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https://math.stackexchange.com/questions/803954/surely-youre-joking-mr-feynman-int-0-infty-frac-sin2xx21x2-dx/803957 | # Surely You're Joking, Mr. Feynman! $\int_0^\infty\frac{\sin^2x}{x^2(1+x^2)}\,dx$ [duplicate]
Prove the following $$$$\int_0^\infty\frac{\sin^2x}{x^2(1+x^2)}\,dx=\frac{\pi}{4}+\frac{\pi}{4e^2}$$$$
I would love to see how Mathematics SE users prove the integral preferably with the Feynman way (other methods are welcome). Thank you. (>‿◠)✌
Original question:
And of course, for the sadist with a background in differential equations, I invite you to try your luck with the last integral of the group.
$$$$\int_0^\infty\frac{\sin^2x}{x^2(1+x^2)}\,dx$$$$
Source: Integration: The Feynman Way
• "The Feynman way" is just a fancy name for the good old differentiation under the integral? – Daniel Fischer May 21 '14 at 11:30
• @DanielFischer I know that Sir but I prefer naming "The Feynman Way" (ô‿ô) – Anastasiya-Romanova 秀 May 21 '14 at 11:33
• @DanielFischer: there seems to be a tendency on this site to label any differentiation of an artificially introduced parameter under the integral sign as "Feynman's <method, way, etc.>." I doubt Feynman invented it, but I see that loads of people read "Surely You're Joking...". – Ron Gordon May 21 '14 at 12:20
• @RonGordon Indeed. It's a good read, by the way. Now, will you have lemon or milk in your tea? – Daniel Fischer May 21 '14 at 12:23
• @V-Moy: Hey, I never said it was wrong; I just think it is a little funny. BTW it would not be the first time that a method or something else is named after its popularizer rather than its inventor. – Ron Gordon May 21 '14 at 12:38
This integral is readily evaluated using Parseval's theorem for Fourier transforms. (I am certain that Feynman had this theorem in his tool belt.) Recall that, for transform pairs $f(x)$ and $F(k)$, and $g(x)$ and $G(k)$, the theorem states that
$$\int_{-\infty}^{\infty} dx \, f(x) g^*(x) = \frac1{2 \pi} \int_{-\infty}^{\infty} dk \, F(k) G^*(k)$$
In this case, $f(x) = \frac{\sin^2{x}}{x^2}$ and $g(x) = 1/(1+x^2)$. Then $F(k) = \pi (1-|k|/2) \theta(2-|k|)$ and $G(k) = \pi \, e^{-|k|}$. ($\theta$ is the Heaviside function, $1$ when its argument is positive, $0$ when negative.) Using the symmetry of the integrand, we may conclude that
\begin{align}\int_0^{\infty} dx \frac{\sin^2{x}}{x^2 (1+x^2)} &= \frac{\pi}{2} \int_0^{2} dk \, \left ( 1-\frac{k}{2} \right ) e^{-k} \\ &= \frac{\pi}{2} \left (1-\frac1{e^2} \right ) - \frac{\pi}{4} \int_0^{2} dk \, k \, e^{-k} \\ &= \frac{\pi}{2} \left (1-\frac1{e^2} \right ) + \frac{\pi}{2 e^2} - \frac{\pi}{4} \left (1-\frac1{e^2} \right )\\ &= \frac{\pi}{4} \left (1+\frac1{e^2} \right )\end{align}
• @V-Moy: Yeah, I'm fast like that. – Ron Gordon May 21 '14 at 12:38
• Surely You're Joking, Mr. Gordon! （‐＾▽＾‐） – Anastasiya-Romanova 秀 May 21 '14 at 12:41
• @RonGordon Why did you answer the same question again since you knew it was a duplicate? You have already answered this question for me : math.stackexchange.com/questions/691798/…. – Jeff Faraci May 22 '14 at 14:15
• @Integrals: because I answered it three months ago and at my age, dementia sets in. I admit that sometimes I just answer the question without looking back through my vast database of answers. So, no, I did not know it was a duplicate. – Ron Gordon May 22 '14 at 14:21
• @Integrals: I would have if I had known. But so what - the worst thing in the world is that duplicate correct (and consistent) answers exist in the site? Vote to close as a duplicate if this bothers you. – Ron Gordon May 22 '14 at 14:27
Here is my attempt: \begin{align} \int_0^\infty\frac{\sin^2x}{x^2(1+x^2)}dx&=\int_0^\infty\left[\frac{\sin^2x}{x^2}-\frac{\sin^2x}{1+x^2}\right]dx\\ &=\int_0^\infty\frac{\sin^2x}{x^2}dx-\frac{1}{2}\int_0^\infty\frac{1-\cos2x}{1+x^2}dx\\ &=\frac{\pi}{2}-\frac{1}{2}\int_0^\infty\frac{1}{1+x^2}dx+\frac{1}{2}\int_0^\infty\frac{\cos2x}{1+x^2}dx\\ &=\frac{\pi}{2}-\frac{1}{2}\frac{\pi}{2}+\frac{1}{2}\frac{\pi}{2e^2}\\ &=\frac{\pi}{4}+\frac{\pi}{4e^2} \end{align} where I use these links: $\displaystyle\int_0^\infty\frac{\sin^2x}{x^2}dx$ and $\displaystyle\int_0^\infty\frac{\cos2x}{1+x^2}dx$ to help me out.
Unfortunately, this is not the Feynman way but I still love this method.
• Thanks for fixing my answer @MycrofD. I'm too excited so I forgot about my English grammar. ≥Ö‿Ö≤ – Anastasiya-Romanova 秀 May 21 '14 at 11:30
• Next time put your attempt in the question. Besides, you got it right (>‿◠)✌. – Shahar May 21 '14 at 11:53
• @Shahar Thanks, +1 for your emo. ≧◠◡◠≦✌ – Anastasiya-Romanova 秀 May 21 '14 at 12:09
• @V-Moy You seem to be an expert in emo's and integrals. ✌ – Sawarnik May 21 '14 at 17:43
• @Sawarnik You might say I'm expert in emo but not in integral, not yet (✿◠‿◠) – Anastasiya-Romanova 秀 May 22 '14 at 8:17
Consider $$I(a)=\int_0^\infty\frac{\sin^2(ax)}{x^2(x^2+1)}dx$$ Differentiate it twice. Since $$\int_0^\infty\frac{\cos(kx)}{x^2+1}dx=\frac{\pi}{2e^k}$$ for $k>0$ we get $I''(a)=\pi e^{-2a}$. Note that $I'(0)=I(0)=0$, so after solving respective IVP we get $$I(a)=\frac{\pi}{4}(-1+2a+e^{-2a})$$ Now it only remains to substitute for $a=1$.
• Wait a second! Why never cross to mind to use Feynman method twice?? +1 Sir! ≥►.◄≤ – Anastasiya-Romanova 秀 May 21 '14 at 12:47
• @V-Moy Because you're not thinking like Feynman! Suppose the Feynman method just transforms your complicated integral into another complicated integral. Well, what tricks do you know for handling complicated integrals? That's right, the Feynman method! :) – David H May 21 '14 at 13:59
• Thanks for your advice Mr. @DavidH. I'll try to think like Feynman. I have a lot of time though. ᕙ(^▽^)ᕗ – Anastasiya-Romanova 秀 May 21 '14 at 16:17
• Why did you post this solution again? You knew this was a duplicate, this is a great solution but still. -1. math.stackexchange.com/questions/691798/… – Jeff Faraci May 22 '14 at 14:14
• @Integrals because it is not forbidden, and this answer perfectly fits to the question even more than the in the link you gave above. – Norbert May 22 '14 at 15:47 | 2020-10-25T22:30:51 | {
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https://math.stackexchange.com/questions/3678612/is-the-sequence-x-n-dfrac1-sqrtn-left1-dfrac1-sqrt2-dfrac1-s | # Is the sequence $x_n=\dfrac{1}{\sqrt{n}}\left(1+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+\ldots+\dfrac{1}{\sqrt{n}}\right)$ monotone?
Observe that $$x_1=1$$ and $$x_2=\dfrac{1}{\sqrt{2}}\left(1+\dfrac{1}{\sqrt{2}}\right)>\dfrac{1}{\sqrt{2}}\left(\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}\right)=1$$.
Thus, $$x_2>x_1$$. In general, we also have $$x_n=\dfrac{1}{\sqrt{n}}\left(1+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+\ldots+\dfrac{1}{\sqrt{n}}\right)>\dfrac{1}{\sqrt{n}}\left(\dfrac{1}{\sqrt{n}}+\dfrac{1}{\sqrt{n}}+\dfrac{1}{\sqrt{n}}+\ldots+\dfrac{1}{\sqrt{n}}\right)=1$$.
Thus, $$x_n\geq 1$$ for all $$n\in \mathbb{N}$$. Also, we have,
$$x_{n+1}=\dfrac{1}{\sqrt{n+1}}\left(\sqrt{n}x_n+\dfrac{1}{\sqrt{n+1}}\right)=\dfrac{\sqrt{n}}{\sqrt{n+1}}x_n+\dfrac{1}{n+1}$$.
Is it true that $$x_{n+1}>x_n$$?
Edit : Thanks to the solution provided by a co-user The73SuperBug. Proving $$x_{n+1}-x_n>0$$ is equivalent to proving $$x_n<1+\dfrac{\sqrt{n}}{\sqrt{n+1}}$$. This is explained below :
\begin{aligned} &x_{n+1}-x_n>0\\ \Leftrightarrow & \dfrac{\sqrt{n}}{\sqrt{n+1}}x_n+\dfrac{1}{n+1}-x_n>0\\ \Leftrightarrow & \left(1-\dfrac{\sqrt{n}}{\sqrt{n+1}}\right)x_n-\dfrac{1}{n+1}<0\\ \Leftrightarrow & x_n<\dfrac{1}{(n+1)\left(1-\dfrac{\sqrt{n}} {\sqrt{n+1}}\right)}\\ \Leftrightarrow & x_n<1+\dfrac{\sqrt{n}}{\sqrt{n+1}}. \end{aligned}
• It's basically a Riemann sum for a convex function. The (good) accepted answer focuses on the specific function you have, but I think a generalization would not harm, so I've added an answer based on this alternative idea. May 17, 2020 at 12:17
The sequence is indeed increasing. Using what you have left off we need to prove: $$x_{n+1} - x_n > 0\iff ...x_n < 1+\dfrac{\sqrt{n}}{\sqrt{n+1}}$$. We prove this by induction on $$n \ge 1$$. Clearly $$x_1 = 1 < 1+ \sqrt{\frac{1}{2}}$$. Assume $$x_n < 1+\dfrac{\sqrt{n}}{\sqrt{n+1}}$$, we show: $$x_{n+1} < 1+\dfrac{\sqrt{n+1}}{\sqrt{n+2}}$$. Using the recursive formula you had above: $$x_{n+1} = \dfrac{\sqrt{n}}{\sqrt{n+1}}x_n+\dfrac{1}{n+1}< \dfrac{\sqrt{n}}{\sqrt{n+1}}\left(1+\dfrac{\sqrt{n}}{\sqrt{n+1}}\right)+\dfrac{1}{n+1}= 1+\dfrac{\sqrt{n}}{\sqrt{n+1}}< 1+\dfrac{\sqrt{n+1}}{\sqrt{n+2}}$$ which is clear because $$n(n+2) < (n+1)^2$$. Thus by induction $$x_n < 1 +\dfrac{\sqrt{n}}{\sqrt{n+1}}$$ and in turn implies $$x_{n+1} > x_n, \forall n \ge 1$$. Thus the sequence is increasing.
A generalization I couldn't pass by. Let $$\color{blue}{S_n=\frac1n\sum_{k=1}^{n-1}f\big(\frac{k}{n}\big)}$$ where $$f:(0,1)\to\mathbb{R}$$ is strictly convex: $$f\big((1-t)a+tb\big)<(1-t)f(a)+tf(b)\quad\impliedby\quad a If we put $$a=k/(n+1),b=(k+1)/(n+1),t=k/n$$ for $$0 here, we obtain $$f\Big(\frac{k}{n}\Big)<\Big(1-\frac{k}{n}\Big)f\Big(\frac{k}{n+1}\Big)+\frac{k}{n}f\Big(\frac{k+1}{n+1}\Big),$$ which, after summing over $$k$$ (to have "$$<$$" still, we must assume $$n>1$$), gives $$\sum_{k=1}^{n-1}f\Big(\frac{k}{n}\Big)<\sum_{k=1}^{\color{red}{n}}\Big(1-\frac{k}{n}\Big)f\Big(\frac{k}{n+1}\Big)+\sum_{k=\color{red}{1}}^{n}\frac{k-1}{n}f\Big(\frac{k}{n+1}\Big)=\frac{n-1}{n}\sum_{k=1}^{n}f\Big(\frac{k}{n+1}\Big),$$ i.e. $$\color{blue}{n^2 S_n<(n^2-1)S_{n+1}}$$. Returning to the question, if we put $$f(x)=1/\sqrt{x}$$, we get $$x_n=S_n+1/n$$ and $$n^2 x_n<(n^2-1)x_{n+1}+1$$; since $$x_{n+1}>1$$, the latter implies the needed $$x_n. | 2022-08-15T01:12:28 | {
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http://mathhelpforum.com/discrete-math/77647-midterm-review.html | # Math Help - Midterm review
1. ## Midterm review
Agh, I've been working on this question for an hour and can't seem to make any headway. I seem to be completely lost on it.
a.) Show that the positive integers less than 11, except 1 and 10 can be split into pairs of integers such that each pair consists of integers that are inverses of each other modulo 11.
b.) use part (a) to show that 10! = -1(mod 11)
Any help would be appreciated
2. Originally Posted by Niotsueq
Agh, I've been working on this question for an hour and can't seem to make any headway. I seem to be completely lost on it.
a.) Show that the positive integers less than 11, except 1 and 10 can be split into pairs of integers such that each pair consists of integers that are inverses of each other modulo 11.
b.) use part (a) to show that 10! = -1(mod 11)
For a), you can find out by trial and error that, for example, $2\times6 = 12\equiv1\!\!\!\pmod{11}$. Therefore 2 and 6 are inverses of each other. So also are 7 and 8, because $7\times8 = 56\equiv1\!\!\!\pmod{11}$.
For b), $10! = 1\times 2\times3\times\cdots\times10 = 1\times(2\times6)\times\cdots\times(7\times8)\time s\cdots\times10$ (pair the numbers off as in part a)).
3. Hello, Niotsueq!
a) Show that the positive integers less than 11, except 1 and 10,
can be split into pairs of integers such that each pair consists of integers
that are inverses of each other modulo 11.
. . . . . $\begin{array}{ccc}2\cdot6 &\equiv & 1\text{ (mod 11)} \\3\cdot4 &\equiv & 1\text{ (mod 11)} \\ 5\cdot9 &\equiv & 1\text{ (mod 11)} \\ 7\cdot8 & \equiv& 1 \text{ (mod 11)} \end{array}$
b) use part (a) to show that 10! = -1 (mod 11)
$10! \;\equiv\;1\cdot2\cdot3\cdot4\cdot5\cdot6\cdot7\cd ot8\cdot9\cdot10\,\text{ (mod 11)}$
. . . $\equiv \;1\cdot(2\cdot6)\cdot(3\cdot4)\cdot(5\cdot9)\cdot (7\cdot8)\cdot10\,\text{ (mod 11)}$
. . . $\equiv\;1\cdot 1\cdot 1\cdot 1\cdot1\cdot10\,\text{ (mod 11)}$
. . . $\equiv\;10\,\text{ (mod 11)}$
. . . $\equiv\;\text{-}1\,\text{ (mod 11)}$
4. Thanks! That's a lot easier than I thought. But, I'm not quite sure why, for instance:
2*6 = 1 (mod 11)
How does this say that 2 and 6 are inverses? I think I may be confused on the definition.
Thanks alot to both of you.
5. Originally Posted by Niotsueq
Thanks! That's a lot easier than I thought. But, I'm not quite sure why, for instance:
2*6 = 1 (mod 11)
How does this say that 2 and 6 are inverses? I think I may be confused on the definition.
Thanks alot to both of you.
$2\cdot 6 \equiv 1 \mod 11$
this shows that they are inverses of each other.
remember what an inverse is in regular multiplication?
$3 \cdot \frac{1}{3}=1$
do you see how the original means they are inverses now?
6. Originally Posted by GaloisTheory1
$2\cdot 6 \equiv 1 \mod 11$
this shows that they are inverses of each other.
remember what an inverse is in regular multiplication?
$3 \cdot \frac{1}{3}=1$
do you see how the original means they are inverses now?
So two numbers are inverses of one another mod(n) if their product is congruent to one mod(n)? | 2014-08-23T05:42:15 | {
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https://math.stackexchange.com/questions/518739/prove-that-the-sequence-a-n-defined-by-a-0-1-a-n1-1-frac-1a-n/518791 | # Prove that the sequence $(a_n)$ defined by $a_0 = 1$, $a_{n+1} = 1 + \frac 1{a_n}$ is convergent in $\mathbb{R}$
I will post the exercise below:
Prove that the sequence $(a_n)$ defined by $a_0 = 1$, $a_{n+1} = 1 + \frac 1{a_n}$ for $n \in \mathbb N$ is convergent in $\mathbb R$ with the Euclidean metric, and determine afterwards is limit. Can you intepret the limit geometrically (hint: Golden ratio)?
So I need to prove that the sequence is convergent in $\mathbb{R}$ with the Euclidean metric, and how do I prove that? The limit must be $1$, but how to interpret it geometrically?
• All I am willing to say is that "hint: Golden ratio" is a very strong hint. Oct 8, 2013 at 9:43
• The limit is not 1. @Duronman, the series is not monotonically increasing either. Oct 8, 2013 at 9:47
• @AdrianRatnapala true! I wrote too fast, sorry, I thought of $a_n + \frac{1}{a_n}$ because of the $1$. I will delete the comment to avoid confusion. Oct 8, 2013 at 9:54
Hint: If the limit $L$ exists, it must satisfy $L = 1 + \frac{1}{L}$, and so it cannot be 1. The solutions are the roots of the equation $L^2 - L - 1 = 0$, and so $L \in \{\frac{1+\sqrt{5}}{2}, \frac{1-\sqrt{5}}{2} \}$. That's where the golden ratio comes into play. Note also that the limit cannot be $\frac{1-\sqrt{5}}{2}$ since $a_n > 0$ for all $n$.
Let $\phi$ be the golden ratio, which is the only positive real such that $\phi = 1 + 1/\phi$. Now consider the sequence $\epsilon_n = a_n - \phi$. We seek to prove the invariant:
$$|\epsilon_n/\epsilon_0| \le \phi^{-n}.$$
For $n=0$, this is clear by inspection. With some algebra we see that:
$$\epsilon_{n+1} = 1 + {1\over a_n} - \phi = -{\epsilon_n\over a_n\phi}.$$
Now our invariant already implies that $a_n = \phi + \epsilon_n \ge \phi - \phi^{-n}\epsilon_0 \ge 1.$ Therefore
$$|\epsilon_{n+1}| \le \left|{\epsilon_n\over \phi}\right| \le \phi^{-(n+1)}.$$
So our invariant remains after $n \rightarrow n+1$. And therefore for all nonnegative integers $n$ by induction. But since $\phi > 1$, that invariant means that $\epsilon_n$ approaches zero or $a_n$ approaches $\phi$ as $n$ goes to infinity.
Writing out the first few terms will yield a very interesting pattern:
$a_0 = \frac{1}{1}, a_1 =\frac{2}{1}, a_2=\frac{3}{2}, a_3=\frac{5}{3} ...$
Do u see it now. Every term is a ratio of fibonacci nos. i.e. if $T_n$ is the nth fibonacci number starting from 1, then $a_n = \frac{T_{n+2}}{T_{n+1}}$. The limit of this is indeed the golden ratio. Try solving the difference equation to see it in case you are unaware of this result.
As for the proof that the observation is correct and not a coincidence, use induction. We verified the base cases above, so assume $$a_n = \frac{T_{n+2}}{T_{n+1}}$$
Now $$a_{n+1} = 1+ \frac{1}{a_n} = 1 + \frac{T_{n+1}}{T_{n+2}}= \frac{T_{n+2} + T_{n+1}}{T_{n+2}} = \frac{T_{n+3}}{T_{n+2}}$$
Hence proved. $\blacksquare$
• But you still need to prove that the ratios of consecutive Fibonnaci numbers actually converge to anything at all. Oct 10, 2013 at 8:44
• I told OP to solve the difference equation for $a_n = a_{n-1} + a_{n-2}$. From that you can evaluate the limit. Oct 10, 2013 at 8:48
• I am missing some knowledge about difference equations. The easiest method to find that limit (if it exists) I know is to find the roots of the iteration rule translated into a polynomial. But I did not know the limit was guaranteed to exist. Oct 14, 2013 at 18:07
• You are referring to the Z Transform method for solving difference equations. With it, you will get the solution for all n. Then you can proceed by my method. Oct 14, 2013 at 18:45
• Adrian, another way is to get the general solution to the recurrence relation and then just take limits. There will invariably be a dominating term which determines the limit. Dec 22, 2013 at 3:30
We will consider decreasing function $f:(0, \infty)\rightarrow(0,\infty), f(x)=1+\frac{1}{x}$. Function $fof:(0, \infty ) \rightarrow(0, \infty)$ is increasing and therefore subsequence $a_{2n},n\geq0$ is increasing, and subsequence $a_{2n +1},n\geq0$ is decreasing. Because $1\leq a_{2n}<a_{2n +1}\leq2$, it follows that both are converging subsequences with common limit the golden ratio, which is the limit sequence in question. | 2022-07-06T14:20:03 | {
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https://math.stackexchange.com/questions/1088313/how-many-different-tsuro-tiles-can-exist | # How many different Tsuro tiles can exist?
