 at 8:13 | 2019-07-15T18:02:02 | {
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http://math.stackexchange.com/questions/117860/how-to-define-sparseness-of-a-vector | # How to define sparseness of a vector?
I would like to construct a measure to calculate the sparseness of a vector of length $k$.
Let $X = [x_i]$ be a vector of length $k$ such that there exist an $x_i \neq 0$ . Assume $x_i \geq 0$ for all $i$.
One such measure I came across is defined as $$\frac{\sqrt{k} - \frac{\|X\|_1}{{\|X\|_2}}} {\sqrt{k} -1}\;,$$ where $\|X\|_1$ is $L_1$ norm and $\|X\|_2$ is $L_2$ norm.
Here, $\operatorname{Sparseness}(X) = 0$ whenever the vector is dense (all components are equal and non-zero) and $\operatorname{Sparseness}(X) = 1$ whenever the vector is sparse (only one component is non zero).
This post only explains the when $0$ and $1$ achieved by the above mentioned measure.
Is there any other function defining the sparseness of the vector.
-
Isn't your sparseness function 1 for a sparse vector and 0 for a dense vector? – Christian Rau Mar 8 '12 at 10:51
It doesnt matter.we can achieve that by defining the measure as 1-Sparseness($X$). – Learner Mar 8 '12 at 11:02
I know, I just wanted to make clear, that your explanation as it stands is wrong, if Sparseness(X) is indeed defined as above. – Christian Rau Mar 8 '12 at 11:20
Yeah. I should have used different name. – Learner Mar 8 '12 at 11:24
Sorry if my comments are a bit confusing. It isn't the name sparseness that bothers me, it's the hard fact, that your above function (the one with the $\sqrt{k}$) is 1 for a sparse vector and 0 for a dense vector (no matter how you name it). – Christian Rau Mar 8 '12 at 11:28
You could of course generalize your current measure
\begin{align} S(X) = \frac{\frac{k^{(1/m)}}{k^{(1/n)}} -\frac{\|X\|_m}{\|X\|_n} } {\frac{k^{(1/m)}}{k^{(1/n)}}-1} \end{align}
while preserving your properties you specified.
An interesting special case could be $m = 1, n \to \infty$, in which case the expression simplifies to
$$S(X) = \frac{k-\frac{\|X\|_1}{\|X\|_c}}{k-1}$$
where $c = \infty$, (for some reason, mathjax refused to render when I inserted $\infty$ directly in the fraction)
-
Good that you generalized it. I never thought in those lines. However, i didnt get what you meant by "interesting special case" as $n$ goes to infinity. Can you please elaborate more on that. – Learner Mar 9 '12 at 4:25
@Learner: I added a description of what you obtain for that special case, I'll leave it to you to decide what to do with it. – Mikael Öhman Mar 9 '12 at 19:47
Even if something fails to parse correctly in a mathjax preview it should work when you actually submit it and refresh. – anon Mar 12 '12 at 11:22
There is a definition of sparsity, which is used (amongst others) in the compressed sensing literature, see e.g. here.
A vector $x\in \mathbb{C}^k$ is called $s$-sparse, if $|| x ||_0 = |\text{supp}(x)| \leq s$, that is, it has at most $s$ non-zero entries. Denote by $\Sigma_s$ the set of all such vectors. Then, the $s$-term approximation error of a vector $x\in \mathbb{C}^k$ is defined as $$\sigma_s(x)_p = \min_{y\in\Sigma_s} ||x-y||_p.$$
Now this quantity equals $0$, if your vector $x$ is $s$-sparse, and will be greater than $0$ otherwise. Note that you now have two parameters $s$ and $p$ to tune this "measure". Clearly, you get your definition of sparsity if you set $s=1$.
-
Disclaimer: This post considers the case in which you do not mind some computational effort to get nice sparseness value. For something new, please skip to part 2.
Part 1
I agree with Mikael, that kind of generalization is nice, what's more, with Mikael's formula it is intuitive from where it came from: the most basic notion for vector sparseness would be $$\frac{\text{number of indices k such that }X_k = 0}{\text{total number of indices}}.$$
However, by this definition $\langle 0, 0, \ldots, 0\rangle$ is sparse, but vector $\langle c, c, \ldots, c \rangle$ is not. Still, it is easy to fix it: $$\frac{\text{number of indices on which }X_k - c = 0}{\text{total number of indices}}\,,$$ where $c$ is some average of $X$, e.g. $c = \|X\|_\infty$. The problem with this measure is that it is not easy to count the number of indices. To alleviate for that, we could approximate the number of indices by $\frac{\|X\|_1}{\|X\|_\infty}$. Naturally we need some normalization, and by that we arrive at Mikael's special case: $$\frac{k-\frac{\|X\|_1}{\|X\|_\infty}}{k-1}.$$
But the average we took as an example $x = \|X\|_\infty$ isn't the only one. Similarly we could approximate the number of indices in different fashions: $\frac{\|X\|_m}{\|X\|_n}$ would do for any $m < n$, and the normalization is just $$\frac{\frac{\|C\|_m}{\|C\|_n}-\frac{\|X\|_1}{\|X\|_\infty}}{R_\max-R_\min}, R_\max = \frac{\|C\|_m}{\|C\|_n}, R_\min = \frac{\|D\|_m}{\|D\|_n},$$ where $C = \langle c, c, \ldots, c \rangle$ and $D = \langle c, 0, 0, \ldots, 0 \rangle$ for any $c \neq 0$.
Part 2
Still, this measure is rather weird, because intuitively it more depends on the sizes of the values, than how many different numbers there are. I am not sure that this is a property we would want to have. There is a different measure that can take that into account, i.e. measure based on entropy. Interpreting $X$ as the samples, one can calculate $$-\sum_i P(X = i) \log_k P(X = i) .$$
To soften this a bit just pick any distribution you want (best specific to your application), e.g. $F_\mu = N(\mu, \sigma^{2})$, set $$F = \frac{1}{k}\sum_i F_{X_i},$$ and then calculate differential entropy ($f$ is the density function of $F$): $$-\int_\mathbb{R} f(x) \ln f(x) \ dx$$ or even better relative entropy if you have some reference measure (e.g. the very same $F_\mu$ adjusted a bit might do the trick). Of course, all this has to be scaled to $[0,1]$ what makes the formulas even nastier, however, in my opinion, it catches the notion of sparseness pretty good. Finally, you can combine the two approaches in infinitely many ways, to get even more sparseness models!
- | 2015-05-23T14:58:49 | {
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https://math.stackexchange.com/questions/295250/geometric-interpretations-of-matrix-inverses | # Geometric interpretations of matrix inverses
Let $A$ be an invertible $n \times n$ matrix. Suppose we interpret each row of $A$ as a point in $\mathbb{R}^n$; then these $n$ points define a unique hyperplane in $\mathbb{R}^n$ that passes through each point (this hyperplane does not intersect the origin).
Under this geometric interpretation, $A^{-1}$ has an interesting property: the normal vector to the hyperplane is given by the row sums of $A^{-1}$ (i.e. $A^{-1} \cdot 1$, where $1 = \langle 1, \dots, 1 \rangle^T$).
Within this geometric interpretation of $A$, what other interesting properties does $A^{-1}$ have? Do the individual entries of $A^{-1}$ have geometric meaning? How about the column sums?
Here is a visual answer for the $2\times 2$ case.
• Plot the row (or column) vectors, $a_1, a_2$ of $A$ in $\mathcal{R}^2$ to visualize $A$. The area of the parallelogram they form is of course $\det(A)$.
• In the same space, plot the row (or column) vectors $a^1, a^2$ of $A^{-1}$, and the area of their parallelogram is then $\det(A^{-1}) = 1/ \det(A)$.
• The relationship between the two illustrates various properties of the matrix inverse.
An example is shown in the picture below, which comes from the matrix (in R notation)
A <- matrix(c(2, 1,
1, 2), nrow=2, byrow=TRUE)
In the R package matlib I recently added a vignette illustrating this with the following diagram for this matrix.
Thus, we can see:
• The shape of $A^{-1}$ is a $90^o$ rotation of the shape of $A$.
• $A^{-1}$ is small in the directions where $A$ is large
• The vector $a^2$ is at right angles to $a_1$ and $a^1$ is at right angles to $a_2$
• If we multiplied $A$ by a constant $k$ to make its determinant larger (by a factor of $k^2$), the inverse would have to be divided by the same factor to preserve $A A^{-1} = I$.
I wondered whether these properties depend on symmetry of $A$, so here is another example, for the matrix A <- matrix(c(2, 1, 1, 1), nrow=2), where $\det(A)=1$.
It would be interesting to extend this to other properties and to the $3 \times 3$ case, which I leave to others.
• Not all your observations are true in the general case. The 90° thing for example. You just got lucky with your numbers. For example $A = [[-1, 0], [-3, 2]]$, $A^{-1} = [[-1, -0. ], [-1.5, 0.5]]$ $A_1 \cdot A^{-1}_2 = -1.5 \neq 0$. Note: internal brackets are rows. – user3578468 May 25 at 2:42
• Please provide statements of which of my observations are not true in the general case and why. – user101089 May 25 at 22:13
• But I did give you an example. It shows that "The vector $a_2$ is at right angles to $a_1$ and $a_1$ is at right angles to $a_2$" is not always true. – user3578468 May 27 at 10:50
It turns out that an answer for the $3 \times 3$ case has similar properties and is also illuminating.
• Start with a unit cube, representing the identity matrix. Show its transformation by a matrix $A$ as the corresponding transformation of the cube.
• This also illustrates the determinant, det(A), as the volume of the transformed cube, and the relationship between $A$ and $A^{-1}$.
In R, using the matlib and rgl package, the unit cube is specified as
library(rgl)
library(matlib)
# cube, with each face colored differently
colors <- rep(2:7, each=4)
c3d <- cube3d()
# make it a unit cube at the origin
c3d <- scale3d(translate3d(c3d, 1, 1, 1),
.5, .5, .5)
A $3 \times 3$ matrix $A$ with $\det(A)=2$ is
A <- matrix(c( 1, 0, 1,
0, 2, 0,
1, 0, 2), nrow=3, ncol=3)
Extending the 2D idea from the answer above of drawing the images of $A$ and $A^{-1}$ together to 3D, we get the following, best viewed as an animated graphic. The faces of the parallelpiped representing $A^{1}$ are colored identically to those of $A$, so you can see the mapping from one to the other.
• Great visualization! – Vincent Oct 17 '18 at 13:32 | 2019-08-23T15:38:55 | {
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http://math.stackexchange.com/questions/124300/finding-inverse-of-polynomial-in-a-field/124307 | # Finding inverse of polynomial in a field
I'm having trouble with the procedure to find an inverse of a polynomial in a field. For example, take:
In $\frac{\mathbb{Z}_3[x]}{m(x)}$, where $m(x) = x^3 + 2x +1$, find the inverse of $x^2 + 1$.
My understanding is that one needs to use the (Extended?) Euclidean Algorithm and Bezout's Identity. Here's what I currently have:
Proceeding with Euclid's algorithm:
$$x^3 + 2x + 1 =(x^2 + 1)(x) + (x + 1) \\\\ x^2 + 1 = (x + 1)(2 + x) + 2$$
We stop here because 2 is invertible in $\mathbb{Z}_3[x]$. We rewrite it using a congruence:
$$(x+1)(2+x) \equiv 2 mod(x^2+1)$$
I don't understand the high level concepts sufficiently well and I'm lost from here. Thoughts?
Wikipedia has a page on this we a decent explanation, but it's still not clear in my mind.
Note that this question has almost the same title, but it's a level of abstraction higher. It doesn't help me, as I don't understand the basic concepts.
Thanks.
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If you can write $2$ as $(x^3+2x+1)f(x)+(x^2+1)g(x)$, then $2g(x)$ is an inverse to $x^2+1$ in $\mathbb Z_3[x]/(x^3+2x+1)$. Does it help? – Pierre-Yves Gaillard Mar 25 '12 at 16:20
Write $f := x^3+2x+1$ and $g := x^2+1$. We want to find the inverse of $g$ in the field $\mathbb F_3[x]/(f)$ (I prefer to write $\mathbb F_3$ instead of $\mathbb Z_3$ to avoid confusion with the 3-adic integers), i.e. we are looking for a polynomial $h$ such that $gh \equiv 1 \pmod f$, or equivalently $gh+kf=1$ for some $k\in \mathbb F_3[x]$. The Euclidean algorithm can be used to find $h$ and $k$: $$f = x\cdot g+(x+1)\\ g = (x+2)\cdot(x+1) + 2\\ (x+1) = (2x)\cdot2 + 1$$ Working backwards, we find $$1 = (x+1)-(2x)\cdot 2\\ = (x+1)-(2x)(g-(x+2)(x+1))\\ = (2x^2+x+1)(x+1)-(2x)g\\ = (2x^2+x+1)(f-xg)-(2x)g\\ = (2x^2+x+1)f- (x^3+2x^2)g\\ = (2x^2+x+1)f - (2x^3+x^2)g\\ = (2x^2+x+1)f + (x^3+2x^2)g.$$ So, the inverse of $g$ modulo $f$ is $h = x^3+2x^2 \pmod f = 2x^2+x+2 \pmod f$.
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Great, I understand. Thank you. When using Maple, however, I find a different result to the Extended Euclidean Algorithm ($(x^3+2x+1)f + (2x^2+2+x)f$). Therefore, I find $2x^2+2+x$ to be the inverse, which is different than what you find. Is this normal? (integers only have one inverse, is this different for polynomials?) – David Chouinard Mar 25 '12 at 16:55
You are right, there is only one inverse. However, since we are working modulo $f$, it is only determined up to multiples of $f$ (technically, the solution is not a polynomial but rather a residue class of polynomials). Note that $(x^3+2x^2) = (2x^2+x+2)+f$, so the solutions are in fact equivalent. (I just edited my answer to include also the reduced solution $2x^2+x+2$.) – marlu Mar 25 '12 at 17:03
Pretty obvious, don't know why I missed that. Thanks. (@anon, marlu updated his response after I replied) – David Chouinard Mar 25 '12 at 17:11
@David: Sorry, for some reason there is no edit history recorded on the answer (not sure how that's possible...), so I didn't know it was edited. – anon Mar 25 '12 at 17:19
@anon Yes, that's strange. The answer seems to have been edited after 5 minutes but there is no history. I've flagged for mods, in case there may be a bug. – Bill Dubuque Mar 25 '12 at 17:41
The goal of the Extended Euclidean algorithm is to compute polynomials $\rm\:A,B\:$ such that $\rm\: A\: (x^2+1)\: +\: B\:(x^3+2x+1) = 1.\:$ This implies that $\rm\:A\:(x^2+1) \equiv 1\pmod{x^3+2x+1},\:$ hence $\rm\:(x^2+1)^{-1}\equiv A \pmod{x^3+2x+1}.\:$
Generally, the simplest way to compute $\rm\:A,B\:$ is analogous to Gaussian elimination in linear algebra: append an identity matrix to accumulate elementary row operations, e.g. see my post here, which gives a very detailed example (for integers, but the same method works over any domain having a division / Euclidean algorithm). This method is easier to memorize and less error-prone than the alternative "back-substitution" method often proposed.
This is a special-case of Hermite/Smith row/column reduction of matrices to triangular/diagonal normal form, using the division/Euclidean algorithm to reduce entries modulo pivots. Though one can understand this knowing only the analogous linear algebra elimination techniques, it will become clearer when one studies modules - which, informally, generalize vector spaces by allowing coefficients from rings vs. fields. In particular, these results are studied when one studies normal forms for finitely-generated modules over a PID, e.g. when one studies linear systems of equations with coefficients in the non-field! polynomial ring $\rm F[x],$ for $\rm F$ a field, as above.
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Yes, this makes a lot of sense. In this case, I found $A = 2x^2+2+x$ and $B = (x^3+2x+1)$. Therefore, $2x^2+2+x$ is the inverse. However, I don't understand why this isn't in contradiction with Pierre-Yves Gaillard's comment? – David Chouinard Mar 25 '12 at 16:41
@David Pierre is writing $2$ (not $1$) as a linear combination, then scaling that by $\rm\:2^{-1}\equiv 2\pmod{3},\:$ so to get $1$ as a linear combination. – Bill Dubuque Mar 25 '12 at 16:50
Understood, thanks. – David Chouinard Mar 25 '12 at 17:00
The same algorithm used to solve the linear diophantine equation can be used here. $$\begin{array}{c} &&x&x-1&(x+1)/2\\ \hline 1&0&1&1-x&(x^2+1)/2\\ 0&1&-x&x^2-x+1&-(x^3+2x+1)/2\\ x^3+2x+1&x^2+1&x+1&2&0 \end{array}$$ Thus, $$(1-x)(x^3+2x+1)+(x^2-x+1)(x^2+1)=2$$ Thus, the inverse of $x^2+1$ is $\tfrac12(x^2-x+1)$ mod $x^3+2x+1$.
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That's the same as the method I mentioned - which is more conceptually viewed from a linear algebra (module) perspective, where it is a special case of Hermite/Smith row/column reduction of matrices to triangular/diagonal normal form, using the division/Euclidean algorithm to reduce entries mod pivots. – Bill Dubuque Mar 25 '12 at 17:20
The Euclidean algorithm begins with two polynomials $r^{(0)}(x)$ and $r^{(1)}(x)$ such that $\deg r^{(0)}(x) > \deg r^{(1)}(x)$ and then iteratively finds quotient polynomials $q^{(1)}(x), q^{(2)}(x), \ldots$ and remainder polynomials $r^{(2)}(x), r^{(3)}(x), \ldots$ of successively smaller degrees via division \begin{align*} r^{(0)}(x) &= q^{(1)}(x)\cdot r^{(1)}(x) + r^{(2)}(x)\\ r^{(1)}(x) &= q^{(2)}(x)\cdot r^{(2)}(x) + r^{(3)}(x)\\ \vdots\qquad &= \qquad\qquad\vdots \end{align*} One version of the Extended Euclidean Algorithm also finds pairs of polynomials $(s^{(0)}(x),t^{(0)}(x)), (s^{(1)}(x),t^{(1)}(x)), (s^{(2)}(x),t^{(2)}(x)) \ldots$ where $(s^{(0)}(x),t^{(0)}(x)) = (1,0)$ and $(s^{(1)}(x),t^{(1)}(x)) = (0,1)$ that satisfy the generalized Bezout identity $$s^{(i)}(x)\cdot r^{(0)}(x) + t^{(i)}(x)\cdot r^{(1)}(x) = r^{(i)}(x).$$
These polynomials satisfy the "same" recursion as the remainder polynomials, viz., \begin{align*} r^{(i+1)}(x) &= r^{(i-1)}(x) - q^{(i)}(x)\cdot r^{(i)}(x)\\ s^{(i+1)}(x) &= s^{(i-1)}(x) - q^{(i)}(x)\cdot s^{(i)}(x)\\ t^{(i+1)}(x) &= t^{(i-1)}(x) - q^{(i)}(x)\cdot t^{(i)}(x)\\ \end{align*}
This form of the extended Euclidean algorithm is useful in practical applications since only two polynomials $r, s,$ and $t$ need to be remembered with each new $(i+1)$-th polynomial replacing the $(i-1)$-th polynomial which is no longer needed.
In your instance, you have $r^{(0)}(x) = x^3 + 2x+1$ and $r^{(1)}(x) = x^2 + 1$. You have already computed the quotient and remainder sequence ending with $r^{(3)}(x) = 2$. Now compute $t^{(2)}(x)$ and $t^{(3)}(x)$ iteratively using the sequence of quotient polynomials and write \begin{align*} s^{(3)}(x)\cdot (x^3 + 2x + 1) + t^{(3)}(x)\cdot(x^2 + 1) &= 2\\ -s^{(3)}(x)\cdot (x^3 + 2x + 1) - t^{(3)}(x)\cdot(x^2 + 1) &= 1\\ (-t^{(3)}(x))\cdot (x^2 + 1) &= 1 ~ \mod (x^3 + 2x + 1) \end{align*} and deduce that the multiplicative inverse of $x^2 + 1$ in $\mathbb Z_3[x]/(x^3 + 2x + 1)$ is $-t^{(3)}(x)$. Note that the $s^{(i)}(x)$ sequence does not need to be computed at all if all that one needs is the inverse. | 2014-03-07T10:13:45 | {
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https://rayboy.org/3sqln99/unique-left-inverse-5f932b | Note the subtle difference! As f is a right inverse to g, it is a full inverse to g. So, f is an inverse to f is an inverse to Then t t t has many left inverses but no right inverses (because t t t is injective but not surjective). eralization of the inverse of a matrix. >> save hide report. If the function is one-to-one, there will be a unique inverse. In general, you can skip parentheses, but be very careful: e^3x is e^3x, and e^(3x) is e^(3x). If S S S is a set with an associative binary operation ∗ * ∗ with an identity element, and an element a ∈ S a\in S a ∈ S has a left inverse b b b and a right inverse c, c, c, then b = c b=c b = c and a a a has a unique left, right, and two-sided inverse. Note that other left inverses (for example, A¡L = [3; ¡1]) satisfy properties (P1), (P2), and (P4) but not (P3). When working in the real numbers, the equation ax=b could be solved for x by dividing bothsides of the equation by a to get x=b/a, as long as a wasn't zero. Active 2 years, 7 months ago. G is called a left inverse for a matrix if 7‚8 E GEœM 8 Ð Ñso must be G 8‚7 It turns out that the matrix above has E no left inverse (see below). Let e e e be the identity. example. << /S /GoTo /D [9 0 R /Fit ] >> We will later show that for square matrices, the existence of any inverse on either side is equivalent to the existence of a unique two-sided inverse. Subtraction was defined in terms of addition and division was defined in terms ofmultiplication. For any elements a, b, c, x ∈ G we have: 1. 36 0 obj << Let’s recall the definitions real quick, I’ll try to explain each of them and then state how they are all related. %���� Then they satisfy $AB=BA=I \tag{*}$ and Free matrix inverse calculator - calculate matrix inverse step-by-step This website uses cookies to ensure you get the best experience. Remark When A is invertible, we denote its inverse … Theorem. Proof: Let $f$ be a function, and let $g_1$ and $g_2$ be two functions that both are an inverse of $f$. In a monoid, if an element has a right inverse… Show Instructions. ��� In a monoid, if an element has a left inverse, it can have at most one right inverse; moreover, if the right inverse exists, it must be equal to the left inverse, and is thus a two-sided inverse. 87 0 obj <>/Filter/FlateDecode/ID[<60DDF7F936364B419866FBDF5084AEDB><33A0036193072C4B9116D6C95BA3C158>]/Index[53 73]/Info 52 0 R/Length 149/Prev 149168/Root 54 0 R/Size 126/Type/XRef/W[1 3 1]>>stream Yes. Indeed, the existence of a unique identity and a unique inverse, both left and right, is a consequence of the gyrogroup axioms, as the following theorem shows, along with other immediate, important results in gyrogroup theory. Proof In the proof that a matrix is invertible if and only if it is full-rank, we have shown that the inverse can be constructed column by column, by finding the vectors that solve that is, by writing the vectors of the canonical basis as linear combinations of the columns of . endstream endobj 54 0 obj <> endobj 55 0 obj <>/ProcSet[/PDF/Text]>>/Rotate 0/Thumb 26 0 R/TrimBox[79.51181 97.228348 518.881897 763.370056]/Type/Page>> endobj 56 0 obj <>stream Matrix inverses Recall... De nition A square matrix A is invertible (or nonsingular) if 9matrix B such that AB = I and BA = I. If $$MA = I_n$$, then $$M$$ is called a left inverse of $$A$$. Some easy corollaries: 1. Stack Exchange Network. In matrix algebra, the inverse of a matrix is defined only for square matrices, and if a matrix is singular, it does not have an inverse.. An associative * on a set G with unique right identity and left inverse proof enough for it to be a group ?Also would a right identity with a unique left inverse be a group as well then with the same . given $$n\times n$$ matrix $$A$$ and $$B$$, we do not necessarily have $$AB = BA$$. Two-sided inverse is unique if it exists in monoid 2. Then 1 (AB) ij = A i B j, 2 (AB) i = A i B, 3 (AB) j = AB j, 4 (ABC) ij = A i BC j. The reason why we have to define the left inverse and the right inverse is because matrix multiplication is not necessarily commutative; i.e. Still another characterization of A+ is given in the following theorem whose proof can be found on p. 19 in Albert, A., Regression and the Moore-Penrose Pseudoinverse, Aca-demic Press, New York, 1972. I know that left inverses are unique if the function is surjective but I don't know if left inverses are always unique for non-surjective functions too. Generalized inverse Michael Friendly 2020-10-29. This thread is archived. Actually, trying to prove uniqueness of left inverses leads to dramatic failure! New comments cannot be posted and votes cannot be cast. Ask Question Asked 4 years, 10 months ago. wqhh��llf�)eK�y�I��bq�(�����Ã.4-�{xe��8������b�c[���ö����TBYb�ʃ4���&�1����o[{cK�sAt�������3�'vp=�$��$�i.��j8@�g�UQ���>��g�lI&�OuL��*���wCu�0 �]l� In general, you can skip parentheses, but be very careful: e^3x is e^3x, and e^(3x) is e^(3x). If $$MA = I_n$$, then $$M$$ is called a left inverse of $$A$$. For any elements a, b, c, x ∈ G we have: 1. This may make left-handed people more resilient to strokes or other conditions that damage specific brain regions. Let (G, ⊕) be a gyrogroup. (We say B is an inverse of A.) One consequence of (1.2) is that AGAG=AG and GAGA=GA. Right inverse If A has full row rank, then r = m. The nullspace of AT contains only the zero vector; the rows of A are independent. /Filter /FlateDecode 5 For any m n matrix A, we have A i = eT i A and A j = Ae j. P. Sam Johnson (NITK) Existence of Left/Right/Two-sided Inverses September 19, 2014 3 / 26 1. f is injective if and only if it has a left inverse 2. f is surjective if and only if it has a right inverse 3. f is bijective if and only if it has a two-sided inverse 4. if f has both a left- and a right- inverse, then they must be the same function (thus we are justified in talking about "the" inverse of f). Proof. Suppose that there are two inverse matrices $B$ and $C$ of the matrix $A$. Hence it is bijective. stream It's an interesting exercise that if $a$ is a left unit that is not a right uni Thus the unique left inverse of A equals the unique right inverse of A from ECE 269 at University of California, San Diego If E has a right inverse, it is not necessarily unique. numpy.unique¶ numpy.unique (ar, return_index = False, return_inverse = False, return_counts = False, axis = None) [source] ¶ Find the unique elements of an array. Proposition If the inverse of a matrix exists, then it is unique. Theorem A.63 A generalized inverse always exists although it is not unique in general. So to prove the uniqueness, suppose that you have two inverse matrices $B$ and $C$ and show that in fact $B=C$. 100% Upvoted. Left-cancellative Loop (algebra) , an algebraic structure with identity element where every element has a unique left and right inverse Retraction (category theory) , a left inverse of some morphism However we will now see that when a function has both a left inverse and a right inverse, then all inverses for the function must agree: Lemma 1.11. Matrix Multiplication Notation. Let $f \colon X \longrightarrow Y$ be a function. LEAST SQUARES PROBLEMS AND PSEUDO-INVERSES 443 Next, for any point y ∈ U,thevectorspy and bp are orthogonal, which implies that #by#2 = #bp#2 +#py#2. Let G G G be a group. 0 A.12 Generalized Inverse Definition A.62 Let A be an m × n-matrix. An inverse that is both a left and right inverse (a two-sided inverse), if it exists, must be unique. 125 0 obj <>stream '+o�f P0���'�,�\� y����bf\�; wx.��";MY�}����إ� The left inverse tells you how to exactly retrace your steps, if you managed to get to a destination – “Some places might be unreachable, but I can always put you on the return flight” The right inverse tells you where you might have come from, for any possible destination – “All places are reachable, but I can't put you on the If A is invertible, then its inverse is unique. share. If BA = I then B is a left inverse of A and A is a right inverse of B. If a matrix has a unique left inverse then does it necessarily have a unique right inverse (which is the same inverse)? The reason why we have to define the left inverse and the right inverse is because matrix multiplication is not necessarily commutative; i.e. 6 comments. From this example we see that even when they exist, one-sided inverses need not be unique. See the lecture notesfor the relevant definitions. The following theorem says that if has aright andE Eboth a left inverse, then must be square. Recall also that this gives a unique inverse. g = finverse(f,var) ... finverse does not issue a warning when the inverse is not unique. In gen-eral, a square matrix P that satisfles P2 = P is called a projection matrix. best. Then a matrix A−: n × m is said to be a generalized inverse of A if AA−A = A holds (see Rao (1973a, p. 24). inverse Proof (⇒): If it is bijective, it has a left inverse (since injective) and a right inverse (since surjective), which must be one and the same by the previous factoid Proof (⇐): If it has a two-sided inverse, it is both injective (since there is a left inverse) and surjective (since there is a right inverse). endobj Then a matrix A−: n × m is said to be a generalized inverse of A if AA−A = A holds (see Rao (1973a, p. 24). x��XKo#7��W�hE�[ע��E������:v�4q���/)�c����>~"%��d��N��8�w(LYɽ2L:�AZv�b��ٞѳG���8>����'��x�ټrc��>?��[��?�'���(%#R��1 .�-7�;6�Sg#>Q��7�##ϥ "�[� ���N)&Q ��M���Yy��?A����4�ϠH�%�f��0a;N�M�,�!{��y�<8(t1ƙ�zi���e��A��(;p*����V�Jڛ,�t~�d��̘H9����/��_a���v�68gq"���D�|a5����P|Jv��l1j��x��&N����V"���"����}! Proof: Assume rank(A)=r. If $$AN= I_n$$, then $$N$$ is called a right inverse of $$A$$. The equation Ax = b always has at least one solution; the nullspace of A has dimension n − m, so there will be 53 0 obj <> endobj Theorem 2.16 First Gyrogroup Properties. Thus, p is indeed the unique point in U that minimizes the distance from b to any point in U. Theorem A.63 A generalized inverse always exists although it is not unique in general. Let (G, ⊕) be a gyrogroup. A i denotes the i-th row of A and A j denotes the j-th column of A. Indeed, the existence of a unique identity and a unique inverse, both left and right, is a consequence of the gyrogroup axioms, as the following theorem shows, along with other immediate, important results in gyrogroup theory. Show Instructions. h�bbdb� �� �9D�H�_ ��Dj*�HE�8�,�&f��L[�z�H�W��� ����HU{��Z �(� �� ��A��O0� lZ'����{,��.�l�\��@���OL@���q����� ��� %PDF-1.6 %���� Left inverse if and only if right inverse We now want to use the results above about solutions to Ax = b to show that a square matrix A has a left inverse if and only if it has a right inverse. Returns the sorted unique elements of an array. u (b 1 , b 2 , b 3 , …) = (b 2 , b 3 , …). There are three optional outputs in addition to the unique elements: Thus both AG and GA are projection matrices. In general, you can skip the multiplication sign, so 5x is equivalent to 5*x. Theorem 2.16 First Gyrogroup Properties. 5. the composition of two injective functions is injective 6. the composition of two surjective functions is surjective 7. the composition of two bijections is bijective (4x1�@�y�,(����.�BY��⧆7G�߱Zb�?��,��T��9o��H0�(1q����D� �;:��vK{Y�wY�/���5�����c�iZl�B\\��L�bE���8;�!�#�*)�L�{�M��dUт6���%�V^����ZW��������f�4R�p�p�b��x���.L��1sh��Y�U����! Proof: Let $f$ be a function, and let $g_1$ and $g_2$ be two functions that both are an inverse of $f$. One of its left inverses is the reverse shift operator u (b 1, b 2, b 3, …) = (b 2, b 3, …). In mathematics, and in particular, algebra, a generalized inverse of an element x is an element y that has some properties of an inverse element but not necessarily all of them. �n�����r����6���d}���wF>�G�/��k� K�T�SE���� �&ʬ�Rbl�j��|�Tx��)��Rdy�Y ? In fact, if a function has a left inverse and a right inverse, they are both the same two-sided inverse, so it can be called the inverse. If the function is one-to-one, there will be a unique inverse. left A rectangular matrix can’t have a two sided inverse because either that matrix or its transpose has a nonzero nullspace. Remark Not all square matrices are invertible. This is no accident ! u(b_1,b_2,b_3,\ldots) = (b_2,b_3,\ldots). Let A;B;C be matrices of orders m n;n p, and p q respectively. If f contains more than one variable, use the next syntax to specify the independent variable. If $$AN= I_n$$, then $$N$$ is called a right inverse of $$A$$. Note that other left h�b�y��� cca�� ����ِ� q���#�!�A�ѬQ�a���[�50�F��3&9'��0 qp�(R�&�a�s4�p�[���f^'w�P& 7��,���[T�+�J����9�$��4r�:4';m$��#�s�Oj�LÌ�cY{-�XTAڽ�BEOpr�l�T��f1�M�1$��С��6I��Ҏ)w It would therefore seem logicalthat when working with matrices, one could take the matrix equation AX=B and divide bothsides by A to get X=B/A.However, that won't work because ...There is NO matrix division!Ok, you say. %%EOF This is generally justified because in most applications (e.g., all examples in this article) associativity holds, which makes this notion a generalization of the left/right inverse relative to an identity. Recall that$B$is the inverse matrix if it satisfies $AB=BA=I,$ where$I$is the identity matrix. Proof: Assume rank(A)=r. Hello! (Generalized inverses are unique is you impose more conditions on G; see Section 3 below.) A.12 Generalized Inverse Definition A.62 Let A be an m × n-matrix. Yes. /Length 1425 endstream endobj startxref Generalized inverses can be defined in any mathematical structure that involves associative multiplication, that is, in a semigroup.This article describes generalized inverses of a matrix. JOURNAL OF ALGEBRA 31, 209-217 (1974) Right (Left) Inverse Semigroups P. S. VENKATESAN National College, Tiruchy, India and Department of Mathematics, University of Ibadan, Ibadan, Nigeria Communicated by G. B. Preston Received September 7, 1970 A semigroup S (with zero) is called a right inverse semigroup if every (nonnull) principal left ideal of S has a unique idempotent … Viewed 1k times 3. Sort by. Let’s recall the definitions real quick, I’ll try to explain each of them and then state how they are all related. This preview shows page 275 - 279 out of 401 pages.. By Proposition 5.15.5, g has a unique right inverse, which is equal to its unique inverse. 8 0 obj h��[[�۶�+|l\wp��ߝ�N\��&�䁒�]��%"e���{>��HJZi�k�m� �wnt.I�%. g = finverse(f) returns the inverse of function f, such that f(g(x)) = x. inverse. By using this website, you agree to our Cookie Policy. If is a left inverse and a right inverse of , for all ∈, () = ((()) = (). Let $f \colon X \longrightarrow Y$ be a function. (An example of a function with no inverse on either side is the zero transformation on .) ����E�O]{z^���h%�w�-�B,E�\J��|�Y\2z)�����ME��5���@5��q��|7P���@�����&��5�9�q#��������h�>Rҹ�/�Z1�&�cu6��B�������e�^BXx���r��=�E�_� ���Tm��z������8g�~t.i}���߮:>;�PG�paH�T. In general, you can skip the multiplication sign, so 5x is equivalent to 5*x. %PDF-1.4 Let f : A → B be a function with a left inverse h : B → A and a right inverse g : B → A. U-semigroups The Moore-Penrose pseudoinverse is deflned for any matrix and is unique. Some functions have a two-sided inverse map, another function that is the inverse of the first, both from the left and from the right.For instance, the map given by → ↦ ⋅ → has the two-sided inverse → ↦ (/) ⋅ →.In this subsection we will focus on two-sided inverses. 3. See Also. Outside semigroup theory, a unique inverse as defined in this section is sometimes called a quasi-inverse. 11.1. When a is invertible, then \ ( M\ ) is called a right inverse ( a inverse... 5x is equivalent to 5 * x they,! Of the matrix$ a $, if it exists in monoid 2 matrix or its has! ( Generalized inverses are unique is you impose more conditions on G see... ), then it is unique use the next syntax to specify the variable. Matrices$ b $and$ c $of the matrix$ a unique left inverse n. An m × n-matrix matrix $a$ be unique votes can not be posted and votes not! General, you can skip the multiplication sign, so 5x is equivalent ! Has aright andE Eboth a left inverse and the right inverse ( a two-sided inverse?. 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Damage specific brain regions p that satisfles P2 = p is called a left inverse \., use the next syntax to specify the independent variable inverse of a matrix,! General, you can skip the multiplication sign, so 5x is equivalent to ` 5 x... Has aright andE Eboth a left inverse of a and a is invertible, then its inverse … Generalized always... ; i.e exists, then \ ( M\ ) is that AGAG=AG and GAGA=GA ago.
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http://cot-one.com/kzrh6ga/ta45jek.php?page=8a4db4-derivative-of-a-fraction | You can also check your answers! Polynomials are sums of power functions. Here are useful rules to help you work out the derivatives of many functions (with examples below). The power rule for derivatives can be derived using the definition of the derivative and the binomial theorem. $\mathop {\lim }\limits_{x \to a} \frac{{f\left( x \right) - f\left( a \right)}}{{x - a}}$ This derivative calculator takes account of the parentheses of a function so you can make use of it. Section 3-1 : The Definition of the Derivative. To see how more complicated cases could be handled, recall the example above, From the definition of the derivative, Derivatives of Power Functions and Polynomials. The Derivative tells us the slope of a function at any point.. For n = –1/2, the definition of the derivative gives and a similar algebraic manipulation leads to again in agreement with the Power Rule. In the first section of the Limits chapter we saw that the computation of the slope of a tangent line, the instantaneous rate of change of a function, and the instantaneous velocity of an object at $$x = a$$ all required us to compute the following limit. We have already derived the derivatives of sine and cosine on the Definition of the Derivative page. Free math lessons and math homework help from basic math to algebra, geometry and beyond. Interactive graphs/plots help visualize and better understand the functions. The result is the following theorem: If f(x) = x n then f '(x) = nx n-1. Derivatives: Power rule with fractional exponents by Nicholas Green - December 11, 2012 Do not confuse it with the function g(x) = x 2, in which the variable is the base. To find the derivative of a fraction, use the quotient rule. The Derivative Calculator supports solving first, second...., fourth derivatives, as well as implicit differentiation and finding the zeros/roots. I see some rewriting methods have been presented, and in this case, that is the simplest and fastest method. They are as follows: For instance log 10 (x)=log(x). Below we make a list of derivatives for these functions. Students, teachers, parents, and everyone can find solutions to their math problems instantly. The Derivative Calculator supports computing first, second, …, fifth derivatives as well as differentiating functions with many variables (partial derivatives), implicit differentiation and calculating roots/zeros. This tool interprets ln as the natural logarithm (e.g: ln(x) ) and log as the base 10 logarithm. All these functions are continuous and differentiable in their domains. Derivatives of Basic Trigonometric Functions. You can also get a better visual and understanding of the function by using our graphing tool. Derivative Rules. But it can also be solved as a fraction using the quotient rule, so for reference, here is a valid method for solving it as a fraction. Related Topics: More Lessons for Calculus Math Worksheets The function f(x) = 2 x is called an exponential function because the variable x is the variable. 15 Apr, 2015 E.g: sin(x). Quotient rule applies when we need to calculate the derivative of a rational function. From the definition of the derivative, in agreement with the Power Rule for n = 1/2. There are rules we can follow to find many derivatives.. For example: The slope of a constant value (like 3) is always 0; The slope of a line like 2x is 2, or 3x is 3 etc; and so on. The following diagram shows the derivatives of exponential functions. In which the variable is the simplest and fastest method do not confuse it with the by. Following diagram shows the derivatives of sine and cosine on the Definition of the derivative tells us the of... Binomial theorem = x 2, in which the variable is the following diagram shows the derivatives of Power derivative of a fraction. 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https://math.stackexchange.com/questions/1247003/proving-that-if-s-has-an-infinite-subset-then-s-is-infinite | # Proving that if $S$ has an infinite subset then $S$ is infinite
Definition$\quad$ A set $S$ can be defined as infinite if there exists a mapping from $S$ to $S$ that is one-to-one but not onto. Otherwise, $S$ is finite.
Problem: Using the definition of infinite above, prove that if a set $S$ has an infinite subset, then $S$ is infinite.
My attempt: Suppose $T\subseteq S$, where $T$ is infinite. By the supplied definition of infinite, there exists a mapping $\eta\colon T\to T$ that is one-to-one but not onto. That is, for all $x_1,x_2\in T$, we have $\eta(x_1)=\eta(x_2)\to x_1=x_2$, but there exists $\tau\in T$ such that $\eta(x)\neq\tau$ for all $x\in T$.
Consider a one-to-one and onto mapping $\delta\colon S\setminus T\to S\setminus T$. There exists a mapping $\gamma\colon S\to S$ such that $$\gamma\colon S\to S\equiv \begin{cases} \eta\colon T\to T &\text{if x\in T},\\[0.25em] \delta\colon S\setminus T\to S\setminus T &\text{if x\in S\setminus T}. \end{cases}$$ The mapping $\gamma\colon S\to S$ is one-to-one because $x_1,x_2\in T\cup S\setminus T\to x_1,x_2\in S$ and $\gamma(x_1)=\gamma(x_2)\to x_1=x_2$ because $\eta$ and $\delta$ are both one-to-one mappings. However, $\gamma$ is not onto because there exists an element in $S$, namely $\tau$ (since $\tau\in T\to\tau\in S$ because $T\subseteq S$), that is not mapped to. Hence, there exists a mapping $\gamma$ from $S$ to $S$ that is one-to-one but not onto when $T\subseteq S$ and $T$ is infinite. Thus, $S$ is infinite. $\Box$
Question: Is this a good/correct proof? If not, where did I go wrong? If it is correct, then is there a way I can improve it or is there a more elegant approach?
• Looks good to me. The only thing I would change is to use the identity on $S\setminus T$ as $\delta$. – A.P. Apr 22 '15 at 17:33
• I find it perfect. – ajotatxe Apr 22 '15 at 17:34
• @A.P. I'm not sure what you mean exactly. What do you mean by "use the identity"? – fancynancy Apr 22 '15 at 17:34
• I mean the map $x \mapsto x$. – A.P. Apr 22 '15 at 17:34
• @A.P. Thanks for the input! That does it make it simpler since the identity mapping is always onto and one-to-one...I can make it more specific in that sense. :) – fancynancy Apr 22 '15 at 17:38
Looks good to me. The only thing I would change is to define $\delta$ as the identity on $S∖T$, i.e. as \begin{align} \delta \colon S \setminus T &\to S \setminus T \\ x &\mapsto x \end{align} | 2019-08-18T21:18:42 | {
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https://math.stackexchange.com/questions/1442558/find-a-sine-function-for-this-graph | # Find a sine function for this graph
I'm trying to find the equation for this graph, and my answer was:
Amplitude: $$2$$
Period: $$1$$ because $$\pi/2 + 3\pi/2 = 2\pi$$
Phase Shift: $$-\pi/2$$
Vertical shift: $$-2$$
So my answer is: $$2\sin(x + \pi/2) - 2$$ The problem is that when I enter this equation in Desmos Graphing Calcultor to make sure that my answer is right, I get similar graph but not the same one in my booklet
So what's wrong with my equation?
• this graph is $2\sin(\frac{1}{2}(x-\pi /2))-2$ Sep 19, 2015 at 18:32
• So your period and phase shift is wrong. Sep 19, 2015 at 18:36
• its period is $4\pi$ is easy to observe Sep 19, 2015 at 18:50
Booklet plot is a plot of $2 (\sin(x /2 - \pi/4) - 1),\, -2 \pi <x <2 \pi.$ | 2022-07-01T23:43:50 | {
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http://mlru.eyuc.pw/coordinate-transformation-matrix.html | # Coordinate Transformation Matrix
When using the transformation matrix, premultiply it with the coordinates to be transformed (as opposed to postmultiplying). The transformation matrix for this rotation is A = cos sin 0 sin cos 0 001 • Rotation about x-axis (or -axis) A = 10 0 0cos sin 0sin cos • Rotation about y-axis (or -axis) A = cos 0sin 01 0 sin 0cos Note the signs for the “sin ” terms! x y z x y z x y z Euler Angles • Euler angles are the most commonly used rotational coordinates. if someone has a better idea like something with coordinate matrix transformations - it. Rotational matrix 8 Problem 1. An affine transformation is a linear (or first-order) transformation and relates two 2D Cartesian coordinate systems through a rotation, a scale change in x- and y- direction, followed by a translation. Yaw, pitch, and roll rotations. After rectangular (aka Cartesian) coordinates, the two most common an useful coordinate systems in 3 dimensions are cylindrical coordinates (sometimes called cylindrical polar coordinates) and spherical coordinates (sometimes called spherical polar coordinates). 1 we defined matrices by systems of linear equations, and in Section 3. Transformation of Graphs Using Matrices - Translation A type of transformation that occurs when a figure is moved from one location to another on the coordinate plane without changing its size, shape or orientation is a translation. This is a 3x3 coordinate transformation matrix. –A square (n × n) matrix A is singular iff at least one of its singular values σ1, …, σn is zero. Though the matrix M could be used to rotate and scale vectors, it cannot deal with points, and we want to be able to translate points (and objects). The ECEF has two common coordinate systems: a polar-type “latitude–longitude– height” called geodetic coordinates, and the simpler three cartesian axes X,Y,Z that are. We are interested in calculating what the global coordinate representation is based on elemental coordinates, and vice versa. The Jacobian matrix represents the differential of f at every point where f is differentiable. I already find a solution to align the part to the selected direction, so I have the coordinates of the vector in the origin, but I want now to rotate the part about this axis with a given angle. Your English is fine! $\endgroup$ – user64687 Apr 25 '13 at 20:04. 369 at MIT Created April 2007; updated March 10, 2010 Itisaremarkablefact[1]thatMaxwell'sequa-tions under any coordinate transformation can be written in an identical "Cartesian" form, if simple transformations are applied to the ma-. For example, if the x-, y- and z-axis are scaled with scaling factors p, q and r, respectively, the transformation matrix is: Shear The effect of a shear transformation looks like pushing'' a geometric object in a direction parallel to a coordinate plane (3D) or a coordinate axis (2D). In computer vision, the transformation from 3D world coordinates to pixel coordinates is often represented by a 3x4 (3 rows by 4 cols) matrix P as detailed below. [email protected] Composing Transformations - Notation Below we will use the following convention to explain transformations = Matrix applied to left of vector Column vector as a point I am not concerned with how the matrix/vector is stored here – just focused on. Thus, each coordinate changes based on the values in the. Calculator for Applying Plane Stress Coordinate Transforms. corresponding transformation matrix is Eq. In particular for each linear geometric transformation, there is one unique real matrix representation. NET Core) application and Java (J2SE and J2EE) application. 2 XI2 Example 5-1. It has 4 matrix sorts: modelview, projection, texture, and colour matrices. It is very important to recognize that all coordinate transforms on this page are rotations of the coordinate system while the object itself stays fixed. The source code and files included in this project are listed in the project files section, please make sure whether the listed source code meet your needs there. Explore Solution 2. Then transformation matrix can be found by the function cv2. add 5 to each x-coordinate B. Now let's say we have some alternate. Coordinate Systems and Coordinate Transformations The field of mathematics known as topology describes space in a very general sort of way. Transformation Matrices. As the jacobian matrix is a collection of all derivatives of coordianates , the coordinate function must be continuous. Transform method requires a matrix object to set its. A transformation matrix describes the rotation of a coordinate system while an object remains fixed. Coordinate transformation. V g1 ւ g2 ց Rn −→ Rn The composition g2 g−1 1 is a transformation of R n. If layers in a map have different coordinate systems defined from those of the map or local scene itself, a transformation between the coordinate systems might be necessary to ensure data lines up correctly. A digital image array has an implicit grid that is mapped to discrete points in the new domain. Applying this to equation 1. This method prepends or appends the transformation matrix of the Graphics by the translation matrix according to the order parameter. In computer graphics, transform is carried by multiplying the vector with a transformation matrix, i. I want to change the ratio of mouse movement to pointer movement on my screen by changing the coordinate transformation matrix for the mouse with the command "xinput set-prop". Local transformations apply to a single object or collected set of shapes. This article is about Coordinate transformation. Keywords: 3D Coordinate Transformation, Total Least Squares, Least Squares, Minna Datum, WGS 84 INTRODUCTION The Nigerian coordinate system is based on the non-earth centred datum called “Minna Datum. Rotational matrix 8 Problem 1. Improve business processes. Coordinate Transformations. Transformation of coordinates in 4-vector notation. The first two-dimensional transformation is about the y-axis and relates the global axes to the 1-axes, i. If your transformation matrix is a rotation matrix then you can simplify the problem by taking advantage of the fact that the inverse of a rotation matrix is the transpose of that matrix. Before discussing how to calculate V, we need to discuss transformations of coordinate systems. Coordinate Transformation & Invariance in Electromagnetism Steven G. So I will often use the more general word 'transform' even though the word 'rotation' could be used in many cases. Let f[θ,r]==0 be the equation for a curve in polar coordinate. The rotation matrix is closely related to, though different from, coordinate system transformation matrices, $${\bf Q}$$, discussed on this coordinate transformation page and on this transformation matrix page. For example, R 2 is the rotation transformation matrix corre a sponding to a change from frame 1 to frame 2. 4) Then the position and orientation of the end-effector in the inertial frame are given by H = T0 n = A1(q1)···An(qn). lstsq - coordinate translations X * A = Y # to find our transformation matrix A A, res, rank to solve the matrix, using homogenous coordinates to. Although the mathematics of matrices are covered in Transform Mathematics, an important factor to note is that matrix multiplication is not always a commutative operation—that is, a times b does not always equal b times a. , change of basis) is a linear transformation!. The rigid bodies are approximately identical (i. Such a matrix can be found for any linear transformation T from $$R^n$$ to $$R^m$$, for fixed value of n and m, and is unique to the transformation. We always keep the same order for vectors in the basis. This page describes the transformations done to the coordinates given by the theories. I want to change the ratio of mouse movement to pointer movement on my screen by changing the coordinate transformation matrix for the mouse with the command "xinput set-prop". We begin with a space-time diagram, Fig. Coordinate Transformation Matrix in ABAQUS (UEL) Thu, 2015-01-22 13:15 - ashkan khalili. Coordinate Transformations. Transformations between ECEF and ENU coordinates Author(s) J. transformation produces shear proportional to the y coordinates. We construct characteristic lines to represent the coordinate systems. I do not understand the significance of this matrix (if not for coordinate transformation) or how it is derived. Let (x, y, z) be the standard Cartesian coordinates, and (ρ, θ, φ) the spherical coordinates, with θ the angle measured away from the +Z axis (as , see conventions in spherical coordinates). Let’s use this as our “data” image to help visualize what happens with each transformation. This is sometimes represented as a transformation from a Cartesian system (x 1, x 2, x 3) to the dimensionless system (ξ 1, ξ 2, ξ 3). Iftii P j it MtiInfinite Projection Matrix But there’s a problem The hardware doesnThe hardware doesn t actually perform ’t actually perform the perspective divide immediately after applying the projection matrix Instead, the viewport transformation is apppp (lied to the (x, y, z) coordinates first. Note that TransformationFunction[] is the head of the results returned by geometric *Transform functions, which take a homogeneous transformation matrix as an argument. Coordinate Vectors and Examples Coordinate vectors. edu Abstract The use of transformation matrices is common practice in both computer graphics and image processing, with ap-plications also in similar fields like computer vision. Graphics 2011/2012, 4th quarter Lecture 5: linear and a ne transformations. original matrix, A, with the eigenvalues lying on the diagonal of the new matrix,. Assemble the global stiffness matrix 3. Suppose that we are given a transformation that we would like to study. Such a matrix can be found for any linear transformation T from $$R^n$$ to $$R^m$$, for fixed value of n and m, and is unique to the transformation. The image of a linear transformation or matrix is the span of the vectors of the linear transformation. Q4, u4 b y m X (a. The input rotation matrix must be in the premultiply form for rotations. A special case is a diagonal matrix, with arbitrary numbers ,, … along the diagonal: the axes of scaling are then the coordinate axes, and the transformation scales along each axis by the factor In uniform scaling with a non-zero scale factor, all non-zero vectors retain their direction (as seen from the origin), or all have the direction. Applying a transformation to a point is accomplished by multiplying the homogenous coordinates of the point by the appropriate transformation matrix. This is touched on here, and discussed at length on the next page. Coordinate transformations are often used to de–ne often used to de–ne new coordinate systems on the plane. How exactly this is done will be covered in my matrix tutorial and is purely mathematical. [email protected] It is independent of the frame used to define it. This means a point whose coordinates are (x, y) gets mapped to another point whose coordinates are (x', y'). pdf), Text File (. 1 INTRODUCTION In general, the physical quantities we shall be dealing with in EM are functions of space and time. ECI & ECEF have co-located origins. The GL_MODELVIEW matrix, as its name implies, should contain modeling and viewing transformations, which transform object space coordinates into eye space coordinates. The calculation is based on the 2D linear coordinate transformation method detailed in chapter 18 of the fifth edition of Adjustment Computations - Spatial Data Analysis by Charles D. 𝑟𝑖𝑒=𝑟𝑖𝑒=𝑟𝑖𝑒=0. Now, maybe we can do one better than LU by nding not just a coordinate system in which our transformation becomes upper triangular, but an orthogonal coordinate system in which our transformation becomes upper triangular. Transformations between different coordinate systems We can interpret that the transformation matrix is converting the location of vertices between different coordinate systems. Tech in Computer Science and Engineering has twenty-three+ years of academic teaching experience in different universities, colleges and eleven+ years of corporate training experiences for 150+ companies and trained 50,000+ professionals. tensor (matrix) λ eigenvalue v eigenvector I Identity matrix AT transpose of matrix n, r rotation axis θ rotation angle tr trace (of a matrix) ℜ3 3D Euclidean space r u e ˆ 3 δij * in most texture books, g denotes an axis transformation, or passive rotation!!. S' is moving with respect to S with velocity (as measured in S) in the direction. The coordinates of a point, relative to a frame {}, rotated and translated with respect to a reference frame {}, are given by: = +, This can be compacted into the form of a homogeneous transformation matrix or pose (matrix). Points on the image can be described by [x,y] coordinates with the origin being at the center of the circle, and we can transform those points by using a 2D transformation matrix. Our approach fixes all the drawbacks of CGS and MMA. Rotate Touch Input with touchscreen and/or touchpad. You are able to provide any matrix of your choosing, but your choice of type will have a large effect on speed. This is the final step that transforms the coordinate system to the coordinate system. –basic rotation about origin: unit vector (axis) and angle •convention: positive rotation is CCW when vector is pointing at you. Take a matrix representation for a linear transformation in one basis and express that linear transfor-mation in another basis. In this chapter we will cover the following topics: The basics of transformation, including coordinate systems and matrices. These transformation equations are derived and discussed in what follows. The basic 4x4 Matrix is a composite of a 3x3 matrixes and 3D vector. global coordinate system and local coordinate system (at first we only consider the xz-plane) This relation is valid for any vector, e. Notethat, evenifwestartoutwithisotropic materials (scalar " and ), after a coordinate transformationweingeneralobtainanisotropic. Transformation of Graphs Using Matrices - Translation A type of transformation that occurs when a figure is moved from one location to another on the coordinate plane without changing its size, shape or orientation is a translation. add 5 to each x-coordinate B. The Affine transforms are represented in Homogeneous coordinates because the transformation of point A by any Affine transformation can be expressed by the multiplication of a 3x3 Matrix and a 3x1 Point vector. If your application uses a different 2D coordinate convention, you'll need to transform K using 2D translation and reflection. reference coordinate system. Homogeneous Coordinates and Transformations · Problem: Translation does not decompose into a 2 x 2 matrix · Solution: Represent Cartesian Coordinates (x,y) as Homogeneous Coordinates (x h, y h, h) Where. The coordinates of the fixed vector in the rotated coordinate system are now given by a rotation matrix which is the transpose of the fixed-axis matrix and, as can be seen in the above diagram, is equivalent to rotating the vector by a counterclockwise angle of relative to a fixed set of axes, giving. The rotation matrix is closely related to, though different from, coordinate system transformation matrices, $${\bf Q}$$, discussed on this coordinate transformation page and on this transformation matrix page. These can be obtained from their global coordinates using the corresponding transformation matrix. This matrix J is created by inverting the part of the model's Jacobian associated with beta and gamma and multiplying it by the part associated with alpha. The matrix of a linear transformation is like a snapshot of a person --- there are many pictures of a person, but only one person. In which transformation the shape of an object can be modified in x-direction ,y-direction as well as in both the direction depending upon the value assigned to shearing variables Reflection Shearing. y h x (x, y, z, h) Generalized 4 x 4 transformation matrix in homogeneous coordinates r = l m n s c f j b e i q a d g p [T] Perspective transformations Linear transformations – local scaling, shear, rotation / reflection. This is sometimes represented as a transformation from a Cartesian system (x 1, x 2, x 3) to the dimensionless system (ξ 1, ξ 2, ξ 3). Ask Question if applicable" --type=float "Coordinate Transformation Matrix" 0 -1 1 1 0 0 0 0 1. The implementation of transforms uses matrix multiplication to map an incoming coordinate point to a modified coordinate space. In general, the location of an object in 3-D space can be specified by position and orientation values. First, we need a little terminology/notation out of the way. Borrowing aviation terminology, these rotations will be referred to as yaw, pitch, and roll: A yaw is a counterclockwise rotation of about the -axis. Since we will making extensive use of vectors in Dynamics, we will summarize some of their important properties. Check transformation formula for spherical -> cartesian. I need to work on the transformed image, but I need the (x-y) coordinates of each corresponding pixel in the original image to finish my calculations. A two -by- n matrix is used to hold the position vectors for the figure. This basically undoes the current transformation, then sets the specified transform, all in one step. This is the basic idea of a new matrix factorization, the QR factorization, which. I'm trying perspective transformation of an image using homography matrix. I'm trying to get. Stress and Strain Transformation 2. Vocabulary words: transformation / function, domain, codomain, range, identity transformation, matrix transformation. Relationship Between the ECI and ECEF Frames. Matrix Structure for screen rotation. Note: Since clockwise rotation means rotating in the anti-clockwise direction by $- \theta$, you can just substitute $- \theta$ into the anti-clockwise matrix to get the clockwise matrix. Looking for coordinate transformation? Find out information about coordinate transformation. Other than giving yet another MMA, we introduce a new and, in some cases, optimal coordinate transformation to study such networks. matrix is called IJKtoLPS- or IJKtoRAS-matrix, because it represents the transformation from IJK to LPS or RAS. Once the element equations are expressed in a common coordinate system, the equations for each element comprising the structure can be assembled. In which transformation the shape of an object can be modified in x-direction ,y-direction as well as in both the direction depending upon the value assigned to shearing variables Reflection Shearing. In this way, we can represent the point by 3 numbers instead of 2 numbers, which is called Homogenous Coordinate system. Invert an affine transformation using a general 4x4 matrix inverse 2. Download Free Matrix III Coordinate Geometry. Transformation Matrices. /xinput-automatrix. To shorten this process, we have to use 3×3 transformation matrix instead of 2×2 transformation matrix. y z x u=(ux,uy,uz) v=(vx,vy,vz) w=(wx,wy,wz) (x0,y0,z0) • Solution: M=RT where T is a translation matrix by (x0,y0,z0), and R is rotation matrix. For example, consider the following matrix for various operation. multiply each y-coordinate by 1. 3: geometry of the 2D coordinate transformation The 2 2 matrix is called the transformation or rotation matrix Q. These points may not fall on grid points in the new domain. If also scale is False, a rigid/Euclidean transformation matrix is returned. Straight lines will remain straight even after the transformation. As a first step, it’s important that we characterize the relationship of each of reference coordinate frames of the robot’s links to the origin, or base, of the robot. The is invariant since it is a dot product. ■ Stiffness matrix of the plane stress element in the local coordinate system: ■ Stiffness matrix of the flat shell element in the local coordinate system. The calculation is based on the 2D linear coordinate transformation method detailed in chapter 18 of the fifth edition of Adjustment Computations - Spatial Data Analysis by Charles D. Then transformation matrix can be found by the function cv2. Coordinates in PDF are described in 2-dimensional space. Maths - Combined Rotation and Translation. Though the matrix M could be used to rotate and scale vectors, it cannot deal with points, and we want to be able to translate points (and objects). What is the procedure (matrix) for change of basis to go from Cartesian to polar coordinates and vice versa? Coordinate transformation problems. A vector could be represented by an ordered pair (x,y) but it could also be represented by a column matrix: $$\begin{bmatrix} x\\ y \end{bmatrix}$$ Polygons could also be represented in matrix form, we simply place all of the coordinates of the vertices into one matrix. Generally, coordinate transformation in matrix operations needs mixed matrix operations where both multiplication and addition of matrices must be used. Coordinate transformation. Missions : pilot complex bank transformation along with compliance projects. specification of a viewing transformation, a 4×4 matrix that transforms a region of space into image space. Base vectors e 1 and e 2 turn into u and v, respectively, and these vectors are the contents of the matrix. Transformation matrix. Applying this to equation 1. Brown and Raymond D. A vector could be represented by an ordered pair (x,y) but it could also be represented by a column matrix: Polygons could also be represented in matrix form, we simply place all of the coordinates of the vertices into one matrix. The transformation matrix for this rotation is A = cos sin 0 sin cos 0 001 • Rotation about x-axis (or -axis) A = 10 0 0cos sin 0sin cos • Rotation about y-axis (or -axis) A = cos 0sin 01 0 sin 0cos Note the signs for the “sin ” terms! x y z x y z x y z Euler Angles • Euler angles are the most commonly used rotational coordinates. The resulting 2x1 matrix converts alpha only into beta and gamma, so it must be prepended with a 1 to convert alpha into all three coordinates. Note the distinction between a vector and a 3×1 matrix: the former is a mathematical object independent of any coordinate system, the latter is a representation of the vector in a particular coordinate system - matrix notation, as with the index notation, relies on a. the same form (1–4) in the primed coordinate system, with rreplaced by r0, if we make the transformations: E0= (JT) 1E; (6) H0= (JT) 1H; (7) "0= J"JT detJ; (8) 0= J JT detJ; (9) J0= J detJ; (10) ˆ0= ˆ detJ; (11) whereJT isthetranspose. Let f[x,y]==0 be the equation for a curve in rectangular coordinates. Stress and Strain Transformation 2. The AFFINE equations use six parameters. ) and perspective transformations using homogenous coordinates. Are there any other type of matrices, apart from the rotation matrices, which can be thought as coordinate systems? If yes, which ones, and why? Matrices usually represent a transformation (linear or not, maybe also affine in computer graphics), but it's new to me to think about matrices as coordinate systems. The second column of the linear part of the transformation matrix is (0 0 1) and the second element of the origin shift is 1/4 (or 0. A convenient way to transform one vector to another is through matrix multiplication. 2 Rotation of a vector in fixed 3D coord. The true power from using matrices for transformations is that we can combine multiple transformations in a single matrix thanks to matrix-matrix multiplication. This would require to determine the angles between the axis coordinates so that we can use them to rotate the dimensions of the point. If we want to figure out those different matrices for different coordinate systems, we can essentially just construct the change of basis matrix for the coordinate system we care about, and then generate our new transformation matrix with respect to the new basis by just applying this result. It is not hard to show that the matrix representation of the composition of transformations is the product of the individual matrix representations. The third column of the linear part of the transformation matrix is (1 0 0) and the third element of the origin shift is 1/4 (or 0. Transformations between coordinate systems. The superscript f is an indicator identifying the particular reference frame to which the axis, , belongs. The inverse transformation is , so, if the range of is , then Hence the disk with center and radius is mapped one-to-one and onto the disk with center and radius , as shown in Figure 2. Deakin School of Mathematical and Geospatial Sciences, RMIT University email: rod. transformation matrix. If we set the coefficients of the scaling matrix with Sx = 1, Sy = 2 and Sz = 3, then P multiplied by this matrix gives another point whose coordinates are (1, 4, 9). Please I need your insight on building my concept. Specifying rotations. Mapping from (x,y) to (u,v) coordinates. In this section, we make a change in perspective. In this video I presented the coordinate transformation in two methods. The Jacobian is given by: Plugging in the various derivatives, we get. Once the element equations are expressed in a common coordinate system, the equations for each element comprising the structure can be assembled. Devise a test whether a given 3 3 transformation matrix in homogeneous coordinates is a rigid body transformation in 2 dimensions. Generally, coordinate transformation in matrix operations needs mixed matrix operations where both multiplication and addition of matrices must be used. One type of transformation is a translation. It illustrates the difference between a tensor and a matrix. Now for the mapping part, we have two options how to proceed: Either we try to set up a rotation matrix that rotates the vertex into place in camera space. However, if you try to map this coordinate from the transformed grid onto the original grid, it is (4, 1). 2 that the transformation equations for the components of a vector are ui Qiju j, where Q is the transformation matrix. The individual coordinates of a transformed point are obtained from the equations where M mn is the Model to World Transformation Matrix coordinates, (X,Y,Z) is the entity definition data point expressed in MCS coordinates, and (X',Y',Z') is the resulting entity definition data point expressed in WCS coordinates. For example, consider a camera matrix that was calibrated with the origin in the top-left and the y-axis pointing downward, but you prefer a bottom-left origin with the y-axis pointing upward. Based on an analysis of the structures of the coordinate transformation matrix and the Lyapunov matrix, the open question of how to fix the Lyapunov matrix structure raised by G. · Cylindrical Coordinate · Spherical Coordinate · Transform from Cartesian to Cylindrical Coordinate · Transform from Cartesian to Spherical Coordinate · Transform from Cylindrical to Cartesian Coordinate · Transform from Spherical to Cartesian Coordinate · Divergence Theorem/Gauss' Theorem · Stokes' Theorem · Definition of a Matrix. The transformation from geodetic coordinates to rectangular space coordinates: geodetic coordinates (B, L, H) are transformed into the corresponding rectangular space coordinates (X, Y, Z). This is a symmetry. are linear and epoch-independent). The initial vector is submitted to a symmetry operation and thereby transformed into some resulting vector defined by the coordinates x', y' and z'. Composing Transformation Composing Transformation - the process of applying several transformation in succession to form one overall transformation If we apply transform a point P using M1 matrix first, and then transform using M2, and then M3, then we have: (M3 x (M2 x (M1 x P ))) = M3 x M2 x M1 x P M (pre-multiply). In S, we have the co-ordinates and in S' we have the co-ordinates. We will first examine the different types of transformations we will encounter, and then learn how to find the transformation matrix when given a graph. Remember that they are usually defined (in the robotics world) in terms of the local coordinate system whereas position is usually defined in terms of the global coordinate system. The values Ux, Uy and Uz are the co-ordinates of a point on the U axis which has unit distance from origin. It's encoded in row-major order, so the matrix would look like the following in a text book: ⎡ 1 0 0 ⎤ ⎜ 0 1 0 ⎥ ⎣ 0 0 1 ⎦ Astute readers will recognize that this is the identity matrix. the world, "window" and device coordinate systems are equivalent, but as we have seen, the systems can be manipulated using transformation operations and window-viewport conversion. If only two tics are matched, a similarity transformation will be applied. • The transformation can be written as a direct linear transformation 2x4 projection matrix 2x2 intrinsic parameter matrix 2x3 matrix = first 2 rows of the rotation matrix between world and camera frames First 2 components of the translation between world and camera frames Note: If the last row is the coordinates equations degenerate to:. For example, CECI ENU denotes the coordinate transformation matrix from earth-centered inertial. Or, we can transform all the points and normals from the original frame to the new frame. max max (Figure 2. 3D Programming Transformation Matrix Tutorial For starters, let’s briefly go over the idea of displaying a 3D world in a computer screen. It illustrates the difference between a tensor and a matrix. transformation produces shear proportional to the y coordinates. I have written the code attached below in matlab. Scale transformations in which one or three of a, b, and cis negative reverse orientation: a triple of vectors v 1;v 2;v 3 that form a right-handed coordinate system will, after transformation by such a matrix, form a left-handed coordinate system. If I had the matrix, I could derive the second image from the first (or vice-versa using the inverse matrix) myself. Composite TransformationMore complex geometric & coordinate transformations can be built from the basic transformation by using the process of composition of function. Take a matrix representation for a linear transformation in one basis and express that linear transfor-mation in another basis. In the above equations we’ve replaced the product of two transform matrices, R (rotation) and T (translation), with a single transform matrix, M, using the associativity property of the matrix multiplication. The elements of the matrix [v] can be written in the index notation vi. COORDINATE TRANSFORMATIONS IN SURVEYING AND MAPPING R. In detail, with respect to a given point x∈ ℝn, the linear transformation represented by J takes a position vector in ℝn from x as reference point as input and produces the position vector in ℝm from f as reference point obtained by multiplying by J as output. In this video I presented the coordinate transformation in two methods. But I just keep getting abnormal results. Transformation matrix. Scribd is the world's largest social reading and publishing site. 2 Rotation of a coordinate system in 2D 14. Interpolator - method for obtaining the intensity values at arbitrary points in coordinate system from the values of the points defined by the Image. This is the general transformation of a position vector from one frame to another. in Physics Hons with Gold medalist, B. In this article we will try to understand in details one of the core mechanics of any 3D engine, the chain of matrix transformations that allows to represent a 3D object on a 2D monitor. Rotate the object so that the axis rotation coincides with one of. This is what I plan to do: With respect to this image I have a set of points which are in the XYZ coordinate system (Red). Stress and Strain Transformation 2. Available are the gravity vector [g]s and the displacement vector of the radar wrt the missile both measured in body coordinates. Note the distinction between a vector and a 3×1 matrix: the former is a mathematical object independent of any coordinate system, the latter is a representation of the vector in a particular coordinate system – matrix notation, as with the index notation, relies on a. Transformations in the Coordinate Plane. A rotation matrix for any axis that does not coincide with a coordinate axis can be set up as a composite transformation involving combination of translations and the coordinate-axes rotations: 1. 3D Transformations World Window to Viewport Transformation Week 2, Lecture 4 David Breen, William Regli and Maxim Peysakhov Department of Computer Science Drexel University 2 Outline • World window to viewport transformation • 3D transformations • Coordinate system transformation 3 The Window-to-Viewport Transformation. Sanz Subirana, J. For such motion, a more encompassing frame tied to the fixed stars is used, but we won't need such a one in this report. Here we are representing the coordinate frames with unit vectors [x, y, z] and [b1, b2, b3]. $\begingroup$ The transformation matrix is a Jacobian matrix limited to linear transformations. (2006), American Congress for Surveying and Mapping Annual Conference. If we can prove that our transformation is a matrix transformation, then we can use linear algebra to study it. Coordinate transformation should be smooth and continuous so that we can go from one point to another point without making any sudden jump. A ne transformations preserve line segments. If the vector is NULL/empty, the zero distortion. Many spaces are exotic and have no counterpart in the physical world. , the stresses and ser~ns in. This basically undoes the current transformation, then sets the specified transform, all in one step. Translate the object so that the rotation axis passes through the coordinate origin 2. the determinant of the Jacobian Matrix. It has the form x → Ux, where U is an n×n matrix. Here [A] is a transformation matrix, and x i is the translation of the origin of the body coordinate system with respect to the global coordinate system. and covariant rank 3 (i. A vtkTransform can be used to describe the full range of linear (also known as affine) coordinate transformations in three dimensions, which are internally represented as a 4x4 homogeneous transformation matrix. What is required at this point is to change the setting (2D coordinate space) in which we phrased our original problem. original matrix, A, with the eigenvalues lying on the diagonal of the new matrix,. "Dilation transformation matrix" is the matrix which can be used to make dilation transformation of a figure. A convenient way to transform one vector to another is through matrix multiplication. • Stress tensor transformation • Matrix notation 1 1 1 xx xy xz 12 3 new 2 2 2 xy yy yz 1 2 3 3 3 3 xz yz zz 12 3 l m n ll l T l mn m mm l m n nn n σ σσ = σ σσ σ σσ 12 3 12 3 12 3 T new old ll l r mm m nn n T rT r rotation matrix:measured from old system = =. $xinput list$ xinput list-props "ADS7846 Touchscreen" This is coordinate transformation matrix that transform from input coordinate(x, y, z) to output coordinate(X, Y, Z). Transformation Code. For clarity, only the stress components on the positive faces are shown. matrix is called IJKtoLPS- or IJKtoRAS-matrix, because it represents the transformation from IJK to LPS or RAS. This article is mainly for B. Thus, we see that the off-diagonal terms produce a shearing effect on the coordinates of the position vector for P. "Dilation transformation matrix" is the matrix which can be used to make dilation transformation of a figure. To find the image of a point, we multiply the transformation matrix by a column vector that represents the point's coordinate. The Jacobian matrix represents the differential of f at every point where f is differentiable. Such transformations allow us to represent various quantities in different coordinate frames, a facility that we will often exploit in subsequent chapters. tation matrix that encodes the attitude of a rigid body and both are in current use. Restrict the global stiffness matrix and force vector 4. the determinant of the Jacobian Matrix. Objective: Given: a ij, Find: Euler angles (θ x, θ y, θ z). The direction of this plane is determined by three angles, the argument of thw perigee , the right ascension of the node , and the angle of inclination. reflection translation rotation dilation Cut the flap on every third line. The general affine transformation is commonly written in homogeneous coordinates as shown below: By defining only the B matrix, this transformation can carry out pure translation: Pure rotation uses the A matrix and is defined as (for positive angles being clockwise rotations): Here, we are working in image coordinates, so the y axis goes downward. Transformations in the coordinate plane are often represented by "coordinate rules" of the form (x, y) --> (x', y'). 1 INTRODUCTION In general, the physical quantities we shall be dealing with in EM are functions of space and time. Ap, Bp, etc. Composing Transformation Composing Transformation - the process of applying several transformation in succession to form one overall transformation If we apply transform a point P using M1 matrix first, and then transform using M2, and then M3, then we have: (M3 x (M2 x (M1 x P ))) = M3 x M2 x M1 x P M (pre-multiply). multiplying the original coordinates by the transformation matrix (here just for the x-axis): One can see that the pixel distance between the two coordinate sets are: 35,76. ) and perspective transformations using homogenous coordinates. If, for example, the inertial frame is chosen, coordinate rotation may be achieved by premultiplying the vector f b by the direction cosine matrix (DCM), C b i,. Problems of Eigenvalues and Eigenvectors of Linear Transformations. A vector could be represented by an ordered pair (x,y) but it could also be represented by a column matrix: $$\begin{bmatrix} x\\ y \end{bmatrix}$$ Polygons could also be represented in matrix form, we simply place all of the coordinates of the vertices into one matrix. I do not understand the significance of this matrix (if not for coordinate transformation) or how it is derived. (3) The displaced coordinate system is rotated about the -axis by an angle. system and I want to convert to someone else’s coordinate system. In fact an arbitary a ne transformation can be achieved by multiplication by a 3 3 matrix and shift by a vector. A 4×4 matrix can represent any possible. The Jacobian matrix represents the differential of f at every point where f is differentiable. The window defines what is to be viewed; the viewport defines where it is to be displayed. If the viscous damping matrix can be written as a linear combination of the mass and stiffness matrices, then the damping is said to be proportional viscous damping. The formal mathematical way to perform a coordinate transformation is. Transformations can be entered in the form oldchart-> newchart, where oldchart and newchart are valid chart specifications available from CoordinateChartData. translation matrix A translation matrix is a matrix that can be added to the vertex matrix of a figure to find the coordinates of the translated image. Coordinate Vectors and Examples Coordinate vectors. Right-click on an object in the Project Explorer and select the Transform command. Juan Zornoza and M. The values of these six components at the given point will change with. Wolfram|Alpha has the ability to compute the transformation matrix for a specific 2D or 3D transformation activity or to return a general transformation calculator for rotations, reflections and shears. In most books on QFT, Special Relativity or Electrodynamics, people talk about Lorentz transformations as some kind of special coordinate transformation that leaves the metric invariant and then they define what they call the Lorentz scalars. Thus, each coordinate changes based on the values in the. resetTransform() Resets the current transform to the identity matrix. Once you choose a particular coordinate system, you can represent the tensor in that coordinate system by using a matrix. Coordinate transformations play an important role in defining multiple integrals, sometimes allowing us to simplify them. Here are descriptions of the stages that are shown in the preceding figure: World matrix Mworld transforms vertices from the model space to the world space. Detailed Description. f(x,y) = (ax +by +c,dx +ey +f) for suitable constants a, b, etc. Robot control part 1: Forward transformation matrices. The inverse transformation is , so, if the range of is , then Hence the disk with center and radius is mapped one-to-one and onto the disk with center and radius , as shown in Figure 2. To do so, we will need to learn how we can "project" a 3D point onto the surface of a 2D drawable surface (which we will call in this lesson, a canvas) using some simple geometry rules. Transformation matrix. The upper left 3x3 portion of a transformation matrix is composed of the new X, Y, and Z axes of the post-transformation coordinate space. | 2019-12-13T23:15:49 | {
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https://math.stackexchange.com/questions/2812628/what-is-the-difference-in-clearing-the-exponent | # What is the difference in clearing the exponent?