The boardgame Tsuro consists of tiles, which each have 8 entry points. Each tile connects each point to exactly one other point. The game manual claims every tile is unique. The game consists of 35 such tiles.
How many unique such tiles could possibly exist?
My reasoning:
• You start with 8 free points. Choose any point, you now have 7 possibilities to connect.
• You now have 6 free points. Choose any point, you now have 5 possibilities to connect.
• You now have 4 free points. Choose any point, you now have 3 possibilities to connect.
• You now have 2 free points. Connect them (no choice possible).
This would lead to 7*5*3 : 105 possibilities. But I wouldn't know how to eliminate "doubles" caused by rotating a tile. Should I divide by 4, since 4 rotations are possible? That would be 26 tiles... but the game itself contains 35 and they are unique.
How should I reason?
• Well, here's why you can't just divide by $4$: Not all the rotations are unique - consider the tile in which each point is connected to the one directly opposite it on the other side - then this tile is rotationally symmetric. Another example in which not all rotations are unique is the bottom left tile in the picture you provided. – Peter Woolfitt Jan 2 '15 at 10:31
You can use Burnside's Lemma. I'll go through the calculation here, though you should read that page for this to make sense.
The set X is all 105 possibilities. The group G that acts on X is $\langle r | r^4 = 1 \rangle$, where r is a 90 degree rotation. Burnside's Lemma is applicable since each tile corresponds exactly to one orbit in X under G. The number of elements of X fixed by each element of G is
• $1: 105$
• $r: 5$
• $r^2: 25$
• $r^3: 5$
So the number of orbits (tiles) is $\frac{1}{4}(105+5+25+5)=35$. The hardest part is computing the $r^2: 25$ entry; here's how I did it:
• Case 1: No pair of antipodal points is connected. Choose any point: you have 6 legal possibilities to connect. Whatever you choose, $r^2$ fixes a second symmetrical connection, so you have 4 points left with 2 legal ways to connect them, for a total of 6*2=12 possibilities.
• Case 2: 2 pairs of antipodal points are connected. There are $\binom{4}{2}=6$ ways to pick the pairs and 2 ways to connect the reamining 4 points, for another 12 possibilities.
• Case 3: All 4 antipodal pairs are connected. There is only 1 such possibility.
• You'd probably also find this page interesting. – Benjamin Cosman Aug 28 '16 at 21:14
• $<abc>$ looks like this: $<abc>$, and $\langle abc\rangle$ looks like this $\langle abc\rangle$. #FriendsDoNotLetFriendsUseTheWrongAnglyThings – Mariano Suárez-Álvarez Aug 28 '16 at 21:19
• Interesting approach, thank you! – Konerak Aug 29 '16 at 7:07
One can reduce further the $105$ possible tiles, down to the minimum possible of $35$ topologically unique tiles via $mod\, 8$ arithmetic, $1+7=(8)=0$.
Modulo $8$ arithmetic (each tile has $8$ ports numbered $0123457$) allows $CW$ and $CCW 2$-dimensional rotations (can not flip a tile, as a flip would represent a $3D$ rotation). To keep this simple, there are no negative integers here, $0-1=7$ is being interpreted as $7+1=0$.
A $45^{\circ}$ rotation is represented by adding/subtracting 1 to/from the port #.
Rotations of $90^{\circ}, 180^{\circ}, 270^{\circ}$ are represented by adding/subtracting $2, 4, 6$ respectively. I just added $2, 4, 6$ used only addition.
Represent a pair of two connected ports of a tile by a $2$-digit number. For example $15$ means port#1 is connected to port#5, i.e. not decimal fifteen.
Represent each tile via a quad set of number pairs, non repeating digits,for example $\{01, 23, 45, 67\}$.
Note $\{01, 14, 26, 57\}$ is not a valid tile as digit "1" -port#1- appears twice.
Example of $180^{\circ}$ rotation:
Tile $\{04, 12, 36, 57\}$ is rotated $180^{\circ}$ by adding (modulo 8) $+4$ to each of its digits.
$$0+4=4$$
$$4+4=(8)=0$$
$$1+4=5$$
$$2+4=6$$
$$3+4=7$$
$$6+4=(10)=2$$
$$5+4=(9)=1$$
$$7+4=(11)=3$$
Thus rotated tile becomes $\{40, 56, 72, 13\}$ and will order/describe/ its ports from low-to-high (as each numbers-pair is commutative, $40=04$ and $72=27$).
Therefore this $180^{\circ}$ rotated tile becomes $\{04, 13, 27, 56\}$.
Rotate all $105$ tiles by $0^{\circ}, 90^{\circ}, 180^{\circ}, 270d^{\circ}$, (find & write all quad sets).
Use Excel, reorder the port# numbers from low-to-high, and eliminate all duplicate tiles.
End result = $35$ unique quad sets representing $35$ unique tiles.
May follow my analysis summary at:
There are 5 ways to make a tile that is invariant under $90^\circ$ rotations: Just connect one pint to any other point except its rotational predecessor or successor, then the rest is fixed. This would correct your count to $\frac{105-5}4+5=29$, which is more, but still too few. You should also count, how many tiles are possible with a $180^\circ$ symmetry. Of course, the $90^\circ$ symmetric tiles are among them, but there are also some "new" such tiles. If there are $k$ such tiles, then the total count gets corrected to $\frac{105-5-k}{4}+5+\frac k2$, so you can make an educated guess what $k$ is ;) | 2019-06-26T22:19:04 | {
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https://math.stackexchange.com/questions/614468/an-extrasensory-perception-strategy/614546 | # An extrasensory perception strategy :-)
Inspired by classical Joseph Banks Rhine experiments demonstrating an extrasensory perception (see, for instance, the beginning of the respective chapter of Jeffrey Mishlove book “The Roots of Consciousness”), I consider the following experiment. A deck of cards is given to a magician John. Then John consecutively takes the cards from the deck, trying to guess suit of the taken card. He looks at the card after the guess for a feedback. The magician wishes to maximize the expected number $E$ of right guesses. For this purpose he devised the following Strategy: at each turn to guess a suit which has a maximal number of card left in the deck. As an easy exercise we can prove that for any sequence of cards in the deck Strategy ensures at least $n$ right guesses, where $n$ is the maximal number of cards with one suit in the deck. But we can consider a more interesting and complicated problem to calculate the expectation $E$ for Strategy (here we are assuming that the deck is so well shuffled such that all sequences of cards have equal probability). By the way, I conjecture that Strategy is the best for maximizing the expectation $E$, that is any other strategy yields not greater value of $E$. Now I wish to evaluate the expectation $E$ for Strategy. For the simplicity we shall consider only a case when there are only two suits ($m\ge 0$ cards of the first suit and $n\ge m$ cards of the second suit). Then $E(0,n)=n$ for each $n$ and we have the following recurrence $$E(m,n)=\frac{n}{m+n}(E(m,n-1)+1)+ \frac{m}{m+n}E(m-1,n)$$
for each $n\ge m\ge 1$.
The rest is true provided I did not a stupid arithmetic mistake.
I was interested mainly in asymptotics for the case $m=n$ and computer calculations suggested that $E(n,n)\sim n+c\sqrt{n}+o(\sqrt{n})$ for $c\approx 0.88\dots$.
Evaluating formulas for $E(m,n)$ for small values of $m\le 6$, I conjectured that there is a general formula
$$E(m,n)=n+m\sum_{i=1}^m\frac {c_{m,i}}{n+i}$$
for each $n\ge m\ge 1$, where $c_{m,i}$ are some integers satisfying the recurrence
$$(m-i)c_{m,i}+ic_{m,i+1}=(m-1)c_{m-1,i}$$
for every $1\le i\le m-1$.
Here are my values for $c_{m,i}$
i\m| 1 2 3 4 5 6
---+------------------------
1 | 1 2 4 8 16 32
2 | -1 -4 -12 -32 -80
3 | 1 6 24 80
4 | -1 -8 -40
5 | 1 10
6 | -1
Then I discovered that for my data $c_{m,i}$ is divisible by $2^{m-i}$. After I did the division, I surprisingly obtained that $$c_{m,i}=(-1)^{i-1}2^{m-i}{m-1 \choose i-1}.$$ I expect that I can easily prove this equality by induction.
But I did not stop at this point because I observed that now the general formula for $E(m,n)$ can be compressed to the form
$$E(m,n)=n+m\int_0^1 x^n(2-x)^{m-1} dx.$$
All of above sounds nice for me and I spent a good time investigating the problem, but I am a professional mathematician, although I am not a specialist in the domain of the above problem. Therefore I care about the following questions. Are the above results new, good and worthy to be published somewhere? What another related problems are worthy to be investigated?
Thanks and merry Holidays.
• – Alex Ravsky Dec 21 '13 at 0:49
• Your expression for $E(m,n)$ looks a lot like Kolmogorov forward equation. – Alex Dec 21 '13 at 1:21
• Is your table of $c_{m,i}$ correct for $i=1$? The triangle and your expression for $c_{m,i}$ looks very like OEIS A013609 – Henry Dec 23 '13 at 22:15
• @Henry Thanks, I corrected the table. – Alex Ravsky Mar 9 '14 at 15:10
• Your integral expression for $E(m,n)$ is closely related to $\displaystyle B_{1/2}(n+1,m) = \int_{0}^{1/2} x^n (1-x)^{m+1} dx$, an incomplete Beta function. – Henry Mar 9 '14 at 16:58
This is to address your question about the asymptotics when $m=n$.
We only need to study the integral
$$I(n) = \int_0^1 x^n (2-x)^{n-1}\,dx$$
or, after we've made the change of variables $x = 1-y$,
\begin{align} I(n) &= \int_0^1 (1-y)^n (1+y)^{n-1}\,dy \\ &= \int_0^1 (1+y)^{-1} \exp\left[n \log\left(1-y^2\right)\right]\,dy. \end{align}
We'll proceed using the Laplace method. The quantity $\log(1-y^2)$ achieves its maximum at $y=0$, and near there we have
$$\log\left(1-y^2\right) \sim -y^2.$$
This motivates us to make the change of variables $\log(1-y^2) = -z^2$, so that
$$I(n) = \int_0^\infty \frac{z e^{-z^2}}{(1+\sqrt{1-e^{-z^2}})\sqrt{1-e^{-z^2}}} e^{-nz^2}\,dz.$$
Near $z=0$ we have
$$\frac{z e^{-z^2}}{(1 + \sqrt{1-e^{-z^2}})\sqrt{1-e^{-z^2}}} = 1 - z + \frac{z^2}{4} + \frac{z^4}{96} - \frac{z^6}{384} + \cdots,$$
and integrating term-by-term we obtain the asymptotic series
$$I(n) \approx \frac{\sqrt{\pi}}{2n^{1/2}} - \frac{1}{2n} + \frac{\sqrt{\pi}}{16n^{3/2}} + \frac{\sqrt{\pi}}{256n^{5/2}} - \frac{5 \sqrt{\pi}}{2048n^{7/2}} + \cdots.$$
Thus
$$E(n,n) \approx n + \frac{\sqrt{\pi}}{2} n^{1/2} - \frac{1}{2} + \frac{\sqrt{\pi}}{16} n^{-1/2} + \frac{\sqrt{\pi}}{256} n^{-3/2} - \frac{5 \sqrt{\pi}}{2048} n^{-5/2} + \cdots.$$
• Thanks, this is very good! :-D This correlates with my computer and analytical calculations. As I already wrote, my old computer calculations suggested that $E(n,n)\sim n+c\sqrt{n}+o(\sqrt{n})$ for $c\approx 0.88\dots$. Now I see that $c\approx \sqrt{\pi}/2$ with the precision up to two significant digits. I derived the integral formula only yesterday so I did not try to derive the asymptotics from it yet. Moreover, I did not know about Laplace method. – Alex Ravsky Dec 21 '13 at 6:55
If I understand right the game, and picturing the evolution as a path in a discrete grid -from $(0,0)$ to ($n,n$)- it's seen that each segment that goes towards the diagonal is a "win"; the other are misses, except for the ones that start from the diagonal itself, half of which are wins. Then, if the path of length $2n$ have $c$ diagonal-touchings (including the start, excluding the end), the total of wins is $n+c/2$.
Hence, the problem is converted to the (probably simpler and already studied) problem of computing the expected numbers of diagonal touchings on a lattice path - or, in a fair ballot counting problem, compute the expected number of ties (or lead changes).
You ask "What (...) related problems are worthy (of investigation)?"
There is the question of what the expected score is if we use that strategy (which is the best strategy) with a Zener pack of 25 cards, namely which contains 5 cards of each of 5 different shapes. I asked that question here: the answer is 8.65.
Then there are misère problems where we try to mimimise the score. I have asked the misère version of the Zener card question here. | 2019-06-17T23:32:08 | {
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https://math.stackexchange.com/questions/804764/why-is-1-frac12-frac13-frac1n-approx-lnn-gamm | Why is $1 + \frac{1}{2} + \frac{1}{3} + … + \frac{1}{n} \approx \ln(n) + \gamma$?
On StackExchange, I read that the harmonic series up to $\frac{1}{n}$ is approximately $\ln(n) + \gamma$, where $\gamma$ is the Euler-Mascheroni constant, which is close to $0.5772$. When I researched the Euler-Mascheroni constant, I only found it defined in terms of the difference between the harmonic series and $\ln(n)$.
Why is the series able to be approximated in this way, and what is the Euler-Mascheroni constant?
• Draw rectangles representing the sum (i.e. having height $1/n$ on the interval $[n, n + 1]$), and the area representing the integral. In the limit, these are almost the same, and the error is $\gamma$. – user61527 May 22 '14 at 0:16
• It's pretty easy to show, using $$\log n = \int_1^n\frac{dx}{x} = \sum_{i=1}^{n-1}\int_i^{i+1}\frac{dx}{x}$$ that $$\left(\sum_{i=1}^n\frac{1}{i}\right)-\log n$$ converges as $n\to\infty$. You can also easily prove that the limit is between $1/2$ and $1$. This limit, or some variant, is often the definition of the Euler-Mascheroni constant. – Thomas Andrews May 22 '14 at 0:21
• @ThomasAndrews I am sure you have a typo there. – chubakueno May 22 '14 at 0:23
• Fixed. @chubakueno – Thomas Andrews May 22 '14 at 0:24
• I like showing $\gamma$ is between one increasing sequence and one decreasing sequence that are also forced to come together; see math.stackexchange.com/questions/306371/… – Will Jagy May 22 '14 at 3:58
$$\log n = \int_1^n \frac{dx}{x} = \sum_{i=1}^{n-1}\int_{i}^{i+1}\frac{dx}{x}$$
So:
$$\left(\sum_{i=1}^n \frac 1 i\right)-\log n = \left(\sum_{i=1}^{n-1}\int_i^{i+1}\left(\frac 1i-\frac1x\right)dx\right) + \frac{1}{n}$$
Now, for $x\in[i,i+1]$, $0\leq\frac{1}i-\frac1 x\leq \frac{1}{i(i+1)}.$
So these terms are positive and $\sum_{i=1}^\infty \frac{1}{i(i+1)} = 1$. So as $n\to\infty$, this means:
$$\left(\sum_{i=1}^n \frac 1 i\right)-\log n$$ converges to a value less than $1$.
It's actually pretty easy to show, since $f(x)=1/x$ is concave, that:
$$\int_i^{i+1}\left(\frac 1i-\frac1x\right)dx>\frac{1}{2i(i+1)}$$
This means that the limit is between $1/2$ and $1$.
This is often the definition of the Euler-Mascheroni constant.
• Don't you need to show the series is increasing? Or have you shown that and I'm missing something. – Alex Zorn May 22 '14 at 0:49
• If $0<a_i<b_i$ and $\sum b_i$ converges, then $\sum a_i$ converges. Basically, $\sum_{i=1}^n a_i$ is increasing as $n\to\infty$ and is bounded above, therefore converges. – Thomas Andrews May 22 '14 at 1:14
As noted in the comments, the "computational" answer is that, comparing areas, we find
$$\int_1^{n+1}\frac{dx}{x}<1+{1\over 2}+\cdots +{1\over n}<1+\int_1^n\frac{dx}{x}$$ Evaluating the integrals gives $$\ln(n+1)<1+{1\over 2}+\cdots +{1\over n}<1+\ln n$$ so we have the very rough approximation that, denoting $$H_n=1+{1\over 2}+\cdots +{1\over n}$$ the sum is around $H_n\approx \ln n$.
A more insightful perspective is to compare the nature of the functions $H_n$ (often called the "harmonic sum") and $\ln x$. The former is discrete, and the latter continuous. In "discrete calculus" you often see $$\Delta f(n)=f(n+1)-f(n)$$ which is the discrete analogue of the derivative. Note that the "derivative" of the harmonic sum is $$\Delta H_n=H_{n+1}-H_n=\left(1+\cdots +{1\over n+1}\right)-\left(1+\cdots +{1\over n}\right)={1\over n+1}$$
Similarly, the derivative of the logarithm is $$\textrm{D}\ln x={1\over x}$$
(The analogy can be drawn further, depending on how much discrete maths you know) There is a sense then, in which the two functions are "companions" of each other - they live in different universes, but within those universes they share many of the same properties. This is a recurring theme in much of mathematics (you may be interested in researching the word "isomorphic", although it does not completely apply here).
• If you average $\ln (n + 1)$ and $\ln n + 1$ you get $\ln n + 0.5$ as an estimate for $H_n$ – vonbrand May 22 '14 at 4:36 | 2021-04-23T05:49:25 | {
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https://math.stackexchange.com/questions/2155401/solving-modulus-equations/2155409 | # Solving Modulus-Equations
So, I've come across this modulus equation in my textbook: $$|2x-1|=|4x+9|$$
I looked at the solution to this equation and understand that in order for both sides to be equal, the quantities inside the brackets must either be the same or the negatives of each other.
The solution then uses the following theorem: if $|p| = b, b>0 \Rightarrow p = -b$ or $p = b$. $$2x-1 = -(4x+9)$$ or $$2x-1 = 4x+9$$ and solves both linear equations to get $x = -\frac{4}{3}$ or $x = -5$
I then asked myself why the solution didn't bother to find $$-(2x-1) = 4x+9$$ or $$-(4x-9) = -(2x-1)$$ and instead only found the two above. I then proceeded to calculate the above linear equations and got the exact same answers as above $x = - \frac{4}{3}$ or $x = -5$
I'd like to know why I achieved the same answers with this.
• Think at why $2x-1 = -(4x+9) \iff -(2x-1) = 4x+9\,$. – dxiv Feb 21 '17 at 23:02
The equation $-(2x-1)=4x+9$ has the same solution as $2x-1=-(4x+9)$ as the only difference between these two equations is it has been multipled by $-1$.
Similarly $2x-1=4x+9$ has the same solution as $-(2x-1)=-(4x+9)$ for the same reason.
So to solve the original equation you need to solve only one out of the pair of equations $-(2x-1)=4x+9$ and $2x-1=-(4x+9)$ and then solve one out of the pair of equations $2x-1=4x+9$ and $-(2x-1)=-(4x+9)$. Which one you do from each pair doesn't matter as they have the same solution.