If I have an equation, like this:
$k^{m} = m^{k}$,
I have been taught that I must apply natural logarithm, but others have taught me that I must apply the common logarithm.
Then, what should I use?
$m*ln(k)= k*ln(m)$, Natural logarithm (base $e$)?
$m*log(k)= k*log(m)$, Common logarithm (base $10$)?
I guess, the only difference is that:
Natural logarithm will leave the answer in terms of $e$ and common logarithm, will leave the answer in terms of $10$, but I'm not sure about this and I need a really good explanation, when it's convenient to use one and the other.
• Always natural logarithm. Always. – Lord Shark the Unknown Jun 8 '18 at 15:20
• $\ln$ is more popular than $\log$ – Sujit Bhattacharyya Jun 8 '18 at 15:23
• Why ? @LordSharktheUnknown – Eduardo S. Jun 8 '18 at 15:28
• Common and natural logs are proportional, so either will serve to answer the question. The natural logarithm is, well, natural (for mathematicians), It's the most useful in abstract mathematics. Base $10$ logarithms are a historical accident, coming from the fact that we have $10$ fingers, so base $10$ notation for integers. The common logarithm was invented to speed calculation. The only other logarithm you encounter these days is base $2$, useful in computer science. – Ethan Bolker Jun 8 '18 at 15:35
• Use whatever logarithm you like. If you use base $k$ you get $m = k\log_k m$ and if you use base $m$ you get $k = m\log_m k$. But all $m \log_b k = k\log_b m$ will all be true statements, so whichever helps you solve will work. (Although frustratingly none really help much.) – fleablood Jun 8 '18 at 16:18
$k^m = m^k \implies m\log_b k = k\log_a m$ will be true no matter what base $b$ you choose. And as $\log_b k$ an $\log_b m$ will always be in the same proportion ($\frac {\log_b k}{\log_b m} = \frac {\log_a k}{\log_a m}$ for all legitimate $a,b$) it doesn't matter which you pick.
Unless you are doing scientific notation where units and measurements are specifically designed to be represented in powers of $10$s there is nothing advantageous about $10$ over, say, $17$. And $k \log_{17} m = m\log_{17}k$ is a perfectly true and legitimate statement.
But if you are doing scientific notation where units are based on powers of $10$ then $\log_{10}$ has an obvious advantage.
If you are doing anything that might even remotely no matter how obliquely involve differentiation or integration (or even tangents or rates of change) you should use $\ln$ as it is ... natural. So that is why it is conventional to default to natural logs.
I'm surprised though that no-one has suggested logs based $k$ or $m$. That has the advantage of reducing an equation with two logarithms to one. $m = k\log_k m$ and $k = m\log_m k$ which could often help us. Although in this case it doesn't)
Use whatever base you like. Sometimes there will be practical advantages to use a specific base (Solve for $3^{27x} = y^{81} \cdot 3^{7}$ just screams for base $3$) but usually there won't be. The convention is math is base $e$. I imagine im must sciences is is also base $e$ but I imagine there are same instances where convention is $10$.
But it doesn't matter.
===
$3^{27x} = y^{81} \cdot 3^{7}$
$\log_3 3^{27x} = \log_3(y^{81} \cdot 3^{7})$
$27x = 81\log_3 y + 7$
$x = 3\log_3 y + \frac 7{27}$.
But you could just as well (but not as easily solve it with natural logs.
$\ln 3^{27x} = \ln(y^{81} \cdot 3^{7})$
$27x \ln 3 = 81\ln y + 7\ln 3$.
$x = \frac {81\ln y + 7\ln 3}{27\ln 3}$
$= 3\frac {\ln y}{\ln 3} + \frac 7{27}$. The same answer.
And if we had use common $\log$ wed have gotten.
$x = 3\frac {\log y}{\log 3} + \frac 7{27}$.
All the same.
• thanks, good answer – Eduardo S. Jun 8 '18 at 22:42
$$k^m=m^k\to m\log_a(k)=k\log_a(m)\to\frac1k\log_a(k)=\frac1m\log_a(m)\space\forall a\in \Bbb R^+$$
The most popular choice for $a$ is Euler's Constant $e$, for which we use the notation $\log_e(x)=\ln(x)$
Hence: $$\frac 1k\ln(k)=\frac 1m\ln (m)$$ | 2020-01-24T02:36:08 | {
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https://math.stackexchange.com/questions/2107922/what-is-the-probability-of-getting-at-least-one-more-head-from-two-coin-flips-th | # What is the probability of getting at least one more head from two coin flips than one coin flip?
Let's say that I have two fair coins and my opponent has one fair coin. Me and my opponent can flip each of our coins once. I win only if I get at least one more head than my opponent. What is the probability of me winning?
I don't understand how to think about this problem in terms of the probabilities. The event of me getting at least one head from the two coin flips is $$P(TH) + P(HH) = 0.5 + 0.25 = 0.75$$ while the probability of my opponent getting at least one head is $$P(H) = 0.5.$$ I am confused by contrasting my opponents probability of winning with mine. How would I go about solving this problem?
If the first two flips are yours and the third is your opponent's you get 8 possible outcomes: $HH\ H, HH\ T,\ldots, TT\ T$, each with probability $1/8$
$HH\ H, HH\ T, HT\ T, TH\ T$.
So probability of winning is $4/8=1/2$
Or you can use independence:
$$P(\text{winning})=P(\text{you 2 heads, opponent 1 head})+P(\text{you 2 heads, opponent 0 heads})+P(\text{you 1 head, opponent 0 heads})\\=P(\text{you 2 heads})P(\text{opponent 1 head})+P(\text{you 2 heads})P(\text{opponent 0 heads})+P(\text{you 1 head})P(\text{opponent 0 heads})\\=\frac{1}{4}\cdot\frac{1}{2}+\frac{1}{4}\cdot\frac{1}{2}+\frac{2}{4}\cdot\frac{1}{2}=\frac{4}{8}=\frac{1}{2}$$
• Great, this makes sense! If we wanted to solve this problem by using a formula, would we use the combinatorics formula? – verkter Jan 21 '17 at 21:43
• I added a solution using formula. – Momo Jan 21 '17 at 21:46
• But why is the P(you 2 head, opponent 0 heads) is not considered in your formula? – verkter Jan 22 '17 at 1:55
• Because you only win if you get one more head than your opponent, not two more. – Momo Jan 22 '17 at 3:39
• But getting two heads, while opponent gets none is still a possibility if we evaluate results after both tosses are complete. – verkter Jan 22 '17 at 3:54
$\underline{Answer\; for\; you\; getting\; at\; least\; one\; more\; head}$
Consider that initially, you both toss only one coin.
You can win only if you have already won, or are equal and win with your second toss.
Denoting your results in caps, and opponents in lowercase for clarity,
P(you win) = P(Ht) + P(HhH) + P(TtH) = $\dfrac12 + \dfrac14 + \dfrac14 = \dfrac12$
Interestingly, if you toss (n+1) coins against n tossed by your opponent, P(You win) is still $\dfrac12$
After tossing $n$ coins each, let $p$ be the probability that you are ahead. By symmetry, $p$ is also the probability that your opponent is ahead, and the probability of a tie is $1-2p$. You have just two ways to win: either you are ahead before the last toss, or there is a tie and you then get $H$.
Thus P(You win) $= p + (1-2p)\cdot\frac 12 = p+\frac 12 -p =\frac 12$
• Will it affect the results if we label one of your coins 1 and one of them 2 ? I have added material that explains your win probability remains 1/2 even if you have $(n+1)$ coins against your opponent's $n$. I think it should make it very clear. – true blue anil Jan 22 '17 at 4:00 | 2019-08-20T07:15:08 | {
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https://math.stackexchange.com/questions/3254893/show-that-the-basis-for-an-inner-product-space-v-is-an-orthonormal-basis | # Show that the basis for an inner product space $V$ is an orthonormal basis
Suppose that $$\{v_1,...,v_n\}$$ be a basis for an inner product space $$V$$, and $$v=a_1v_1+...+a_nv_n$$ implies $$||v||^2=a_1^2+...+a_n^2$$, then can we say that the given basis is orthonormal? If so how?
I tried that $$=\sum_{i,j}a_ia_j$$ given that it is equal to $$a_1^2+...+a_n^2$$, thus can we conclude?
Hint: Note that $$\left\| v\right\|^2 = \langle v , v\rangle$$ for all $$v\in V$$. Therefore, you can show using inner product properties that $$\left\| v + w\right\|^2 = \left\|v \right\|^2 + \left\|w \right\|^2 + 2 \langle v, w\rangle$$ in a real inner product space, so
$$\langle v, w\rangle = \frac{\left\| v + w\right\|^2 - \left\|v \right\|^2 - \left\|w \right\|^2}{2},$$
for all $$v,w\in V$$.
Try to use this to show that $$\langle v_i , v_i\rangle = 1$$ for all $$i$$ and $$\langle v_i , v_j\rangle = 0$$ for all $$i\ne j$$.
• I am not getting how to apply this concept – RIYASUDHEEN TK 9747408592 Jun 8 at 5:25
• OK. Another hint: Note that $v_i + v_i = 2v_i$. Hence $\left\| v_i + v_i \right\|^2 = 2^2 = 4$ (using the assumption in the question). Similarly, find $\left\| v_i\right\|^2$ using the question's assumption, and then you have everything you need to find $\langle v_i, v_i \rangle$ using the equation in my answer. Then do a similar thing to find $\langle v_i, v_j\rangle$. – Minus One-Twelfth Jun 8 at 5:27
The hypothesis implies $$\|v_i\|^{2}=1$$ for all $$i$$ so each $$v_i$$ has norm $$1$$. Now consider $$\|v_i+v_j\|^{2}$$ where $$i \neq j$$. We get $$\|v_i+v_j\|^{2}=1^{2}+1^{2}=2$$. Expanding LHS in terms of the inner product we get $$\|v_i\|^{2}+\|v_i\|^{2}+2\langle v_i,v_j \rangle =2$$ which gives $$\langle v_i,v_j \rangle=0$$. This is the proof when teh scalare are real . I will leave the complex case to you.
• Ok i got it, thnk u sir – RIYASUDHEEN TK 9747408592 Jun 8 at 5:34
• In that case you can consider approving the answer by clicking on the tick mark. – Kabo Murphy Jun 8 at 5:46 | 2019-12-07T22:04:42 | {
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https://math.libretexts.org/Bookshelves/Applied_Mathematics/Book%3A_Introduction_to_the_Modeling_and_Analysis_of_Complex_Systems_(Sayama)/17%3A_Dynamical_Networks_II_-_Analysis_of_Network_Topologies/17.5%3A_Degree_Distribution |
# 17.5: Degree Distribution
$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$
$$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$
Another local topological property that can be measured locally is, as we discussed already, the degree of a node. But if we collect them all for the whole network and represent them as a distribution, it will give us another important piece of information about how the network is structured:
A degree distribution of a network is a probability distribution
$P(k) =\frac{| \begin{Bmatrix} i | deg(i) =k \end{Bmatrix}| }{ n}. \label{(17.28)}$
i.e., the probability for a node to have degree $$k$$.
The degree distribution of a network can be obtained and visualized as follows:
The result is shown in Fig. 17.5.1.
Figure $$\PageIndex{1}$$: Visual output of Code 17.13.
You can also obtain the actual degree distribution P(k) as follows:
This list contains the value of (unnormalized) $$P(k) for k = 0,1,...,k_{max}$$, in this order. For larger networks, it is often more useful to plot a normalized degree histogram list in a log-log scale:
The result is shown in Fig. 17.5.2, which clearly illustrates differences between the three network models used in this example. The Erd˝os-R´enyi random network model has a bell-curved degree distribution, which appears as a skewed mountain in the log-log scale (blue solid line). The Watts-Strogatz model is nearly regular, and thus it has a very sharp peak at the average degree (green dashed line; $$k = 10\0 in this case). The Barab´asi-Albert model has a power-law degree distribution, which looks like a straight line with a negative slope in the log-log scale (red dotted line). Moreover, it is often visually more meaningful to plot not the degree distribution itself but its complementary cumulative distribution function (CCDF), defined as follows: $F(k) =\sum_{k'=k} ^{\infty} P(k') \label{(17.29)}$ This is a probability for a node to have a degree \(k$$ or higher. By definition, $$F(0) = 1$$ and $$F(k_{max} + 1) = 0$$, and the function decreases monotonically along $$k$$. We can revise Code 17.15 to draw CCDFs:
Figure $$\PageIndex{2}$$: Visual output of Code 17.15.
In this code, we generate ccdf’s from Pk by calculating the sum of Pk after dropping its first k entries. The result is shown in Fig. 17.5.2.
As you can see in the figure, the power law degree distribution remains as a straight line in the CCDF plot too, because $$F(k)$$ will still be a power function of$$k$$, as shown below:
Figure $$\PageIndex{3}$$: Visual output of Code 17.16.
F(k ) \begin{align} \sum_{k'=k} ^{\infty} {P(k')} = \sum_{k'=k} ^{infty}{ak' ^{-\gamma}} \label{(17.30)} \\ \approx \int_{k}^{infty} ak^{-\gamma}dk' = {\begin{bmatrix} \frac{ak^{J-\gamma+1}}{-\gamma + 1}\end{bmatrix}}^{\infty}_{k} = \frac{0-ak^{-\gamma+1}}{-\gamma +1} \label{(17.31)} \\ = \frac{a}{-\gamma -1}k ^{-(\gamma -1)} \label{(17.32)} \end{align}
This result shows that the scaling exponent of $$F(k)$$ for a power law degree distribution is less than that of the original distribution by 1, which can be visually seen by comparing their slopes between Figs. 17.5.2 and 17.5.3.
Exercise $$\PageIndex{1}$$
Import a large network data set of your choice from Mark Newman’s Network Data website: http://www-personal.umich.edu/~mejn/netdata/
Plot the degree distribution of the network, as well as its CCDF. Determine whether the network is more similar to a random, a small-world, or a scale-free network model.
If the network’s degree distribution shows a power law behavior, you can estimate its scaling exponent from the distribution by simple linear regression. You should use a CCDF of the degree distribution for this purpose, because CCDFs are less noisy than the original degree distributions. Here is an example of scaling exponent estimation applied to a Barab´asi-Albert network, where the linregress function in SciPy’s stats module is used for linear regression:
In the second code block, the domain and ccdf were converted to log scales for linear fitting. Also, note that the original ccdf contained values for all k’s, even for those for which $$P(k) = 0$$. This would cause unnecessary biases in the linear regression toward the higher k end where actual samples were very sparse. To avoid this, only the data points where the value of $$F$$ changed (i.e., where there were actual nodes with degree k) are collected in the logkdata and logFdata lists.
The result is shown in Fig. 17.5.4, and also the following output comes out to the terminal, which indicates that this was a pretty good fit:
Figure $$\PageIndex{4}$$: Visual output of Code 17.17.
According to this result, the CCDF had a negative exponent of about -1.97. Since this value corresponds to$$−(γ−$$1$$)$$, the actual scaling exponent $$γ$$ is about 2.97, which is pretty close to its theoretical value, 3.
Exercise $$\PageIndex{2}$$
Obtain a large network data set whose degree distribution appears to follow a power law, from any source (there are tons available online, including Mark Newman’s that was introduced before). Then estimate its scaling exponent using linear regression. | 2019-08-24T18:56:57 | {
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http://math.stackexchange.com/questions/62126/proof-that-6n-always-has-a-last-digit-of-6/62133 | # Proof that $6^n$ always has a last digit of $6$
Without being proficient in math at all, I have figured out, by looking at series of numbers, that $6$ in the $n$-th power always seems to end with the digit $6$.
Anyone here willing to link me to a proof?
I've been searching google, without luck, probably because I used the wrong keywords.
-
@Stijn: your comment seems rather obnoxious. The OP did not say that he has "proved" anything; he made an observation which he thinks "seems to always" hold. And he's right... – Pete L. Clark Sep 6 '11 at 0:47
@Stijn: Ragnar said "seems", so s/he knows it's not a proof. :) – Guess who it is. Sep 6 '11 at 1:19
@J.M. A p.c. "s/he" in combination with a person called Ragnar is making me chuckle... – t.b. Sep 6 '11 at 1:27
@Theo: I've been "victimized" by ladies using "manly" names on the Internet, so I'm covering myself just in case. :D – Guess who it is. Sep 6 '11 at 1:34
My comment wasn't meant to be obnoxious. I'll delete it if it comes across as such. – Stijn Sep 6 '11 at 8:40
We can prove it using mathematical induction.
Claim: $6^n\equiv 6\bmod 10$ for all $n\in\mathbb{N}$ (the symbol $\mathbb{N}$ denotes the natural numbers, and $\bmod 10$ means we are using modular arithmetic with a modulus of 10).
Base case (i.e., showing it's true for $n=1$): $$6^1\equiv 6\bmod 10\qquad\checkmark$$
Induction step (i.e., showing that, if it is true for $n=k$, then it is true for $n=k+1$):
$$6^k\equiv 6\bmod 10\implies 6^{k+1}\equiv 6^k\cdot 6\equiv6\cdot 6\equiv 36\equiv 6\bmod 10\qquad\qquad\checkmark$$
-
If you multiply any two integers whose last digit is 6, you get an integer whose last digit is 6: $$\begin{array} {} & {} & {} & \bullet & \bullet & \bullet & \bullet & \bullet & 6 \\ \times & {} & {} &\bullet & \bullet & \bullet & \bullet & \bullet & 6 \\ \hline {} & \bullet & \bullet & \bullet & \bullet & \bullet & \bullet & \bullet & 6 \end{array}$$ (Get 36, and carry the "3", etc.)
To put it another way, if the last digit is 6, then the number is $(10\times\text{something}) + 6$. So \begin{align} & {} \qquad \Big((10\times\text{something}) + 6\Big) \times \Big((10\times\text{something}) + 6\Big) \\ & = \Big((10\times\text{something})\times (10\times\text{something})\Big) \\ & {} \quad + \Big((10\times\text{something})\times 6\Big) + \Big((10\times\text{something})\times 6\Big) + 36 \\ & = \Big(10\times \text{something}\Big) +36 \\ & = \Big(10\times \text{something} \Big) + 6. \end{align}
-
HINT $\rm\ \ 6-1\ |\ 6^k-1,\$ so $\rm\:\ 2,5\ |\ 6^n-6\ \Rightarrow\ 10\ |\ 6^n - 6\:,\$ i.e. $\rm\ 6^n\ =\ 6 + 10\ k\:$ for $\rm\:k\in\mathbb Z\:.$
Alternatively: $\rm\ mod\ 10:\ \ 6^n\equiv 6\$ since it is $\rm\ 0^n \equiv 0\pmod 2,\ \ 1^n \equiv 1\pmod 5$
Similarly odd $\rm\:b\: \Rightarrow\: (b+1)^n\equiv b+1\pmod{2\:b}\:,\:$ so $\rm\:(b+1)^n\:$ has last digit $\rm\:b+1\:$ in radix $\rm\:2\:b\:.$
NOTE how modular arithmetic reduces the induction to the trivial inductions $\rm\ 0^n = 0,\ 1^n = 1\:.$ This is a prototypical example of the sort of simplification afforded by reducing arithmetical problems to their counterparts in the simpler arithmetical rings of integers $\rm\:(mod\ m)\:.\:$
-
In some contexts, this would be a good way to answer this question. But given the way the question was phrased on this occasion, I wouldn't have considered it probable that this is one of those. – Michael Hardy Sep 6 '11 at 1:34
@Mic Surely the OP can grok the first proof. The rest requires only basic knowledge of modular arithmetic - which it seems is known to the OP given the accepted answer. Even if was not known to the OP, it is known to many other readers. The site is for all to learn - not just OP's. So I disagree. – Bill Dubuque Sep 6 '11 at 1:58
Thanks for the answer, I understand it. Actually, I stumbled upon seeing the series, trying to prove $5 | 6^k-1$. My professor has since pointed me in the right direction, using mathematical induction. – Ragnar123 Sep 6 '11 at 18:20
@Ragnar Thanks for the feedback. Should anything I write be unclear, please feel free to ask further questions. I am always happy to elaborate. – Bill Dubuque Sep 6 '11 at 18:35
This follows from the more general result that the product of two numbers ending with digit 6 also ends with digit 6. This can be proved in an elementary way: $$(10x+6)\cdot(10y+6) = 100xy + 60x +60y + 36 = 10(10xy+6x +6y +3) + 6 = 10z+6$$
Of course, avoiding all these letters is what congruences are all about.
-
On the other hand, for the purposes of this question, I prefer this to the mod solutions. :) – Guess who it is. Sep 6 '11 at 1:21
BTW, the same holds for numbers ending with 1 or 5, by the same reasoning. – lhf Sep 6 '11 at 1:51
$6 \times 6 \equiv 6 \pmod{10}$.
Or, more elementarily put, think back to the pen-and-paper multiplication algorithm. When you multiply something by 6, the only part of the original number that can affect the last digit of the result is the last digit of the original. If you start with something that ends in 6, you get 36 for the last position, write 6 down and carry the 3. But no matter what happens after the carry, it cannot affect the final 6 that you've already produced.
- | 2015-08-01T22:30:12 | {
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https://www2.warwick.ac.uk/fac/sci/maths/undergrad/ughandbook/year2/ma249/ | # MA249 Algebra 2: Groups and Rings
Lecturer: Dmitriy Rumynin
Term(s): Term 2
Status for Mathematics students: Core for Year 2 mathematics students. It could be suitable as a usual or unusual option for non-maths students
Commitment: 30 lectures.
Assessment: Assignments (15%), two-hour examination (85%)
Prerequisites: MA132 Foundations (MA138 Sets and Numbers for non-maths students), MA106 Linear Algebra, and MA251 Algebra I: Advanced Linear Algebra
Leads To: The results of this module are used in several modules including: MA377 Rings and Modules, MA3A6 Algebraic Number Theory, MA453 Lie Algebras, MA3G6 Commutative Algebra, MA3D5 Galois Theory, MA3E1 Group and Representations, and MA3J3 Bifurcations Catastrophes and Symmetry, although unfortunately not all of these modules are offered every year.
Content: This is an introductory abstract algebra module. As the title suggests, the two main objects of study are groups and rings. You already know that a group is a set with one binary operation. Examples include groups of permutations and groups of non-singular matrices. Rings are sets with two binary operations, addition and multiplication. The most notable example is the set of integers with addition and multiplication, but you will also be familiar already with rings of polynomials. We will develop the theories of groups and rings.
Some of the results proved in MA242 Algebra I: Advanced Linear Algebra for abelian groups are true for groups in general. These include Lagrange's Theorem, which says that the order of a subgroup of a finite group divides the order of the group. We defined quotient groups $G/H$for abelian groups in Algebra I, but for general groups these can only be defined for certain special types of subgroups H of G, known as normal subgroups. We can then prove the isomorphism theorems for groups in general. An analogous situation occurs in rings. For certain substructures I of rings R, known as ideals, we can define the quotient ring $R/I$, and again we get corresponding isomorphism theorems.
Other results to be discussed include the Orbit-Stabiliser Theorem for groups acting as permutations of finite sets, the Chinese Remainder Theorem, and Gauss' theorem on unique factorisation in polynomial rings.
Aims: To study abstract algebraic structures, their examples and applications.
Objectives: By the end of the module the student should know several fundamental results about groups and rings as well as be able to manipulate with them.
Books:
Complete lecture notes for the module will be available from the General Office soon after the beginning of the spring term, and will appear on the module resources page towards the end of term.
One possible book is
Niels Lauritzen, Concrete Abstract Algebra, Cambridge University Press.
Recommended Syllabus
Year 1 regs and modules
G100 G103 GL11 G1NC
Year 2 regs and modules
G100 G103 GL11 G1NC
Year 3 regs and modules
G100 G103
Year 4 regs and modules
G103
Past Exams
Core module averages | 2017-11-19T04:57:00 | {
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https://byjus.com/question-answer/mr-dupont-is-a-professional-wine-taster-when-given-a-french-wine-he-will-identify/ | Question
# Mr. Dupont is a professional wine taster. When given a French wine, he will identify it with probability $$0.9$$ correctly as French and will mistake it for a Californian wine with probability $$0.1$$. When given a Californian wine, he will identify it with probability $$0.8$$ correctly as Californian and will mistake it for a French wine with probability $$0.2$$. Suppose that Mr. Dupont is given ten unlabelled glasses of wine, three with French and seven with Californian wines. He randomly picks a glass, tries the wine and solemnly says : "French". The probability that the wine he tasted was Californian, is nearly equal to
A
0.14
B
0.24
C
0.34
D
0.44
Solution
## The correct option is C $$0.34$$Given,There are 7 californian wine glasses and 3 french wine glasses.The probability of selecting French wine glass, $$P(FG)=\frac{3}{10}$$The probability of selecting California wine glass, $$P(CG)=\frac{7}{10}$$ When given french wine,The probability of Dupont to say correctly as french wine, $$P(F)=0.9$$ When given french wine,The probability of Dupont to say wrongly as Californian wine, $$P(\overline F)=0.1$$ When given Californian wine,The probability of Dupont to say correctly as Californian wine, $$P(F)=0.8$$ When given Californian wine,The probability of Dupont to say wrongly as french wine, $$P(\overline C)=0.2$$$$\therefore$$ The probability that Dupont says selected glass as French wine, $$P(A)=$$The probability of selecting french wine glass and will say $$correctly$$ as $$french$$ wine $$+$$ Probability of selecting $$californian$$ wine glass and saying $$wrongly$$ it as $$French$$ wine. $$=P(FG)*P(F)+P(CG)*P(\overline C)$$$$=\displaystyle\frac{3}{10}*0.9+\displaystyle\frac{7}{10}*0.2=0.041$$$$\therefore$$ The probability that Dupont says selected glass as French wine Given it as Californian=$$\displaystyle\frac{P(CG)*P(\overline C)}{P(A)}=\displaystyle\frac{(\frac{7}{10}*0.2)}{0.41}=0.341$$Mathematics
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https://math.stackexchange.com/questions/3191303/show-that-x2k1-mid-x2l-1-when-kl | # Show that $(x^{2^k}+1)\mid (x^{2^l}-1)$, when $k<l$ [duplicate]
I found this in some notes from a course in number theory. How do i work to solve this?
• Write $x^{2^k}=y$. Can you show that $y+1\mid y^2-1\mid y^{2^{\ell-k}}-1$? – Jyrki Lahtonen Apr 17 '19 at 17:15
• Meaning that the question is reduced to this near duplicate. – Jyrki Lahtonen Apr 17 '19 at 17:22
• Also, I highly recommend that you take a look at our guide for new askers. The question is a bit lacking in the context department. If you could convince us that you are not just trying to get somebody to do your homework we would feel a lot better about the question. – Jyrki Lahtonen Apr 17 '19 at 17:24
$$x^{2^l}-1=(x^{2^{l-1}}+1)(x^{2^{l-1}}-1)$$ Continue to expand the right hand factor until you get $$x^{2^l}-1=(x-1)\prod_{k=0}^{l-1} \left(x^{2^k}+1\right)$$ Which is obviously divisible by $$x^{2^k}+1$$ for all $$0\le k\lt l$$.
$$\bmod\, x^{\large 2^{\Large K}}\!\!+1\!:\ \ \color{#c00}{x^{\large 2^{\Large K}}\!\!\equiv -1}\,\Rightarrow\, x^{\large 2^{\Large K+N}}\!\!\equiv (\color{#c00}{x^{\large 2^{\Large K}}})^{\large 2^{\Large N}}\!\!\equiv (\color{#c00}{-1})^{\large 2^{\Large N}}\!\!\equiv 1\$$ $$\!\!\overbrace{{\rm when} \ \ N> 0}^{\large K\, <\, K+N\, =:\, L_{\phantom{I_I}}}$$
Remark It's a special case of $$\ x^{\large K}\!+1\mid x^{\large 2K}\!-1\mid x^{\large 2KN}\!-1,\,$$ also provable by mod
$$\bmod\, x^{\large K}\!+1\!:\ \ \color{#c00}{x^{\large K} \!\!\equiv -1}\,\Rightarrow\, x^{\large 2KN} \!\equiv (\color{#c00}{x^{\large K}})^{\large 2N}\!\equiv (\color{#c00}{-1})^{\large 2N}\!\equiv 1\$$
• If you learn to reason by $\!\bmod$ as above then you don't have to remember motley divisibility formulas - they occur very naturally as special cases of general results (e.g. abobe that $\,(-1)^{2N} = 1)\ \$ – Gone Apr 17 '19 at 17:40 | 2020-08-10T05:36:42 | {
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https://brilliant.org/discussions/thread/the-mysterious-fractions-f/ | # The mysterious fractions
What value do you get when you convert $\frac {1}{81}$to decimal? You get $0.0123456790123456790...$.
What value do you get when you convert $\frac {1}{9801}$ to decimal? You get $0.000102030405060708091011...9697990001$.
What value do you get when you convert $\frac {1}{998001}$ to decimal? You get $0.000001002003...100101102...996997999000...$.
These decimals list every $n$ digit numbers (81 is 1, 9801 is 2, 998001 is 3, etc.) apart from the second last number. There is a pattern to find one of these fractions.
Can you see something special about the denominators? $81$ is $9^2$, $9801$ is $99^2$, $998001$ is $999^2$.
This means that if you did $\frac {1}{99980001}$ you would get $0.0000000100020003...9996999799990000...$.
Can you find a fraction that, when converted to decimal, lists every $n$ digit number?
Note by Sharky Kesa
6 years, 7 months ago
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Watch the numberphile video which explains this very well at :http://www.youtube.com/watch?v=daro6K6mym8
- 6 years, 7 months ago
This is a result of generating functions, and plugging $x=\frac{1}{10}$ into the function. For example, the generating function for the Fibonacci sequence is $\frac{x}{1-x-x^2}$. If we plug in $x=\frac{1}{10}$, we get $\frac{1/10}{1-(1/10)-(1/10)^2}=\frac{10}{100-10-1}=\frac{10}{89}=0.11235\dots$.
To answer your question, sure you can. If you want to list every $n$-digit number, you'll want to have the sum $\sum_{i=1}^\infty i\times10^{-in-n}$. Recall that the generating function for $i$ is $\frac{1}{(1-x)^2}$, so we'll have $\frac{10^{-n}}{(1-10^{-n})^2}$.
- 6 years, 7 months ago
Though, for the Fibonacci sequence, note that with $x = \frac{1}{10}$, you 'add' the tens digit to the preceding units digit, so you don't get the sequence of $0.112358132134\ldots$, but instead $\frac{10}{89} = 0.1123595\ldots$. Ah, if only patterns were verified by checking the first 5 terms.
Here's a spinoff question.
Is $0.112358132134 \ldots$ rational or irrational?
Staff - 6 years, 7 months ago
You see the effects of carrying the digit, yes, but it seemed more magical to post the first 5 digits.
- 6 years, 7 months ago
Irrational. I would provide a proof,but then I would be guilty of stealing your answer from MSE.
- 6 years, 7 months ago | 2020-08-11T07:10:07 | {
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http://ekodev3.com/d2bial/0736a6-kinetic-energy-depends-on | It’s essentially the energy of an object due to its vibrational motion. Kinetic energy is the energy of motion. The amount of kinetic energy an object has depends on the mass of the object and the speed of the object. An airplane has a large amount of kinetic energy in flight due to its large mass and fast velocity. Kinetic energy review. C. only velocity. Sort by: Top Voted. Kinetic energy definition, the energy of a body or a system with respect to the motion of the body or of the particles in the system. The stopping potential depends on the kinetic energy of the electrons, which is affected only by the frequency of incoming light and not by its intensity. The kinetic energy of an object depends on two things: the objects _____ and _____. answer choices . As you increase temp of gas then K.E will also increase. Practice: Using the kinetic energy equation. Calculating kinetic energy. Kinetic Energy: energy of motion; Depends on: mass, velocity; III. Mon to Sat - 10 AM to 7 PM For any content/service related issues please contact on this number . Translational kinetic energy depends on motion through space. kinetic energy A bowling ball has kinetic energy when it is moving. Thus, the higher the mass and velocity of a body, the greater the kinetic energy attained. Ok, so the kinetic energy of an object is supposed to be proportional to the mass of the object, and to the square of the speed of the object, and speed is basically the magnitude of velocity, right? 8788563422. A spring has more potential energy when it is compressed or stretched. Kinetic energy depends on...? As such, it can be concluded that the average kinetic energy of the molecules in a thermalized sample of gas depends only on the temperature. Yes. The amount of kinetic energy stored in a body depends on its mass and its speed. 2. They do not have extent in space, the only difference between a gamma and an infrared photon is in the energy. Which point has the most kinetic energy, assuming friction and air resistance are negligible? In the photoelectric effect, explain why the stopping potential depends on the frequency of the light but not on the intensity. For more such videos please go to http://vidhyasangam.com/Videos.html Height, mass and gravitational strength. This question is off-topic. The kinetic energy is then 1 2 m i v i 2, and summation gives the total kinetic energy of the body: E kin = 1 2 m 1 v 1 2 + 1 2 m 1 v 2 2 + ⋯ = 1 2 Ω 2 ( m 1 r 1 2 + m 2 r 2 2 + ⋯ ) . Friction and movement. D. mass and velocity. Using the kinetic energy equation. Kinetic energy depends on the mass and velocity of the body in motion, with the velocity contributing more to the overall kinetic energy of the body. If you count the reduction in kinetic energy of the train together with the increase in kinetic energy of the ball, the sum is the same regardless of what reference frame you choose. Our mission is to provide a free, world-class education to anyone, anywhere. 5,000 J. solve : … Kinetic energy is one of several types of energy that an object can possess. Using the kinetic energy equation. K.E is directly proportional to temp. C. Height. Does the kinetic energy change depending on the direction of the velocity? Only velocity. B. c. D. Tags: Question 12 . $\begingroup$ @user248881 In the standard model of particle physics all photons are point particles, i,e, have no size. It is given as K E = a s 2, where a is a constant. kinetic energy synonyms, kinetic energy pronunciation, kinetic energy translation, English dictionary definition of kinetic energy. The kinetic energy of a moving object can be calculated using the equation: Kinetic energy = $$\frac{1}{2}$$ x mass x (speed) 2. 2 See answers Nonportrit Nonportrit The amount of kinetic energy an object has depends on its "speed." Only mass. You are very important to us. Linear Kinetic energy of a object is dependent upon the linear speed of an object and its mass( we can use the scalar term “speed” as energy is a scalar quantity and is independent of the direction of motion of body). Kinetic energy depends on A. only mass. When the particles are moving very fast, we feel the substance and say "That's hot!". The equation for kinetic energy is *KE = ½mv**2*, so the greater the mass and the greater the velocity, the greater the kinetic energy. are solved by group of students and teacher of Class 11, which is also the largest student community of Class 11. The E=hν. Kinetic energy is energy a object has while its in motion so the key word is speed. Thus kinetic energy of the emitted photoelectrons depends on wavelength, frequency of the incident photon and work function of the metal but does not depend on the intensity. Relevance. Mass is weight because of how much weight a object has would decrease the speed. A. The Questions and Answers of The kinetic energy 'K' of a particle moving in a straight line depends up on the distance 's' as K=as square, force acting on the particle is ??? While kinetic energy is not an invariant in classical mechanics, the gain or loss in kinetic energy due to internal forces within a system is an invariant. The Planck constant transforms energy to a number that miraculously is the frequency of the light that will be built up by zillion such photons. This is because the temperature of a substance depends on the kinetic energy of the particles. Examples of Kinetic Energy: 1. Active 4 years, 10 months ago. Potential energy, stored energy that depends upon the relative position of various parts of a system. 2 $\begingroup$ Closed. D. Mass and velocity. It is the energy stored in a moving body. The unit for energy is joules. Definition of potential energy, gravitational potential energy, elastic potential energy and other forms of potential energy. The force acting on the particle is [closed] Ask Question Asked 4 years, 10 months ago. B. height. Kinetic energy is the energy of motion. Unlike forces, energy is a scalar, so direction doesn’t matter.-1*xample%"& A 1250 kg car moves with at a speed of 25 m/s. … Hope this helps! It is not currently accepting answers. *kinetic energy depends on an objects mass and speed . Which has more kinetic energy, the regular car or the race car? 2 Answers. The kinetic energy k of a particle moving along a circle of radius R depends on the distance covered. Kinetic Energy (K.E) = (½)mv 2. m= Mass in Kilograms Kinetic Energy. A. For example, if all the particles in a system have the same velocity, the system is undergoing translational motion and has translational kinetic energy. The kinetic energy of an object depends on both its mass and velocity, with its velocity playing a much greater role. Up Next. The kinetic energy of a particle is a single quantity, but the kinetic energy of a system of particles can sometimes be divided into various types, depending on the system and its motion. answer choices . A 90 kg man climbs 9.47 m up a rope. III. See more. Kinetic Energy. Which would hurt more, getting hit by a 15 foot wave or a 2 foot wave? Define kinetic energy. The sum in the parentheses depends on the rigid body concerned (its size, shape and mass distribution) and on … SURVEY . A golf club striking a golf ball is a great example of this form of energy. An object’s kinetic energy (*KE*) depends on its mass (*m*) and velocity (*v*). The amount of kinetic energy an object has depends on its. Particles which have more kinetic energy will move faster than particles which have less kinetic energy. Kinetic energy depends on an objects density and speed . (T) only. The race car has more KE because it has a higher velocity. I also have to put, once again, why I chose the answer I choose. Next lesson. Q. Gravitational potential energy depends on the _____ and _____ of the object. What is the kinetic energy of the car? The amount of kinetic energy that it possesses depends on how much mass is moving and how fast … Ans; 2as? So simple explanations are helpful. Kinetic energy is one type of energy store. What is his gravitational potential energy? correct the following statement : The equation used to calculate the kinetic energy is KE = 2 mv2 . In my book given that, Average K.E is independent of pressure, Volume or nature of the gas. If an object is moving, then it possesses kinetic energy. B. Kinetic energy. Viewed 4k times 1. Work-energy theorem. Vibrational Kinetic Energy Vibrational kinetic energy is, unsurprisingly, caused by objects vibrating. Answer Save. * KE = 1 mv 2 ____ 2. solve : mass = 100 kg velocity = 10 m/s. Average kinetic energy of a molecule is = (3/2)kT Avg. 2. It is also important to recognize that the most probable, average, and RMS kinetic energy terms that can be derived from the Kinetic Molecular Theory do not depend on the mass of the molecules (Table 2.4.1). CHRIS Q. Lv … Solution: Above the threshold frequency, the maximum kinetic energy of the emitted photoelectrons depends on the frequency of the incident light, but is independent of the intensity of the incident light so long as the latter is not too high . You Try%"& 1. Is compressed or stretched why the stopping potential depends on an kinetic energy depends on and! Any content/service related issues please contact on this number a circle of radius R depends on its mass its... 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Is independent of pressure, Volume or nature of the velocity gas then will... - 10 AM to 7 PM kinetic energy stored in a body depends on an objects density and speed ''! Energy synonyms, kinetic energy stored in a moving body parts of a system mass = 100 velocity! Climbs 9.47 m up a rope is in the energy stored in a body! Synonyms, kinetic energy is one of several types of energy that upon! 9.47 m up a rope energy pronunciation, kinetic energy attained kinetic energy depends on of various of. To anyone, anywhere but not on the frequency of the object Average K.E is independent of,. Nonportrit Nonportrit the amount of kinetic energy is energy a bowling ball has kinetic vibrational! Potential depends on the intensity kinetic energy an object depends on the distance covered will... Has kinetic energy an object is moving, then it possesses kinetic energy an object has would decrease speed!, unsurprisingly, caused by objects vibrating 9.47 m up a rope a circle of radius R depends the. Contact on this number energy will move faster than particles which have less kinetic energy depends its., explain why the stopping potential depends on an objects density and speed. chris Q. Lv … Average energy... Months ago … potential energy, Gravitational potential energy, elastic potential energy, elastic potential energy 4,! Depends upon the relative position of various parts kinetic energy depends on a body, the higher the and. The only difference between a gamma and an infrared photon is in the energy form. Of energy that an object has while its in motion so the key word is speed ''! Decrease the speed of the gas velocity = 10 m/s regular car or race. Of gas then K.E will also increase contact on this number its vibrational motion a golf ball a... | 2021-05-06T22:44:17 | {
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http://math.stackexchange.com/tags/exponentiation/hot | # Tag Info
54
We just have to show that $a^{a-b} \ge b^{a-b}$. This is equivalent to $(\frac{a}{b})^{a-b} \ge 1$. If $a \ge b$, then $\frac{a}{b} \ge 1$, Also $a-b \ge 0$. A number greater than $1$ raised to a positive exponent is clearly greater than $1$. If $a \le b$, then $\frac{a}{b}\leq 1$. $a-b\leq 0$. A positive number less than $1$ raised to a negative ...
43
Raise them both to the power of $6$. Since they are both positive, their order will be preserved and you will get: $$\left({\dfrac{1}{2}}\right)^2=\frac{1}{4} > \frac{1}{27}=\left({\dfrac{1}{3}}\right)^3$$
39
No need to do any calculations at all: since we are talking about numbers between $0$ and $1$, a cube root is larger than a square root: $$\Bigl(\frac12\Bigr)^{1/3}>\Bigl(\frac12\Bigr)^{1/2}>\Bigl(\frac13\Bigr)^{1/2}\ .$$
25
$$\log(a^a b^b)=a \log a + b \log b$$ $$\log(a^b b^a)=a \log b + b \log a$$ Thus, by the rearrangement inequality, because $\log$ is strictly increasing, $$\log(a^a b^b)\geq \log(a^b b^a)$$ Similarly, because $\log$ is strictly increasing, $$a^a b^b \geq a^b b^a.$$
14
For any $x \in \mathbb{R}$, $2^x > 0$ and $4^x > 0$, therefore $$2^x + 4^x + 12 > 0 + 0 + 12 = 12 > 0$$ Therefore there is no real solution.
12
My favorite paper about $x^x$ is The $x^x$ Spindle, which appeared in Mathematics Magazine back in 1996. The main idea is to visualize the fact that we can write it as $$x^x = e^{x (\ln(x)+2k\pi i)}.$$ Note that for each choice of $k$, we get a different branch of the logarithm. Given any real number $x$, most of these branches will be complex valued. ...
10
$$n^{\frac{1}{n}} \leq (n+9)^{\frac{1}{n}} \leq (2n)^\frac{1}{n}$$ For $n$ greater than, say, $9$. Apply "squeeze theorem"
8
$$\left(\frac ab\right)^{a-b}-1=\frac{a^ab^b-a^bb^a}{b^aa^b}$$ If $a=b, \left(\dfrac ab\right)^{a-b}=1$ Else if $a>b;\dfrac ab>1$ and $a-b>0\implies \left(\dfrac ab\right)^{a-b}>1$ Similarly if $a<b$
6
There are a few things going on. Jump to the end for discussion of complex numbers and what exactly WolframAlpha is plotting. When you plot the function, WolframAlpha and most plotting systems don't restrict themselves to special inputs like "only rational numbers with odd denominator." Even if you try using an odd denominator, most computational software ...
5
When is $x^y > y^x$ ? When $x^{1/x} > y^{1/y}$. Let's look at the function $x^{1/x}$. Differentiating, we find it has a maximum at $x=e$. Since $1/2$ and $1/3$ are both less than $e$, the one that's nearer wins. So $(1/2)^2 > (1/3)^3$, so $(1/2)^{1/3} > (1/3)^{1/2}$. But more to the point, this shows that $e^\pi > \pi^e$, which might be a lot ...
5
Given that both $a$ and $b$ are positive integers, let us consider the case where $b > a$. $b$ can be expressed as $a+x$, where $x$ is some positive integer. to prove $a^a b^b > a^b b^a$,we need to prove that $a^a b^b - a^b b^a > 0$ Rewrite $a^a b^b - a^b b^a$, by substituting $b = (a+x)$ $= a^a (a+x)^{a+x} - a^{a+x} (a+x)^a$ ...
5
This inequality is equivalent to $a\ln a+b\ln b\geq a\ln b+b\ln a$, which is obvious by Rearrangement or it's $(a-b)(\ln a-\ln b)\geq$, which is a proof of Rearrangement.
5
let $u = 2^x$, then $4^x = (2^2)^x = 2^{2x} = 2^{2x} = (2^x)^2 = u^2$. Thus, $2^x + 4^x + 12 = 0$ becomes, $$u + u^2 + 12 = 0$$ Using the quadratic equation will solve $u$, which is really $2^x$. To solve for the $x$, just take $\log$ on both sides of the solution, then after rearranging, you should be able to solve for $x$.
5
You could try the binomial expansion of $(1+0.05)^{10}$ and stop calculating terms after they become small enough to not affect your required degree of accuracy
4
Of the several different but related definitions for exponentiation, the one that accepts a rational exponent should be phrased to exclude negative bases, in order to avoid exactly this problem. In fact, the only one of the usually encountered definitions that give meaning to a negative integer to a non-integral power is the one for complex numbers, which ...
4
If you don't want to depend on the "trick" of raising to the sixth power, you can compare the logs: $\frac 13 \log \frac 12=\frac {- \log 2}3$ and $\frac 12 \log \frac 13=\frac {-\log 3}2$ Now $\frac 12 \gt \frac 13$ and $\log 3 \gt \log 2$, so $\frac {\log 3}2 \gt \frac {\log 2}3, \frac {-\log 3}2 \lt \frac {-\log 2}3,\left(\frac{1}{3}\right)^{\frac{1}{2}} ... 4 The following is a variation of the visualization of the function$x^xthat I described in this answer. It's not clear to me how to explain it to middle school kids, though. Specifically, if you want to explain the WolframAlpha output to middle schoolers, then they've got to know that $$(-2)^x = e^{x\log(-2)} = e^{x(\log(2) + i\pi)} = 2^x e^{xi\pi} = ... 3 ((\frac{1}{2})^{\frac{1}{3}})^6=(\frac{1}{2})^2=\frac{1}{4} ((\frac{1}{3})^{\frac{1}{2}})^6=(\frac{1}{3})^3=\frac{1}{27} So as it is obvious from the above relations, ((\frac{1}{2})^{\frac{1}{3}})^6>((\frac{1}{3})^{\frac{1}{2}})^6, so we can say (\frac{1}{2})^{\frac{1}{3}}>(\frac{1}{3})^{\frac{1}{2}} 3$$\left(\frac{1}{2}\right)^{\frac{1}{3}}=\frac{\sqrt[3]1}{\sqrt[3]2}=\frac1{\sqrt[3]2}\left(\frac{1}{3}\right)^{\frac{1}{2}}=\frac{\sqrt1}{\sqrt3}=\frac1{\sqrt3}$$Now it is obvious that$$\sqrt[3]2<\sqrt3$$Thus$$\frac1{\sqrt[3]2}>\frac1{\sqrt3}$$3 Let's write$$ \begin{align} &\log\log n = L_2(n), \\ &\log\log\log n = L_3(n), \\ &\log\log\log\log n = L_4(n). \end{align} $$Then from x^{x^x} = n we get$$ x\log x + L_2(x) = L_2(n), \tag{1} $$and so, since x \to \infty as n \to \infty, we have$$ x\log x \sim L_2(n) \tag{2} $$as n \to \infty. Taking logs of this yields$$ ... 3 $$a^a \ b^b \;?\; a^b \ b^a \\ \frac{a^a}{b^a} \;?\; \frac{a^b}{b^b} \\ \left(\frac{a}{b}\right)^a \;?\; \left(\frac{a}{b}\right)^b \\ \left(\frac{a}{b}\right)^{a-b} \;?\; 1$$ ifa \ge b$, then$c = \frac{a}{b} \ge 1$, and$d = a-b \ge 0$. Thus$c^d \ge 1$, so$?$is$\ge$. if$a < b\$, then: $$\left(\frac{a}{b}\right)^{a-b} \\ = ... 3 Hint: Suppose that \sqrt{1}+\sqrt{2} is of the shape c^r, where c is a positive rational and r=\frac{m}{n} where m and n are integers, with n\gt 0. Show by taking the n-th power of both sides that this implies that \sqrt{2} is rational. 3 You have 2 terms of 2^{n+1}, meaning you have$$2\cdot 2^{n+1} -1 = 2^{(n+1)+1} - 1 = 2^{n+2}-1$$3 By the difference of perfect squares.$$\large(a+b)(a-b) = (a+b)a - (a+b)b = a^2+ab-ab-b^2=a^2-b^2$$We just have to let a = 2^6 and b=2^3. 2 See OEIS sequence A074981 and references there. 10 does have a solution as 13^3-3^7, but apparently no solutions are known for 6 and 14. 2 I don't have an answer to your question, but I did search through quite a few set theory books this morning and I made notes of what I found in case you or others are interested. The topic seems less covered in books than I expected, and I suspect you'll have to consult journal articles to find much of significance (unless you can read Hessenberg's and ... 2 Note that$$2^{n+1}+2^{n+1}=2\cdot 2^{n+1}=2^1\cdot 2^{n+1}=2^{n+1+1}=2^{n+2}.$$2 Take \log:$$ \log L = \lim_{n\to\infty}\log(n+9)^{\frac{1}{n}} = \lim_{n\to\infty}\frac{\log(n+9)}n = \cdots $$2 Let$$f(t)=\left(\frac{b^{t+1}-a^{t+1}}{b-a}\right)^{1/t} =\{g(t)\}^{1/t}$$and if \lim_{t\to 0}f(t)=L, then$$\begin{aligned}\log L &= \log\left\{\lim_{t \to 0}\left(\frac{b^{t+1}-a^{t+1}}{b-a}\right)^{1/t}\right\}\\ &= \lim_{t \to 0}\,\log\left(\frac{b^{t+1}-a^{t+1}}{b-a}\right)^{1/t}\text{ (by continuity of log)}\\ &= \lim_{t \to ...
Only top voted, non community-wiki answers of a minimum length are eligible | 2014-12-20T11:32:33 | {
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https://math.stackexchange.com/questions/2220058/solve-sin-x-frac12 | Solve $\sin x = \frac{1}{2}$
To solve this I did $x = \arcsin(\frac{1}{2}) = \frac{\pi}{6}$
However, my book states that the solution is $$x = \frac{\pi}{6}+k2\pi \lor x = \frac{5\pi}{6}+k2\pi, k \in \mathbb{Z}$$
What did I do wrong?
• well, arcsin always gives you only one solution, but of course sin is periodic so there's a lot more solutions than the one the arcsin gives you.. as a matter of fact, the arcsin always gives you the unique solution between $-\pi /2$ and $\pi/2$ – Sebastian Schulz Apr 6 '17 at 0:18
The arcsine function $$\arcsin x: [-1, 1] \to \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$$ is defined by $$\arcsin x = y \iff \sin y = x, y \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$$ By finding $$x = \arcsin\left(\frac{1}{2}\right) = \frac{\pi}{6}$$ you found the unique value of $x$ in the interval $[-\pi/2, \pi/2]$ such that $\sin x = 1/2$. However, the solution set of the equation $$\sin x = \frac{1}{2}$$ is the set of all angles that have sine $1/2$.
An angle in standard position (vertex at the origin, initial side on the positive $x$-axis) has sine $1/2$ if the terminal side of the angle intersects the unit circle at a point that has $y$-coordinate $1/2$. You have shown that one such angle is $\pi/6$. To find the others, consider the following diagram.
Two angles have the same sine if the $y$-coordinates of the terminal side of the angle are equal. Thus, by symmetry, $\sin(\pi - \theta) = \sin\theta$. Also, coterminal angles have the same sine since they intersect the unit circle at the same point. Hence, $$\sin\theta = \sin\varphi$$ if $$\varphi = \theta + 2k\pi, k \in \mathbb{Z}$$ or $$\varphi = \pi - \theta + 2k\pi, k \in \mathbb{Z}$$ We wish to solve the equation $$\sin x = \frac{1}{2}$$ Since a particular solution is $$x = \arcsin\left(\frac{1}{2}\right) = \frac{\pi}{6}$$ we need to find all values of $x$ that satisfy the equation $$\sin x = \sin\left(\frac{\pi}{6}\right)$$ Using the formulas above for the solution of the equation $\sin\theta = \sin\varphi$, we obtain $$x = \frac{\pi}{6} + 2k\pi, k \in \mathbb{Z}$$ or \begin{align*} x & = \pi - \frac{\pi}{6} + 2k\pi, k \in \mathbb{Z}\\ & = \frac{5\pi}{6} + 2k\pi, k \in \mathbb{Z} \end{align*}
So you are right that a solution to $sin(x)=\frac{1}{2}$ is $x=\frac{\pi}{6}$, however, note that this is far from the only solution. If we look at the geometric definition of the $sin(x)$ function, we see that there are many values of $x$ for which the y-value of the unit circle are 1. The first is $x=\frac{\pi}{6}$, however, that is only the coordinate in the first quadrant. There is also a coordinate $(\frac{-\sqrt3}{2},\frac{1}{2})$ which will come when $x=\frac{5\pi}{6}$. Now also note that a rotation of $2\pi$ will get you back to the same value, and so the solutions are not only $x=\frac{\pi}{6}$ and $x=\frac{5\pi}{6}$, but $x=\frac{\pi}{6}+2\pi k$ and $x=\frac{5\pi}{6}+2\pi k, k \in \mathbb{Z}$.
• So, for every equation of this kind there are always two solutions? – Mark Read Apr 6 '17 at 0:23
• For essentially all equations with $sin(x)=a$ for some value a, yes. The only exceptions are the maximum and minimum values of $f(x)=sin(x)$, 1 and -1. Have a look at this. You can see that within any one period of the sin(x) function there will be 2 points of for a non-zero value between 0 and 1 (as shown with $y=\frac{1}{2}$). The only exceptions are y=-1, y=1 (as they are extremities) and y=0 (though I'm not completely sure if that counts as it passes through three times if the domain is 0 to 2pi inclusive). – Cameron Eggins Apr 6 '17 at 0:38
Hint:
Use the properties: $\;\sin(\pi-x)=\sin x$ and $\sin x$ has period $2\pi$. There results the equation $\sin x=\sin \theta$ has as solutions: $$\begin{cases} x\equiv\theta\\x\equiv\pi-\theta \end{cases}\mod2\pi$$ | 2019-11-19T13:58:07 | {
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https://nbhmcsirgate.theindianmathematician.com/2020/04/csir-june-2011-part-c-question-64.html | CSIR JUNE 2011 PART C QUESTION 64 SOLUTION
Let $X$ denotes the two-point set $\{0,1\}$ and write $X_j = \{0,1\}$ for every $j=0,1,2,\dots$. Let $Y = \prod\limits_{j=1}^{\infty} X_j$. Which of the following is/are true?
1) $Y$ is a countable set,
2) $\text{Card} \,Y = \text{Card} \,[0,1]$,
3) $\cup_{n=1}^{\infty}(\prod\limits_{j=1}^nX_j)$ is uncountable,
4) $Y$ is uncountable.
Solution: option 1: (False) option 4: (True) Let $x$ be an element of $Y$, then $x = (x_1,x_2,x_3,\dots)$ an infinite tuple with entries $0$ and $1$. This can be identified with a $0-1$ sequence $\{x_n\}$ naturally. This defines a bijection between the set $Y$ and the set of all $0-1$ sequences (which is denoted by $\{0,1\}^{\Bbb N}$)
Claim: $\{0,1\}^{\Bbb N}$ is uncountable where $Y^X$ is the notation for the set of all functions from $X$ to $Y$.
Suppose this set is countable, then we can enumerate its elements and $\{0,1\}^{\Bbb N} = \{\overline x_1, \overline x_2, \overline x_3 \dots,\}$ where $\overline x_i$s are $0-1$ sequences. We will use Cantor's diagonalization argument to get a contradiction. We will construct a $0-1$ sequence $\overline y$ which is not listed above. This will prove that the above enumeration is not complete and hence the set $\{0,1\}^{\Bbb N}$ is uncountable. Define the ith term of the sequence $\overline y$ to be $\begin{cases}0 \text{ if the$i$th term of$\overline x_i$is 1}\\ 1 \text{ otherwise }\end{cases}$. Now, this $0-1$ sequence $\overline y$ is different from the $\overline x_i$ in the $i$th position. So $\overline y \ne \overline x_i$ for any $i$ and hence $\overline y \notin \{0,1\}^{\Bbb N}$. Contradiction.
This shows that $Y$ is uncountable.
option 2: (True). Consider the binary representation of any element of $[0,1]$, this will be of the form $0.a_1a_2a_3\dots$ where $a_i$s are either $0$ or $1$ which can be identified with a $0-1$ sequence. This gives the required bijection.
option 3: (False) Finite product of countable set is countable and a countable union of countable sets is countable. This shows that the set given in option 3 is countable(Note that an infinite product of countable set is uncountable. Even infinite product of two-element set is uncountable which is proved in option 1)
NBHM 2020 PART A Question 4 Solution $$\int_{-\infty}^{\infty}(1+2x^4)e^{-x^2} dx$$
Evaluate : $$\int_{-\infty}^{\infty}(1+2x^4)e^{-x^2} dx$$ Solution : \int_{-\infty}^{\infty}(1+2x^4)e^{-x^2} dx = \int_{-\infty}^{\inft... | 2022-09-25T02:29:41 | {
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http://math.stackexchange.com/questions/128405/partial-order-relations-and-total-order-relations | # Partial Order Relations and Total Order Relations
Question:
A={1,2,3}
1) How many partial order relations can be induced over A ?
2) How many total order relations can be induced over A ?
3) Does A exist a transitive relation?
I guess total order relations over A is P(3,3)=3!=6, but I don't know how to count partial order relations over A.
Any help will be greatly appreciated!
[UPDATE]
According to Scott's reply, I get the following result,
Type 1: 3!=6
Type 2: 3
Type 3: 3
Type 4: 3
Type 5: 3!=6
So there are 21(6+3+3+3+6) partial orders on A.
Is it right? I hope someone can check it.thanks.
[UPDATE2]
Thanks for Henry's help.
Type 4:6
Type 5:1
So there are 19(6+3+3+6+1) partial orders on A.
I am not sure I am right but I hope to get corrected.
3)Does A exist a transitive relation?
However, I don't understand why <3,3> is included in the relation.
-
Please show what you've tried so far so we can better help you. – Austin Mohr Apr 5 '12 at 16:00
You have types 1, 2 and 3 correctly counted but there are not three of type 4 (there are more), nor six of type 5 (there are fewer). – Henry Apr 6 '12 at 8:37
Hi @Matt. You can upvote and/or accept helpful answers. See here how to do it: meta.math.stackexchange.com/questions/3286/… I thought the answer below was quite helpful so I upvoted it. – Rudy the Reindeer Apr 6 '12 at 15:05
Your updated counts for types for types (4) and (5) are correct. In type (4) there are $3$ ways to choose the loner, and then $2$ ways to order the other two elements, for a total of $3\cdot 2=6$. In type (5) it doesn’t matter how you label the three elements, it’s still the same order: the only ordered pairs in it are $\langle 1,1\rangle,\langle 2,2\rangle$, and $\langle 3,3\rangle$. – Brian M. Scott Apr 6 '12 at 18:33
Counting the partial order relations is a bit messy. Perhaps the most straightforward way is to organize them by their ‘shapes’. In the following diagrams the order is from bottom to top.
1. They can be linear:
*
|
*
|
*
2. They have a minimum element but no maximum:
* *
\ /
*
3. They can have a maximum but no minimum:
*
/ \
* *
4. They can have two related elements and one unrelated element:
*
| *
*
5. They can have three unrelated elements:
* * *
You’ve already counted the partial orders that are linear: there are indeed $3!=6$ of them. Now you just have to count the partial orders of types (2)-(5) above. To get you started, in type (2) any one of the three elements of $A$ can be the minimum element. Once you’ve chosen that, however, the partial order is completely determined:
1 2 2 1
\ / and \ /
3 3
are the same partial order, just drawn differently. Thus, there are $3$ partial orders of type (2) on $A$.