I drew a graph of $$y = |4x+9| - |2x-1|.$$ Not to scale, for $x$ I made four squares equal $1,$ because the interesting $x$ values are $1/2$ and $-9/4.$ The graph is three lines joining up, but with different slopes (four times what I depicted). Had I checked the answer first, I would have moved it over a bit so as to see more of the negative $x$ axis. | 2019-12-14T08:25:28 | {
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Step 1: When we multiplying 0 and 0, we get 0. It consists of only 0, 1 digit and rules for addition, subtraction, and multiplication are the same as decimal numbers. INTEGER BINARY 0 000 1 001 2 010 3 011 4 100 5 101 6 110 7 111. Applying the rules from the previous section, the multiplication follows these steps: The first two numbers (5. To do it we set the value range from 0 to 1, and choose the symmetric matrix type. The fastest way of multiplying two big integers. Algorithm: To multiply two decimal numbers x and y, write them next to each other, as in the example below. Converting from hex to binary is a lot like converting binary to hex. However, each bit of implies different algorithmic path during each iteration, that is, if , only a point doubling is necessary. , digits) is performed in a manner similar to decimal multiplication. The multiplication of an n-bit binary number with an m-bit binary number results in a product that is up to m + n bits in length for both signed and unsigned words. mlt": nn = na if n = 2 then f$= "b" + ". That is all the way from 1 x 1 to 12 x 12. How to use this calculator: In the calculator, there are two input fields intended for entry of binary. This online calculator for addition and subtraction multiplication and division of binary numbers online. For matrix multiplication, the number of columns in the first matrix must be equal to the number of rows in the second matrix. This is the first step in this step the least significant bit or the right most bit of B is multiplied with all the digits of A from the right side and the result is written. Each cell should contain a zero or a one. Multiplication of binary numbers can be achieved either combinationally or sequentially. Therefore, multiplication must be explicitly denoted by using the * operator, as in a * b. Two dots in the right-most box is worth one dot in the next box to the left. This gives signal of overflow-7+7=0 00111 binary 7 01001 two's complement of 7 11110 carries 10000 result of addition 16 - but fifth bit can be ignored, real result is 0. There are various number systems in the world. In fourth case, a binary addition is creating a sum of 1 + 1 = 10 i. They follow their own set of simple rules and trade on their own special exchange that has been set up just for them. abstract syntax 2. Lattice multiplication is also known as Italian multiplication, Gelosia multiplication, sieve multiplication, shabakh, Venetian squares, or the Hindu lattice. The magnitude can be interpreted as a +01 Id, the remainder from a carry out of I 28d. Learn multiplication table with easy to memorize, helpful times tables are simple to read. Watch this illustration and play with the code. These rules are applied for each operation, such that nested calculations imply the precision of each component. en: binary numbers adding substraction multiplication division. Message fields can be one of the following For example, when an old binary parses data sent by a new binary with new fields, those new fields become unknown fields in the old. (Where there is only one number above, you just carry down the 1. • More general rule: – The principle of inclusion and exclusion. Let's illustrate the multiplication rules for 3-bit binary numbers A = 111 and B = 101, beginning with the lowest and highest digits of the multiplier (Figure 2. For all a b b a ba. reflexive: if for every x X, xRx holds, i. Commute= travel (move) If you move swap the numbers ina multiplication and addition the anwser will still be the same! Rule: a+b = b+a Sample: 1+2= 3, 2+1=3. Same rule does not hold for natural numbers because if we take two numbers such as x and y and perform binary. 1 The Multiplication Rule. 1 * 1 = 1 2. According to the binary multiplication rules, the numbers in the bracket give the decimal equivalents of the binary numbers. Observations on Multiply Version 3. MUL Multiply; DEC VAX; signed multiplication of scalar quantities (8, 16, or 32 bit integer) in general purpose registers or memory, available in two operand (first operand multiplied by second operand with result replacing second operand) and three operand (first operand multiplied by second operand with result placed in third operand) (MULB2 multiply byte 2 operand, MULB3 multiply byte 3 operand, MULW2 multiply word 2 operand, MULW3 multiply word 3 operand, MULL2 multiply long 2 operand. The examples of valid hexadecimal numbers are 4E1or CD3A01. (25 ends up with 0010 0101) Now another thing to remember, there is a rule established in the multiplication of the values as written in the book, Cryptography and Network Security, that multiplication of a value by x (ie. However, it is not guaranteed to be compiled using efficient routines, and thus we recommend the use of scipy. Note that in each subsequent row, placeholder 0's need … As an example of binary multiplication we have 101 times 11, 101 x 1 1. Here's the demo for subtraction. 0 is written in the given. Only valid for independent events. Binary Conversion Practice! ! ! !Binary Places: 32, 16, 8, 4, 2, 1 Convert these binary numbers to decimal: 1 10 11 100 101 1000 1011 1100 10101 11111 Convert these decimal numbers to binary:. It can be calculated easily if we know the following rules. I'd be happy to share it with anyone once I figure out how to place it on this site. Sponsored Links. Two's complement converter calculator is used to calculate the 2's complement of a binary or a decimal number. So, the result became 0. Quantities with unlike units may sometimes be multiplied, resulting in such units as foot-pounds, gram-centimeters, and kilowatt-hours. Binary literals can make relationships among data more apparent than they would be in hexadecimal or octal. To multiply in binary, you multiply the first number by each of the digits of the second number in turn starting from the right-hand side (in the same way that you would do multiplication in decimal). This is illustrated below. This means 3 groups of 2 equals 6 in all. C program to convert decimal number to roman. Existing Algorithms: • The naive binary multiplication algorithm has a time complexity O(n^2) where n is the number of bits of the numbers being multiplied. Arrays, multiplication and division Jennie Pennant, with the help of Jenni Way and Mike Askew, explores how the array can be used as a thinking tool to help children develop an in-depth understanding of multiplication and division. Other :: Binary Classification. ) Keep going, always adding pairs of numbers from the previous row. Additionally, the output is restricted to a 10-bit word with binary-point-only scaling of 2-4. Binary Card Game Explained ; Binary Card Game, the computer plays against. Instead of needing to know your multiplication tables you only really need to know three things. Power rule with positive exponents; Power rule with negative exponents: Problem type 1; Simplifying a sum or difference of three univariate polynomials; Product rule with positive exponents; Multiplying binomials with leading coefficients of 1; Solving a two-step equation with integers; Solving an equation to find the value of an expression. binary calculator as a 1 and 0. Binary Multiplication Calculator is an online tool for digital computation to perform the multiplication between the two binary numbers. }\) We use symbols to represent binary operations instead of function names, just as we do with addition and multiplication of integers. A multiplication algorithm is an algorithm (or method) to multiply two numbers. Converting Decimal Numbers to Binary Numbers. It is enough to show only one rule-break to prove that {-1, 1} is not a group with respect to addition. 24 Addition and Multiplication in Q The following rules define binary operations on Q. So in the example above: Relative uncertainty = (0. You will be able to use it both as a binary to decimal converter and as a decimal to binary calculator. Binary multiplication is implemented using the same basic longhand algorithm that you learned in grade school. 2 Unary Operations Up: 1 Binary Operations Previous: 1 Addition and subtraction Index 2 Matrix multiplication Matrices are conformable for multiplication if and only if the number of columns in the first matrix equals the number of rows in the second matrix. A binary operator; the result is $$\binary{1}$$ if at least one of the two operands are $$\binary{1}\text{,}$$ otherwise the result is \(\binary{0}\text{. The general rule is that if the OV flag is set to I, then complement the sign bit. • Binary multiplication of unsigned integers reduces to "shift and add". The process in the binary system is easier than it is in the Binary multiplication calculator is a simple method to multiply binary values without using manual mehtods. When a string contains single or double quote characters, however, use Function annotations should use the normal rules for colons and always have spaces around the. basis is used in all matrix and matrix-vector multiplication algorithms, which are considered in this and the following sections. Rule 3: Lastly, perform all additions and subtractions, working from left to right. Binary Addition, Subtraction , Multiplication and division. + 01 0 01 1 10 · 01 0 00 1 01 Table C. Binary integer multiplier 210 first receives binary coefficients (C1 and C2) of two input operands and performs a binary multiplication, C' = C1 · C2. Definition: Binary operation. If we know the rules of binary multiplication & division, the solution of t. Note that binary operators work on vectors and matrices as well as scalars. 1415; …) converges to a unique. That means that every piece of binary code in a computer must be converted into a physical object or state. Binary Multiplication: Binary multiplication is exactly as it is in decimal, i. When a string contains single or double quote characters, however, use Function annotations should use the normal rules for colons and always have spaces around the. The rules for binary multiplication are: In truth table form, the multiplication of two bits, a x b is: Observe that a x b is identical to the logical and operation. Multiplication by an Integer Constant 7 The average time complexity seems to be exponential. The rule for addition of two binary coded decimal (BCD) numbers is given below. MUL Multiply; DEC VAX; signed multiplication of scalar quantities (8, 16, or 32 bit integer) in general purpose registers or memory, available in two operand (first operand multiplied by second operand with result replacing second operand) and three operand (first operand multiplied by second operand with result placed in third operand) (MULB2 multiply byte 2 operand, MULB3 multiply byte 3 operand, MULW2 multiply word 2 operand, MULW3 multiply word 3 operand, MULL2 multiply long 2 operand. Addition Worksheet A :. Binary Options Trading the simplest form of Best Trading Platform in the financial sector, that has become the traders best choice recently in Option Trading. Starting from the right, the first "A" represents the "ones" place, or 16 0. C program for fractional decimal to binary fraction conversion. (c) G= (N,+), the natural numbers in Z with addition as the binary operation. The evaluation using the and and or operators follow these rules: and and or evaluates expression Bitwise operators are used to compare integers in their binary formats. This means you are either multiplying each digit by 0 or 1, which will give you either a 0 or 1 as the answer. Test your method out on these numbers, and find a way to check your work. Binary Multiplication Binary multiplication is performed in same way as decimal numbers. The group operation is multiplication of complex numbers. E A B C E E A B C A A B C E B B C E A C C E A B T Combination order is "top" then "side"; e. 4 cm) × 100% = 5. See full list on byjus. Binary multiplication is arguably simpler than its decimal counterpart. An example is the unary and binary minus, (). Let’s add two binary numbers to understand the binary addition. If you want to build up one for yourself, you can look at this scheme of 2 bits times 2 bits multiplier -. What is the result of multiplying the binary number 10010 by 101? 1011010. Then R is called a commutative ring with respect to these operations if the following properties hold: (i) Closure: If a,b R, then the sum a+b and the product a·b are uniquely defined and belong to R. What if the digits in each number are not even, or the same? What is the rule in dividing it into two parts? Example: x = 12345 y = 2478 or. Translate the number 22810 в binary like this: the Integer part of the number is divided by the base of the new. In mathematical operations involving significant figures, the answer is. Binary Numbers Toggle the 1s and 0s by clicking on them to reveal dots and make binary numbers. semantic algebras 3. Multiplying a binary term (multiplicand m) by 2 simply shifts mone power level up, for example, 2 · K = P or 2 · P = T. Generally you won’t need to use this operator since it’s redundant. The first two numbers (5. The Multiplying Binary Numbers (Base 2) (A) Math Worksheet from the Multiplication Worksheets Page at Math-Drills. For example, an int data type can hold 10 digits, is stored in 4 bytes, and doesn't accept decimal points. Let A = {1, 3, 5, 9, 11, 13} and let$\odot$define the binary operation of multiplication modulo 14. Note: In the Chain Rule, we work from the outside to the inside. There are four rules of binary addition. (Where there is only one number above, you just carry down the 1. C program to convert decimal number to roman. Booth's Multiplication Algorithm is used to multiplication of two signed binary numbers. So basically all you need to do is multiply the powers. unambiguouslywrite a⋆b⋆c to denote either of the iterated products. 1 The Number of Elements in a List. FrankenPC makes a fair start showing binary addition. Let's illustrate the multiplication rules for 3-bit binary numbers A = 111 and B = 101, beginning with the lowest and highest digits of the multiplier (Figure 2. If we only want to invest in a single N-bit adder,. Complex multiplication is a more difficult operation to understand In the above formula for multiplication, if v is zero, then you get a formula for multiplying a complex. Binary division and multiplication are both pretty easy operations. for overloading binary operators, addition, subtraction, multiplication, division and comparison". " The Multiplicand is the number taken or multiplied. Fast multiplication. For each number of nodes, n, there is a certain number of possible binary tree configurations. Note that in each subsequent row, placeholder 0's need … As an example of binary multiplication we have 101 times 11, 101 x 1 1. We see what are the rules of binary multiplication and division and how to solve. Two last carries are 11. How to use this calculator: In the calculator, there are two input fields intended for entry of binary numbers. The unary minus operator returns the operand multiplied by -1. The Standard Multiplication Algorithm. There are four rules of binary addition. abstract syntax 2. Example 7. For example, an int data type can hold 10 digits, is stored in 4 bytes, and doesn't accept decimal points. Enter the two numbers that you want to implement the. Just as we get a number when two numbers are either added or subtracted or multiplied or are divided. Binary options are a different kind of option than any you're likely familiar with. When performing a binary. 1 Lecture 8: Binary Multiplication & Division Today s topics: Addition/Subtraction Multiplication Division Reminder: get started early on if signs disagree These rules fulfil the equation above 19. For example, in mathematics and most computer languages, multiplication is granted a higher precedence than addition, and it has been this way since the. 1415; …) converges to a unique. If neither value is a double but one is a float, then Java treats both values as floats. 1 The Number of Elements in a List. Binary numbers can represent two states: 0, 1, and 2-bit binary numbers can represent (2=) 4 states: 00, 01, 10, and 11; by analogy, 7-bit binary numbers can represent (2=) 128 species. The arithmetic of binary numbers means the operation of addition, subtraction, multiplication and division. Beginners introduction to binary, hexadecimal and octal numbers. It's still a base-10 representation, but the algorithm doesn't need the digits to be between 0 and 9. Home Work. Stanford University. The relative uncertainty gives the uncertainty as a percentage of the original value. C program for addition of binary numbers. Conversion rules for converting decimal to other number system First step is to identify the base of target number system i. The Manchester Baby, operational in 1948, had no multiplication hardware. Post navigation. b) Whenever b(i,j) =1 , a(i,t) should also be 1, but for both alias sets i, t ; alias (i,j); Binary Variables b(i,j. For a node's world transformation: W = P. A simplistic way to perform multiplication is by repeated addition. There are four basic operations for binary addition, as mentioned above. It uses "engine" of Mathematical calculator. There are no ads, popups or other garbage. The magnitude can be interpreted as a +01 Id, the remainder from a carry out of I 28d. See General Rules for Operator Overloading for more information. 2 * 10 = 20 and 3 * 100 = 300. Floating Point Addition. Let us take two binary numbers A = 1001 and B = 101 we want to find out A × B. The 1←2 Rule Whenever there are two dots in single box, they “explode,” disappear, and become one dot in the box to the left. Dividing Binary Numbers. Robot Multiplication for Kids the 11s Master your Multiplication Facts by byrih. The general rule is: (x m) n = x mn. + Addition. The precedence order for arithmetic operators places multiplication and division above addition and. There are many applications of matrices in computer programming; to represent a graph data structure, in solving a system of linear equations and more. L = local transformation matrix. 2nd: Change the division sign to a multiplication sign. Binary Multiply - Repeated Shift and Add. The binary multiplication is much easier as it contains only 0s and 1s. The number can be converted to decimal by multiplying out as follows: 1*1 + 0*2 + 1*4 = 5. For example, multiplication of two 4-bit numbers requires a ROM having eight address lines, four of them, X 4 X 3 X 2 X 1 being allocated to the multiplier, and the remaining four, Y 4 Y 3 Y 2 Y 1 to the multiplicand. Then, b is called inverse of a. abstract syntax 2. The 3 basic. More Topics Related. More Examples:. Binary addition/subtraction is similar to regular (daily life) addition/subtraction, but here addition/subtraction performs only two digits those are 0 and 1, these are binary digits hence such kind of addition/subtraction is called binary addition/subtraction. Find the missing multiple or product for any multiplication fact You can use the printable multiplication table below to check your answers or as a study guide. With three or more, it is also straightforward, but you use the Even-Odd Rule. 741 Op-Amp Circuit Binary Addition Binary Multiplication Section 4. Last but not the least, a major reason computers use the binary system is that the two-state system is the number system best suited to the optical and magnetic storage components of the computer. How to do Addition, Subtraction, Multiplication and Division in Android Studio. However, it is not guaranteed to be compiled using efficient routines, and thus we recommend the use of scipy. Point multiplication is usually computed by a sequence of point doublings (triplings, halvings) and additions, whose numbers depend on the length and the number of non-zero digits of k. Binary literals can make relationships among data more apparent than they would be in hexadecimal or octal. Example − Addition Binary Subtraction. To start practising, just click on any link. To convert 0xDEADBEEF (a commonly used code to indicate a system crash), begin by sorting the digits into "bins":. A lower precedence number means tighter binding Other numeric operations are similar: subtraction (-), and multiplication (*). binary calculator as a 1 and 0. In Binary Arithmetic, 1 + 1 = 10, as per the rules of binary addition. Leibniz pointed out in 1703 that to do simple arithmetic in binary, such as addition and multiplication, you don’t need to memorize such rules as 5 + 4 = 9, or 6 × 7 = 42. But we do not know much more. Assume the multiplicand (A) has N bits and the multiplier (B) has M bits. How to convert letters (ASCII characters) to binary and vice versa. eve Engels, 2006 Slide 7 of 10 –Example: two-digit multiplication a 1 a 0 x b 1 b 0 a 1 b 0 a 0 b 0 a 1 b 1 a 0 b 1 + p 2 p 1 p 0 p 0 = a 0 b 0 p 1 = a 1 b 0 + a 0 b 1 p 2 = a 1 b 1 + a 0 b 1 a 1 b 0 (addition here, not OR). Binary Addition, Subtraction , Multiplication and division. Multiplication And Division of Octal Numbers Calculator, Octal Multiplication And Division: Required Data Entry Enter A Octal Value. Addition, subtraction, multiplication, and division are all binary operators with which we are familiar from grade school. (b) G= (N,·), the natural numbers in Z with multiplication as the binary oper-ation. flipped over). For example, the real numbers form a field, with ‘+’ and ‘·’denoting ordinary addition and multiplication. " Every digit must be either 1 or 0. Binary Arithmetic 1: Adding binary numbers. Multiplication done algebraically. All it includes are addition of binary. Binary arithmetic is essential part of all the digital computers and many other digital system. What is the associative property of Multiplication? Associative property of multiplication is the rule that says that the way that factors are grouped in a multiplication problem will not affect the answer. In modern terms it employs the decomposition of one number into its binary components, addition to produce "doubles" or multiples of the other number, and computing a total of doubles identified by odd divisors or the powers of two. Overflow rule:- If two number are added, and they are both positive or negative than overflow occurs if and only if the result has opposite sign. Let us understand the binary multiplication on natural numbers and real numbers. 0 + 0 = 0 0 + 1 = 1 1 + 0 = 1 1 + 1 = 0. Such models can be linearized quite easily and are in most cases better solved as a linear. Lesson 1: Introduction to Multiplication. The rules of differentiation (product rule, quotient rule, chain rule, …) have been implemented in JavaScript code. Multiplication Tables L All possible binary combinations of symmetry operations can be summarized in a multiplication table. number system to which we have to convert the decimal number system. In mathematics, a binary operation or dyadic operation is a calculation that combines two elements (called operands) to produce another element. 