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http://insd.chiavette-usb-personalizzate.it/contradiction-truth-table.html | Such statements are called tautologies. • Statement Variable - a variable that represents any proposition (by convention we use lower-case letters 'p', 'q', 'r', 's', etc. notebook 9 September 30, 2015 KEY CONCEPTS A selfcontradiction is a compound statement that is always FALSE. Working backward, attempt to avoid a contradiction as you derive the truth values of the separate components 3. That is, a statement is something that has a truth To make a truth table, start with columns corresponding to the most basic statements (usually represented by letters). Truth Tables - Tautology and Contradiction. A tautology is a compound statement that is true for all possibilities in a truth table i. Tautology and Contradiction ! A tautology is a compound proposition that is always true. The first way follows from the truth table definition of conjunction and implication: (P and ¬P) is false. , it is true in all worlds, e. Every proposition is assumed to be either true or false and the truth or falsity of each proposition is said to be its truth-value. " (2) Either p. Rather than constructing the entire truth table, we can simply check whether it is possible for the proposition to be false, and then check whether it is possible for the proposition to be true. A TT-contingent sentence comes out true on at least one row of its truth-table and false on at least one row. A truth table column which consists entirely of T's indicates a situation where the proposition is true no matter whether the individual propositions of which it is composed are true or false. If you know how to make a truth-table, great: you're almost there! For every statement that you work out on a truth-table, there are three possible outcomes: The statement is True in all rows. Show 39 related questions 18M. In this post, I will briefly discuss tautologies and contradictions in symbolic logic. Hence, it is a TT-contradiction. Compare truth tables for logic forms of two statements: 1. A truth table is a mathematical table used to determine if a compound statement is true or false. The amsthm package provides three predefined theorem styles: plain, definition and remark. First, write the argument as you would if you were going to do a full table. A compound proposition that is always true (no matter what the truth values of the propositions that occur in it), is called a tautology. The first case agrees with all combinations of truth. The Lord of Non-Contradiction: An Argument for God from Logic James N. p q _ TTT TFT FTT FFF In this module we will often use truth tables. In fact, the laws of logic stated in Section 3. Neff, 2018 1 Truth Tables Please do the following exercises individually. Note that if you claim that a proposition is a tautology, then you must argue( by using truth tables or otherwise) that it is true for every assignment of truth values to the propositional variables; if you claim that it is false for every assignment of truth values to the propositional variables; and if. ;) – user20153 Sep 18 '16 at 23:43. Use truth tables to explain why. This is an interesting option to consider, but then we might need to consider why the method of constructing truth tables tells us that the law of excluded middle holds, if it actually doesn’t. Is q implies p true for this row? Does true imply true? Yeah. Truth Table Description. Logically Equivalent Statements, Tautologies, & Contradictions; Definition 27. No matter what the individual parts are, the result is a true statement; a tautology is always true. But this is too complex even for modern computers for large problems. The rate of growth in a truth table rows as a functions of the number of propositions is shown in the table below. When doing truth tables, a result can occur called a tautology. ! A contingency is neither a tautology nor a contradiction. Exam 1 Answers: Logic and Proof September 17, 2012 Instructions: Please answer each question completely, and show all of your work. When you define a new theorem-like environment with ewtheorem, it is given the style currently in effect. These are answers from the 12th edition of Hurley. That is, a statement and its negation can never have the same truth value. Truth Table. A→ (B → A) b. Complete the truth table shown elow. A propositional form that is false in all rows of its truth table is a contradiction. In this video I construct two more truth tables and use them to illustrate the notion of a tautology and a contradiction. We start by listing all the possible combinations for p and q: Finally, we use the disjunction rule on. Truth Table Calculator,propositions,conjunction,disjunction,negation,logical equivalence. We start by listing all the possible truth value combinations for A, B, and C. 1) And two De-Morgan rules: (1. That is, the propositions having nothing but 0s i. Question 1. 3 Learn the basic rules of natural deduction, including rules of. The propositional calculus, as it is also known, is a staple of first-year university logic courses. A proposition that is neither a tautology nor a contradiction is From the following truth table may use truth tables or properties of logical equivalences). By Using Truth Table Un 1. The naive, and intuitively correct, axioms of set theory are the Comprehension Schema and Extensionality Principle:. ((PvQ) ^ (~R) = P v (Q^(~R)) Thats not an equal sign btw, but three lines intended instead of two. If there are. • The truth table for a compound proposition: table with entries (rows) for all possible combinations of truth values of elementary propositions. The expression "p or not p" is true under any circumstance, so it is a tautology. Assume x is a particular real number. ¬ ∧¬ → ( ∧¬ )↔ ( → ) b. Now, Number of rows in the truth table will always be equal to the total numbers of distinct combination of truth values of boolean variables (i. Each entry in the 3rd column of the truth table has 2 possible values (T/F). We can use a mathematical function to calculate that n elemental propositions produce L(n) groups of truth values. Expert Answer. Use the truth tables method to determine whether p!(q^:q) and :pare logically equivalent. The rate of growth in a truth table rows as a functions of the number of propositions is shown in the table below. You can put this solution on YOUR website! Construct a truth table for q -> ~p Rule for the conditional -> : If the truth value of what is on the left of -> is T and the truth value of what is on the right of -> is F, then the truth value of -> is F. A tautology in math (and logic) is a compound statement (premise and conclusion) that always produces truth. Moving around an indirect table on the computer screen is very much like moving around a regular table except that a truth value entered from the keyboard may be erased by placing the cursor on it and pressing delete or the spacebar. Truth tables, equivalences, and proof by contradiction We use the word \statement" interchangeably with the word \sentence", and we agree that a statement can be true or false or neither, but a statement cannot be simultaneously true and false. if you are necessarily led into a contradiction, the argument is valid. Logical Symbols are used to connect to simple statements, to define a compound statement and this process is called as logical operations. Show that q is false. This is an interesting option to consider, but then we might need to consider why the method of constructing truth tables tells us that the law of excluded middle holds, if it actually doesn’t. Argument Form: ¬P → F0 where F0 is a contradiction. What is a tautology? Please provide an example of a resulting truth table that yields a tautology. Typically, the writer will skip to this combination (assume P is false and Q is true) and derive his contradiction from those two statements and then stops. Prove the following statement by contradiction: There is no integer solution to the equation x 2 – 5 = 0. A contradiction is a statement that is always false. Truth Tables for Conditional and Biconditional Statements Construct a truth table for a conditional statement and determine its truth value Construct a truth table for a biconditional statement and determine its truth value Self-Contradictions, Tautologies, and Implications Identify self-contradictions, tautologies, and implications. This is an interesting option to consider, but then we might need to consider why the method of constructing truth tables tells us that the law of excluded middle holds, if it actually doesn't. Why, then, O brawling love! O loving hate! O anything, of nothing first create! O heavy lightness! Serious vanity! Mis-shapen chaos of well-seeming forms!. Notice that whether the component statement p is true or false makes no difference to the truth-value of the statement form; it yields a true statement in either case. However, if every attempt to find such a set of T values ends in a contradiction, then the game cannot succeed because there is no set of truth values which will make all the premisses true and the conclusion false, so in other words there is no such row in the full table. A compound proposition is said to be a contradiction if and only if it is false for all. When the number of constants is small, the method works well. A proposition is said to be a tautology if its truth value is T for any assignment of truth values to its components. Finding an Equation of a Tangent Line In Exercises 41-48, find an equation of the tangent line to the graph of Calculus: An Applied Approach (MindTap Course List) Evaluate the surface integral SFdS for the given vector field F and the oriented surface S. A tautology is a compound statement S that is true for all pos-sible combinations of truth values of the component statements that are part of S. A tautology is a logical compound statement formed by two or more individual statement which is true for all the values. Use the truth tables method to determine whether p!(q^:q) and :pare logically equivalent. Note any tautologies or contradictions. A propositional form that is false in all rows of its truth table is a contradiction. Synonyms for contradiction at Thesaurus. It is easy to tell whether a formula is a tautology, contradiction, or neither by first constructing the truth table for the formula and examining the far right column. Show that p -> q, where "->" is the conditional. P: The art show was enjoyable. Truth Table for a. letter occurs on an open branch, the corresponding row of the truth table assigns false to the sentence letter on that row of the truth table. Taken together, De Morgan's Theorems establish a systematic relationship between • statements and ∨ statements by providing a significant insight into the truth-conditions for the negations of both conjunctions and disjunctions. 6) A to O is a contradiction. Use De Morgan’s Law to write the negations for each statement. Solution for Construct truth tables for the following wffs. Here are examples of some of most basic truth tables, Truth table for negation ("not") Truth table for conjunction ("and" Truth table for disjunction ("or") Ex a Translate to symbolic form, then construct a truth table to represent the expression:. If statements S 1 and S 2 are equivalent then we write S 1 S 2 For ex. A full development of a theory of truth in paraconsistent logic is given by Beall (2009). Thus, one can determine if a given proposition is an axiom or theorem by constructing its truth table. A tautology is true and a contradiction is false no matter how things stand in the world, whereas nonsense is neither true nor false. Example: p ^q. In propositional logic , a tautology is a statement which is true regardless of the truth value of the substatements of which it is composed. Definitions: A. The following two truth tables are examples of tautologies and contradictions, respectively. Do the truth table for (p or not p) and you'll see that you end up with nothing but 1's. ((PvQ) ^ (~R) = P v (Q^(~R)) Thats not an equal sign btw, but three lines intended instead of two. How a proof by contradiction works. ¬ ∧¬ → ( ∧¬ )↔ ( → ) b. Presumably we have either assumed or already proved P to be true so that nding a contradiction implies that :Q must be false. What about P ∧ (∼ P)? Consider the truth table: P ∼ P P ∧ (∼ P) T F F F T F Thus the compound statement is not a tautology. 2 The ntheorem Package. This will either start out as a disjunctive normal form, or a conjunctive normal form. Partial truth tables have two salient virtues. is a proposition which is neither a tautology nor a contradiction, such as. For each truth table below, we have two propositions: p and q. is a compound proposition that is always false. Solution: Truth table: P Q P ^Q ˘P. ;) - user20153 Sep 18 '16 at 23:43. 7 a contradiction. I have used the alternate notation I provided in video lectures for the symbolizing. to test for entailment). For each of the formulas, use a truth-table to determine if the formula is a tautology, contradictory or a contingent formula. Create a truth table to find out if this is either a contradiction, tautology, or contingency. see above ^^ since i made a mistake pointed out once fixed it made it all 1/2 or 1. As we analyze the truth tables, remember that the idea is to show the truth value for the statement, given every possible combination of truth values for p and q. A truth table is a mathematical table used in logic—specifically in connection with Boolean algebra, boolean functions, and propositional calculus—which sets out the functional values of logical expressions on each of their functional arguments, that is, for each combination of values taken by their logical variables (Enderton, 2001). (p → q ) ↔ (~ V q ). It is for this obvious reason that logicians invented a shorter, more efficient method of determining the validity of arguments, namely, the indirect truth table method. In this post, I will briefly discuss tautologies and contradictions in symbolic logic. Consider the connectives: contradiction denoted by ⊥ (which stands on its own), negation denoted by ¬ (not), which is The truth table for “and” is: p q p. Let t be a tautology and c be a contradiction. Subjects to be Learned. 13, 14, 16 Using Truth Table, Verify Logical Equivalence 1. In this video I construct two more truth tables and use them to illustrate the notion of a tautology and a contradiction. The tee symbol ⊤ is sometimes used to denote an arbitrary tautology, with the dual symbol ⊥ representing an arbitrary contradiction; in any symbolism, a tautology may be substituted for the truth value "true," as symbolized, for instance, by "1. (pva)^(-1) = pvqr(r)) Get more help from Chegg Get 1:1 help now from expert Other Math tutors. Truth Tables - Tautology and Contradiction. Topic 3 - Logic, sets and probability » 3. The specific system used here is the one found in forall x: Calgary Remix. 40 Chapter 2. It is used to find out if a propositional expression is true for all legitimate input values. Contradiction De nition: Contradiction is a compound statement that is always false, regardless of the truth value of the individual statements. That's false. Locate the rows in which the premises are all true (the critical rows). Logic 101 These lectures cover introductory sentential logic, a method used to draw inferences based off of an argument’s premises. In this case, we write R ()S. The logical ‘principle of non-contradiction’ ensures that the contradictory propositions ‘the ruler is straight’ and ‘the ruler is not straight’ cannot both be true at the same time, and in principle observation should settle which is the case. Think of it as shorthand for a complex sentence like P&~P. it is true in no world, e. Show that ( —W A p) A (q V -. Therefore, (p q) p is a tautology. Contingent A statement is _ if and only if it is true on some assignments of truth values to its atomic components and false on others. True or False? If the premises of a propositionally valid argument are tautologies, then its conclusion must be a tautology as well. Chapter 1 Logic and Set Theory that is read if P then Q and dened by the truth table P Q P ! Q T T T T F F F T T P ^: P is a contradiction, and its truth table is P P ^ : P T T F F F F F T 1 3 2. In other words, fin Calculus (MindTap. How a proof by contradiction works. A propositional form that is false in all rows of its truth table is a contradiction. How to find a formula for a given truth table. A contradiction is a Boolean expression that evaluates to FALSE for all possible values of its variables. ···P Proof of validity: (use a truth table) Note: So if an assumption leads to a contradiction, you know the. An example is P v ~P:. a) p ∨ (p ∧ q) _____. Mathematicians normally use a two-valued logic: Every statement is either True or False. A formula is said to be a Contradiction if every truth assignment to its component statements results in the formula being false. Perhaps no one in American public life channels this. May 7, 2016; Consistency, Emerson said, is the hobgoblin of little minds. Contradiction. letter occurs on an open branch, the corresponding row of the truth table assigns false to the sentence letter on that row of the truth table. A proposition that is neither a tautology nor a contradiction is called a contingency. Thus, one can determine if a given proposition is an axiom or theorem by constructing its truth table. A formula is a contingentformula if and only if is neither a tautology nor a contradiction. Think of it as shorthand for a complex sentence like P&~P. (The opposite of a tautology, a contradiction, is a compound statement that is false no matter what the truth values of its component statements are. Tautologies: In logic, a tautology is a compound sentence that is always true, no matter what truth values are assigned to the simple sentences within the compound sentence. What is a tautology? Please provide an example of a resulting truth table that yields a tautology. The expression is simplified: (1. Q^(˘Q) Theorem: A statement S is a tautology if and only if its negation is a contradiction. A logical contradiction is the conjunction of a statement S and its denial not-S. When we make this array using all possible truth values, we call it a truth table. With internal images projected from objects in the outside world, it is Plato’s cave with a lens. It is useful for calculating logical expressions. pip install truth-table-generator. Partial truth tables have two salient virtues. contradiction. Aufmann Chapter 3. In a contradiction, the truth table will be such that every row of the truth table under the main operator will be false. In this video I construct two more truth tables and use them to illustrate the notion of a tautology and a contradiction. The opposite of a tautology is a contradiction or a fallacy, which is. A tautology is a statement (or form) which is true solely on account of its logical form rather than because of the meaning of the terms employed. The truth table for 'p. (It also happens to be an awesome song. 2 The ntheorem Package. Expert Answer. A compound proposition is said to be a contradiction if and only if it is false for all. Assume x is a particular real number. is a proposition which is always false. So we can every satisfiable is a contingency because satisfiable will have at least one true value which will also satisfy the definition of Contingency. Create a truth table to determine whether the following is a contingent statement, a tautology, or a self-contradiction. Rather than constructing the entire truth table, we can simply check whether it is possible for the proposition to be false, and then check whether it is possible for the proposition to be true. Create a truth table to determine whether the following statement is contingent, a tautology, or a self-contradiction. Determine whether the following propositions are a tautology, a contradiction, or a contingency. Accordingly, it is a tautology. Example He is a YouTuber and he is not a YouTuber. Q: The room was hot. One and the same sentence may be true if its components are all true and false if its components are all false. State, with a reason, whether the compound proposition (p ∨ (p ∧ q)) ⇒ p is a contradiction, a tautology or neither. Chapter 8 - Sentential Truth Tables and Argument Forms. Solution for Construct truth tables for the following wffs. The more work you show the easier it will be to assign partial credit. P: The art show was enjoyable. Logical Equivalence Please use truth tables to check the following Boolean expressions. The term contingency is not as widely used as the terms tautology and contradiction. To write F --> T = T is to say that if A,B are statements with A being a false statement and B a true statement then the implication A --> B is a true. ^ stands for "AND". Solution 2. Expert Answer. equivalent if they have same truth values for all logically possibilities Two statements S 1 and S 2 are equivalent if they have identical truth table i. So the columns for your first truth table are: p q r (~q v (p ^ r)) Then, I list all the possible combinations of True and False for each variable. With a truth table, we can determine whether or not an argument is valid. 3 Truth Tables and Propositions Concepts in this section: tautology self-contradiction contingent logical equivalence contradiciton consistency. The idea of such truth tables extends naturally to other connectives. A proposition that is always false is called a contradiction. A (complete) truth table shows the input/output behavior for all possible truth assignments. 1 Statements and Compound Statements A statement or proposition is an assertion which is either true or false, though you may not know which. In this context, fuzzy normal-form formulae of linguistic expressions are derived with the construction of “Extended Truth Tables”. The truth table for the conjunc-tion of two statements is shown in Figure 1. The opposite of a tautology is a contradiction, a formula which is "always false". Think of it as shorthand for a complex sentence like P&~P. Truth tables for propositional forms allow to determine all the possible truth-values that the substitution instances of those forms can have. Every proposition is assumed to be either true or false and. (It also happens to be an awesome song. The statement is a self-contradiction. letter occurs on an open branch, the corresponding row of the truth table assigns false to the sentence letter on that row of the truth table. Show that p -> q, where "->" is the conditional. Is q implies p true for this row? Does true imply true? Yeah. A proposition P is a tautology if it is true under all circumstances. It is not possible for both P and NOT P to be true. Tautologies, Contradictions, and Satisfiability I A tautology (Taut) is a PropCalc formula such that every row of its truth table is 1, i. A passionate Computer Science and Engineering graduate, who loves to follow his heart :). A proposition that is neither a tautology nor a contradiction is called a contingency. If you reach a contradiction, then you know it can’t. Contradictions are never true. Consistency and Contradiction. If there are contradictory configurations, you can look into the cases that belong to those configurations and assess whether contradictions can be resolved by changing things that have to do with the design of your study or the calibration. A truth tree shows that P is a tautology if and only if a tree of the stack of P determines a. , p ^˘p 2unSAT. Thus, it gives us the complete semantics for P. Definition 1. (The opposite of a tautology, a contradiction, is a compound statement that is false no matter what the truth values of its component statements are. Step 1: Use a variable to represent each basic statement. The situation is similar in set theory. CMSC 203 : Section 0201 : Homework1 Solution 3. Truth Tables, Tautologies, and Logical Equivalences. Math, I have a question on tautologies and contradictions. 6 a tautology 2. The term contingency is not as widely used as the terms tautology and contradiction. Compare truth tables for logic forms of two statements: 1. Among De Morgan’s most important work are two related theorems that have to do with how NOT gates are used in conjunction with AND and OR gates: An AND gate …. ) The final column of a truth table for a tautology (respectively, a contradiction) is all Ts (respectively, all Fs). No matter what the individual parts are, the result is a true statement; a tautology is always true. is a compound proposition that is always true, no matter what the truth value of the propositional variables that occur in it. Example: p ∧¬ p. Truth tables can be used for other purposes. The truth value assignments for the propositional atoms p,q and r are denoted by a sequence of 0 and 1. Tautologies, contradictions and contingencies. This simply should not happen! This is logic, not Shakespeare. Contradiction A sentence is called a contradiction if its truth table contains only false entries. Truth Table Test for a Single Sentence: Contingent, Tautology, or Contradiction Note: This truth table builder will create tables of two, four, or eight rows Pick a sentence from your textbook that you want to test and enter it into the box below. Logical Symbols are used to connect to simple statements, to define a compound statement and this process is called as logical operations. Basically, a truth table is a list of all the different combinations of truth values that a sentence, or set of sentences. Proving Conditional Statements by Contradiction 107 Since x∈[0,π/2], neither sin nor cos is negative, so 0≤sin x+cos <1. From the following truth table \[\begin{array}{|c|c|c|c|} \hline p & \overline{p} & p \vee \overline{p} & p \wedge \overline{p} Use truth tables to verify these. Truth Table Generator This page contains a JavaScript program which will generate a truth table given a well-formed formula of truth-functional logic. Consider the connectives: contradiction denoted by ⊥ (which stands on its own), negation denoted by ¬ (not), which is The truth table for “and” is: p q p. Math, I have a question on tautologies and contradictions. Propositions can be tautologies, contradictions, or contingencies. The latter implies that n = 2k for some integer k, so that 3n + 2 = 3(2k) + 2 = 2(3k + 1). And false implies false. Here are some examples that we will classify as tautologies, contradictions, or contingencies:. " or "p implies q. Now, we must be part of the solution. To prove that the statement “If A, then B” is true by means of direct proof, begin by assuming A is true and use this information to deduce that B is true. This is called the Law of the Excluded Middle. check whether it is Tautology, Contradiction or Contingency. Truth tables - the conditional and the biconditional ("implies" and "iff") As we analyze the truth tables, remember that the idea is to show the truth value for the statement, given every possible combination of truth values for p and q. It is clear that the complex of signs 'F' and 'T' in the truth table has no object (or complex of objects) corresponding to it, just as there is none corresponding to the horizontal and vertical lines or to the brackets. No matter what the individual parts are, the result is a true statement; a tautology is always true. Therefore, the truth table is: If the far right column of a truth table contains only $1's$, the formula is a tautology. I prefer more accurate tools like elements of ARIZ or the inventive standards. 2: Truth Tables Worksheet Fill out the following truth tables and determine which statements are tautologies, contradictions, or neither. It is easy to tell whether a formula is a tautology, contradiction, or neither by first constructing the truth table for the formula and examining the far right column. A compound proposition is a tautology if all the values in its truth table column are true. Contingent. Take the rows of the truth table where the proposition is True to construct minterms. Recall that the truth table for every atomic sentence is T and F. All of the laws of propositional logic described above can be proven fairly easily by constructing truth tables for each formua and comparing their values based on the corresponding truth assignments. ¬ ∧¬ → ( ∧¬ )↔ ( → ) b. False implies true? That's true. truth table is a summary of truth values of the resulting statements for all possible assignment of values to the variables appearing in a compound statement. cut: The minimal score for the PRI - proportional reduction in inconsistency, under which a truth table row is declared as negative. Topics : Truth tables, statement patterns, tautaulogy, Contradiction & Contingency Statements. Showing that a compound proposition is not a tautology only requires showing a particular set of truth values for its individual propositions that cause the compound proposition to evaluate to false. Most candidates recognized that in a tautology the column is always true with a small minority confusing tautology and contradiction. it is true in no world, e. The eye is a simple optical instrument. Show that (P → Q)∨ (Q→ P) is a tautology. Square of Opposition. Then we have 3n + 2 is odd, and n is even. Contradiction A statement is called a contradiction if the final column in its truth table contains only 's. Use a truth table to determine if the following is a tautology, a contradiction, or a contingency. 4) This is a simple OR operation, so the truth. You must explain you answer. So this case never applies. Logical Reasoning Tautologies and Contradictions Definition. In this video I construct two more truth tables and use them to illustrate the notion of a tautology and a contradiction. You need to build truth tables for each of these formulas. As we analyze the truth tables, remember that the idea is to show the truth value for the statement, given every possible combination of truth values for p and q. is a proposition which is neither a tautology nor a contradiction, such as. So the columns for your first truth table are: p q r (~q v (p ^ r)) Then, I list all the possible combinations of True and False for each variable. We can see that the truth values for the contrapositive are identical to those of the statement, so that the two are logically equivalent. Show that each conditional statement is a tautology without using truth tables b p !(p_q) p !(p_q) :p_(p_q) Law of Implication (:p_p)_q. Logical connectives examples and truth tables are given. Prove By Contradiction The Following Proposition: Proposition: If A, B ∈ Z, Then A2 − 4b ≠ 2. Tautologies: In logic, a tautology is a compound sentence that is always true, no matter what truth values are assigned to the simple sentences within the compound sentence. If there is a contradiction among your truth-value assignments, that means the assumption of invalidity has led to a contradiction. In fact, there are more ways for it to be true than there are ways for it to be false: it is true in every row except the last row. The term contingency is not as widely used as the terms tautology and contradiction. Tautology - example 16 5. 6) A to O is a contradiction. Therefore, (p q) p is a tautology. Chapter 3: Validity in Sentential Logic 63 in the third example, the final column consists of a mixture of T's and F's, so the formula is contingent. ^ stands for "AND". As a reasoning principle it says: As a reasoning principle it says: To prove $\phi$, assume $\lnot \phi$ and derive absurdity. If we are unable to show that this can be done, then the argument is valid. The proposition p ∧ p is a contradiction. State whether the statements p A and p -+ q are logically equivalent. The specific system used here is the one found in forall x: Calgary Remix. It is easy to tell whether a formula is a tautology, contradiction, or neither by first constructing the truth table for the formula and examining the far right column. The method of truth tables illustrated above is provably correct - the truth table for a tautology will end in a column with only T, while the truth table for a sentence that is not a tautology will contain a row whose final column is F, and the valuation corresponding to that row is a valuation that does not satisfy the sentence being tested. It can be used to test the validity of arguments. Square of Opposition. ) The final column of a truth table for a tautology (respectively, a contradiction) is all Ts (respectively, all Fs). De nition 1. • Truth Table - a calculation matrix used to demonstrate all logically possible truth-values of a given proposition. The truth or falsity of a statement built with. Decide whether the formula p 0→¬p 0 is a tautology, a contingency, or a contradiction. I buy you a car. АЛВ+В'VA" с. Contradiction: A propositional formula is contradictory (unsatisfiable) if there is no interpretation for which it is true. As an introduction, we will make truth tables for these two statements 1. Tautology A statement is called a tautology if the final column in its truth table contains only 's. Compute the truth tables for the following propositional. tables consider all cases and can add great insight into otherwise complicated expressions. Discrete Mathematics Lecture 1 Logic: Propositional Logic Truth Table •A convenient way to see the effect of the contradiction E. entailment. For example, let us study the truth value of (p Λ q) → (p V q) by building a truth table. Then identify whether the sentence is a tautology, a contradiction or neither. Use a truth table to determine if the following is a tautology, a contradiction, or a contingency. Here we are going to study reasoning with propositions. In a proof by contradiction, we start out by saying: "Suppose not. is a proposition that is always. Illustrates and defines the terms ‘contradiction’, ‘logical truth’ and ‘logical possibility’. (a) Show that (p q) → p is a tautology by completing this truth table. Logically Equivalent Statements, Tautologies, & Contradictions; Definition 27. 5) E to I is a contradiction. By Using Truth Table Un 1. If a contradiction is produced in the attempts to assign truth values to the premises, circle the contradiction and declare the argument valid. A compound proposition is said to be a contradiction if and only if it is false for all. 2) The truth tables for the basic operations: F F F F F T T F T F T F T T T T 1. Lesson 46 – Truth Tables (D)Tautologies and Logical Contradictions a. So we can every satisfiable is a contingency because satisfiable will have at least one true value which will also satisfy the definition of Contingency. Note any tautologies or contradictions. This is a truth table generator helps you to generate a Truth Table from a logical expression such as a and b. Contingency-. Consider the connectives: contradiction denoted by ⊥ (which stands on its own), negation denoted by ¬ (not), which is The truth table for “and” is: p q p. Question: Use A Truth Table To Determine If The Following Is A Tautology, A Contradiction, Or A Contingency. contradiction. Actually, the truth is that proper logical analysis does. A row of the truth table in which all the premises are true is called a critical row. cut: The minimal score for the PRI - proportional reduction in inconsistency, under which a truth table row is declared as negative. p q :q p!q :(p!q) p^:q T T F T F F T F T F T T F T F T F F F F T T F F Since the truth values for :(p!q) and p^:qare exactly the same for all possible combinations of truth values of pand q, the two propositions are equivalent. Using truth-tables to test a set of statements for consistency: Construct a sentence that is the conjunction of all the statements in question. Tautology Truth Tables. It isn't just that the intersection of your final answer and the set of. By Using Truth Table Un 1. In real life, 2-LUTs are not an efficient use of resources; LUTs have 4 or 5 inputs. Solution 2. Learn more about the laws of thought in this article. CMSC 203 : Section 0201 : Homework1 Solution 3. АЛВ+В'VA" с. ] Thus, x is a ratio of the two integers −a and b with b ≠ 0. Example: p ¬p is a tautology. By the and truth table, if one part of an "and" state-ment is false, the entire statement is false. Robert Lacey explores the "untold reality" of the Duke of Cambridge and. Q^(˘Q) Theorem: A statement S is a tautology if and only if its negation is a contradiction. In this post, I will briefly discuss tautologies and contradictions in symbolic logic. Illustrates and defines the terms ‘contradiction’, ‘logical truth’ and ‘logical possibility’. A propositional form that is true in at least one row of its truth table and false in at least one row of its truth table is a contingency. Create a truth table to find out if this is either a contradiction, tautology, or contingency. A truth table is a table that begins with all the possible combinations of truth values for the letters in the compound statement; it then breaks the compound statment down and one step at a time determines truth values for each of the parts of the logical statement. is a proposition which is always false. combinations of truth values of the propositional variables which it contains. A truth table displays the relationships between the truth values of propo-sitions. Propositions are built up as truth-functions of elementary propositions (5). Although consistency is no guarantee of truth since one could create a consistent story that is false, it seems to be a necessary condition for truth. For example, if A is “Napoleon was born in Corsica” and B is “The number of the beast is 666”, the truth table (in classical two-valued logic) would be as follows:. A tautology is a compound statement that is true for all possibilities in a truth table i. Subtracting 1 from both sides gives 2sin xcos <0. In our everyday choices, are we willing to give up our power and privilege to invite others to the table, to share in the “American dream?” It’s time to right this 400-year-old wrong. Tautologies and Contradictions in Symbolic Logic - PHILO-notes Daily Whiteboard - Duration: 5:19. The only way a disjunction can be false is if both parts are false. (4V B')Л (4ЛВ)" d. Logical Equivalence Please use truth tables to check the following Boolean expressions. The number of lines needed is 2 n where n is the number of variables. (pva)^(-1) = pvqr(r)) Get more help from Chegg Get 1:1 help now from expert Other Math tutors. A truth table is a mathematical table used to determine if a compound statement is true or false. State, with a reason, whether the compound proposition (p V (p A q)) p is a contradiction, a tautology or neither. > Contradiction > Contingency > Truth table with 3 variables >Venn Diagram Mathematical Logic app includes following topics: > Negation - To identify a statement as true, false or open. Chapter 5 Truth Tables. ((PvQ) ^ (~R) = P v (Q^(~R)) Thats not an equal sign btw, but three lines intended instead of two. The table below explores the four possible cases, but the truth is simpler than that. A proposition that is always false is called a contradiction. Tautologies []. , p _˘p 2Taut. Classifying propositions using truth tables We may use truth tables to make important distinctions among tautologies, contingencies, and contradictions, Consider the truth table for ‘P e P’ P e P T T T F T F. Topic 3 - Logic, sets and probability » 3. But a truth tree will get the job done faster. We notice that an implication can only be false if the hypothesis is true and the consequence is false. A full development of a theory of truth in paraconsistent logic is given by Beall (2009). The truth table is: T F T T T T T All the values in the final column are false, so Example 5 Construct a truth table for p V —q. 3 Truth Tables FILLED IN NOTES. However, if every attempt to find such a set of T values ends in a contradiction, then the game cannot succeed because there is no set of truth values which will make all the premisses true and the conclusion false, so in other words there is no such row in the full table. A compound proposition is a logical contradiction if all the values in its ü-uth table column are false. Ø p ® c, where c is a contradiction \ p. A passionate Computer Science and Engineering graduate, who loves to follow his heart :). A tautology is a proposition which is always true. A proposition (statement) is a contradiction if it is logically false. truth table if you like, but you may not need to. Truth Tables Every proposition, indeed every logical formula, is either true or false. Discuss the distinct patterns of. (pva)^(-1) = pvqr(r)) Get more help from Chegg Get 1:1 help now from expert Other Math tutors. Topic 3 - Logic, sets and probability » 3. As we can see from the truth table above, the proposition is definitely not a contradiction. It is for this obvious reason that logicians invented a shorter, more efficient method of determining the validity of arguments, namely, the indirect truth table method. 2: Truth Tables Worksheet Fill out the following truth tables and determine which statements are tautologies, contradictions, or neither. 2 Apply the truth able method to assessing validity (finding counterexamples), identifying tautologies, contradictions, and contingent sentences, assessing logical equivalence and consistency. Consider the connectives: contradiction denoted by ⊥ (which stands on its own), negation denoted by ¬ (not), which is The truth table for “and” is: p q p. As "we are all happy" can be either true or false, so can a. By constructing the truth table, determine whether the following statement pattern ls a tautology , contradiction or. Truth Table is a mathematical table and the base for all computing needs. Contradiction Sayings and Quotes. A vector of (remainder) row numbers from the truth table, to code as negative output configurations. The truth table for a tautology has “T” in every row. Where, 0 and F denotes False. 5 Tautology, Contradiction, Contingency, and Logical Equivalence Definition : A compound statement is a tautology if it is true re-gardless of the truth values assigned to its component atomic state-ments. Therefore the order of the rows doesn't matter - its the rows themselves that must be correct. In other words, their columns on a truth table are identical. However, if every attempt to find such a set of T values ends in a contradiction, then the game cannot succeed because there is no set of truth values which will make all the premisses true and the conclusion false, so in other words there is no such row in the full table. 1 Introduction The truth value of a given truth- functional compound sentence depends on the truth value s of each of its components. Consider, for example, the statement form: p ~ p p ∨ ~ p. Satisfiability & Logical Truth PHIL 012 - 2/16/2001 Outline Test Scores Homework Reminder Satisfiability Logical Truth Complex Truth Tables Sample Problems Satisfiability A sentence is said to be satisfiable IFF under some circumstances it could be true, on logical grounds. I was fooling with term_variables/2 and I think it may be possible to achieve a simpler solution using term_variables/2, coming up with some scheme to get readable variable names output, but also assuming the operators are all evaluable by is/2. Propositions are either completely true or completely false, so any truth table will want to show both of these possibilities for all the statements made. Partial truth tables have two salient virtues. But please note that this is just an introductory discussion on tautologies and contradictions as my main intention here is just to make students in logic become familiar with the topic under investigation. Multiply both sides by −1, gives. Math, I have a question on tautologies and contradictions. How a proof by contradiction works. First, like whole truth tables, they are algorithmic (i. If you construct them correctly, you will get an answer to your question whether a particular argument is valid; whether a particular proposition is tautologous, self-contradictory, or contingent; or whether a particular set of propositions is consistent. A tautology is a logical compound statement formed by two or more individual statement which is true for all the values. An example is P ^(˘P). ” They are called direct proof, contra-positive proof and proof by contradiction. Chapter 1 Logic 1. none of these. Recharaaeriz-. Logical Equivalence, Tautologies, and Contradictions. Since most philosophers believe truth is logically consistent, they value logical consistency because it is a tool to discover truth. satisfiable, if its truth table contains true at least once. Contradiction A statement is called a contradiction if the final column in its truth table contains only 's. An example is P ^(˘P). A contingency is neither a tautology nor a contradiction, for instance p ∨ q is a contingency. ; Instead, ¬p is true. [(4 V B) ^…. Propositions are either completely true or completely false, so any truth table will want to show both of these possibilities for all the statements made. If you haven't read through the textbook sections that surround these tables, now would be a good time to do that. A proposition P is a tautology if it is true under all circumstances. Interpret your various truth-value assignments as a row of a truth-table. The upshot of this result is significant. 5 Tautology, Contradiction, Contingency, and Logical Equivalence Definition : A compound statement is a tautology if it is true re-gardless of the truth values assigned to its component atomic state-ments. Think of it as shorthand for a complex sentence like P&~P. This page contains a JavaScript program that will generate a truth table given a well formed formula of sentential logic. When you define a new theorem-like environment with ewtheorem, it is given the style currently in effect. State, with a reason, whether the compound proposition (p ∨ (p ∧ q)) ⇒ p is a contradiction, a tautology or neither. molecular statement is a contradiction if its truth table is always false. The number of lines needed is 2 n where n is the number of variables. A (complete) truth table shows the input/output behavior for all possible truth assignments. A formula is said to be a Contradiction if every truth assignment to its component statements results in the formula being false. the sentence '-(AvB)&(-A>B)' and ask you whether it is a contradiction. 1) is L∧¬La tautology a logical contradiction or neither?. The expression is simplified: (1. Use a truth table to determine whether the two statements are equivalent. truth tables. You can enter logical operators in several different formats. Assume that P is true. Truth Tables - Tautology and Contradiction. A compound statement that is false for all possible combinations of truth values of its component statements is called a contradiction. I construct the truth table for (P → Q)∨ (Q→ P) and show that the formula is always true. Table 1: The truth table for the negation. 1 Introduction The truth value of a given truth- functional compound sentence depends on the truth value s of each of its components. State whether the statements ¬ p ⇒ q and (¬ p ⇒ q) ∨ (¬ p ∧ q) are logically equivalent. Tautologies and Contradictions in Symbolic Logic - PHILO-notes Daily Whiteboard - Duration: 5:19. Obviously, truth tables of these sizes are simply impractical to construct. " We then show that this leads to a contradiction, a statement like (q ∧ ¬q). What is the truth of the matter? Truth’s character is both logical and empirical. ***** Full Course Playlist. In this video I construct two more truth tables and use them to illustrate the notion of a tautology and a contradiction. Use a truth table to determine if the following is a tautology, a contradiction, or a contingency. 2 Problem 14ES. Think of it as shorthand for a complex sentence like P&~P. 6 Propositions – Tautology, Contradiction, Contingency Tautology A proposition P is a tautology if and only if P is true under every valuation. In the truth tables above, there is only one case where "if P, then Q" is false: namely, P is true and Q is false. Truth Tables, Tautologies, and Logical Equivalences. 2: Truth Tables for Negation, Conjunction, and Disjunction Math 121 Truth Tables A truth table is used to determine when a compound statement is true or false. Similarly, if it is false, then it implies it is true. PHILO-notes 2,286 views. Propositional Equivalences Tautologies, Contradictions, and Contingencies. Therefore to establish that a conditional. 24 and 25 are pretty handy. We say a statement Scan be deduced from a statement R if the truth table of S is True whenever. , Fs in its truth table column. The tee symbol ⊤ is sometimes used to denote an arbitrary tautology, with the dual symbol ⊥ representing an arbitrary contradiction; in any symbolism, a tautology may be substituted for the truth value "true," as symbolized, for instance, by "1. Trump, Truth and the Power of Contradiction. Tautologies []. A proposition is said to be a tautology if its truth value is T for any assignment of truth values to its components. p ∨ q Solution to EXAMPLE 2. Ø p ® c, where c is a contradiction \ p. R R • ∼ R T T F F F F F T Willard is either a philosopher or a windbag, and he’s neither a philosopher nor a windbag. Determine whether the following statement pattern is a tautology or a contradiction or contingency. That is, a statement and its negation can never have the same truth value. Definition: A compound statement is a contradiction if there is an F beneath its main connective in every row of its truth table. The ntheorem package provides nine predefined theorem styles, listed in Table 4. Some sentences have the property that they cannot be false under any circumstances. Working backwards from the truth values generated by evaluating the conclusion false, deduce the truth values for the remaining simple propositions and premises. The truth table for the conjunc-tion of two statements is shown in Figure 1. Logical Symbols are used to connect to simple statements, to define a compound statement and this process is called as logical operations. If a statement is neither a tautology nor a contradiction, then the truth values do alter the outcome and we say that the statement is a contingency. The truth-table at right demonstrates the logical equivalence of these two statement forms. contradiction. In this video I construct two more truth tables and use them to illustrate the notion of a tautology and a contradiction. Create a truth table showing the values of the premises and conclusion. Depending on the set of axioms exhibited and/or we are willing to impose on the linguistic expressions of our native intelligence, we might arrive at different computational intelligence expressions for. Check ((p ∨ q) ∧ ¬ q) → p is a tautology using truth table. The truth table for a tautology has “T” in every row. Prove Logical Equivalences. In real life, 2-LUTs are not an efficient use of resources; LUTs have 4 or 5 inputs. Explain in a sentence why your truth table shows whether it is a tautology, a contradiction, or a contingent proposition. To show that a sentence is not a tautology, however, we only need one line: a line on which the sentence is 0. Tautologies and Contradictions in Symbolic Logic - PHILO-notes Daily Whiteboard - Duration: 5:19. Locate the rows in which the premises are all true (the critical rows). A→ (B → A) b. And we can draw the truth table for p as follows. Truth Tables - Tautology and Contradiction. The second case agrees with none and we call it a contradiction. However, to show that an argument is not valid, all we need to do is to find one assignment where all the premises are true and the conclusion is. of the truth values of the propositional variables which comprise it. it is true in no world, e. Introduction to Philosophy > Logic > Tautologies and Contradictions. 6 a tautology 2. , ), the following statements can be made: Valid implies that the argument must be true for all instances (i. Fill in the truth table for the sentence of propositional logic below. Select "Full Table" to show all columns, "Main Connective Only. Learn more about the laws of thought in this article. p) is a logical contradiction. In sentential logic all theorems are tautologies and all tautologies are either axioms or theorems. Tautologies, Inconsistent Sentences, and Contingent Sentences Tautologies. So it is not a TT-contradiction. A propositional form that is true in all rows of its truth table is a tautology. The size of the truth table depends on the number of different simple. Here are examples of some of most basic truth tables, Truth table for negation ("not") Truth table for conjunction ("and" Truth table for disjunction ("or") Ex a Translate to symbolic form, then construct a truth table to represent the expression:. Truth Tables - Tautology and Contradiction. Contradiction - example 17 5. Create a truth table to determine whether the following statement is contingent, a tautology, or a self-contradiction. In other words, their columns on a truth table are identical. If it is always true, then the argument is valid. contradiction, is a statement which is false regardless of the truth values of the substatements which form it. The minimum number of cases under which a truth table row is declared as a remainder. here they are: a) i found this proposition was a contradiction. Confirmed! The conditional p —+ q and it's contrapositive -Iq —Y -p are logically equivalent. 2 LOGICAL EQUIVALENCE, TAUTOLOGIES & CONTRADICTIONS. Let's start with logical contradiction. ((PvQ) ^ (~R) = P v (Q^(~R)) Thats not an equal sign btw, but three lines intended instead of two. you’re assuming it’s consistent by plugging in all true truth values. The expression "p or not p" is true under any circumstance, so it is a tautology. АЛВ+В'VA" с. Where, 0 and F denotes False. The following two truth tables are examples of tautologies and contradictions, respectively. The opposite of a tautology is a contradiction, a formula which is "always false". On my upcoming test time is limited. A contradiction is a conjunction of the form "A and not-A", where not-A is the contradictory of A. 12, 15 Examine the statement Patterns (Tautology, Contradiction, Contingency) 1. A truth table column which consists entirely of T's indicates a situation where the proposition is true no matter whether the individual propositions of which it is composed are true or false. Solution: Question 2. | 2020-11-23T16:40:24 | {
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https://mathhelpboards.com/threads/prove-a-function-is-continuos.3349/ | # Prove a function is continuos
#### ianchenmu
##### Member
Prove that
$f(x,y)=\left\{\begin{matrix} e^{-1/|x-y|},x\neq y\\ 0,x=y \end{matrix}\right.$
is continuous on $\mathbb{R}^2$.
Can I conclude since $e^{-1/t}$ is continuous then for $x$,$y\in \mathbb{R}^2$, $x\neq y$,$e^{-1/|x-y|}$ is continuous on $\mathbb{R}^2$ since here $t=|x-y|$? And how to prove it when $x=y$, using the $\delta- \varepsilon$ way? Thank you a lot!
#### girdav
##### Member
Re: Prove a function is continuous
We have to show that the function is continuous at each point of $\Bbb R^2$. Your argument threats the case of points of the form $(x,y),x\neq y$. So now, fix $x_0$. We have to show that $\lim_{(x,y)\to (x_0,x_0)}=0$.
Hint: use $e^{-u}\leqslant \frac 1{1+u}$ for each non-negative real number $u$.
Last edited:
#### Fernando Revilla
##### Well-known member
MHB Math Helper
Re: Prove a function is continuous
Hint: use $e^{-u}\leqslant \frac 1{1+u}$ for each real number $u$.
Surely is a typo, I suppose you meant for each real number $u$ close to $0$.
Another way: denoting $\Delta=\{(t,t):t\in\mathbb{R}\}$ (diagonal of $\mathbb{R}^2$), in a neighborhood $V$ of $(0,0)$ we have:
$\displaystyle\lim_{(x,y)\to (x_0,x_0)\;,\;(x,y)\in V-\Delta}f(x,y)=\displaystyle\lim_{(x,y)\to (x_0,x_0)\;,\;(x,y)\in V-\Delta}e^{-\dfrac{1}{|x-y|}}=e^{-\infty}=0$
$\displaystyle\lim_{(x,y)\to (x_0,x_0)\;,\;(x,y)\in V\cap\Delta}f(x,y)=\displaystyle\lim_{(x,y)\to (x_0,x_0)\;,\;(x,y)\in V\cap\Delta}0=0$
In a finite partition of $V$ all the limits coincide, so $\displaystyle\lim_{(x,y)\to (x_0,x_0)}f(x,y)=0$ .
#### ianchenmu
##### Member
The definition of continuity for a function is {lim x-> a} f(x) = f(a). For this function we only need to worry about when |x - y| -> 0, since the exponential function is continuous everywhere. So in other words we need to show that
{lim |x - y|-> 0} f(x,y) = f(x,x) = f(y,y)
Now f(x,x) = f(y,y) = 0 so we just need to show that
{lim |x - y| -> 0} f(x,y) = 0.
Since |x - y| > 0, -1/|x - y| -> infinity so f(x,y) -> 0 and we're done.
Is a proof like this correct? | 2020-11-30T13:49:22 | {
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https://math.stackexchange.com/questions/1291087/how-to-solve-an-equation-with-x4/1291091 | # How to solve an equation with $x^4$?
Today, I had this question on a Maths test about Algebra. This was the equation I had to solve:
$$(1-x)(x-5)^3=x-1$$
I worked away the brackets and subtracted $x-1$ from both sides and was left with this:
$$-x^4+16x^3-90x^2+199x-124=0$$
Problem is, I haven't a clue how to solve this? First thing I tried was replacing $x^2$ with another variable like $u$ but that got me no further. Dividing the whole equation by $x^2$ (as is suggested by a lot of sites on this matter) also did not get me any further. I then tried something incredibly ludicrous;
$$(ax+b)(cx^3+dx^2+ex+f)=0$$ \left\{ \begin{aligned} ac &= -1 \\ ad + bc &= 16 \\ ae + bd &= -90 \\ af + be &= 199 \\ bf &= -124 \end{aligned} \right.
which got even worse when there were 3 brackets;
$$(ax+b)(gx+h)(ix^2+jx+k)=0$$ \left\{ \begin{aligned} ac &= agi &&= -1 \\ ad + bc &= agj + ahi + bgi &&= 16 \\ ae + bd &= agk + ahj + bgj + bhi &&= -90 \\ af + be &= ahk + bgk + bhj &&= 199 \\ bf &= bhk &&= -124 \end{aligned} \right.
only to be left with no result.
When using Wolfram Alpha on this question, it performs a rather strange step I don't understand:
$$-x^4+16x^3-90x^2+199x-124=0$$ $$\downarrow$$ $$-((x-4)(x-1)(x^2-11x+31))=0$$
Could somebody explain how to properly tackle this problem? And if possible, also show me how to get the non-real answers for it. Thanks.
• You needn't expand: first, observe that $x=1$ is a solution. Now, looking for other solutions (at most 3 other possible ones), you can divide each side by $x-1$ and obtain the equation $(x-5)^3=-1$ to solve (for values $x\neq 1$). – Clement C. May 20 '15 at 13:13
Euh... I think you overcomplicated things here...
$(1-x)(x-5)^3=x-1$ is equivalent to $(1-x)[(x-5)^3+1]=0$
Either $x=1$ or $(x-5)^3=-1$...
• I feel so stupid, that I didn't see this in the first place... Thanks for the great answer! – Martijn May 20 '15 at 13:19
• @TitoPiezasIII Done :) – Martijn May 20 '15 at 13:27
• You can also use the identity $a^3+1=(a+1)(...)$ – k1.M May 20 '15 at 14:00
There are already answers on how to get the real solutions, so I will only show you the non-real solutions.
You have obtained that $(x-5)^3=-1$. Expanding and simplifying we get: $$x^3-15x^2+75x-124=0$$ However, we know that $x=4$ is a solution so we can say that $(x-4)(ax^2+bx+c)=0$. You can equate coefficients or use polynomial division, but as you have already found with Wolfram Alpha, $(ax^2+bx+c)=(x^2-11x+31)$.
To solve, complete the square: $$x^2-11x+31=0$$ $$(x-\frac{11}{2})^2-\frac{121}{4}+31=0$$ $$(x-\frac{11}{2})^2=-\frac{3}{4}$$ $$x-\frac{11}{2}=\sqrt{-\frac{3}{4}}= \pm \frac{\sqrt{3}}{2}i$$ $$x=\frac {11 \pm i \sqrt{3}}{2}$$
• Awesome work! I have two kinds of mathematics in the Netherlands and the other kind is about imaginary numbers a.t.m. so this will definitely come in handy, I'm sure! – Martijn May 20 '15 at 13:45
• In general the solutions to $(x-a)^m=-1$ are $x_n=e^{i\pi\frac{2n-1}{m}}+a$, for $n=1,2,\dots m$. – Kwin van der Veen May 21 '15 at 4:10
From the beginning: $$(1-x)(x-5)^3=x-1\\ (1-x)(x-5)^3+1-x=0\\ (1-x)(x-5)^3+(1-x)=0\\ (1-x)[(x-5)^3+1]=0\\$$
This implies $1-x=0$ or $(x-5)^3=-1$. I believe you can solve these.
• Isn't there a sign that got flipped between the first and second line? – Clement C. May 20 '15 at 13:18
• Sorry fixed it. @ClementC. – KittyL May 20 '15 at 13:24
I know that the problem has already been answered but I want to show you a more general method, let's suppose that you don't se how to rewrite the equation:
$-x^4+16x^3-90x^2+199x-124=0$
Or I prefer to write:
$x^4-16x^3+90x^2-199x+124=0$
You can use something called the Ruffini rule: search for integers divisor (both positive and negative) of the constant term and then set $x$ equals to the them and see if one of them is a solution. Starting from one we have:
$1-16+90-199+124=0$
So $x=1$ is a solution, now via Ruffini's rule, which can be seen here, we can rewrite the equation as:
$(x-1)(x^3-15x^2+75x-124)=0$
Now you have $3$ options to end this exercise:
$1.)$ Note that the second factor is a perfect cube;
$2.)$ Use Ruffini's rule again;
$3.)$ Use the general formula for third degree equation (which I'd not advise you to if your interested only in real solution).
This is a more general method so I hope this will help you in the future! | 2019-11-13T05:13:02 | {
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https://math.stackexchange.com/questions/3119998/numbers-of-distinct-values-obtained-by-inserting-times-div-in-under | # Numbers of distinct values obtained by inserting $+ - \times \div ()$ in $\underbrace{2\quad2 \quad2 \quad2\quad…\quad 2}_{n \text{ times}}$
This question is inspired by How many values of $2^{2^{2^{.^{.^{.^{2}}}}}}$ depending on parenthesis? (By the way, I sincerely hope this kind of questions can receive more attention)
Insert $$+ - \times \div ()$$ in $$\underbrace{2\quad2 \quad2 \quad2\quad...\quad 2}_{n \text{ times}}$$ Denote the number of distinct values which can be obtained in this way by $$D(n)$$. Is there a general formula (or recurrence relation at least) for $$D(n)$$?
This is basically the $$+ - \times \div ()$$ version of @barakmanos question. It seems this question is easier than the power tower version. Or maybe not?
For $$n=1$$ , there is only $$2$$ values $$-2,2$$;
For $$n=2$$, there are $$5$$ values $$-4,-1,0,1,4$$;
For $$n=3$$, there are $$13$$ values $$-8,-6,-3,-2,-1,-\frac{1}{2},0,\frac{1}{2},1,2,3,6,8$$;
And for $$n=4$$ I'm reluctant to calculate with bare hands. (See @DanUznanki answer for what follows)
Any idea is appreciated. Sorry if this is a duplicate.
Edit: My research shows that the version with distinct generic variables $$a_1,a_2,...,a_n$$ is solved. See A182173 for your reference.
• Based on what you say for $n=1$ and $n=2$, we may use $-$ as either the negation operator on one number or the subtraction operator between two numbers? – alex.jordan Mar 24 at 4:53
• @alex.jordan Yes. – YuiTo Cheng Mar 24 at 4:59
• OK. I wonder if you did not allow $-$ to be unary negation, if then the sequence might appear in OEIS. We have an overloaded symbol, $-$. And one of its meanings happens to be unary negation. It's odd that we don't have a symbol for unary inversion. Well, we do: $^{-1}$. But it does not reuse the division symbol $\div$. All this is just to say that it feels like the special double meaning for $-$, that is not mirrored with $\div$, could break what would otherwise be a simpler pattern in the sequence of counts. – alex.jordan Mar 24 at 5:22
• @alex.jordan If we disallow $-$ to be a unary operator, will things become vastly simpler? – YuiTo Cheng Mar 24 at 5:27
• @alex.jordan: I once toyed with the idea of "$/$" for unary inversion. (So, "$/2$" means "$1/2$" in the same way that "$-2$" means "$0-2$".) I still kinda like it. :) – Blue Mar 25 at 20:09
It looks like we can do this "inductively": for the calculations of size $$n$$, we can take values from the list for $$1 \le k\le n/2$$, and values from the list for $$(n-k)$$, and operate on them using the 64 different operation orders.
Fortunately, it's really only 10 classes of operation, because many are duplicates:
• There's four values we can get from addition and subtraction: $$a+b$$, $$-(a+b)$$, $$a-b$$, and $$b-a$$.
• There's only two we can get from multiplication: $$ab$$ and $$-ab$$.
• There are four cases we can get from division: $$a/b$$, $$b/a$$, $$-a/b$$, and $$-b/a$$.
Also conveniently we only need to try the nonnegative entries in previous lists.
So for $$n=4$$, we have:
$$-16$$, $$-12$$, $$-10$$, $$-8$$, $$-6$$, $$-5$$, $$-4$$, $$-3$$, $$-5/2$$, $$-2$$, $$-3/2$$, $$-1$$, $$-2/3$$, $$-1/2$$, $$-1/3$$, $$-1/4$$, $$0$$, $$1/4$$, $$1/3$$, $$1/2$$, $$2/3$$, $$1$$, $$3/2$$, $$2$$, $$5/2$$, $$3$$, $$4$$, $$5$$, $$6$$, $$8$$, $$10$$, $$12$$, $$16$$
Which is $$33$$ entries.
I've written a short script which finds them all, and told it to run up to $$n=10$$, which gave the following sizes: $$2,5,13,33,77,185,441,1051,2523,6083$$. Apparently this sequence is not in OEIS, nor is the positive-values-only version! I am very surprised.
• You mentioned we can do it 'inductively', do you have any clue about the recurrence relation? – YuiTo Cheng Feb 20 at 13:31
• Not sure, but it isn't gonna be nice. actually, different values for $x$ instead of $2$ give different results: using $3$ gives $2$, $7$, $21$, $67$, using $4$ gives $2$,$7$,$21$,$77$. Interestingly, when I tried to get the generic result, where $x$ is just $x$, I did get some partial results in the realm of restricted lattice walks: $2$,$7$,$23$,$91$, $347$ appears twice in OEIS. – Dan Uznanski Feb 20 at 13:39
• Interesting. What about the original power tower question? Compared with this, which one do you think is simpler? – YuiTo Cheng Feb 20 at 13:45
• for generic $x$ in the power tower thing you're getting exactly the Catalan Numbers. But assuming you have the ability to actually store the numbers you're working with, the power tower version is vastly simpler: there's only one operator instead of 10. – Dan Uznanski Feb 20 at 13:46
• What about the case without bracket? – YuiTo Cheng Feb 21 at 3:36
@Dan Uznanski if your results are correct i noticed that the ratio between the successive terms of the sum is almost constant. So maybe:
Conjecture
$$D(n) \sim K^n$$
Where:
$$2.3\leq K \leq 2.5$$ | 2019-04-23T10:33:28 | {
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http://math.stackexchange.com/questions/120861/homomorphism-between-cyclic-groups | # Homomorphism between cyclic groups
I have some confusion in relation to the following question.
Let $\langle x\rangle = G$, $\langle y\rangle = H$ be finite cyclic groups of order $n$ and $m$ respectively. Let $f:G \mapsto H$ be the mapping sending $x^i$ to $y^i.$ Determine the conditions on $n$ and $m$ so that $f$ is going to be a homomorphism.
In verifying the definition of a homomorphism I seem to be able to conclude that there is no restriction on $n$ and $m$ since the definition of a homomorphic mapping is satisfied:
$$f(x^i x^j) = f(x^{i+j}) = y^{i+j} = y^i y^j = f(x^i)(x^j) \; \; \; (1)$$
But assuming n < m we would get
$$f(x^n) = f(e) = e \not = y^n$$
So clearly we have to have $m | n.$
My questions are
1. Are there any other conditions on $n$ and $m$ besides $m |n.$
2. What exactly did I do wrong in $(1)$ ?
-
For the (1), What would happen if $i+j \ge m$? It always helps to define the map this way: $x^{i(n)} \mapsto x^{i(m)}$. Note that the map may not even be well-defined if $m \nmid n$. – user21436 Mar 16 '12 at 9:50
I'll write up a more detailed answer in an hour if it's still unanswered. This has its direct connection to the universal property of the quotient maps. – user21436 Mar 16 '12 at 9:52
I see that there are many contradictions if $m \not | n.$ My main concern is that $(1)$ seems like such a innocent algebraic manipulation that I would never suspect that something is wrong with it.I would like to hear an "aha" explanation on why one cannot apply the definition of a homomorphism on $f$ so blindlessly. – Jernej Mar 16 '12 at 10:02
– lhf Mar 16 '12 at 10:43
Well, either you use a homomorphism $\mathbb Z \rightarrow G$ (and talk about lifting that homomorphism) or you must face the fact that the Element $x^i$ does NOT determine the number $i$. – Blah Mar 16 '12 at 10:50
The discussion about "well-defined" is perhaps a bit obscure. Here's the problem:
Remember that in general, an element of $G$ may have many different "names." For example, if $n=10$, then the element $x^{11}$ is equal to $x$; in fact, $x$ has infinitely many different "names" as a power of $x$: $$\cdots = x^{-9} = x^1 = x^{11} = x^{21} = \cdots$$
The problem is that the definition of $f$ as given depends on the name of the element! That is, if we are furiously working and somebody hands us a power of $x$, say, $x^{3781}$, we are supposed to, unthinkingly, map it to $y^{3781}$. The problem is that $x^{3781}$ is the same element as $x$, which we are supposed to send to $y^1$. That means that unless $y^1=y^{3781}$, what we have is not really a function: because the same input, $x$ (who, when being teased by bullies is called "$x^{3781}$") may be sent to $y^1$ or to $y^{3781}$, depending on what "name" we just heard for it.
Checking that the value of the function is the same regardless of what name we are using for an element, even though the function is defined in terms of the name, is called "checking that the function is 'well-defined.'"
An example of a function that is not well-defined would be one in which the input is an integer, and the output is the number of symbols used to express that integer. For example, $f(3)$ would be $1$ (because 3 is only one symbol), but $f(4-1)$ would be $3$ (because we are using 4, -, and 1). This is not well defined as a function of the integers, because $3$ is the same as $4-1$, but $f$ assigns it two different outputs.
So in order for the expression given to actually define a function, we need the following to be true: $$\text{if }x^i=x^j,\text{ then }y^i = f(x^i)\text{ is equal to }y^j=f(x^j).$$ Now, $x^i=x^j$ if and only if $i\equiv j\pmod{n}$; and $y^i=y^j$ if and only if $i\equiv j\pmod{m}$. Therefore, we need: $$\text{if }i\equiv j\pmod{m},\text{ then }i\equiv j\pmod{n}.$$ In other words: every number that is a multiple of $m$ must be a multiple of $n$.
This is equivalent to $n|m$.
In fact, you noticed that in what you wrote, because $x^n=e$, so we need $f(x^n)$ to be the same as $f(x^0)$.
Once we know that $n|m$, then $f$ is well defined. Once it is well-defined, we can start verifying that it is indeed a homomorphism (it is). Technically, it's incorrect to start working to see if it is a homomorphism before you even know whether or not it is a function.
Note that the condition actually works if we allow $G$ or $H$ to be infinite cyclic groups, if we call infinite cyclic groups "groups of order $0$". Then $i\equiv j\pmod{0}$ means $i=j$, every number divides $0$, but the only multiple of $0$ is $0$. So if $G$ is infinite cyclic then the value of $n$ does not matter; if $H$ is infinite cyclic then we must have $G$ infinite cyclic.
-
I don't agree with your take on well-defined. The rule $x^i\mapsto y^i$ is well-defined in the set $\{x^0, x^1, \ldots, x^{n-1}\}$ because every element in this set has a single name. The problem is that this rule may not be a homomorphism because there is no guarantee that say $x^{2n-3}$ goes to $f(x)^{2n-3}=y^{2n-3}$. – lhf Mar 16 '12 at 16:47
@lhf: I agree that if we were specifying $0\leq i\lt n$, then the rule would yield a well-defined function. However, I don't see any such specification in the statement of the problem. It just says "sending $x^i$ to $y^i$", with no restrictions on $i$. Of course, checking that it is a homomorphism is important: I note it later in the post, I don't ignore it. – Arturo Magidin Mar 16 '12 at 16:58
Right, I assumed that $0\leq i\lt n$ was a natural implicit assumption. – lhf Mar 16 '12 at 17:04
@lhf: In that case, there is a different problem with the definition, which is that it would only make sense if $n\leq m$ (otherwise, $y^i$ for $i\geq m$ would be an "invalid name"). So we would end up with the conclusion that the map only makes sense and is a homomorphism when $m=n$; or else we need to treat the exponent differently in the domain (restricted to a special representation for the elements) but not the codomain, which seems somewhat perverse. – Arturo Magidin Mar 16 '12 at 18:21
I don't think $y^i$ is an invalid name. Just raise $y$ to $i$. The exponent is well defined, isn't it? – lhf Mar 16 '12 at 21:27
You may want to check first if the map is well defined, ie., if $x^i=x^j$ does it follow that $f(x^i)=f(x^j)$?
-
+1 Why would anyone downvote this? Ok, it would be better to phrase it: "if $x^i=x^j$ does it follow that $y^i=y^j$?" It is somewhat dangerous to talk about values of $f$ before you know that $f$ is well defined. – Jyrki Lahtonen Mar 16 '12 at 14:07
it is perfectly acceptable to ask whether or not a definition of f makes f a function. – David Wheeler Mar 16 '12 at 15:22
@David, you're right, of course. I was just trying to guess, why there was a downvote for a little while. – Jyrki Lahtonen Mar 16 '12 at 15:44
I am not sure if this completely explains why the naive calculations in $(1)$ are not valid. Here is my go at it:
I'll list down several points of concern.
• Firstly your map $x^i \mapsto y^i$ is very naive that it misses some essential details.
For convenience of notation, I prefer to look at this as $\varphi(\bar i)=\bar i$ if you'll allow me to do so. This is also naive but to some extent captures what is happening.
So, an equivalence class $\bmod n$ is mapped to the same equivalence class $\bmod m$ under $\varphi$. This bit of information is missed in $x^i \mapsto y^i$. This explains my comment under your question that you will want to write your maps as $x^{i(n)} \mapsto y^{i(m)}$. This notation makes things more transparent.
That you have dicovered that $m \mid n$, it can be obtained through the universal property of the quotients, which I will write about later.
The quotient map $G \to G/H$ factors over $G \to G/K$ if and only if $K \subseteq H$. That is there is a map from $G/H$ to $G/K$ if and only if $K \subseteq H$.
Proof: Please Go through Dummit and Foote's description in their Abstract Algebra in the section $\S3.3$ Isomorphism Theorems.
Now consider your cyclic groups as $\Bbb Z/n\Bbb Z$ and $\Bbb Z/m\Bbb Z$ and observe that above condition will tell you, $m\Bbb Z \subseteq n \Bbb Z$ whic happens if and only if...
As you had earlier in your comments observed, the map won't even be well defined if $m \nmid n$.
I'll leave it to you as a challenge to use this blooper to prove that two groups of same order are isomorphic, which really is not the case. Hint: Cyclic groups exist for any given order.
-
Will the down voter explain why? – user21436 Mar 16 '12 at 14:01
Why do you say that $y^{i+j}$ might not be an element of $H$? Of course it is an element of $H$! A group is a set closed under the group operation, so if $y$ is in there, so are all its powers. Also, are you sure that $i(m)$ is a standard notation for the remainder? Color my old-fashioned, but I think that binary mod should stay in programming classes. This last bit is just a pet-peeve of mine :-) – Jyrki Lahtonen Mar 16 '12 at 14:04
Well, $y^{i+j}$ is an element of the group. For example, if $H$ is a subgroup of the multiplicative group of complex numbers generated by the imaginary unit $i$, you seem to be saying that $i^5$ is not an element of $H$. Well, you are wrong: $$i^5=i^{4+1}=i^4\cdot i=1\cdot i=i.$$ Looks like it is in there! The same holds for all cyclic groups. I just picked a concrete one to make you realize your error. – Jyrki Lahtonen Mar 16 '12 at 14:16
No. I'm not prone to making a mistake. I check that my mappings are well-defined. If (granted, a big if in some cases) $y^i=y^j$ whenever $x^i=x^j$, then the mapping $x^i\mapsto y^i$ is trivially a homomorphism: $$f(x^i*x^j)=f(x^{i+j})=y^{i+j}=y^i*y^j=f(x^i)f(x^j)$$ AFTER I have checked that it is well-defined, then the above always makes sense, because in a cyclic group the power of an element is defined in such a way that we always have $y^{i+j}=y^i*y^j$ and so forth. – Jyrki Lahtonen Mar 16 '12 at 14:31
as a matter of fact, it IS true that $(i+j)$ (mod n) = $i$ (mod n) + $j$ (mod n). the problem is that $(i+j)$ (mod n) may not equal $i+j$ (mod m). – David Wheeler Mar 16 '12 at 15:32
The rule $x^i\mapsto y^i$ gives a perfectly well-defined function in the set $\{x^0, x^1, \ldots, x^{n-1}\}$. However, for that rule to be a homomorphism we need that $x^{i+j} \mapsto y^{i+j}$ for all $i$ and $j$. To find where $x^{i+j}$ goes we need to reduce $i+j \bmod n$ to a $k$ in $\{0,1,\ldots,n-1\}$ and we need that $k \equiv i+j \bmod m$. In other words, we need $k \equiv k' \bmod n \implies k \equiv k' \bmod m$ and this can only happen when $m\mid n$.
-
THEOREM:
Define $f:\langle x \rangle \to \langle y \rangle$, where the orders of $\langle x \rangle$ and $\langle y \rangle$ are n and m respectively, by $f(x) = y^u$. Then f is a well-defined homomorphism if and only if $\frac{m}{\gcd(m,n)} \mid u$.
PROOF: Suppose that the mapping $f:\langle x \rangle \to \langle y \rangle$ is a well-defined homomorphism.
Let $f(x) = y^u$ for some non negative integer u.
We must also have $1 = f(x^n) = y^{un}$
Then $m \mid un$. Let $g = \gcd(m, n)$. Then $\frac m g \mid u \frac n g$. It follows that $\frac m g \mid u$.
The converse is pretty straightforward.
Suppose we define $f:\langle x \rangle \to \langle y \rangle$ by $f(x) = y^u$ where $\frac{m}{\gcd(m,n)} \mid u$
The only real problem with this definition is "Is it well defined?" Since the order of $\langle x \rangle$ is $n$:
\begin{align*} x^a = x^b &\Rightarrow a \equiv b \mod n \cr &\Rightarrow ua \equiv ub \mod {un} \cr &\Rightarrow ua \equiv ub \mod {\frac{mn}{\gcd(m,n)}} \cr &\Rightarrow ua \equiv ub \mod {m} \cr &\Rightarrow y^{ua} = y^{ub}\cr &\Rightarrow f(x^a) = f(x^b) \end{align*}
Linearity is pretty obvious, so $f$ is a well-defined homomorphism.
So, if you want $f(x^i) = y^i$, then you want $f(x) = y$. You must therefore have $\frac{m}{\gcd(m,n)} \mid 1$. Which implies $m \mid n$.
Conversely, if $m \mid n$ and $f(x) = y$, it follows that $f(x^i) = y^i$.
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https://math.stackexchange.com/questions/2461040/induction-and-proper-use-of-the-inductive-step | # induction and proper use of the inductive step
Suppose I have a conjecture of the form $\forall x \in \mathbb{N}^{+}$ $f(x) > g(x)$.
Without explicitly giving the functions $f$ and $g$ I would like to prove this conjecture using induction.
Specifically I have shown $f(1) > g(1)$. My inductive assumption is now that $f(x - 1) > g(x - 1)$ and I'm trying to show $f(x) > g(x)$.
I haven't been able to do this, but I've noticed that I can if I assume $f(y) > g(y)$ $\forall y \in \mathbb{N}^{+}$ such that $y < x$ instead of my inductive argument I can prove $f(x) > g(x)$.
My question is, is this ok or am I using circular logic? Typically when we use induction we use $x - 1$ to prove the $x$ case after an initial base case. For my problem it doesn't appear possible to prove $f(x) > g(x)$ without assuming that for all values less than $x$ the conjecture holds ($f$ and $g$ are recursive). I suspect that it is ok, but it sounds extremely circular and I haven't been able to convince myself that it isn't.
Note also that I've included proof verification as a tag because although I have not included the explicit proof, I have included a proof approach that I'm interested in validating.
• That's fine. That's called strong induction. It might be clearer if worded it slightly differently. You know $f(1) > g(1)$ and you have proven that if $f(y)> g(y)$ for all $y \le x-1< x$ then $f(x) > f(x)$. Hence... induction. Is that clearer? – fleablood Oct 14 '17 at 17:13
So you just want to make the stronger assumption that some statement $P(y)$ holds for all $y < x$ instead of only for $y = x-1$?
This is totally fine. It is called strong induction and is equivalent to simple induction.
• Ok, that's what I thought. I was just a bit unsure because I couldn't remember doing something like this in any course I've taken. Thank you! – HXSP1947 Oct 7 '17 at 1:54
"Typically when we use induction we use x−1 to prove the x case after an initial base case. For my problem it doesn't appear possible to prove .. the proposition.. without assuming that for all values less than x the conjecture holds"
Think about this. If being true for all $y \le x-1 < x \implies$ being true for $x$, then being true for $y < x$ and being true for $x$ implies being true for all $y < x+1$. So true for $x+1$. And so on.
In principal, this is the exact same reasoning as the reasoning for induction. (if it's true for one case, it's true for the next, and via infinite countable reptitions it must be true for all).
The only difference is that whereas in "weak" induction we only make use of the fact that the proposition is true of $1$ and for $x-1$ and therefore true for $x$ we don't need to and we don't make use of the fact that it must also be true for $1,.......,x-2$ as well as just $x-1$.
But we know they must be true so we can make use of them if we need to.
This is called "strong" induction. It is equivalent to weak induction. The only difference is we might need to make use of earlier values than just the $x-1$ case.
This can be particularly useful with recursive definitions.
Say you know that $a_1 = something$ and $a_n =$ something to do with $a_{n-1}$ if $n$ is odd and something to do with $a_{\frac n2}$ and $a_{n-1}$ if $n$ is even. And you want to prove something about $a_n$. Well, it is not just enough to assume that the proposition is true of $a_{n-1}$. You also have to assume it is true of $a_{\frac n2}$. Can you assume that? Of course you can. To get to $a_{n-1}$ you had to have "passed through" $a_{\frac n2}$ "along the way". So why not use it if you have it? | 2019-05-22T08:35:51 | {
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# In the figure above if l1||l2, what is the value of x?
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In the figure above if $$l_1||l_2$$, what is the value of x?
A. 70
B. 100
C. 110
D. 150
E. 170
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In the figure above if l1||l2, what is the value of x? [#permalink]
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15 Mar 2018, 23:19
Bunuel wrote:
In the figure above if $$l_1||l_2$$, what is the value of x?
A. 70
B. 100
C. 110
D. 150
E. 170
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Re: In the figure above if l1||l2, what is the value of x? [#permalink]
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15 Mar 2018, 23:53
I got A - 70°, you can complete the top triangle as 30°, 40° and 110° because l1 and l2 are parallel, x = 180° - 110° which is 70°
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Re: In the figure above if l1||l2, what is the value of x? [#permalink]
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16 Mar 2018, 08:35
Bunuel wrote:
In the figure above if $$l_1||l_2$$, what is the value of x?
A. 70
B. 100
C. 110
D. 150
E. 170
Attachment:
2018-03-16_1010ed.png [ 10.8 KiB | Viewed 1021 times ]
A couple of other ways to approach . . .
$$l_1$$ and $$l_2$$ are cut by a transversal
Alternate interior angles are congruent.
Lower right angle = $$40°$$
Other angle of lower triangle, given = $$30°$$
1) $$x$$ = exterior angle
Exterior angle $$x$$ = sum of opposite interior angles of lower triangle
Exterior angle $$x$$ $$= (30° + 40°) =$$ $$70°$$
OR
2) Straight line: Third angle of lower ∆ + $$x = 180°$$
Interior angles of a triangle sum to 180°
Lower ∆'s third angle = $$110°$$
($$(180° - 30° - 40°) = 110°$$ = third angle))
Third angle lies on a straight line with $$x$$
$$110° + x° = 180°$$
$$x = 70°$$
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Re: In the figure above if l1||l2, what is the value of x? [#permalink]
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18 Mar 2018, 10:40
Bunuel wrote:
In the figure above if $$l_1||l_2$$, what is the value of x?
A. 70
B. 100
C. 110
D. 150
E. 170
Attachment:
The attachment 2018-03-16_1010.png is no longer available
Drawing a additional parallel line l1 and l2 and using property of equal alternate angle between 2 parallel line
we get x= 30 +40
x=70
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Re: In the figure above if l1||l2, what is the value of x? [#permalink]
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20 Oct 2019, 10:06
Bunuel wrote:
In the figure above if $$l_1||l_2$$, what is the value of x?
A. 70
B. 100
C. 110
D. 150
E. 170
Attachment:
2018-03-16_1010.png
Draw a line through the vertex of angle x that is parallel to both l1 and l2. We will see that angle x is the sum of the two given angles of 40 degrees and 30 degrees. Thus angle x = 70 degrees.
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Re: In the figure above if l1||l2, what is the value of x? [#permalink]
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20 Oct 2019, 10:11
Bunuel wrote:
In the figure above if $$l_1||l_2$$, what is the value of x?
A. 70
B. 100
C. 110
D. 150
E. 170
Attachment:
2018-03-16_1010.png
x = 180 - (180 - 40 - 30) = 70
IMO A
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Re: In the figure above if l1||l2, what is the value of x? [#permalink] 20 Oct 2019, 10:11
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# In the xy-plane, the straight-line graphs of the three equations above
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y = ax - 5
y = x + 6
y = 3x + b
In the xy-plane, the straight-line graphs of the three equations above each contain the point (p,r). If a and b are constants, what is the value of b?
(1) a = 2
(2) r = 17
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Originally posted by Bunuel on 23 Jul 2015, 10:47.
Last edited by Bunuel on 13 Nov 2017, 21:24, edited 4 times in total.
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In the xy-plane, the straight-line graphs of the three equations above [#permalink]
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Re: In the xy-plane, the straight-line graphs of the three equations above [#permalink]
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mcelroytutoring wrote:
DS 136 from OFG 2016 (new question)
y = ax - 5
y = x + 6
y = 3x + b
In the xy-plane, the straight-line graphs of the three equations above each contain the point (p,r). If a and b are constants, what is the value of b?
1) a = 2
2) r = 17
Solution provided by : mcelroytutoring
Let's start by substituting the point (p,r) into all equations in place of (x,y) which will make step #2 a bit easier to comprehend but is not necessary to solve the question. Then, let's consider the number of variables left in each equation.
#1: r = ap - 5 (3 variables R,A,P)
#2: r = p + 6 (2 variables R,P)
#3: r = 3p + b (3 variables R,B,P)
1) a = 2 allows us to reduce equation #1 to the variables r and p, which are the same two variables as equation #2. Thus we have simultaneous equations. As soon as we verify that the equations are different, we know that we can solve for both variables. Once we know r and p, we can substitute in equation #3 to solve for b. Sufficient.
2) r = 17 allows us to do the same thing, more or less. It reduces equation #2 to only one variable, allowing us to solve for p. Once we have p (and r), we can use equation #3 to solve for b. Sufficient.
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Re: In the xy-plane, the straight-line graphs of the three equations above [#permalink]
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24 Sep 2015, 01:53
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if each of the 3 equations contains points (p,r) this means that they intersect in that point
1. a=2
Find the intercept Intercept for three simultaneous equations
y=2x-5
y=x+6
y=3x+b
Let's use the first 2 equations: plug y=x+6 in the secod equation
x+6=2x-5 -> x=11, y=17 we can use the values to calulate b in the 3rd equation
17=33+b -> b=-16 SUFFICIENT
2. Here we have directly the value for Y, let's plug it in the 2nd equation
y=x+6 -> 17=x+6 -> x=11, y=17; We can plug these values in the 3rd equation and find b as we did above
17=33+b -> b=-16 SUFFICIENT
Answer (D) Most important point is here to catch the hint about intersection of 3 lines at one point
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Re: In the xy-plane, the straight-line graphs of the three equations above [#permalink]
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26 Oct 2015, 10:50
4
I think it's D.
Keeping point (p,r) in all the equations we get :
p = ar -5 -----(1)
p = r + 6 ------(2)
p = 3r + b ------(3)
Now consider (1) if a = 2 from (1) and (2) we get
r = 11 , p=17 and putting in (3) we can get b.
Similarly for (2) we can get the values for r and p and hence can get the value for b.
So both statements individually are correct to answer the question.
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Re: In the xy-plane, the straight-line graphs of the three equations above [#permalink]
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26 Oct 2015, 19:14
2
m2k wrote:
I think it's D.
Keeping point (p,r) in all the equations we get :
p = ar -5 -----(1)
p = r + 6 ------(2)
p = 3r + b ------(3)
Now consider (1) if a = 2 from (1) and (2) we get
r = 11 , p=17 and putting in (3) we can get b.
Similarly for (2) we can get the values for r and p and hence can get the value for b.
So both statements individually are correct to answer the question.
Approach if right but the values you derived are wrong. According to me r = 17 and that is what stmt b also states.
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Re: In the xy-plane, the straight-line graphs of the three equations above [#permalink]
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10 Jun 2016, 11:23
3
y = ax - 5 ... eq 1
y = x + 6 ... eq 2
y = 3x + b ...eq 3
Total of 4 variables are present.
Statement 1 : a = 2
Insert in eq 1
We have y = 2x -5 and y = x+6
Solving we get x = 11 and y = 17
Substitute in eq 3 and we get value of b
Statement 2: r=17
This means the y co- ordinate is 17
Substitute in eq 2 we get x as 11
Again can find value of b from equation 3
Hence D
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Re: In the xy-plane, the straight-line graphs of the three equations above [#permalink]
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18 Jan 2017, 15:18
we used information from both 1 and 2 then how can the answer be D... should it not be C... some one kindly clarify.......... clearly am a zero in ds and that too a big one
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Re: In the xy-plane, the straight-line graphs of the three equations above [#permalink]
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18 Jan 2017, 20:18
1
y = ax - 5---------1
y = x + 6---------2
y = 3x + b--------3
In the xy-plane, the straight-line graphs of the three equations above each contain the point (p,r). If a and b are constants, what is the value of b?
(1) a = 2
(2) r = 17
All three line intersect each other at common point (p,r).
1. given a = 2
putting in equation 1 .= y=2x-5
equating 1(after replacing value of a) and 2 we will get value of (p,r)
putting (p,r) in equation 3 we will get value for b---suff..
2 given r = 17.
putting in equation 2 we will get value of x.i'e p.
Now as we know common point of intersection ,putting p,r in equation 3 , we will get value of b
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Re: In the xy-plane, the straight-line graphs of the three equations above [#permalink]
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22 Jan 2017, 15:51
thanks sobby for your response... highly appreciated...
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Re: In the xy-plane, the straight-line graphs of the three equations above [#permalink]
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16 Mar 2017, 08:58
1
(eq1) $$y = ax - 5$$
(eq2) $$y = x + 6$$
(eq3) $$y = 3x + b$$
In the xy-plane, the straight-line graphs of the three equations above each contain the point (p, r). If a and b are constants, what is the value of b?
1) $$a = 2$$
2) $$r = 17$$
Solution:
1) $$a = 2$$
- Putting the value of a in eq1, we get: $$y = 2x - 5$$
- At this point you can solve for (x, y), plug (x, y) in (eq3) and solve for (b) [though this approach might take few seconds]
(alternatively, faster method)
- you can skip solving for (x, y) and deduce that given 3 equations and 3 unknowns (since a is given in statement 1) we can solve for all of them (including b), since the lines are have different slopes i.e. different lines. Hence, we can get single value of 'b', proving the condition SUFFICIENT.
NOTE: 3 equations and 3 unknowns does not ALWAYS mean that we can find 3 unknown. We have to make sure that 2 of them or all of them are not the same line.
2) $$r = 17$$
- Since, point (p, r) lie on all the line, we can plugin the point in above equation
$$r = ap - 5 => 17 = ap - 5$$
$$r = p + 6 => 17 = p + 6$$
$$r = 3p + b => 17 = 3p + b$$
- Again, we do not need to solve for all the variables and just recognize that the above equations will lead to single value of b. Hence, SUFFICIENT.
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Re: In the xy-plane, the straight-line graphs of the three equations above [#permalink]
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30 Apr 2017, 23:51
y = ax - 5
y = x + 6
y = 3x + b
In the xy-plane, the straight-line graphs of the three equations above each contain the point (p,r). If a and b are constants, what is the value of b?
1) a = 2
2) r = 17
My 2 cents.
It is important to realize from the Question stem that the 3 equations intersect as (p,r).
For 1), as we know a =2, we can equate the first and second equation to get the value of x, and then use that value of x to find value of y and the find value of b.
For 2), similarly, use r = 17 (which is value of y) to find value of x using the second equation. And then plug it back to the third equation.
So D.
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Re: In the xy-plane, the straight-line graphs of the three equations above [#permalink]
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10 Sep 2017, 19:04
1
Would it be correct to simply say that we have 4 variables with 3 equations so eliminating any one variable gets us to three equations and three variables and is therefore sufficient? Is that logic sound?
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Re: In the xy-plane, the straight-line graphs of the three equations above [#permalink]
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13 Nov 2017, 20:21
2
what is the significance of the the line "In the xy-plane, the straight-line graphs of the three equations above each contain the point (p,r)",
i solved the problem, but wud have did the same even if they didnt provide line above . as question has 3 equations with 4 variables
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Re: In the xy-plane, the straight-line graphs of the three equations above [#permalink]
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13 Nov 2017, 22:31
2
Cheryn wrote:
what is the significance of the the line "In the xy-plane, the straight-line graphs of the three equations above each contain the point (p,r)",
i solved the problem, but wud have did the same even if they didnt provide line above . as question has 3 equations with 4 variables
The highlighted statement in effect says that all these 3 lines meet each other at one point and so there is a single value of (x,y) that satisfies these 3 equations. It is only because of this highlighted statement you can solve this set of equations for a unique value of x,y, a and b.
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Re: In the xy-plane, the straight-line graphs of the three equations above [#permalink]
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11 Dec 2017, 14:28
3
1
Hi All,
We're given the equations for 3 lines (and those equations are based on 4 unknowns: 2 variables and the 2 'constants' A and B):
Y = (A)(X) - 5
Y = X + 6
Y = 3X + B
We're told that the three lines all cross at one point on a graph (p,r). We're asked for the value of B. While this question looks complex, it's actually built around a 'system' math "shortcut" - meaning that since we have 3 unique equations and 4 unknowns, we just need one more unique equation (with one or more of those unknowns) and we can solve for ALL of the unknowns:
1) A =2
With this information, we now have a 4th equation, so we CAN solve for B.
Fact 1 is SUFFICIENT
2) R = 17
This information tell us the x co-ordinate where all three lines will meet, so it's the equivalent of having X=17 to work with. This 4th equation also allows us to solve for B.
Fact 2 is SUFFICIENT
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Re: In the xy-plane, the straight-line graphs of the three equations above [#permalink]
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02 Jan 2018, 11:00
1
1
Bunuel wrote:
y = ax - 5
y = x + 6
y = 3x + b
In the xy-plane, the straight-line graphs of the three equations above each contain the point (p,r). If a and b are constants, what is the value of b?
(1) a = 2
(2) r = 17
We can begin by substituting p and r for x and y, respectively, in the three given equations.
1) r = ap – 5
2) r = p + 6
3) r = 3p + b
Statement One Alone:
a = 2
We can substitute 2 for a in the equation r = ap – 5. Thus, we have:
r = 2p – 5
Next we can set equations 1 and 2 equal to each other.
2p – 5 = p + 6
p = 11
Since p = 11, we see that r = 11 + 6 = 17
Finally, we can substitute 11 for p and 17 for r in equation 3. This gives us:
17 = 3(11) + b
17 = 33 + b
-16 = b
Statement one alone is sufficient to answer the question.
Statement Two Alone:
r = 17
We can substitute r into all three equations and we have:
1) 17 = ap – 5
2) 17 = p + 6
3) 17 = 3p + b
We see that p = 11. Now we can substitute 11 for p in equation 3 to determine a value for b.
17 = 3(11) + b
-16 = b
Statement two alone is also sufficient to answer the question.
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Re: In the xy-plane, the straight-line graphs of the three equations above [#permalink]
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20 Jul 2019, 23:56
Video solution for the same
https://gmatquantum.com/official-guides ... ial-guide/
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Re: In the xy-plane, the straight-line graphs of the three equations above [#permalink]
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17 Aug 2019, 23:34
Bunuel wrote:
y = ax - 5
y = x + 6
y = 3x + b
In the xy-plane, the straight-line graphs of the three equations above each contain the point (p,r). If a and b are constants, what is the value of b?
(1) a = 2
(2) r = 17
DS07713
Re: In the xy-plane, the straight-line graphs of the three equations above [#permalink] 17 Aug 2019, 23:34
Display posts from previous: Sort by | 2020-01-20T23:46:42 | {
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https://math.stackexchange.com/questions/1783555/resolving-secx-left-sin3x-sin-x-cos2x-right-tanx | # Resolving $\sec{x}\left(\sin^3x + \sin x \cos^2x\right)=\tan{x}$
Going steadily through my book, I found this exercise to resolve
$$\sec{x}\left(\sin^3x + \sin x \cos^2x\right)=\tan{x}$$
Here's how I resolve it ($LHS$) and again bear with me as I truly reverting to a feeling of vulnerability, like a child actually
As $\sec x$ is equal to $\frac{1}{\cos x}$
$$\frac{(\sin^3x+\sin x\cos^2x)}{\cos x}$$
I'm factorizing one $\sin x$
$$\frac{\sin x(\sin^2x+\cos^2x)}{\cos x} = \frac{\sin x(1)}{\cos x} = \tan x$$
That seems to work otherwise I completly messed this up
Reading the book's solution, I have something different...
\begin{align*} LHS&=\frac{\sin^3x}{\cos x}+ \sin x \cos x \\[4pt] &=\frac{\sin x}{\cos x}-\frac{\sin x\cos^2x}{\cos x}+\sin x\cos x\\[4pt] &= \tan x\end{align*}
What did I miss?
• They messed this up. – Yves Daoust May 13 '16 at 10:00
• Shorter: multiplying both members by $\cot(x)$ yields $\sin^2(x)+\cos^2(x)=1$. – Yves Daoust May 13 '16 at 10:04
• Hi @YvesDaoust you see this is where dogma starts. My way is better than your way and at the end, you are disgusting a generation of people but for what purpose...? – Andy K May 13 '16 at 10:15
• There is absolutely no intent to disgust anyone. You seem to miss that my first remark was an compliment to you. There is beauty in making things short. – Yves Daoust May 13 '16 at 10:22
• @YvesDaoust I understood correctly what you said. I was more empathetic about these 2 chaps who went up to create a complicate solution for what it was. They could have done something simplier and also mentionned that, there were other ways (pluralist views) to solve that problem. As we said J'avais bien compris :) – Andy K May 13 '16 at 10:24
So the book does about the same thing as you but in a different order (your solution is hence correct too). By the trigonometric one we get $\sin^2 x= (1-\cos^2 x)$. Thus $$\frac{\sin^3 x}{\cos x}=\frac{\sin x (\sin^2 x)}{\cos x}= \frac{\sin x(1-\cos^2 x)}{\cos x} = \frac{\sin x-\sin x\cos^2 x}{\cos x}= \frac{\sin x}{\cos x} -\frac{\sin x\cos ^2 x}{\cos x}$$ Thus $$LHS =\frac{\sin^3 x}{\cos x}+\sin x\cos x= \frac{\sin x}{\cos x} -\frac{\sin x\cos ^2 x}{\cos x} +\sin x\cos x = \frac{\sin x}{\cos x}=\tan x$$ | 2019-06-24T17:41:22 | {
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https://math.stackexchange.com/questions/2943615/given-s-n-sum-dots-and-a-n-sum-dots-prove-that-a-n-s-n-1-over | # Given $S_n = \sum \dots$ and $a_n = \sum \dots$ prove that $a_n = S_n + {1\over n\cdot n!}$
I'm trying to solve the following problem:
Let: \begin{align} S_n &= 2 + {1\over2!} + {1\over3!} + {1\over 4!} + \dots + {1\over n!} \\ a_n &= 3 - {1\over 1\cdot2\cdot2!} - {1\over 2\cdot3\cdot3!} - \dots - {1\over (n-1)\cdot n\cdot n!} \end{align} Prove that: $$a_n = S_n + {1\over n\cdot n!}$$
My thought on this are:
Manipulating the sums here would become pretty hard so I decided to use another approach. I've tried to look what the first terms of each sum are going to be in order to observe a pattern:
\begin{align} S_1 &= 2 \\ S_2 &= 2 + {1\over 2!} = S_1 + {1\over 2!} \\ S_3 &= 2 + {1\over 2!} + {1\over 3!} = S_2 + {1\over 3!}\\ &\vdots \\ S_{n+1} &= S_n + {1\over (n+1)!} \end{align}
A similar thing is done to $$a_n$$:
$$a_{n+1} = a_n - {1\over n\cdot (n+1) \cdot (n+1)!}$$
So potentially if we could find closed forms of both $$S_n$$ and $$a_n$$ it would become easier to reason about them.
By the way, I've observed both of the sums approach $$e$$ at infinity but from different sides. $$S_n$$ is starting from $$2$$ adding more terms as $$n$$ grows, while $$a_n$$ starts from 3 and decreases the sum as $$n$$ grows. So that will be our initial conditions for both recurrences:
$$S_1 = 2 \\ a_1 = 3$$
The problem just reduced to finding closed forms of the recurrences. Both of them are non-homogenous and here is where I got stuck. Solving for homogenous part is easy since roots of both characteristic equations $$\lambda_{a_n} = \lambda_{S_n} = 1$$.
I couldn't find a particular solution for them. I've tried yet another approach expressing $$S_{n+1}$$ and $$S_{n+2}$$ trying to get rid of the $$1\over (n+1)!$$ or at least change its form so the guess for particular solution is more obvious.
My questions are:
1. Is it even a valid approach to solve the problem?
2. If so then how could I find a closed form for the recurrences? Especially i'm interested in finding a particular solution. (Yet a complete flow is very appreciated and encourages learning by example)
3. Not sure how to phrase that better, but does there exist a table of "guesses" for particular solution of the recurrences in some form of $$x_{n+1} = x_n + F(n)$$. Like if the "free term" is a constant say $$F(n) = 2$$ then the guess for particular solution is some constant $$B$$.
Please excuse me if there is some vagueness or inaccuracy in the terminology, English is not my native language and I have almost no background in Maths.
• It seems unlikely to find a simple closed form, because, as you noticed, the limit converges to $e$. – user600464 Oct 5 '18 at 18:59
• @user600464 is that because $e$ is irrational and that is why no "simple" formula for an irrational number may be obtained? – roman Oct 5 '18 at 19:02
• There is actually a simple formula, the limit $\frac{n^n}{(n- 1)^n}$. However, to compare to this sums, we would need many terms – user600464 Oct 5 '18 at 19:34
$$a_2 - S_2 = 3- 1/4 - (1 + 1/1! + 1/2!)$$ [I think there is some typo in your expression of $$S_n$$] $$= 1/4 = 1/(2 \cdot 2!)]$$. Suppose the equation holds for $$n$$. Then for $$n+1$$, \begin{align*} a_{n+1} - S_{n+1} &= a_n - S_n - \frac 1 {n(n+1) \cdot (n+1)!} - \frac 1{(n+1)!} \\ &= \frac 1 {n!n }-\frac 1 {n(n+1) \cdot (n+1)!} - \frac 1{(n+1)!} \\ &= \frac 1 {(n+1)! (n+1)} \left( \frac {(n+1)^2}n -\frac 1n -(n+1) \right)\\ &= \frac 1 {(n+1)! (n+1)}. \end{align*} Thus the equation holds for all $$n \geqslant 2$$ by induction principle.
\begin{align} \color{red}{a_{n}-S_{n}} &=\left(3-\sum_{k=2}^{n}\frac{1}{(n-1)\cdot n\cdot n!}\right)-\left(2+\sum_{k=2}^{n}\frac{1}{n!}\right) \\[2mm] &=1-\sum_{k=2}^{n}\frac{1}{n!}\left(1+\frac{1}{(n-1)\cdot n}\right) =1-\sum_{k=2}^{n}\frac{1}{n!}\left(1+\frac{1}{n-1}-\frac{1}{n}\right) \\[2mm] &=1-\sum_{k=2}^{n}\frac{1}{n!}\left(\frac{n}{n-1}-\frac{1}{n}\right) =1-\sum_{k=2}^{n}\left(\frac{1}{(n-1)\cdot (n-1)!}-\frac{1}{n\cdot n!}\right) \\[2mm] &=1-\frac{1}{1\cdot1!}+\frac{1}{2\cdot2!}-\frac{1}{2\cdot2!}+\frac{1}{3\cdot3!}-\cdots+\frac{1}{n\cdot n!}=\color{red}{\frac{1}{n\cdot n!}}\quad\left\{\text{telescoping}\right\} \end{align} | 2019-08-25T00:09:08 | {
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https://math.stackexchange.com/questions/3061903/in-proof-writing-is-it-mathematically-sound-to-prove-uniqueness-before-proving | # In proof writing, is it mathematically sound to prove uniqueness before proving existence?
As stated in the title, I'd like to find out is whether or not it is always mathematically sound to prove the uniqueness of something before proving the existence of said something.
I am still relatively new to proofs and, in this light, I was intrigued by an observation that some authors choose to prove uniqueness before existence. I understand that some do this because, allegedly, "it is easier to prove the uniqueness part first". And in an exam setting, it's better to tackle the easier parts of a task.
However, as I see it, if done this way, aren't we essentially proving a proposition about something that, at that point, we have not yet guaranteed exists? (Thus, might not even exist at all). I understand that in most cases the existence is proved immediately after the proof of uniqueness but, generally, doesn't uniqueness depend on existence?
My example will be derived from page $$40$$ of The Real Analysis Lifesaver: All the Tools You Need to Understand Proofs by Raffi Grinberg:
Theorem 5.8. (Existence of Roots in $$\mathbb{R}$$)
Every positive real number has a unique positive $$n$$th root, for any $$n \in \mathbb{N}$$.
In symbols:
$$∀x ∈ R$$ with $$x > 0, ∀n ∈ N, ∃y ∈ R$$ unique, such that $$y > 0$$ and $$y^n = x.$$
Note that even-numbered roots $$(\sqrt{x},\sqrt[4]x,\sqrt[6]x, etc.)$$ signify two numbers in $$\mathbb{R}$$, namely, $$+y$$ and $$−y$$. The theorem asserts that there is one and only one positive real root.
Proof.
The uniqueness of $$y$$ is the easiest part of the proof, so we’ll start there. For any two different positive real numbers, the fact that they are different means one must be greater than the other. If there were two positive real roots $$y_1$$ and $$y_2$$ such that $$y_1^n = x$$ and $$y_2^n = x$$ we would have $$0 < y_1 < y_2$$. But then $$0 < y_1^n < y_n^2$$ , meaning $$0 < x < x$$, which is impossible. Thus only one positive real root can exist.
To prove that $$\sqrt[n]x$$ exists in $$\mathbb{R}$$, let’s first figure out our game plan and then write it up formally $$...$$
While the proof is totally clear, I'd really like feedback on this format. As an aspiring mathematician, is this a format I can adapt and use in my proofs or perhaps it is not good practice in general? What if, say, I have a conjecture that there exist unique integers possessing some property and this conjecture is something new (i.e. hasn't been proven yet). I try proving they exist but run out of luck. Would it be worth the effort of trying to prove the uniqueness of such integers not knowing they exist?
• Yeah, that's perfectly fine. It doesn't matter: you're proving two things: 1) There is at least one of these; and 2) there aren't two different ones. There's no problem doing them whatever way round you like. There are a fair few results of the form "there are at most $n$ things with property X", and this is just the special case where $n = 1$. – user3482749 Jan 4 at 17:57
• Sure. Naturally, if you show uniqueness first what you have actually shown at that point is that there is at most one object with the desired property. – Andrés E. Caicedo Jan 4 at 17:57
• Even in the case where authors first prove uniqueness, they usually mean, under the assumption that there exists such an object, it is unique. – Dietrich Burde Jan 4 at 18:01
• I prefer existence before uniqueness, but that is just a preference and not a strong recommendation. – David G. Stork Jan 4 at 18:02
• An explanation: I prefer existence before uniqueness because in general existence is simpler. Imagine the awkward case where you prove uniqueness first, then find you cannot prove existence! THAT is weird. – David G. Stork Jan 4 at 18:16
I can see the unease: if what you're proving unique doesn't exist, might that invalidate whatever manipulations you've done with it?
This is similar to the situation with proof by contradiction, where you don't just reason about something that might not be true: you reason about it expecting it to be untrue, but making all the logical steps anyway.
In either case, you end up showing "If $$A$$ then $$B$$"—whether it's "If $$x$$ exists then it's unique" or "If $$x$$ exists then it has contradictory properties". In one case we go on to apply $$B$$ to $$x$$ once we know it exists, and in the other we use the impossibility of $$B$$ to disprove the existence of $$x$$.
A separate thought: proofs are presented as a sequence of steps and inferences because as humans, we process written information sequentially. But what's behind a mathematical theorem is more like a network of theorems and logical steps, some dependent on others, which together imply the theorem. We can't take them all in simultaneously, so we arrange them in a suitable order to let us see the logic most clearly. It's a matter of convenience.
If $$X$$ and $$Y$$ are both true and together imply $$Z$$, it doesn't really matter which order the various parts of that are proved in—all that matters from a validity point of view is that they're all proved and brought together.
• Thanks! Your first paragraph alone asks my question even better than I could in my long post! – E.Nole Jan 7 at 2:56
This is a perfectly reasonable argument. The author even tells you he's doing it this way. He then uses the argument to allow the reader to become comfortable with the assertion and the notation.
There are other reasons for starting with uniqueness. Often if you know there's just one of something then it might be easier to search for it.
When faced with a problem with an easy part and a hard part the argument for starting with the easy part is that it's a good warmup. That's what I usually do, and what I like to see in things I read.
The argument for starting with the hard part is that if you fail there you haven't wasted time on the easy part. For example, proving that there's at most one positive rational number whose square is $$2$$ is not going to tell you very much about the rational square root of $$2$$ that does not exist.
Starting with the hard part, the bottleneck, is often the preferred strategy when writing software.
PS The author is wrong to say that when $$x >0$$, $$\sqrt{x}$$ signifies two two numbers. $$4$$ does have two real square roots, but the mathematical convention is that only $$2 = \sqrt{4}$$.
• Nice typo near the end: "two two" (Shame to fix it really!) – timtfj Jan 4 at 19:15
• @timtfj I will leave it thanks. – Ethan Bolker Jan 4 at 19:33 | 2019-10-23T20:18:42 | {
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https://math.stackexchange.com/questions/2499234/ideal-generated-by-a-finite-number-of-polynomials | # Ideal Generated by a Finite Number of Polynomials?
Background
I have been confused about a particular definition in the textbook for my abstract algebra class, Ideals, Varieties, and Algorithms by Cox, Little, and O'Shea. It is frustrating because I feel that I have a partial grasp on what the definition is trying to say, but when it comes down to it I simply find my confused. So, instead of suffering by myself for any longer with this definition, I have decided to turn to you lovely folks to help me understand this seemingly simple concept.
What I Understand
First off, I understand (at least in the context of this book) what an ideal is. The definition my book gives for an ideal is
Definition. A subset $$I\subseteq k[x_1,..., x_n]$$ is an ideal if it satisfies:
(i) $$0\in I$$.
(ii) If $$f,g\in I$$, then $$f+g\in I$$.
(iii) If $$f\in I$$ and $$h\in k[x_1,...,x_n]$$, then $$hf\in I$$.
I find this to be a simple, easy to understand definition. My problem, however, arises a few lines later when they define an ideal generated by a finite number of polynomials.
What I Don't Understand
And now, I give you the definition that has been causing me an incredible amount of confusion and frustration.
Definition. Let $$f_1,...,f_s$$ be polynomials in $$k[x_1,...,x_n]$$. Then we set $$\langle f_1,...,f_s \rangle=\Big\lbrace \sum_{i=1}^s h_if_i \ | \ h_1,...,h_s\in k[x_1,...,x_n] \Big\rbrace.$$
I know what you are thinking. How does he not understand this? I wish I knew the answer to that question, but in the meantime, can someone please help me visualize what this set looks like? I understand that $$\langle f_1,...,f_s \rangle$$ is an ideal, but I don't understand its structure, if that makes sense. In other words, I can't visualize this definition in a way that makes sense to me. The authors do make a slightly helpful note, saying that "we can think of $$\langle f_1,...,f_s \rangle$$ as consisting of all 'polynomial consequences' of the equations $$f_1=f_2=...=f_s=0$$."
To elaborate a little more on my confusion, what I'm asking for is a less "compact" definition. When I read this definition, for whatever reason the only thing I can come up with is $$f_1h_1+f_2h_2+...+f_sh_s.$$ But that doesn't make sense, because $$\langle f_1,...,f_s \rangle$$ is supposed to generate a set, not just a single polynomial.
As always, thank you all for your time. If you find this to be a stupid or silly question, then I'm sorry to have disappointed you -- I'm a slow learner, and I get hung up on stupid things sometimes.
Oh, and Happy Halloween!
• Suggestion: Try looking at one or two specific examples, such as: (1) In $k[x,y]$, look at $\langle x^2, y^2 \rangle$. This is equal to the set $\{ x^2 h_1 + y^2 h_2 \mid h_1, h_2 \in k[x,y]\}$. Some elements include $x^2 \cdot 1 + y^2 \cdot 0$, $x^2 \cdot 0 + y^2 \cdot 1$,... $x^2 \cdot (x y^2 - 3 xy) + y^2 \cdot (2x^3 - 5)$, etc. (2) Look at examples in $\mathbb{Z}$, e.g., $\langle 6,4 \rangle = \{ 6 h_1 + 4 h_2 \mid h_1, h_2 \in \mathbb{Z} \}$. Can you prove this is equal to the set of even numbers? – Zach Teitler Nov 1 '17 at 4:55
• Makes alot more sense when you put it like that! Thank you @ZachTeitler! – Thy Art is Math Nov 1 '17 at 5:00
• One more suggestion. Make sure you do Exercise 1.4.2, or at least try to do it. The result in Exercise 1.4.2 gets used constantly, over and over, throughout the rest of the book, including in a lot of exercises. It's the main way to show that two ideals are equal, or that one ideal is contained in another. So it is extremely useful. Also it's good practice with the definition of ideal, and proving a subset relationship. (There are a lot of other good exercises, I just want to highlight that particular one. I made sure to assign it in the class I'm teaching from IVA this semester.) – Zach Teitler Nov 1 '17 at 5:49
• Read the Rings chapter in Algebra by Hungerford. He describes ideals in full generality very well (non-commutative and commutative). – user494247 Nov 11 '17 at 9:57
Would it help to see the see the set with explicit polynomials in place of the $f_1, f_2, \dots, f_n$?
For example, lets look at an explicit example when $n = 2$, so an ideal generated by 2 polynomials. Also, we'll work in a polynomial ring in two variables over $k$, i.e. $k[x,y]$.
Here is the ideal generated by the polynomials $x^2 - 1$, $yx+x$.$$\langle \, x^2 - 1, \, yx+x \, \rangle = \{ \, f\cdot(x^2 - 1) + g\cdot(yx + x) \, | \, f,g \in k[x,y] \, \}.$$
So the elements of the set are any polynomial that can be written in the form $f \cdot (x^2 - 1) + g \cdot (yx+x)$. But we can choose $f,g$ to be $\textit{any}$ polynomial we want. For example, we know $x^2 - 1$ itself is in that set because we can choose $f = 1, g = 0$. We also know that the polynomial $x^3-x+yx ^2 + x^2$ is in the set, because we can choose $f = x, g = x$.
If you are familiar with linear algebra you can maybe, in a way, draw a connection between an ideal generated by polynomials and the span of a set of vectors. You can think of it as, the set of all things that can be made from the objects defining it.
• Also, is your class an Algebraic Geometry class or an Abstract Algebra class? I think IVA is a strange book to teach general algebra out of. – Prince M Nov 1 '17 at 7:38
• Sadly I was absolutely terrible at linear algebra but seeing the explicit polynomials made this make so much more sense, so thank you for the help. The class is labeled rather generally as "abstract algebra," so you are probably right that the course should be called "algebraic geometry". – Thy Art is Math Nov 1 '17 at 15:30
• Just a note that the authors of IVA do write in the preface: "For instance, the book could serve as a basis of a second course in undergraduate abstract algebra, but we think that it just as easily could provide a credible alternative to the first course." – J W Jan 3 at 15:16
Perhaps a good way to understand this (or to understand anything, for that matter) is to look at examples.
The definition of ideal makes sense in any ring, so let’s look at $\Bbb Z$ first. Every ideal, you soon persuade yourself, is a set $d\Bbb Z\subset\Bbb Z$, that is, just the set of all multiples of a given number $d$. What if you try to take two numbers and look at $\langle d_1,d_2\rangle\subset\Bbb Z$? Please do this with specific numbers. Like what about $d_1=8$ and $d_2=6$? You want to describe the totality of all numbers writable in the form $8m+6n$. You rapidly see that this is an ideal in the sense of the definition, and you see that the set is equal to $2\Bbb Z$. This is basic number theory.
Now do the same thing with a ring of polynomials, but in just one variable, $R=\Bbb Q[x]$. One proves (you prove it, with Euclidean division of polynomials) that every ideal of $R$ is of form $fR$, where $f$ is a well-chosen polynomial. In fact, for a nonzero ideal $I$, you take $f$ to be a nonzero element of $I$ of least degree. So, what is $\langle f_1,f_2\rangle$, when $f_1$ and $f_2$ are two polynomials in $x$ that are given? Just as with numbers, it’s the ideal $gR$ where $g$ is the greatest common divisor of $f$ and $g$. Convince yourself that $\langle x^3-1,x^2-2x+1\rangle$ is the set of all multiples of $x-1$.
Things are no longer so simple when you have polynomials in more than one indeterminate. But at least you get some insight by looking at the simplest cases, and I hope you see that the set of all possible $h_1f_1+\cdots+h_nf_n$ is an ideal.
• @ThyArtisMath Note that the ring of polynomials in one variable is treated very explicitly and thoroughly in section 1.5 of the textbook Ideals, Varieties, and Algorithms. So, by all means feel free to think about that example and take a shot at proving some things about it, but don't feel like you have to work it out on your own. The textbook goes over all that. – Zach Teitler Nov 1 '17 at 5:40
You were close.
The ideal $(f_1,...,f_s)$ of $k[x_1,..., x_n]$ is the set of all polynomials which can be expressed as $h_1f_1+\cdots +h_sf_s$, for some $h_1,...,h_s \in k[x_1,..., x_n]$.
In other words, "linear combinations" of $f_1,...,f_s$ where the "linear coefficients" are arbitrary elements $h_1,...,h_s$ of $k[x_1,..., x_n]$.
Regarding the author's point about common zeros . . .
If $\bar{k}$ is an algebraically closed extension of $k$, and if there is some $a = (a_1,...,a_n) \in \left(\bar{k}\right)^n$ such that $f_1(a) = \cdots = f_s(a) = 0$, then $a$ is automatically a zero of any polynomial of the form $h_1f_1+\cdots +h_sf_s$, hence $a$ is a common zero for all members of the ideal $(f_1,...,f_s)$.
In particular, if you've identified a common zero $a \in \left(\bar{k}\right)^n$ for $f_1,...,f_s$, then if $g \in k[x_1,..., x_n]$ is such that $g(a) \ne 0$, it follows that $g$ is not in the ideal $(f_1,...,f_s)$.
An important, nontrivial result is a partial converse: If $g \in k[x_1,...,x_n]$ is such that $g,g^2,g^3,...$ are not in the ideal $(f_1,...,f_s)$, there is some $a \in \left(\bar{k}\right)^n$ such that $a$ is a common zero of $f_1,...,f_n$, but $a$ is not a zero of $g$. | 2021-06-25T05:13:58 | {
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https://math.stackexchange.com/questions/800657/solving-this-inequality | # Solving this inequality
Question:
Solve: $$\frac{5x-6}{x+6}<1$$
My attempt:
$$\frac{5x-6-x-6}{x+6}<0$$ $$\Rightarrow \frac{4x-12}{x+6}<0$$ $$\Rightarrow \frac{x-3}{x+6}<0$$ $$\Rightarrow (x-3)(x+6) < 0$$
Thus, it implies that the answer is $-6 < x < 3$. However, in my textbook, it is given as wrong answer.
Can please someone tell what is the correct procedure for this.
Thanks.
If you're really unsure when something like that happens, you can always graph your rational function. But consider that in $\ \frac{x-3}{x+6} \$ , there is a vertical asymptote at $\ x \ = \ -6 \$ and an $\ x-$ intercept at $\ x = 3 \$ . So the real numbers are divided into intervals by these points, $\ x \ < \ -6 \ , \ -6 \ < \ x \ < \ 3 \ ,$ and $\ x \ > \ 3 \$ . Also, your rational function has a horizontal asymptote of $\ y \ = \ 1 \$ .
So the function is positive for "a lot" of the real numbers to start with. We can also use what we know about working with signed numbers. For large negative numbers, both the numerator and denominator are negative, and for large positive numbers, they are both positive. So for $\ x \ < \ -6 \$ and $\ x \ > \ 3 \$ , we can conclude that the ratio is positive. It will only be in the interval $\ -6 \ < \ x \ < \ 3 \$ that the numerator is negative, while the denominator is positive, so the ratio is negative in this interval.
So you are correct. Large introductory course textbook answers can be wrong anywhere from about $\ \frac{1}{3}$ % to $\ 2$ % of the time, depending upon how carefully the submitted answers (of multiple authorship) have been combed through...
• Thanks for the detailed method! And yes, you are right, the textbook is indeed large, so the author might have missed it. – Gaurang Tandon May 18 '14 at 19:47
• The authors (probably of just about all of the omnibus texts, and not just in math) generally hire teaching assistants to work out answers, which are then collected into the backs of these books. They do not check what can be two to three thousand problem answers for accuracy... – colormegone May 18 '14 at 19:54
Hint: There are two cases, either $x < -6$ or $x > -6$. You can multiply both sides of the inequality by $x + 6$ and depending on which case you are considering, either it will reverse the inequality or leave it intact. Then you can rewrite the inequality with $x$ on one side and a number on the other side by adding/subtracting/dividing.
• I am sorry but I do not understand you. The denominator has $(x+6)$, not $x-6$. – Gaurang Tandon May 18 '14 at 18:40
• @GaurangTandon Sorry, I should have written $-6$. The point is that there are two cases, either $x+6$ is positive or negative and it will flip the inequality or not depending on what case you're in. – user2566092 May 18 '14 at 18:59 | 2019-12-08T05:45:56 | {
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http://otvedaem.com/rw8q0t2p/compute-combinations-r-cb7d5c | Hi again, I am exploring if R can help me to get all possible combinations of members in a group. We use the combn() function for finding all possibilities: To calculate the number of combinations the binomial coefficient is used: To give you some intuition consider the above example: you have possibilities for choosing the first ball, for the second, for the third and so on up to the sixth ball. If argument FUN is not NULL, applies a function given by the argument to each point.If simplify is FALSE, returns a list; otherwise returns an array, typically a matrix. Venables, Bill. Write A Program To Compute The Number Of Combinations Of 'r Items From A Given Set Of 'N' Items. The following C function comb requires a two-dimensional array to store the intermediate results. / r! The row names are ‘automatic’. Show transcribed image text. While I’m at it, I will examine combinations and permutations in R. As you may recall from school, a combination does not take into account the order, whereas a permutation does. / ((n - r)! Theorem 3. For factorial watch this video https://youtu.be/IBlnyh9hPwA Combination : C(n,r) = n!/(r! Permutations are specific selections of elements within a set where the order in which the elements are arranged is important, while combinations involve the selection of elements without regard for order. : Proof. Combinations of a,b,c,d,e,f,g that have at least 2 of a,b or c . C++ Program to Compute Combinations using Factorials C++ Programming Server Side Programming The following is an example to compute combinations using factorials. The Combinations Calculator will find the number of possible combinations that can be obtained by taking a sample of items from a larger set. Search the stuart package. 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In English we use the word "combination" loosely, without thinking if the order of things is important. e.g. Permutations and combinations have uses in math classes and in daily life. Without repetition simply means that when one has drawn an element it cannot be drawn again, so with repetition implies that it is replaced and can be drawn again. The number of r-combinations of a set with n elements, where n is a nonnegative integer and r is an integer with 0 r n, equals C(n;r) = nCr = n r = n! r = 5. and. combinations enumerates the possible combinations of a See the expression argument to the options command for details on how to do this. The first factors vary fastest. Combinations vs. Permutations. R/compute.combinations.R defines the following functions: compute.combinations. Questionnaire. The number of combinations of r objects is n C r = n! edit close. Combinations of a,b,c,d,e,f,g that have at least 2 of a,b or c . Thankfully, they are easy to calculate once you know how. / (64! Let us see this in action, as an example we’ll see how many different ways there are of four runners reaching the finishing line: After this rather complicated function the calculation of the number of ways is simple, it is just the factorial function (it should again be obvious why): As you will see when solving real world problems with R the above functions often come in handy, so you should add them to your ever growing tool set – have fun and stay tuned! Command (⌘)-R: Start up from the built-in macOS Recovery system. Compute the combinations of choosing r items from n elements. Computes all combinations of r elements from n. GitHub Gist: instantly share code, notes, and snippets. Generate All Combinations of n Elements, Taken m at a Time. Of course, when the values are large enough, a possible stack overflow will occur when recursion depths become large. Where, N! However, mathematicians are focused on how many elements will exist within a Combinatorics problem, and have little interest in actually going through the work of creati… For that we need to use the itertools package. The number of r-combinations of a set with n elements, where n is a nonnegative integer and r is an integer with 0 r n, equals C(n;r) = nCr = n r = n! Permutation and combination. Recursive Combination Algorithm Implementation in C++ The above is simple and handy if you want to list all combinations given n and b. I start with a list of vectors and run the function below, which loops through 1:n where n is the number of sets and then uses combn to generate all combinations of my sets taken m at a time.. Syntax: Rules In Detail The "has" Rule. The order in which you combine them doesn't matter, as you will buy the two you selected anyways. After you’ve entered the required information, the nCr calculator automatically generates the number of Combinations and the Combinations with Repetitions. Combinations are used in a large number of game type problems. Let's do a little experiment in R. We'll toss two fair dice, just as we did in an earlier post, and see if the results of the two dice are independent. A data frame containing one row for each combination of the supplied factors. This type of activity is required in a mathematics discipline that is known as combinatorics; i.e., the study of counting. This makes computations feasible for very large numbers of combinations. Fortunately, the science behind it has been studied by mathematicians for centuries, and is well understood and well documented. The columns are labelled by the factors if these are supplied as named arguments or named components of a list. link brightness_4 code # A Python program to print all # combinations of given length . Theorem 3. End Example Combinations tell you how many ways there are to combine a given number of items in a group. Two different methods can be employed to count r objects within n elements: combinations and permutations. We are … combn() function in R Language is used to generate all combinations of the elements of x taken m at a time. Rules In Detail The "has" Rule. Computer Glossary; Who is Who; Permutation and Combination in Python? One of the key advantage of python over other programming language is that it comes with huge set of libraries with it. Jan. 2001. http://cran.r-project.org/doc/Rnews, combinations(n, r, v=1:n, set=TRUE, repeats.allowed=FALSE) Basically, it shows how many different possible subsets can be made from the larger set. Then a comma and a list of items separated by commas. enumerates the possible permutations. permutations (n r)! This means that for the example of the combination lock above, this calculator does not compute the case where the combination lock can have repeated values, for example 3-3-3. Now, there are possible positions for the first ball that is drawn, for the second… and so on and because the order doesn’t matter we have to divide by , which gives the binomial coefficient. Syntax: combn(x, m) Parameters: x: total number of elements taken r: number of elements taken at a time out of “x” elements Example 1: Will this result in a fractional number? The first factors vary fastest. The combinations were formed from 3 letters (A, B, and C), so n = 3; and each combination consisted of 2 letters, so r = 2. Vignettes . C (n,r): is the total number of combinations. Generates the combinations for choosing r items from a set of n items. We will perhaps cover those in a later post. If you're working with combinatorics and probability, you may need to find the number of permutations possible for an ordered set of items. All these combinations are emitted in lexicographical order. R/compute.combinations.R defines the following functions: compute.combinations. Python Server Side Programming Programming. Description. stuart Subtests Using Algorithmic Rummaging Techniques. I assume that your rank starts at $0$, as this simplifies the code (for me).. In this section, we are going to learn how to find permutation and combination of a given sequence using python programming language. - omegahat/Combinations macOS Recovery installs different versions of macOS, depending on the key combination you use while starting up. My goal is to compute the intersections of several vectors (sets of identifiers, gene-names to be specific). The number says how many (minimum) from the list are needed for that result to be allowed. filter_none. Permutations and combinations are part of a branch of mathematics called combinatorics, which involves studying finite, discrete structures. If your Mac is using a firmware password, you're prompted to enter the password. If x is a positive integer, returns all combinations of the elements of seq(x) taken m at a time. Vignettes . to access the Math PROB menu or press [ALPHA][WINDOW] to access the shortcut menu. r: number of elements chosen from the set for sampling! This video describes how to use the TI-30 to compute combinations Remember to use the second function button in order to access combinations. r!) This type of activity is required in a mathematics discipline that is known as combinatorics; i.e., the study of counting. Mathematics and statistics disciplines require us to count. Compute the combinations of choosing r items from n elements. Mathematics and statistics disciplines require us to count. nCm: Compute the binomial coefficient ("n choose m"), where n is any real number and m is any integer. It returns r length subsequences of elements from the input iterable. For example, you have a urn with a red, blue and black ball. How to calculate combination. We have 4 choices (A, C, G and T) a… play_arrow. Package index. "Programmers Note", R-News, Vol 1/1, Computing with combinations in SAS/IML. 10^3 ## [1] 1000 nrow (P_wi) ## [1] 1000. How many combinations are there for selecting four? permutations if length of input sequence is n and input parameter is r. Combination This method takes a list and a input r as a input and return a object list of tuples which contain all possible combination of length r in a list form. Combinatorics has many applications within computer science for solving complex problems. Our last case is permutations (of all elements) without repetitions which is also the most demanding one because there is no readily available function in base R. So, we have to write our own: As you can see it is a recursive function, to understand recursion read my post: To understand Recursion you have to understand Recursion…. which will be of the form n(n-1)...(n-r+1)/1.2...r. Similar to factorial, we initialize the result as 1 and multiply by n-i and divide by i+1. In R we use the choose() function to calculate it: So, you see that the probability of winning the lottery are about the same, no matter whether you play it… or not. If x is a positive integer, returns all combinations of the elements of seq(x) taken m at a time. = 11,238,513. Algorithms Begin function CalCombination(): Arguments: n, r. Body of the function: Calculate combination by using the formula: n! Example has 1,a,b,c. Let's take a more straightforward example where you choose three balls called R(red), B(blue), G(green). As far as I know there are no very convenient formulae for $r$ in between. For example, a deck of (n = 52) cards of which a (k = 5) card hand is drawn. Generate all combinations of the elements of x taken m at a time. * (n-r)!. So that gives . A permutation is an arrangement of objects in which the order is important (unlike combinations, which are groups of items where order doesn't matter).You can use a simple mathematical formula to find the number of different possible ways to order the items. rdrr.io Find an R package R language docs Run R in your browser R Notebooks. * (n-r)!) For the example, you can calculate 10! Exactly one of arguments "x" and "n" should be given; no provisions for function evaluation. No. Permutations . If you choose two balls with replacement/repetition, there are permutations: {red, red}, {red, blue}, {red, black}, {blue, red}, {blue, blue}, {blue, black}, {black, red}, {black, blue}, and {black, black}. When you think about it this is the same as because all the coefficients smaller than can be eliminated by reducing the fraction! To evaluate a permutation or combination, follow these steps: There are two ways to access the nPr and nCr templates: Press. See the expression argument to the Getting all possible combinations. options command for details on how to do this. where you have three positions with the numbers zero to nine each. Then we force the program to backtrack and find the next combination by evaluating the always failing ~. Only 1 Powerball number is picked from 26 choices, so there are only 26 ways of doing this. There are several notations for an r-combination from a set of n distinct elements: C(n;r), nCr (n, choose r), and n r, the binomial coe cient, which is the topic of the next section. Combin… A permutation is calculated n P r. Start on 'n' and count backwards 'r' numbers, multiplying them together. This is particularly important when completing probability problems.Let's say we are provided with n distinct objects from which we wish to select r elements. This is a C++ program to compute Combinations using Recurrence Relation for nCr. in a lottery it normally does not matter in which order the numbers are drawn). A combination is a way to select a part of a collection, or a set of things in which the order does not matterand it is exactly these cases in which our combination calculator can help you. Search the stuart package. Variations In this section, we will show you how it’s done. In some cases, you can also refer to combinations as “r-combinations,” “binomial coefficient” or “n choose r.” In some references, they use “k” instead of “r”, so don’t get confused when you see combinations referred to as “n choose k” or “k-combinations.” How do you calculate combinations in Excel? rdrr.io Find an R package R language docs Run R in your browser R Notebooks. The combntns function provides the combinatorial subsets of a set of numbers. This number will go in the denominator of our probability formula, since it is the number of possible outcomes. That was simple! It generate nCr * r! Each row of C contains a combination of k items chosen from v. The elements in each row of C are listed in the same order as they appear in v. If k > numel(v), then C is an empty matrix. = 69! Caution: The number of combinations and permutations increases rapidly We will solve this problem in python using itertools.combinations() module.. What does itertools.combinations() do ? Note that AB and BA are considered to be one combination, because the order in which objects are selected does not matter. So there are 11,238,513 possible ways of picking 5 numbers from a choice of 69 numbers. For factorial watch this video https://youtu.be/IBlnyh9hPwA Combination : C(n,r) = n!/(r! All combinations of v, returned as a matrix of the same type as v. Matrix C has k columns and n!/((n –k)! Another way of thinking about it is how many ways are there to, from a pool of six items, people in this example, how many ways are there to choose four of them. specified size from the elements of a vector. n: total number of elements in the given set. 10^3 ## [1] 1000 nrow (P_wi) ## [1] 1000. https://www.mathsisfun.com/combinatorics/combinations-permutations.html We will solve this problem in python using itertools.combinations() module.. What does itertools.combinations() do ? To use values of n above about 45, you will need to increase Generate all combinations of the elements of x taken m at a time. For p = 5 and k = 3, the problem is: “For each observation of the 5 variables, find the largest product among any 3 values.” In the SAS/IML language, you can solve problems like this by using the ALLCOMB function to generate all combinations of size k from the index set {1,2,…,p}. FAQ. Or use Option-Command-R or Shift-Option-Command-R to start up from macOS Recovery over the Internet. So I would like for each set of line with the same symbol calculate the average (or median) of the lines. with (10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1), which gives you 3,628,800. See the answer. * (n - r)!, where n represents the total number of items, and r represents the number of items being chosen at a time. In other words: "My fruit salad is a combination of apples, grapes and bananas" We don't care what order the fruits are in, they could also be "bananas, grapes and apples" or "grapes, apples and bananas", its the same fruit salad. The core question you must be able to answer is how many elements there are in a substructure of yours. To use values of n above about 45, you will need to increase R's recursion limit. Posted on June 3, 2019 by Learning Machines in R bloggers | 0 Comments, The area of combinatorics, the art of systematic counting, is dreaded territory for many people so let us bring some light into the matter: in this post we will explain the difference between permutations and combinations, with and without repetitions, will calculate the number of possibilities and present efficient R code to enumerate all of them, so read on…. Expert Answer . This is because first, we multiply by n and divide by 1. In python, we can find out the combination of the items of any iterable. Press the number on the menu that corresponds to the template you want to insert. : factorial . Recall that we need to find n!/r!(n-r)! !arg:(?m. Thank you in advance. A data frame containing one row for each combination of the supplied factors. The number says how many (minimum) from the list are needed for that result to be allowed. The number of permutations with repetition (or with replacement) is simply calculated by: where n is the number of things to choose from, r number of times. Here are the steps to follow when using this combination formula calculator: On the left side, enter the values for the Number of Objects (n) and the Sample Size (r). 5!) And then one would need some form of inclusion/exclusion to count those choices where some item is … When n gets large, the package provides a mechanism for dealing with each combination as it is generated so that one does not have to hold the entire collection around and operate on them after creating the entire collection. rows, where n is length(v). The word "has" followed by a space and a number. Now, either n or n-1 have to be even (as they are consecutive numbers). For example, if you want a new laptop, a new smartphone and a new suit, but you can only afford two of them, there are three possible combinations to choose from: laptop + smartphone, smartphone + suit, and laptop + suit. The formula for a combination is: nCr = (n!)/(r!(n-r)!). Home / R Documentation / base / expand.grid: Create a Data Frame from All Combinations of Factor Variables expand.grid: Create a Data Frame from All Combinations of Factor Variables Description Usage Arguments Value Note References See Also Examples Description. (n r)! 5!) Calculates a table of the number of combinations of n things taken r at a time. Before that, let me quickly show you how we can use one formula to find out the total number of combinations. all combinations of 1:n taken two at a time (that is, the indices of x that would give all combinations of the elements of x if x with length n had been given). For the numerator, we need the number of ways to draw one Ace and four other cards (none of them Aces) from the deck. Combination formula : If we have n distinct elements and if we are taking r elements at a time, we can have the below amount of combinations : nCr. Example has 1,a,b,c. This is the key distinction between a combination … In R: A biological example of this are all the possible codon combinations. The columns are labelled by the factors if these are supplied as named arguments or named components of a list. Next, we multiply by n-1 and divide by 2. In all cases, you can imagine somebody drawing elements from a set and the different ways to do so. : Proof. We won’t cover permutations without repetition of only a subset nor combinations with repetition here because they are more complicated and would be beyond the scope of this post. Permutation implies that the order does matter, with combinations it does not (e.g. See the shortcut menu in the second screen. Generate All Combinations of n Elements, Taken m at a Time Description. This problem has existing recursive solution please refer Print all possible combinations of r elements in a given array of size n link. The idea is to fix one element after the other [for (i in 1:n) and cbind(v[i], ...)] and permute the remaining elements [perm(v[-i])] down to the base case when only one element remains [if (n == 1) v], which cannot be permuted any further. combos = combntns(set,subset) returns a matrix whose rows are the various combinations that can be taken of the elements of the vector set of length subset.Many combinatorial applications can make use of a vector 1:n for the input set to return generalized, indexed combination subsets.. * (n-r)!) Package index. Let us start with permutations with repetitions: as an example take a combination lock (should be permutation lock really!) Let us now move on to calculating the number of combinations given n and r What does this algorithm do? r! n = 69. and. For this calculator, the order of the items chosen in the subset does not matter. Using the TI-84 Plus, you must enter n, insert the command, and then enter r. See the PROB menu in the first screen. What makes matters a little bit more complicated is that the recursive call is within a for loop. - omegahat/Combinations 10 P 7 = 10 x 9 x 8 x 7 x 6 x 5 x 4 (start on 10 and count down 7) Your program would start off with a variable 'x' assigned a value of 1. So in your example, we're ordering combinations lexicographically so we can use the binomial coeffecient to find how many elements there are of our substructures. The word "has" followed by a space and a number. We all know that the total number of solution to pick combination of n items out of m items is C(m, n), and sometimes denoted as $C_m^n$ or $(_n^m)$. I will have only a single line by gene in the end. stuart Subtests Using Algorithmic Rummaging Techniques. This is particularly important when completing probability problems.. Let's say we are provided with n distinct objects from which we wish to select r elements. The row names are ‘automatic’. number of things n ≦300 \) Customer Voice. Combinations and Permutations What's the Difference? To calculate combination, all you need is the formula, that too, in case you want to determine it manually. r! It returns r length subsequences of elements from the input iterable. Limitations. rdrr.io Find an R package R language docs Run R in your browser R Notebooks. n C r = 69 C 5 = 69! Taking $r=1$ gives $(1+x)^n = \sum_{k=0}^n \binom{n}{k}x^k$ and letting $r$ tend to infinity one gets $1/(1-x)^n = \sum_{k=0}^\infty \binom{-n}{k}(-x)^k = \sum_{k=0}^\infty \binom{k+n-1}{k}x^k$, the two formulae in the question. There are several notations for an r-combination from a set of n distinct elements: C(n;r), nCr (n, choose r), and n r, the binomial coe cient, which is the topic of the next section. (comb= bvar combination combinations list m n pat pvar var. Unlike permutations, where group order matters, in combinations, the order doesn't matter. k!) to access the probability menu where you will find the permutations and combinations commands. This problem has been solved! We use the expand.grid() function for enumerating all possibilities: The formula for calculating the number of permutations is simple for obvious reasons ( is the number of elements to choose from, is the number of actually chosen elements): The next is combinations without repetitions: the classic example is a lottery where six out of 49 balls are chosen. If you have to solve by hand, keep in mind that for each factorial, you start with the main number given and then multiply it by the next smallest number, and so on until you get down to 0. To calculate combinations, we will use the formula nCr = n! I start with a list of vectors and run the function below, which loops through 1:n where n is the number of sets and then uses combn to generate all combinations of my sets taken m at a time.. However, it is under-represented in libraries since there is little application of Combinatorics in business applications. This problem has existing recursive solution please refer Print all possible combinations of r elements in a given array of size n link. Collect all sets on the respective higher level [X ] and return the whole matrix X. Rather than type in the formula each time, it should be (a lot) easier to use the permutation and combination commands. with n and r!. Mathematically This Is Denoted By: N! Caution: The number of combinations and permutations increases rapidly with n and r!. Similarly, next whe… R's recursion limit. All the combinations emitted are of length ‘r’ and ‘r’ is a necessary argument here. How many combinations if I'm starting with a pool of six, how many combinations are there? Imagine you've got the same bag filled with colorful balls as in the example in the previous section.Again, you pick five balls at random, but this time, the order is important - it does matter whether you pick the red ball as first or third. / ( (69 - 5)! My goal is to compute the intersections of several vectors (sets of identifiers, gene-names to be specific). We can easily write an iterative function to compute the value. Denotes The Factorial Of N. If N . permutations(n, r, v=1:n, set=TRUE, repeats.allowed=FALSE), the of this package were written by Gregory R. Warnes. The formula for calculating the number of permutations is simple for obvious reasons ( is the number of elements to choose from, is the number of actually chosen elements): In R: 10^3. We first roll the dice 100,000 times, and then compute the joint distribution of the results of the rolls from the two dice. 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Are of length ‘ r ’ represents the size of different combinations that can be made from the macOS! Many applications within computer science for solving complex problems ( ⌘ ) -R: start up the! Cover those in a group use values of n things taken r at a time, when values. Are to combine a given number of elements from a set of n! Store the intermediate results ( minimum ) from the elements of a given set of combinations! A large number of combinations of the rolls from the list of combinations one row each! Hi again, I am exploring if r can help me to all. Button in order to access the nPr and nCr templates: press of... Example to compute combinations using Recurrence Relation for nCr of size n link sets. Are consecutive numbers compute combinations r order to access the shortcut menu calculate combinations, we multiply by n-1 and divide 2! What makes matters a little bit more complicated is that the recursive call is within for! Large enough, a possible stack overflow will occur when recursion depths become large labelled compute combinations r factors! All # combinations of r elements in the subset does not matter in which you combine them does matter. Run r in your browser r Notebooks formula to find out the combination of the items of any.. R ’ as input here ‘ r ’ and ‘ r ’ represents the size of combinations! An iterative function to compute combinations Remember to use values of n elements taken. The combntns function provides the combinatorial subsets of a set of libraries with it that it comes huge... Can use one formula to find permutation and combination of the rolls from the input iterable the factors if are. Large numbers of combinations and the combinations emitted are of length ‘ r ’ and ‘ ’., depending on the menu that corresponds to the template you want to insert ( ⌘ ):! | 2021-02-27T09:15:31 | {
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https://amsi.org.au/ESA_Senior_Years/SeniorTopic4/4b/4b_2content_8.html | ## Content
### Sampling from the continuous uniform distribution
Now we consider samples from the continuous uniform distribution.
Recall from the module Continuous probability distributions that, if the random variable $$U$$ has a uniform distribution on the interval $$(a, b)$$, which we write as $$U \stackrel{\mathrm{d}}{=} \mathrm{U}(a,b)$$, then $$U$$ has the following pdf:
$f_U(u) = \begin{cases} \dfrac{1}{b-a} & \text{if $$a < u < b$$,}\\ 0 &\text{otherwise.} \end{cases}$
In particular, if $$U \stackrel{\mathrm{d}}{=} \mathrm{U}(0,1)$$, then
$f_U(u) = \begin{cases} 1 &\text{if $$0 < u < 1$$,}\\ 0 &\text{otherwise.} \end{cases}$
We saw this distribution in the section Mechanisms for generating random samples, where we discussed obtaining a random sample in Excel. The Excel function $$\sf\text{RAND()}$$ gives observations from this distribution.
Figures 21 to 23 show three independent random samples from the $$\mathrm{U}(0,1)$$ distribution, each of size $$n=10$$. Note how they vary.
Figure 21: First random sample of size $$n=10$$ from the $$\mathrm{U}(0,1)$$ distribution.
Figure 22: Second random sample of size $$n=10$$ from the $$\mathrm{U}(0,1)$$ distribution.
Figure 23: Third random sample of size $$n=10$$ from the $$\mathrm{U}(0,1)$$ distribution.
Figure 24 shows the distribution of many independent random samples of size $$n=10$$ from the $$\mathrm{U}(0,1)$$ distribution. Remember that this means that any single observation in any of the samples is equally likely to be observed at any point along the number line between 0 and 1.
Figure 24: Dot plots of 100 random samples of size $$n=10$$ from the $$\mathrm{U}(0,1)$$ distribution.
As we did for the Normal and exponential distributions, we now look at random samples of size $$n=20$$ (figure 25) and $$n=100$$ (figure 26); again, we see that the larger samples conform more closely to the parent distribution.
Figure 25: Histograms of 100 random samples of size $$n=20$$ from the $$\mathrm{U}(0,1)$$ distribution.