17 * 5 = 85 / (Division) By using this operator we can perform division of numbers. Binary Triple Octal Decimal Hexadecimal Binary-decimal. If ⋆ is associative we can. Assigning a precedence to grammar rules. If you skip parentheses or a multiplication sign, type at least a whitespace, i. No strict rules are defined but just easy to read but precise. Use as an example 18 or 10010: 18 = 16 + 2 = 2 4 + 2 1. (2) In each part below, a rule is given that determines a binary operation on Z. Sponsored Links. Enter the two numbers that you want to implement the. A carry-save adder's propagation time is constant, while traditional a 2 input adder's propagation time is proportional the the width of the operands. C program to convert decimal number to roman. One technique for performing multiplication in any number base is by repeated addition; for example, 6×4=6+6+6+6=24 in decimal (similarly, 4×6=4+4+4+4+4+4=24). When a multiplication problem is given abstractly, there is no such distinction, so we prefer to use the symmetrical term "factor. This is the currently selected item. mlt": nn = nb open f$ for binary as #1 for n2 = 1 to nn randomize timer x$= ltrim$ (str$(int (rnd * 10))) seek #1, n2: put #1, n2, x$ next n2 seek #1, n2 close (1) next n ' open "a" + ". This is, of course, 10 times the starting value - that is how you multiply things by 10. Decitrig: A rule with the trigonometric scales with decimal degrees (S, ST, T) [0. So, the result became 0. multiplication of any element by 0 yields 0; 3. The number 1010110 is represented by 7 bits. Therefore, multiplication must be explicitly denoted by using the * operator, as in a * b. The value of a particular N-bit binary number x in a U(a,b)representation is given by the expression x. , E A B C E EE = E EA = A EB = B EC = C A AE = A AA = B AB = C AC = E B BE = B BA = C BB = E BC = A C CE = C CA = E CB = A CC = B. Commercial applications like computers, mobiles, high speed calculators and some general purpose processors require […]. Binary Multiplication Calculator is an online tool for digital computation to perform the multiplication between the two binary numbers. In Karatsuba algorithm for multiplying two numbers, we divide each number into two. A binary multiplier is a combinational logic circuit used in digital systems to perform the multiplication of two binary numbers. Unit-2: Binary Multiplication-Booth's Algorithm. As an example of binary multiplication we have 101 times 11, 101 x 1 1. eve Engels, 2006 Slide 7 of 10 –Example: two-digit multiplication a 1 a 0 x b 1 b 0 a 1 b 0 a 0 b 0 a 1 b 1 a 0 b 1 + p 2 p 1 p 0 p 0 = a 0 b 0 p 1 = a 1 b 0 + a 0 b 1 p 2 = a 1 b 1 + a 0 b 1 a 1 b 0 (addition here, not OR). Read about Binary Addition (Binary Arithmetic) in our free Electronics Textbook. The unary minus operator returns the operand multiplied by -1. Multiplication of a binary number n by the binary number 100 (=4) is done by adding two bits 0 to the right of the binary representation of n. The exact same rule exists in binary. The binary number system uses only two digits 0 and 1 due to which their addition is simple. A, 1 is entered in its position that yields 1000 since 2 3 = 8. The interaction of the parts makes the dinner range from good to great. There are four basic operations for binary addition, as mentioned above. The form calculates the bitwise exclusive or using the function gmp_xor. One technique for performing multiplication in any number base is by repeated addition; for example, 6×4=6+6+6+6=24 in decimal (similarly, 4×6=4+4+4+4+4+4=24). to multiply and divide polynomials, use conv and deconv on. The result of binary arithmetic for A with and B with gives the worst case p:. The multiplication method is sometimes called doubling and halving or called Russian peasant multiplication. Arithmetic operations in the binary system are extremely simple. (Like multiplying by 10 in our normal notation. Binary Option trading platforms allows e-Traders to make investments by predicting the future direction of an asset and make up to 85% profits in a span of few minutes. 0026), with the final result having 8 decimal places (0. Instead, you need. To convert fraction to binary, start with the fraction in question and multiply it by 2 keeping notice of the resulting integer and fractional part. Multiplying and dividing positive and negative numbers is a simple operation with two numbers. " The Multiplicand is the number taken or multiplied. As a result of multiplication you will get a new matrix that has the same quantity of rows as the 1st one has and the same quantity of columns as the 2nd one. 11 ⋅ 2 − 3 1 01. FREE holiday, seasonal, and themed multiplication worksheets to help teach the times tables. Starting from the right, the first "A" represents the "ones" place, or 16 0. Specifying Field Rules. To perform addition, subtraction, multiplication and division in python, you have to ask from user to enter any two number, and then ask again to enter the operator to perform the desired operation. Binary Card Game Explained ; Binary Card Game, the computer plays against. Binary code, as it turns out, is easy to convert from electronic information (e. The numbers we are multiplying together are called factors, and the result of multiplication is called the. multiply numbers right to left and multiply each digit of one number to every digit of the other number, them sum them up. Draws binary random numbers (0 or 1) from a Bernoulli distribution. Each row and each column is unique and contains as many zeros as ones. (b) Using binary representa-tion 10011 1101 10011 10011 1011111 10011 11110111 (c) Adding immediately Figure 7. The magnitude can be interpreted as a +01 Id, the remainder from a carry out of I 28d. them sum them up. (LL 0, LL 1,LL 01, etc. For a node's world transformation: W = P. Most processors perform n-bit by n-bit multiplication and produce a 2n-bit result (double bits) assuming there is no overflow condition. Here Binary number is a number that can be represented using only two numeric symbols - 0 and 1. " Every digit must be either 1 or 0. Binary, or base 2, is one of many possible Number Systems. Multiplication of pure imaginary numbers by non-finite numbers might not match MATLAB. Suppose for instance that we want to multiply 13 11, or in binary notation, x= 1101 and y= 1011. In mathematics when you perform computational actions, you must have in mind that there is a sequence that need to be respected in. Distributivity of Binary Operations. Learn Java program multiplication starting from its overview, How to write, How to set environment , How to run, Example like Addition, Subtraction , Division, Multiplication etc. C++ Program to Multiply Two Matrix Using Multi-dimensional Arrays This program takes two matrices of order r1*c1 and r2*c2 respectively. Note that in each subsequent row, placeholder 0's need … As an example of binary multiplication we have 101 times 11, 101 x 1 1. The representations of the multiplicand and product are not specified; typically, these are both also in two's complement representation, like the multiplier, but any number system that supports addition and subtraction will work as well. Matrix multiplication falls into two general categories: Scalar: in which a single number is multiplied with every entry of a matrix. A general rule when multiplying a Qm format number by a Qn format number, is that the product will be a Q(m+n) number. In binary, 2 * 3 = 6 is 10 * 11 = 110, and 4 * 3 = 12 is 100 * 11 = 1100. For example: x= 1234 y= 2456 Then a = 12, b = 34, c = 24 , d = 56. That is binary numbers can be represented in general as having p binary digits and q fractional digits. Multiplying and dividing positive and negative numbers is a simple operation with two numbers. * (Multiplication) By using this operator we can perform multiplication of numbers. Other arithmetic operators work only with numbers and always convert their operands to numbers. Two dots in the right-most box is worth one dot in the next box to the left. resample(rule[, axis, closed, label, …]) Resample time-series data. Convert number. The balanced ternary multiplication table is depicted in Fig. If the array isn't sorted, you must sort it using a sorting technique such as merge sort. By using this website, you agree to our Cookie Policy. Note that since binary operates in base 2, the multiplication rules we need to remember are those that involve 0 and 1 only. The question is the extent to which we can unify addition and multiplication, realizing them as terms in a single underlying binary operation. Multiplication by an Integer Constant 7 The average time complexity seems to be exponential. A 4-bit, 2's complement example:. Binary multiplication can be achieved by using a ROM as a look-up’ table. Browse other questions tagged rsa algorithm-design montgomery-multiplication or ask your own question. Note that in each subsequent row, placeholder 0's need to be added, and the value shifted to the left, just like in decimal multiplication. (25 ends up with 0010 0101) Now another thing to remember, there is a rule established in the multiplication of the values as written in the book, Cryptography and Network Security, that multiplication of a value by x (ie. Our source code has been written in characters we can read, but it needs to be. To find the probability of the two dependent events, we use a modified version of Multiplication Rule 1, which was presented in the last lesson. In the case of decimal multiplication, we need to remember 3 x 9 = 27, 7 x 8 = 56, and so on. Linear algebra. com binary converter online for free. It gives a range of activations, so it is not binary activation. This online calculator for addition and subtraction multiplication and division of binary numbers online. So in the example above: Relative uncertainty = (0. In this tutorial, we will create a custom pipe so that the currency format can run well. According to all the rules above, x will be converted to int first. Leibniz pointed out in 1703 that to do simple arithmetic in binary, such as addition and multiplication, you don’t need to memorize such rules as 5 + 4 = 9, or 6 × 7 = 42. If an operation consists of k steps and. Also, we'll perform addition and subtraction on them. The first two numbers (5. To help me remember I wrote some fully commented code for an 8 x 8 bit unsigned multiplier. Stats: Probability Rules. The code works fine when I give an integer input, but doesn't give correct result for floating point input. 5^2)^25 (25 multiplication steps) = ((3. For K-12 kids, teachers and parents. 0015 to 1,000,000), two binary scales for adding and subtracting fractions, a scale of drill sizes, a scale of thread sizes, and millimeters. 70 × 10-1 with 9. The arithmetic unit performs parallel one-step addition (subtraction), multiplication and division. Converting Decimal Numbers to Binary Numbers. Binary Addition, Subtraction , Multiplication and division. In binary multiplication, we only need to remember the following. Similarly, whenever we would like to sum two binary numbers, only we will have a carry if the product is bigger than 1 because, in binary numbers, 1 is the highest number. To find the probability of the two dependent events, we use a modified version of Multiplication Rule 1, which was presented in the last lesson. What is the rules of binary in addition? As with decimal numbers, you start by adding the bits (digits) one column, or place. Other arithmetic operators work only with numbers and always convert their operands to numbers. It is enough to show only one rule-break to prove that {-1, 1} is not a group with respect to addition. Test yourself on multiplication facts. The difference being the "point" bookkeeping required which is the same as addition. Take some parts away or add new ones in, and you get a different outcome, but not a binary, win/lose one. The "Commutative Laws" say we can swap numbers over and still get the same answer. ) Vector : A rule with hyperbolic functions. There are four rules for binary multiplication: Multiplication is always 0, whenever at least one input is 0. Here is a new (harder) goal: come up with a general method to find the binary number related to any number of dots without actually going through the “exploding dot” process. Now multiplication is going to occur like this. Just binary utilities that work right in your browser. binary arithmetic rules. Multiplication rules. Mutually Exclusive Events. Leave a Comment on 540. ⌚ September 25, 2014 Binary multiplication is performed the same as decimal multiplication. numbers are multiplied and divided through a process called shifting. Similarly, whenever we would like to sum two binary numbers, only we will have a carry if the product is bigger than 1 because, in binary numbers, 1 is the highest number. The matrix product is designed for representing the composition of linear maps that are represented by matrices. Though binary division not too difficult, it can initially be a bit harder to understand than the other binary operations as they shared similarities. Then we multiply the entire top number by each individual digit of the bottom number. General binary operations that follow these patterns are studied in abstract algebra. Multiplication Example Multiplicand 1000ten Multiplier x 1001ten-----1000 0000 0000 1000-----Product 1001000ten In every step • multiplicand is shifted • next bit of multiplier is examined (also a shifting step) • if this bit is 1, shifted multiplicand is added to the product. However, it is not guaranteed to be compiled using efficient routines, and thus we recommend the use of scipy. The method used for ancient Egyptian multiplication is also closely related to binary numbers. This is a simple online tool to convert English into binary. Binary Multiplication is generally simpler than decimal multiplication. The binary number system uses only two digits 0 and 1 due to which their addition is simple. Enter expression with binary numbers and get the result. This online calculator for addition and subtraction multiplication and division of binary numbers online. In this tutorial, we will create a custom pipe so that the currency format can run well. The Binary Calculator is used to perform addition, subtraction, multiplication and division on two In mathematics and computer science, binary is a positional numeral system with a base of 2. Let’s covert the same binary number to an octal number: $100100010101111_{2}=100 100 010 101 111$ $100=4$ $010=2$ \$101=5. It shows you how the product is generated in real-time, step-by-step, and allows you to highlight the individual multiplication steps used to get the answer. To figure the decimal value of a binary number, you multiply each bit by its corresponding power of two and then add the results. And all utilities work exactly the same way — load. Generally you won’t need to use this operator since it’s redundant. This follows Chapter 4 of Schmidt’s book. There are four rules of binary addition. The general rule is: (x m) n = x mn. To read binary numbers, and convert them to their decimal equivalent, you have two options: you can either use the Binary to Decimal Converter at ConvertBinary. binary: A binary operator operates on two operands. The multiplication of a 16 bit binary "a" with a 24 bit binary "b and the complementary defence of both the general interest and the rules of the competitive. In the modern world, multiplication, division, addition, and subtraction are estimated by the binary calculator within a second, same rule as applied in the decimal system. A general rule when multiplying a Qm format number by a Qn format number, is that the product will be a Q(m+n) number. 2 steps per bit because Multiplier & Product combined How can you make it faster? What about signed multiplication? • trivial solution: make both positive. The magnitude can be interpreted as a +01 Id, the remainder from a carry out of I 28d. Browse other questions tagged rsa algorithm-design montgomery-multiplication or ask your own question. We have already discussed the binary addition and binary subtraction in detail in the previous articles now we are going to discuss binary multiplication in a detailed manner. If we take the number 237 and shift each digit 1 place to the left, adding a zero to the right hand side, we get 2370. How to use this calculator: In the calculator, there are two input fields intended for entry of binary numbers. Binary Multiplication •Sizing •In binary addition –we are generally representing something that ultimately is to be executed in hardware •Our hardware cannot change the number of bits (wires) it can hold •We must establish a maximum number size •For multiplication the size of the result must be the sum of the. Binary Arithmetic 1: Adding binary numbers. Take some parts away or add new ones in, and you get a different outcome, but not a binary, win/lose one. Convert number. Note that since binary operates in base 2, the multiplication rules we need to remember. (countable) A calculation involving multiplication. 16 + 0 + 0 + 2 + 1 = 19. Binary Multiplication •Sizing •In binary addition –we are generally representing something that ultimately is to be executed in hardware •Our hardware cannot change the number of bits (wires) it can hold •We must establish a maximum number size •For multiplication the size of the result must be the sum of the. Leibniz pointed out in 1703 that to do simple arithmetic in binary, such as addition and multiplication, you don’t need to memorize such rules as 5 + 4 = 9, or 6 × 7 = 42. You just take a regular number (called a "scalar") and multiply it on. Unicode Character 'MULTIPLICATION SIGN' (U+00D7). Consider the simple problem of multiplying 110 2 by 10 2. A, 1 is entered in its position that yields 1000 since 2 3 = 8. Multiplying Complex Numbers Calculator. Binary addition is the simplest method to add any of the binary numbers. Addition Worksheet A :. Let’s add two binary numbers to understand the binary addition. So a binary matrix is such an array of 0's and 1's. A, 1 is entered in its position that yields 1000 since 2 3 = 8. Align the numbers as an ordinary subtraction. The binary number system uses 0s and 1s to represent numbers. No problem with unsigned (always positive) numbers, just use the same Multiplication in Two's complement cannot be accomplished with the standard technique since, as. The number 1010110 is represented by 7 bits. Binary multiplication is actually much simpler to calculate than decimal multiplication. Learn binary conversions and arithmetic with interactive demonstrations and explanations. ), with steps shown. If given condition is FALSE (not TRUE), expr_set2 will get executed. The procedure for binary multiplication is similar to that in decimal system. Computer method: Computer method is used by digital machines to multiply the binary numbers. There are four rules for binary subtraction: Borrow 1 is required from the next higher order bit to subtract 1 from 0. The binary number system uses only two digits 0 and 1 due to which their addition is simple. Sponsored Links. | 2021-03-02T10:58:54 | {
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https://forum.math.toronto.edu/index.php?PHPSESSID=k441o32850lkospornhudrbhs7&action=printpage;topic=1558.0 | # Toronto Math Forum
## MAT334-2018F => MAT334--Tests => Term Test 2 => Topic started by: Victor Ivrii on November 24, 2018, 04:54:06 AM
Title: TT2A Problem 1
Post by: Victor Ivrii on November 24, 2018, 04:54:06 AM
Using Cauchy's integral formula calculate
$$\int_\Gamma \frac{z\,dz}{z^2-4z+5},$$
where $\Gamma$ is a counter-clockwise oriented simple contour, not passing through eiter
of $2\pm i$ in the following cases
(a) The point $2+i$ is inside $\Gamma$ and $2-i$ is outside it;
(b) The point $2-i$ is inside $\Gamma$ and $2+i$ is outside it;
(c) Both points $2\pm i$ are inside $\Gamma$.
Title: Re: TT2A Problem 1
Post by: ZhenDi Pan on November 24, 2018, 05:30:25 AM
We have
\int_\Gamma \frac{zdz}{z^2-4z+5}
Let
f(z) = \frac{z}{z^2-4z+5} = \frac{z}{(z-(2-i))(z-(2+i))}
Question a:
The point $2-i$ is outside of the contour $\Gamma$ and the point $2+i$ is inside of the contour $\Gamma$. Then let
g(z) =\frac{z}{z-2+i} \\
g(2+i) = \frac{2+i}{2i} \\
\int_\Gamma f(z)dz = \int_\Gamma \frac{g(z)}{(z-(2+i))}dz = 2\pi i g(z_0) = 2\pi i g(2+i) = 2\pi i \cdot \frac{2+i}{2i}= \pi(2+i)
Question b:
The point $2+i$ is outside of the contour $\Gamma$ and the point $2-i$ is inside of the contour $\Gamma$. Then let
g(z) =\frac{z}{z-2-i} \\
g(2-i) = -\frac{2-i}{2i} \\
\int_\Gamma f(z)dz = \int_\Gamma \frac{g(z)}{(z-(2-i))}dz = 2\pi i g(z_0) = 2\pi i g(2-i) = 2\pi i \cdot -\frac{2-i}{2i}= -\pi(2-i)
Question c:
Both points $2+i$ and $2-i$ are inside of the coutour $\Gamma$. Then we have
z_0 = 2+i \\
z_1 = 2-i \\
\left.Res(f;2+i) = \frac{z}{z-2+i} \right\vert_{z=2+i} = \frac{2+i}{2i} \\
\left.Res(f;2-i) = \frac{z}{z-2-i} \right\vert_{z=2-i} = - \frac{2-i}{2i}
So the Residue Theorem gives us
\int_\Gamma f(z)dz = 2\pi i(\frac{2+i}{2i}-\frac{2-i}{2i}) = 2\pi i \cdot 1= 2\pi i
Title: Re: TT2A Problem 1
Post by: Yifei Wang on November 24, 2018, 05:34:25 AM
We can rewrite the fraction as:
$let f(z) =\frac{z}{z^2-4z+5}$
as
$\frac{z}{(z-(2+i))(z-(2-i))}$
a. When $2+i$ is inside
$f(z) = \int \frac{\frac{z}{z-2+i}}{z-2-i}~dz$
By Cauchy's thm
$f(z) = \int \frac{\frac{z}{z-2+i}}{z-2-i}~dz= 2i\pi *f(2+i) = \pi * (2+i)$
b. When $2-i$ is inside
$f(z) = \int \frac{\frac{z}{z-2-i}}{z-2+i}~dz$
By Cauchy's thm
$f(z) = \int \frac{\frac{z}{z-2-i}}{z-2+i}~dz= 2i\pi *f(2-i) = -\pi * (2-i)$
c. When both points are inside
$f(z) = \int \frac{z}{(z-(2+i))(z-(2-i))}~dz = 2i\pi (Res(f, 2+i) + Res(f, 2-i)) = 2\pi i$
Title: Re: TT2A Problem 1
Post by: Yifei Wang on November 24, 2018, 05:36:15 AM
ZhenDi Pan
I think you are missing the $z$ on the numerator.
Title: Re: TT2A Problem 1
Post by: ZhenDi Pan on November 24, 2018, 06:23:00 AM
Yes thank you I corrected it. Still our answers are different, I don't know where went wrong though.
Title: Re: TT2A Problem 1
Post by: Zhuoer Sun on November 24, 2018, 12:50:28 PM
Yifei Wang
I think you did the last part wrong. For part (c), you don't need to multiply by 2ipi again. The answer should be the sum of what you got from part (a) and (b), as you've already included 2ipi in previous parts. The final answer should just be pi∗(2+i)−pi∗(2−i).
Title: Re: TT2A Problem 1
Post by: Yifei Wang on November 25, 2018, 02:38:21 PM
Thank you for the correction!
Title: Re: TT2A Problem 1
Post by: Victor Ivrii on November 29, 2018, 07:57:21 AM
Remark: Since integrand is $\frac{1}{z} +O(\frac{1}{z^2}$ the residue at $\infty$ is $-1$ and answer to (c) is $2\pi i$ (independently from(b),(c)). Yet another solution to (c) | 2022-11-26T10:22:06 | {
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http://math.stackexchange.com/questions/219693/finding-a-closed-formula-for-1-cdot2-cdot3-cdots-k-dots-nn1n2-cdotsk | # Finding a closed formula for $1\cdot2\cdot3\cdots k +\dots + n(n+1)(n+2)\cdots(k+n-1)$
Considering the following formulae:
(i) $1+2+3+..+n = n(n+1)/2$
(ii) $1\cdot2+2\cdot3+3\cdot4+...+n(n+1) = n(n+1)(n+2)/3$
(iii) $1\cdot2\cdot3+2\cdot3\cdot4+...+n(n+1)(n+2) = n(n+1)(n+2)(n+3)/4$
Find and prove a 'closed formula' for the sum
$1\cdot2\cdot3\cdot...\cdot k + 2\cdot3\cdot4\cdot...\cdot(k+1) + ... + n(n+1)(n+2)\cdot...\cdot (k+n-1)$
generalizing the formulae above.
I have attempted to 'put' the first 3 formulae together but i am getting know where and wondered where to even start to finding a closed formula.
-
Do you know how to prove by induction? – Clive Newstead Oct 23 '12 at 22:27
Yes, using the basic step n=1 and then induction step n+1 showing it follows. But don't i need to 'find' the formula before proving it using induction? – Matt Oct 23 '12 at 22:29
The pattern looks pretty clear: you have
\begin{align*} &\sum_{i=1}^ni=\frac12n(n+1)\\ &\sum_{i=1}^ni(i+1)=\frac13n(n+1)(n+2)\\ &\sum_{i=1}^ni(i+1)(i+2)=\frac14n(n+1)(n+2)(n+3)\;, \end{align*}\tag{1}
where the righthand sides are closed formulas for the lefthand sides. Now you want
$$\sum_{i=1}^ni(i+1)(i+2)\dots(i+k-1)\;;$$
what’s the obvious extension of the pattern of $(1)$? Once you write it down, the proof will be by induction on $n$.
Added: The general result, of which the three in $(1)$ are special cases, is $$\sum_{i=1}^ni(i+1)(i+2)\dots(i+k-1)=\frac1{k+1}n(n+1)(n+2)\dots(n+k)\;.\tag{2}$$ For $n=1$ this is $$k!=\frac1{k+1}(k+1)!\;,$$ which is certainly true. Now suppose that $(2)$ holds. Then
\begin{align*}\sum_{i=1}^{n+1}i(i+1)&(i+2)\dots(i+k-1)\\ &\overset{(1)}=(n+1)(n+2)\dots(n+k)+\sum_{i=1}^ni(i+1)(i+2)\dots(i+k-1)\\ &\overset{(2)}=(n+1)(n+2)\dots(n+k)+\frac1{k+1}n(n+1)(n+2)\dots(n+k)\\ &\overset{(3)}=\left(1+\frac{n}{k+1}\right)(n+1)(n+2)\dots(n+k)\\ &=\frac{n+k+1}{k+1}(n+1)(n+2)\dots(n+k)\\ &=\frac1{k+1}(n+1)(n+2)\dots(n+k)(n+k+1)\;, \end{align*}
exactly what we wanted, giving us the induction step. Here $(1)$ is just separating the last term of the summation from the first $n$, $(2)$ is applying the induction hypothesis, $(3)$ is pulling out the common factor of $(n+1)(n+2)\dots(n+k)$, and the rest is just algebra.