Figure 26: Histograms of 100 random samples of size $$n=100$$ from the $$\mathrm{U}(0,1)$$ distribution.
Finally, we consider increasing the sample size further, over a wider range; each row in figure 27 has a different sample size, shown by the label at the left of the row. There is more lack of uniformity in the histograms of the smaller samples than in the larger samples: to enable a fair comparison, the same bin widths have been used throughout.
Figure 27: Histograms of random samples of varying size from the $$\mathrm{U}(0,1)$$ distribution. The same sample size, indicated at left, has been used for the 10 histograms in each row.
Next page - Content - Sampling from the binomial distribution | 2022-09-26T10:01:10 | {
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Cobbs approach allows students to build a deep understanding of statistical concepts over time as they analyze and design experiments. Have a good day. Our solutions are written by Chegg experts so you can be assured of the highest quality! 11, Ex. Microelectronics Circuit Analysis and Design Donald Neamen 4th Solutions For different square roots, the eigenvectors may also be different. that following holds well: Sinceis Our interactive player makes it easy to find solutions to Understanding Analysis (Undergraduate Texts In Mathematics) 1st Edition problems you're working on - just go to the chapter for your book. It's easier to figure out tough problems faster using Chegg Study. Linear Algebra Done Right 3rd ed. Amusing ourselves to describe slader homework and digital and cyber sheet, postman. This terrific book will become the text of choice for the single-variable introductory analysis course; take a … Do not just copy these solutions. 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Thus we know that R and R1 are identical, only their presentation in the matrix form may be slightly different. that the following condition holds well: Add This terrific book will become the text of choice for the single-variable introductory analysis course; take a … a natural number andis Access Understanding Analysis 2nd Edition Chapter 1.4 Problem 3E solution now. http://linear.axler.net. The textbook is "Understanding Analysis" by Stephen Abbott. Access Understanding Analysis (Undergraduate Texts in Mathematics) 1st Edition Chapter 1.4 solutions now. Now is the time to redefine your true self using Slader’s Understanding Analysis answers. Understanding Analysis is so well-written and the development of the theory so well-motivated that exposing students to it could well lead them to expect such excellence in all their textbooks. JavaScript is required to view textbook solutions. The problem asks to show the square root is unique. Please complete all the solutions of Hoffman Kunze Linear algebra specially chapter 7 8 9 10. YOU are the protagonist of your own life. and the eigenspace is orthogonal to each other (since they are spanned by distinct orthonormal vectors). © 2003-2020 Chegg Inc. All rights reserved. 459 verified solutions. Please only read these solutions after thinking about the problems carefully. See Solutions for Homework #1. Oh, I thought you were the author of LADR. Interaction and grammar from the measuring slader purely online courses slader help calculus homework sheet content of the bureau of justice calculus large. on both the sides of above inequality: Therefore, there exists a rational numbersuch Then there exists a natural numbersuch Consider the case when, Can you find your fundamental truth using Slader as a Understanding Analysis solutions manual? I just wanted to let you know that this blog means so much to me; it helped me go through Hoffman-Kunze (which is a great book, I definitely recommend it.) View the primary ISBN for: Understanding Analysis 0th Edition Textbook Solutions. where each $u_j$ is in $E(\lambda_j,T)$. Bookmark it to easily review again before an exam. Solved: Free step-by-step solutions to exercise 1 on page 11 in Understanding Analysis (9781493927128) - Slader Solutions to Understanding Analysis (9781493927128), Pg. A nal goal I have for these notes is to illustrate by example how the form and grammar of a written argument are intimately connected to the Solutions. Below, you can find links to the solutions of linear algebra done right 3rd edition by Axler. 458 verified solutions. Introduction to Design and Analysis of Experiments explains how to choose sound and suitable design structures and engages students in understanding the interpretive and constructive natures of data analysis and experimental design. Hit a particularly tricky question? I must be very tired, but how does 1.A:2 work? … Understanding Analysis is perfectly titled; if your students read it, that’s what’s going to happen. By the same logic b1^n,....,bn^2 are eigenvalues of T. Thus the eigenspaces of R and R1 with respect to same eigenvalue is identical. Everyone's understand to a certain problem might be different. 1 :: Homework Help and Answers :: Slader Solutions. Keep going! NOW is the time to make today the first day of the rest of your life. Circuit analysis is the process of finding all the currents and voltages in a network of connected components. Spectrum Monitors : this window displays a list of active spectrum monitors and hybrid APs streaming data to your client, the radio band the device is monitoring, and the date and time the SM or hybrid AP was connected to your client. Now is the time to redefine your true self using Slader’s Understanding Analysis answers. Nor to the square computation or cube=1... What am I missing?https://www.wolframalpha.com/input/?i=((-1%2Bsqrt(3i))%2F2)%5E2%3D(-1-sqrt(3i))%2F2. Introductory Circuit Analysis, the number one acclaimed text in the field for over three decades, is a clear and interesting information source on a complex topic. You are correct. Understanding Analysis is so well-written and the development of the theory so well-motivated that exposing students to it could well lead them to expect such excellence in all their textbooks. Chapter 1 The Real Numbers 1.1 Discussion: The Irrationality of $\\sqrt 2$ (no exercises) 1.2 Some Preliminaries 1.3 The Axiom of Completeness 1.4 Consequences of … See Solutions for Homework #2. Without doing too much work, show how to prove Theorem 1.4.3 in the case where a < 0 by converting this case into the one already proven. You can get free manual solution 1- click on the name of the book 2- following the open link of http://libgen.io Therefore, V can have an orthonormal basis consisting of eigenvectors of R. let them be e1, ...., en. Under expressed in the required skill set required for the knowledge and understanding in history. Do you have a TeX file or pdf of all the solutions compiled? Please only read these solutions after thinking about the problems carefully. Tomorrow's answer's today! By the way,One question about baby rudin, P.23 Chapter 1, Exercise 10. I am working through Axler's Measure, Integration and Real Analysis. Below, you can find links to the solutions of linear algebra done right 3rd edition by Axler. What a wonderful blog for those who want to explore linear algebra! First of all, thank you. The Thirteenth Edition contains updated insights on the highly technical subject, providing readers with the most current information in circuit analysis. since R is a square root of T. T ei = ai^2 * ei. I got of to a horrible start trying to dust off my linear algebra skills, haha! Save my name, email, and website in this browser for the next time I comment. I think you are doing the same (almost) thing as the textbook in a different form. Thank you so much! First Exam will be on Wednesday, October 8, 2014. The problems are. Where can I donate to you? Wait a minute... That i is not under the radical sign! I'm no longer having classes at univ, so it will be nice to have some help for learning this out of curiosity. Slader teaches you how to learn with step-by-step textbook solutions written by subject matter experts. Thanks again. Understanding Rapid Systems Of salder As Slader expands, it should win over interest (and pocketbooks) of highschool students to convince them to move their homework activities to its online group. Not even Wolfram Alpha agrees. Shed the societal and I'm going to use this for self-study. Never mind. The best part? Solutions. Take the one which suits you. also a rational number. It's really wonderful that this kind of exercises was being shared that it can be able to help a lot of people in testing their knowledge of algebra. We look at the basic elements used to build circuits, and find out what happens when elements are connected together into a circuit. Thanks once again. Do not just copy th … LOL, Here is the official site of this book. waiting for help:I cannot find the list of mistakes in Linear Algebra Done Right（3E）. Understanding Elementary Analysis 2nd Edition homework has never been easier than with Chegg Study. A modern introduction to probability and statistics : understanding why and how / F.M. Hence there might have many different choices of square roots. Download free Textbook PDF or purchase low-cost hardcopy Welcome. troductory student rmly in mind. In my teaching of analysis, I have come to understand the strong correlation between how students learn analysis and how they write it. This is immensely helpful to those of us who can't afford school and choose to self-study. INTEGRATION Problems with Solutions Solutions Manual for: Understanding Analysis, Second Edition Real Analysis Description of Analysis Problems and Solutions in EAL AND COMPLEX ANALYSIS Math 431 - Real ... is the time to redefine your true self using Slader’s Introduction to Real Analysis answers. let a1,..., an be the according eigenvalues. Thanks so much :), Thanks. See Solutions for Homework #4 and #5. Semester 1. Shed the societal and cultural narratives holding you back and let step-by-step Introduction to Probability and Statistics textbook solutions reorient your old paradigms. Solutions. Without you I would never hat made it, Thank you very much. The proof of 7.36 is awful. Can you find your fundamental truth using Slader as a Understanding Analysis solutions manual? This is an alternate ISBN. for those who are taking an introductory course in complex analysis. … Understanding Analysis is perfectly titled; if your students read it, that’s what’s going to happen. Let R be the positive square root of T. Then from the definition of positive operators, R is self-adjoint. ;). Assume that there exist another positive square root, let's call it R1. Thus, it's also nice that they can have some thing that would help them to learn besides from school. Second, I am studying for my qualifying exam and I am using Axler's book. let v1,....,vn be a orthonormal basis consisting of eigenvectors. 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Holding you back and let step-by-step Understanding Analysis solutions manual step-by-step Introduction to Probability and Statistics solutions... One question about baby rudin, understanding analysis slader Chapter 1, Exercise 10 hardcopy! Of positive operators, R is self-adjoint find links to the solutions compiled be e1,.... vn... Required for the knowledge and Understanding in history many different choices of square roots, eigenvectors. Exercise 10 you are doing the same ( almost ) thing as textbook... Please complete all the solutions of Hoffman Kunze linear algebra skills, haha your students read it, that s. Qualifying exam and I am studying for my qualifying exam and I wish to contribute something back your..., October 8, 2014 and # 5 cobbs approach allows students to build circuits, website! For help: I can not find the list of mistakes in linear algebra skills haha... Are spanned by distinct orthonormal vectors ) 3E solution now together into a circuit Session,. Covid-19 Relief Fund is not under the radical sign be very tired, but does... Are written by subject matter experts Analysis '' by Stephen Abbott say is... Cyber sheet understanding analysis slader postman to donate, please donate to US PayPal COVID-19 Relief Fund 's it... Thus, it 's also nice that they can have an orthonormal basis consisting eigenvectors... Day of the rest of your life student as an interactive self-study supplement to the website companion circuit. Thing as the textbook in a network of connected components by Stephen Abbott now is the official site of book! Out of curiosity therefore, V can have an orthonormal basis consisting of of. Not under the radical sign Analysis 2nd Edition PDF solution manuals over time as they and! Out what happens when elements are connected together into a circuit to contribute something back minute... that is. 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Might do the same for Axler 's Measure, Integration and Real Analysis from there might have many choices! Calculus large or PDF of all the solutions of linear algebra done 3rd! Supplement to the solutions of linear algebra specially Chapter 7 8 9 10 Te_1. Longer having classes at univ, so it will be nice to have some for! The problem asks to show the square root is unique student as an interactive self-study to. Blog for those who want to explore linear algebra skills, haha so it be! Analysis PDF ( Profound Dynamic Fulfillment ) today square root of T. from! S Understanding Analysis answers want to explore linear algebra done right 3rd Edition by Axler Slader ’ s Analysis! 'Ve greatly enhanced an already strong book 's educational value find out what happens when elements are connected into... To redefine your true self using Slader as a Understanding Analysis answers email, and the condition. Is immensely helpful to those of US who ca n't afford school and choose to self-study solutions, you greatly. Off my linear algebra done right 3rd Edition by Axler or PDF of all solutions! Figure out tough problems faster using Chegg Study better than downloaded Elementary Analysis 2nd Edition Chapter 1.4 now! Negative eigenvalues, if b1^2 = a1^2 Then we know for sure b1 = a1 ) online courses Slader understanding analysis slader! Analysis textbook solutions understanding analysis slader your old paradigms certain problem might be different is Study... Used to build circuits, and the eigenspace is orthogonal to each other ( since they are by! Please complete all the solutions of linear algebra done right 3rd Edition Axler. To serve the student as an interactive self-study supplement to the solutions of linear algebra done Right（3E） can find to! Currents and voltages in a different form links to the website companion of Analysis! Univ, so it will be on Wednesday, October 8, 2014 skills, haha access Analysis. You are doing the same ( almost ) thing as the textbook in a different form,!... Be very tired, but how does 1.A:2 work of circuit Analysis is perfectly titled ; your. | 2021-07-30T14:42:54 | {
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http://math.stackexchange.com/questions/95598/complex-conjugate-of-function | Complex conjugate of function
I have a wavefunction $\psi(x,t)=Ae^{i(kx-\omega t)}+ Be^{-i(kx+\omega t)}$. $A$ and $B$ are complex constants.
I am trying to find the probability density, so I need to find the product of $\psi$ with it's complex conjugate. The problem is, im not sure what is it's complex conjugate, I know the complex conjugate of $5+4i$ is $5-4i$, but what would be the complex conjugate of $\psi$? Is it just $-Ae^{i(kx-\omega t)}-Be^{-i(kx+\omega t)}$?
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The complex conjugation factors through sums and products. So you can take the complex conjugate of the factor with A and B separately. The constant A and B form know problem, this goes according to the usual rules. This leaves something of the form $e^{(a+bi)}$. Now note that $e^{(a+bi)}= e^a(\cos(b)+i \sin(b))$ Taking the complex conjugate now and using $\cos(b)=-\cos(b)$ and $-\sin(b)=\sin(-b)$, you find the complex conjugate $e^{a+i(-b)}$.
This means: $\bar{\psi} = \bar{A} \mathrm{e}^{-i (k x - \omega t)} + \bar{B} \mathrm{e}^{i (k x + \omega t)}$
Note the use of the minus sign to compactly write the complex conjugate of $e^{a+ib}$. In computation this is what write, but you might want to keep the explanation of the $\cos$ and $\sin$ in the back of your head.
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It's easier to note once and for all that $\overline{e^z}=e^{\overline z}$, because the exponential function can be defined uniquely without committing to a choice between $i$ and $-i$ (e.g., by power series). – Henning Makholm Jan 1 '12 at 18:42
Thanks, so I want to find $\psi\bar{\psi}$. Do I just multiply it out like this? $(Ae^{i(kx-\omega t)}+ Be^{-i(kx+\omega t)})(\bar{A} \mathrm{e}^{-i (k x - \omega t)} + \bar{B} \mathrm{e}^{i (k x + \omega t)})$, and so i get: $A\bar A+B\bar B+A\bar Be^{i(kx-\omega t)+i(kx+\omega t)}+B\bar Ae^{-i(kx+\omega t)-i(kx-\omega t)}$? Is that correct? – Thomas Jan 1 '12 at 18:46
Yes, if needed you can further simply to: $A\bar A+B\bar B+A\bar Be^{2ikx}+B\bar Ae^{-2ikx}$ – MrOperator Jan 1 '12 at 18:52
The complex conjugation will map $A \to \bar{A}$ and $B \to \bar{B}$.
If, say $A= 5 + 4 i$, then $\bar{A} = 5 - 4 i$, as you noted. So $$\bar{\psi} = \bar{A} \mathrm{e}^{-i (k x - \omega t)} + \bar{B} \mathrm{e}^{i (k x + \omega t)}$$
- | 2014-08-29T05:18:19 | {
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https://math.stackexchange.com/questions/3107650/show-that-the-sequence-a-n-left1-frac13-right-cdot-left1-frac1/3107660 | # Show, that the sequence $a_n = \left(1+\frac{1}{3}\right)\cdot\left(1+\frac{1}{9}\right)\cdot\ldots\cdot\left(1+\frac{1}{3^n}\right)$ converges.
I need to show, that the following sequence converges. I think I can somehow do it by Riemann Integral, but I cannot figure out a way to extract $$\frac{1}{n}$$ from it. I also cannot find two sequences which could let me show that it converges by the sandwich theorem. How to approach it?
$$a_n = \left(1+\frac{1}{3}\right)\cdot\left(1+\frac{1}{9}\right)\cdot\ldots\cdot\left(1+\frac{1}{3^n}\right)$$
• What about taking the logarithm ?
– user65203
Feb 10 '19 at 16:54
• The sequence is monotone. Is it bounded? Feb 10 '19 at 16:55
If you aplied logarithm you get that $$\log{a_n} = \sum_{n=1}^{\infty} \log{(1+\frac{1}{3^n})}$$. Now using that $$\log{x} \leq x-1$$ for $$x>0$$ you get that $$\log{a_n} \leq \sum_{n=1}^{\infty} \frac{1}{3^n}$$ that is a geometric series that converges, so as $$\log{a_n}$$ converges, then $$a_n$$ converges.
• How did you get from $\log{a_n} = \sum_{n=1}^{\infty} \log{1+\frac{1}{3^n}}$ to $\log{a_n} \leq \sum_{n=1}^{\infty} \frac{1}{3^n}$ ? I don't understand that part Feb 10 '19 at 17:10
• $\log{x} \leq x-1$ is the same as $\log{(1+x)}\leq x$, so $\log{(1+3^{-n})} \leq 3^{-n}$ Feb 10 '19 at 17:12
• Could you please use brackets? I don't know if you mean $log(1)+x$ or $log(1+x)$ Feb 10 '19 at 17:14
• Already edit, I meant $\log{(1+x)}$ Feb 10 '19 at 17:16
• Thank you very much. Could you please show me the path from $log(x) \leq x - 1$ to $log(1+x) \leq x$? Feb 10 '19 at 17:19
If you are ok with using GM-AM (inequality between geometric and arithmetic mean) and with using the well known limit
• $$\left(1+ \frac{x}{n}\right)^n\stackrel{n\to \infty}{\longrightarrow}e^x$$
then you can reason directly as follows: $$\begin{eqnarray*} \prod_{k=1}^n \left(1+\frac{1}{3^k}\right) & \leq & \left(\frac{\sum_{k=1}^n \left(1+\frac{1}{3^k}\right) }{n} \right)^n\\ & = & \left(\frac{n + \sum_{k=1}^n \frac{1}{3^k}}{n} \right)^n\\ & \leq & \left(1 + \frac{\sum_{k=1}^{\infty} \frac{1}{3^k}}{n} \right)^n \\ & = & \left(1 + \frac{\frac{1}{2}}{n} \right)^n \\ & \stackrel{n \to \infty}{\longrightarrow} & \sqrt{e} \\ \end{eqnarray*}$$
Since $$a_n$$ is obviously increasing and according to above calculation also bounded, it follows that $$a_n$$ is also convergent.
More generally, let $$a_n =\prod_{k=1}^n (1+c^k)$$ where $$0 < c < 1$$. This problem is $$c = \frac13$$.
Let $$b_n = \ln(a_n) =\sum_{k=1}^n \ln(1+c^k)$$.
Then $$b_n$$ is increasing and, since $$\ln(1+x) < x$$ for $$x > 0$$, $$b_n \lt \sum_{k=1}^n c^k =\dfrac{c-c^{n+1}}{1-c} \lt\dfrac{c}{1-c}$$ for all $$n$$.
Therefore $$b_n$$ is a bounded, monotone increasing sequence and so converges.
Therefore $$a_n = e^{b_n}$$ also converges.
Another proof can be given using Cauchy's criterion.
If $$n > m$$, $$b_n-b_m =\sum_{k=m+1}^n \ln(1+c^k) \lt\sum_{k=m+1}^n c^k =\dfrac{c^{m+1}-c^n}{1-c} \lt\dfrac{c^{m+1}}{1-c}$$.
This last can be made as small as desired by choosing $$m$$ large enough.
Note: To show that $$\ln(1+x) < x$$ for $$x > 0$$:
$$\ln(1+x) =\int_1^{1+x}\dfrac{dt}{t} =\int_0^{x}\dfrac{dt}{1+t} \lt\int_0^{x}dt =x$$. | 2022-01-20T13:40:49 | {
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http://mathhelpforum.com/algebra/25703-urgent-algebra-homework-xmas-stuff-d.html | # Math Help - Urgent algebra homework! Xmas stuff... :D
1. ## Urgent algebra homework! Xmas stuff... :D
Despite the season being over I am still left puzzled over this problem.
The following questions are related to the popular Christmas song "The Twelve Days of Christmas".
We all know it's fairly simple to figure out the amount of presents found over the twelve days of Christmas. It is 364 presents.
How many ways did you find to solve for 364 presents?
Tough what about Twelve Days of Christmas? How many presents would we receive then?
And the toughest of them all THE nth day of Christmas? How on Earth are we supposed to explain and work out the nth day of Christmas?
Help needed.
Thanks!
2. Originally Posted by chocole
Despite the season being over I am still left puzzled over this problem.
The following questions are related to the popular Christmas song "The Twelve Days of Christmas".
We all know it's fairly simple to figure out the amount of presents found over the twelve days of Christmas. It is 364 presents.
How many ways did you find to solve for 364 presents?
Tough what about Twelve Days of Christmas? How many presents would we receive then?
And the toughest of them all THE nth day of Christmas? How on Earth are we supposed to explain and work out the nth day of Christmas?
Help needed.
Thanks!
Tiina
Maybe I am not familiar with the version you know. If you get 1 present on the first day, 2 presents on the 2nd day, and n presents on the nth day, then your total is:
$\sum_{i=0}^n i = \frac{n(n+1)}{2}$
3. Originally Posted by colby2152
Maybe I am not familiar with the version you know. If you get 1 present on the first day, 2 presents on the 2nd day, and n presents on the nth day, then your total is:
$\sum_0^n i = \frac{n(n+1)}{2}$
The problem states how many presents would be received over any other number (n) of days.
I am brain dead. I've never been this stuck... on this kind of problem. XD
4. Originally Posted by chocole
The problem states how many presents would be received over any other number (n) of days.
I am brain dead. I've never been this stuck... on this kind of problem. XD
The formula for the sum of an arithmetic series is:
$S_{n} = \frac{n}{2} \left( 2a + (n - 1)d \right)$
a = 1 and d = 1
So:
$S_{n} = \frac{n}{2} \left( 2 + (n - 1) \right)$
$S_{n} = \frac{n}{2} \left( n + 1 \right)$
$S_{n} = \frac{n^2 + n}{2}$
Where n = the number of days.
So on the 1st day you would have 1 present, on the 2nd day, you would receive 3 presents, etc...
5. Hello, chocole!
The following questions are related to the Christmas song "The Twelve Days of Christmas".
We all know how to figure out the number of presents given over the 12 days.
It is 364 presents.
How many ways did you find to solve for 364 presents?
$\begin{array}{cccc}\text{Day} & & \text{Presents} \\
1 &1 & 1 \\
2 & 1+2 & 3 \\
3 & 1+2+3 & 6 \\
4 & 1+2+3+4 & 10 \\
\vdots & \vdots & \vdots
\end{array}$
. These are "triangular numbers."
On the $n^{th}$ day, my true love gave to me: . $\frac{n(n+1)}{2}$ gifts.
By the $n^{th}$ day, my true love gave to me a total of:
. . $\sum^n_{k=1}\frac{k(k+1)}{2} \;=\;\frac{n(n+1)(n+2)}{6}$ gifts. | 2014-12-29T01:36:49 | {
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https://math.stackexchange.com/questions/3251458/conditional-probability-that-standard-gaussian-random-variable-is-larger-than-an | # Conditional probability that standard gaussian random variable is larger than another
Say there are two standard Gaussian random variables $$X$$ and $$Y$$. I am trying to evaluate the probability that the larger of the two is selected, given that it is known whether $$X$$ is positive or negative (the strategy is selecting $$X$$ if $$X$$ is positive and selecting $$Y$$ if $$X$$ is negative). In equation form this is $$Pr(X-Y>0|X>0) + Pr(X-Y<0|X<0)$$
How can this expression be evaluated? Numerically it appears to be $$\frac{3}{4}$$, and intuitively this makes sense.
I am also interested in this probability in the more general case, where the strategy involves selecting $$X$$ if $$X-S>0$$ and $$Y$$ otherwise, where $$S$$ is another independent Gaussian random variable.
• I don't understand your question. Did you mean $\text{Pr}(X-Y>0|X>0)\text{Pr}(X>0)+\text{Pr}(X-Y<0|X<0)\text{Pr}(X<0)$? – Angela Pretorius Jun 5 '19 at 4:32
• Yes, that's right! – tankerjeel Jun 5 '19 at 20:07
The following assumes $$X$$ and $$Y$$ are independent.
As Angela Richardson pointed out, you probably actually want to compute $$P(X-Y > 0 \mid X > 0) P(X>0) + P(X-Y <0 \mid X< 0) P(X<0)$$ which is $$3/4$$. (The quantity in your post is not $$3/4$$.)
For the first term, it suffices to compute $$P(X-Y > 0, X>0)$$. (Why?)
Consider the region of the plane that contains $$(x,y)$$ pairs satisfying $$x-y>0$$ and $$x>0$$. Then use rotational symmetry of the vector $$(X,Y)$$ to compute the probability.
The other term can be handled similarly.
• Thanks. Is there a way to evaluate the more general case as well? I'm guessing symmetry no longer holds once $S$ is introduced. – tankerjeel Jun 5 '19 at 20:09
• In case anyone is curious, the stackexchange post in this link gives the solution as to how to compute the more general case. When $S$ is the standard normal, the probability turns out to be exactly $\frac{2}{3}$, and approaches $\frac{3}{4}$ as the variance of $S$ approaches $0$. – tankerjeel Jun 5 '19 at 23:57 | 2020-10-24T15:34:28 | {
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http://mathhelpforum.com/calculus/154765-volume-sphere-hole.html | # Thread: Volume of sphere with hole
1. ## Volume of sphere with hole
A hole of diameter d is drilled through a sphere of radius r in such a way that the axis of the hole passes through the centre of the sphere. Find the volume of the solid that remains.
I drew a picture of a sphere centred on the cartesian plane, with a hole going through its centre (in the shape of a cylinder).
I already knew that the volume of the sphere was $\frac {4}{3}\pi r^{3}$. I then worked out the volume of the hole (cylinder):
$
\int_{-r}^{r} \frac{1}{4}\pi d^{2}dy$
$= \frac{1}{2}\pi d^{2} r
$
Volume of solid remaining = $\frac {4}{3}\pi r^{3} - \frac{1}{2}\pi d^{2} r$
but the correct answer was $\frac{1}{6}\pi (4r^{2} - d^{2})^{\frac{3}{2}}$
I cant see where I went wrong cause surely the objective of this problem is to minus the hole (cylinder) from a sphere to get the remaining volume?
2. The volume subtracted out by drilling the hole is not precisely a cylinder, because a cylinder usually has flat ends. The material removed has curved ends (matches up with the sphere). So what do you get when you change this?
3. Very intresting problem. It sounds like Ackbeet is right in the fact that the figure removed is not
Spoiler:
only
a cylinder.
Good Luck!
4. Rather than subtracting the round-ended cylinder from the volume of the sphere, I think it's probably easier to set up the integral for the volume. You can use cylindrical shell elements of width $dx$, circumference $2 \pi x$, and height $2 \sqrt {R^2 - x^2}$, and integrate from x = d/2 to x = R.
5. I got the integral $V=-2\pi\int_{r^{2}-\frac{d^{2}}{4}}^{0} \sqrt{u} du$ after making the substitution of $u=r^{2} - x^{2}$ then my final answer was:
$\frac{4}{3}\pi(r^{2}-\frac{d^{2}}{4})^{\frac{3}{2}}$. Did I do something wrong or is there any way to simplify it to $\frac{1}{6}\pi(4r^{2} -d^{2)^\frac{3}{2}$ ?
6. They are the same answer. Proof:
$\displaystyle{\frac{4}{3}\pi(r^{2}-\frac{d^{2}}{4})^{\frac{3}{2}}}$
$\displaystyle{=\frac{4}{3}\pi(4r^{2}-d^{2})^{\frac{3}{2}}\left(\frac{1}{4}\right)^{\fra c{3}{2}}}$
$\displaystyle{=\frac{4}{3}\pi(4r^{2}-d^{2})^{\frac{3}{2}}\,\frac{1}{4^{3/2}}}$
$\displaystyle{=\frac{1}{8}\,\frac{4}{3}\pi(4r^{2}-d^{2})^{\frac{3}{2}}}$
$\displaystyle{=\frac{1}{6}\pi(4r^{2}-d^{2})^{\frac{3}{2}}.}$ | 2017-02-19T21:44:56 | {
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https://mathstats.uncg.edu/sites/pauli/112/HTML/secstatements.html | ## Section1.2Statements
Statements are declarative sentences that are either true or false. The statements are formulated in such a way that any reader, who knows what all the words mean, can understand them.
1. “Victoria likes cookies.” is a declarative sentence, and it is either true or false, so it is a statement.
2. “Broccoli is green.” is a declarative sentence and it is true, so it is a statement.
3. “Broccoli is pink.” is a declarative sentence and it is false, so it is a statement.
4. “Cookies!” is not a declarative sentence, so it is not a statement.
From now on we concentrate on statements about the integers.
Consider the following:
1. “2 is equal to 3.” is a statement. It is false.
2. “2 plus 3 is equal to 5.” is a statement. It is true.
3. “2 plus 3” is not a statement, as it is not a declarative sentence; it is not even a sentence, as it does not contain a verb.
When we write a statement using the symbols $$=\text{,}$$ $$\ne\text{,}$$ $$\lt\text{,}$$ $$\le\text{,}$$ $$>\text{,}$$ or $$\ge\text{,}$$ the comparison symbol takes the place of the verb. A mathematical statement always has a verb or a symbol that takes the place of the verb, just as a sentence does.
A mathematical expression consists of objects and operations. The objects can be numbers or variables (see the next section) and the operations can be, for example $$+\text{,}$$ $$\cdot\text{,}$$ or $$-\text{.}$$ Unlike a statement, an expression has no comparison symbol, that means it has no “verb.” So expressions by themselves are not true or false, but expressions can be used in statements, as in Example 1.1.6.
We formulate Example 1.2.2 using symbols.
1. “2 = 3” is a statement. It is false.
2. “2 + 3 = 5” is a statement. It is true.
3. “2 + 3” is an expression. As it does not have a verb it is not a statement.
We identify whether statements about integers are true or false. Notice that all these examples when read out have a verb in them, namely “is”.
1. $$2=2$$ is read “2 is equal to 2”. This is a true statement.
2. $$2=3$$ is read “2 is equal to 3”. This is a false statement.
3. $$2\gt 3$$ is read “2 is greater than”. This is a false statement.
4. $$2\ne 3$$ is read “2 is not equal to 3”. This is a true statement.
5. $$2\le 2$$ is read “2 is less than or equal to 2”. This is a true statement.
We give some examples of expressions and statements and identify them.
1. $$2 + 3$$” is an expression.
2. $$2+3 = 5$$” is a statement.
3. $$2 + 1 + 5$$” is an expression.
4. $$2+ 1 + 5 \lt 10$$” is a statement.
Decide whether the following are statements or not. If they are statements decide whether they are true or false.
1. “Sunflower”
2. “Stop signs are red.”
3. $$2$$ is equal to $$3\text{.}$$
4. $$\displaystyle (1+2)-4687$$
5. $$\displaystyle 2+3=7$$
6. $$\displaystyle 3 > -100$$
Solution.
1. “Sunflower” is not a sentence, so it is not a statement.
2. “Stop signs are red.” is a declarative sentence, so it is a statement. It is true.
3. “2 is equal to 3” is a declarative sentence, so it is a statement. As $$2\ne 3$$ the statement is false.
4. $$(1+2)-4687$$ is not a statement as it has no verb.
5. $$2+3=7$$ is a statement, the verb is ‘=’ (is equal to). As $$2+3=5$$ it is a false statement.
6. $$3 > -100$$ is a statement, the verb is “$$>$$” (is greater than). It is a true statement.
When a statement is true, we usually do not write “is true.” When a statement is false, always write “is false”.
In Checkpoint 1.2.7 recognize statements and for statements decide whether they are true or false.
Determine which of the following are mathematical statements.
For the statements decide whether they are true or false.
1. $$\displaystyle 13\le 38$$
2. $$\displaystyle -11-38$$
3. $$\displaystyle 13\cdot 38$$
4. $$\displaystyle 13\ne 38$$
### Subsection1.2.2Compound Statements
In mathematics we often deal with multiple statements that overlap. In these cases instead of writing each statement separately, we often write them as one string of statements. This allows us to connect the statements directly.
Instead of writing “$$2 + 3 = 5$$” and “$$5 = 1+4\text{,}$$” we write “$$2 + 3 = 5 = 1+4\text{.}$$
We can also do this with inequalities.
Writing “$$2+5 = 7 \lt 10$$” means both “2+5 = 7” and “$$7 \lt 10\text{.}$$” In words, “$$2$$ plus $$5$$ is $$7$$ and $$7$$ is less than $$10\text{.}$$
Compound statements are often used to prove identities, that is, when proving that two expressions are equal. The proofs of Theorem 1.4.6 and Theorem 1.4.7 and Theorem 1.4.12 in the next chapter is written that way. | 2022-09-28T21:56:53 | {
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https://math.stackexchange.com/questions/1877296/derivation-of-y-ax-h2k-from-y-ax2bxc-given-a-vertex-and-a-point | Derivation of $y=a(x-h)^2+k$ from $y=ax^2+bx+c$ given a vertex and a point
Derive $y=a(x-h)^2+k$ from $y=ax^2+bx+c$ given a vertex and a point.
Recently I have been solving a problem to which I could not find a solution. Google search of "quadratic equation given vertex and a point," yielded what I have been looking for. However, although I have already solved the problem, I am still wondering how to derive the equation I was looking for before I googled it.
I was given a vertex $V(-3, -2)$ and a point $P(-4, 0)$ of a parabola. Using $y=ax^2+bx+c$, I have derived an equation for finding the vertex for each parabola with $V(\frac{-b}{2a}, -a(\frac{b}{2a})^2+c)$.
I knew that given the same vertex, the parabola $y=x^2+6x+7$ was close to what I've been looking for. However, this parabola didn't go through $P(-4, 0)$, because it was too wide.
At this point it seemed as if I had enough information to derive an equation, that given a vertex and a point I would obtain a parabola according to the restrictions. However, from this point on I didn't know how to proceed further.
• Note that by symmetry, $(-2,0)$ is also a point on the parabola. – user137731 Jul 31 '16 at 23:53
• Is your goal to determine an equation of the parabola that has vertex $V(-3, -2)$ and passes through the point $P(-4, 0)$? – N. F. Taussig Jul 31 '16 at 23:56
• @N.F.Taussig Yes. – user270346 Jul 31 '16 at 23:56
• I was looking for a general equation, that given the vertex and a point of a parabola I could determine it's coefficients $a, b, c$ as in $y=ax^2+bx+c$. – user270346 Jul 31 '16 at 23:57
• Use the two points you're given and the one you get from symmetry. Plug each in to get a system of linear equations that you can hopefully then solve. – user137731 Jul 31 '16 at 23:58
$$y = ax^2 + bx + c$$
$$y - c = a\left(x^2 + \frac{b}{a}x\right)$$
$$y - c + a\frac{b^2}{4a^2}= a\left(x^2 + \frac{b}{a}x + \frac{b^2}{4a^2}\right)$$
$$y - c + \frac{b^2}{4a}= a\left(x+\frac{b}{2a}\right)^2$$
$$y = a\left(x-\left(-\frac{b}{2a}\right)\right)^2 + \left(c- \frac{b^2}{4a}\right)$$
$$y = a\left(x-h\right)^2 + k$$
where $h = -\frac{b}{2a}$ and $k = c- \frac{b^2}{4a}$
If you're going in reverse (going from vertex $(h,k)$ and point $(x,y)$ to quadratic parameters $a,b,c$), then you can take the last few equations, isolate $a,b,c$, and translate them into terms of $h,k,x,y$:
$$a = \frac{y-k}{(h-x)^2}$$
$$b = -2ha = \frac{-2h(y-k)}{(h-x)^2}$$
$$c = k + \frac{b^2}{4a} = k + \frac{h^2(y-k)}{(h-x)^2}$$
• This answer followed by the @N. F. Taussig answer is the order what makes most sense to me after picking up on this problem from where I left it. Thanks! – user270346 Aug 1 '16 at 0:35
• @Marcus Stuhr, Regarding your edit, note that the parabola $y=x^2+6x+7$ doesn't have roots $x=-4$ and $x=-2$ as required. – user270346 Aug 1 '16 at 0:47
• I've done as you've requested. Please, take your time. Thanks. – user270346 Aug 1 '16 at 0:55
• @Matt I believe you're looking for $2x^2 + 12x + 16 = 0$, which goes through $(-4,0)$ and $(-2,0)$ and has vertex $(-3,-2)$. – Marcus Andrews Aug 1 '16 at 1:15
Since you know that the parabola has vertex $V(-3, -2)$, the vertex form of its equation is $$y = a[x - (-3)]^2 + (-2) = a(x + 3)^2 - 2$$ Since the parabola passes through the point $P(-4, 0)$, we can substitute $-4$ for $x$ and $0$ for $y$ to determine $a$. \begin{align*} 0 & = a(-4 + 3)^2 - 2\\ 2 & = a(-1)^2\\ 2 & = a \end{align*} Hence, $y = 2(x + 3)^2 - 2$. You can expand this expression to obtain the standard form of the equation.
Addendum: In general, if you know the vertex $V(h, k)$ and a point $P(u, v) \neq V(h, k)$ on the parabola, you can write $$y = a(x - h)^2 + k$$ then substitute $u$ for $x$ and $v$ for $y$ to determine $a$. \begin{align*} v & = a(u - h)^2 + k\\ v - k & = a(u - h)^2\\ \frac{v - k}{(u - h)^2} & = a \end{align*} Then $$y = a(x - h)^2 + k = \frac{v - k}{(u - h)^2}(x - h)^2 + k$$ Again, expanding the expression to obtain its standard form enables you to determine the coefficients $a$, $b$, and $c$.
• Thanks for your answer, however that's not my question. Please forgive me if I was not clear enough. I was looking for a general equation, that given the vertex and a point of a parabola I could determine it's coefficients $a, b, c$ as in $y=ax^2+bx+c$, based on what I knew. So, it's more about how do you derive the general equation for solving this type of problems, rather than solving the problem itself. – user270346 Aug 1 '16 at 0:04
• This is especially useful due to the other point suggested by the symmetry. Thanks for your answer. – user270346 Aug 1 '16 at 0:37
• On a side note, do you have any idea as why the vertex equation for this parabola $y=2(x+3)^2-2$ and $y=x^2+6x+7$ following from it, don't have same roots? – user270346 Aug 1 '16 at 0:51
• If you expand $y = 2(x + 3)^2 - 2$, you obtain \begin{align} y & = 2(x + 3)^2 - 2\\ & = 2(x^2 + 6x + 9) - 2\\ & = 2x^2 + 12x + 18 - 2\\ & = 2x^2 + 12x + 16 \end{align} You forgot to multiply the terms in the parentheses by $2$. As you can verify, both $y = 2(x + 3)^2 - 2$ and $y = 2x^2 + 12x + 16$ have the roots $-4$ and $-2$. – N. F. Taussig Aug 1 '16 at 1:00
Generalized Form:
Given $y=ax^2+bx+c$, we can move the loose number $c$ to the other side and try completing the square! So from that, we get $$y-c=ax^2+bx\tag{1}$$
Factoring out $a$ gives us $y-c=a\left(x^2+\frac bax\right)$. Completing the square by dividing the coefficient of the $x$ term by $2$ and squaring it gives us $$y-c+a\left(\frac {b^2}{4a^2}\right)=a\left(x+\frac {b}{2a}\right)^2\tag{2}$$
Simplifying the left hand side and move it to the right hand side to obtain $$y=a\left(x-\left(-\frac {b}{2a}\right)^2\right)+\left(\frac {4ac-b^2}{4a}\right)\tag{3}$$
And since the Vertex follows the formula $(h,k)=\left(-\frac {b}{2a},\frac {4ac-b^2}{4a}\right)$, you get the Vertex formula by plugging it in. | 2020-01-19T21:58:54 | {
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https://math.stackexchange.com/questions/1409368/sum-of-sum-n-1-infty-1n1-fracx2n-12n-1 | Sum of $\sum_{n=1}^{\infty}(-1)^{n+1}\frac{x^{2n-1}}{2n-1}$
I've been working with the series:
$$\sum_{n=1}^{\infty}(-1)^{n+1}\frac{x^{2n-1}}{2n-1}$$
From the ratio test it is clear that the series converges for $|x| < 1$, but I'm unable to obtain the sum of the series.
I'm looking for any hint of how to obtain the sum.
• Differentiate by x, use the geometric sum, integrate, and use that f(0)=0. Result is arctan(x) – Ákos Somogyi Aug 25 '15 at 17:11
Let $f(x)=\sum_{n=1}^{\infty}\frac{(-1)^{n+1}x^{2n-1}}{2n-1}$. Then we have
\begin{align} f'(x)=\sum_{n=1}^{\infty} (-1)^{n+1}x^{2n-2}&=\frac{-1}{x^2}\sum_{n=1}^{\infty} (-x^2)^{n}\\\\ &=\frac{1}{1+x^2} \tag 1 \end{align}
Integrating $(1)$ and using $f(0)=0$ reveals that
$$\bbox[5px,border:2px solid #C0A000]{f(x)=\arctan(x)}$$
• (+1) Similar to the approach I was going to take. I was going to note that it was the same as $$\sum_{n=0}^\infty(-1)^n\frac{x^{2n+1}}{2n+1}$$ and then show that the derivative of that was $\frac1{1+x^2}$ – robjohn Aug 25 '15 at 17:46
• @robjohn Thanks! I skipped any discussion of interval of convergence and also omitted discussing the legitimacy of differentiating term by term since it appeared that the OP was already aware of at least the first of these issued. – Mark Viola Aug 25 '15 at 17:50
Let $y$ be our sum so:
$$y=\sum_{n=1}^{\infty}{{(-1)}^{n+1}\frac{{x}^{2n-1}}{2n-1}}$$
Let's differentiate it to get:
$$y'=\sum_{n=1}^{\infty}{{(-1)}^{n+1}{x}^{2n-2}}\\ y'=\sum_{n=1}^{\infty}{{(-1)}^{n-1}{x}^{2n-2}}\\ y'=\sum_{n=1}^{\infty}{{(-1)}^{n+1}{(-x^2)}^{n-1}}\\ y'=\frac{1}{1+x^2}$$
Now integrate to get the original sum to get:
$$y=\arctan{x} +C$$
It is easy to see that for $x=0$ we have $y=0$, so $C=0$, hence the sum equals:
$$\sum_{n=1}^{\infty}{{(-1)}^{n+1}\frac{{x}^{2n-1}}{2n-1}}=\arctan{x}$$
• Isn't that just a repeat of what's already posted? – Macavity Aug 25 '15 at 17:40
• I was typing my answer at the same time that's why I didn't notice the other answer – Oussama Boussif Aug 25 '15 at 17:43
• (+1) good answer. I've had it happen many times that a similar answer has been posted while I am typing up mine. In fact, it happened here, but I happened to see Dr. MV's answer before I posted. – robjohn Aug 25 '15 at 17:49
$$(-1)^{n+1}\dfrac{x^{2n-1}}{2n-1}=i^{2(n+1)}\cdot\dfrac{x^{2n-1}}{2n-1}=-i\cdot\dfrac{(ix)^{2n-1}}{2n-1}$$
If $S=\sum_{n=1}^\infty(-1)^{n+1}\dfrac{x^{2n-1}}{2n-1},$
$$i S=\sum_{n=1}^\infty(-1)^{n+1}\dfrac{(ix)^{2n-1}}{2n-1}$$
Now for $-1<y\le1,\ln(1+y)=y-\dfrac{y^2}2+\dfrac{y^3}3-\dfrac{y^4}4+\cdots$
$\ln(1-y)=-y-\dfrac{y^2}2-\dfrac{y^3}3-\dfrac{y^4}4-\cdots$
$\ln(1+y)-\ln(1-y)=?$
$$\implies2i S=\ln(1+ix)-\ln(1-ix)=\ln\dfrac{1+ix}{1-ix}$$
Let $1=r\cos A,x=r\sin A, x=\tan A$
Now, $$\ln\dfrac{1+ix}{1-ix}=\ln(e^{2iA})=2iA=2i\arctan x$$
• I know what you're trying to show, but the appearance of $x$ and $y$ and the fact that I think the sum for $iS$ looks like it has too many $i^{2n-1}$ in it, makes things a bit hard to follow. – robjohn Aug 25 '15 at 18:02 | 2019-10-23T20:25:49 | {
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https://ww2.mathworks.cn/help/stats/quantile.html | # quantile
Quantiles of a data set
## Syntax
``Y = quantile(X,p)``
``Y = quantile(X,N)``
``Y = quantile(___,'all')``
``Y = quantile(___,dim)``
``Y = quantile(___,vecdim)``
``Y = quantile(___,'Method',method)``
## Description
example
````Y = quantile(X,p)` returns quantiles of the elements in data vector or array `X` for the cumulative probability or probabilities `p` in the interval [0,1]. If `X` is a vector, then `Y` is a scalar or a vector having the same length as `p`.If `X` is a matrix, then `Y` is a row vector or a matrix where the number of rows of `Y` is equal to the length of `p`. For multidimensional arrays, `quantile` operates along the first nonsingleton dimension of `X`. ```
example
````Y = quantile(X,N)` returns quantiles for `N` evenly spaced cumulative probabilities (1/(`N` + 1), 2/(`N` + 1), ..., `N`/(`N` + 1)) for integer `N`>1. If `X` is a vector, then `Y` is a scalar or a vector with length `N`. If `X` is a matrix, then `Y` is a matrix where the number of rows of `Y` is equal to `N`.For multidimensional arrays, `quantile` operates along the first nonsingleton dimension of `X`. ```
example
````Y = quantile(___,'all')` returns quantiles of all the elements of `X` for either of the first two syntaxes.```
example
````Y = quantile(___,dim)` returns quantiles along the operating dimension `dim` for either of the first two syntaxes.```
example
````Y = quantile(___,vecdim)` returns quantiles over the dimensions specified in the vector `vecdim` for either of the first two syntaxes. For example, if `X` is a matrix, then `quantile(X,0.5,[1 2])` returns the 0.5 quantile of all the elements of `X` because every element of a matrix is contained in the array slice defined by dimensions 1 and 2.```
example
````Y = quantile(___,'Method',method)` returns either exact or approximate quantiles based on the value of `method`, using any of the input argument combinations in the previous syntaxes.```
## Examples
collapse all
Calculate the quantiles of a data set for specified probabilities.
Generate a data set of size 10.
```rng('default'); % for reproducibility x = normrnd(0,1,1,10)```
```x = 1×10 0.5377 1.8339 -2.2588 0.8622 0.3188 -1.3077 -0.4336 0.3426 3.5784 2.7694 ```
Calculate the 0.3 quantile.
`y = quantile(x,0.30)`
```y = -0.0574 ```
Calculate the quantiles for the cumulative probabilities 0.025, 0.25, 0.5, 0.75, and 0.975.
`y = quantile(x,[0.025 0.25 0.50 0.75 0.975])`
```y = 1×5 -2.2588 -0.4336 0.4401 1.8339 3.5784 ```
Calculate the quantiles of a data set for a given number of quantiles.
Generate a data set of size 10.
```rng('default'); % for reproducibility x = normrnd(0,1,1,10)```
```x = 1×10 0.5377 1.8339 -2.2588 0.8622 0.3188 -1.3077 -0.4336 0.3426 3.5784 2.7694 ```
Calculate four evenly spaced quantiles.
`y = quantile(x,4)`
```y = 1×4 -0.8706 0.3307 0.6999 2.3017 ```
Using `y = quantile(x,[0.2,0.4,0.6,0.8])` is another way to return the four evenly spaced quantiles.
Calculate the quantiles along the columns and rows of a data matrix for specified probabilities.
Generate a 4-by-6 data matrix.
```rng default % For reproducibility X = normrnd(0,1,4,6)```
```X = 4×6 0.5377 0.3188 3.5784 0.7254 -0.1241 0.6715 1.8339 -1.3077 2.7694 -0.0631 1.4897 -1.2075 -2.2588 -0.4336 -1.3499 0.7147 1.4090 0.7172 0.8622 0.3426 3.0349 -0.2050 1.4172 1.6302 ```
Calculate the 0.3 quantile for each column of `X` (`dim` = 1).
`y = quantile(X,0.3,1)`
```y = 1×6 -0.3013 -0.6958 1.5336 -0.1056 0.9491 0.1078 ```
`quantile` returns a row vector `y` when calculating one quantile for each column of a matrix. For example, `-0.3013` is the 0.3 quantile of the first column of `X` with elements (0.5377, 1.8339, -2.2588, 0.8622). Because the default value of `dim` is 1, you can return the same result with `y = quantile(X,0.3)`.
Calculate the 0.3 quantile for each row of `X` (`dim` = 2).
`y = quantile(X,0.3,2)`
```y = 4×1 0.3844 -0.8642 -1.0750 0.4985 ```
`quantile` returns a column vector `y` when calculating one quantile for each row of a matrix. For example `0.3844` is the 0.3 quantile of the first row of `X` with elements (0.5377, 0.3188, 3.5784, 0.7254, -0.1241, 0.6715).
Calculate the $N$ evenly spaced quantiles along the columns and rows of a data matrix.
Generate a 6-by-10 data matrix.
```rng('default'); % for reproducibility X = unidrnd(10,6,7)```
```X = 6×7 9 3 10 8 7 8 7 10 6 5 10 8 1 4 2 10 9 7 8 3 10 10 10 2 1 4 1 1 7 2 5 9 7 1 5 1 10 10 10 2 9 4 ```
Calculate three evenly spaced quantiles for each column of `X` (`dim` = 1).
`y = quantile(X,3,1)`
```y = 3×7 2.0000 3.0000 5.0000 7.0000 4.0000 1.0000 4.0000 8.0000 8.0000 7.0000 8.5000 7.0000 2.0000 4.5000 10.0000 10.0000 10.0000 10.0000 8.0000 8.0000 7.0000 ```
Each column of matrix `y` corresponds to the three evenly spaced quantiles of each column of matrix `X`. For example, the first column of `y` with elements (2, 8, 10) has the quantiles for the first column of `X` with elements (9, 10, 2, 10, 7, 1). `y = quantile(X,3)` returns the same answer because the default value of `dim` is 1.
Calculate three evenly spaced quantiles for each row of `X` (`dim` = 2).
`y = quantile(X,3,2)`
```y = 6×3 7.0000 8.0000 8.7500 4.2500 6.0000 9.5000 4.0000 8.0000 9.7500 1.0000 2.0000 8.5000 2.7500 5.0000 7.0000 2.5000 9.0000 10.0000 ```
Each row of matrix `y` corresponds to the three evenly spaced quantiles of each row of matrix `X`. For example, the first row of `y` with elements (7, 8, 8.75) has the quantiles for the first row of `X` with elements (9, 3, 10, 8, 7, 8, 7).
Calculate the quantiles of a multidimensional array for specified probabilities by using the `'all'` and `vecdim` input arguments.
Create a 3-by-5-by-2 array `X`. Specify the vector of probabilities `p`.
`X = reshape(1:30,[3 5 2])`
```X = X(:,:,1) = 1 4 7 10 13 2 5 8 11 14 3 6 9 12 15 X(:,:,2) = 16 19 22 25 28 17 20 23 26 29 18 21 24 27 30 ```
`p = [0.25 0.75];`
Calculate the 0.25 and 0.75 quantiles of all the elements in `X`.
`Yall = quantile(X,p,'all')`
```Yall = 2×1 8 23 ```
`Yall(1)` is the 0.25 quantile of `X`, and `Yall(2)` is the 0.75 quantile of `X`.
Calculate the 0.25 and 0.75 quantiles for each page of `X` by specifying dimensions 1 and 2 as the operating dimensions.
`Ypage = quantile(X,p,[1 2])`
```Ypage = Ypage(:,:,1) = 4.2500 11.7500 Ypage(:,:,2) = 19.2500 26.7500 ```
For example, `Ypage(1,1,1)` is the 0.25 quantile of the first page of `X`, and `Ypage(2,1,1)` is the 0.75 quantile of the first page of `X`.
Calculate the 0.25 and 0.75 quantiles of the elements in each `X(i,:,:)` slice by specifying dimensions 2 and 3 as the operating dimensions.
`Yrow = quantile(X,p,[2 3])`
```Yrow = 3×2 7 22 8 23 9 24 ```
For example, `Yrow(3,1)` is the 0.25 quantile of the elements in `X(3,:,:)`, and `Yrow(3,2)` is the 0.75 quantile of the elements in `X(3,:,:)`.
Find median and quartiles of a vector, `x`, with even number of elements.
Enter the data.
`x = [2 5 6 10 11 13]`
```x = 1×6 2 5 6 10 11 13 ```
Calculate the median of `x`.
`y = quantile(x,0.50)`
```y = 8 ```
Calculate the quartiles of `x`.
`y = quantile(x,[0.25, 0.5, 0.75])`
```y = 1×3 5 8 11 ```
Using `y = quantile(x,3)` is another way to compute the quartiles of `x`.
These results might be different than the textbook definitions because `quantile` uses Linear Interpolation to find the median and quartiles.
Find median and quartiles of a vector, `x`, with odd number of elements.
Enter the data.
`x = [2 4 6 8 10 12 14]`
```x = 1×7 2 4 6 8 10 12 14 ```
Find the median of `x`.
`y = quantile(x,0.50)`
```y = 8 ```
Find the quartiles of `x`.
`y = quantile(x,[0.25, 0.5, 0.75])`
```y = 1×3 4.5000 8.0000 11.5000 ```
Using `y = quantile(x,3)` is another way to compute the quartiles of `x`.
These results might be different than the textbook definitions because `quantile` uses Linear Interpolation to find the median and quartiles.
Calculate exact and approximate quantiles of a tall column vector for a given probability.
When you perform calculations on tall arrays, MATLAB® uses either a parallel pool (default if you have Parallel Computing Toolbox™) or the local MATLAB session. To run the example using the local MATLAB session when you have Parallel Computing Toolbox, change the global execution environment by using the `mapreducer` function.
`mapreducer(0)`
Create a datastore for the `airlinesmall` data set. Treat `'NA'` values as missing data so that `datastore` replaces them with `NaN` values. Specify to work with the `ArrTime` variable.
```ds = datastore('airlinesmall.csv','TreatAsMissing','NA',... 'SelectedVariableNames','ArrTime');```
Create a tall table on top of the datastore, and extract the data from the tall table into a tall vector.
`t = tall(ds) % Tall table`
```t = Mx1 tall table ArrTime _______ 735 1124 2218 1431 746 1547 1052 1134 : : ```
`x = t{:,:} % Tall vector`
```x = Mx1 tall double column vector 735 1124 2218 1431 746 1547 1052 1134 : : ```
Calculate the exact quantile of x for `p` = 0.5. Because `X` is a tall column vector and `p` is a scalar, `quantile` returns the exact quantile value by default.
```p = 0.5; % Cumulative probability yExact = quantile(x,p)```
```yExact = tall double ? ```
Calculate the approximate quantile of x for `p` = 0.5. Specify `'Method','approximate'` to use an approximation algorithm based on T-Digest for computing the quantiles.
`yApprox = quantile(x,p,'Method','approximate')`
```yApprox = MxNx... tall double array ? ? ? ... ? ? ? ... ? ? ? ... : : : : : : ```
Evaluate the tall arrays and bring the results into memory by using `gather`.
`[yExact,yApprox] = gather(yExact,yApprox)`
```Evaluating tall expression using the Local MATLAB Session: - Pass 1 of 4: Completed in 1.4 sec - Pass 2 of 4: Completed in 0.6 sec - Pass 3 of 4: Completed in 0.74 sec - Pass 4 of 4: Completed in 0.76 sec Evaluation completed in 4.8 sec ```
```yExact = 1522 ```
```yApprox = 1.5220e+03 ```
The values of the approximate quantile and the exact quantile are the same to the four digits shown.
Calculate exact and approximate quantiles of a tall matrix for specified cumulative probabilities along different dimensions.
When you perform calculations on tall arrays, MATLAB® uses either a parallel pool (default if you have Parallel Computing Toolbox™) or the local MATLAB session. To run the example using the local MATLAB session when you have Parallel Computing Toolbox, change the global execution environment by using the `mapreducer` function.
`mapreducer(0)`
Create a tall matrix `X` containing a subset of variables from the `airlinesmall` data set. See Quantiles of Tall Vector for Given Probability for details about the steps to extract data from a tall array.
```varnames = {'ArrDelay','ArrTime','DepTime','ActualElapsedTime'}; % Subset of variables in the data set ds = datastore('airlinesmall.csv','TreatAsMissing','NA',... 'SelectedVariableNames',varnames); % Datastore t = tall(ds); % Tall table X = t{:,varnames} % Tall matrix```
```X = Mx4 tall double matrix 8 735 642 53 8 1124 1021 63 21 2218 2055 83 13 1431 1332 59 4 746 629 77 59 1547 1446 61 3 1052 928 84 11 1134 859 155 : : : : : : : : ```
When operating along a dimension that is not 1, the `quantile` function calculates the exact quantiles only, so that it can perform the computation efficiently using a sorting-based algorithm (see Algorithms) instead of an approximation algorithm based on T-Digest.
Calculate the exact quantiles of `X` along the second dimension for the cumulative probabilities 0.25, 0.5, and 0.75.
```p = [0.25 0.50 0.75]; % Vector of cumulative probabilities Yexact = quantile(X,p,2)```
```Yexact = MxNx... tall double array ? ? ? ... ? ? ? ... ? ? ? ... : : : : : : ```
When the function operates along the first dimension and `p` is a vector of cumulative probabilities, you must use the approximation algorithm based on t-digest to compute the quantiles. Using the sorting-based algorithm to find the quantiles along the first dimension of a tall array is computationally intensive.
Calculate the approximate quantiles of `X` along the first dimension for the cumulative probabilities 0.25, 0.5, and 0.75. Because the default dimension is 1, you do not need to specify a value for `dim`.
`Yapprox = quantile(X,p,'Method','approximate')`
```Yapprox = MxNx... tall double array ? ? ? ... ? ? ? ... ? ? ? ... : : : : : : ```
Evaluate the tall arrays and bring the results into memory by using `gather`.
`[Yexact,Yapprox] = gather(Yexact,Yapprox);`
```Evaluating tall expression using the Local MATLAB Session: - Pass 1 of 1: Completed in 2.9 sec Evaluation completed in 3.8 sec ```
Show the first five rows of the exact quantiles of `X` (along the second dimension) for the cumulative probabilities 0.25, 0.5, and 0.75.
`Yexact(1:5,:)`
```ans = 5×3 103 × 0.0305 0.3475 0.6885 0.0355 0.5420 1.0725 0.0520 1.0690 2.1365 0.0360 0.6955 1.3815 0.0405 0.3530 0.6875 ```
Each row of the matrix `Yexact` contains the three quantiles of the corresponding row in `X`. For example, `30.5`, `347.5`, and `688.5` are the 0.25, 0.5, and 0.75 quantiles, respectively, of the first row in `X`.
Show the approximate quantiles of `X` (along the first dimension) for the cumulative probabilities 0.25, 0.5, and 0.75.
`Yapprox`
```Yapprox = 3×4 103 × -0.0070 1.1148 0.9321 0.0700 0 1.5220 1.3350 0.1020 0.0110 1.9180 1.7400 0.1510 ```
Each column of the matrix `Yapprox` corresponds to the three quantiles for each column of the matrix `X`. For example, the first column of `Yapprox` with elements (–7, 0, 11) contains the quantiles for the first column of `X`.
Calculate exact and approximate quantiles along different dimensions of a tall matrix for `N` evenly spaced cumulative probabilities.
When you perform calculations on tall arrays, MATLAB® uses either a parallel pool (default if you have Parallel Computing Toolbox™) or the local MATLAB session. To run the example using the local MATLAB session when you have Parallel Computing Toolbox, change the global execution environment by using the `mapreducer` function.
`mapreducer(0)`
Create a tall matrix `X` containing a subset of variables from the `airlinesmall` data set. See Quantiles of Tall Vector for Given Probability for details about the steps to extract data from a tall array.
```varnames = {'ArrDelay','ArrTime','DepTime','ActualElapsedTime'}; % Subset of variables in the data set ds = datastore('airlinesmall.csv','TreatAsMissing','NA',... 'SelectedVariableNames',varnames); % Datastore t = tall(ds); % Tall table X = t{:,varnames}```
```X = Mx4 tall double matrix 8 735 642 53 8 1124 1021 63 21 2218 2055 83 13 1431 1332 59 4 746 629 77 59 1547 1446 61 3 1052 928 84 11 1134 859 155 : : : : : : : : ```
To find evenly spaced quantiles along the first dimension, you must use the approximation algorithm based on T-Digest. Using the sorting-based algorithm (see Algorithms) to find quantiles along the first dimension of a tall array is computationally intensive.
Calculate three evenly spaced quantiles along the first dimension of `X`. Because the default dimension is 1, you do not need to specify a value for `dim`. Specify `'Method','approximate'` to use the approximation algorithm.
```N = 3; % Number of quantiles Yapprox = quantile(X,N,'Method','approximate')```
```Yapprox = MxNx... tall double array ? ? ? ... ? ? ? ... ? ? ? ... : : : : : : ```
To find evenly spaced quantiles along any other dimension (`dim` is not `1`), `quantile` calculates the exact quantiles only, so that it can perform the computation efficiently by using the sorting-based algorithm.
Calculate three evenly spaced quantiles along the second dimension of `X`. Because `dim` is not 1, `quantile` returns the exact quantiles by default.
`Yexact = quantile(X,N,2)`
```Yexact = MxNx... tall double array ? ? ? ... ? ? ? ... ? ? ? ... : : : : : : ```
Evaluate the tall arrays and bring the results into memory by using `gather`.
`[Yapprox,Yexact] = gather(Yapprox,Yexact);`
```Evaluating tall expression using the Local MATLAB Session: - Pass 1 of 1: Completed in 4.2 sec Evaluation completed in 5.1 sec ```
Show the approximate quantiles of `X` (along the first dimension) for the three evenly spaced cumulative probabilities.
`Yapprox`
```Yapprox = 3×4 103 × -0.0070 1.1149 0.9322 0.0700 0 1.5220 1.3350 0.1020 0.0110 1.9180 1.7400 0.1510 ```
Each column of the matrix `Yapprox` corresponds to the three evenly spaced quantiles for each column of the matrix `X`. For example, the first column of `Yapprox` with elements (–7, 0, 11) contains the quantiles for the first column of `X`.
Show the first five rows of the exact quantiles of `X` (along the second dimension) for the three evenly spaced cumulative probabilities.
`Yexact(1:5,:)`
```ans = 5×3 103 × 0.0305 0.3475 0.6885 0.0355 0.5420 1.0725 0.0520 1.0690 2.1365 0.0360 0.6955 1.3815 0.0405 0.3530 0.6875 ```
Each row of the matrix `Yexact` contains the three evenly spaced quantiles of the corresponding row in `X`. For example, `30.5`, `347.5`, and `688.5` are the 0.25, 0.5, and 0.75 quantiles, respectively, of the first row in `X`.
## Input Arguments
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Input data, specified as a vector or array.
Data Types: `double` | `single`
Cumulative probabilities for which to compute the quantiles, specified as a scalar or vector of scalars from 0 to 1.
Example: 0.3
Example: [0.25, 0.5, 0.75]
Example: (0:0.25:1)
Data Types: `double` | `single`
Number of quantiles to compute, specified as a positive integer. `quantile` returns `N` quantiles that divide the data set into evenly distributed `N`+1 segments.
Data Types: `double` | `single`
Dimension along which the quantiles of a matrix `X` are requested, specified as a positive integer. For example, for a matrix `X`, when `dim` = 1, `quantile` returns the quantile(s) of the columns of `X`; when `dim` = 2, `quantile` returns the quantile(s) of the rows of `X`. For a multidimensional array `X`, the length of the `dim`th dimension of `Y` is the same as the length of `p`.
Data Types: `single` | `double`
Vector of dimensions, specified as a positive integer vector. Each element of `vecdim` represents a dimension of the input array `X`. In the smallest specified operating dimension (that is, dimension `min(vecdim)`), the output `Y` has length equal to the number of quantiles requested (either `N` or `length(p)`). In each of the remaining operating dimensions, `Y` has length 1. The other dimension lengths are the same for `X` and `Y`.
For example, consider a 2-by-3-by-3 array `X` with ```p = [0.2 0.4 0.6 0.8]```. In this case, `quantile(X,p,[1 2])` returns an array, where each page of the array contains the 0.2, 0.4, 0.6, and 0.8 quantiles of the elements on the corresponding page of `X`. Because 1 and 2 are the operating dimensions, with `min([1 2]) = 1` and `length(p) = 4`, the output is a 4-by-1-by-3 array.
Data Types: `single` | `double`
Method for calculating quantiles, specified as `'exact'` or `'approximate'`. By default, `quantile` returns the exact quantiles by implementing an algorithm that uses sorting. You can specify `'method','approximate'` for `quantile` to return approximate quantiles by implementing an algorithm that uses T-Digest.
Data Types: `char` | `string`
## Output Arguments
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Quantiles of a data vector or array, returned as a scalar or array for one or multiple values of cumulative probabilities.
• If `X` is a vector, then `Y` is a scalar or a vector with the same length as the number of quantiles requested (`N` or `length(p)`). `Y(i)` contains the `p(i)` quantile.
• If `X` is an array of dimension d, then `Y` is an array with the length of the smallest operating dimension equal to the number of quantiles requested (`N` or `length(p)`).
collapse all
### Multidimensional Array
A multidimensional array is an array with more than two dimensions. For example, if `X` is a 1-by-3-by-4 array, then `X` is a 3-D array.
### Nonsingleton Dimension
A nonsingleton dimension of an array is a dimension whose size is not equal to 1. A first nonsingleton dimension of an array is the first dimension that satisfies the nonsingleton condition. For example, if `X` is a 1-by-1-by-2-by-4 array, then the third dimension is the first nonsingleton dimension of `X`.
### Linear Interpolation
Linear interpolation uses linear polynomials to find yi = f(xi), the values of the underlying function Y = f(X) at the points in the vector or array x. Given the data points (x1, y1) and (x2, y2), where y1 = f(x1) and y2 = f(x2), linear interpolation finds y = f(x) for a given x between x1 and x2 as follows:
`$y=f\left(x\right)={y}_{1}+\frac{\left(x-{x}_{1}\right)}{\left({x}_{2}-{x}_{1}\right)}\left({y}_{2}-{y}_{1}\right).$`
Similarly, if the 1.5/n quantile is y1.5/n and the 2.5/n quantile is y2.5/n, then linear interpolation finds the 2.3/n quantile y2.3/n as
`${y}_{\frac{2.3}{n}}={y}_{\frac{1.5}{n}}+\frac{\left(\frac{2.3}{n}-\frac{1.5}{n}\right)}{\left(\frac{2.5}{n}-\frac{1.5}{n}\right)}\left({y}_{\frac{2.5}{n}}-{y}_{\frac{1.5}{n}}\right).$`
### T-Digest
T-digest[2] is a probabilistic data structure that is a sparse representation of the empirical cumulative distribution function (CDF) of a data set. T-digest is useful for computing approximations of rank-based statistics (such as percentiles and quantiles) from online or distributed data in a way that allows for controllable accuracy, particularly near the tails of the data distribution.
For data that is distributed in different partitions, t-digest computes quantile estimates (and percentile estimates) for each data partition separately, and then combines the estimates while maintaining a constant-memory bound and constant relative accuracy of computation ($q\left(1-q\right)$ for the qth quantile). For these reasons, t-digest is practical for working with tall arrays.
To estimate quantiles of an array that is distributed in different partitions, first build a t-digest in each partition of the data. A t-digest clusters the data in the partition and summarizes each cluster by a centroid value and an accumulated weight that represents the number of samples contributing to the cluster. T-digest uses large clusters (widely spaced centroids) to represent areas of the CDF that are near ```q = 0.5``` and uses small clusters (tightly spaced centroids) to represent areas of the CDF that are near `q = 0` or `q = 1`.
T-digest controls the cluster size by using a scaling function that maps a quantile q to an index k with a compression parameter $\delta$. That is,
`$k\left(q,\delta \right)=\delta \cdot \left(\frac{{\mathrm{sin}}^{-1}\left(2q-1\right)}{\pi }+\frac{1}{2}\right),$`
where the mapping k is monotonic with minimum value k(0,δ) = 0 and maximum value k(1,δ) = δ. The following figure shows the scaling function for δ = 10.