-
For each extra bracket (n+1) then (n+2) you're adding +1 on the bottom and then an addition bracket. im thinking the closed formula for $\sum_{k=1}^nn(n+1)(n+2)\dots(n+k-1)\;;$ would be $\frac1{k+1}(n+k-1)$ im not too sure – Matt Oct 23 '12 at 22:40
@Matt: Look at the closed forms in $(1)$ again: your $\frac1{k+1}$ is fine, but you should have $k$ more factors, not just one. – Brian M. Scott Oct 23 '12 at 22:47
Im thinking something $\frac1{k+1}n(n+k)$ but not sure how to 'repeat' brackets for more terms of $k$. so for $k=1$ it works as it gives $\frac12n(n+1)$ but then for $k=2$ all i get it $\frac13n(n+2)$ rather than $\frac13n(n+1)(n+2)$ – Matt Oct 23 '12 at 22:52
@Matt: When $k=1$ the RHS has two factors involving $n$; when $k=2$ it has three factors involving $n$; and when $k=3$ it has four factors involving $n$. In each case the smallest is $n$ and the largest is $n+k$. In the general case, then you should expect to have $k+1$ factors involving $n$, running from $n$ up through $n+k$. – Brian M. Scott Oct 23 '12 at 22:57
So it is simply $\frac1{k+1}(n+k)$ – Matt Oct 23 '12 at 23:01
If you divide both sides by $k!$ you will get binomial coefficients and you are in fact trying to prove $$\binom kk + \binom{k+1}k + \dots + \binom{k+n-1}k = \binom{k+n}{k+1}.$$ This is precisely the identity from this question.
The same argument for $k=3$ was used here.
Or you can look at your problem the other way round: If you prove this result about finite sums $$\sum_{j=1}^n j(j+1)\dots(j+k-1)= \frac{n(n+1)\dots{n+k-1}}{k+1},$$ you also get a proof of the identity about binomial coefficients.
-
From (i), (ii) and (iii) it is reasonable to guess that your sum will be $$n(n+1)\cdot...\cdot(n+k)/(k+1)$$ Try to prove this by induction.
-
For a fixed non-negative $k$, let $$f(i)=\frac{1}{k+1}i(i+1)\ldots(i+k).$$ Then $$f(i)-f(i-1)=i(i+1)\ldots(i+k-1).$$ By telescoping,
$$\sum_{i=1}^ni(i+1)(i+2)\dots(i+k-1)=\sum_{i=1}^n\left(f(i)-f(i-1)\right)=f(n)-f(0)=f(n)$$ | 2014-03-13T17:37:26 | {
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http://mdnu.schuetzen-wuerm.de/nth-term-test-for-divergence-calculator.html | Nth Term Test For Divergence Calculator
nth root calculator. = cA , where c is a constant. If you're ever in doubt and would like to We made this SAT® score calculator because we saw that everyone else simply replicated the tables when creating what they called a "calculator". com Test for Divergence This test, according to Wikipedia, is one of the easiest tests to apply; hence it is the first “test” we check when trying to determine whether a series converges or diverges. Derivatives. Many authors do not name this test or give it a shorter name. nth Term Test for Divergence (ONLY) If lim 0n n. if the limit is equal to a number different then zero, or if the limit is infinity, or if the limit does not exist) then the series diverge. 5n² = 5,20,45,80,125. So the nth term tending to zero is only a necessary condition for convergence and is not sufficient. X n are the observation, then the G. nth term test, divergence test, zero test (KristaKingMath). Using the nth-Term Test for Divergence 613 The series in Example STUDY TIP 5(c) will play an important role in this chapter. Uncertainties are assessed by re-sampling the distributions and re-computing divergence estimates 100 times. geometric series, nth term test (Test for Divergence). HOWEVER, just because a series’ terms approach 0, it is NOT convergent!!! This is a test for divergence, not a test for convergence. Author: Juan Carlos Ponce Campuzano. 9: Use Divergence Test to determine whether r 1 ln. If , then the series is convergent or; 2. Dierentiate term by term (new power series is for f0(x)). nth-term-test-for-divergence. In mathematics, the nth-term test for divergence[1] is a simple test for the divergence of an infinite series When testing if a series converges or diverges, this test is often checked first due to its ease of use. Geometric series test. 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On this page, we explain how to use it and how to avoid one of the most common pitfalls associated with this test. Do not sell my info. Nearly every calculus book begins with the same example, and it’s so darn fine that I will bow to peer pressure and use it as well. The key idea is to apply the classical inequality x>=log(1+x) (valid for x>-1) with x=1/k and sum over k, 1<=k<=n-1. What is important to point out is that there is an nth-term test for sequences and an nth-term test for series. In particular, the converse to the test is not true; instead all one can say is. terms tend to zero. If you have a table of values, see Riemann sum calculator for a table. They also crop up frequently in real analysis. Your project deserves the perfect stock photo. Step-by-step Solutions » Walk through homework problems step-by-step from beginning to end. MIT License. 005 when n= 2, so we must add only two terms, 1 1 33! 11. 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This test, according to Wikipedia, is one of the easiest tests to apply; hence it is the first "test" we check when trying to determine whether a series converges or diverges. Recall: Divergence Test If #lim_{n to infty}a_n ne 0#, then #sum_{n=1}^{infty}a_n# diverges. Hints help you try the next step on your own. A function that calls itself is known as a recursive function. If P b n diverges, and a n b n for all n beyond a certain value, then P a n also diverges. Incorrect Test Choice ITC Student chooses an incorrect test, such as an nth term test, or a geometric test. Click the "Calculate" button to calculate the Student's t-critical value. Substitute the values of and in above formula. Use the NEB Tm Calculator to estimate an appropriate annealing temperature when using NEB PCR products. So in this case we would have X10 k=1 2k +1 = 3+5+7++21, and in this case the sum of the series is equal to 120. Testing for Convergence or Divergence of a Series (continued). Converges to 4 63 by Geo Series Test 2b. The nth term test for divergence The first, and usually simplest, test for divergence of a series is the nth term test. Telescoping Series Pdf. Inn Preliminary Question True Of False? Given The Series An, If N=1 Lim An = Oo Then, The Series Must Diverge. By using this website, you agree to our Cookie Policy. Calculus tutoring on Chegg Tutors Learn about Calculus terms like Tests For Convergence And Divergence on Chegg Tutors. Since 1/5 is not equal to 0, the series diverges. Derivatives. Why is my calculator giving me a huge number for sin(3. Test for convergence or divergence. Before the empirical analysis, we made a simple graph of the collected sample data from 1980 to 2015, as shown in Figs. The key idea is to apply the classical inequality x>=log(1+x) (valid for x>-1) with x=1/k and sum over k, 1<=k<=n-1. ; Many authors do not name this test or give it a shorter name. so tuis series fails the nth term test. First of all, enter the nth term. Step 3: Next, substitute the number 1 to 5 into 5n². The infinite series converges (i. If , then the series is convergent or; 2. A summary of the Nth term test for divergence. Step-by-step Solutions » Walk through homework problems step-by-step from beginning to end. This series converges because the series. Instead, we use the nth term test for divergence. Take the limit of An as n approaches zero. 4 Comparison tests - The comparison test - The limit comparison test. The calculator will generate all the work with detailed explanation. The calculator is useful in helping students explore convergence and divergence and guess the limit of sequences. If the limit of sequence {a n} doesn’t equal 0, then the series ∑ a n is divergent. This simplifies finding say the 42nd term. Limit Comparison Test If lim (n-->) (a n / b n) = L, where a n, b n > 0 and L is finite and positive, then the series a n and b n either both converge or both diverge. Free Practice Tests for learners of English. The full name of the test is "nth term test for divergence. The limit of the nth term, [n+2]/ [5n+17], is lim n→∞ n+2 5n+17 =lim n→∞ 1+ 2 n 5+ 17 n = 1 5. For this example we should use the integral test since many of the other tests seem complicated. (-1)^(n+1)*(n/n^2+4) 8. or if the limit does not exist, then. Use the nth term test to determine whether the following series converges or diverges. If nl!im• an = 0, we can't conclude anything. How old are you? A. If not, we can use the divergence test to conclude the series diverges. See full list on gauravtiwari. Maclaurin Series Taylor and Maclaurin Series interactive applet 3. For the biases: Where the sampled b corresponds to the biases used on the linear transformation for the ith layer on the nth sample. I Geometric series. nth-Term Test for Divergence. This simple t-test calculator, provides full details of the t-test calculation, including sample mean, sum of A t-test is used when you're looking at a numerical variable - for example, height - and then comparing the averages of two separate populations or. ( , , ) Leave empty, if you don't need the divergence at a specific point. The Integral Calculator lets you calculate integrals and antiderivatives of functions online — for free! The calculator lacks the mathematical intuition that is very useful for finding an antiderivative, but on the other hand it can try a large number of possibilities within a short amount of time. The limit limn→∞ sn is called the value of the series P k xk and is written as P∞ k=1 xk Finally, P k xk. As with all risk calculators, calculated risk numbers are +/- 5% at best. I have seen so many tweets of late with people getting their heads frazzled with nth-child and nth-of-type. A summary of the Nth term test for divergence. A prerequisite of convergence for an infinite series is that its terms approach 0. geometric series, nth term test (Test for Divergence). T-Test Calculator for 2 Independent Means. Pay Close Attention to the Next Two Exercises. In mathematics, the nth-term test for divergence is a simple test for the divergence of an infinite series: If or if the limit does not exist, then diverges. It means you have to enter, for example, a3, a5, a6, etc. Thus, there are sequences that can be defined recursively, analytically, and those that can. Tests for convergence and divergence are methods to determine the convergence or divergence of infinite series. Return to the Series, Convergence, and Series Tests starting page. The sum is. Root Test: Suppose that the terms of the sequence in question are non-negative, and that there exists r such. Free Practice Tests for learners of English. Evaluate integral theorems of Green, Gauss & Stokes to find lines, surfaces & volumes. By using this website, you agree to our Cookie Policy. The infinite series converges (i. As with the Ratio Test this test will also tell whether a series is absolutely convergent or not rather than simple convergence. You can make the process of transfering the application to your calculator sweet and simple with Texas Instrument’s handy TI connect software. Let $\displaystyle \sum_{n \mathop = 1}^\infty a_n$ be a series of real numbers $\R$ or complex numbers $\C$. 6 Alternating series, absolute and. The harmonic series is as follows: H 1 = 1 H 2 = H 1 + 1/2 H 3 = H 2 + 1/3 H 4 = H 3 + 1/4 H n = H n-1 + 1/n. or if the limit does not exist, then. Above is what is commonly referred to as the “nth Term Test. an Dr an1 or an an1 Dr: Again, in this case it is relatively easy to find a formula for the nth term: an Da0rn. USED: When you suspect the terms of the given series do not approach zero. Calculator for double sums, the connections of two sums, an inner and an outer sum. Limit Comparison Test Calculator. 1A6: In addition to examining the limit of the sequence of partial sums of the series, methods for determining whether a series of numbers converges or diverges are the nth term test, the comparison test, the limit comparison test, the integral test, the ratio test, and the alternating series test. • The nth-Term Test for Divergence • The Integral Test and its relationship to improper integrals and areas of rectangles • Use of the Integral Test to introduce the test for p-series • Comparisons of series • Alternating series and the Alternating Series Remainder • The Ratio and Root Tests. • Determining convergence or divergence of sequences. This calculator demonstrates the application of the Hardy-Weinberg equations to loci with more than two alleles. If this test is inconclusive, that is, if the limit of a_n IS equal to zero (a_n=0), then you need to use another test to determine the behavior. Please choose one option for each question then click Test Result to obtain your result and level (50 questions). The key idea is to apply the classical inequality x>=log(1+x) (valid for x>-1) with x=1/k and sum over k, 1<=k<=n-1. Nth term in a geometric series So for this I have calculated what I believe to be the correct few terms of Q and P Q0=75, Q1=84, Q2=94. n n ∞ ∑= + diverges. The n and the 2 cancel from the numerator and the denominator. It is possible to optimize our trainable weights. It says: Problems. Divergence test; Alternating series test; Integral test; Ratio test; nth root test; Evaluating a Taylor series for a particular value of x in order to calculate a. This test contains grammar and vocabulary questions and your test result will help you choose a level to practise at. If , then the series is divergent or; 3. The Nth Term Test is one of the most simple tests for convergence. On average it takes 2,200 steps to walk a mile, but this varies based on the length of a person's stride, which is determined by their height and pace. How old are you? A. A summary of the Nth term test for divergence. - Geometric series, convergent/divergent series, nth-term test for divergence - Combining series, adding or deleting terms, reindexing. So in fact every series we have written down is now seen to converge, including the one we started with. INTRODUCTION TO SEQUENCES AND SERIES, NTH TERM DIVERGENCE TEST A sequence is basically a list of numbers based on some defining rule. If you're behind a web filter, please make sure that the domains *. nth term test, divergence test, zero test (KristaKingMath). When we have the model for our data under both hypotheses we simply plug our data into Equation 1 to calculate the likelihood ratio and choose H1 if its value is positive or H0 otherwise (assuming equal priors). Roots are often written using the radical symbol √, with √Y, denoting the square. How to use The … Continue reading →. The nth term test or the Test for Divergence is used to show that the limit of the terms is not zero so the series must diverge. use the ratio test to determine which of the following series converges and which diverges X∞ n=1 2n +5 3n, ∞ n=1 (2n)! (n!)2, X∞ n=1 n! nn;. Ifthe test was inconclusive. (a) • Â n=1 nn nn2 (b) • Â n=0 (1)n n2 +1+2n n2 10 " §en it §yn2 a-, because n-it E -2 fo all n 72. Home What's New Blog About Privacy Terms Popular Problems Help. Loading Found a content error?. Function Grapher and Calculator Description:: All Functions. A summary of the Nth term test for divergence. Learners analyze geometric series in detail. Use only the Divergence Test to determine if the statement is true, false, or can't be decided yet. % Progress. Then use your descrip- tion to write a formula for the nth term of each sequence. # Optionally ignore the kl divergence term. 5n² = 5,20,45,80,125. The most common way to find the gcd is the Euclidean algorithm. Note: This test only determines the divergence of a series. How Does a Calculator Work? ~ A series that has no last. However, the equation above can be used to calculate the number of genotypes for a locus with any number alleles. Let the sequence $\sequence {a_n}$ be such that the limit superior $\displaystyle \limsup_{n \mathop \to \infty} \size {a_n}^{1/n} = l$. If ($\displaystyle \lim_{n \rightarrow \infty} a_n eq 0$), then the series ($\displaystyle \sum_{n=0}^\infty a_n$) diverges. Helps you plan out daily resin use by letting you know exactly what you need! Having trouble deciding between Artifacts? Use our trusty damage calculator to see which artifact has the edge!. ) (b ax Cos y 8. The limit of the nth term, [n+2]/ [5n+17], is lim n→∞ n+2 5n+17 =lim n→∞ 1+ 2 n 5+ 17 n = 1 5. The n th-term test for divergence is a very important test, as it enables you to identify lots of series as divergent. This calculator allows test solutions to calculus exercises. Nth term test for testing the series whether it is convergent or divergent Msc mathematics Mca entrance exam du And many more. Solving slope, simplify exponential expressions, multiply and divide monomials free worksheets, complex numbers solver, linear algebra for dummies, fractions math. 64 with lots of other numbers after the decimal point. The harmonic series is as follows: H 1 = 1 H 2 = H 1 + 1/2 H 3 = H 2 + 1/3 H 4 = H 3 + 1/4 H n = H n-1 + 1/n. Helps you plan out daily resin use by letting you know exactly what you need! Having trouble deciding between Artifacts? Use our trusty damage calculator to see which artifact has the edge!. 1 Nth Term Test for Divergence. Nth term test for divergence in hindi. Note that this is only a test for divergence. Calculator of Hardy-Weinberg equilibrium. Using the formula above you can quickly calculate any nth term. As part of our spreadcheats, today we will learn how to calculate moving average using excel formulas. Recall: Divergence Test If #lim_{n to infty}a_n ne 0#, then #sum_{n=1}^{infty}a_n# diverges. The Art of Convergence Tests. If that is the case, you may conclude that the series diverges by Divergence (Nth Term) Test. By following the above equation to reach an outcome where you are effectively showing that the terms of the target series will become less than. By using this website, you agree to our Cookie Policy. geometric series, nth term test (Test for Divergence). If desired, the precise definition of limit can be carefully explained; and students may even be made to memorize it, but it should not be emphasized. Please choose one option for each question then click Test Result to obtain your result and level (50 questions). Finding The nth Term Of A Geometric Sequence. Why is my calculator giving me a huge number for sin(3. An activity to calculate the nth term of a linear sequence. The task is to find the Nth Harmonic Number. com Test for Divergence This test, according to Wikipedia, is one of the easiest tests to apply; hence it is the first “test” we check when trying to determine whether a series converges or diverges. The formula for the nth term of the arithmetic sequence (also called the general term) is: A n = a + (n -1)d. We have ve tests for convergence: 1) the Divergence Test, 2) the Alternating Series Test, 3) the Ratio Test, 4) the Integral (comparison) Test, and 5) the Comparison Test. If the series is an alternating series, state whether it converges absolutely or conditionally. A summary of the Nth term test for divergence. This simplifies finding say the 42nd term. 3n+5/(n*2^n) As. An explanation of how to use and how not to use the nth term test for divergence. Geometric series test. How do you use the Nth term test on the infinite series ∞ ∑ n=1 ln(2n + 1 n + 1) ? By the nth term test (Divergence Test), we can conclude that the posted series diverges. Then use your descrip- tion to write a formula for the nth term of each sequence. One test calculator to answer all your pre- and post-test analysis questions. Free series convergence calculator - test infinite series for convergence step-by-step This website uses cookies to ensure you get the best experience. Answer Save. Even if the terms are approaching zero the series could still diverge, though, hence the Harmonic Series, so this test isn't super-useful on its own. In mathematics, the nth-term test for divergence[1] is a simple test for the divergence of an infinite series When testing if a series converges or diverges, this test is often checked first due to its ease of use. Then its sum is. Determining Convergence or Divergence of an Infinite Series. The sum is. The nth term test for divergence The first, and usually simplest, test for divergence of a series is the nth term test. the true conversion rates (the numbers you would get if you ran the test forever) for Variation A and Variation B fall somewhere within their respective bell curves above. The next test for convergence or divergence of series works especially well for series involving powers. The task is to find the Nth Harmonic Number. Note: The Nth Term Test for Divergence is the special name given to the contrapositive of Theorem 2. com/yt/4758819/?ref=ytd. This tends to zero, showing that even if the nth term tends to zero, the series may still be divergent. This test cannot be used for convergence. Thank you for your questionnaire. Free series convergence calculator - test infinite series for convergence step-by-step This website uses cookies to ensure you get the best experience. ) (b ax Log y e 4. First Term in the Series. Let $\displaystyle \sum_{n \mathop = 1}^\infty a_n$ be a series of real numbers $\R$ or complex numbers $\C$. The series might converge. 2) fails to provide. Example: the sequence {3, 5, 7, 9, } starts at 3 and jumps 2 every time: As a FormulaSaying "starts at 3 and jumps 2 every time" is fine, but it doesnt help us calculate the: 10th term, 100th term, or nth term, where n could be any term number we want. Displaying the steps of calculation is a bit more involved, because the Derivative Calculator can't completely depend on Maxima for this task. The most common way to find the gcd is the Euclidean algorithm. They also crop up frequently in real analysis. ifthe series converges absolutely. We offer free personalized SAT test prep in partnership with the test developer, the College Board. If the calculator did not compute something or you have identified an error, please write it in comments below. com Test for Divergence This test, according to Wikipedia, is one of the easiest tests to apply; hence it is the first “test” we check when trying to determine whether a series converges or diverges. It states that if the limit of a_n is not zero, or. There are plenty of tests to knwo the conv/div of a series. Integral Test and p-Series. Is the nth-term test not applicable in this case? Can you clarify?. The Integral Test. Sending completion. For example, Abel’s Test allows you to define convergence or divergence by the types of functions contained in the series. 