The scaling function translates the quantile q to the scaling factor k in order to give variable size steps in q. As a result, cluster sizes are unequal (larger around the center quantiles and smaller near `q = 0` or ```q = 1```). The smaller clusters allow for better accuracy near the edges of the data.
To update a t-digest with a new observation that has a weight and location, find the cluster closest to the new observation. Then, add the weight and update the centroid of the cluster based on the weighted average, provided that the updated weight of the cluster does not exceed the size limitation.
You can combine independent t-digests from each partition of the data by taking a union of the t-digests and merging their centroids. To combine t-digests, first sort the clusters from all the independent t-digests in decreasing order of cluster weights. Then, merge neighboring clusters, when they meet the size limitation, to form a new t-digest.
Once you form a t-digest that represents the complete data set, you can estimate the end-points (or boundaries) of each cluster in the t-digest and then use interpolation between the end-points of each cluster to find accurate quantile estimates.
## Algorithms
For an n-element vector `X`, `quantile` computes quantiles by using a sorting-based algorithm as follows:
1. The sorted elements in `X` are taken as the (0.5/n), (1.5/n), ..., ([n – 0.5]/n) quantiles. For example:
• For a data vector of five elements such as {6, 3, 2, 10, 1}, the sorted elements {1, 2, 3, 6, 10} respectively correspond to the 0.1, 0.3, 0.5, 0.7, 0.9 quantiles.
• For a data vector of six elements such as {6, 3, 2, 10, 8, 1}, the sorted elements {1, 2, 3, 6, 8, 10} respectively correspond to the (0.5/6), (1.5/6), (2.5/6), (3.5/6), (4.5/6), (5.5/6) quantiles.
2. `quantile` uses Linear Interpolation to compute quantiles for probabilities between (0.5/n) and ([n – 0.5]/n).
3. For the quantiles corresponding to the probabilities outside that range, `quantile` assigns the minimum or maximum values of the elements in `X`.
`quantile` treats `NaN`s as missing values and removes them.
## References
[1] Langford, E. “Quartiles in Elementary Statistics”, Journal of Statistics Education. Vol. 14, No. 3, 2006.
[2] Dunning, T., and O. Ertl. “Computing Extremely Accurate Quantiles Using T-Digests.” August 2017. | 2022-01-28T16:14:48 | {
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"url": "https://ww2.mathworks.cn/help/stats/quantile.html",
"openwebmath_score": 0.7601086497306824,
"openwebmath_perplexity": 1108.302201282896,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9814534316905262,
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https://math.stackexchange.com/questions/3044937/method-to-factor-an-expression | # Method to factor an expression
As the title says, i want to factorize an expression, but i don't have any clue how to proceed.
Here is the expression :
$$2x² -7x +3$$
And here is the factorized form :
$$(x-3)(2x - 1)$$
My question is, which method or rule to use to go from first to second ?
please note that I am a beginner, and the only question that i found which is closer to mine is this post.
Thank you for your help !
Rewrite the expression into the form:
$$2x^2-6x-x+3$$ ,
then group the first two terms together and the last two terms together:
$$(2x^2-6x)-(x-3)=2x(x-3)-(x-3)$$ ,
next extract the common factor:
$$(x-3)(2x-1)$$ ,
and you are done.
• I like the simplicity of your method, thank's :) – ganzo db Dec 18 '18 at 16:41
• @ganzodb You are welcome – Matko Dec 18 '18 at 18:24
If it concerns quadratic polynomial $$ax^2+bx+c$$ then start with calculating discriminant: $$D:=b^2-4ac$$
If $$D$$ is negative then give up (unless you are familiar with complex numbers already).
If $$D$$ is nonnegative then: $$ax^2+bx+c=a(x-x_1)(x-x_2)$$ where $$x_1=\frac{-b+\sqrt D}{2a}$$ and $$x_2=\frac{-b-\sqrt D}{2a}$$.
Especially if $$D$$ is a perfect square (as in your case, where $$D=25$$) then there is reason to cheer.
• As you supposed, i'm not good with complex numbers :) ! – ganzo db Dec 18 '18 at 16:54
It relies on the following property of quadratic polynomials:
If the quadratic polynomial $$\;ax^2+bx+c\;(a\ne 0)$$ has roots $$\xi_0$$ and $$\xi_1$$ (real or complex, distinct ot not), it can factored as $$ax^2+bx+c=a(x-\xi_0)(x-\xi_1).$$
This property is a consequence of a more general property of polynomials (of any degree) and the ring of polynomials over a field being a euclidean domain:
If a polynomial $$p(x)$$ has root $$\xi$$, it is divisible by $$x-\xi$$.
In the present case, the discriminat of $$2x² -7x +3$$ is $$\;\Delta=49-4\cdot 2\cdot 3=25$$, so its roots are $$\;\frac{7\pm 5}4==\bigl\{3,\frac 12\bigr\}$$, and the factorisation is $$2(x-3)\Bigl(x-\frac12\Bigr)=(x-3)(2x-1).$$
• i wish i could understand the last one ( \frac{7\pm 5}4 ). – ganzo db Dec 18 '18 at 16:57
• @ganzodb: It's just the formula $\frac{-b\pm\sqrt\Delta}{2a}.$. – Bernard Dec 18 '18 at 18:22
In general ,if $$x_1$$ and $$x_2$$ are roots of $$\underbrace{a}_{\neq 0}x^2+bx+c=0$$ then $$ax^2+bx+c=k(x-x_1)(x-x_2)=k[x^2-(x_1+x_2)x+x_1x_2]$$ comparing the coefficients, $$a=k,b=-k(x_1+x_2),c=kx_1x_2$$ Consequently $$\text{sum of the roots}=-\frac{b}{a}\;\;\&\;\;\text{product of the roots}=\frac{c}{a}$$
So your case, $$x_1+x_2= \frac{7}{2}$$ and $$x_1x_2=\frac{3}{2}$$
Now solve these to get $$x_1$$ and $$x_2$$ to finish your conclusion
To find $$x_1$$ and $$x_2$$, $$2x^2-7x+3=0$$ implies $$x^2-\frac{7}{2}x+\frac{3}{2}=0$$ which means $$x^2-2\left(\frac{7}{4}\right)x=-\frac{3}{2}$$ which is same as $$x^2-2\left(\frac{7}{4}\right)x+\frac{49}{16}=-\frac{3}{2}+\frac{49}{16}=\frac{25}{16}$$ so $$\left(x-\frac{7}{4}\right)^2=\frac{25}{16}$$ and so $$x-\frac{7}{4}=\pm \sqrt{\frac{25}{16}}=\pm \frac{5}{4}$$ so $$x=\frac{7}{4} \pm \frac{5}{4}=\frac{7\pm 5}{4}$$
• Thank you for the time spent in the explanation, i'm still unfamiliar with some mathematical concepts. – ganzo db Dec 18 '18 at 16:53 | 2021-07-31T18:38:37 | {
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https://s12762.gridserver.com/1gb40i0t/43d397-master-electrician-test | 1 ? the This is the reason I like best. While we can use other methods to solve such a problem, if we know the multiplicative inverse of our coefficient matrix, then we can easily solve the problem by simply multiplying both sides by the inverse. A second-order matrix can be represented by . 's' : ''}}. Step 2:. a number multiplied by it’s multiplicative inverse gives the multiplicative identity. Here are three ways to find the inverse of a matrix: 1. See Also . link to the specific question (not just the name of the question) that contains the content and a description of Track your scores, create tests, and take your learning to the next level! If we multiply matrix A by the inverse of matrix A, we will get the identity matrix, I. From the top row, we get 1(11) + -2(5) = 11 - 10 = 1. All other trademarks and copyrights are the property of their respective owners. Reduce the left matrix to row echelon form using elementary row operations for the whole matrix (including the right one). F) Find A(BC). The concept of solving systems using matrices is similar to the concept of solving simple equations. The theory, as usual, is below the calculator as The multiplicative inverse of a matrix is similar in concept, except that the product of matrix $$A$$ and its inverse $$A^{−1}$$ equals the identity matrix. By using this website, you agree to our Cookie Policy. COMEDK 2012: The multiplicative inverse of (3 + 4i/4 - 5 i) is (A) ((-8/25) , (31/25) ) (B) ((-8/25) , (-31/25) ) (C) ((8/25) , (-31/25) ) (D) ((8/25 So, for the number 2, it is 1/2. There are a couple of ways to do this. Inverse of a Matrix. The term inverse matrix generally implies the multiplicative inverse of a matrix. Create an account to start this course today. In general, the inverse of n X n matrix A can be found using this simple formula: where, Adj(A) denotes the adjoint of a matrix and, Det(A) is Determinant of matrix A. Yes, we write the inverse with a superscript of -1. This calculator finds the modular inverse of a matrix using the adjugate matrix and modular multiplicative inverse. 5/7 minus 4/7. The multiplicative inverse of a modulo m exists if and only if a and m are coprime (i.e., if gcd(a, m) = 1). Send your complaint to our designated agent at: Charles Cohn This website uses cookies to ensure you get the best experience. Hence, I is known as the identity matrix under multiplication. I don't … first two years of college and save thousands off your degree. f(g(x)) = g(f(x)) = x. Find the value of . The multiplicative inverse of a matrix is similar in concept, except that the product of matrix $A$ and its inverse ${A}^{-1}$ equals the identity matrix. You see, it is useful to learn about the multiplicative inverse of a matrix because if we know it, then we can use it to help us solve equations with matrices in them. If Varsity Tutors takes action in response to AX - BX + D = C. Use A - 1 to find the solution, (x_1, x_2), to the given system. C) Find AB. Your name, address, telephone number and email address; and The multiplicative inverse of a nonsingular matrixis its matrix inverse. Start Here; Our Story; Hire a Tutor; Upgrade to Math Mastery. Multiplying Matrices. Multiplicative Inverse Property Calculator. multiplicative inverse synonyms, multiplicative inverse pronunciation, multiplicative ... may use this activity to consolidate their students' learning of certain concepts of matrices such as the algorithm for matrix multiplication and the concept of the multiplicative inverse of a matrix. B) Find 2A + 3B. If you've found an issue with this question, please let us know. 3. is the multiplicative inverse of . The identity matrix is a square matrix containing ones down the main diagonal and zeros everywhere else. St. Louis, MO 63105. Thus, if you are not sure content located With this knowledge, we have the following: 101 S. Hanley Rd, Suite 300 Create your account. To learn more, visit our Earning Credit Page. Coolmath privacy policy. The multiplicative inverse of a matrix is the matrix that gives you the identity matrix when multiplied by the original matrix. Define multiplicative inverse. Note the first and the last columns are equal. After you multiply, you can then easily find the answer by translating back to equation form. Matrix Multiplication Calculator Here you can perform matrix multiplication with complex numbers online for free. Zero … MULTIPLICATIVE INVERSES For every nonzero real number a, there is a multiplicative inverse l/a such that. Attempt to find inverse of cross multiplication using skew symmetric matrix. I will use the determinant method. Plug the value in the formula then simplify to get the inverse of matrix C. Step 3:. or more of your copyrights, please notify us by providing a written notice (“Infringement Notice”) containing and career path that can help you find the school that's right for you. Inverse Matrices: The inverse of a matrix, when multiplied to the matrix, in both orders must produce an identity matrix. Same thing when the inverse comes first: ( 1/8) × 8 = 1. All operations on residue matrices are performed the same as for the integer matrices except that the operations are done in modular arithmetic. either the copyright owner or a person authorized to act on their behalf. The following formula is used to calculate the inverse matrix value of the original 2×2 matrix. information contained in your Infringement Notice is accurate, and (c) under penalty of perjury, that you are Amy has a master's degree in secondary education and has taught math at a public charter high school. Study.com has thousands of articles about every I find the modular multiplicative inverse (of the matrix determinant, which is $1×4-3×5=-11$) with the extended Euclid algorithm (it is $-7 \equiv 19 \pmod{26}$). your copyright is not authorized by law, or by the copyright owner or such owner’s agent; (b) that all of the The multiplicative inverse of a matrix is similar in concept, except that the product of matrix A and its inverse A –1 equals the identity matrix. I explain that today we will find the multiplicative inverse of a matrix. A coin is continually flipped until it comes up tails, at which time that coin is put aside and the other co, Given the matrices A = \begin{bmatrix} 4 & -2 & 3\\ -2 & 1 & 3\\ 1 & 2 & 2 \end{bmatrix}, \quad C =\begin{bmatrix} 1 & -3 & 0\\ -3 & 1 & 0\\ 0 & 0 & -2 \end{bmatrix} , find C^TA, Compute Let A= \begin{bmatrix}3&7&8&9&-60&2&-5& 7&30&0& 1&5&0 0&0&2&4&10&0&0&-2&0\end{bmatrix}, Determine whether the following statements are True or False. 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The identity matrix is a square matrix containing ones down the main diagonal and zeros everywhere else. Let's review what we've learned. (v) Existence of multiplicative inverse : If A is a square matrix of order n, and if there exists a square matrix B of the same order n, such that AB = BA = I. where I is the unit matrix of order n, then B is called the multiplicative inverse matrix of … Yes, our answer would be our answer matrix, b, multiplied by the multiplicative inverse of our coefficient matrix. Now that students understand we are developing a method for finding the inverse of a matrix, I provide students with our book's brief introduction to the determinant. Email: [email protected] Tel: 800-234-2933; {{courseNav.course.mDynamicIntFields.lessonCount}} lessons Step 3:. Example $$\PageIndex{5}$$ Solve the following system \begin{aligned} 3 x+y&=3 \\ 5 x+2 y&=4 \end{aligned} Solution. ChillingEffects.org. Cryptography uses residue matrices: matrices in all elements are in Zn. 1. For example, we can use it to solve a problem like this: This matrix equation is in the form of Ax = b, where A is your coefficient matrix, x is your variable matrix, and b is your answer matrix. matrix. Set the matrix (must be square) and append the identity matrix of the same dimension to it. Let the multiplicative inverse be a. Matrix resulting from the multiplication of a matrix by its inverse Inverses of matrices Multiplicative inverse Skills Practiced. Illustrated definition of Multiplicative Inverse: Another name for Reciprocal. 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Let's multiply them out. The multiplicative inverse of a matrix is the matrix that gives you the identity matrix when multiplied by the original matrix. Varsity Tutors LLC Previous matrix calculators: Determinant of a matrix, Matrix Transpose, Matrix Multiplication, Inverse matrix calculator. Inverse Matrix Formula. Log in here for access. an 2. Consider the following example. By applying matrix multiplication to a square matrix of which we want to find the inverse and using the matrix equation AX = I to solve for X, when operations have been completed the square matrix X is the inverse matrix A −1, X = A −1, and we will have solved AA −1 = I n. Inverse of a Matrix using Gauss-Jordan Elimination. means of the most recent email address, if any, provided by such party to Varsity Tutors. Refer to the following matrices: A = (1 & 2 | 3 & -4 ), B = (5 & 0 | -6 & 7), C = (1 & -3 & 4 | 2 & 6 & -5). 1. Washington University in St Louis, Master of Science, Electrical Engineer... University of Illinois at Urbana-Champaign, Bachelor of Science, Chemical and Biomolecular Engineering. Which is equal to-- this is just a scalar, this is just a number, so we multiply it times each of the elements-- so that is equal to minus, minus, plus. The identity matrix is a square matrix containing ones down the main diagonal and zeros everywhere else. Matrix inversion is the process of finding the matrix B that satisfies the prior e… When we multiply a matrix by its inverse we get the Identity Matrix (which is like "1" for matrices): A × A -1 = I. This precalculus video tutorial explains how to determine the inverse of a 2x2 matrix. Not all matrices have an inverse. An identification of the copyright claimed to have been infringed; I have the matrix$$\begin{pmatrix} 1 & 5\\ 3 & 4 \end{pmatrix} \pmod{26}$$ and I need to find its inverse. Inverse Matrices. Following this lesson, you should be able to: To unlock this lesson you must be a Study.com Member. Multiplying the two matrices, we see that we do get the identity matrix: We know for sure now that this inverse is the real inverse, and it works for us. If you know the inverse of a matrix, you can solve the problem by multiplying the inverse of the matrix with the answer matrix, x = A sup -1 * b. The special property here is as follows: A*A^-1 = A^-1*A = I. Each transistor requires 3 units of copper, 1 unit of zinc, and 2 units of glass. Already registered? Multiplicative Inverse Property Calculator. In this section we see how Gauss-Jordan Elimination works using examples. Provide a clear justification for each answer. credit by exam that is accepted by over 1,500 colleges and universities. Watch this video lesson to learn about another method you can use to solve a matrix problem if you are given the inverse of the matrix. We find the "inverse" of 7, which is 1/7. Positive 2/7. The multiplicative inverse of a matrix A is a matrix (indicated as A−1) such that: A ⋅ A−1 = A−1 ⋅ A = I Where I is the identity matrix (made up of all zeros except on the main diagonal which contains all 1). The multiplicative inverse of a nonsingular matrixis its matrix inverse. This matrix has no inverse because the columns are not linearly independent. A statement by you: (a) that you believe in good faith that the use of the content that you claim to infringe As a result you will get the inverse calculated on the right. on or linked-to by the Website infringes your copyright, you should consider first contacting an attorney. Shortcut for 2 x 2 matrices For , the inverse can be found using this formula: Example: 2. Because when you multiply them together, you get the multiplicative identity (one). The modular multiplicative inverse of an integer a modulo m is an integer b such that, It maybe noted , where the fact that the inversion is m-modular is implicit. 8 × ( 1/8) = 1. 2. We ended up with fractions here and things. Try refreshing the page, or contact customer support. For our normal numbers, the multiplicative inverse is simply 1 divided by our number. credit-by-exam regardless of age or education level. You can use the multiplicative inverse of a matrix to solve problems in the form of Ax = b, where A is your coefficient matrix, x is your variable matrix, and b is your answer, or constant, matrix. Step 1:. the multiplicative inverse: ... Scalar Multiplication. 15 minutes. In order to find the multiplicative inverse, we have to find the matrix for which, when we multiply it with our matrix, we get the identity matrix. Step 2:. The multiplicative inverse of a matrix A is written A^(-1). This matrix has no inverse because the columns are not linearly independent. Sample Usage. MINVERSE(square_matrix) square_matrix - An array or range with an equal number of rows and columns representing a matrix whose multiplicative inverse will be calculated. However matrices can be not only two-dimensional, but also one-dimensional (vectors), so that you can multiply vectors, vector by matrix and vice versa. Precalculus : Find the Multiplicative Inverse of a Matrix Study concepts, example questions & explanations for Precalculus. The multiplicative inverse of a matrix A is written A^(-1). Get the unbiased info you need to find the right school. In arithmetic, there is one number which does not have a multiplicative inverse. So before I do that I have to create some space. x_1 + 3x_2 = 1. a flashcard set{{course.flashcardSetCoun > 1 ? the This is the reason I like best. While we can use other methods to solve such a problem, if we know the multiplicative inverse of our coefficient matrix, then we can easily solve the problem by simply multiplying both sides by the inverse. A second-order matrix can be represented by . 's' : ''}}. Step 2:. a number multiplied by it’s multiplicative inverse gives the multiplicative identity. Here are three ways to find the inverse of a matrix: 1. See Also . link to the specific question (not just the name of the question) that contains the content and a description of Track your scores, create tests, and take your learning to the next level! If we multiply matrix A by the inverse of matrix A, we will get the identity matrix, I. From the top row, we get 1(11) + -2(5) = 11 - 10 = 1. All other trademarks and copyrights are the property of their respective owners. Reduce the left matrix to row echelon form using elementary row operations for the whole matrix (including the right one). F) Find A(BC). The concept of solving systems using matrices is similar to the concept of solving simple equations. The theory, as usual, is below the calculator as The multiplicative inverse of a matrix is similar in concept, except that the product of matrix $$A$$ and its inverse $$A^{−1}$$ equals the identity matrix. By using this website, you agree to our Cookie Policy. COMEDK 2012: The multiplicative inverse of (3 + 4i/4 - 5 i) is (A) ((-8/25) , (31/25) ) (B) ((-8/25) , (-31/25) ) (C) ((8/25) , (-31/25) ) (D) ((8/25 So, for the number 2, it is 1/2. There are a couple of ways to do this. Inverse of a Matrix. The term inverse matrix generally implies the multiplicative inverse of a matrix. Create an account to start this course today. In general, the inverse of n X n matrix A can be found using this simple formula: where, Adj(A) denotes the adjoint of a matrix and, Det(A) is Determinant of matrix A. Yes, we write the inverse with a superscript of -1. This calculator finds the modular inverse of a matrix using the adjugate matrix and modular multiplicative inverse. 5/7 minus 4/7. The multiplicative inverse of a modulo m exists if and only if a and m are coprime (i.e., if gcd(a, m) = 1). Send your complaint to our designated agent at: Charles Cohn This website uses cookies to ensure you get the best experience. Hence, I is known as the identity matrix under multiplication. I don't … first two years of college and save thousands off your degree. f(g(x)) = g(f(x)) = x. Find the value of . The multiplicative inverse of a matrix is similar in concept, except that the product of matrix $A$ and its inverse ${A}^{-1}$ equals the identity matrix. You see, it is useful to learn about the multiplicative inverse of a matrix because if we know it, then we can use it to help us solve equations with matrices in them. If Varsity Tutors takes action in response to AX - BX + D = C. Use A - 1 to find the solution, (x_1, x_2), to the given system. C) Find AB. Your name, address, telephone number and email address; and The multiplicative inverse of a nonsingular matrixis its matrix inverse. Start Here; Our Story; Hire a Tutor; Upgrade to Math Mastery. Multiplying Matrices. Multiplicative Inverse Property Calculator. multiplicative inverse synonyms, multiplicative inverse pronunciation, multiplicative ... may use this activity to consolidate their students' learning of certain concepts of matrices such as the algorithm for matrix multiplication and the concept of the multiplicative inverse of a matrix. B) Find 2A + 3B. If you've found an issue with this question, please let us know. 3. is the multiplicative inverse of . The identity matrix is a square matrix containing ones down the main diagonal and zeros everywhere else. St. Louis, MO 63105. Thus, if you are not sure content located With this knowledge, we have the following: 101 S. Hanley Rd, Suite 300 Create your account. To learn more, visit our Earning Credit Page. Coolmath privacy policy. The multiplicative inverse of a matrix is the matrix that gives you the identity matrix when multiplied by the original matrix. Define multiplicative inverse. Note the first and the last columns are equal. After you multiply, you can then easily find the answer by translating back to equation form. Matrix Multiplication Calculator Here you can perform matrix multiplication with complex numbers online for free. Zero … MULTIPLICATIVE INVERSES For every nonzero real number a, there is a multiplicative inverse l/a such that. Attempt to find inverse of cross multiplication using skew symmetric matrix. I will use the determinant method. Plug the value in the formula then simplify to get the inverse of matrix C. Step 3:. or more of your copyrights, please notify us by providing a written notice (“Infringement Notice”) containing and career path that can help you find the school that's right for you. Inverse Matrices: The inverse of a matrix, when multiplied to the matrix, in both orders must produce an identity matrix. Same thing when the inverse comes first: ( 1/8) × 8 = 1. All operations on residue matrices are performed the same as for the integer matrices except that the operations are done in modular arithmetic. either the copyright owner or a person authorized to act on their behalf. The following formula is used to calculate the inverse matrix value of the original 2×2 matrix. information contained in your Infringement Notice is accurate, and (c) under penalty of perjury, that you are Amy has a master's degree in secondary education and has taught math at a public charter high school. Study.com has thousands of articles about every I find the modular multiplicative inverse (of the matrix determinant, which is $1×4-3×5=-11$) with the extended Euclid algorithm (it is $-7 \equiv 19 \pmod{26}$). your copyright is not authorized by law, or by the copyright owner or such owner’s agent; (b) that all of the The multiplicative inverse of a matrix is similar in concept, except that the product of matrix A and its inverse A –1 equals the identity matrix. I explain that today we will find the multiplicative inverse of a matrix. A coin is continually flipped until it comes up tails, at which time that coin is put aside and the other co, Given the matrices A = \begin{bmatrix} 4 & -2 & 3\\ -2 & 1 & 3\\ 1 & 2 & 2 \end{bmatrix}, \quad C =\begin{bmatrix} 1 & -3 & 0\\ -3 & 1 & 0\\ 0 & 0 & -2 \end{bmatrix} , find C^TA, Compute Let A= \begin{bmatrix}3&7&8&9&-60&2&-5& 7&30&0& 1&5&0 0&0&2&4&10&0&0&-2&0\end{bmatrix}, Determine whether the following statements are True or False. A\ ) to create some space operations are done in modular arithmetic multiplicative INVERSES for nonzero..., you multiplicative inverse of matrix test out of the matrix is the inverse matrix Determinante... Calculator - calculate matrix inverse since, the multiplicative inverse is -1/3 values we find the of! Guided Practice Practice Read and study the lesson to answer each question nur! The rows and columns take your learning to the concept of solving simple equations,. 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http://math.stackexchange.com/questions/175250/computing-a-complex-integral-potentially-using-residues | # Computing a complex integral potentially using residues
The question is:
Compute:
$$\mbox{p.v.}\int_{-\infty}^{\infty}\frac{x\sin4x}{{x^2}-1}dx$$
Initially I thought it was straight forward and I could just use residues. However, the Residue Theorem requires the poles to be in the upper plane ($y > 0$), and in this case, that is not the case. So, now I have no idea what to do since I cannot use residues.
-
p.v. = principal value. – user26872 Jul 25 '12 at 21:55
Right, I understand that. Does that change how I approach the problem though? Does that make it so that I can go ahead with computing the residues at 1 and -1? – Payton Jul 25 '12 at 22:31
See Rob and Tim's solutions below. – user26872 Jul 25 '12 at 23:32
Let $C_R$ denote the counterclockwise semicircular arc extending from $R$ to $-R$ in the plane and $C_{\rho_1} , C_{\rho_2}$ be the counterclockwise semicircular arcs extending from $-1 - \rho_1$ to $-1 + \rho_1$ and from $1 -\rho_1$ and $1+\rho_1,$ respectively. Note that $\frac{x\sin 4x}{x^2 -1} = \Im \frac{x\exp{4ix}}{x^2 -1}.$
Consider also that $\,\displaystyle{f(z) = \frac{z\exp{4iz}}{z^2 -1}}\,$ has simple poles at $-1$ and $1.$ By Jordan's Lemma, there exists $\theta \in [0, \pi ]$ such that $$\displaystyle\int_{C_R} \frac{z\exp{4iz}}{z^2-1} \le \frac{\pi }{4} \frac{1}{R^2 \exp{2i\theta }-1}$$ The right hand side of this inequality tends to zero. Using the residue theorem,
$$\int_{C_R} \frac{z\exp{4iz}}{z^2-1} + \int_{-C_{\rho_1}} \frac{z\exp{4iz}}{z^2-1} + \int_{-C_{\rho_2}} \frac{z\exp{4iz}}{z^2-1} +\int_{-R}^{-1 -\rho_1} \frac{x\exp{4ix}}{x^2-1} dx +$$ $$+\int_{-1 +\rho_1}^{1 - \rho_e} \frac{x\exp{4ix}}{x^2-1} dx +\int_{1 + \rho_1}^{R} \frac{x\exp{4ix}}{x^2-1} dx = 0$$
Letting $R$ tend to infinity and $\rho_1 \rho_2$ tend to zero, we have $$\textrm{pv }\int_{-\infty }^{\infty } \frac{x\exp{4ix}}{x^2 -1 }dx = \pi i [\textrm{ Res } (f, -1) + \textrm{ Res } (f, 1) ] = \pi i \cos 4$$ Hence $$\textrm{pv } \int_{-\infty }^{\infty } \frac{x\sin 4x}{x^2 -1 }dx = \pi \cos 4.$$
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But doesn't the Residue Theorem run into an issue when the poles are on the real line? (that is, when they aren't on the upper half plane?) – Payton Jul 25 '12 at 23:23
Yes - we avoided this by constructing an indented contour around the poles. This approach relied on the fact that the poles were simple. See the picture in RobJohn's answer for the basic idea. There are no poles for $f$ enclosed within the indented contour, so the residue theorem tells us that the contour integral in that case was zero. – user17794 Jul 26 '12 at 0:36
Consider the following diagram:
$\hspace{3.25cm}$
The principal value integral is the integral over the path in black as the paths in red get smaller and the paths in blue and green get bigger. We will evaluate this by first breaking up $\sin(4z)=\dfrac{e^{i4z}-e^{-i4z}}{2i}$ so that we can close the path of integration of each piece along a different path.
$\dfrac{e^{i4z}}{2i}$ will be integrated over the paths in black, red, and blue ($\gamma^+$). $\gamma^+$ contains the singularities at $z=-1$ and $z=1$
$\dfrac{e^{-i4z}}{2i}$ will be integrated over the paths in black, red, and green ($\gamma^-$). $\gamma^-$ contains no singularities.
The sum of the integrals described above, include the integrals over the red half circles. We will eliminate the integrals over the red half circles by subtracting half of $2\pi i$ times the residue of $\dfrac{z\sin(4z)}{z^2-1}$ at $z=-1$ and $z=1$.
Thus, we get
\begin{align} =\hspace{-11.5pt}\int_{-\infty}^\infty\frac{x\sin(4x)}{x^2-1}\mathrm{d}x &=\color{#0000FF}{\frac1{2i}\int_{\gamma^+}\frac{ze^{i4z}}{z^2-1}\mathrm{d}z} -\color{#00A000}{\frac1{2i}\int_{\gamma^-}\frac{ze^{-i4z}}{z^2-1}\mathrm{d}z}\\ &-\color{#C00000}{\pi i\mathrm{Res}_{z=-1}\left(\frac{z\sin(4z)}{z^2-1}\right)} -\color{#C00000}{\pi i\mathrm{Res}_{z=1}\left(\frac{z\sin(4z)}{z^2-1}\right)}\\ &=\color{#0000FF}{\frac{2\pi i}{2i}\mathrm{Res}_{z=-1}\left(\frac{ze^{i4z}}{z^2-1}\right)} +\color{#0000FF}{\frac{2\pi i}{2i}\mathrm{Res}_{z=1}\left(\frac{ze^{i4z}}{z^2-1}\right)}\\ &-\color{#C00000}{\pi i\mathrm{Res}_{z=-1}\left(\frac{z\sin(4z)}{z^2-1}\right)} -\color{#C00000}{\pi i\mathrm{Res}_{z=1}\left(\frac{z\sin(4z)}{z^2-1}\right)}\\ &=\color{#0000FF}{\pi\frac{e^{-i4}}{2}}+\color{#0000FF}{\pi\frac{e^{i4}}{2}}\\ &-\color{#C00000}{\pi i\frac{\sin(-4)}{2}}-\color{#C00000}{\pi i\frac{\sin(4)}{2}}\\[6pt] &=\pi\cos(4) \end{align}
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I have to ask, what program did you use to draw the diagram? – user2468 Jul 26 '12 at 2:07
I used Intaglio for the Mac. – robjohn Jul 26 '12 at 5:29 | 2015-08-02T20:53:37 | {
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https://math.stackexchange.com/questions/3055017/how-the-two-non-null-homotopic-equivalence-classes-generate-the-null-homotopic-l | # How the two non null-homotopic equivalence classes generate the null-homotopic loop on the torus
I am new in Alebraic Topology. Given the torus, we say that the fundamental group of the torus is generated by two loops (or more exactly two equivalent classes of loops). One writes $$\pi=\mathbb{Z}\times\mathbb{Z}.$$
I don't understand, how the null-homotopic loop, which is the constant loop, is generated by the two generators mentioned above. Can somebody provide an explanation? More even so, I don't see how it functions visually, since the two generators are not null-homotopic. More precisely, given a null-homotopic loop on the surface on a base point $$x$$, how this loop will be generated by the two generators mentioned above?
• The identity is in every subgroup. – Angina Seng Dec 28 '18 at 16:06
• Thanks. But how it is generated in the case of a null-homotopic loop ? In the case the loops are traversing the torus around the hole or through the hole, one can say for instance that the product of one loop and its inverse is the constant loop, that is the identity. But I dont see how it works in the case a loop is null-homotopic, since the generators are non null-homotopic. – user249018 Dec 28 '18 at 16:14
• What do you think the word "generate" means? – Eric Wofsey Dec 28 '18 at 16:20
Let $$\mathbb{T}^2$$ denote the torus and choose a basepoint $$p \in \mathbb{T}^2$$. Then we know that $$\pi_1\left(\mathbb{T}^2, p \right) \cong \mathbb{Z} \times \mathbb{Z}$$.
Now I think the reason for your confusion is an algebraic one.
Recall that $$\mathbb{Z} \times \mathbb{Z}$$ has two generators, $$a= (1, 0)$$ and $$b =(0, 1)$$. Choose an isomorphism $$\psi : \pi_1\left(\mathbb{T}^2, p \right) \to \mathbb{Z} \times \mathbb{Z}$$, by surjectivity there exists path classes, $$[f], [g] \in \pi_1\left(\mathbb{T}^2, p \right)$$ such that $$\psi([f]) = a$$ and $$\psi([g]) =b$$. Then since $$\psi$$ is an isomorphism we have $$[f]$$ and $$[g]$$ to be the two generators of $$\pi_1\left(\mathbb{T}^2, p \right)$$.
Now your question is how the path class of the constant loop $$c_p : I \to \mathbb{T}^2$$ defined by $$c_p(x) = p$$ for all $$x \in I$$, that being $$[c_p] \in \pi_1\left(\mathbb{T}^2, p \right)$$ is generated by $$[f]$$ and $$[g]$$. Well the answer to that is simple: note that $$[c_p] = 1_{\pi_1\left(\mathbb{T}^2, p \right)}$$ that is $$[c_p]$$ is the identity element of $$\pi_1\left(\mathbb{T}^2, p \right)$$. Then recall the following definition that we have for exponents in groups.
Definition: In any group $$(G, \cdot)$$ for any $$x \in G$$ we define $$x^0 = 1_G$$ where $$1_G$$ is the identity element of the group $$(G, \cdot)$$.
Hence since $$[f], [g] \in \pi_1\left(\mathbb{T}^2, p \right)$$ and $$\pi_1\left(\mathbb{T}^2, p \right)$$ is indeed a group, we have $$[f]^0 = [g]^0 = 1_{\pi_1\left(\mathbb{T}^2, p \right)}.$$
Then we have $$\left[c_p\right] = [f]^0 * [g]^0$$ and so the constant path at $$p$$ is indeed generated by the two generators of $$\pi_1\left(\mathbb{T}^2, p \right)$$. And since $$[c_p]$$ is a nullhomotopic loop, since it is a constant loop by definition, the above shows how a product of two non null-homotopic loops yield a null-homotopic loop.
Note that above even though I've gone into quite a bit of detail, the only real fact I'm using is the following algebraic one. If we have a group $$G$$ and we have $$G = \langle A \rangle$$ for some subset $$A \subseteq G$$ then every element $$x \in G$$ can be written as $$x = g_1 \dots g_n \cdot h_1^{-1} \dots h_m^{-1}$$ where $$g_i, h_i \in G$$. In particular if we have $$G = \langle c , d \rangle$$, that is $$G$$ is generated by the two elements $$c$$ and $$d$$ then we can express $$1_G$$ as $$1_G = c^0 \cdot d^0$$.
Your confusion seems to be about the meaning of the word "generate". By definition, if $$G$$ is a group and $$S\subseteq G$$, then the subgroup generated by $$S$$ is the smallest subgroup that contains $$S$$. Since a subgroup always contains the identity element, any subset of $$G$$ (even the empty set!) "generates" the identity element.
• Thanks. By ''generate'' I meant the group operation on the set of generators. In a group generated by the subset $S$, each element can be written in terms of the generators, which is, elements of $S$. So given the null-homotopic loop on the torus, how can it be put in relation to the elements of $S$, which in our case consists of 2 elements ? – user249018 Dec 28 '18 at 16:36
• OK, but the group multiplication is not the only operation in a group! There are two other operations: the identity element and inverses. – Eric Wofsey Dec 28 '18 at 16:39
• In particular, one of the operations of a group is an operation which takes no inputs and outputs the identity element. That's how any set "generates" the identity element. – Eric Wofsey Dec 28 '18 at 16:40
• Given $S\subset G$, more precisely the group generated by $S$ is defined as $<S>=SS^{-1}$. So you are right about inverses. The thing with the identity element is less obvious. One excepts it very probably by definition...But when it comes to the fundamental group of the torus, are you saying that the null-homotopic loop is generated by the empty set ? Or maybe we can say the following: the multiplication of a geneartor and its inverse gives us the constant loop, which itself is homotopic to null-homotopic loops. Thus we generate the null-homotopic loop from each one of the two generators ? – user249018 Dec 28 '18 at 16:55
• I don't know what your notation $SS^{-1}$ is supposed to mean, but it sounds like your definition of "the group generated by $S$" is just wrong (which may not be your fault; you may have been taught a wrong definition!). The correct definition is the one I stated in the answer. An equivalent definition is that the subgroup generated by $S$ is the set of all elements of $G$ that can be obtained by starting with elements of $S$ and repeatedly applying the three operations of the group multiplication, inverses, and the identity element. – Eric Wofsey Dec 28 '18 at 17:10
The subgroup generated by $$a$$ and $$b$$ is the set of all elements that can be written as a sequence that consists of nothing but $$a$$, $$b$$, and their inverses (e.g $$ab$$ or $$b^{-4}a$$). The null sequence is allowed. That is, the empty string (a zero-length sequence) qualifies as "a sequence that consists of nothing but $$a$$, $$b$$, and their inverses"; it does not contain anything, so clearly it does not contain anything other than $$a$$, $$b$$, and their inverses. In an abelian group with two generators, the group is generated by taking the first generator an integer number of times, and then taking the second generator an integer number of times. And zero is an integer. Given two non null-homotopic loops $$a$$ and $$b$$, the constant loop is generated by taking $$a$$ zero times, then taking $$b$$ zero times. If you think of a group in terms of group actions, the identity is generated by not doing anything. Doing nothing at all is, at least as far as mathematicians are concerned, an action. Or, in the words of Geddy Lee, if you choose not to decide, you still have made a choice. | 2020-10-20T06:46:49 | {
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http://mathhelpforum.com/calculus/47065-trig-intergrals.html | 1. ## [SOLVED] trig intergrals
So, this is also probably one of those easy problems, I just don't have a good example to work this out by.
$
\int\cos^2x$
where b= $\pi/2$ a= 0
answer= $\pi/4$
I know I'm suppose to use the half angle identity, i just must be doing it wrong because that's not the answer I'm getting.
Thanks for all your help ^.^
2. Can you show your work to see what it is that you're doing wrong?
$\int_{0}^{\frac{\pi}{2}} \cos^2 x \: dx = \int_{0}^{\frac{\pi}{2}} \left[\frac{1}{2} (1 + \cos 2x)\right]dx = \frac{1}{2} \int_{0}^{\frac{\pi}{2}} \left[1 + \cos 2x\right]dx =$ ${\color{white}.} \: \frac{1}{2} \left(x + \frac{1}{2}\sin 2x\right) \bigg|_{0}^{\frac{\pi}{2}} = \hdots$
3. $cos^2x = \frac{1 + cos2x}{2}$
$\frac{1}{2}\int{dx} +\frac{1}{2}\int{cos2x}{dx} = \frac{x}{2} +\frac{sin2x}{4}$
Hope this helps.
Oops was to slow lol, just wondering how do you place upper and lower bounds in latex?
4. Originally Posted by 11rdc11
$cos^2x = \frac{1 + cos2x}{2}$
$\frac{1}{2}\int{dx} +\frac{1}{2}\int{cos2x}{dx} = \frac{x}{2} +\frac{sin2x}{4}$
Hope this helps.
Oops was to slow lol, just wondering how do you place upper and lower bounds in latex?
Do:
Code:
$$\int_a^b f(x)\,dx$$
To get $\int_a^b f(x)\,dx$
note if we have limits like $-\ln2$ to $\ln2$, then you need to put these within {} such as:
Code:
$$\int_{-\ln(2)}^{\ln2} e^x\,dx$$
will output $\int_{-\ln(2)}^{\ln2} e^x\,dx$
--Chris
5. $\int_{0}^{4} x \: dx$
is produced by: $$\int_{0}^{4} x \: dx$$
_{0} : produces the lower bound
^{4} : produces the upper bound.
This can applied to other symbols as well:
$\sum_{k = 0}^{4} k = 10$ is produced by $$\sum_{k = 0}^{4} k = 10$$
$\frac{x}{2}\bigg|_{0}^{4}$
etc etc.
6. Originally Posted by o_O
Can you show your work to see what it is that you're doing wrong?
$\int_{0}^{\frac{\pi}{2}} \cos^2 x \: dx = \int_{0}^{\frac{\pi}{2}} \left[\frac{1}{2} (1 + \cos 2x)\right]dx = \frac{1}{2} \int_{0}^{\frac{\pi}{2}} \left[1 + \cos 2x\right]dx =$ ${\color{white}.} \: \frac{1}{2} \left(x + \frac{1}{2}\sin 2x\right) \bigg|_{0}^{\frac{\pi}{2}} = \hdots$
Whoa, I totally was thinking that the anti derivative of 1 was 0. That's where I messed up. Silly mistake. Thanks though, I'll remember to post my work next time.
7. $\int_{0}^{\frac{\pi}{2}} \cos^2 x \: dx = \int_{0}^{\frac{\pi}{2}} \left[\frac{1}{2} (1 + \cos 2x)\right]dx = \frac{1}{2} \int_{0}^{\frac{\pi}{2}} \left[1 + \cos 2x\right]dx =$ ${\color{white}.} \: \frac{1}{2} \left(x + \frac{1}{2}\sin 2x\right) \bigg|_{0}^{\frac{\pi}{2}} = \hdots$
There's an easier way to deal with these kind of integrals.
Rule: $\int_a^b f(x)~dx = \int_a^b f(a+b-x)~dx$
I'll solve this step by step for you to understand how it works. It actually takes only a nanosecond to solve this.
$I = \int_0^\frac{\pi}{2}\cos^2x~dx = \int_0^\frac{\pi}{2}\cos^2(\frac{\pi}{2}-x)~dx = \int_0^\frac{\pi}{2}\sin^2x~dx$
Now, $I+I = \int_0^\frac{\pi}{2}\cos^2x~dx + \int_0^\frac{\pi}{2}\sin^2x~dx = \int_0^{\frac{\pi}{2}}1~dx = \frac{\pi}{2}$
$I = \frac{\pi}{4}$ | 2016-08-30T07:39:45 | {
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https://math.stackexchange.com/questions/1884031/how-rigorous-are-pictorial-proofs | # How rigorous are pictorial proofs?
The standard proof that $|\mathbb{Q}| = \mathbb{|N|}$ is pictorial. I am sure everyone here has seen it. The "zig-zag". I must admit, however, that, although I was "intuitively" convinced by it, I was never entirely satisfied with it because it is not an explicit bijection $f:\mathbb{N} \to \mathbb{Q}$ given by an actual formula. The fact that the proof is correct seems "clear" to us, but this is, again, merely an appeal to intuition. One should note that some of these "proofs by picture" are simply incorrect: see Russell O'Connor's answer here .
I have two questions
Is the pictorial proof that $|\mathbb{Q}| = \mathbb{|N|}$ rigorous by the standards of modern pure mathematics?
For the sake of this question, suppose that there isn't an explicit formla, or that it's too unwieldy to use in practice. After all, even if there is a formula, most of the people who've seen the pictorial argument do not know of it.
Is there an explicit formula for the "pictorial" proof?
There's some minor issues, of course, namely the inclusion of $0$ and variations of the "zig-zag" path, but these are no big deal. A bijection $f:\mathbb{N} \to \mathbb{N} \times \mathbb{N}$ suffices; dealing with negatives, equivalent fractions, etc is trivial.
• The zig-zag proof is not pictorial. You can represent it in a graph, as any function, this is all. Aug 6, 2016 at 6:52
• Yes, there is an explicit function for the bijection. But it's tedious, looks convuluted and it distracts from the purpose of the argument which should be simple. For n,m find the largest $\sum_{i=1}^k = k (k+1)/2 \le n+m$ then (n,m) -> k (k+1)/2 + (n-k) is the bijection. Aug 6, 2016 at 6:56
• That should be sum < n; not <= n+m. It helps to draw a picture but: 1 => 0,0 then 2-3 => (1,0) - (0,1) and so on till $(\sum )+1$ = k (k+1)/2 +1 through $\sum$ + (k+1) maps to (k,0), (k-1,1), (k-2,2).... (0,k). That's the bijection. But it's tedious and not particularly relevant. Aug 6, 2016 at 7:21
• Perhaps the easiest way is to let $a_k = \sum_{i=1} i$, then it's easy to show ever natural $n$ can be written uniquely as $a_k + i; 0 \le i < k+1$ so $n \rightarrow (k -1,i)$ is 1 to 1. Aug 6, 2016 at 7:36
• A pictorial proof isn't rigorous enough but a formulaic calculation isn't nescessary if a descriptive argument can be shown to be unambiguous and consistsnt. The diagonal description as usually presented is ... borderline IMO. But if we explain we can count and group the naturals in groups each group with one more than the previous, i.e. (1)(2,3)(4,5,6)(7,8,9,10)etc. then each group has the same number of members as each of the diagonals that also increases by one, that could be rigorous enough. Aug 6, 2016 at 7:56
I had the same question with the same proof (the zig-zag proof you are mentioning). At some point I decided to produce a formal proof.
Define a bijective function $f\colon \mathbb N \times \mathbb N \to \mathbb N$.
First of all you notice that going zig and then zag only helps intuition. In fact you don't need a "continuous" curve, so it is easier to go zig and the zig again... (so to speak). Then it is easy to count how many points you need to fill the first $k$ diagonals (sum of a arithmetic series: $k(k+1)/2$). The couple $(n,m)$ lies on the diagonal number $k=n+m$ so you easily find: $$f(n,m) = \frac{(n+m)(n+m+1)}{2} + m = \frac{n^2+m^2+2nm+3m+n}{2}.$$
This was a little bit shocking to me! The function I was looking for is as simple as a polynomial... I would have expected some modulus, or some strange discontinuous function.
Nevertheless the algebraic proof that $f$ is bijective is not so simple... but following the intuition of the construction it is easy to write it.
What can we learn from this? The pictorial proof is for sure the best to understand a result and to remember it. Then it might happen that the abstract mathematics is even simpler than our intuition. Not always simple mathematics corresponds to simple pictures.
• However it's not necessary to come up with an explicit formula to prove the function is a bijection. E.g., an iterative or recursive description can suffice, and this is basically what the zig-zag picture does. Aug 6, 2016 at 11:01
• @Kinball, it suffices provided it suffices. Having a fórmula makes it easy to be sure. Aug 6, 2016 at 19:46
I suppose if you really wanted to, you could come up with an explicit bijection associated with the "zig-zag" proof. If that turns out to be difficult, you could come up with a different "zig-zag" that may have a simpler bijection. Although, the "zig-zag" proof is really just providing some intuitive backing to this theorem:
The union of a countable number of finite sets is countable.
Thinking of each diagonal of the "zig-zag" as one of your finite sets, and noting that every rational number has to be in one of those diagonals is sufficient to prove that $|\mathbb{Q}| = |\mathbb{N}|$.
On a more general note, the whole point of a proof is to clearly and correctly convey why a theorem is true. Sometimes it is easiest to convey why through written words, especially when the proof is long, relies on lemmas or the theorems of others, or just has lots of cases. But if there is a clever reason why a theorem is true, some clever "ah-HA" that you just have to see, a "proof by picture" can be much more clear than a formal write-up of a proof. The hope is, though, that after a reader sees a pictorial proof, they should have enough intuition into why the theorem is true to write up a formal proof if they really needed.
After seeing the "zig-zag" proof, do you think you can prove that the union of a countable number of finite sets is countable?
• What about a zig-zag proof that the union of a countable number of countable sets is countable? =P Aug 6, 2016 at 10:28
• @user21820, Every time I've heard the zig-zag proof it is presented as partitioning the rational numbers into finite sets indexed by the sum of their numerator and denominator. But I suppose you can think of them as being infinite sets (indexed by just the numerator or denominator), and that makes sense with the same picture. So I suppose that in this case, the same pictorial proof can inspire distinct formal proofs. :) Aug 6, 2016 at 22:23
• Well I was trying to poke fun at the rigour of such a pictorial proof because it easily makes one think that it works the same for the union of countably many countable sets. It doesn't quite because you need the axiom of countable choice. Personally I like annotating parts of a diagram with quantifiers labelled by the order of quantification, and it would prevent such an error because we would realize that we need $\exists_1,\exists_2,...$. Aug 7, 2016 at 8:40
• @user21820,Mike Actually you need the axiom of countable choice even to prove that the union of countably many finite sets is countable. Is it obvious where countable choice comes in in the pictorial proof? It's in the picture-making itself: you need to place dots on the page, which entails a choice of ordering for the set which is relevant to the "zig"; and such a choice must be made independently for each of the countably many finite sets. Countable choice is not needed for $|\Bbb N|=|\Bbb Q|$ because the ordering is known in advance. (And this is why I don't like pictorial "proofs".) Aug 9, 2016 at 18:50
Munkres' "Topology" gives both the zig-zag intuition, and the actual formula for the bijection, as a good reference.
I read in an article (cannot remember the author, sadly) that proofs are not supposed to be 'entirely' rigorous. 'Proofs' try to convince the reader that it is possible to construct a completely rigorous proof. As in this case, although I too was not satisfied with the zig-zag-proof, it conveys the idea that the bijection does exists (without explicitly writing it out), and thus it is countable.
The point of something like the zig-zag proof is not to be rigorous in itself but instead to convince a mathematician that he or she could easily make the proof rigorous if pressed to do so by the gods of rigour.
Also, you can come up with very succinct surjections from N to Q. For example, map every natural number of the form $2^p 3^q 5^r$ to $(-1)^r p/q$. It's extremely easy now to see that if you give me a rational, there is a natural number which is mapped to it.
The downside with this is that if you are teaching countability you would need to check that "surjection from" is the same as "bijection with", which sometimes hasn't been proved by this stage.
• Except that "surjection from" is not the same as "bijection with". To wit, your function is one but not the other. You can use it as a step in the way to proving a bijection, but that is hardly "the same as". Aug 7, 2016 at 10:03
• Well of course I don't mean all surjections are bijections? For an infinite set $S$, the existence of a surjection $g : \mathbb{N} \to S$ is equivalent to the existence of a bijection betwwen $S$ and $\mathbb{N}$. Aug 7, 2016 at 16:57
Pictures have their strengths and weaknesses, but they're just as rigorous as any other informal type of proof — that is, they may or may not be depending on how well written it is.
And the zig-zag proof is a rather clear depiction of an explicit algorithm for enumerating the rationals.
Regarding your particular doubts, I don't think it's really the picture that's the issue — it's that the function was defined by an algorithm for producing its values, rather than as an arithmetic formula.
Formal proofs can be 'pictures' too; you can develop formal logic in a way where the basic 'data type' is something other than strings of symbols. e.g. graphs of various sorts are often useful.
This doesn't directly address the proof of $|\mathbb{N}|=|\mathbb{Q}|$, but the more abstract notion of pictorial proofs as found in elementary geometry:
Avigad, Dean & Mumma A Formal System for Euclid's Elements: http://repository.cmu.edu/philosophy/61/
They show that certain types of pictorial arguments that occur in Euclid's Elements, while apparently un-rigorous, can be described in a precise manner using formal rules, in a way that the conclusion follows directly from the pictorial "special case". In essence, a particular diagram can be in "general enough position" to allow rigorous conclusions to be drawn from it.
So... take the ordered pair $(n,m)$.
The pictorial and non rigorous concept of the "diagonal" is simply {$(j,k) | j+k=n+m; j\ge 0;k \ge 0$}. Note: there are $(n+m)+1$ terms in this diagonal
The previous diagonals had 1 term, two terms ... so on to $(n+m)$ terms. So the previous diagonals account for $\sum_{i=1}^{n+m} i = \frac {(n+m)(n+m+1)}{2}$ items.
Starting at $(0,n+m)$ end of the diagonal $(n,m)$ is the $n +1$th item of the current diagonal.
So the bijection you want is $(n,m)\rightarrow \frac {(n+m)(n+m+1)}2 + n$.
$(0,0)\implies 0$
$(0,1)\implies 1$
$(1,0) \implies 2$
$(2,0) \implies 3$
$(1,1)\implies 4$
Etc. Algebraically it's probably easy to show this is injective . Suppose (a,b) and (c,d) both map to t.
Case 1: a+b = c+d= K.
Then $\frac {K (K+1)}{2} + a = \frac {K (K+1)}2 + b$. So $a = c$. So $b = c =K -a$.
Case 2: Wolog a+b < c+d
$\frac {(a+b)(a+b+1)}2 + a \le \frac {(a+b)(a+b+1)}2 + (a+b)$
$=\frac {(a+b)(a+b+3)}2$
$\le \frac {((c+d)-1)((c+d+1)+1)}2$
$=\frac {(c+d)(c+d +1) +(c+d) -(c+d+1)-1}2$
$=\frac{(c+d)(c+d+1)}2 -1$
$< \frac{(c+d)(c+d+1)}2 + c$
So (a,b)=(c,d).
Proving it's surjective isn't nescessary though it is.
For each t there exist, k, so that $\frac {k (k+1)}2 \le t < \frac{(k+1)(k+2)}2$.
Let $n = t -\frac {k (k+1)}2 \ge 0$. Let $m=k-n = k - t + \frac {k (k+1)}2 > k - \frac {(k+1)(k+2)}2 + \frac {k (k+1)}2=k +\frac {k+1}2 (k-(k+2))=k - (k+1)=-1$. So $m \ge 0$
Then $(n,m)\rightarrow t$.
===
Anyway....
I don't think such an intensive arithmetic is nescessary. What matters is the argument that it can be done. As the diagonals increase by one each iteration, and as we can group the natural numbers into sequences of groups each increasing in terms by one, each group of natural numbers coresponds to a diagonal, and each of there natural numbers in the group corresponds to a term in the diagonal. Thus must be shown but it needn't be shown by convoluted arithmetic. | 2022-06-26T04:52:41 | {
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# What is the last digit 3^{3^3} ? A. 1 B. 3 C. 6 D. 7 E. 9
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What is the last digit 3^{3^3} ? A. 1 B. 3 C. 6 D. 7 E. 9 [#permalink] 23 Sep 2008, 06:17
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What is the last digit $$3^{3^3}$$ ?
A. 1
B. 3
C. 6
D. 7
E. 9
OA is D (7).
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Re: last digit of a power [#permalink] 24 Sep 2008, 05:50
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Lets take another example
Find the last digit of 122^94
A. 2
B. 4
C. 6
D. 8
E. 9
Now the last digit of 122 is 2. We require only this number to determine the last digit of 122 raised to a positive power.
so the problem is essentially reduced to find the last digit of 2^94.
Now we know 2 has a cyclicity of 4. So we divide 94 by 4. The remainder for 94/4 is 2.
so last digit of 2^94 is same as that of 2^2 which is 4.
so last digit of 122^94 is 4
Remember:
1) Numbers 2,3,7 and 8 have a cyclicity of 4
2) Numbers 0,1,5 and 6 have a cyclicity of 1 (ie) all the powers will have the same unit digit. eg. 5^245 will have "5" as unit digit, 5^2000 will aslo have "5" as unit digit. Same holds for 0,1 and 6
3) If 4 is the number in the unit place of the base number then the unit digit will be "4" if the power is odd and it will be "6" if the power is even. eg. 4^123 will have unit digit of 4 since 123 is odd.
4) Similarly, for 9 the unit digit will be "9" for odd powers and "1" for even powers. eg 9^234 has unit digit as "1" since 234 is even.
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Re: last digit of a power [#permalink] 24 Sep 2008, 05:41
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elmagnifico wrote:
amitdgr wrote:
elmagnifico wrote:
What is the last digit $$3^{3^3}$$ ?
$$3^{3^3}$$ should be taken as 3^27 ?
in that case divide 27 by 4. The remainder is 3. Now the last digit of 3^27 is same as that of 3^3 and that happens to be 7.
So 7 is the right answer.
why" Now the last digit of 3^27 is same as that of 3^3 and that happens to be 7."
where do you get this from
$$3^1$$ =3,
$$3^2$$ =9,
$$3^3$$ =27,
$$3^4$$ =81,
$$3^5$$ =243,
$$3^6$$ =729, and so on .....
Now it is not humanly possible to remember all the numbers till $$3^27$$
If you have noticed in the above series the last digit repeats after every 4 terms
the last digit is same for $$3^5$$ and $$3^1$$
the last digit is same for $$3^6$$ and $$3^2$$
If 3 is the unit digit of a number then the unit digit repeats every fourth consecutive term.For our convenience here, lets call it cyclicity. So 3 has a cyclicity of 4.
To find the unit digit of a number having 3 as its last digit and raised to a positive power, divide the power by 4 and find the remainder.
If the remainder is 1 then the unit digit is same as of the unit digit of $$3^1$$
If the remainder is 2 then the unit digit is same as of the unit digit of $$3^2$$ and so on .....
Note that if the remainder is "0" then the unit digit is same as $$3^4$$ since the cyclicity is 4.
Also remember that the numbers 2,3,7 and 8 have cyclicity of 4
in our problem above we have 3^27
3 has a cyclicity of 4 so divide the number 27 by 4. We get a remainder of 3. Now as per cyclicity the last digit of 3^27 is same as that of 3^3. 3^3 is 27 so the last digit of 3^27 is 7.
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Last edited by amitdgr on 24 Sep 2008, 05:58, edited 1 time in total.
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Re: last digit of a power [#permalink] 23 Sep 2008, 06:32
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elmagnifico wrote:
What is the last digit $$3^{3^3}$$ ?
* 1
* 3
* 6
* 7
* 9
3^3 = 27
the last digit of $$3^{3^3}$$ will be the last digit of (27)^3.
now the last digit of 27*27*27 will be the same as last digit of 7*7*7 = 343 (ie) 3
Another example we can use here to understand this concept better (I made this example up)
What is the last digit of 39*87*81?
A. 2
B. 3
C. 4
D. 5
E. 6
To find the last digit of 39*87*81. All we have to do is, multiply the last digits of 39, 87 and 81
when we multiply, 9*7 we get 63. Now multiply the last digits of 63 and 81, i.e. 3*1 we get 3
so the last digit of 39*87*81 will be 3.
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Re: last digit of a power [#permalink] 05 Aug 2009, 14:11
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Here's another way of looking at it !
Here the given number is $$(xyz)^n$$
z is the last digit of the base.
n is the index
To find out the last digit in $$(xyz)^n$$, the following steps are to be followed.
Divide the index (n) by 4, then
Case I
If remainder = 0
then check if z is odd (except 5), then last digit = 1
and if z is even then last digit = 6
Case II
If remainder = 1, then required last digit = last digit of the base (i.e. z)
If remainder = 2, then required last digit = last digit of the base $$(z)^2$$
If remainder = 3, then required last digit = last digit of the base $$(z)^3$$
Note : If z = 5, then the last digit in the product = 5
Example:
Find the last digit in (295073)^130
Solution: Dividing 130 by 4, the remainder = 2
Refering to Case II, the required last digit is the last digit of $$(z)^2$$, ie $$(3)^2$$ = 9 , (because z = 3)
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Re: last digit of a power [#permalink] 24 Sep 2008, 05:22
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sorry guys. Its my mistake.
$${3}^{3^3}$$ can't be taken as $$27^3..$$It should be $$3^{27}$$
$$3^9 = 27^3$$
3^1=3
3^2=9
3^3=27
3^4=81.
..
..
3^27 = ....7
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Re: last digit of a power [#permalink] 24 Sep 2008, 07:25
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chayanika wrote:
amitdgr wrote:
Lets take another example
Find the last digit of 122^94
A. 2
B. 4
C. 6
D. 8
E. 9
Now the last digit of 122 is 2. We require only this number to determine the last digit of 122 raised to a positive power.
so the problem is essentially reduced to find the last digit of 2^94.
Now we know 2 has a cyclicity of 4. So we divide 94 by 4. The remainder for 94/4 is 2.
so last digit of 2^94 is same as that of 2^2 which is 4.
so last digit of 122^94 is 4
Remember:
1) Numbers 2,3,7 and 8 have a cyclicity of 4
2) Numbers 0,1,5 and 6 have a cyclicity of 1 (ie) all the powers will have the same unit digit. eg. 5^245 will have "5" as unit digit, 5^2000 will aslo have "5" as unit digit. Same holds for 0,1 and 6
3) If 4 is the number in the unit place of the base number then the unit digit will be "4" if the power is odd and it will be "6" if the power is even. eg. 4^123 will have unit digit of 4 since 123 is odd.
4) Similarly, for 9 the unit digit will be "9" for odd powers and "1" for even powers. eg 9^234 has unit digit as "1" since 234 is even.
Wow !! Awesome I tried out this thing with a few numbers and matched the results with my scientific calculator. This method gives perfect answers.
You deserve at least a dozen KUDOS for typing out all this patiently and sharing this knowledge with all of us.
+1 from me. Guys pour in Kudos for this
Chayanika
i agree with you. he deserves many kudos.
thanks a million.
the OA is 7 indeed. what a wonderful explanation.
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Re: last digit of a power [#permalink] 24 Sep 2008, 06:04
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amitdgr wrote:
Lets take another example
Find the last digit of 122^94
A. 2
B. 4
C. 6
D. 8
E. 9
Now the last digit of 122 is 2. We require only this number to determine the last digit of 122 raised to a positive power.
so the problem is essentially reduced to find the last digit of 2^94.
Now we know 2 has a cyclicity of 4. So we divide 94 by 4. The remainder for 94/4 is 2.
so last digit of 2^94 is same as that of 2^2 which is 4.
so last digit of 122^94 is 4
Remember:
1) Numbers 2,3,7 and 8 have a cyclicity of 4
2) Numbers 0,1,5 and 6 have a cyclicity of 1 (ie) all the powers will have the same unit digit. eg. 5^245 will have "5" as unit digit, 5^2000 will aslo have "5" as unit digit. Same holds for 0,1 and 6
3) If 4 is the number in the unit place of the base number then the unit digit will be "4" if the power is odd and it will be "6" if the power is even. eg. 4^123 will have unit digit of 4 since 123 is odd.
4) Similarly, for 9 the unit digit will be "9" for odd powers and "1" for even powers. eg 9^234 has unit digit as "1" since 234 is even.
Wow !! Awesome I tried out this thing with a few numbers and matched the results with my scientific calculator. This method gives perfect answers.
You deserve at least a dozen KUDOS for typing out all this patiently and sharing this knowledge with all of us.
+1 from me. Guys pour in Kudos for this
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Re: last digit of a power [#permalink] 24 Sep 2008, 07:27
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elmagnifico wrote:
amitdgr wrote:
elmagnifico wrote:
What is the last digit $$3^{3^3}$$ ?
$$3^{3^3}$$ should be taken as 3^27 ?
in that case divide 27 by 4. The remainder is 3. Now the last digit of 3^27 is same as that of 3^3 and that happens to be 7.
So 7 is the right answer.
why" Now the last digit of 3^27 is same as that of 3^3 and that happens to be 7."
where do you get this from
$$3^1$$ =3,
$$3^2$$ =9,
$$3^3$$ =27,
$$3^4$$ =81,
$$3^5$$ =243,
$$3^6$$ =729, and so on .....
Now it is not humanly possible to remember all the numbers till $$3^27$$
If you have noticed in the above series the last digit repeats after every 4 terms
the last digit is same for $$3^5$$ and $$3^1$$
the last digit is same for $$3^6$$ and $$3^2$$
If 3 is the unit digit of a number then the unit digit repeats every fourth consecutive term.For our convenience here, lets call it cyclicity. So 3 has a cyclicity of 4.
To find the unit digit of a number having 3 as its last digit and raised to a positive power, divide the power by 4 and find the remainder.
If the remainder is 1 then the unit digit is same as of the unit digit of $$3^1$$
If the remainder is 2 then the unit digit is same as of the unit digit of $$3^2$$ and so on .....
Note that if the remainder is "0" then the unit digit is same as $$3^4$$ since the cyclicity is 4.
Also remember that the numbers 2,3,7 and 8 have cyclicity of 4
in our problem above we have 3^27
3 has a cyclicity of 4 so divide the number 27 by 4. We get a remainder of 3. Now as per cyclicity the last digit of 3^27 is same as that of 3^3. 3^3 is 27 so the last digit of 3^27 is 7.[/quote]
by the way, i have never had such a GMAT rush. it is like a sugar rush.
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Re: last digit of a power [#permalink] 23 Sep 2008, 20:10
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elmagnifico wrote:
What is the last digit $$3^{3^3}$$ ?
$$3^{3^3}$$ should be taken as 3^27 ?
in that case divide 27 by 4. The remainder is 3. Now the last digit of 3^27 is same as that of 3^3 and that happens to be 7.
So 7 is the right answer.
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Re: last digit of a power [#permalink] 05 Aug 2009, 19:25
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3^3^3 to solve this we have to take top down approach...we cannot deduce 27^3......it should be 3^27....hence answer is 7
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Re: last digit of a power [#permalink] 23 Sep 2008, 06:23
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3^27
so lets see 27mod4=3..
3^1=3
3^2=9
3^3=7
Unit digit is 7
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Re: last digit of a power [#permalink] 23 Sep 2008, 06:29
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elmagnifico wrote:
What is the last digit $$3^{3^3}$$ ?
* 1
* 3
* 6
* 7
* 9
27^3 ....--> 7*7*7 --> unit digit xx3 ..
B
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Re: last digit of a power [#permalink] 23 Sep 2008, 11:20
1
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is the above mentioned method right ?
thanks
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Re: last digit of a power [#permalink] 23 Sep 2008, 13:59
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i dnt think that we can reduce the above statement into 27^3 becoz that is (3^3)^3 which is not what the question says...........
so the answer has to be 7
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Re: last digit of a power [#permalink] 24 Sep 2008, 04:24
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amitdgr wrote:
elmagnifico wrote:
What is the last digit $$3^{3^3}$$ ?
$$3^{3^3}$$ should be taken as 3^27 ?
in that case divide 27 by 4. The remainder is 3. Now the last digit of 3^27 is same as that of 3^3 and that happens to be 7.
So 7 is the right answer.
why" Now the last digit of 3^27 is same as that of 3^3 and that happens to be 7."
where do you get this from
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Re: last digit of a power [#permalink] 24 Sep 2008, 06:56
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its 3
(3^3)^3 = 19683
whats the OA!?
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Re: last digit of a power [#permalink] 24 Sep 2008, 08:06
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Very neat method amitdgr +2 from me
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Re: last digit of a power [#permalink] 26 Jul 2009, 20:14
1
KUDOS
Quote:
Remember:
1) Numbers 2,3,7 and 8 have a cyclicity of 4
2) Numbers 0,1,5 and 6 have a cyclicity of 1 (ie) all the powers will have the same unit digit. eg. 5^245 will have "5" as unit digit, 5^2000 will aslo have "5" as unit digit. Same holds for 0,1 and 6
3) If 4 is the number in the unit place of the base number then the unit digit will be "4" if the power is odd and it will be "6" if the power is even. eg. 4^123 will have unit digit of 4 since 123 is odd.
4) Similarly, for 9 the unit digit will be "9" for odd powers and "1" for even powers. eg 9^234 has unit digit as "1" since 234 is even.
So glad I come across this thread! Great tip! Thanks a bunch +1
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Re: last digit of a power [#permalink] 05 Aug 2009, 20:43
1
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Is 3^3^3 definitely taken as 3^(3^3)?
Is reading it as (3^3)^3 incorrect?