2b) Diverges (nth Term Test) 2c) Diverges (Integral Test) 2d) Diverges by the nth Term Test n n lim 1 n 1 → = + 2e) Converges, Geometric 2 r 1 3 = 2f) Diverges (nth Term Test) 2g) Diverges (nth Term Test) 2h) Converges (Integral Test) 3a) Converges to 3 by Telescoping 3b) Diverges by Integral Test 3c) Converges to 25 23 by Geometric Series. App Downloads. The sum of the rst n terms of an A. 1A6: In addition to examining the limit of the sequence of partial sums of the series, methods for determining whether a series of numbers converges or diverges are the nth term test, the comparison test, the limit comparison test, the integral test, the ratio test, and the alternating series test. If ($\displaystyle \lim_{n \rightarrow \infty} a_n eq 0$), then the series ($\displaystyle \sum_{n=0}^\infty a_n$) diverges. If the in nite series X1 k=1 a k converges, then a k!0 as k !1. However, the equation above can be used to calculate the number of genotypes for a locus with any number alleles. Displaying the steps of calculation is a bit more involved, because the Derivative Calculator can't completely depend on Maxima for this task. This leaves us with an expression with nothing in the denominator. Step 2: The infinite geometric series is. Root Test: Suppose that the terms of the sequence in question are non-negative, and that there exists r such. Converges to 4 63 by Geo Series Test 2b. nth-Term Test for Divergence The contra-positive of Theorem 8 provides a useful test for divergence. The theorem then says ∂P (4) P k · n dS = dV. nth term test. so tuis series fails the nth term test. Online Integral Calculator » Solve integrals with Wolfram|Alpha. Limit Comparison Test Calculator. In mathematics, the nth-term test for divergence[1] is a simple test for the divergence of an infinite series When testing if a series converges or diverges, this test is often checked first due to its ease of use. Step-by-step Solutions » Walk through homework problems step-by-step from beginning to end. Proof of the divergence theorem. I hope that this was helpful. (-1)^(n+1)*(n/n^2+4) 8. One of the newer topics covered in the Maths GCSE as well as other qualifications is how to find the nth term of a quadratic sequence, but did you know there is a method of finding the nth term on the Casio Classwiz. The first term of the series is. com/yt/4758819/?ref=ytd. Free online beam calculator that calculates the reactions, deflection and draws bending moment and shear force diagrams for cantilever or simply supported beams. A summary of the Nth term test for divergence. The leading term in the inner expansion for the normal gravitational field gives the Bouguer formula. For example, the sum of the series n={1,1,1,1. Question 1 : Find the nth term of the following sequences. This is the special symbol that means "nth root", it is the "radical" symbol (used for square roots) with a little n to mean nth root. Practice using the nth term test to determine sequence divergence. Khan Academy has been translated into dozens of languages, and 100 million people use our platform worldwide. Example Cont. The Integral Test. The calculator does not go beyond 5 alleles and 15 possible genotypes. After reading through the example question below check out the worksheets and practice questions. So the 5-th term of a sequence starting with 1 and with a difference (step) of 2, will be: 1 + 2 x (5 - 1) = 1 + 2 x 4 = 9. Share yours for free!. The infinite series converges (i. The divergence of the harmonic series is proved by direct comparison with a series whose nth partial sum telescopes to the natural logarithm of n. In this situation the nth term test for divergence (Theorem 14. The integral test for convergence is only valid for series that are 1) Positive : all of the terms in the series are positive, 2) Decreasing : every term is less than the one before it, a_(n-1)> a_n, and 3) Continuous : the series is defined everywhere in its domain. There are divergent series whose nth terms approach 0. T-Test Calculator for 2 Independent Means. Testing convergence/divergence of a series using the nth term test for divergence, the geometric series test, and the telescoping series test. Thus, f_x(x,y) = Cx for some constant C = log y and f_y(x,y) = Clogy for some constant C = x. In this video we discuss the nth Term Test for Divergence, the conditions the test has to meet, how to perform it, and its limitations. nth term test, divergence test, zero test (KristaKingMath). Diverges because. • Determining convergence or divergence of sequences. Section 4-11 : Root Test. In other words, we do not have a definite conclusion. In mathematics, the nth-term test for divergence is a simple test for the divergence of an infinite series: If or if the limit does not exist, then diverges. Your budget deserves straightforward royalty-free pricing that lets you use an image just about anywhere, as often as you want. The following rule is a corollary of the comparison test: if. Step 4: Now, take these values (5n²) from the numbers in the original number sequence and work out the nth term of these numbers that form a linear sequence. The nth term is. It is difficult to explicitly calculate the sum of most alternating series, so. Get an answer for 'sum_(n=2)^oo n/ln(n) Determine the convergence or divergence of the series. Preliminary Question True of False? Given the series > an, if lim an = 0 then, the series must converge. S D ∂z The closed surface S projects into a region R in the xy-plane. Although many documents discuss excess returns caused by medium-term momentum and long-term reversal in the US market, we empirically test this conclusion using the method of Jegadeesh and Titman. In mathematics, the nth-term test for divergence is a simple test for the divergence of an infinite series Unlike stronger convergence tests, the term test cannot prove by itself that a series converges. 9: Use Divergence Test to determine whether r 1 ln. Example: the sequence {3, 5, 7, 9, } starts at 3 and jumps 2 every time: As a FormulaSaying "starts at 3 and jumps 2 every time" is fine, but it doesnt help us calculate the: 10th term, 100th term, or nth term, where n could be any term number we want. App Downloads. Simplifying terms. If the in nite series X1 k=1 a k converges, then a k!0 as k !1. Integral Test Let X1 n=1 a nbe a series with positive terms, and let f(x) be the function that results when n is replaced by xin the formula for a n. To determine if the series is convergent or divergent, apply the nth-Term Test for Divergence. A prerequisite of convergence for an infinite series is that its terms approach 0. You will not be able to see the correct answers to the questions. # Every Nth epoch, save weights: # Loop across test chunks # Calculate number of batches:. Justify your answer. This test helps determine if a series diverges. (1 point) The nth term test for Divergence of Series. The nth term test for divergence The first, and usually simplest, test for divergence of a series is the nth term test. ( , , ) Leave empty, if you don't need the divergence at a specific point. nth term = dn + (a - d). We Want To Check For Divergence Of The Series N > --- And We Have Decided To Try The N=2 Nth Term Test For Divergence. This video provides two examples of how to apply the nth term divergent test to determine if a infinite series is divergent. 6, 11, 16, 21, For this sequence d = 5, a = 6. you could try to find the general formula for the Nth Opt-in alpha test for a. You can operate the calculator directly from your keyboard, as well as. In mathematics, the nth-term test for divergence is a simple test for the divergence of an infinite series: If → ∞ ≠ or if the limit does not exist, then ∑ = ∞ diverges. Also, it can identify if the sequence is arithmetic or geometric. Find the nth term of a geometric sequence Contact Us If you are in need of technical support, have a question about advertising opportunities, or have a general question, please contact us by phone or submit a message through the form below. An activity to calculate the nth term of a linear sequence. Telescoping Series Pdf. Integral konvergence divergence. for Series TEST nth-term Geometric series p-series SERIES ∑ an ∞ ∑ ar (i) Converges with sum S = n −1 n =1 ∞ Useful for the comparison tests if the The comparison series ∑ bn is often a geometric series of a pseries. By following the above equation to reach an outcome where you are effectively showing that the terms of the target series will become less than. In mathematics, the nth-term test for divergence is a simple test for the divergence of an infinite series:. , the nth term is less than 0. and decreasing to show convergence, find a larger series. Sending completion. So in this case we would have X10 k=1 2k +1 = 3+5+7++21, and in this case the sum of the series is equal to 120. Helps you plan out daily resin use by letting you know exactly what you need! Having trouble deciding between Artifacts? Use our trusty damage calculator to see which artifact has the edge!. Diverges by Geo Series Test or nth Term Test 2g. Test Review 25 multiple choice formula sheet provided Calculator active *be able to do explicit rule/nth term for both arithmetic and geometric sequences *use and create a recursive formula *find finite and infinite sums *find a specific term *use sigma notation *divergence and convergence. This utility helps solve equations with respect to given variables. with the rst term a and the common difference d is n called an arithmetic series. How Does a Calculator Work? ~ A series that has no last. S D ∂z The closed surface S projects into a region R in the xy-plane. Only the variables i and j may occur in the sum term. ( , , ) Leave empty, if you don't need the divergence at a specific point. An activity to calculate the nth term of a linear sequence. Guidelines: Calculate cube root of 27 For example, use the square root calculator below to find the square root of 7. This is an excellent test to start with because the limit is often easy to calculate. diverges if (a) is not 0 know nothing if (a) equals 0. Many authors do not name this test or give it a shorter name. A function that calls itself is known as a recursive function. = , where ai = Then, = Therefore, by the Nth Term Test for Divergence, Exercise 6: Discuss the convergence or divergence of the series. Exercise 5: Discuss the convergence or divergence of the series. If the limit of sequence { an } doesn’t equal 0, then the series ∑ an is divergent. MIT License. If , then the series is convergent or; 2. Complete Solution. Thus, the harmonic series is a demonstration that the nth term test is a test for divergence only and cannot be used to show a series converges. fibonaciexp2. The n th-term test for divergence is a very important test, as it enables you to identify lots of series as divergent. The list of online calculators for sequences and series. an bn cancels the dominant terms in. This test helps determine if a series diverges. The divergence of the harmonic series is proved by direct comparison with a series whose nth partial sum telescopes to the natural logarithm of n. 12 3 45 The height of each bounce is three-fourths the height of the preceding bounce. In this section, we face the problem of deciding which method to use to test a series for convergence or divergence. The infinite series often contain an infinite number of terms and its nth term represents the nth term of a sequence. It can be used to determine that a series diverges, but it doesn't tell you when a series converges. If r = 1, the ratio test is inconclusive, and the series may converge or diverge. If that Show bx e y ax, sin 0) (2 2 1 2 y b a ay y 10. n"' term test for Divergence Geometric Series Test p-Series Test Telesco in Series Test Inte ral Test Direct Comparison Test Limit Com rison Test Alternating Series Test Ratio Test Root Test 10. 140625)? 12 answers. For instance, the terms of. The nth-term test for divergence is a very important test, as it enables you to identify lots of series as divergent. 3 Newton’s Newton Method Nature and Nature’s laws lay hid in night:. Test for Convergence v1. Using the fx-991EX and the Simul. Welcome to Clip from. Nth Term Test for Divergence - (3 Helpful Examples!) Nth term, georgia algebra 1 eoct online prep test, calculate square root fraction. Just because a(n) of a series approaches 0 does not mean it converges. See full list on gauravtiwari. In this radical calculator Y is a positive real number and x is a Nth root or radical power. Diverges because. This means our calculations are accurate and up-to-date to the practice materials shared from the test maker. Excluding the initial 1, this series is geometric with constant ratio r = 4/9. To use this calculator, enter the values in the given input boxes. Quick question: Doesn't the nth-term test for divergence state that if the limit of a function DOES NOT EXIST or is any number other than zero then it diverges? 1 and DNE both fit the criteria for divergence based from the nth-term test. Definition of Convergence and Divergence in Series. It states that if the limit of a_n is not zero, or. Then use your descrip- tion to write a formula for the nth term of each sequence. Unlike the nth term test for divergence from the last module, this module gives several tests which, if successfully applied, give a definitive answer of whether a series converges or not. terms tend to zero. We are Cambridge Assessment English. MEMORY METER. Simple Present, Present Progressive, Present Perfect, Simple Past, If-Satz Type I Level: lower intermediate. Is each term of the series positive? If so, a comparison test (page 417) may be used. M) of a series containing n observations is the nth root of the product of the values. The Secant Method, when it is working well, which is most of the time, is fast. 3n+5/(n*2^n) As. educreations. The main purpose of this calculator is to find expression for the n th term of a given sequence. You should start with a rm knowledge of each test and the ability to recall quickly the details of each test. If you're ever in doubt and would like to We made this SAT® score calculator because we saw that everyone else simply replicated the tables when creating what they called a "calculator". This online BMI Calculator is delivered to you at no charge and will allow you to obtain your Body Mass Index (BMI) to see whether you are underweight, of normal weight, overweight, or obese according to the classification employed by the World Health Organization (WHO). Is the first test test tests that be will lookingover. In exercise 22-28, test for convergence or divergence using each test at least once. use the ratio test to determine which of the following series converges and which diverges X∞ n=1 2n +5 3n, ∞ n=1 (2n)! (n!)2, X∞ n=1 n! nn;. Many authors do not name this test or give it a shorter name. = , where ai = Then, = Therefore, by the Nth Term Test for Divergence, Exercise 6: Discuss the convergence or divergence of the series. So I decided to knock this together so people can play around and test their nth's out!. If P b n diverges, and a n b n for all n beyond a certain value, then P a n also diverges. Since , the series is converges. Where: a n is the nth term of the sequence, a is the first term, d is the common difference. Here you can calculate a determinant of a matrix with complex numbers online for free with a very detailed solution. Basically if r = 1, then the ratio test fails and would require a different test to determine the convergence or divergence of the series. Converges to 4 63 by Geo Series Test 2b. The n th-term test for divergence is a very important test, as it enables you to identify lots of series as divergent. If the value of when n goes to inifinity the summation of the series will just continue to add up values, and. Get the free "Convergence Test" widget for your website, blog, Wordpress, Blogger, or iGoogle. It is possible to optimize our trainable weights. So the first term of the nth term is 5n². If the limit of sequence {a n} doesn’t equal 0, then the series ∑ a n is divergent. Geometric series test. I have seen so many tweets of late with people getting their heads frazzled with nth-child and nth-of-type. Since the series diverges. The divergence of the harmonic series is proved by direct comparison with a series whose nth partial sum telescopes to the natural logarithm of n. Consider a series S a n such that a n > 0 and a n > a n+1 We can plot the points (n,a n) on a graph and construct rectangles whose bases are of length 1 and whose heights are of length a n. The last two tests that we looked at for series convergence have required that all the terms in the series be positive. to show that this diverges should I use the $n^{th}$ term test? So far I have substituted infinity for $n$. The key idea is to apply the classical inequality x>=log(1+x) (valid for x>-1) with x=1/k and sum over k, 1<=k<=n-1. The series may or may not. Nth Root (Radical) calculator is a simple tool used to calculate the N th Root of the given real number which has the N th radical value. Test for Convergence v1. fibonaciexp2. 4 Series Tests for Convergence and Divergence. Converges by p-series Test 2f. Determining convergence or divergence of the sequence with the nth term. - Nth Term Test for Divergence - Geometric Series: Examining the ratio to determine Numerically calculate the value of the derivative of a function at a. One of the newer topics covered in the Maths GCSE as well as other qualifications is how to find the nth term of a quadratic sequence, but did you know there is a method of finding the nth term on the Casio Classwiz. Use the Standard Deviation Calculator to calculate your sample's standard deviation and mean. For example, sum_(n=1)^(infty)(-1)^n does not converge by the limit test. find a formula for the nth term directly. Sending completion. Abstract: Based on the probability generating function of stuttering Poisson distribution (SPD), this paper considers some equivalent propositions of SPD. this proves the P-series test. n"' term test for Divergence Geometric Series Test p-Series Test Telesco in Series Test Inte ral Test Direct Comparison Test Limit Com rison Test Alternating Series Test Ratio Test Root Test 10. Looking at this function closely we see that f(x) presents an improper behavior at 0 and only. you could try to find the general formula for the Nth Opt-in alpha test for a. If the terms of a rather conditionally convergent series are suitably arranged, the series may be made to converge to any desirable value or even to diverge according to the. Incorrect Test Choice ITC Student chooses an incorrect test, such as an nth term test, or a geometric test. Suppose that there exists r such that. Fortunately, it’s also very easy to use. If the limit of sequence {a n} doesn’t equal 0, then the series ∑ a n is divergent. Test infinite series for convergence step-by-step. One of the newer topics covered in the Maths GCSE as well as other qualifications is how to find the nth term of a quadratic sequence, but did you know there is a method of finding the nth term on the Casio Classwiz. The calculator will approximate the definite integral using the Riemann sum and sample points of your choice: left endpoints, right endpoints, midpoints, and trapezoids. According to divergent test rules, we can't conclude anything about it. In mathematics, the nth-term test for divergence is a simple test for the divergence of an infinite series Unlike stronger convergence tests, the term test cannot prove by itself that a series converges. (1 point) Assume we are trying to determine the. To perform the test, first you need to calculate a measure known as U for each sample. Many authors do not name this test or give it a shorter name. Use the Standard Deviation Calculator to calculate your sample's standard deviation and mean. nth-Term Test for Divergence. The calculation. Step 4: Now, take these values (5n²) from the numbers in the original number sequence and work out the nth term of these numbers that form a linear sequence. Consider, if x 1, x 2 …. A series contain terms whose order matters a lot. The Nth Term Test is one of the most simple tests for convergence. 3 The nth Term Test for Divergence. Khan Academy has been translated into dozens of languages, and 100 million people use our platform worldwide. 60 – 75 min 75 min 75 min 75 min 60 – 75 min 75 min • grid paper • scientific calculator • graphing calculator (optional) Dynamic Activity; Animation • scientific calculator • graphing calculator (optional) Animations. Always check the Nth Term Test for Divergence. If → ∞ ≠ or if the limit does not exist, then ∑ = ∞ diverges. Although many documents discuss excess returns caused by medium-term momentum and long-term reversal in the US market, we empirically test this conclusion using the method of Jegadeesh and Titman. If you have a table of values, see Riemann sum calculator for a table. The limit test is inconclusive when the limit is zero. com online calculator provides basic and advanced mathematical functions useful for school or college. Share yours for free!. The nth term test is only a test for divergence. Determinant is calculated by reducing a matrix to row echelon form and multiplying its main diagonal elements. It's quick, free and gives an instant score. Always check first. (2 votes). or if the limit does not exist, then. Let s n(p)be the nth partial sum of the p-series ∞ k=1 1/k p. This means our calculations are accurate and up-to-date to the practice materials shared from the test maker. Alternating Series Test - Proof. Using the nth-Term Test for Divergence 613 The series in Example STUDY TIP 5(c) will play an important role in this chapter. so Etfe converges. The test that we are going to look into in this section will be a test for alternating series. This video explains how to apply the nth term divergence test to an infinite series. Find more Mathematics widgets in Wolfram|Alpha. Also, it can identify if the sequence is arithmetic or geometric. If a series converges, its nth term approaches 0. com/yt/4758819/?ref=ytd. (n^10 + 10)/n! 7. ParticipantsConversions. Equations and terms. 4 Comparison tests: the comparison test, limit comparison test. Suppose that the in nite series X1 k=1 a k converges. A function that calls itself is known as a recursive function. The Integral Test. Test for Divergence for Series, Two Examples. If , then the series is divergent or; 3. Nth Term Test. Learners analyze geometric series in detail. The ratio test is especially useful, but the integral test is one i dread to use. By the nth term test (Divergence Test), we can conclude that the posted series diverges. an Dr an1 or an an1 Dr: Again, in this case it is relatively easy to find a formula for the nth term: an Da0rn. 1 Equation 2 Use 3 Explanation 4 Video Explanation Calculate the limit as stated above. The n and the 2 cancel from the numerator and the denominator. Example Cont. For example, students can be. n=1 n 2 + n. To each series we associate the sequence of nth terms of the series and also the sequence of partial sums of the series. • The nth-term test for divergence • The integral test and its relationship to improper integrals and areas of rectangles • Use of the integral test to introduce the test for p-series • Comparisons of series • Alternating series and the alternating series remainder • The ratio and root tests • Taylor polynomials and approximations. Nth term test for testing the series whether it is convergent or divergent Msc mathematics Mca entrance exam du And many more. First Term in the Series. If the th Term Divergence Test is inconclusive, the next step is to examine the corresponding series. This is used by this calculator. # Every Nth epoch, save weights: # Loop across test chunks # Calculate number of batches:. Function Grapher and Calculator Description:: All Functions. Use the NEB Tm Calculator to estimate an appropriate annealing temperature when using NEB PCR products. terms tend to zero. We use the product rule for f and the quotient rule for V as. Usually we need about 45 percent more iterations than with the Newton Method to get the same accuracy, but each iteration is cheaper. pdf Due December 2: Page 515: 3,4,24, 26, 49,50 Due December 4: Page 528: 7-12;14-18 (use LCT, DCT, nth term lor geometric series). Identify the test used. The calculation. 2b) Diverges (nth Term Test) 2c) Diverges (Integral Test) 2d) Diverges by the nth Term Test n n lim 1 n 1 → = + 2e) Converges, Geometric 2 r 1 3 = 2f) Diverges (nth Term Test) 2g) Diverges (nth Term Test) 2h) Converges (Integral Test) 3a) Converges to 3 by Telescoping 3b) Diverges by Integral Test 3c) Converges to 25 23 by Geometric Series. Nth Term Test For Convergence This test basically tells you that if your terms aren't approaching zero, there's no way the series converges. | 2021-03-02T13:46:01 | {
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https://math.stackexchange.com/questions/3261187/what-is-dfraci4-%CF%80-int-z-4-dfracdzz-cos-z | What is $\dfrac{i}{4-π} \int_{|z|=4} \dfrac{dz}{z \cos z}$?
How to evaluate the integral $$\dfrac{i}{4-π} \int_{|z|=4} \dfrac{dz}{z \cos z}?$$
Here $$f(z)=\dfrac{1}{z \cos z}$$ has poles at $$0$$ and $$\frac{\pm π}{2}$$ .
Residue at $$z=0$$ is 1 and residues at remaining poles add up to give 0. So the integral using Cauchy integral formula is $$2π(4-π)$$.
I think I am wrong. How to get the integral?
It's not true that residues add up to $$0$$. We have $$\cos z = \sin(\frac{\pi}{2}-z) = \sin(\frac{\pi}{2}+z)$$ so $${\rm Res}_{z=0} \frac{1}{z\cos z} = \lim_{z\rightarrow 0}\frac{1}{\cos z} = 1$$ $${\rm Res}_{z=\frac{\pi}{2}} \frac{1}{z\cos z} = \lim_{z\rightarrow \frac{\pi}{2}}\frac{(z-\frac{\pi}{2})}{z \sin(\frac{\pi}{2}-z)} = -\frac2\pi$$ $${\rm Res}_{z=-\frac{\pi}{2}} \frac{1}{z\cos z} = \lim_{z\rightarrow -\frac{\pi}{2}}\frac{(z+\frac{\pi}{2})}{z \sin(\frac{\pi}{2}+z)} = -\frac2\pi$$ and $$\sum {\rm Res} = 1 - \frac{4}{\pi} = \frac{\pi-4}{\pi}$$ so the final result is $$\frac{i}{4-\pi}\cdot 2\pi i\frac{\pi-4}{\pi} = 2$$
You made a mistake in your residues. The residues of $$1/(z\cos z)$$ at $$z=\pm\frac\pi2$$ are both $$-\frac2\pi$$.
• ohh...yes....thank you Jun 13 '19 at 16:20
The function $$f(z)=\frac{1}{z\cos(z)}$$ has simple poles at $$z=0$$ and $$z= (2n-1)\pi/2$$ for $$n\in \mathbb{Z}$$. The poles that are inside the circle $$|z|=4$$ are at $$z=0$$ and $$z=\pm \pi/2$$.
The residues of $$f$$ at the implicated poles are
\begin{align} \text{Res}\left(\frac{1}{z\cos(z)}, z=0\right)&=1\\\\ \text{Res}\left(\frac{1}{z\cos(z)}, z=\pi/2\right)&=-\frac2\pi\\\\ \text{Res}\left(\frac{1}{z\cos(z)}, z=-\pi/2\right)&=-\frac2\pi \end{align}
Therefore we have
$$\frac{i}{4-\pi}\oint_{|z|=4}f(z)\,dz=\frac{i}{4-\pi}\times 2\pi i\times\left(1-\frac{4}{\pi}\right)=2$$ | 2021-10-21T07:00:03 | {
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http://math.stackexchange.com/questions/4261/proving-a-binomial-sum-identity/247057 | # Proving a binomial sum identity
Mathematica tells me that
$$\sum _{k=0}^n { n \choose k} \frac{(-1)^k}{2k+1} = \frac{(2n)!!}{(2n+1)!!}.$$
Although I have not been able to come up with a proof.
Proofs, hints, or references are all welcome.
-
You could consider the integral $$\int_{0}^{1} (1-x^2)^n dx .$$
-
Aha this is the integral I was searching for! Using the integral $\int_0^1 (1-t)^n \, dt$ I was able to prove $\sum _{k=0}^n {n \choose k}\frac{(-1)^k}{k+1} = \frac{1}{n+1}$ I thought there should be a corresponding integral to my problem and after seeing your answer the integral I was searching for is obvious now! – yjj Sep 8 '10 at 11:02
Sums of the form $$\sum_{k=0}^n(-1)^k{n\choose k}f(k)$$ can often be attacked via the Calculus of finite differences.
-
$$S_n=\sum\limits_{k=0}^n \dfrac{(-1)^k \binom{n}{k}}{2k+1}$$
We have:
\begin{align}S_n&=\dfrac{(-1)^n}{2n+1}+\sum\limits_{k=0}^{n-1} \dfrac{(-1)^k \binom{n}{k}}{2k+1}\\ &=\dfrac{(-1)^n}{2n+1}+\sum\limits_{k=0}^{n-1}\left[\dfrac{n}{n-k}.\dfrac{(-1)^k \binom{n-1}{k}}{(2k+1)}\right]\\ &=\dfrac{(-1)^n}{2n+1}+\dfrac{1}{2n+1}\sum\limits_{k=0}^{n-1}\left[\dfrac{n(2n+1)}{n-k}.\dfrac{(-1)^k \binom{n-1}{k}}{(2k+1)}\right]\\ &=\dfrac{(-1)^n}{2n+1}+\dfrac{1}{2n+1}\sum\limits_{k=0}^{n-1}\left[\dfrac{2n^2-2nk+2nk+n}{n-k}.\dfrac{(-1)^k \binom{n-1}{k}}{(2k+1)}\right]\\ &=\dfrac{(-1)^n}{2n+1}+\dfrac{1}{2n+1}\sum\limits_{k=0}^{n-1}\left[\dfrac{2n^2-2nk}{n-k}.\dfrac{(-1)^k \binom{n-1}{k}}{(2k+1)}\right]\\&+\dfrac{1}{2n+1}\sum\limits_{k=0}^{n-1}\left[\dfrac{2nk+n}{n-k}.\dfrac{(-1)^k \binom{n-1}{k}}{(2k+1)}\right]\\ &=\dfrac{(-1)^n}{2n+1}+\dfrac{2n}{2n+1}\sum\limits_{k=0}^{n-1}\dfrac{(-1)^k \binom{n-1}{k}}{(2k+1)}+\dfrac{1}{2n+1}\sum\limits_{k=0}^{n-1} \dfrac{n(-1)^k \binom{n-1}{k}}{n-k}\\ &=\dfrac{2n}{2n+1}\sum\limits_{k=0}^{n-1}\dfrac{(-1)^k \binom{n-1}{k}}{(2k+1)}+\dfrac{1}{2n+1}\sum\limits_{k=0}^{n-1} \left[(-1)^k \binom{n}{k}\right]+\dfrac{(-1)^n}{2n+1}\\ &=\dfrac{2n}{2n+1}\sum\limits_{k=0}^{n-1}\dfrac{(-1)^k \binom{n-1}{k}}{(2k+1)}+\dfrac{1}{2n+1}\sum\limits_{k=0}^n \left[(-1)^k \binom{n}{k}\right]\end{align} Therefore, $$S_n=\dfrac{2n}{2n+1}S_{n-1}+0 \Rightarrow S_{n-1}=\dfrac{2n-2}{2n-1}S_{n-2} ... \Rightarrow S_1=\dfrac{2}{3}S_0$$ and $S_0=1$
Hence, $$S_n=\dfrac{2n(2n-2)...2}{(2n+1)(2n-1)...3.1}=\dfrac{(2n)!!}{(2n+1)!!}$$
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As stated, you divide by zero since you sum to $k=n$ and $n-k$ is in the denominator. – Julian Kuelshammer Nov 29 '12 at 7:36
Wah! Thanks for pointing out errors in my solution. – hxthanh Dec 1 '12 at 12:12
Fix this problem as follows sum for $k=0$ to $n-1$ and $k=n$ is calculated separately – hxthanh Dec 1 '12 at 12:34
You can edit your post accordingly. – Julian Kuelshammer Dec 1 '12 at 12:37
Your identity is a special case of the Chu-Vandermonde identity.
${}_2 F_1(-n,b;c;1)=\frac{(c-b)_n}{(c)_n}$
with $b=\frac12$ and $c=\frac32$. More info on it is in A=B, as mentioned already by John.
-
It's on page 181. – Ｊ. Ｍ. Sep 8 '10 at 11:05
I can't offer specific help, but I'd recommend thumbing through Concrete Mathematics looking for techniques. It has many sums that look similar, though of course the difficulty is in the details.
There's also the book A=B, but Concrete Mathematics gives an introduction to the content of A=B and in my opinion is easier to read, so I'd start with Concrete Mathematics.
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There are a powerful algorithms generalizing telescopy (Gosper, Zeilberger et al.) that easily tackle this case of the Chu-Vandermonde identity and much more complicated sums. For example, see this paper which gives as an application a very interesting q-analogy - namely that L. J. Rogers' classical finite version of Euler's pentagonal number theorem is simply the dual of a special case of a q-Chu-Vandermonde.
- | 2014-10-20T09:41:46 | {
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https://math.stackexchange.com/questions/1822068/how-many-permutations-of-1-2-3-n-there-are-with-no-2-consecutive-numbe | # How many permutations of $\{1,2,3,…,n\}$ there are with no 2 consecutive numbers?
How many permutations of $$\{1,2,3,...,n\}$$ are there with no 2 consecutive numbers?
For example:
$$n=4$$, $$2143$$, $$3214$$, $$1324$$ are the permutations we look for and $$1234$$, $$1243$$, $$2134$$ are what we DON'T look for.
My solution:
we will sub from all the permutations the bads. All permutations= $$n!$$
bads:
Lets define $$A_i$$ to be a group of all permutations containing $$i,i+1$$ in it. ($$1\leq i\leq n-1$$)
My problem is to count to group of :$$|A_i \cap A_j|$$ where $$1 \leq i < j \leq n-1$$.
Because there is a big difference between $$A_i \cap A_{i+1}$$ and $$A_i \cap A_{i+5}$$ (for instance), we can't say that both are the same. So we need to divide $$|A_i \cap A_j|$$ to 2 options.
1. when $$i+1 = j$$ then $$|A_i \cap A_j|=(n-2)!$$
2. when $$i+1 < j$$ then $$|A_i \cap A_j|=(n-2)!$$
So $$|A_i \cap A_j|=(n-2)!+(n-2)!$$
This is a big mistake but I have no clue why.
(the rest of the solution is the same).
Any help would be welcome.
Stav
• It makes no sense to add your two figures of $(n-2)!$ together: the first applies only when $i+1=j$, and the second applies only when $i+1<j$, so they never both to the same pair of $i$ and $j$. – Brian M. Scott Jun 11 '16 at 19:07
• @BrianM.Scott Studying the subject and wasn't sure if the following statement was true.$N(A_iA_j)$ was same for all $i\neq j$(even if $i+1=j$) by calculation through inclusion principle. thus there was no contradiction? – user416486 Apr 23 '18 at 15:49
## 3 Answers
If, as appears to be the case, you’re planning to use an inclusion-exclusion argument, you’ll need $\left|\bigcap_{k\in I}A_k\right|$ for every non-empty $I\subseteq[n-1]$. The trick is to realize that no matter how the integers in $I$ are spaced, the cardinality is going to be the same. When $|I|=2$, the case that you’re discussing in the question, there are two possibilities: the two members of $I$ are consecutive, or they are not. But in both cases it turns out that the cardinality of the intersection is $(n-2)!$: there are different calculations for the two cases, but they lead to the same result. Note that there is no reason at all to add these calculations: no $I$ belongs to both cases.
The set $I$ can be divided into blocks of consecutive integers. For example, $I=\{2,3,5,6,7,9\}$ has $3$ blocks: $\{2,3\},\{5,6,7\}$, and $\{9\}$. Suppose that $I$ has a block $\{k,k+1,\ldots,k+r\}$. Then every permutation in $\bigcap_{k\in I}A_k$ must contain the subsequence $\langle k,k+1,\ldots,k+r,k+r+1\rangle$; call this an extended block. If $I$ has $b$ blocks altogether, each permutation of $[n]$ that belongs to $\bigcap_{k\in I}A_k$ must contain all $b$ extended blocks as subsequences, but it can contain them in any order. Those extended blocks contain altogether $|I|+b$ integers: the blocks themselves contain $|I|$ integers, and each extended block has one extra integer on the righthand end. That leaves $n-|I|-b$ members of $[n]$ that can be permuted arbitrarily, because they aren’t in any extended block of $I$. Thus, the permutations in $\bigcap_{k\in I}A_k$ are really permutations of
$$b+(n-|I|-b)=n-|I|$$
things: $b$ extended blocks, and the $n-|I|-b$ single elements that are not part of any extended block. It follows that
$$\left|\bigcap_{k\in I}A_k\right|=(n-|I|)!$$
whenever $\varnothing\ne I\subseteq[n-1]$.
Now the inclusion-exclusion principle tells you that
\begin{align*} \left|\bigcup_{k\in[n-1]}A_k\right|&=\sum_{\varnothing\ne I\subseteq[n-1]}(-1)^{|I|-1}\left|\bigcap_{k\in I}A_k\right|\\ &=\sum_{\varnothing\ne I\subseteq[n-1]}(-1)^{|I|-1}(n-|I|)!\\ &=\sum_{k=1}^{n-1}\binom{n-1}k(-1)^{k-1}(n-k)!\;, \end{align*}
and I’ll leave the rest for you to finish off.
• First of all thank you very much for your time. Secondly Im having hard time to undestand how I={2,3,5,6,7,9}I={2,3,5,6,7,9} has 3 blocks: {1,2},{5,6,7}{1,2},{5,6,7}, and {9}. how and why did you choose those blcoks specifically? what does a block mean? – Stav Alfi Jun 11 '16 at 19:53
• @Stav: By block I simply mean a maximal set of consecutive integers. In the set $\{2,3,5,6,7,9\}$ the sets $\{2,3\},\{5,6,7\}$, and $\{9\}$ are the longest strings of consecutive integers, so they’re the blocks. The blocks of $\{1,5,6,7,8,12,13,14,20\}$ are $\{1\},\{5,6,7,8\},\{12,13,14\}$, and $\{20\}$: in each block the numbers are consecutive, and the blocks are as long as possible. Note that in the case $|I|=2$, your two cases correspond to one block, when $i+1=j$, and two blocks, when $i+1<j$. – Brian M. Scott Jun 11 '16 at 19:58
• No, Thank you! I could not get a better answer! The second paragraph provided the answer to my question in the best way. Combining all was perfectlly clear. Thank you agian and good night from isreal! :) – Stav Alfi Jun 11 '16 at 20:24
• @Stav: You’re welcome! (And have a good night.) – Brian M. Scott Jun 11 '16 at 20:26
• @MarkoRiedel I think Brian's expression counts the permutations that do have 2 consecutive numbers. If you take $n!$ minus his sum, it reproduces your sequence. – David Zhang Jun 11 '16 at 22:05
Consider the following recurrence: the desired count $Q_n$ is $1$ for $n=1$ and $1$ for $n=2.$ For $n\gt 2$ we obtain an admissible permutation either by placing the value $n$ anywhere at $n-1$ possible positions of an admissible permutation from $Q_{n-1}$ (this is not $n$ because we may not place $n$ next and to the right of $n-1$) or we construct a permutation having exactly one pair of consecutive numbers and place the value $n$ between these two. This can be done by taking a permutation from $Q_{n-2}$ and replacing one of the $n-2$ values by a fused pair containing the value and its successor and incrementing the values that are larger than the first element of the fused pair.
We get the recurrence
$$Q_n = (n-1) Q_{n-1} + (n-2) Q_{n-2}$$
and $Q_1= Q_2= 1.$ This yields the sequence
$$1, 3, 11, 53, 309, 2119, 16687, 148329, 1468457, 16019531, 190899411, \\ 2467007773, 34361893981, 513137616783,\ldots$$
which is OEIS A000255, where a detailed entry may be found.
The Maple code for this was as follows.
with(combinat);
C :=
proc(n)
option remember;
local perm, pos, res;
res := 0;
perm := firstperm(n);
while type(perm, list) do
for pos to n-1 do
if perm[pos] + 1 = perm[pos+1] then
break;
fi;
od;
if pos = n then
res := res + 1;
fi;
perm := nextperm(perm);
od;
res;
end;
Q :=
proc(n)
option remember;
if n = 1 or n = 2 then return 1 end if;
(n - 1)*Q(n - 1) + (n - 2)*Q(n - 2)
end;
Addendum. Here is my perspective on the inclusion-exclusion approach. We take as the underlying partially ordered set the set $P$ of subsets (these are the nodes of the poset) of $\{1,2,\ldots,n-1\}$ where a subset $S\in P$ represents permutations where the elements of $S$ are next to their successors, plus possibly some other elements also next to their successors. The partially ordered set is ordered by set inclusion. To compute the cardinality of the permutations corresponding to $S$ suppose that the elements of $S$ listed in order form $m$ blocks. We first remove these from $[n].$ We must also remove the elements that are consecutive with the rightmost element of each block, so we have now removed $|S|+m$ elements. We then put the augmented and fused blocks back into the permutation and permute them. We have added in $m$ blocks, therefore the net change is $-|S|-m +m = -|S|.$ Hence by inclusion-exclusion we compute the quantity
$$\sum_{S\in P, S\ne\emptyset} (-1)^{|S|} (n-|S|)!.$$
Now this depends only on the number $q$ of elements in $S$ so we get
$$\sum_{q=1}^{n-1} {n-1\choose q} (-1)^q (n-q)!.$$
We must now ask about the weight assigned to a permutation with $p$ elements (call this set $T$) next to their successor. This permutation is included in or rather represented by all sets $S$ that are subsets of the set $T$, which is the poset spanned by the singletons and $T$ being the topmost node. We obtain
$$\sum_{q=1}^p (-1)^q {p\choose q} = -1 + (1-1)^p = -1.$$
The count assigns the weight minus one to the permutation. It follows that when we add $n!$ exactly those permutations remain that do not contain consecutive adjacent values, for a result of
$$n! + \sum_{q=1}^{n-1} {n-1\choose q} (-1)^q (n-q)!.$$
We may simplify this to
$$\sum_{q=0}^{n-1} {n-1\choose q} (-1)^q (n-q)!.$$
In both cases, there are $(n-2)!$ permutations.
In the first case, there is the triple $(i,i+1,i+2)$ and $n-3$ other numbers.
In the second case, there are two pairs, and $n-4$ other numbers.
• You say that becouse in both cases we will permutate n-2 numbers , the answer should be (n-2)!. but my quastion is why you neglect the big deffrence between those 2 cases? – Stav Alfi Jun 11 '16 at 14:25
• They are different; but they both have the same number of permutations. – Empy2 Jun 11 '16 at 14:27
• Then how do we know when to sum 2 options instead of taking only one as an answer like now? – Stav Alfi Jun 11 '16 at 14:28
• @Stav: For each fixed pair of $i$ and $j$ with $1\le i<j\le n-1$ there are $(n-2)!$ permutations in $A_i\cap A_j$; the reasoning explaining this is different when $j=i+1$ from when $j>i+1$, but in both cases the result of the reasoning is $(n-2)!$. – Brian M. Scott Jun 11 '16 at 19:06
• Thank you for your comment. Just to clarify the rest of the prove, I see that |Ai∩Aj| for all cases is (n-2)! but very hard to guess for the |Ai1∩Ai2∩....Aik|=(n-k)!, how do you know that with out proving it? – Stav Alfi Jun 11 '16 at 19:21 | 2019-07-15T17:58:47 | {
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https://math.stackexchange.com/questions/3991510/converting-n-is-an-odd-negative-integer-into-formal-logic/3991560 | # Converting “n is an odd negative integer” Into Formal Logic
I am learning how to think about mathematical truth at the level of formal logic and I am tasked with converting the following basic statement into something formal: "n is a negative integer that is odd"
My first attempt at this was the following: Let $$n \in \mathbb{Z}$$ such that $$2n + 1 < 0$$
From my point of view this statement is saying that "n is an integer and it's values are restricted by the inequality $$2n + 1 < 0$$". But how this actually translated to a mathematician was "$$n$$ is an integer satisfying the inequality $$2n + 1<0$$" and after some thought, I concluded my initial statement was indeed wrong.
So I went back to the drawing board and came up with this:
$$\exists k \in \mathbb{N} \quad n = -2k + 1$$
But I feel that this formulation is still missing something, and I just can't put my finger on it. Should something be said about $$n$$ as well? Another way I thought of writing this was:
$$\exists k \in \mathbb{N} \quad (n = -2k + 1) \Rightarrow n \in \mathbb{Z}$$
$$\exists n \in \mathbb{Z} \quad \exists k \in \mathbb{N} \quad n = -2k + 1$$ Any help thinking through this would be appreciated.
• “n is an odd negative integer” is a predicate; thus, the formula must be of form $P(n)$ with $n$ free. – Mauro ALLEGRANZA Jan 19 at 14:41
• @MauroALLEGRANZA Could you perhaps provide an example demonstrating what you mean? It doesn't need to be related precisely to my question, but just to give a bit more insight. I understand that $P(n)$ is a statement about $n$, but I am missing what you mean by "$n$ free". – GrayLiterature Jan 19 at 14:48
• $\text {OddNegInt}(n) \leftrightarrow [n \in \mathbb Z \land (n < 0) \land \exists k \in \mathbb N (n=2k+1)]$ – Mauro ALLEGRANZA Jan 19 at 14:56
• @MauroALLEGRANZA I see now. "N is an odd negative integer" can be translated as "n is an integer AND n is odd AND n is negative" – GrayLiterature Jan 19 at 15:02
• @MauroALLEGRANZA Although, I think that the last part in your formulation for $k$ would require that $k \in \mathbb{Z}$. Since if $n = 2k + 1$ then n is strictly positive for any $k \in \mathbb{N}$. Would you agree with this observation? – GrayLiterature Jan 19 at 15:15
We can write "$$n$$ is a negative integer" as $$n\in\mathbb{Z}^-$$ And "$$n$$ is odd" as $$\exists k\in\mathbb{Z}.\ n=2k-1$$ Combine both: $$\exists k\in\mathbb{Z}.\ n=2k-1 \wedge n\in\mathbb{Z}^-$$ Or alternatively, $$\exists k\in\mathbb{Z}.\ \mathbb{Z}^-\ni n=2k-1$$
Hope this helps. :)
• No, $\mathbb{Z}$ also contains all positive integers. "$n$ is a negative integer" is writen $n \in \mathbb{Z} \wedge n < 0$. – Olivier Roche Jan 19 at 15:29
• @OlivierRoche: As far as the popular convention says, $\mathbb {Z}^+$ is the set of positive integers, and $\mathbb {Z}^-$ is the set of negative integers. – ultralegend5385 Jan 19 at 23:40
• Oh sorry I didn't see the superscript! – Olivier Roche Jan 20 at 0:50
This answer follows the comment provided by Mauro ALLEGRANZA in the comments:
The statement "$$n$$ is an odd negative integer" can be translated to "$$n$$ is an integer AND $$n$$ is odd AND $$n$$ is negative". Converting this to formal logic we can write:
• "$$n$$ is an integer" == $$n \in \mathbb{Z}$$
• "$$n$$ is negative" == $$n < 0$$
• "$$n$$ is odd" == $$\exists k \in \mathbb{Z} \quad (n = 2k +1)$$ (edit : if one takes $$k \in \mathbb{N}$$, then $$n > 0$$)
and combining these statements about $$n$$:
• "n is an odd negative integer" == ($$n \in \mathbb{Z}) \quad \land \quad (n<0) \quad \land \quad \exists k \in \mathbb{Z} \thinspace (n = 2k +1)$$
edit : Since $$\exists k \in \mathbb{Z} \thinspace (n = 2k +1)$$ already implies that $$n \in \mathbb{Z}$$, one can omit it and state :
$$(n<0) \quad \land \quad \exists k \in \mathbb{Z} \thinspace (n = 2k +1)$$
• Thank you for the edit. – GrayLiterature Jan 19 at 16:01 | 2021-03-02T15:13:01 | {
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https://web.sas.upenn.edu/ancoop/2018/10/10/the-elements-eye-view/ | ## The Element’s-Eye View
There are three things to keep in mind when it comes to set-theory proofs (which most of our proofs hereafter will be):
1. Sets are defined by their membership. Because of this, we nearly always take an element’s-eye view of any proof involving sets. Notice that in the proofs we presented above, the first thing we did to get going was say something like “Let $x\in$. . .”.