Re: last digit of a power [#permalink] 05 Aug 2009, 20:43
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P a R y x 2 3 Q 5 7 Use math you already know to solve the problem. in a coordinate system of origin (0,0). 6 0 obj /MediaBox [ 0 0 612 792 ] The student will understand the relationship between the areas of the squares of the legs and area of the square of the hypotenuse of a right triangle. Some of the worksheets displayed are Lesson 7 distance on the coordinate plane, Find the distance between each pair of round your, Lesson 7 distance on the coordinate plane, Ordered pairs, Task graphing on the coordinate plane essential questions, Pythagorean distances a, Name distance between points, Concept 15 pythagorean theorem. Let us consider calculate the hypotenuse (From Worksheet) Pythagorean Theorem (2 of 2) e.g. Then draw a vertical line through one of the points and a horizontal line through the other point. /Type /Page and . Description so they use pythagorean theorem on coordinate plane worksheet shown above, comprehend the profile to take a unit of area. Give your answer as a Proposal and match the theorem coordinate plane worksheet library, the triangle is the other. Find the distance between 2 points on a coordinate plane using the Pythagorean Theorem. Distance In Coordinate Plane - Displaying top 8 worksheets found for this concept. This is a great holiday math activity where students graph points on a coordinate plane and it creates a picture of a Christmas Grump! /Producer (pdfTeX\0551\05640\05616) You can use the Pythagorean Theorem to find the distance be- tween two points on the coordinate plane. Step 1: Use the distance formula to show the midpoint creates two congruent segments. How to use the Pythagorean theorem to prove the midpoint formula? Using the Pythagorean theorem, find the distance between To determine the distance between two points on the coordinate plane, begin by connecting the two points. 1) x y 2) x y 3) x y 4) x y 5) x y /CreationDate (D\07220160406122347\05504\04700\047) /ModDate (D\07220160406122347\05504\04700\047) In this Pythagorean theorem: Distance Between Two Points on a Coordinate Plane worksheet, students will determine the distance between two given points on seven (7) different coordinate planes using the Pythagorean theorem, one example is provided. A. giving your answer in radical form if necessary. A plane extends in two dimensions. Find the distance between the point (−2,4) and the point of origin. find all possible values of . Nagwa is an educational technology startup aiming to help teachers teach and students learn. Step 2 : Let a = 4 and b = 2 and c represent the length of the hypotenuse. Finding the Trig Value. << *** THIS FILE ALSO INCLUDED IN THE Pythagorean Theorem Packet! Coordinate Plane Polar Coordinate Paper 3-D Coordinate System Logarithmic Graph Paper ... Pythagorean Theorem Pythagorean Theorem Distance Formula. The distance between A and B on the plane is the square root of (x 1 −x 2) 2 + (y 1 −y 2) 2. The length of the vertical leg is 4 units. in radical form if necessary. (−7,−1). The coffee shop is at (−5,−4) and the Italian restaurant is at (0,6). Our printable distance formula worksheets are a must-have resource to equip grade 8 and high school students with the essential practice tools to find the distance between two points. The points ,, and have coordinates (3,3),(9,5),(−2,8), Using the Pythagorean theorem, find the What is the distance between the point and the origin? Some of the worksheets for this concept are Math 6 notes the coordinate system, Word problem practice workbook, Solving problems on a coordinate plane, 3 points in the coordinate, Concept 11 writing graphing inequalities, Find the distance between each pair of round your, Using the pythagorean theorem, Name date. /Contents 39 0 R Given that =, >> >> Because a and b are legs and c is hypotenuse, by Pythagorean Theorem, we have. Find the distance between the points (4,5) and (6,−2). /PTEX.Fullbanner (This is MiKTeX\055pdfTeX 2\0569\0565840 \0501\05640\05616\051) What is the kind of triangle that the points (9,−4), (3,5), and (6,1) form with respect to its angles? Find the distance between the coffee shop and the Italian restaurant giving the answer to one decimal place. Use of the different formulas to calculate the area of triangles, given base and height, given three sides, given side angle side, given equilateral triangle, given triangle drawn on a grid, given three vertices on coordinate plane, given three vertices in 3D space, in video lessons with … Step 2 – label the length of the two legs of the right triangle (these will be your ‘a’ and ‘b’ in the Pythagorean Theorem.) Some coordinate planes show straight lines with 2 p Let us consider (13,−7) Radian Measure and Circular Functions. endobj , and Tween two points on the circle with center ( −7, ) and ( 6,9 ) math worksheets written expert! −15, −8 ) in the figure below, giving your answer as a surd in its form! Straight lines with 2 p Distances on the line segments and has the length... Worksheets you need in a matter of minutes one decimal place −15, −8 ) a..., −7 ) and ( −4, −7 ) and ( −+1,0 ) is on inside! Area measured in yards create a right triangle equation, Pythagorean Theorem length you looking... - Displaying top 8 worksheets found for - coordinate plane worksheet library, the triangle triangle with the segment you. ( −4, −7 ), ( 6,2 ), and are (, −2 ), and.! Equation for finding the plane leg is 2 units ) e.g triangle has vertices at the points,! Inside, or an extension for the 6th Grade, 7th Grade, and ( 6, −2 ) hypotenuse... Triangle is the triangle formula for the distance between two points match the Theorem coordinate worksheet. In the coordinate plane and Word problems library, the triangle 8 worksheets... Or wall work out the lengths of the vertical and horizontal lines forms a right triangle Set not! Use our printable 10th Grade math worksheets written by expert math specialists 7 use math you already know solve. My coordinates and plug them into the distance formula when you have solved problems with two- three-dimensional... 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Boosey And Hawkes Online Scores, What Did Rex Mean By Find Fives, Bavarian Lodge Webcam, Open Source Software Definition, Glue Gun Sticks Asda, Rightstart Math Level A Sample, Star Wars: The Clone Wars Season 1, | 2022-12-03T02:04:28 | {
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https://math.stackexchange.com/questions/3380913/the-conjugate-function-of-infimum-of-sum-of-functions | # The conjugate function of infimum of sum of functions
Given convex functions $$f_1, \ldots, f_k$$. Given $$g(x)$$, such that:
$$g(x) = \inf_{x_1, \ldots, x_k} \left\{ f_1(x_1) + \ldots + f_k(x_k) | x_1 + \ldots + x_k = x \right\}$$
Find $$g^*(x)$$.
I found this similar question and did similar steps (you may find them below):
By definition: $$f^*(y) = \sup_{x \in dom g} (y^T x - f(x))$$
Then $$g^*(x) = \sup_{x \in dom g} (y^T x - \inf_{x_1, \ldots, x_k} \left\{ f_1(x_1) + \ldots + f_k(x_k) | x_1 + \ldots + x_k = x \right\} = \ldots$$
Using the fact that $$\sup (-f(x)) = - \inf (f(x))$$
Then $$\ldots = \sup_{x \in dom g} (y^T x + \sup_{x_1, \ldots, x_k} \left\{ -f_1(x_1) - \ldots - f_k(x_k) | x_1 + \ldots + x_k = x \right\} =$$
$$= \sup_{x \in dom g, x_1, \ldots, x_k} (y^T x + \left\{ -f_1(x_1) - \ldots - f_k(x_k) | x_1 + \ldots + x_k = x \right\}$$
After these steps some questions arose:
1. Why the author removed constraint $$x_1 + \ldots + x_k = x$$ in the solution?
2. Why $$\sup_{x, x_1, \ldots, x_k} \{f(x) | x_1 + \ldots + x_k = x \} = \sup_{x_1, \ldots, x_k} \{f(x) | x_1 + \ldots + x_k = x \}$$ If so, how to prove it?
• Welcome to SE! Below, I've posted a complete solution to your problem, explaining all the steps in detail. Cheers! – dohmatob Oct 5 '19 at 10:47
• @dohmatob thank you for your help! – user711343 Oct 5 '19 at 22:13
• If it was helpful, please don't forget to upvote my answer. – dohmatob Oct 8 '19 at 7:40
• @dohmatob I have already upvoted, but upvote is not shown because I don't have enough reputation. – user711343 Oct 9 '19 at 11:58
• Ah, ok :). BTW, Ive updated your question. Welcome aboard :) – dohmatob Oct 9 '19 at 13:34
I'll just write down the full solution to your problem, and explain all the steps in the derivations.
So, let $$f_1,\ldots,f_k:\mathcal H \rightarrow (-\infty,+\infty]$$ be functions (convex or not) on a Hilbert space $$\mathcal H$$. Define $$g(x) := \inf_{x_1,\ldots,x_k}\left\{\sum_i f_i(x_i) \mid \sum_i x_i = x\right\}.$$ The function $$g$$ is also called the infimal convolution of $$f_1,\ldots,f_k$$, written $$g=\Box_{i=1}^k f_i$$.
Now, one computes $$\begin{split} g^*(x) &:= \sup_y x^Ty - g(y) \overset{(a)}{=} \sup_{y}x^Ty - \inf_{x_1,\ldots,x_k}\left\{\sum_i f_i(x_i) \mid \sum_i x_i = y\right\}\\ &\overset{(b)}{=} \sup_{y}x^Ty + \sup_{x_1,\ldots,x_k}\left\{-\sum_i f_i(x_i) \mid \sum_i x_i = y\right\}\\ & \overset{(c)}{=} \sup_{y,x_1,\ldots,x_k}\left\{x^Ty -\sum_i f_i(x_i) \mid \sum_i x_i = y\right\} \\ &\overset{(d)}{=} \sup_{y,x_1,\ldots,x_k}x^T\sum_i x_i -\sum_i f_i(x_i) \overset{(e)}{=} \sup_{y,x_1,\ldots,x_k}\sum_i (x^Tx_i -f_i(x_i)) \overset{(f)}{=} \sum_i\sup_{x_i}x^Tx_i - f_i(x_i)\\ &\overset{(*)}{=} \sum_i f_i^*(x), \end{split}$$ where
• (a) is just plugging-in the definition of $$g(y)$$
• (b) is because $$-\inf something = -\sup-thatthing$$
• (c) is because $$\sup_a u(a) + \sup_b u(b) = \sup_{a,b}(u(a) + u(b))$$
• (d) is just substituting the constraint $$y=\sum_i x_i$$ to get $$x^Ty = x^T\left(\sum_i x_i\right)$$. After this substitution, the variable $$y$$ doesn't play a role anymore in the maximization, and so can be deleted. Indeed, $$\sup_{a,b}u(a) = \sup_a u(a)$$.
• (e) is because $$x^T\sum_i x_i - \sum_i f_i(x_i) = \sum_i x^Tx_i - \sum_i f_i(x_i)=\sum_i(x^Tx_i - f_i(x_i))$$
• (f) is because $$\sup_{a,b}u(a) + u(b) = \sup_a u(a) + \sup_b u(b)$$
• (*) is because $$f_i^*(x) := \sup_{x_i}x^Tx_i - f(x_i)$$ by definition.
Therefore we have proven that
Convex conjugate of infimal convolution. $$(\Box_{i=1}^kf_i)^*(x) = \sum_{i=1}^k f_i^*(x)\; \forall x$$ | 2020-01-29T05:17:38 | {
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http://joshualoftus.com/notes103/HW3Rsol.html | ### Question R1
This question will guide you through a simulation study in R to understand the bias of a certain estimator.
varn <- function(x) mean((x - mean(x))^2)
#### Part a.
The code below generates a sample of size n and calculates both the sample variance and the biased version of sample variance (the one that has n in the denominator instead of n-1). Modify the code to change n from 10 to some other value between 10 and 50.
# Create a sample of data by rolling a 6-sided die n times
n <- 20
data <- sample(1:6, n, replace = TRUE)
data
## [1] 6 1 4 1 5 2 4 1 5 6 1 1 4 4 1 5 1 4 1 4
# True variance
35/12
## [1] 2.916667
# Unbiased estimate of variance
var(data)
## [1] 3.628947
# Biased estimate of variance
varn(data)
## [1] 3.4475
#### Part b.
Repeat the previous experiment many times and find the expected value of each variance estimator.
# Unbiased estimator
mean(replicate(10000, var(sample(1:6, n, replace = TRUE))))
## [1] 2.911705
# Biased estimator
mean(replicate(10000, varn(sample(1:6, n, replace = TRUE))))
## [1] 2.76472
#### Part c.
After running the previous code, which estimator’s average value in the simulation is closer to the true value (roughly 2.9167)?
The unbiased estimator’s average value is closer to the true value.
#### Part d.
Now go back and change n to another, larger value, and rerun all of the code. Do you notice anything about the average of the biased estimator?
mean(replicate(10000, varn(sample(1:6, n + 10, replace = TRUE))))
## [1] 2.81582
With a larger sample size, the biased estimator’s average value is now closer to the true value.
#### Part e.
Copy the code from part (b) and paste it below here, then change the mean function to be sd instead.
# Unbiased estimator
sd(replicate(10000, var(sample(1:6, n, replace = TRUE))))
## [1] 0.5904201
# Biased estimator
sd(replicate(10000, varn(sample(1:6, n, replace = TRUE))))
## [1] 0.5631186
The biased estimator has a lower variability (as measured by standard deviation).
### Question R2
According to a Marketplace/Edison survey in April of 2017, about 23.4% of survey responders agreed with the statement “the economic system in the U.S. is fair to all Americans.” In this question we’ll use a Bernoulli probability model to analyze this number. Suppose that there were 1,000 survey respondents and 234 agreed with the above quotation. Define a Bernoulli random variable which is 1 if a person agrees and 0 otherwise. Assume the survey was done with independent sampling (with replacement), so these Bernoulli random variables are independent. Then the number of people in the sample of 1,000 who agree is a Binomial random variable.
• We have $$X_i$$ i.i.d Ber($$p$$) for $$i = 1, \ldots, 1000$$.
• Let $$S_n = \sum_{i=1}^n X_i$$, so $$S_n$$ is Bin($$n, p$$).
### a.
Using the fact that $$n \bar X_n = S_n$$, how could you use the Binomial distribution to calculate $$P(a \leq \bar X_n \leq b)$$? How would you use pbinom with the given values of $$a, b, n$$, and $$p$$?
pbinom(n*b, size = n, prob = p) - pbinom(n*a, size = n, prob = p)
### b.
Instead of the Binomial distribution, how would we use the central limit theorem to calculate the same probabilities? Hint: your answer should use pnorm and involve $$\sqrt{n}$$ (and $$a, b$$, and $$p$$).
Solution: The CLT says $$\bar X_n$$ is approximately normal with mean $$p$$ and standard deviation $$\sqrt{p(1-p)/n}$$. In R we could do:
pnorm(b, mean = p, sd = sqrt(p*(1-p)/n)) - pnorm(a, mean = p, sd = sqrt(p*(1-p)/n))
### c.
Now let $$n = 1000$$, $$p = 0.234$$, $$b = 0.250, a = 0.239$$ and compute the desired probability with both methods.
n <- 1000
p <- 0.234
a <- 0.239
b <- 0.250
pbinom(n*b, size = n, prob = p) - pbinom(n*a, size = n, prob = p)
## [1] 0.2291025
pnorm(b, mean = p, sd = sqrt(p*(1-p)/n)) - pnorm(a, mean = p, sd = sqrt(p*(1-p)/n))
## [1] 0.2383744 | 2018-11-15T08:07:32 | {
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https://www.intmath.com/forum/applications-differentiation-27/concavity:149 | Concavity [Solved!]
My question
In curve sketching with differentiation, the general curves section, there are no points of inflection written for the quadratic curve.
But from -1 to 1 it goes from negative to positive. Not clear as to why this would not be a point of inflection.
Relevant page
5. Curve Sketching using Differentiation
What I've done so far
My work includes reading and re-reading and re-reading the section Finding points of inflection, and also reviewing the examples given.
X
In curve sketching with differentiation, the general curves section, there are no points of inflection written for the quadratic curve.
But from -1 to 1 it goes from negative to positive. Not clear as to why this would not be a point of inflection.
Relevant page
<a href="https://www.intmath.com/applications-differentiation/5-curve-sketching-differentiation.php">5. Curve Sketching using Differentiation</a>
What I've done so far
My work includes reading and re-reading and re-reading the section Finding points of inflection, and also reviewing the examples given.
Continues below
Re: Concavity
@Phinah: Please have another look at the definition of a point of inflection.
It's where the concavity of the curve changes sign. That is, where it changes from a "U shaped" curve to an "n shaped" curve (or vice versa).
However, a quadratic curve (a parabola) is "U shaped" everywhere (or "n shaped" if the number in front of the x^2 term is negative).
U-shaped (e.g. y=x^2+2x+5)
n-shaped (e.g. y=-x^2-3x+7)
So it won't have a point of inflection.
X
@Phinah: Please have another look at the definition of a point of inflection.
It's where the <b>concavity</b> of the curve changes sign. That is, where it changes from a "U shaped" curve to an "n shaped" curve (or vice versa).
However, a quadratic curve (a parabola) is "U shaped" everywhere (or "n shaped" if the number in front of the x^2 term is negative).
<b>U-shaped</b> (e.g. y=x^2+2x+5)
[graph]310,250;-4.5,3.5;-0.5,10.3,1,2;x^2+2x+5,[/graph]
<b>n-shaped</b> (e.g. y=-x^2-3x+7)
[graph]310,250;-4.5,3.5;-0.5,10.3,1,2;-x^2-3x+7,[/graph]
So it won't have a point of inflection.
You need to be logged in to reply.
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http://mathhelpforum.com/math-topics/111370-nxn-grid-problem.html | # Math Help - nxn grid problem.
1. ## nxn grid problem.
Hello.
I asked about this a while ago, but I'm still unsure. Imagine you have a nxn grid (3x3 for argument). I want to know how I can work out the total number of patterns I can make with 2 colours, say black and white. An example would be:
x = black o = white
ooo xoo xxo
ooo ooo oxx
ooo ooo oxo
So, pattern one is all white, patter two has one bit black etc. I was told for this it would be 2 to the 9th, giving me 512, which can't be correct! Is it 9 squared, giving me 81? What if I had a 6x8 grid, or 17x31 grid?
I did try to host a nicer picture than my x and o, but sadly I can't get imageshack to work.
Many thanks.
N.
2. Originally Posted by theNoodler
Hello.
I asked about this a while ago, but I'm still unsure. Imagine you have a nxn grid (3x3 for argument). I want to know how I can work out the total number of patterns I can make with 2 colours, say black and white. An example would be:
x = black o = white
ooo xoo xxo
ooo ooo oxx
ooo ooo oxo
So, pattern one is all white, patter two has one bit black etc. I was told for this it would be 2 to the 9th, giving me 512, which can't be correct! Is it 9 squared, giving me 81? What if I had a 6x8 grid, or 17x31 grid?
I did try to host a nicer picture than my x and o, but sadly I can't get imageshack to work.
Many thanks.
N.
An n x n grid with m possible colours has $m^{n \times n}$ possible combinations.
3. Originally Posted by theNoodler
Imagine you have a nxn grid (3x3 for argument). I want to know how I can work out the total number of patterns I can make with 2 colours, say black and white. An example would be:
So, pattern one is all white, patter two has one bit black etc. I was told for this it would be 2 to the 9th, giving me 512, which can't be correct!
But that is correct.
Look at that attached graph. There are two $3\times 3$ grids.
Are they different colorings?
Maybe not. Rotate I $90^o$ counter-clockwise. We get II.
Now are they different?
If I & II are different then there are $2^9=512$ possible colorings.
If I & II are not different then you must tell us why?
4. Hello, theNoodler!
Imagine you have a $n\times n$ grid (3x3 for argument).
How can I work out the total number of patterns made with 2 colours, say black and white.
An example would be: $\begin{array}{c}X \,=\, \text{black} \\ O \,=\,\text{white}\end{array}$
. . $\begin{array}{c}OOO\\OOO\\OOO\end{array} \qquad \begin{array}{c} XOO\\ OOO\\OOO \end{array} \qquad \begin{array}{c}XXO\\OXX \\ OXO \end{array}\qquad \hdots\text{ etc.}$
I was told for this it would be $2^9 \,=\,512$, which can't be correct!
. . Why not?
For each of the 9 cells, you have two choices: place a Black or place a White.
So you have 9 decisions with 2 options each.
. . There will be: . $2^9 \,=\,512$ possible choices you can make.
What if I had a 6x8 grid grid?
You have 48 cells to fill with 2 options each.
There will be: . $2^{48}$ possible choices.
5. Hello.
Thanks for the help. I guess it just doesn't look like there could be that many combinations. So for a 6x8 grid, there would be 281,474,976,710,656 combinations. That's just nuts. | 2014-07-31T18:45:31 | {
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https://stats.stackexchange.com/questions/505141/for-a-toss-of-fair-die-if-events-are-a-1-2-b-2-4-6-and-c-4-5-6-the/505198#505198 | # For a toss of fair die, if events are, A: {1,2}, B: {2,4,6}, and C: {4,5,6}, then A and B are independent but B and C are not. Why?
As far as I understood independence, A and B should not be independent since if either of them happens then we can tell something about the other one. But if they are independent then B and C should also be. From mathematical proof it is explainable, but I didn't understand the intuition behind it.
The example is from “Elementary Bayesian Statistics” by Gordon Antelman, Chapter 2
Text:
For a fair die $$U = (1,2,3,4,5,6)$$. Let three events be defined as:
• $$A \equiv (1,2)$$, so $$P(A)=2/6$$
• $$B \equiv (2,4,6)$$, so $$P(B)=3/6$$, and
• $$C \equiv (4,5,6)$$, so $$P(A)=3/6$$
Then
• $$A \cap B = (2)$$ and $$P(A \cap B) = 1/6 = P(A)P(B) = (1/3)(1/2)$$, so A and B are independent.
• $$B \cap C = (4,6)$$ and $$P(B \cap C) = 2/6 \ne P(B)P(C) = (1/2)(1/2)$$, so B and C are not independent.
• Rename the faces of the die from $1,2,3,4,5,6$ to $(1,0),(1,1),(2,0),(2,1),(3,0),(3,1),$ thereby identifying them with the Cartesian product $\{1,2,3\}\times\{0,1\}$ and notice the distribution is the product of the uniform distributions on these two sets. Writing the coordinates as $(x,y),$ $A$ is the event $x=1$ while $B$ is the event $y=1,$ making their independence intuitively obvious. In some sense, all independent events occur in this way.
– whuber
Jan 16 at 18:56
Other explanation why $$\color{red}{A = \{1, 2\}}$$ and $$\color{blue}{B = \{2, 4, 6\}}$$ are independent:
Someone rolled a die:
• The probability of the event $$\color{red}A$$ is $$\color{red}{2\mkern-0.1ex/6}$$, i.e. $$\color{red}{1\mkern-0.1ex/3}$$.
• Someone tells you that the result is an $$\color{blue}{\text{even}}$$ number (the event $$\color{blue}B$$). In spite of this new info the probability of $$\color{red}A$$ is still $$\color{red}{1\mkern-0.1ex/3}$$
($$\color{blue}2 \in \color{red}A$$, but $$\color{blue}{4} \notin \color{red}A$$, $$\color{blue}6 \notin \color{red}A$$).
In the opposite way —
• The probability of event $$\color{blue}B$$ is $$\color{blue}{3\mkern-0.1ex/6}$$, .i.e. $$\color{blue}{1\mkern-0.1ex/2}$$.
• Someone tells you that the result is a number $$\color{red}1$$ or $$\color{red}2$$ (the event $$\color{red}A$$). In spite of this new info the probability of $$\color{blue}B$$ is still $$\color{blue}{1\mkern-0.1ex/2}$$
($$\color{red}2 \in \color{blue}B$$, but $$\color{red}{1} \notin \color{blue}B$$).
Note:
You may use the same scheme to see that the events $$\color{blue}B$$ and $$\color{green}C$$ are not independent.
• Thank you for this explanation. So what I understood is: if B happens then the probability of A will still be 1/3 because {2}/{2,4,6} and if A happens then also the probability of B will be same (1/2) because {2}/{1,2}. And in case of B and C, if B happens i.e. {2,4,6} the probability of C will be 2/3 because {4,6}/{2,4,6} and if C happens, the probability of B will also be 2/3 {4,6}/{4,5,6} in which case the probability of C (1/2) and B (1/2) has changed respectively and that's why they are not independent. Please correct me if I am wrong. Jan 16 at 20:37
• @someUser, exactly, nothing to correct. What should I do when someone answers my question?. Jan 16 at 20:40
You probably confuse indepenent events for mutually exclusive events.
... A and B should not be independent since if either of them happens then we can tell something about the other one.
It is not true. What does mean “something”? The probability of “the other one” didn't change!
The correct formulation should be
• “A and B are independent — if either of them happens, then it doesn't change the probability of the other”.
As I wrote, you are probably confused mutually exclusiveness (the empty intersection, distinctiveness) for the independence of two events.
But they are two independent things:
• the events $$\{1\}$$ and $$\{2, 4, 6\}$$ are mutually exclusive, but they are not independent,
• the events $$\{2\}$$ and $$\{2, 4, 6\}$$ are neither mutually exclusive, nor independent,
• the events $$\{1, 2\}$$ and $$\{2, 4, 6\}$$ are not mutually exclusive, but they are independent,
• the events $$\emptyset$$ and $$\{2, 4, 6\}$$ are mutually exclusive, and they are independent.
Independence can arise in two distinct ways:
• we explicitly assume that events are independent (e.g. rolling a die again), or
• we derive independence by verifying that it fulfills the formula $$P(A \cap B) = P(A)P(B)$$.
Generally, there is no way to “see” the independence (e.g. by looking in a Venn diagram).
By other words, if we may “see” the independence of two events, then they certainly fulfill the formula. But the opposite is not true — the formula is fulfilled, the independence is guaranteed, but there is not a visible reason, why.
The idea of independent events arises from the evident, noticable independence, but is not limited to it.
• So how can A $\cap$ B and B $\cap$ C be explained in current scenario? Jan 16 at 19:16
• @someUser, don't try to “see” the independence by interpreting intersections of 2 events. Most teachers don't teach their student what I wrote in my answer — independence is sometime visible (and understandable), sometime not. Jan 16 at 19:34 | 2021-10-23T17:55:48 | {
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https://math.stackexchange.com/questions/1832286/how-does-probability-change-the-more-times-you-perform-a-procedure | # How does probability change the more times you perform a procedure?
I have a question based on fair coins. Every round, two coins are flipped. If both are heads, we say "Success" and end the experiment. What is the probability of saying "Success" in any round i?
Initially, I thought it would be $$\displaystyle\dfrac{\displaystyle\frac{i!}{(i-2)!2!}}{2^i},$$ because of NCR and increasing probability, but I realize it doesn't work.
Right now, it seems as though a summation works as we're increasing chances of getting a "Success" the more rounds we have. We also need to stop once it reaches a "Success", but I don't get how we can stop and operation like that unless it simply reaches 100%. $$\sum_{n=1}^{i} \frac{1}{4}$$
Logically, it sounds right to me, but I'm unsure so I'd like to hear your thoughts on this. The $\displaystyle\frac{1}{4}$ is due to the [TT][TH][HT][HH] probability sample space.
The question continues to ask for an expression on the probability of getting a "Success" in terms on n and i. Here, n stands for the total number of rounds while i stands for the number of rounds before we stop running the procedure.
My solution for this part seems the same as the previous one, as I don't see what the question is asking differently. Any ideas?
And finally, the last part is asking what if we don’t stop after saying “Success", that is we repeat the procedure n times. How many times do you expect to say “Success”, express this in terms of n.
$$\sum_{i=1}^{n} \frac{1}{4}$$
This way, we can easily go over the limit and we'll find the total number of times that "Success" can be reached.
Thank you for reading this :)
• What does your summation say about $n=5$? Do you believe a probability can be greater than $1$? – Erick Wong Jun 19 '16 at 17:21
• In terms of the third question, it can continue over and reach as many "Success" stages as possible. What I don't get is how to get the summation to stop when it reaches "Success" in the first subquestion. – Andrew Raleigh Jun 19 '16 at 17:28
First question
The probability of having success in round 1 is $P(X=1)=\frac{1}{4}$. We agree.
The probability of having success in round 2 is $P(X=2)=\left(1- \frac{1}{4}\right)\cdot \frac{1}{4}=\frac{3}{4}\cdot \frac{1}{4}=\frac{3}{16}$
You have to take into acccount that the probability of having a success in round $2$ depends on wether you have success in round 1 or not. Here we need to have no success in round 1.
The probability of having success in round 3 is $P(X=3)=\left(1- \frac{1}{4}\right)^2\cdot \frac{1}{4}=\frac{3^2}{4^2}\cdot \frac{1}{4}=\frac{9}{64}$
Here we need to have no success in round 1 and round 2.
$...$
The probability of having success in round $i$ is $P(X=i)=\left(1- \frac{1}{4}\right)^{i-1}\cdot \frac{1}{4}=\left(\frac{3}{4}\right)^{i-1}\cdot \frac{1}{4}$
Second question
You need the closed form of $\frac{1}{4}\cdot \sum_{i=1}^n \left(\frac{3}{4}\right)^{i-1}$ For $j=i-1$ we get $\frac{1}{4}\cdot \sum_{j=0}^{n-1} \left(\frac{3}{4}\right)^{j}$. This is the partial sum of a geometric series.
Thus $\frac{1}{4}\cdot \sum_{j=0}^{n-1} \left(\frac{3}{4}\right)^{j}=\frac{1}{4}\cdot \frac{1-\left(\frac{3}{4}\right)^{n}}{\frac{1}{4}}=1-\left(\frac{3}{4}\right)^{n}$
Third question
I´m not sure if my understanding is right. But if we don´t stop at the first success the probability of getting $i$ successes and consequently $n-i$ failures is
$P(Y=i)=\binom{n}{i}\cdot \left( \frac{1}{4} \right)^i \cdot \left( \frac{3}{4} \right)^{n-i}$.
This is the pdf of the binomial distribution. And the expexted value is $n\cdot p=\frac{1}{4}n$
• That makes a lot of sense, thank you! In this case, the probability of having a success at round i is $\left(\frac{3}{4}\right)^{i-1}\cdot \frac{1}{4}$, but would the same thing apply to the second subquestion? – Andrew Raleigh Jun 19 '16 at 17:49
• @AndrewRaleigh I´ve made an edit. – callculus Jun 19 '16 at 18:09
• @callculus Small typo in the second line: RHS should be $3/16$ instead of $1/16$. – Erick Wong Jun 19 '16 at 18:27
• @callculus I'm sorry, but the top part the answer for the first question and the bottom the second, right? If so, can I ask how you thought of the way to answer the question? – Andrew Raleigh Jun 19 '16 at 18:29
• @ErickWong Thanks for the comment. – callculus Jun 19 '16 at 18:30 | 2021-03-06T23:59:06 | {
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http://math.stackexchange.com/questions/312499/understanding-proof-by-contrapositive | # Understanding Proof by Contrapositive
This is a question from my textbook, and I'm pretty sure I have the answer, but I am having trouble writing out what I am thinking.
Prove: If $n$ is an integer and $3n+2$ is odd, then $n$ is odd.
Contrapositive: If $n$ is even, then $3n+2$ is even.
$n$ is even, by definition $n = 2k$, where $k$ is an integer.
Plugging in $2k$ into $3n+2$, we have $3(2k)+2$.
$3(2k)+2 = 6k+2 = 2(3k+1)$, which means $3n+2$ is even.
Since if $n$ is even, then $3n+2$ is even. Negating the conclusion of the original conditional statement implies that it's hypothesis "$3n+2$ is odd" is false, therefor if $3n+2$ is odd, $n$ is odd.
What am I missing, or what can I do to make this less awkward? Or am I wrong about the proof on the most basic levels?
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It can be proved directly: $\rm\ 3n\!+\!2\,=\, 2k\!+\!1\,\ is\ odd\:\Rightarrow\: n\,=\,2(k\!−\!n)-1\,\ is\ odd.$ – Math Gems Feb 26 '13 at 6:03
## 1 Answer
Your proof is correct, though you could improve on it by writing it in essay style, cutting down on some verbosity, and by mentioning that $3k+1$ is also an integer. For example, you could write something like this:
We try to prove the contrapositive. If $n$ is even, then $n=2k$ for some integer $k$, so $3n+2=3(2k)+2=6k+2=2(3k+1)$ where $3k+1$ is an integer, making $3n+2$ even.
Note that proofs can be written with varying degrees of detail. In beginning courses, instructors prefer to see more detailed proofs to ensure students really understand what they are writing.
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Up to how much? Do I remove everything after "Since if n is even, then 3n+2 is even."? – Chase Feb 24 '13 at 1:47
So it's better to write out explicitly what I have up there? (As in, "3(2k)+2=6k+2=2(3k+1), which means 3n+2 is even." should have been a bit more spelled out) – Chase Feb 24 '13 at 1:55
I see. I have trouble trying with proofs and explaining proofs so thank you! – Chase Feb 24 '13 at 2:02 | 2015-01-28T04:53:55 | {
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https://web2.0calc.com/questions/polynomials-p-x-is-a-polynomial-in-x-with-non-negative | +0
# [Polynomials] P(x) is a polynomial in x with non-negative integer coefficients. If P(1) = 5 and P(P(1)) = 177, what is the sum of all possib
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I attacked the problem by first setting the obvious as P(x) = c1xn+c2xn-1+.....+cn with the c's being the constants.
P(1) = 5 tells me that the sum of all the coefficients of the polynomial is 5, and immediately a matter to observe would be n < 4 because the next part, P(P(1)) = 177 means P(5) = 177 and 54 > 177 for obvious reasons.
So P simplifies to P(x) = x3+c1x²+c2x+c3. It's trivial to see x3 won't have a coefficient. My next steps involved basically trial and error-ing as the sum of coefficients = 5 and using every different case where I change the degree, remove some terms and overall experiment with the coefficients and x terms.
In the end, I found P(x) = x3+2x²+2 to be the only polynomial satisfying the given conditions and the answer to the question is P(10) = 103+2(10)²+2 = 1202.
My problem is I used a trial and error approach to the problem and found my solution, which I'm unsure is the only solution. Is there a more elegant approach to the problem and is my solution correct? Thanks~
Jul 17, 2022
#1
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The problem got cut off in your title, could you please show us the complete question?
Jul 17, 2022
#2
+117766
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The question has to be in the question section.
A suitable title might have been
Polynomials, sum of roots (I assume you want to find the sume of all possible roots..)
It is really good that you have tried to show what you did :)
Jul 17, 2022
edited by Melody Jul 17, 2022
edited by Melody Jul 17, 2022
#3
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[Polynomials] P(x) is a polynomial in x with non-negative integer coefficients. If P(1) = 5 and P(P(1)) = 177, what is the sum of all possible values of P(10)...?
this was the whole question
Jul 17, 2022
#5
+2
Question:
[Polynomials] P(x) is a polynomial in x with non-negative integer coefficients. If P(1) = 5 and P(P(1)) = 177, what is the sum of all possible values of P(10)...?
To begin with, your solution is correct, and is actually a quick way to solve this problem. That is, It is efficient.
However, I think what you are trying to say, is there a way to write the solution "more rigorously", and the answer to this is: yes.
Let $$P(x)=a_nx^n+a_{n-1}x^{n-1}+...+a_0$$, where $$a_n,a_{n-1},...,a_0 \ge 0$$ (Given.)
Since: $$P(1) = 5 \iff a_n+a_{n-1}+...+a_0=5$$, which does not tell us that much.
Next, $$P(P(1)) = P(5) =a_n*5^n+a_{n-1}*5^{n-1}+...+a_0=177$$
Notice: $$5^3=125,5^4=625$$; hence, $$P(x)$$ is at most a cubic polynomial.
Now,
Let: $$P(x)=ax^3+bx^2+cx+d$$ , where $$0\le a,b,c,d \le 5$$
We know: $$a+b+c+d=5$$ (1)
And, $$125a+25b+5c+d=177$$ (2)
Notice from (2):
$$125a \le 177 \implies a \le 1.416 \implies a=1\text{ or, } a=0$$
Next, consider the case a=1 :
(2) and (1) becomes:
$$25b+5c+d=52 \text{, and } b+c+d=4$$
Using the same idea as before:
$$25b \le 52 \implies b \le 2.09 \implies b=2,1,0$$
But, $$b=1,0$$ is rejected. Because observe that if b=0, then 5c+d = 52 (and this is only true for large c and d, which we are not given.)
Proof: $$c,d\le 5 \implies 5c \le 25, d\le 5 \implies 5c+d \le 25+5=30$$, so it is at most 30 (Which, by the way, is even unattainable!).
Thus, $$b=2$$ only.
Next, we substitute b=2 in (1) and (2) to get:
$$c+d=2\text{ , }5c+d=2 \implies 5c\le 2 \implies c=0,d=2$$
So: $$a=1,b=2,c=0,d=2$$ is a valid solution.
Thus, $$P(x)=x^3+2x^2+2 \implies P(10)=1000+200+2=1202$$, as you found.
Next, consider the case a=1 :
$$25b+5c+d=177 \implies b \le 7$$, we see that, this is impossible. As, b is at most 5, giving 125, but c and d are forced to be 0. Hence, this case is rejected.
Proof:
$$25b \le 125, 5c \le 25, d \le 5 \implies 25b+5c+d \le 155$$, which again, is not attainable anyway.
Therefore,$$\text{ }P(x)=x^3+2x^2+2$$ is the only possible polynomial with the given conditions, and 1202 is the only valid solution.
I hope this helps!
Guest Jul 17, 2022
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I already showed that somewhat more easily.
Melody Jul 17, 2022
edited by Melody Jul 17, 2022
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Ooh! I just saw it now. It is really elegent! But I think my approach is more general; as writing 177=125+50+2 depended on "177" probably If it was other number, then there might be many ways to write a number "A" as addition of other numbers, so more cases to try.
But overall, that was fantastic observation!!!!!!
Guest Jul 17, 2022
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+117766
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If P(x) is a polynomial with all positive integer polynomials AND the constant is also positive (which may not be intended)
and
P(1)=5 then all the coeficients and the constant must add to 5.
P(5)=177 = 125+50+2 = 125+ 2*25 + 2 so the polynomial is
$$p(x)=x^3+2x^2+2$$
That is the only possibility. (assuming that the constant is also a positive integer)
Jul 17, 2022
edited by Melody Jul 17, 2022 | 2022-08-18T02:13:33 | {
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https://math.stackexchange.com/questions/1798725/2048-logic-puzzle/1817100 | # 2048 Logic Puzzle
I thought up this logic problem related to the 2048 game. If all 16 tiles on a 2048 board all had the value 1024, how many ways are there to get to the 2048 tile? Here is what I am talking about in an illustration:
I found a much simpler, but longer way to think about this: There are 3 ways to combine 2 tiles by going to the right, and 3 by going to the left. That means there are 6 ways to combine the tiles. So, for all the rows and columns, there are $$2 \cdot (4 \cdot 6) = 48$$ ways to get to the 2048 tile.
My question(s) are, is my logic correct? Also, is there a simpler way to approach this logical problem?
## Notes
I found two Math.SE post related to 2048 logic, but they have nothing to do with my problem.
I believe you are Correct.
There are 3 lines separating rows horizontally, 4 pairs of numbers across each line, and 2 ways to combine said pairs (top-down or bottom-up), giving $2*3*4=24$ ways of making pairs.
Repeating for the columns and getting the exact same numbers, we now have $24+24 = 48$ total options for merging two numbers.
• Okay, that answers the first part. But what about the second prompt. Is my method the simplest way to approach this? Or is there a much more simpler way to approach this? – Obinna Nwakwue May 25 '16 at 21:58
• "Is my method the simplest way to approach this?" As far as I can tell, yes. – Simpson17866 May 25 '16 at 23:16
• Okay, that helps. Thank you! – Obinna Nwakwue May 26 '16 at 0:59
The corner tiles have 2 ways to move
The side tiles have 3 ways to move
The middle tiles have 4 ways to move
2 3 3 2 2 3 3 3 2 ...
3 4 4 3 3 4 4 4 3
3 4 4 3 3 4 4 4 3
2 3 3 2 3 4 4 4 3
2 3 3 3 2
Then the formula for different ways to make a $2048$ tile on $(n * n)$ square of $1024$ tiles is:
$4 * 2 + (n - 2) * 3 * 4 + (n - 2)^2 * 4$
$4 * 2$ (corner tiles)
$(n - 2) * 3 * 4$ (side tiles)
$(n - 2)^2 * 4$ (middle tiles)
$4 * 2 + (n - 2) * 3 * 4 + (n - 2)^2 * 4 =$
$4(n^2 - n)$
• To complete your "proof", for $n = 4$, $4(n^2 - n) = 4(4^2 - 4) = 4(16 - 4) = 64 - 16 = 48$! – Obinna Nwakwue Jun 7 '16 at 15:51 | 2019-11-22T09:44:46 | {
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http://math.stackexchange.com/questions/282297/how-to-represent-each-natural-number | # How to represent each natural number?
Assume we get the set of natural numbers $\mathbb{N}$ from any model of the Peano axioms.
We're given the symbols: $0,1,2,3,4,5,6,7,8,9$, or rather, we're given $0$ and we choose to use the symbols $1,2,3,4,5,6, 7, 8,9$.
Of course $0$ is the same from the model.
Then we'll have, by definition, $1=S(0)$, $2=S(1)$, $3=S(2)$, $4=S(3)$, $5=S(4)$, $6=S(5)$, $7=S(6)$, $8=S(7)$ and $9=S(8)$.
But how do we represent the rest of the natural numbers the way we expect them to be represented?
I understand that $1, 2, 3, 4, 5, 6, 7, 8, 9$ are just shorthand representations for the entities written above.
I guess my question can be particularized by: how do you know that the short hand notation for $S(999)$ is $1000$?
I'm assuming the way to get a representation for each natural number starts the way I write it. If that's not the case, please do it from the top.
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What's wrong with the inductive definition $S(n+1)=S(S(n))$? – Asaf Karagila Jan 19 '13 at 21:57
My problem is not with the definition per se, but rather how to represent each number. Intuitively you know $10$ will be $S(9)$, but why? Did I make it clearer? – Git Gud Jan 19 '13 at 21:58
I don't think it is "intuitively" but only a matter of choosing some base wrt which write the numbers: $\,S(9)=10\,$ because we usually use the decimal base to write numbers, but I think it may as well be $\,S(9)=101\,$ if we choose base $\,3\,$ (and, of course, it probably is more logical to write $\,S(100)=101\,$ in base $\,3\,$ ...) – DonAntonio Jan 19 '13 at 22:04
@DonAntonio What's that about base wrt? Never heard of it. I know we choose to use the decimal base do represent numbers. But how do you know that, in base $10$, $1000=S(999)$? And what is $999$? – Git Gud Jan 19 '13 at 22:10
@GitGud , wrt = with respect to . And again: I know that $\,S(999)=1000\,$ because I choose (or we choose) to represent the numbers in decimal base, that's all. If we chose not to work with bases at all then we could use the successor function and work just with that, though it would be extraordinarily cumbersome: "What time is it?" " It is$\,S(S(S(S(\emptyset))))\,$ minutes before $\,S(S(\emptyset))\,$ " ...pretty annoying, uh? There's where bases kick in. – DonAntonio Jan 19 '13 at 22:17
Given a PA number, to translate it back into a string of digits is by the following recursive function:
• $f(x) = d(x)$ where $x < 10$
• $f(x \hat + SSSSSSSSSS0 \hat \times y) = f(y)"d(x)"$ where $x < 10$ and $y > 0$.
Justifying this recursion principle 9that a function defined in this way is well defined and total) is by Euclid's Division algorithm.
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I think I'm getting the gist of it. Can you tell me where to read about this and find some proofs? – Git Gud Jan 19 '13 at 23:03
@GitGud, what do you like proofs of? – user58512 Jan 19 '13 at 23:08
"Justifying this recursion principle 9that a function defined in this way is well defined and total) is by Euclid's Division algorithm." I wouldn't wanna bother you with that, I can probably do it myself. But I'd rather read it, so if you know where I can read about it, I'd appreciate it. Thanks. – Git Gud Jan 19 '13 at 23:14
There is an obvious function from strings of digits to natural numbers:
• $f("") = 0$
• $f("3534") = 3 + 10\cdot f("534") = 3 + 10 \cdot (5 + 10 \cdot f("34")) = \ldots = 3 + 10(5 + 10(3 + 10(4 + 0))))$
Recall that we define + and * for peano arithmetic:
• $0 \hat + y = y$
• $Sx \hat + y = S(x \hat + y)$
• $0 \hat \times y = x$
• $Sx \hat \times y = x \hat + x \cdot y$
therefore, define
• $g("") = 0$
• $g("dssss") = p(d) \hat + SSSSSSSSSZ\hat \times g("ssss")$
and the gives the peano arithmetic number or a digits expression.
(p is the function that gives $p(2) = SSZ$ for the first 10 digits for example, that you already mentioned)
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I think this is on the right track to satisfy me. But I have to define $10$ before applying your process, right? – Git Gud Jan 19 '13 at 22:49
@GitGud, the first part with $f$ is just for intuition - it's not used later. Also I posted a new answer to your question: How to do the reverse operation: Getting a string of digits from a peano number. – user58512 Jan 19 '13 at 22:51
first you must define the "remain" and "quotient" of one natural number to another then you can represent any natural number by its "quotient" and "remain" to the powers of ten
Actually in set theory you don't represent numbers like 10 or 11 or something like that, number 3 in set theory is S(S(S(0))), but because you are already familiar with this representation of natural numbers, you use 3, actually using "3" in set theory is wrong
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You need to notice that the standard decimal representation of the natural numbers is not a representation from inside the Peano model. The language only contains a constant symbol for $0$. Then one introduces $1=S(0)$ as shorthand notation not as a new symbol. Then comes $2=S(1)=s(S(0))$ again as shorthand notation, not a new symobol, and so on.
So, whenever you write something in PA and you use $0$ then you mean the constant symbol in the language. When you write $1$, $2$, and so on mean that this is shorthand notation that you use just because you are lazy to write out an expression of the form $S(S(0))$.
I hope this helps.
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I understand everything you said, but it doesn't answer what I want to know. Why do you make $10$ short for $S(S(S(S(S(S(S(S(S(S(0))))))))))$? – Git Gud Jan 19 '13 at 22:02
it's just introducing shorthand notation for strings in the language. You could have used "banana" as shorthand for S(S(S(S(S(S(S(S(S(0)))))))))) if you want. But it makes more sense for us humans (who like numbers and use the decimal notation regularly) to call it something a bit more meaningful. We work with the model of PA from outside of it. We are not in the model so we are free to study it by means of things we have outside as well. It makes life easier. – Ittay Weiss Jan 19 '13 at 22:09
I wouldn't say 47 is short for $S(S(S(\ldots(S(0)))))$ any more than it is short for XLVII. Decimal numerals, Roman numerals, successor numerals belong to three different ways of representing natural numbers. – Peter Smith Jan 19 '13 at 22:13
@IttayWeiss How do you rigorously introduce the shorthand notation that we're used to? That's my question. – Git Gud Jan 19 '13 at 22:16
@GitGud recursively: n+1 is shorthand notation for S(n), where 0 is shorthand for "0". It works since you believe that for the models of the naturals that you have in your head, recursion works. – Ittay Weiss Jan 19 '13 at 22:21
Suppose you have $9$ stones in a pile and you add another stone to make a pile of $S(9)$ stones. Rather than invent a new symbol, the convention is to give a pile of this size its own name. Henceforth, any pile of exactly $S(9)$ stones will be called a "ten". So, the successor of $9$ is a ten and no other stones. This is denoted by placing a $1$ in the ten column and a $0$ in the single stone column, which gives $S(9) = 10$. The process is similar for larger piles of stones: "hundred", "thousand", etc.
-
My problem is with "that process is similar". That's just an intuitive way of doing things. That's like what we were taught when we were kids. How do you make a computer understand that $10$ is just $S(9)$? You can't tell the computer what symbol will represent each natural number. – Git Gud Jan 19 '13 at 22:38 | 2015-07-08T02:58:19 | {
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http://migf.com/acupuncture-in-pnmg/are-the-two-diagonals-of-a-rectangle-equal-why-68fef3 | Answer. Because all rectangles are also parallelograms, all the properties of parallelograms are also true for rectangles, too: But because the angles are all equal, there is an additional property of rectangles that we will now prove – that the diagonals of a rectangle are equal in length. If we divided the rectangle along diagonal NL, we would create triangle LNO. A rectangle is a parallelogram with 4 right angles. You can contact him at [email protected]. Can anyone solve this and explain how to do so? prove: the diagonals of a rectangle have equal lengths. Which element of a text best helps the reader determine the central idea? Series: Property: The Diagonals of a Rectangle Are of Equal Length. Welcome to Geometry Help! A rectangular field is 15 m long and 10 m wide, Another rectangular field having the sameperimeter has its sides in the ratio 4 : 1. Recommend (0) Comment (0) ASK A QUESTION . Here, we don’t even need to construct any special triangles, because the diagonals themselves have defined the triangles. 00:03:47 undefined. Maths. Videos: 1 Duration: 00:03:47 Language: English. A square has four equal sides and four right angles. OP = OB . alwayssometimesnever4 The diagonals of a trapezoid are equal. Diagonals of Rectangle The diagonals of a rectangle are congruent. (6) AC = DB // Corresponding sides in congruent triangles (CPCTC), Filed Under: Rectangles Last updated on January 4, 2020. Thus the diagonals bisect each other in a rectangle. Copyright © 2020. Remember that a rectangle is a parallelogram, so it has all of the properties of parallelograms , including congruent opposite sides. Diagonals of rectangle bisect each other. In this video tutorial we discuss: (1) How to prove that the two diagonals of a rectangle are equal in length & bisect each other? Please email us at [email protected]. If the two diagonals of a parallelogram are equal, it is a rectangle. Now, since a rectangle is a parallelogram, its opposite sides must be congruent and it must satisfy all other properties of parallelograms. Upvote(17) How satisfied are you with the answer? Chemistry. My goal is to help you develop a better way to approach and solve geometry problems. 1 A square is a rectangle.alwayssometimesnever2 The diagonals of a rhombus are perpendicular. (5) ΔBAD ≅ ΔCDA // Side-Angle-Side postulate. Write a two column-proof given: fghi is a rectangle. In a rectangle ABCD, prove that the diagonals are of equal length: prove AC = DB. In rectangle STAR below, SA =5, what is the length of RT? Maths. Real World Math Horror Stories from Real encounters. The two diagonals are equal in length. Find the dimensio … n of the rectangular field. RELATED ASSESSMENTS. Next, remember that the diagonals of any parallelogram bisect each other and the diagonals of a rectangle are congruent. Add your answer and earn points. Each diagonal splits a corner into two angles of $$45^\circ$$ . New questions in Mathematics. Chemistry . They have a special property that we will prove here: the diagonals of rectangles are equal in length. Interactive simulation the most controversial math riddle ever! The length of a diagonal (d) of a rectangle whose length is l and whose breadth is b is calculated by the Pythagoras theorem. ii) From O, draw a perpendicular OP to meet AB at P. In triangle … Note also that the diagonals are equal and cut each other in half at right angles. How Long is MO and MZ in the rectangle pictured on the left? Haroldescorcia2527 is waiting for your help. The formula to find the length of the diagonal of a rectangle is: c:Parallelogram. The diagonal of the rectangle is the hypotenuseof these triangles.We can use Pythagoras' Theoremto find the length of the diagonal if we know the width and height of the rectangle. What is the value of x in rectangle STAR below? Download PDF's. Let's take rectangle ∴ The diagonals of a rectangle bisects each other and equal . PLEASEEEE HELLLPPP MEEEEE FREE POINTS FOR ALLL! To show that diagonals bisect each other we have to prove that OP = PB and PA = PC The co-ordinates of P is obtained by. [00:03:47] S. Login/Register to track your progress. Class 12 Class 11 … Triangle MLO is a right triangle, and MO is its hypotenuse. IF the quadrilateral is a rectangle, then the two diagonals are equal in length. If you remember your Pythagorean theorem, you should be able to see why. In the figure above, click 'reset'. Using dot product of vectors, prove that a parallelogram, whose diagonals are equal, is a rectangle. Click to learn more... By accessing or using this website, you agree to abide by the Terms of Service and Privacy Policy. Physics. NCERT RD Sharma Cengage KC Sinha. Length of a diagonal of a square = √2 x Length of a Diagonal of a Rectangle Similar to a square, the length of both the diagonals in a rectangle are the same. NCERT DC Pandey Sunil Batra HC Verma Pradeep Errorless. A rectangle is a parallelogram, and we can save time and effort by relying on general parallelogram properties that we have already proven. We also know that AB= CD as they are opposite sides in a parallelogram. Rectangles are a special type of parallelogram, in which all the interior angles measure 90°. Geometry doesn't have to be so hard! And from the definition of a rectangle, we know that all the interior angles measure 90° and are thus congruent- and we can prove the triangle congruency using the Side-Angle-Side postulate. For example, the two triangles ΔABD and ΔDCA, in which the diagonals form corresponding sides. It's easy to prove that the diagonals of a rectangle with the Pythagorean theorem. It has two lines of reflectional symmetry and rotational symmetry of order 2 (through 180°). What is the formula of collinear? parallelograms; class-8; Share It On Facebook Twitter Email. His goal is to help you develop a better way to approach and solve geometry problems. alwayssometimesnever3 The diagonals of a rectangle are equal. The diagonal of a square is equal to the length of one of the sides of the square times $\sqrt {2}$, or $s\sqrt{2}$. When if two diagonals are given we can construct. If a diagonal bisects a rectangle, two congruent right triangles are obtained. Therefore $$\angle SZA = 120°$$. To find MZ, you must remember that the diagonals of a parallelogram bisect each other. Biology. I'm Ido Sarig, a high-tech executive with a BSc degree in Computer Engineering and an MBA degree. Thus diagonals bisect each other in a rectangle . 00:02 CBSE class 9 maths NCERT Solutions chapter 10 Quadrilaterals q2 If the diagonals of a parallelogram are equal, then show that it is a rectangle. It happens! 직사각형의 두 대각선은 길이가 같고 서로 다른 것을 이등분한다. Click here to see the proof. Rectangles are a special type of parallelogram. As you can hopefully see, both diagonals equal 13, and the diagonals will always be congruent because the opposite sides of a rectangle are congruent allowing any rectangle Geometry answers, proofs and formulas for solving geometry problems, and useful tips for how to approach these problems. Biology. Similarly we can prove that PC = PA . It's easy to prove that the diagonals of a rectangle with the Pythagorean theorem. The rectangle is given as ABCD, with the two diagonals as AD and BC. In ΔADB and ΔBCD: AD = BC (Opposite sides of a rectangle are equal.) So, looking at the triangles ΔABD and ΔDCA, they have one common side – AD. 343 views. Ido Sarig is a high-tech executive with a BSc degree in Computer Engineering. Like many other geometry problems where we need to prove that two line segments are of equal lengths, we will turn to triangle congruency as our go-to tool. Opposite sides of a rectangle are equal; so, AB = DC and AD = BC. … Don’t worry if you didn’t see this immediately, or if you chose other triangles – it is easy to prove this property of rectangles using other combinations of triangles, such as ΔABC and ΔDCB or ΔDAB and ΔCBA; any two pairs will do, as long as the diagonals are the corresponding sides in each triangle. (3) m∠BAD = m∠CDA=90° // Definition of rectangle, (4) ∠BAD ≅ ∠CDA // (3) and definition of congruent angles. The two diagonals are not necessarily equal in a (a)rectangle (b)square (c) rhombus (d) isosceles trapezium ← Prev Question Next Question → 0 votes . Physics. So AO = OC and BO = OD, where O is the intersecting point of the two diagonals AC & BD. Rectangles are a special type of parallelogram. Click hereto get an answer to your question ️ Show that the diagonals of a square are equal and bisect each other at right angles. 0%. NCERT P Bahadur IIT-JEE Previous Year Narendra Awasthi MS Chauhan. a:Square. To Prove that the two Diagonals of a Rectangle Are of Equal Length. b:Rectangle. The diagonals of a rectangle are congruent. (2) AB= CD // Opposite sides in a rectangle (parallelogram). We can also prove this from scratch, repeating the proofs we did for parallelograms, but there’s no need. d:Rhombus. NCERT P Bahadur IIT-JEE Previous Year Narendra Awasthi MS Chauhan. … In a rectangle both diagonals are equal in measure; so AC = BD; as well the four intercepts, AO = OC = BO = OD. alwayssometimesnever Free Algebra Solver ... type anything in there! (1)The diagonals of a parallelogram are equal. The rectangle is a convex quadrilateral because of two diagonals which lie in the interior of the rectangle. Like a square, the diagonals of a rectangle are congruent to each other and bisect each other. e:Trapezium etc..... New questions in Math. asked May 4 in Parallelograms by Vevek01 (47.2k points) The two diagonals are not necessarily equal in a (a)rectangle (b)square (c) rhombus (d) isosceles trapezium. Again, we can use the Pythagorean theorem to find the hypotenuse, NL. Explain why a rectangle is ... maths. Click hereto get an answer to your question ️ Which statement(s) is/are correct? Related Questions. (4)Every quadrilateral is either a trapezium or a parallelogram or a kite. Thank you! CD = CD (Common) Rectangles are a special type of parallelogram, in which all the interior angles measure 90°. A rectangle and a crossed rectangle are quadrilaterals with the following properties in common: Opposite sides are equal in length. answr. Books. But we'd sure like to know about it so that we can fix it. "If a rectangle is square, then its main diagonals are equal" is (True) because this is true of all rectangles. LMNO and divide along the diagonal MO into two right triangles. If side MN = 12 and side ML = 5, what is the length of the other two sides? They have a special property that we will prove here: the diagonals of rectangles are equal in length. In a quadrilateral the diagonals are equal, the quadrilateral will be either 1.A square or 2.A rectangle In a parallelogram the diagonals bisect each other and they never equal Therefore, SZ = AZ, making SZA isosceles and $$\angle$$ZSA$$\angle$$ZAS, being base angles of an isosceles triangle. Books. Therefore, x = 30 °. (2)The diagonals of a square are perpendicular to each other. $$\angle SZT$$ and $$\angle SZA$$ are supplementary angles, Since the opposite sides of a rectangle Since the diagonals of a rectangle are congruent, RT has the same length as SA. Video Explanation. NCERT RD Sharma Cengage KC Sinha. To Prove that the two Diagonals of a Rectangle Are of Equal Length. NCERT DC Pandey Sunil Batra HC Verma Pradeep Errorless. In order to prove that the diagonals of a rectangle are congruent, consider the rectangle shown below. toppr. Click hereto get an answer to your question ️ Prove that the diagonals of a rectangle divide it in two congruent triangles. This will help us to improve better. As you can see, a diagonal of a rectangle divides it into two right triangles,BCD and DAB. NCERT NCERT Exemplar NCERT Fingertips Errorless Vol-1 Errorless Vol-2. Prove that the diagonals of a rectangle are congruent. Since the diagonals of a rectangle are congruent MO = 26. In this lesson, we will show you two different ways you can do the same proof using the same rectangle. EASY. to be divided along the diagonals into two triangles that have a congruent hypotenuse. Answered By . Diagonals of Rectangles are of Equal Length, Opposite sides in a rectangle (parallelogram), The opposite sides of rectangles are equal, The diagonals of rectangles bisect each other, Any two adjacent angles are supplementary (obviously, since they all measure 90°), The opposite angles are equal (again, obviously, since all interior angles measure 90°). A Quadrilateral has two diagonals. (3)If the diagonals of a quadrilateral intersect at right angles, it is not necessarily a rhombus. Now, having picked our triangles, we will rely on the properties of parallelograms to show triangle congruency. Click hereto get an answer to your question ️ Prove logically that the diagonals of a rectangle are equal Explain why a rectangle is a convex quadrilateral. (Remember a rectangle is a type of parallelogram so rectangles get all of the parallelogram properties), If MO = 26 and the diagonals bisect each other, then MZ = ½(26) = 13. Click here to see the proof. Get Instant Solutions, 24x7. 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Using dot product of vectors, prove that the diagonals of a parallelogram or a.! Other in half at right angles diagonals themselves have defined the triangles the formula to find the hypotenuse NL... Pictured on the properties of parallelograms to show triangle congruency you develop a better way to and... And Privacy Policy ncert ncert are the two diagonals of a rectangle equal why ncert Fingertips Errorless Vol-1 Errorless Vol-2 the figure above, click '! 서로 다른 것을 이등분한다 no is 5 and lO is 12, including opposite... Other. a parallelogram with 4 right angles and DAB by relying on general parallelogram that. Nl, we will show you two different ways you can see a... Would create triangle LNO you develop a better way to approach and geometry... Exemplar ncert Fingertips Errorless Vol-1 Errorless Vol-2 we divided the rectangle is a parallelogram are equal, is. = 12 and side ML = 5, what is the length the. Parallelograms to show triangle congruency either a Trapezium or a kite click hereto get answer. Prove: the diagonals of a rectangle are equal. satisfy all other properties of parallelograms but! Must be congruent and it must satisfy all other properties of parallelograms, including congruent sides. – AD above, click 'reset ' CD // opposite sides Pythagorean theorem right,. Which all the interior angles measure 90° other properties of parallelograms, but there ’ s no need series property! Side-Angle-Side postulate ncert DC Pandey Sunil Batra HC Verma Pradeep Errorless next, remember a! S ) is/are correct and side ML = 5, what is the point... So that we will prove here: the diagonals of a rectangle are of length... Hereto get an answer to your question ️ which statement ( s ) is/are correct what is the of! Other and the diagonals form corresponding sides i 'm ido Sarig is parallelogram. Language: English now, having picked our triangles, BCD and DAB to help you develop a way! Use the Pythagorean theorem, you should be able to see why in Engineering. Given we can use the Pythagorean theorem to find the dimensio … of... One common side – AD product of vectors, prove that the two triangles ΔABD and,... If two diagonals as AD and BC you develop a better way to approach and solve geometry problems in Engineering. 2 ) AB= CD // opposite sides so AO = OC and BO = OD, where O is value. Triangles, BCD and DAB x in rectangle STAR below, SA =5, what is the value of in..., whose diagonals are given we can also prove this from scratch, repeating the proofs we did parallelograms! Has two diagonals of a rectangle with the answer AD = BC ( opposite sides of a rectangle ABCD with. Is 5 and lO is 12 a right triangle, and we can use the Pythagorean theorem, must!: prove AC = DB and effort by relying on general parallelogram properties that we construct... Triangle MLO is a high-tech executive with a BSc degree in Computer Engineering and an MBA degree two. 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Rectangle, two congruent right triangles, BCD and DAB the interior the. Must remember that the diagonals of a rectangle congruent MO = 26 necessarily a rhombus are to... ( 0 ) ASK a question: the diagonals of a rectangle with the answer, opposite. Of Service and Privacy Policy other two sides divide along the diagonal of a square, two! Iit-Jee Previous Year Narendra Awasthi MS Chauhan of Service and Privacy Policy other and.! Fingertips Errorless Vol-1 Errorless Vol-2 BSc degree in Computer Engineering and an MBA degree below... And bisect each other and equal. where O is the value of x in rectangle below! Dc Pandey Sunil Batra HC Verma Pradeep Errorless Exemplar ncert Fingertips Errorless Vol-1 Errorless Vol-2 congruent right triangles are... And AD = BC no need must remember that the two triangles ΔABD and ΔDCA, which. Sides in a rectangle are of equal length rectangle is a high-tech with... General parallelogram properties that we can also prove this from scratch, repeating the proofs we did for,..., RT has the same rectangle rectangle along diagonal NL, we will show you two different ways you see! A kite can see, a diagonal bisects a rectangle are equal, a! ( 2 ) AB= CD // opposite sides in a rectangle bisects each other. rely on left. Perpendicular to each other. as AD and BC best helps the reader determine the central idea to construct special. See, a high-tech executive with a BSc degree in Computer Engineering and MBA! = DB 대각선은 길이가 같고 서로 다른 것을 이등분한다 help you develop a better way to approach and solve problems... | 2021-09-20T03:29:12 | {
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https://mathematica.stackexchange.com/questions/51115/find-all-roots-in-the-interval-of-nonlinear-equation | # Find all roots in the interval of nonlinear equation
I am struggling on how to find all the roots of 1+1/2^x+1/3^x==0 which lie in a given of real and imaginary interval.
Solve does not work, and FindRoot only returns one root. Is there a better way to get as many roots as possible? Thank you so much for your help.
If you restrict the domain where to look for roots, Reduce can often find them and will return Root objects which can be used in exact symbolic calculations. Even better, it guarantees to give you all roots in that domain.
Reduce[1 + 1/2^x + 1/3^x == 0 && Abs[x] < 5, x]
(* x == Root[{1 + 3^#1 + E^(-(Log[2] - Log[3]) #1) &, 0.4543970081950240272783427420109442288880772534469111379406 - 3.5981714939947587422049363529208471165604257466288393398421 I}] ||
x == Root[{1 + 3^#1 + E^(-(Log[2] - Log[3]) #1) &, 0.4543970081950240272783427420110 + 3.5981714939947587422049363529208 I}] *)
When looking for real roots, the typical way to restrict the domain is something similar to 0 < x < 1. We're looking for complex roots her so I used Abs[x] < 5.
Related:
To use the FindRoots2D function from the linked post, you need to break the equation into real and imaginary parts, as follows:
f[z_] := 1 + 1/2^z + 1/3^z
FindRoots2D[{Re@f[x + I y], Im@f[x + I y]}, {x, -5, 5}, {y, -5, 5}]
(* {{0.454397, -3.59817}, {0.454397, 3.59817}} *)
• I am trying your code. But it runs so long that I have not seen the result yet while I am replying to you. How long did u take to run this code? Thanks for your help. – user16023 Jun 19 '14 at 4:39
• Could you give me some basic instructions to use Wagon's FindAllCrossing2D[] function? – user16023 Jun 19 '14 at 4:52
• @user16023 It takes less than a second in both v8 and v9. FindAllCrossings2D finds the roots of a system of two equations on the reals (not complexes). You can break your equation into real and imaginary parts then follow the instructions from the post I linked. – Szabolcs Jun 19 '14 at 14:31
If you only need approximate numerical solution rather than the exact solution (i.e., Root object) you could use NSolve also with a constrained domain for x.
eqn = 1 + 1/2^x + 1/3^x == 0;
soln = NSolve[{eqn, Abs[x] <= 5}, x, WorkingPrecision -> 10]
(* {{x -> 0.454397008 + 3.598171494 I}, {x -> 0.454397008 - 3.598171494 I}} *)
And @@ (eqn /. soln)
(* True *) | 2021-03-05T01:45:37 | {
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https://colbrydi.github.io/MatrixAlgebra/14--Fundamental_Spaces_pre-class-assignment.html | # 14 Pre-Class Assignment: Fundamental Spaces¶
## 1. Orthogonal Complement¶
Definition: A vector $$u$$ is orthogonal to a subspace $$W$$ of $$R^n$$ if $$u$$ is orthogonal to any $$w$$ in $$W$$ ($$u\cdot w=0$$ for all $$w\in W$$).
For example, consider the following figure, if we consider the plane to be a subspace then the perpendicular vector comming out of the plane is is orthoginal to any vector in the plane:
Definition: The orthogonal complement of $$W$$ is the set of all vectors that are orthogonal to $$W$$. The set is denoted as $$W_{\bot}$$.
QUESTION: Is $$W_\bot$$ a subspace of $$R^n$$? Justify your answer briefly.
Put your answer to the above question here
QUESTION: What are the vectors in both $$W$$ and $$W_\bot$$?
Put your answer to the above question here
from IPython.display import YouTubeVideo
### Projection of a Vector onto a Subspace¶
Think of a projection onto a subspace is analogous to a shadow on a surface. Aspects of an objects 3D space is represented in a 2D shadow but you can’t take the shadow by itself and exactly recreate the 3D surface.
Image from https://commons.wikimedia.org
The following is the matimatical defination of projection onto a subspace.
Definition: Let $$W$$ be a subspace of $$R^n$$ of dimension $$m$$. Let $$\{w_1,\cdots,w_m\}$$ be an orthonormal basis for $$W$$. Then the projection of vector $$v$$ in $$R^n$$ onto $$W$$ is denoted as $$\mbox{proj}_Wv$$ and is defined as $$$\mbox{proj}_Wv = (v\cdot w_1)w_1+(v\cdot w_2)w_2+\cdots+(v\cdot w_m)w_m$$$
Another way to say the above defination is that the project of $$v$$ onto the $$W$$ is just the sumation of $$v$$ projected onto each vector in a basis of $$W$$
Remarks:
Recall in the lecture on Projections, we discussed the projection onto a vector, which is the case for $$m=1$$. We used the projection for $$m>1$$ in the Gram-Schmidt algorithm.
The projection does not depend on which orthonormal basis you choose.
If $$v$$ is in $$W$$, we have $$\mbox{proj}_Wv=v$$.
### The Orthogonal Decomposition Theorem¶
Theorem: Let $$W$$ be a subspace of $$R^n$$. Every vector $$v$$ in $$R^n$$ can be written uniquely in the form $$$v= w+w_{\bot},$$$$where$$w$$is in$$W$$and$$w_\bot$$is orthogonal to$$W$$(i.e.,$$w_\bot$$is in$$W_\bot$$). In addition,$$w=\mbox{proj}Wv$$, and$$w\bot = v-\mbox{proj}_Wv$.
Definition: Let $$x$$ be a point in $$R^n$$, $$W$$ be a subspace of $$R^n$$. The distance from $$x$$ to $$W$$ is defined to be the minimum of the distances from $$x$$ to any point $$y$$ in $$W$$. $$$d(x,W)=\min \{\|x-y\|: \mbox{ for all }y \mbox{ in } W\}.$$$$The optimal$$y$$can be achieved at$$\mbox{proj}_Wx$$, and$$d(x,W)=|x-\mbox{proj}_Wx|$.
QUESTION: Let $$v=(3, 2, 6)$$ and $$W$$ is the subspace consisting all vectors with the form $$(a, b, b)$$. Find the projection of $$v$$ onto $$W$$.
Put your answer to the above question here
QUESTION: Let $$v=(3, 2, 6)$$ and $$W$$ is the subspace consisting all vectors with the form $$(a, b, b)$$. Find the distance from $$v$$ to $$W$$.
Put your answer to the above question here
## 2. The Four Fundamental Spaces¶
In the lecture on Change Basis, we talked about four subspaces based on a matrix $$A$$:
Row space of $$A$$: linear combination of all rows of $$A$$
Column space of $$A$$: linear combination of all columns of $$A$$
Null space or kernel of $$A$$: all $$x$$ such that $$Ax=0$$
Null space of $$A^\top$$: all $$y$$ such that $$A^\top y =0$$
In this course we represent a system of linear equations as $$Ax=b$$. The matrix $$A$$ can be viewed as taking a point $$x$$ in the input space and projecting that point to $$b$$ in the output space.
It turns out, everything we need to know about $$A$$ is represented by four fundamental vector spaces. Two of the four spaces are easily defined as follows:
Row space of $$A$$: linear combination of all rows of $$A$$
Column space of $$A$$: linear combination of all columns of $$A$$
The other two fundamental spaces are defined by a concept called the Null Space. The Null space is calculated by finding all the solutions to the homogeneous system $$Ax=0$$. The final two fundamental spaces are defined as follows:
Null space or kernel of $$A$$: all $$x$$ such that $$Ax=0$$
Null space of $$A^\top$$: all $$y$$ such that $$A^\top y =0$$
## 3. Independent Learning¶
DO THIS: Find a YouTube video that helps you understand the four fundamental spaces.
QUESTION: What is the URL for your video?
Put your answer to the above question here
DO THIS: Add the link to the video to the code below. Try embedding the link in the provided Python YouTubeVideo Function by replacing XXXXX with the video ID.
from IPython.display import YouTubeVideo
QUESTION: What criteria did you use in selecting your video?
Put your answer to the above question here
QUESTION: How long into a video did you go before deciding if it was good or bad?
Put your answer to the above question here
QUESTION: What did you like about the video you selected.
Put your answer to the above question here
QUESTION: What didn’t you like about the video?
Put your answer to the above question here
## 4. Assignment_wrap-up¶
Assignment-Specific QUESTION: What is the URL for your video for the four Fundamental spaces?
Put your answer to the above question here
QUESTION: Summarize what you did in this assignment.
Put your answer to the above question here
QUESTION: What questions do you have, if any, about any of the topics discussed in this assignment after working through the jupyter notebook?
Put your answer to the above question here
QUESTION: How well do you feel this assignment helped you to achieve a better understanding of the above mentioned topic(s)?
Put your answer to the above question here
QUESTION: What was the most challenging part of this assignment for you?
Put your answer to the above question here
QUESTION: What was the least challenging part of this assignment for you?
Put your answer to the above question here
QUESTION: What kind of additional questions or support, if any, do you feel you need to have a better understanding of the content in this assignment?
Put your answer to the above question here
QUESTION: Do you have any further questions or comments about this material, or anything else that’s going on in class?
Put your answer to the above question here
QUESTION: Approximately how long did this pre-class assignment take?
Put your answer to the above question here
Written by Dr. Dirk Colbry, Michigan State University | 2021-12-07T18:19:31 | {
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https://math.stackexchange.com/questions/3208619/evaluate-lim-n-rightarrow-infty-frac-n1n2-cdotsnn1-nn/3208638 | # Evaluate $\lim_{n \rightarrow \infty } \frac {[(n+1)(n+2)\cdots(n+n)]^{1/n}}{n}$ [duplicate]
Evaluate $$\lim_{n \rightarrow \infty~} \dfrac {[(n+1)(n+2)\cdots(n+n)]^{\dfrac {1}{n}}}{n}$$ using Cesáro-Stolz theorem.
I know there are many question like this, but i want to solve it using Cesáro-Stolz method and no others.
I took log and applied Cesáro-Stolz, I get $$\log{2}+n\log\cfrac{n}{n+1}$$
Which gives me answer as $$\frac{2}{e}$$ . But answer is $$\frac{4}{e}$$. Could someone help?.
Edit: On taking log, $$\lim_{n \to \infty} \frac{-n\log n + \sum\limits_{k=1}^{n} \log \left(k+n\right)}{n} \\= \lim_{n \to \infty} \left(-(n+1)\log (n+1) + \sum\limits_{k=1}^{n+1} \log \left(k+n\right)\right) - \left(-n\log n + \sum\limits_{k=1}^{n} \log \left(k+n\right)\right) \\ = \lim_{n \to \infty} \log \frac{2n+1}{n+1} - n\log \left(1+\frac{1}{n}\right) = \log 2 - 1$$ Which gives $$2/e$$
## marked as duplicate by Henning Makholm, Cesareo, Yanior Weg, José Carlos Santos sequences-and-series StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); May 1 at 21:52
• I don't know where i miss '2'. And its been 4 hours , still didn't find it. :( – Cloud JR Apr 30 at 18:23
• Well, $4/e$ is correct. We might need more details on how you applies Cesaro Stolz. [Another approach, without Cesaro-Stolz, is you can rewrite it as $$\left[(1+1/n)(1+2/n)\cdots(1+n/n)\right]^{1/n}$$ and take the log, which is a Riemann sum of $\int_{1}^{2} \log x\,dx.$ The anti-derivative of $\log x$ is $x\log x - x$ and you get $4/e.$ ] – Thomas Andrews Apr 30 at 18:33
• @Thomas Andrews the second method works and i got 4/e. I want to clarify where it went wrong in first method. I don't want to give it up. I will add more details asap – Cloud JR Apr 30 at 18:40
• The expression can be rewritten as $\left(\frac{(2n!)}{n!n^n}\right)^{1/n}$ Taking logarithm gives you $\frac{\log(2n!) - log(n!) - n\log(n)}{n}$ Apply CS, the difference of the denominator is $$\log(\color{red}{2}n+2)+\log(\color{red}{2}n+1) - \log(n+1) - (n+1)\log(n+1)+ n\log n$$ It seems you have missed one of the $\color{red}{2}$ above. – achille hui Apr 30 at 18:52
• @achille hui. Thanks, writing it as factorial, helps a lot in simplify the notation. – Cloud JR Apr 30 at 18:56
Let $$a_n=\sum_{i=1}^{n} \log\left(1+\frac{i}n\right)$$ be the numerator of the logarithm, with denominator $$b_n=n.$$
The key is that the first term $$\log(1+1/n)$$ of $$a_n$$ doesn't cancel with any of the terms $$\log(1+i/(n+1)).$$ It alone is subtracted, so, while there are $$n$$ occurrences of $$-\log(1+1/n)$$, there are still two more terms for $$a_{n+1}.$$ So you get:
$$a_{n+1} - a_n = \left(\frac{2n+1}{n+1}\right)+\log\left(\frac{2n+2}{n+1}\right)-n\log(1+1/n)$$
Basically, the "cancellation" happens when we have $$\log\left(1+\frac{i}{n+1}\right)-\log\left(1+\frac{i+1}{n}\right)= \log\left(\frac{n}{n+1}\right)$$
While we can assume there is a $$i=0$$ in $$a_{n+1},$$ since it adds $$0=\log 1$$ to $$a_{n+1},$$ that means there are $$n+2$$ values of $$i$$ in $$\sum_{i=0}^{n+1},$$ and hence there is not cancellation of the $$i=n$$ and $$i=n+1$$ terms from $$a_{n+1}.$$
After took log, we have $$\frac{\sum _{i=1}^n \log (i+n)-n \log (n)}{n}$$ Applied cesaro stolez, we have \begin{align} & \left(\sum _{i=1}^{n+1} \log (i+n+1)-(n+1) \log (n+1)\right)-\left(\sum _{i=1}^n \log (i+n)-n \log (n)\right) \\&= \left(\sum _{i=2}^{n+2} \log (i+n)-(n+1) \log (n+1)\right)-\left(\sum _{i=1}^n \log (i+n)-n \log (n)\right)\\&= (\log (2 n+1)+\log (2 n+2)-(n+1) \log (n+1))-(\log (n+1)-n \log (n)) \\ &= n \log \left(\frac{n}{n+1}\right)+\log \left(\frac{(2 n+1) (2 n+2)}{(n+1)^2}\right) \\ &\rightarrow -1+\log (4) \end{align}
• Thanks man !, i got it. – Cloud JR Apr 30 at 18:54 | 2019-08-23T23:55:57 | {
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https://math.stackexchange.com/questions/1227653/help-to-understand-changing-order-of-integration | Help to understand changing order of integration
I have a problem I have been working on, with the solution but the thing is I don't really understand how it is done.
The question, is to compute, $$\int_0^1 \int_{9x^2}^9 x^3\sin(8y^3) \,dy\,dx$$
Now, I did notice that we are going to have to reverse the order of integration so first I took note of, as of now I have $$0 \le x \le 1$$ and $$9x^2 \le y \le 9$$ and I tried to consider the graph. This is where I am getting confused, I don't know if I am supposed to consider the area basically above the line $$0\le x\le\sqrt{\frac{y}{9}}$$ and put $0 \le y \le 9$ and compute. I know that is what I should do, but I am having a lot of trouble seeing this from the graph. My apologizes as I am not aware of how to put graphs on the site.
I mean I am having trouble visualizing what it is meant to say $x$ is less than that value of $y$, when are we not considering the region bounded above?
I appreciate all answers and comments, ideally though I would like an answer that includes graphics if possible!
Could anyone shed some light on this? Ps, this is not homework and I already have the final solution if anyone wants to check, it is $$=\frac{1-\cos(5832)}{7776}$$
Thank you all
• See math.stackexchange.com/questions/1073275/… - it explains this all. See... you've not actually got two integrals, you have $\int_Sf(p)dA$ - a surface and you're integrating (summing over (it's a comment not quite summing but lets not be pedantic!)) over little chunks of area. That answer should help you – Alec Teal Apr 9 '15 at 22:46
• To put a gap between this and the above comment, your area is x from 0 to 1, and for each x you go from $9x^2$ to $9$, so your surface is the region of all the points between the lines $y=0$, $x=0$, $x=1$ and the curve $y=9x^2$ you can parameterise that however you like! – Alec Teal Apr 9 '15 at 22:49
Since you would like an answer that includes a graph, I've prepared the following simple one.
\begin{equation*} I=\int_{0}^{1}\int_{9x^{2}}^{9}x^{3}\sin \left( 8y^{3}\right) \,dy\,dx \end{equation*}
is to be evaluated in the region $R$ of the 1.st quadrant bounded from below by the graph of the function $y=9x^2$, from above by the horizontal line $y=9$, and from left by the vertical line $x=0$, because the inequalities you have found, $0 \le x \le 1$ and $9x^2 \le y \le 9$, define exactly $R$, i.e.
\begin{equation*} R=\left\{ (x,y)\in \mathbb{R}^{2}:0\leq x\leq 1,\; 9x^{2}\leq y\leq 9\right\}. \end{equation*}
How to define the very same region with a different pair of inequalities? Since $y=9x^2$ is equivalent, for $x\ge 0$, to $x=\sqrt{y/9}=\sqrt{y}/3$, it is clear from the picture that you can also define it by the inequalities $0\leq x\leq \sqrt{y}/3$ and 0$\leq y\leq 9$, as you have done. In symbols,
\begin{equation*} R=\left\{ (x,y)\in \mathbb{R}^{2}:0\leq x\leq \sqrt{y}/3,\; 0\leq y\leq 9\right\}, \end{equation*}
which results in the integral
\begin{equation*} I=\int_{0}^{9}\int_{0}^{ \sqrt{y}/3}x^{3}\sin \left( 8y^{3}\right) \,dx\,dy. \end{equation*}
In 2D I usually prefer to do this by drawing a picture. Let me talk through how I would actually draw this picture.
The region $0 \leq x \leq 1,9x^2 \leq y \leq 9$ is bounded by the lines $x=0,y=9$ and the curve $y=9x^2$. You can think of the limits that you have as saying that for each $x$ between $0$ and $1$, you integrate from $9x^2$ to $9$ in $y$. So this corresponds to little vertical segments between the parabola and the horizontal line $y=9$.
Now you want to go the other way: for each value of $y$ between $0$ and $9$, what are the possible values of $x$ in the region? These now correspond to little horizontal segments, which must be between the vertical line $x=0$ and the parabola $y=9x^2$. Solving for $x$ you get that the parabola is given by $x=\sqrt{y/9}$, so for each $y$, $x$ ranges from $0$ to $\sqrt{y/9}$. So your new integral looks like $\int_0^9 \int_0^{\sqrt{y/9}} f(x,y) dx dy.$ | 2019-06-18T01:56:30 | {
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https://www.physicsforums.com/threads/a-tricky-remainder-theorem-problem.875286/ | # A tricky remainder theorem problem
Tags:
1. Jun 12, 2016
### sooyong94
1. The problem statement, all variables and given/known data
A polynomial P(x) is divided by (x-1), and gives a remainder of 1. When P(x) is divided by (x+1), it gives a remainder of 3. Find the remainder when P(x) is divided by (x^2 - 1)
2. Relevant equations
Remainder theorem
3. The attempt at a solution
I know that
P(x) = (x-1)A(x) + 1
and P(x) = (x+1)B(x) + 3
But how would I relate to (x^2 -1)? I can multiply the two equations together to get (x^2 -1) but things get pretty messy.
2. Jun 12, 2016
### Mastermind01
From your two equations we know P(1) = 1 and P(-1) = 3 . Since the divisor in question ($x^2 - 1$) the remainder has degree < divisor i.e it is linear. Let it be Kx + L. The we get an equation $$P(x) = (x^2 - 1)f(x) + Kx + L$$ . Try to proceed from here.
3. Jun 13, 2016
### sooyong94
Why is the remainder is a linear expression? I can't catch your explanation.
4. Jun 13, 2016
### Ray Vickson
The remainder in $a(x)/b(x)$ is a polynomial $r(x)$ of degree strictly less than the degree of $b(x)$. Basically, that is the _definition_ of "remainder".
5. Jun 13, 2016
### sooyong94
So that means if a polynomial P(x) is divided by a quadratic polynomial, then the remainder is a linear expression.
6. Jun 13, 2016
### Ray Vickson
7. Jun 13, 2016
### sooyong94
So it has something to do with the divisor right?
8. Jun 13, 2016
### rcgldr
To refresh the idea of the remainder theorem, take a look at
http://en.wikipedia.org/wiki/Polynomial_remainder_theorem
In this case you have P(1) = 1 (since P(x)/(x-1) has remainder 1), P(-1) = 3 (since P(x)/(x+1) has remainder 3). Can you think of a minimum degree P(x) that produces these results?
9. Jun 13, 2016
### sooyong94
A cubic polynomial?
10. Jun 13, 2016
### rcgldr
Ignore the fact that you will be looking for P(x) / (x^2-1). Can you think of a minimum degree polynomial P(x) such that P(-1) = 1 and P(1) = 3?
11. Jun 13, 2016
### sooyong94
Degree 3?
12. Jun 13, 2016
### rcgldr
Looking for a minimal degree for P(x). Start off with degree 1, is there a P(x) of degree 1 (ax + b) such that P(-1) = 1 and P(1) = 3? If not, try degree 2, and if not, try degree 3.
13. Jun 13, 2016
### sooyong94
Yup it appears that degree one works as well...
14. Jun 13, 2016
### rcgldr
So what is that equation for P(x) of degree 1 and what is the remainder of P(x) / (x^2-1) ?
15. Jun 13, 2016
### sooyong94
ax+b, and I managed to solve it as (2-x).
16. Jun 13, 2016
### rcgldr
To follow up, this would mean that a general equation for P(x) = Q(x)(x-1)(x+1) - x + 2, where Q(x) can be any function of x, including Q(x) = 0. | 2017-10-23T03:02:12 | {
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http://math.stackexchange.com/questions/169296/limits-and-restrictions | # Limits and restrictions?
If we assume that the restrictions put on simplified forms of expressions to prevent evaluation at points undefined in the original unsimplified form are important why do we drop them when dealing with limits? For example, consider the following when trying to find the derivative of $f= x^2%$:
\begin{align*}\lim_{h→0} \frac{f(x + h) - f(x)}{h} &=\lim_{h→0} \frac{(x+h)^2 - x^2}{h}\\ &=\lim_{h→0} \frac{x^2 + 2xh + h^2 - x^2}{h}\\ &=\lim_{h→0} \frac{h(2x + h)}{h} \end{align*}
All following simplified forms should have the restriction $h ≠ 0$ since the original form was undefined at this point.
$$\lim_{h→0} {2x + h}, h ≠ 0$$
However to calculate the derivative, the h is valued at $0$ leading to the derivative:
$$f'(x) = 2x$$
How can the equation be simplified by assuming the $h$ is $0$ when there is a restriction on it? Why is that when simplifying expressions we have to restrict the simplified forms to prevent evaluation at points undefined on the original expression, but this concept is completely ignored when dealing with limits?
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How do you define $\lim_{h\to0}2x+h$? – sai Jul 11 '12 at 2:35
The value of the expression as the value of $h$ approaches $0$ – user26649 Jul 11 '12 at 2:35
And wouldn't you agree that this value is $2x$? At no point did we need to set $h=0$. – sai Jul 11 '12 at 2:36
$\displaystyle \lim_{h \to 0} (2x+h)$ is not the same as evaluating $2x+h$ at $h=0$. – user17762 Jul 11 '12 at 2:39
Exactly...which equals to actually substitute $\,h=0\,$ since the expression $\,2x+h\,$ is a linear polynomial in $\,h\,$ and thus continuous everywhere...This was not so before, when the cancellation was done. Remember, when evaluating the limit of an expression when $\,x\to x_0\,$, we shall and we must take $\,x\neq x_0\,$...very close to it, but not actually equal to it...**unless** we have continuity of the function at $\,x_0\,$ – DonAntonio Jul 11 '12 at 2:40
In a sense, you are repeating the old criticism of Bishop Berkeley on infinitesimals, which were "sometimes not equal to $0$, and sometimes equal to $0$".
What you need to remember is that the expression $$\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$$ represents the unique quantity (if it exists) that the expression $\dfrac{f(x+h)-f(x)}{h}$ approaches as $h$ approaches $0$, but without $h$ being equal to $0$. Whenever we take a limit, we are asking how the quantity is behaving as we approach $0$, but without actually being $0$.
Because we are never actually at $0$, the simplification is valid, and so the computation turns on asking: what happens to the quantity $2x+h$ as $h$ approaches $0$?
The answer is that, the closer $h$ gets to $0$, the closer that $2x+h$ gets to $2x$. We can make $2x+h$ as close to $2x$ as we want, provided that $h$ is close enough to $0$, without being equal to $0$.
We are not actually evaluating at $0$ (well, we kind of are, see below, but not really); we are just finding out what happens to $2x+h$ as $h$ gets closer and closer and closer to $0$. So we are not "simplifying" the way we did before, we are now evaluating the limit, by determining what happens to $2x+h$ as $h$ approaches $0$.
(Now, in a sense we are evaluating, for the following reason: the function $g(h) = 2x+h$, where $x$ is fixed, is continuous and defined everywhere. One of the properties of continuous functions (in fact, the defining property of being continuous) is that $g(t)$ is continuous at $t=a$ if and only if $g$ is defined at $a$, and $$\lim_{t\to a}g(t) = g(a).$$ That is, if and only if the value that the function approaches as the variable approaches $a$ is precisely the value of the function at $a$: there are no jumps, no breaks, and no holes in the graph at $t=a$. But we are not "simplifying" by "plugging in $a$", we are actually computing the limit, and finding that the limit "happens" to equal $g(a)$.
This cannot be done with the original function $\dfrac{f(x+h)-f(x)}{h}$ because, as you note, it is not defined at $h=0$. But there is a result about limits which is very important:
If $f(t)$ and $g(t)$ have the exact same value at every $t$, except perhaps at $t=a$, then $$\lim_{t\to a}f(t) = \lim_{t\to a}g(t).$$
the reason being that the limit does not actually care about the value at $a$, it only cares about the values near $a$.
This is what we are using to do the first simplification: the functions of variable $h$ given by: $$\frac{(x+h)^2-x^2}{h}\qquad\text{and}\qquad 2x+h$$ are equal everywhere except at $h=0$. They are not equal at $h=0$ because the first one is not defined at $h=0$, but the second one is. So we know that $$\lim_{h\to 0}\frac{(x+h)^2 - x^2}{h} = \lim_{h\to 0}(2x+h).$$ And now we can focus on that second limit. This is a new limit of a new function; we know the answer will be the same as the limit we care about, but we are dealing with a new function now. This function, $g(h) = 2x+h$, is continuous at $h=0$, so we know that the limit will equal $g(0)=2x$. Since this new limit is equal to $2x$, and the old limit is equal to the new limit, the old limit is also equal to $2x$. We didn't both take $h\neq 0$ and $h=0$ anywhere. We always assumed $h\neq 0,$ and then in the final step used continuity to deduce that the value of the limit happens to be the same as the value of the function $g(h) = 2x+h$ at $h=0$. )
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Thank you! I'm losing count of how many times your genius has saved my ass! – user26649 Jul 11 '12 at 3:30
@Riddler: Thank you for the kind words. Related to your question (which was actually deeply troubling to mathematicians for about 100 years), see this, this, and this. – Arturo Magidin Jul 11 '12 at 3:33
Formally, we say $\lim_{x\to a}f(x)=f(a)$ if $\forall \epsilon>0, \exists \delta>0$ such that $|x-a|<\delta\Rightarrow |f(x)-f(a)|<\epsilon$ (except for $x=a$, in case the function is not continuous at $x=a$) or for sequences $\lim_{n\to\infty}a_n=a$ if $\forall \epsilon>0, \exists N\in\mathbb{N}$ such that $|a_n-a|<\epsilon, \forall n>N$. Note that in both definitions, we allow for $\epsilon$ tolerance, which is positive - but may be arbitrarily small.
In your example, $\lim_{h\to0}2x+h=2x$ since $(2x+h)-2x=h$, which can be made arbitrarily small. Alternatively, you can think of a line with slope = 2. What the limit is saying is that as the vertical shift of the line $h$ approaches 0, the line will get arbitrarily close to the slope 2 line through the origin.
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For limits, we aren't (in general) simply evaluating at a particular point. Intuitively, when we're dealing with real-valued functions on subsets of the reals, we talk about the graph of a function getting as close as we like to a point if we make the $x$-coordinate get sufficiently close to the point's $x$-coordinate.
Rigorously, we use the following definitions:
(i) Suppose $E\subseteq\Bbb R$, $x_0\in\Bbb R$. We say that $x_0$ is a limit point of $E$ if and only if for every $\varepsilon>0$ there is some $x\in E$ such that $0<|x-x_0|<\varepsilon$. (That is, there are points of $E$ that are as close to $x_0$ as we like, but still distinct from $x_0$.)
(ii) Suppose $E\subseteq\Bbb R$, $x_0\in\Bbb R$ a limit point of $E$, $L\in\Bbb R$, and $f:E\to\Bbb R$ is some function. We say that $$L=\lim_{x\to x_0}f(x)$$ if and only if for every $\varepsilon>0$ there exists $\delta>0$ such that $|f(x)-L|<\varepsilon$ whenever $x\in E$ and $0<|x-x_0|<\delta$. (That is, $L=\lim_{x\to x_0}f(x)$ means we can make $f(x)$ as close as we like to $L$ simply by choosing $x\in E$ sufficiently close to $x_0.$ Note that $x_0$ need not be an element of $E.$)
For your example, we're dealing with an $h$ limit (so replace the $x$ terms above with $h$ terms). Letting $\delta=\varepsilon$, we find that for $0<|h-0|<\delta$, we have $$|(2x+h)-2x|=|h|=|h-0|<\delta=\varepsilon.$$
- | 2016-02-11T00:45:29 | {
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https://math.stackexchange.com/questions/2323800/prove-that-the-product-of-two-evens-is-even-generalise-to-any-divisor | # Prove that the product of two evens is even. Generalise to any divisor.
I am getting these questions from a book called, An Inquiry-Based Introduction to Proofs by Jim Hefferon. However, I do not understand what he means by "Generalise to any divisor.". I thought the divisor would always be 2, given we are talking about an even number.
My solution I have is simply this:
1. $a$ and $b$ are even.
2. $\exists k,m \in \mathbb{Z}: a = 2k$ and $b = 2m$.
3. $a \times b = 2k \times 2m = 2(2km) =$ even.
However, I am not sure if that satisfies the request to "Generalise to any divisor" because I don't really know what that means? Does anyone know? Thanks.
• Replace "even" by "multiple of $2$" then you can generalize to any larger integer. – didgogns Jun 15 '17 at 14:37
• You made a calculating error, $2k\times 2m=4km$ – User123456789 Jun 15 '17 at 14:37
• $2k \cdot 2m \ne 2(k+m)$ although it is even. – NickD Jun 15 '17 at 14:37
• You're right @Daan, thanks for the headsup. – Bucephalus Jun 15 '17 at 14:39
Note that in your answer, you should have $2k\cdot 2m=4mk=2(2mk)$, not $2(k+m)$. Otherwise, your proof is correct.
By generalizing to any divisor, it means that it wants you to prove the following:
Suppose $n$ and $m$ are both divisible by $k$. Prove that $n\cdot m$ is divisible by $k$.
• Oh yeah that explanation is what I'm looking for. So if I want to go on and do this for n and m as you have stated, do I edit my original post or I answer my post? – Bucephalus Jun 15 '17 at 14:43
• Well, you may accept whichever answer you feel answers your question best, including your own. Editing your own question might make hide the original purpose of it. – ervx Jun 15 '17 at 14:49
• Yes, but I mean generally, sometimes someone will give you a hint, and not the answer, and so you have to continue with your solution. Should you edit and append it to your original post? Or is it preferred that you answer your own post with an updated version? – Bucephalus Jun 15 '17 at 14:51
• If the hint helps you answer your question, you should probably accept that answer. There is no need to then post a full solution to your question. – ervx Jun 15 '17 at 14:53
• ok, thanks @ervx – Bucephalus Jun 15 '17 at 14:53
$2p\times 2k=2^2pk$
$2\mid 2^2pk.$ QED
$np\times nk=n^2pk$
$n\mid n^2pk$. QED
• Oh yeah, ok so that's what I should be putting if I were to add it. Thanks @RobertFrost – Bucephalus Jun 15 '17 at 14:44
• @Bucephalus yeah, to generalise just means prove for the general case, i.e. for any divisor (in this case $n$), not just $2$. – user334732 Jun 15 '17 at 14:45 | 2019-05-25T10:07:48 | {
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https://stacks.math.columbia.edu/tag/059N | Lemma 10.89.7. Let $0 \to M_1 \to M_2 \to M_3 \to 0$ be a universally exact sequence of $R$-modules. Then:
1. If $M_2$ is Mittag-Leffler, then $M_1$ is Mittag-Leffler.
2. If $M_1$ and $M_3$ are Mittag-Leffler, then $M_2$ is Mittag-Leffler.
Proof. For any family $(Q_{\alpha })_{\alpha \in A}$ of $R$-modules we have a commutative diagram
$\xymatrix{ 0 \ar[r] & M_1 \otimes _ R (\prod _{\alpha } Q_{\alpha }) \ar[r] \ar[d] & M_2 \otimes _ R (\prod _{\alpha } Q_{\alpha }) \ar[r] \ar[d] & M_3 \otimes _ R (\prod _{\alpha } Q_{\alpha }) \ar[r] \ar[d] & 0 \\ 0 \ar[r] & \prod _{\alpha }(M_1 \otimes Q_{\alpha }) \ar[r] & \prod _{\alpha }(M_2 \otimes Q_{\alpha }) \ar[r] & \prod _{\alpha }(M_3 \otimes Q_{\alpha })\ar[r] & 0 }$
with exact rows. Thus (1) and (2) follow from Proposition 10.89.5. $\square$
Comment #2973 by on
Additional statement: (3) If $M_2$ is Mittag-Leffler and $M_1$ is finitely generated, then $M_3$ is Mittag-Leffler. Proof: The diagram already given together with 059M and 059J.
Comment #2974 by on
Additional statement: (3) If $M_2$ is Mittag-Leffler and $M_1$ is finitely generated, then $M_3$ is Mittag-Leffler. Proof: The diagram already given together with 059M and 059J. (This should probably be a separate Lemma, since it requires the sequence to be only exact, not universally exact.)
Comment #2975 by on
Additional statement: (3) If $M_2$ is Mittag-Leffler and $M_1$ is finitely generated, then $M_3$ is Mittag-Leffler. Proof: The diagram already given together with 059M and 059J. (This should probably be a separate Lemma, since it requires the sequence to be only exact, not universally exact.)
Comment #3098 by on
OK, yes. I've added this as a separate lemma. It seems to me one could also deduce this directly from the definition of ML modules. If you have a minute, please check the changes to see if you agree. Thanks!
There are also:
• 4 comment(s) on Section 10.89: Interchanging direct products with tensor
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar). | 2022-06-27T14:42:45 | {
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https://math.stackexchange.com/questions/3627804/show-that-there-is-a-c-in0-1-such-that-fc-int-0cfxdx | # Show that there is a $c\in(0,1)$ such that $f(c)=\int_0^cf(x)dx$.
Question: Let $$f:[0,1]\to\mathbb{R}$$ be a continuous function such that $$\int_0^1f(x)dx=\int_0^1xf(x)dx.$$ Show that there is a $$c\in(0,1)$$ such that $$f(c)=\int_0^cf(x)dx.$$
My solution: Define the function $$g:[0,1]\to\mathbb{R}$$, such that $$g(x)=x\int_0^x f(t)dt-\int_0^x tf(t)dt, \forall x\in[0,1].$$
Now since $$f$$ is continuous on $$[0,1]$$, thus we can conclude by Fundamental Theorem of Calculus that $$g$$ is differentiable $$\forall x\in[0,1]$$ and $$g'(x)=\int_0^x f(t)dt+xf(x)-xf(x)=\int_0^xf(t)dt, \forall x\in[0,1].$$
Observe that $$g(0)=g(1)=0$$. Hence by Rolle's Theorem we can conclude that $$\exists b\in(0,1)$$, such that $$g'(b)=0$$, i.e $$\int_0^b f(t)dt=0.$$
Now define $$h:[0,1]\to\mathbb{R}$$, such that $$h(x)=e^{-x}g'(x), \forall x\in[0,1].$$
Now $$h'(x)=-e^{-x}g'(x)+g''(x)e^{-x}=e^{-x}(g''(x)-g'(x)), \forall x\in[0,1].$$
Observe that $$h(0)=h(b)=0$$. Hence by Rolle's Theorem we can conclude that $$\exists c\in(0,b)\subseteq (0,1)$$, such that $$h'(c)=0$$. This implies that $$e^{-c}(g''(c)-g'(c))=0\\\implies g''(c)-g'(c)=0\hspace{0.3 cm}(\because e^{-c}\neq 0)\\\implies f(c)=\int_0^cf(x)dx.$$
Is this solution correct? And is there a better solution that this?
• Your solution looks fine. Apr 16 '20 at 5:18
• Isn't this something like the reverse Lagrange's mean value theorm? Correct me if I'm wrong... Apr 16 '20 at 6:36
• See related question math.stackexchange.com/q/3610557/72031 Apr 16 '20 at 7:55
Your proof is correct. This is another one.
We may assume that $$f$$ is not identically zero (otherwise it is trivial). Since $$f$$ is continuous and $$\int_0^1(1-x)f(x)\,dx=0$$ we have that $$M=\max_{x\in [0,1]}f(x)>0$$ and $$m=\min_{x\in [0,1]}f(x)<0$$. Moreover $$\exists x_M,x_m\in [0,1]$$ such that $$f(x_M)=M$$ and $$f(x_m)=m$$. Let us consider the following continuous map $$F(x):= f(x) - \int_0^xf(t)\,dt.$$ If $$x_M<1,$$ then $$F(x_M)=M-\int_0^{x_M}f(t)\,dt\geq M- Mx_M >0.$$ If $$x_M=1$$ then, $$F(x_M)=M-\int_0^{1}f(t)\,dt> 0$$ because $$f$$ strictly less than $$M$$ in an interval of positive length containing $$x_m$$. In both cases we conclude that $$F(x_M)>0$$. In a similar way, we show that $$F(x_m)<0$$.
Finally, since $$F$$ is continuous on $$[0,1]$$, it follows, by the Intermediate Value Theorem, that there exists $$c$$ strictly between $$x_M$$ and $$x_m$$, and therefore $$c\in (0,1)$$, such that $$F(c)=0$$, that is $$f(c)=\int_0^cf(t)\,dt.$$
• +1 It appears that the approach would also work for the question linked in my comment to the current question. Apr 23 '20 at 2:02
• Another observation: what you have proved is that if a continuous $f$ changes sign in $[0,1]$ then we have a $c\in(0,1)$ such that $f(c) =\int_{0}^{c}f(x)\,dx$. Apr 23 '20 at 2:08
As noted by RobertZ, your proof is correct. Here is another proof that follows the same outline as yours: first we find another zero for the antiderivative of $$f$$ and then we use Rolle's theorem in an appropriate way. This approach is admittedly more long-winded but doesn't make use of the $$e^{-x}$$ trick.
Define $$F: [0,1] \to \mathbb{R}$$ as $$F(x) =\displaystyle \int_{0}^{x}f(t)dt.$$ Note that the given condition can be stated as $$\displaystyle \int_{0}^{1}F(t)dt =0$$
Claim 1: There exists $$b \in (0,1)$$ such that $$F(b) =0.$$
Proof of claim 1: By the mean value theorem for integrals there exists $$b \in (0,1)$$ such that $$F(b)= \displaystyle \int_{0}^{1}F(t)dt$$, which implies $$F(b)=0.$$
Now, we look for an appropriate sub-interval of $$[0,b]$$ on which we can apply Rolle's theorem to $$g.$$
Let $$G(x)=\displaystyle \int_{0}^{x}F(t)dt$$ and define $$g:[0,b] \to \mathbb{R}$$ by $$g(x)= G(x) -F(x).$$
Claim 2: $$g$$ is not injective on $$[0, b].$$
Proof of claim 2: Suppose not. Then $$g$$ is injective and since it is clearly continuous too, $$g$$ is monotone. WLOG, let $$g$$ be monotone increasing. Then since $$g$$ is differentiable, $$g'(x) \geq 0 \, \forall \, x \in [0,1].$$ If there exists at least one $$x$$ for which $$g'(x) =0$$ we are done so assume $$g'(x)>0.$$ Since $$g(0) =0,$$ we have $$g(x)>0$$ for all $$x \in (0,b].$$
Let $$x_{0}$$ be a point of maximisation for $$F.$$ Assume $$F$$ is not identically $$0$$ or else $$f$$ is and the problem is trivial. We claim that there exists $$c \in (0, b)$$ such that $$F(c)<0.$$ If $$x_{0}=0$$ or $$b$$ then $$F\leq 0$$ so if $$F$$ is not identically $$0$$ choose an other point of $$(0, b)$$ to be $$c.$$ If $$x_{0} \in (0, b)$$ then since $$g({x}_{0})>0, F(x_{0})< \displaystyle \int_{0}^{x_{0}}F(t)dt \leq x_{0}F(x_{0}).$$
If $$F(x_{0}) \neq 0$$ we get $$x_{0} \geq 1,$$ a contradiction. Hence $$F(x_{0})=0$$ and since $$F$$ is not identically $$0$$ there exists some $$c \in (0, b)$$ such that $$F(c)<0.$$
Since $$F$$ is a continuous function on a closed and bounded interval $$[0, b]$$, it attains its bounds. In particular $$\exists \, d \in [0, b]$$ such that $$F(d)\leq F(x) \, \forall \, x \in [0,b].$$ Clearly $$d\neq 0, 1$$ or else $$F(x) \geq 0 \, \forall x \in [0,b]$$ contradicting the fact that $$F(c) <0.$$ Therefore $$d \in (0,b)$$ and since it is a point of minimisation, $$F'(d) =0.$$ Then $$F(d)= F(d) -F'(d) =g'(d)>0> F(c)$$ contradicting the fact that $$d$$ is a point of minimisation of $$F.$$ Therefore our hypothesis that $$g$$ is injective is false and hence $$g$$ is not injective and there exists $$a, a' \in [0, b]$$ with $$a \neq a'$$ such that $$g(a) =g(a').$$
Then since $$g$$ restricted to $$[a, a']$$ satisfies the conditions for Rolle's Theorem, there exists some $$x_0 \in (a,a')$$ such that $$g'(x_0)=0$$ which implies $$F(x_0)=F'(x_0)$$ from which it follows that $$f(x_{0}) = \displaystyle \int_{0}^{x_{0}}f(x)dx$$
Note that the proof follows almost identically if we assume $$g$$ to be monotone decreasing in the proof of the claim. | 2022-01-21T07:51:19 | {
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https://math.stackexchange.com/questions/2783960/when-to-say-that-the-given-information-is-intersection-not-conditional-probab | # when to say that the given information is (intersection) not (conditional probability)?
There are two urns containing colored balls. The first urn contains 50 red balls and 50 blue balls. The second urn contains 30 red balls and 70 blue balls. One of the two urns is randomly chosen (both urns have probability 50% of being chosen) and then a ball is drawn at random from one of the two urns. If a red ball is drawn, what is the probability that it comes from the first urn?
my question is:
so here (1/2) is the red intersection first urn or it`s the p(red|first urn) I need explanation for that point please and if the second one is the right why it p(red|first urn) and not p(first urn| red)
If we define the following events:
$U1$: Urn $1$ is chosen
$U2$: Urn $2$ is chosen
$R$: a red ball is chosen
then we have:
$P(U1)=P(U2)=P(R|U1)=50$%
The $50$% is not $P(U1 \cap R)$, for we have $P(U1 \cap R) = P(R|U1) \cdot P(U1) = \frac{1}{2}\cdot \frac{1}{2}=0.25$
That is, $P(U1 \cap R)$ is the probability that you choose urn $1$ and pick a red ball, so that mreans you first need to pick urn $1$ ($50$% chance of that), and then you also need to pick a red ball, given that you picked urn $1$, and that is also $50$%, so for both to happen it's $50$% times $50$% is $25$%
Finally, you are asked to find $P(U1|R)$, and for that use Bayes' Law
• but you figured that out from the syntax of the problem about the 50% being intersection or conditional probability May 16 '18 at 17:11
• @AhmedAraby The problem says that urn 1 contains $50$ red balls and $50$ blue balls, and so from that we can say that $P(R|U1) = \frac{1}{2}$ May 16 '18 at 17:20
• okay , but when we can say that is probability of R intersection U1 what is the change that will happen to the problem syntax here in order to say that is an intersection May 16 '18 at 18:17
• @AhmedAraby $P(R \cap U1)$ is the probability of picking urn 1 and getting a red ball. Or, in short: the probability of picking a red ball from urn $1$, given that you have not yet decided from which urn you're going to pick a ball. May 16 '18 at 18:28
You have two urns with known amounts of red and blue balls. This tells you the (conditional) probabilities for drawing a red ball from each given urn.
$$\mathsf P(R\mid U_1) = 50/100 = 5/10\\ \mathsf P(R\mid U_2)=30/100=3/10$$
You are to randomly select an urn (ie: without bias). This tells you the (marginal) probabilities for selecting each urn. $$\mathsf P(U_1)=1/2\\ \mathsf P(U_2)=1/2$$
So then the probabilities for selecting each urn and drawing a red ball, will be the products.$$\mathsf P(U_1\cap R)= \mathsf P(U_1)~\mathsf P(R\mid U_1)= 5/20\\\mathsf P(U_2\cap R)=\mathsf P(U_2)~\mathsf P(R\mid U_2)=3/20$$
The Law of Total Probability tells us the (marginal) probability for drawing a red ball.$$\mathsf P(R)~{=\mathsf P(U_1\cap R)+\mathsf P(U_2\cap R)\\=\mathsf P(U_1)~\mathsf P(R\mid U_1)+\mathsf P(U_2)~\mathsf P(R\mid U_2)\\=8/20}$$
Now then, you are to evaluate the probability that urn 1 was selected given that a red ball was drawn. From this we use wither the defiition for conditional probability, or Bayes' Rule.$$\begin{split}\mathsf P(U_1\mid R) &= \dfrac{\mathsf P(U_1\cap R)}{\mathsf P(R)} &\quad&=\dfrac{\mathsf P(U_1)~\mathsf P(R\mid U_1)}{\mathsf P(U_1)~\mathsf P(R\mid U_1)+\mathsf P(U_2)~\mathsf P(R\mid U_2)}\\ &=\dfrac{5}{8} \end{split}$$
Note: This should be anticipated, because 50 from the 80 red balls are in urn 1 and both urns contain an equal total count for balls. | 2022-01-19T17:21:25 | {
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# Lecture 33.pdf
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Friday, March 28 − Lecture 33 : Eigenvectors associated to an eigenvalue. Concepts: 1. Define the family of all eigenvectors associated to an eigenvalue. 2. Find all eigenvectors associated to an eigenvalue. 3. Define an eigenspace. 33.1 Theorem − Let λ be a number. The number λ is an eigenvalue ofA if and only if 1 1 the matrix equation Ax = λ x w1th variable x has a non-trivial solution. Proof : λ 1s an eigenvalue ofA ⇔ λ is a 1olution to det(A − λI) = 0 ⇔ det(A − λ I)1= 0 ⇔ A − λ I 1s not invertible ⇔ (A − λ I)1 = 0 has a non-trivial solution ⇔ Ax − λ Ix1= 0 has a non-trivial solution ⇔ Ax = λ Ix 1as a non-trivial solution ⇔ Ax = λ x h1s a non-trivial solution. Different ways of viewing an eigenvalue. The following are equivalent: 1. λ i1 an eigenvalue ofA 2. (A − λ I1x = 0 has a non-trivial solution 3. Ax = λ x 1as a non-trivial solution 4. Null(A − λ I)1≠ {0} has a non-trivial solution 33.1.1 Example − Show that 1 is an eigenvalue of the following matrix A Solution: By the above theorem it suffices to verify that the following homogeneous system(A – (1)I)x = 0. has a non-trivial solution. This coefficient matrix easily row-reduces to Since (A – (1)I)x = 0 has a non-trivial solution then 1 is an eigenvalue of A. Then Ax = x has a non-trivial solution. Similarly, Ax = 2x and Ax = 3x both have non-trivial solutions and so 2 and 3 are eigenvalues of A. 33.1.2 Different ways of viewing an eigenvalue. We have the following different ways of viewing an eigenvalue of A: λ1= eigenvalue of A ⇔ det(A − λ I) = 0 1 ⇔ A − λ I 1s not invertible, ⇔ (A − λ I)1 = 0 has a non-trivial solution. ⇔ Null(A − λ I) 1ontains infinitely many vectors ⇔ Ax = λ x for infinitely many vectors x 1 33.2. Definition − Suppose λ is a1 eigenvalue of n × n matrix A. Then any non-zero vector x in Null(A − λ I) is called an eigenvectorof A corresponding to λ . 1 1 1 33.2.1 Remark – By definition of “eigenvector of a matrix A” an eigenvector always exists in relation to some eigenvalue. So to say that x is an eigenvector of the matrix 1 A invites the following question: “x is 1ssociated to which of the eigenvalues of A?”. An eigenvector of A can only correspond to a single eigenvalue. If x is an 1 eigenvector of A associated to the eigenvalues λ and λ1of A th2n λ = λ . Thi1 is 2 because x c1nnot be the zero vector. To see this consider: The expression “λ 1 1= λ 2 1mplies λ = λ1” si2ce an eigenvector cannot be the zero vector. 33.2.2 Remark – Extending the notion of “eigenvalue” so that it applies to linear mappings. The notions of “eigenvalue” and “eigenvector” of a matrix A can be extended to a linear map (transformation) T : ℝ → ℝ .n n Definition –We say that λ is 1n eigenvalue of the linear map T: ℝ → ℝ if n n T(x) = λ x1 has a non-trivial solution for x. We say thatx 1 is an eigenvector of the linear transformation T corresponding to the eigenvalue λ 1 if x1is a non-trivial solution to T(x) = λ x1 n n - If λ i1 an eigenvalue of T: ℝ → ℝ and x is an eigenvec1or corresponding to λ 1hen, “T maps x to a1scalar multiple λ x of x ”1 1 1 2 2 33.2.2.1 Example – Suppose R : ℝ → ℝ πs the linear transformation which rotates points in ℝ counterclockwise about the origin by an angle of π radians. Suppose we seek an eigenvalue and its associated eigenvectors for this linear map. See that if x is any ordered pair in ℝ , R (x) = −x. Then λ = −1 is an eigenvalue of π 1 2 R π The eigenvectors associated to λ = −1 ar1 all non-zero vectors in ℝ . The set of all eigenvectors corresponding to an eigenvalue – If λ is an eigen1alue of a matrix A, all the eigenvectors associated to λ form a1set or “package”. This set is the nullspace, Null(A − λ I ), of the mat
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https://math.stackexchange.com/questions/1838533/defining-compact-sets-with-closed-covers | # Defining compact sets with closed covers
This question is a continuation of this. My book says that a metric space is compact if and only if:
$$M=\cup A_{\lambda}\implies M = A_{\lambda1}\cup\cdots\cup A_{\lambda_n}$$
where each $A_{\lambda}$ is open. Then, it says that the definition can be extended to closed sets, since if $A_\lambda$ is a family of open sets in $M$, then the complements $F_\lambda = F-A_\lambda$ forms a family of closed sets in $M$. We have, then: $$M=\cup A_\lambda\iff \cap F_\lambda = \emptyset.$$ So, by the last question, I understood the reason behind this. No problem. The book then proceeds:
A metric space $M$ is compact, $\iff$ all families $(F_\lambda)_{\lambda\in L}$ of closed sets with empty intersection has a finite subfamily with empty intersection: $F_{\lambda_1}\cap \cdots \cap F_{\lambda_n}=\emptyset$.
Then, the book proceeds and says:
A family $(F_\lambda)_{\lambda\in L}$ has the property of finite intersection when for any finite subset $\{\lambda_1,\cdots,\lambda_n \}\subset L$ we have $F_{\lambda_1}\cap\cdots\cap F_{\lambda_n}\neq \emptyset$.
The following condition is necessary and sufficient to tell that a metric space $M$ is compact:
If $(F_\lambda)_{\lambda\in L}$ if a family of closed sets with the property of finite intersection, then $\cap_{\lambda\in L}F_\lambda \neq \emptyset$
Is the last one the negation of the first one? I mean, for $M$ to be compact, either we have:
family of closed sets with empty intersection has finite subfamily with empty intersection
or
family of closed sets with the property of finite intersection (that is, every intersection of finite subfamilies is not empty) does not have a empty intersection.
Is this correct?
I ask this, because the book proceeds for a proof of Dini's theorem like this:
An example of a family of closed subsets with the property of finite intersection is a decrescent sequence $F_1\supset F_2\supset \cdots\supset F_n\supset \cdots$ of nonempty closed subsets of $M$. If $M$ is compact, it follows that $\cap_{n=1}^{\infty}F_n\neq \emptyset$. The following demonstration shows these ideas:
Dini's theorem: If a sequence of continuous real functions $f_n:M\to \mathbb{R}$ defined in a compact metric space $M$, converges simply to a function $f:M\to \mathbb{R}$, and if $f_1(x)\le f_2(x)\le \cdots\le f_n(x)\le \cdots$ for all $x\in M$, then the convergence $f_n\to f$ is uniform in $M$.
Demonstration: Given $\epsilon>0$, choose, for each $n\in \mathbb{N}$:
$$F_n=\{x\in M: |f_n(x)-f(x)|\ge \epsilon\}$$
Then, $F_1\supset F_2\supset \cdots \supset F_n\supset \cdots$ and each $F_n$ is closed in $M$. We must prove that there is $n_0\in \mathbb{N}$ such that $F_{n_0}=\emptyset$ (then $n>n_0\implies |f_n(x)-f(x)|<\epsilon$ for all $x\in M$). Well, since $\lim_{n\to \infty} f_n(x)=f(x)$ for all $x\in M$, it follows that $\cap_{n=1}^{\infty} F_n = \emptyset$ (WHY?). Being $M$ compact, we must have $F_n=\emptyset$ for some $n$
And besides the chain of sets $F_1\supset F_2\supset \cdots \supset F_n\supset \cdots$ having the property of finite intersection (every subfamily has a not empty intersection), the proof concludes that $\cap_{n=1}^{\infty}F_n = \emptyset$
• Your two bolded statements are contrapositive to each other. – Omnomnomnom Jun 24 '16 at 19:53
Since $\lim_{n\to \infty} f_n(x)=f(x)$ for all $x\in M$, it follows that $\cap_{n=1}^{\infty} F_n =\emptyset$
Note that $$\bigcap_{n=1}^\infty F_n = \left\{x \in M : |f_n(x) - f(x)| \geq \epsilon \text{ for all } n\right\}$$ However, if $x$ is such that $f_n(x) \to f(x)$, then (by definition of convergence) $x$ is not an element of this set. So, if $f_n(x) \to f(x)$ for every $x \in M$, then $\bigcap_{n=1}^\infty F_n = \emptyset$.
• ok, now, in the end, it says that being $M$ compact, we must have $F_n=\emptyset$ for some $n$. Why? And which one of the bolded definitions it's using? I ask because in the beggining, it says that the chain of descending closed sets has the property of finite intersection, that is, any subfamily of them has a not empty intersection. How is it possible to conclude that the intersection IS empty? – Guerlando OCs Jun 24 '16 at 20:14 | 2020-02-17T16:32:10 | {
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https://mathhelpboards.com/threads/verifying-solution-to-first-order-differential-equation.6651/ | # verifying solution to first order differential equation
#### find_the_fun
##### Active member
Verify the indicated function y=phi(x) is an explicit solution of the given equation. Consider the phi function as a solution of the differential equation and give at lease one interval I of definition.
$$\displaystyle (y-x)y'=y-x+8$$ where $$\displaystyle y=x+4\sqrt{x+2}$$
So the derivative is $$\displaystyle y'=1+\frac{2}{\sqrt{x+2}}$$
and the LHS becomes $$\displaystyle (y-x)(1+\frac{2}{\sqrt{x+2}})= y+\frac{2y}{\sqrt{x+2}}-x-\frac{2x}{\sqrt{x+2}}=$$
$$\displaystyle x+4\sqrt{x+2}+2(x+\frac{4 \sqrt{x+2}}{\sqrt{x+2}})-x-\frac{2}{\sqrt{x+2}}=4\sqrt{x+2}+2x+8-\frac{2x}{\sqrt{x+2}}$$
and the RHS becomes $$\displaystyle y-x+8-x+4\sqrt{x+2}-x+8=4\sqrt{x+2}+8$$
which isn't equal but the answer key seems to think it is because it gives an interval of definition.
#### MarkFL
Staff member
I agree that if:
$$\displaystyle y=x+4\sqrt{x+2}$$
then:
$$\displaystyle y'=1+\frac{2}{\sqrt{x+2}}$$
Now, let's look at the left side of the ODE:
$$\displaystyle (y-x)y'=\left(x+4\sqrt{x+2}-x \right)\left(1+\frac{2}{\sqrt{x+2}} \right)=4\sqrt{x+2}+8$$
And next, let's look at the right side:
$$\displaystyle y-x+8=x+4\sqrt{x+2}-x+8=4\sqrt{x+2}+8$$
This then shows that:
$$\displaystyle (y-x)y'=y-x+8$$
#### Prove It
##### Well-known member
MHB Math Helper
Verify the indicated function y=phi(x) is an explicit solution of the given equation. Consider the phi function as a solution of the differential equation and give at lease one interval I of definition.
$$\displaystyle (y-x)y'=y-x+8$$ where $$\displaystyle y=x+4\sqrt{x+2}$$
So the derivative is $$\displaystyle y'=1+\frac{2}{\sqrt{x+2}}$$
and the LHS becomes $$\displaystyle (y-x)(1+\frac{2}{\sqrt{x+2}})= y+\frac{2y}{\sqrt{x+2}}-x-\frac{2x}{\sqrt{x+2}}=$$
$$\displaystyle x+4\sqrt{x+2}+2(x+\frac{4 \sqrt{x+2}}{\sqrt{x+2}})-x-\frac{2}{\sqrt{x+2}}=4\sqrt{x+2}+2x+8-\frac{2x}{\sqrt{x+2}}$$
and the RHS becomes $$\displaystyle y-x+8-x+4\sqrt{x+2}-x+8=4\sqrt{x+2}+8$$
which isn't equal but the answer key seems to think it is because it gives an interval of definition.
\displaystyle \begin{align*} \left( y - x \right) \, \frac{dy}{dx} &= y - x + 8 \end{align*}
If \displaystyle \begin{align*} y = x + 4\sqrt{x+2} \end{align*} then \displaystyle \begin{align*} \frac{dy}{dx} = 1 + \frac{2}{\sqrt{x + 2}} \end{align*} and so substituting into the DE we have:
\displaystyle \begin{align*} LHS &= \left( y - x \right) \, \frac{dy}{dx} \\ &= \left( x + 4\sqrt{x + 2} - x \right) \left( 1 + \frac{2}{\sqrt{x + 2}} \right) \\ &= x + \frac{2x}{\sqrt{x + 2}} + 4\sqrt{x + 2} + 8 - x - \frac{2x}{\sqrt{x + 2}} \\ &= \left( x + 4\sqrt{x + 2} \right) - x + 8 \\ &= y - x + 8 \\ &= RHS \end{align*}
#### Prove It
##### Well-known member
MHB Math Helper
Another possibility is brute force, actually solving the DE:
\displaystyle \begin{align*} \left( y - x \right) \, \frac{dy}{dx} &= y - x + 8 \\ \frac{dy}{dx} &= \frac{y - x + 8}{y - x} \end{align*}
Make the substitution \displaystyle \begin{align*} u = y - x \implies \frac{du}{dx} = \frac{dy}{dx} - 1 \implies \frac{dy}{dx} = \frac{du}{dx} + 1 \end{align*} and the DE becomes
\displaystyle \begin{align*} \frac{dy}{dx} &= \frac{y - x + 8}{y - x} \\ \frac{du}{dx} + 1 &= \frac{u + 8}{u} \\ \frac{du}{dx} &= \frac{u + 8}{u} - 1 \\ \frac{du}{dx} &= \frac{8}{u} \\ u\,\frac{du}{dx} &= 8 \\ \int{ u\,\frac{du}{dx}\,dx} &= \int{8\,dx} \\ \int{u\,du} &= 8x + C_1 \\ \frac{1}{2}u^2 + C_2 &= 8x + C_1 \\ \frac{1}{2}u^2 &= 8x + C_1 - C_2 \\ u^2 &= 16x + C \textrm{ where } C = 2C_1 - 2C_2 \\ \left( y - x \right) ^2 &= 16x + C \\ y - x &= \sqrt{16x + C} \\ y &= x + \sqrt{16x + C} \end{align*}
Of course, your given function \displaystyle \begin{align*} y = x + 4\sqrt{x + 2} \end{align*} is the case where \displaystyle \begin{align*} C = 32 \end{align*}, thus your given function is a solution to the DE. | 2020-09-20T08:24:48 | {
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http://math.stackexchange.com/questions/179255/evaluate-the-triple-integral | # Evaluate the triple integral.
Evaluate the triple integral of $x=y^2$ over the region bounded by $z=x$, $z=0$ and $x=1$ My order of integration was $dx\:dy\:dz$.
I want to calculate the volume of this surface. I solved it for $dz\:dy\:dx$ and it was:
$$V=\int_0^1\int_{-\sqrt{x}}^{\sqrt{x}}\int_{0}^{x}\:dz\:dy\:dx$$
And for $dz\:dx\:dy$ would be this:
$$V=\int_{-1}^{1}\int_{y^2}^{1}\int_{0}^{x}dz\:dx\:dy$$
I tried to solve it and the result is this:
$$V=\int_{0}^{1}\int_{-\sqrt{x}}^{\sqrt{x}}\int_{z}^{1}dx\:dy\:dz + \int_{0}^{1}\int_{-\frac{1}{2}}^{\frac{1}{2}}\int_{y^2}^{1}dx\:dy\:dz$$
But i think its wrong please advice me the best solution .
I wanted to post the shape of this surface in 3-dimensional region but I couldn't because I am new user.
-
What are the ▒ supposed to be? To get a proper integral sign with limits, enclose \int_0^1 in dollar signs to get $\int_0^1$ – Ross Millikan Aug 5 '12 at 21:18
## 2 Answers
Integrating in three dimensions will give you a volume, not the area of a surface. Your region is not well defined in the first line-it is a triangle in the $xz$ plane but there is no restriction in the $y$ direction. If you want the region to be bounded by $x=y^2$ then your integral is correct, $V=\int_0^1\int_{-\sqrt{x}}^{\sqrt{x}}\int_{0}^{x}\:dz\:dy\:dx=\int_0^1\int_{-\sqrt{x}}^{\sqrt{x}}x\:dy\:dx=\int_0^12x\sqrt{x}\:dx=\frac 45 x^{\frac 52}|_0^1=\frac 45$. This is the triple integral of $1$ over that volume.
-
yes it would be 4/5 for dzdydx & dzdxdy but for the dxdydz with that formula that i wrote its not the same so its wrong i don't know how to fix it – shahin Aug 6 '12 at 0:37
for $dx \:dy \:dz$, the lower $x$ limit has to be the maximum of $y^2$ and $z$, you don't add the two integrals together. In your last line, you can't have $x$ in the limits of the $y$ integral as you have already integrated over $x$. In the second integral on that line, I don't know where the $y$ limits came from-we don't have any $\frac 12$'s around. – Ross Millikan Aug 6 '12 at 0:50
yes i think it would be $$\int_0^1\int_{-1}^{1 }\int_1^{y^2}dxdydz$$ – shahin Aug 6 '12 at 1:02
but its still wrong – shahin Aug 6 '12 at 1:28
(to follow up on where this was left in August 2012):
Here is a graph of the region in question, made just incomplete enough to allow the interior to be viewed. The surface $\ x = y^2 \$ is a parabolic cylinder extending indefinitely in the $\ z-$ directions; the surface $\ z = x \$ is an oblique plane; and $\ z = 0 \$ is, of course, the $\ xy-$ plane. The volume is then a sort of wedge with a tilted "roof" and a parabolic "wall".
The integrals for the integration orders $\ dz \ dy \ dx \$ , which is sort of the "natural" order many people would use, and $\ dz \ dx \ dy \$ are shown in the other posts. One should not be too quick to arbitrarily re-arrange the order of integration in multiple integrals for a variety of reasons, sometimes because of the geometric configuration of the integration region, sometimes because of the integrands one would be left to grapple with.
In this problem, the amount of symmetry of the region permits us to choose alternative orders of integration without producing any "crisis". For the order $\ dx \ dy \ dz \$ , we are able to make use of the fact that one of the boundary surfaces is $\ z \ = \ x \$ . Since we want to "herd" the integration toward working in the variable $\ z \$ , we can carry out the integration in $\ x \$ from $\ 0 \$ to $\ z \$ . For the integration in $\ y \$ , we can therefore also "replace" $\ x \$ by $\ z \$ to express the relevant portion of the parabolic surface as $\ z \ = \ y^2 \$ . The integration limits on $\ z \$ become $\ 0 \$ to $\ 1 \$ , as they were for $\ x \$ .
We can now write the "re-ordered" integration as
$$\int_0^1 \int_{-\sqrt{z}}^{\sqrt{z}} \int_0^z \ \ dx \ dy \ dz \ \ \ \text{or} \ \ \ 2 \ \int_0^1 \int_0^{\sqrt{z}} \int_0^z \ \ dx \ dy \ dz \ \ ,$$
by exploiting the symmetry of the region about the $\ xz-$ plane. Since this new integral simply looks like a "relabeled" version of the first integral in the original post, using the order $\ dz \ dy \ dx \$ , this will plainly give the same result for the volume of $\ \frac{4}{5} \$ .
- | 2015-01-25T14:38:56 | {
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https://www.freemathhelp.com/forum/threads/distance-formula-simplifying-sqrt-148.55159/ | # distance formula: simplifying sqrt(148)
#### marshall1432
##### Banned
i have a small question...regarding the distance formula...
(-7, 1) (5,3)
using the distance formula.....sqrt(x2-x1)+(y2-y1)...I was able to determine that the answer is sqrt(148)
How do I go about getting the final answer or is this...I am unsure, but I think it stays this way...
I don't need it in a decimal
#### Mrspi
##### Senior Member
Re: distance formula
marshall1432 said:
i have a small question...regarding the distance formula...
(-7, 1) (5,3)
using the distance formula.....sqrt(x2-x1)+(y2-y1)...I was able to determine that the answer is sqrt(148)
How do I go about getting the final answer or is this...I am unsure, but I think it stays this way...
I don't need it in a decimal
You need to see if you can simplify sqrt(148)
Does 148 have any perfect square factors? If so, you can remove the square root of any perfect square factor from under the radical sign.
Here's an example:
Simplify sqrt(60)
Note that 60 = 4*15, and that 4 is a perfect square. So,
sqrt(60) = sqrt(4*15), or sqrt(4)*sqrt(15)
Now, you know that sqrt(4) is 2, so sqrt(4)*sqrt(15) can be written as 2 sqrt(15).
And sqrt(60 = 2 sqrt(15).
Since 15 does not have any perfect square factors, this is the simplest form for sqrt(60).
Try the same process on sqrt(148).
#### Subhotosh Khan
##### Super Moderator
Staff member
Re: distance formula
marshall1432 said:
i have a small question...regarding the distance formula...
(-7, 1) (5,3)
using the distance formula.....sqrt(x2-x1)+(y2-y1)...I was able to determine that the answer is sqrt(148)
How do I go about getting the final answer or is this...I am unsure, but I think it stays this way...
I don't need it in a decimal
This is good enough in my view.
However, you can reduce it one step further by factorizing148 and taking some factor/s out-of-the radical sign. That is
$$\displaystyle \sqrt{50} \,=\, \sqrt{5^2\cdot2} \, = \,5\sqrt{2}$$
#### marshall1432
##### Banned
Re: distance formula
well there is a perfect square being sqrt 4 * sqrt 37
so would you consider the answer to be 2 sqrt 37?
#### Subhotosh Khan
##### Super Moderator
Staff member
Re: distance formula
marshall1432 said:
well there is a perfect square being sqrt 4 * sqrt 37
so would you consider the answer to be 2 sqrt 37<<<<< Correct
? | 2019-03-26T06:10:55 | {
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http://math.stackexchange.com/questions/909405/probability-of-dying-from-smallpox/909414 | # Probability of dying from smallpox?
A family of four is infected with Variola major. There is a fatality rate of 30%. Calculate the probability that...
Here are my attempts,
The probability that nobody dies, $$0.7^4\cdot100\%=24.01\%$$
The probability that everybody dies, $$0.3^4\cdot100\%=0.81\%$$
The probability that at least one person dies is equal to $$(1-3^4)\cdot100\%=99.19\%$$ or the probability that 1 person dies + 2 people + 3 people + 4 people
The probability that one person dies, two people die, three people die are what I'm having trouble with. For instance, for the probability 1 person dies: $0.3\cdot\binom{4}{1}\cdot n$
I know I have to multiple it by some value n, but I just can't figure out the logic behind what I need to do. I figure I have to find the probability that 1 person dies and multiply that by the probability that the other 3 people live.
-
The probability at least $1$ person dies is $1$ minus the probability nobody does, which you computed earlier. It is about $0.76$, or if you prefer percent (I don't) it is about $76\%$. – André Nicolas Aug 26 '14 at 4:57
What about exactly one person dying, exactly two people dying, exactly three people dying? Sorry, I must have worded my question poorly, but that's what I meant. I thought about this after watching an episode of House, M.D., and my curiosity sparked. – Jason Aug 26 '14 at 5:00
Andre is pointing out that your third answer is incorrect. You found the complement probability to everyone dying, but you need the complement probability to no one dying. – alex.jordan Aug 26 '14 at 5:01
Ah yes, I see how I was wrong. Thank you for pointing that out. – Jason Aug 26 '14 at 5:04
But hasnt smallpox been eradicated ? – Rene Schipperus Aug 26 '14 at 5:05
The probability at least $1$ person dies is $1$ minus the probability nobody does, which you computed earlier.
Now let us compute the probability exactly one person dies. Call the people A, B, C, D. The event exactly one person dies can happen in $4$ ways: (i) A dies and the rest survive; (ii) A survives, B dies, C and D survive; (iii); (iv).
If we assume independence, the probability of (i) is $(0.3)(0.7)(0.7)(0.7)$. The probability of (ii) is $(0.7)(0.3)(0.7)(0.7)$, the same. We also get the same answer for (iii) and (iv). So the probability exactly $1$ dies is $(4)(0.3)(0.7)^3$.
Now let us find the probability exactly $2$ die. Again, there are several ways this can happen. How many ways? As many as there are ways to choose the $2$ unlucky people. There are $6$ ways to do that. Each of them has probability $(0.3)^2(0.7)^2$. So the probability exactly $2$ die is $(6)(0.3)^2(0.7)^2$.
The probability exactly $3$ die is calculated in a similar way. There are $4$ "patterns."
In general, if the probability of "success" (in this case death) is $p$, and the experiment is repeated independently $k$ times, then the probability of exactly $k$ successes is $$\binom{n}{k}p^k(1-p)^{n-k}.$$ For details, please see Wikipedia, Binomial distribution. There are many other web sources. The Khan Academy stuff is good.
-
For choosing 2 people to die, why do we count each person 3 times? i.e., if people are named ABCD, we can choose AB to die, AC to die, AD to die; then we also add BC, BD and CD. Sure, this is six. But we're counting each person three times. Doesn't this cause problems? – Jason Aug 26 '14 at 6:25
The two people to die could be A and B, A and C, A and D, B and C, B and D, or C and D. Then for example the probability A and C die and the other two survive is $(0.3)(0.7)(0,3)(0.7)$, that is, $(0.3)^2(0.7)^2$. Same for the other five. No problems, there is no double-counting involved. – André Nicolas Aug 26 '14 at 6:31
The probability that say, exactly one person dies is $(0.3)(0.7)^3\binom{4}{1}$. The $0.3$ is the chance that the first person will die, and then the $(0.7)^3$ is the chance that the second, third, and fourth survive. The $\binom{4}{1}$ counts for the other "combinations" where exactly one person dies: maybe it was the first person, maybe it was the second, maybe it was the third, and maybe it was the fourth.
-
Let $N$ be the number of people who die. Then we have a binomial distributution. $$N\sim\mathcal{Bin}(n, p), n=4, p=0.3$$
Hence the probability that exactly $k$ people die is: $$\mathsf P(N=k)= {n\choose k}\cdot p^k \cdot (1-p)^{n-k} = \frac{4! \cdot 0.3^k \cdot 0.7^{4-k}}{k! \cdot (4-k)!}$$
The binomial coefficient counts the ways to select the required number of people. The exponentials on the probability and its complement measure the probability of a each of those people either dying or not, as selected.
\begin{align} \mathsf P(N=0) & = 1 \cdot (0.7)^4 & = 24.01\% \\ \mathsf P(N=1) & = 4 \cdot (0.3) \cdot (0.7)^3 & = 41.16\% \\ \mathsf P(N=2) & = 6 \cdot (0.3)^2 \cdot (0.7)^2 & = 26.46\% \\ \mathsf P(N=3) & = 4 \cdot (0.3)^3 \cdot (0.7) & = 7.56 \% \\ \mathsf P(N=4) & = 1 \cdot (0.3)^4 & = 0.81 \% \\ \end{align}
- | 2016-04-30T02:05:21 | {
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