2. Generally, the way to go to prove two sets are equal is to show (separately) that each is a subset of the other.
3. Keep in mind what your goal is. That is, pay attention to the consequent. Your goal is to translate from the language of the consequent into the language of the antecedent.
These three principles are not only good advice for writing a proof, but they are in fact good advice for {\em figuring out} a proof as well. The following examples will illustrate principles 1 and 3.
Theorem. If $A$ and $B$ are sets with $2^A\subseteq 2^B$, then $A\subseteq B$.
Proof. Let $A$ and $B$ be set with $2^A\subseteq 2^B$. Since we have to show $A\subseteq B$, let $x\in A$. Then $\{x\}\subseteq A$, so $\{x\}\in 2^A$. Since $2^A\subseteq 2^B$, we have $\{x\}\in 2^B$. But this means $\{x\}\subseteq B$. Since $x\in\{x\}$, we see that $x\in B$.
So we’ve shown every $x\in A$ also has $x\in B$, i.e. $A\subseteq B$. $\Box$
Theorem. If $A$ and $B$ are sets with $A\subseteq B$, then $2^A\subseteq 2^B$.
Proof. Let $A$ and $B$ be sets with $A\subseteq B$. Since we have to show $2^A\subseteq 2^B$, let $x\in 2^A$. Then $x\subseteq A$. Since $A\subseteq B$, and set inclusion is transitive, we have $x\subseteq B$. But this is exactly the statement that $x\in 2^B$.
So we’ve show that every $x\in 2^A$ also has $x\in 2^B$, i.e. $2^A\subseteq 2^B$. $\Box$
Notice that in both cases, we were concerned with the {\em consequent} of the conditional we’d be asked to show (principle III), and that we proceeded by following elements (principle I). In the first proof, we translated from the language of elements of $A$ and $B$ to the language of subsets of $A$ and $B$ to the language of elements of $2^A$ and $2^B$, applied our assumption, and translated back. In the second proof, we translated from the language of elements of $2^A$ and $2^B$ to the language of subsets of $A$ and $B$, applied our assumptions, and translated back.
Activity Consider the following couplet:
Heffalumps and woozles are very confuzle.
A heffalump or woozle’s very sly.
Explain why both lines are talking about the same thing as follows. First, give a mathematical name to what they describe. Then explain why each line takes a different perspective on that mathematical object, and why those perspectives are not contradictory, the difference between “and” and “or” notwithstanding.
## Meta Data
Title: The Element’s-Eye View
Date Posted: October 10, 2018
Posted By:
Category: sets
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https://puzzling.stackexchange.com/questions/60536/largest-odd-factors-summing-to-a-square | Largest odd factors summing to a square
I just found this awesome puzzle from the Tournament of the Towns (though I'm sure it's appeared other places too). The connection between odd factors and square is surprising, and the proof has a lovely 'aha' moment. Enjoy!
For any natural number $x$, prove that $$\sum_{y=x+1}^{2x}(\text{largest odd factor of y})=x^2.$$
• I'm never around when the good math puzzles are posted :( – Quintec Feb 14 '18 at 1:00
• dude same :( aiyaaa – NL628 Feb 14 '18 at 4:42
Let $H(x) = (\text{largest odd factor of x})$. We can make two observations about this function:
1. $H(x) = H(\frac{x}{2})$, if x is even
2. $H(x) = x$, if x is odd
Let $$F(x) = \sum_{y=x+1}^{2x}H(x).$$
$F(1) = 1 = 1^2$. By induction, if $F(x) = x^2$, then
\begin{align*} F(x+1) &= F(x) + H(2x+2) + H(2x+1) - H(x+1) \\&= F(x) + H(x+1) + (2x+1) - H(x+1) \\&= F(x) + 2x+1 \\&= x^2 + 2x + 1 \\&= (x+1)^2 \end{align*}
• Whoa! Nice proof! What I had in mind was more like Gareth's proof, but this is really neat and fast. – Rand al'Thor Feb 14 '18 at 12:11
ffao posted his answer while I was writing this up. I like my solution better, though in some sense it's equivalent to ffao's, so I'm posting mine too. I will of course be entirely unoffended if Rand gives the mighty green checkmark to ffao, who after all got there first.
It is well known that
the sum of all odd numbers below $2x$ is equal to $x^2$. And there are $x$ of these, from $2\cdot1-1$ to $2x-1$.
So presumably the sum we have here,
which has the right number of terms and consists entirely of odd numbers, will simply be a permutation of this.
Well,
take any odd number $t$ between $1$ and $2x$ inclusive. Keep doubling it until the next doubling would go beyond $2x$. You will end up with a number from $x+1$ to $2x$ inclusive, and clearly each such number appears just once. And the largest odd divisor of this number will be exactly $t$. We're done.
Or, to put it a bit more formally (possibly harder to follow quickly but easier to convince yourself it's definitely correct):
let $A=\{\,y\,:\,x+1\leq y\leq2x\,\}$ and $B=\{\,z\,:1\leq z\leq2x\,\&\,z \textrm{ odd}\,\}$;
define $f:A\rightarrow B$ and $g:B\rightarrow A$ by
$f(y)=\textrm{largest odd divisor of$y$}$ and
$g(z)=\textrm{largest$2^kz$that's$\leq2x$}$;
then $f,g$ are inverses and therefore our sum, which is $\sum_{y\in A}f(y)$,
equals $\sum_{z\in B}z$, which famously equals $x^2$.
• Yes, in some sense this is actually equivalent to what I ended up doing. The good side to my solution is that there is no actual thinking required to reach it, but it ends up being less clean. – ffao Feb 14 '18 at 0:10
• You can also word it this way: Let: $x+1 = O_{x+1}*2^{E_{x+1}}$ ... $2x = O_{2x}*2^{E_{2x}}$ None of the numbers in the range can be a multiple of another, but if the $O$s of two numbers are identical, one of the numbers will be the multiple of other, so all $O$s must be different, which are the odd numbers up to $2x-1$. The sum will be $x^2$. – Nautilus Feb 14 '18 at 9:30
• "clearly each such number appears just once" - I think this could do with just a little more explanation, as it's really the crux of this proof. – Rand al'Thor Feb 14 '18 at 12:13
• This is the proof I had in mind when I posted the question (the 'aha' moment being what's in your first two spoilertags). But ffao's proof is so unexpectedly short and slick! I'm not sure where to put the checkmark now, but knowing me, it'll probably be a while until I award it anyway ;-) – Rand al'Thor Feb 14 '18 at 12:14
• One of the reasons for the more-formal spoilered paragraph was to be more explicit about that "clearly". (I claim it really is obvious that those functions f and g are inverses, and that's all you need.) – Gareth McCaughan Feb 14 '18 at 13:40 | 2020-05-26T10:49:40 | {
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http://mathhelpforum.com/statistics/224720-dependent-probability-question.html | Math Help - dependent probability question?
1. dependent probability question?
Hello, I had a question about how to calculate dependent probability. Calculating independent probability is easy, simply multiply the probability of each event: P1 * P2 * P3.
Therefore, if 3 different resources are working on 3 independent and unrelated tasks, and each resource has an 80% probability of completing his task, then the total probability of all resources completing all tasks, successfully, would be .80 * .80 * .80 = 51.2%
However, I need to figure out the probability formula, for a more advanced scenario. Consider a scenario, where 1 resource is working on 3 tasks in sequence, and the probability of the resource completing each individual task is 80%.
In this scenario, the first task would be independent, but the 2nd task would be dependent on completion of the 1st task, and the 3rd task would be dependent on completion of the 1st and 2nd task. So can you tell me the formula I would use to calculate total probability in this scenario?
2. Re: dependent probability question?
Let's look at failure rather than success. Then the overall probability of success is 1 - probability of failure.
You can fail at step 1, step 2, or step 3.
Pr[fail at step 1] = 1 - .8 = .2
Pr[fail at step 2] = Pr[fail at step 2 | success at step 1]
Now assuming failure at step 2 is independent of success at step 1 we get
Pr[fail at step 2] = Pr[fail at step 2] Pr[success at step 1] = .2 * .8 = .16
Pr[fail at step 3] = Pr[fail at step 3 | success at 1 and 2] = .2 * .8 * .8 = .128
The total probabily of failure is then the sum of these (.2 + .16 + .128) = .488
So Pr[Success] = 1 - .488 = .512
This is identical to the non sequenced case. Why? Because in order to succeed all 3 steps have to be succeeded at in either case.
Where this methodology becomes very useful is when there are multiple paths to success as well as failure.
3. Re: dependent probability question?
My probability abilities are not advanced but i can get a surprising number of problems correct by using probability trees and other basic probability techniques.
When I learn more advanced probability theories they are more likely to make sense because I can 'see' that the formula is logical.
Learn to use trees effectively and the answer to these problems becomes much easier to discover. (You don't always need to draw the whole tree)
That's what I think anyway.
4. Re: dependent probability question?
Originally Posted by Melody2
My probability abilities are not advanced but i can get a surprising number of problems correct by using probability trees and other basic probability techniques.
When I learn more advanced probability theories they are more likely to make sense because I can 'see' that the formula is logical.
Learn to use trees effectively and the answer to these problems becomes much easier to discover. (You don't always need to draw the whole tree)
That's what I think anyway.
In industry probability trees are used EXTENSIVELY.
5. Re: dependent probability question?
romsec - thanks for your feedback. However, I think the total probability should be lower, if 1 resource works on 3 tasks in sequence. Here's why: If the resource fails at task 1, then he will never get to task 2 or task 3. If the resource fails at task 2, then he will never get to task 3.
On the other hand, if 3 different resources work on these 3 different tasks independently, then this approach will completely sidestep the dependencies/risks that I described with the single resource above, so it seems like the realistic probability of success should be greater.
It seems like probability theory should have a formula that factors in the dependencies I described above. Can you provide an updated probability formula based on the additional context I provided above?
6. Re: dependent probability question?
I tell you what. You look at my derivation and tell me specifically what bits you think are incorrect. I took into account all the dependencies you put in by requiring serial operation.
7. Re: dependent probability question?
@romsek - I'm only a beginner at statistics. Maybe there's some type of statistics concept that we're not accounting for here?
I was under the impression that the total probability for independent events was calculated differently than the total probability for dependent events, and the total probability for dependent events would be lower. Maybe it will help if I simplify my example?:
Scenario 1: 3 independent events, each event has a probability of 80%: .80 * .80 * .80 = 51.2%
Scenario 2: 3 events with sequential dependency, each distinct event has a probability of 80%: ???
Maybe part of the problem here, is that when I say each event has a probability of 80%, that implies that the dependency relationship has no impact, compared to scenario 1. I was assuming that the total probability formula for dependent events would have a built-in function to lower the total probability.
Does my simplified example change anything? Or do you still assert that the total probability for these scenarios should be identical?
8. Re: dependent probability question?
Originally Posted by random512
@romsek - I'm only a beginner at statistics. Maybe there's some type of statistics concept that we're not accounting for here?
I was under the impression that the total probability for independent events was calculated differently than the total probability for dependent events, and the total probability for dependent events would be lower. Maybe it will help if I simplify my example?:
Scenario 1: 3 independent events, each event has a probability of 80%: .80 * .80 * .80 = 51.2%
Scenario 2: 3 events with sequential dependency, each distinct event has a probability of 80%: ???
Maybe part of the problem here, is that when I say each event has a probability of 80%, that implies that the dependency relationship has no impact, compared to scenario 1. I was assuming that the total probability formula for dependent events would have a built-in function to lower the total probability.
Does my simplified example change anything? Or do you still assert that the total probability for these scenarios should be identical?
In both cases there is a single outcome that denotes success of the entire venture. All 3 tasks must be completed successfully. There is only a single way this can be achieved and that is that all 3 tasks are done successfully.
The probability that your 3 independent folks each succeeding is .8. The probability that all 3 independently succeed is (.8)^3 = .512
Task 1 succeeds with pr. .8. Only then does it proceed to Task 2. The probability that we get to Task 2 is thus .8. We don't care what happens if we fail at Task 1, if we do we fail.
We have arrived at Task 2 with pr .8. Now we have .8 pr of completing Task 2. There's no other dependency of our success on Task 1, it got here, that's all we need to know. So our pr of success of Task 2 is again .8 but also times the pr. that it got here at all, which we know to be .8. So the overall pr of completing Task 2 is (.8)^2 = .64
Now we make it to Task 3 with pr .64. Once again there is no dependency on how well things went at task 1 or 2, the thing got here and that's all we know and we have pr .8 of success at Task 3. So the overall pr of completing Task 3 is .8 times the pr it got here at all .64, or (.8)^3 = .512.
There are all sorts of dependencies you could put between your tasks that would show the sort of conditional probability effects you are thinking of. But you haven't done that here.
9. Re: dependent probability question?
@Romsek - Thanks for your knowledge and patience. I think I figured out what I was missing: total probability of overrun.
3 independent events can execute concurrently. Therefore, if each event has an overrun probability of 50% then the total probability of overrun can be kept to 50%. However, if the 3 events must execute in sequence, then the probability of overrun is:
1 - (.50 * .50 * .50) = 87.5%
So that's how the concept of total probability can be used to prove the cost/impact of dependent events, as opposed to independent events.
10. Re: dependent probability question?
If you want to set up an example demonstrating what you have in mind consider this.
Let's say we're making toy dolls since it's the holidays. We have to cut the parts, assemble the parts, paint the doll.
We have 3 elves and they have different abilities regarding making these dolls. For elf A his pr success at those tasks is say {0.85, .6, .4}
elf B could be {0.6, 0.9, 0.6}, and elf C could be {0.5, 0.5, 0.8}
You could have each elf build the entire doll at their station but it's pretty clear that there's a better strategy and that's the pair the elf's task with their best ability.
That's where your sequential operation comes in since you have to build the dolls in a sequence.
If you really want to get fancy you can have the probability of success at a station depend on the quality that the previous station produced. It's harder to assemble poorly cut parts. It's harder to paint a poorly assembled doll, and incorporate all that in to further show how important your strategy of matching worker and task is.
All this will certainly show the effect of sequential vs parallel operation.
11. Re: dependent probability question?
Originally Posted by random512
@Romsek - Thanks for your knowledge and patience. I think I figured out what I was missing: total probability of overrun.
3 independent events can execute concurrently. Therefore, if each event has an overrun probability of 50% then the total probability of overrun can be kept to 50%. However, if the 3 events must execute in sequence, then the probability of overrun is:
1 - (.50 * .50 * .50) = 87.5%
So that's how the concept of total probability can be used to prove the cost/impact of dependent events, as opposed to independent events.
ok I'm glad you figured it out.
12. Re: dependent probability question?
One more related question. What's the proper way to describe the probability multiplication factor? For example:
* the probability of a single risk event occurring is 50%
* the probability of at least 1/3 risk events occurring is:
1 - ( .50 * .50 * .50) = 87.5%
Is it valid to say that additional risk factors create an "exponential" probability that at least 1 risk event will occur? "Exponential" is a good word, because it's descriptive, and powerful. However, I want to make sure that I use the word appropriately. So can I describe this multiplication effect as "exponential" or should I choose a different word? | 2015-05-03T12:19:03 | {
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https://math.stackexchange.com/questions/3212103/find-smallest-contraction-coefficient | # Find Smallest contraction coefficient
I have been given the following function $$f:[-1,1]\to \mathbb{R}$$: $$f(x)=\ln(x+2)-x$$ And I have been asked whether it is a contraction or not, and if it is, I have to find the smallest contraction coefficient, such that $$0.
Attempt
Since $$0<|f'(x)|<1$$ for $$(-1,1)$$ we must have that the function is a contraction, and i would intuitively say that $$\left|-\frac{2}{3}\right|=\frac{2}{3}$$ is the smallest contraction coefficient.
Doubts
I have no concrete theorem or example to support my claim, and therefore I am skeptical.
Since $$f$$ is differentiable $$\forall x \in [-1,1], |f'(x)|= |\cfrac{1}{x+2} - 1| = |\cfrac{-x-1}{x+2}| = \cfrac{1+x}{2+x}$$ and $$\forall x \in [-1,1], f''(x) = \cfrac{1}{(x+2)^2} \ge 0$$ so $$f'$$ is growing. And we have $$| f'(1)| = \cfrac{2}{3}$$ So $$\forall x \in [-1,1], |f'(x)| \le \cfrac{2}{3}$$ Therefore $$\cfrac{2}{3}$$ is indeed the smallest contraction coefficient.
Hint: Because of the mean value inequality, you want to maximize $$|f'(x)|$$ for $$x \in [-1,1]$$.
I think that Rasmus' realized that $$L=\frac 23$$ is the maximum value for $$f'$$ and therefore will work as a contraction constant. This was probably not the question. The question, I think, was about the possibility of getting a contraction constant $$\tilde L < L$$. The answer is no, it is not possible. If you consider a smaller constant you immediately get an interval of the form $$[\xi, 1]$$ where the contraction inequality does not hold. | 2019-06-25T11:51:08 | {
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http://eigen.tuxfamily.org/dox/group__LeastSquares.html | Please, help us to better know about our user community by answering the following short survey: https://forms.gle/wpyrxWi18ox9Z5ae9
Eigen 3.3.9
Solving linear least squares systems
This page describes how to solve linear least squares systems using Eigen. An overdetermined system of equations, say Ax = b, has no solutions. In this case, it makes sense to search for the vector x which is closest to being a solution, in the sense that the difference Ax - b is as small as possible. This x is called the least square solution (if the Euclidean norm is used).
The three methods discussed on this page are the SVD decomposition, the QR decomposition and normal equations. Of these, the SVD decomposition is generally the most accurate but the slowest, normal equations is the fastest but least accurate, and the QR decomposition is in between.
# Using the SVD decomposition
The solve() method in the BDCSVD class can be directly used to solve linear squares systems. It is not enough to compute only the singular values (the default for this class); you also need the singular vectors but the thin SVD decomposition suffices for computing least squares solutions:
Example:Output:
#include <iostream>
#include <Eigen/Dense>
using namespace std;
using namespace Eigen;
int main()
{
cout << "Here is the matrix A:\n" << A << endl;
cout << "Here is the right hand side b:\n" << b << endl;
cout << "The least-squares solution is:\n"
<< A.bdcSvd(ComputeThinU | ComputeThinV).solve(b) << endl;
}
Here is the matrix A:
-1 -0.0827
-0.737 0.0655
0.511 -0.562
Here is the right hand side b:
-0.906
0.358
0.359
The least-squares solution is:
0.464
0.043
This is example from the page Linear algebra and decompositions .
# Using the QR decomposition
The solve() method in QR decomposition classes also computes the least squares solution. There are three QR decomposition classes: HouseholderQR (no pivoting, so fast but unstable), ColPivHouseholderQR (column pivoting, thus a bit slower but more accurate) and FullPivHouseholderQR (full pivoting, so slowest and most stable). Here is an example with column pivoting:
Example:Output:
MatrixXf A = MatrixXf::Random(3, 2);
VectorXf b = VectorXf::Random(3);
cout << "The solution using the QR decomposition is:\n"
<< A.colPivHouseholderQr().solve(b) << endl;
The solution using the QR decomposition is:
0.464
0.043
# Using normal equations
Finding the least squares solution of Ax = b is equivalent to solving the normal equation ATAx = ATb. This leads to the following code
Example:Output:
MatrixXf A = MatrixXf::Random(3, 2);
VectorXf b = VectorXf::Random(3);
cout << "The solution using normal equations is:\n"
<< (A.transpose() * A).ldlt().solve(A.transpose() * b) << endl;
The solution using normal equations is:
0.464
0.043
If the matrix A is ill-conditioned, then this is not a good method, because the condition number of ATA is the square of the condition number of A. This means that you lose twice as many digits using normal equation than if you use the other methods.
Eigen
Namespace containing all symbols from the Eigen library.
Definition: Core:309
Eigen::ComputeThinU
@ ComputeThinU
Definition: Constants.h:385
Eigen::DenseBase::Random
static const RandomReturnType Random()
Definition: Random.h:113
Eigen::MatrixBase::bdcSvd
BDCSVD< PlainObject > bdcSvd(unsigned int computationOptions=0) const
Definition: BDCSVD.h:1269
Eigen::ComputeThinV
@ ComputeThinV
Definition: Constants.h:389
Eigen::Matrix
The matrix class, also used for vectors and row-vectors.
Definition: Matrix.h:180 | 2021-01-20T08:04:49 | {
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https://math.stackexchange.com/questions/2134031/is-this-formula-for-frace2-3e21-known-how-to-prove-it | # Is this formula for $\frac{e^2-3}{e^2+1}$ known? How to prove it?
I found an interesting infinite sequence recently in the form of a 'two storey continued fraction' with natural number entries:
$$\frac{e^2-3}{e^2+1}=\cfrac{2-\cfrac{3-\cfrac{4-\cdots}{4+\cdots}}{3+\cfrac{4-\cdots}{4+\cdots}}}{2+\cfrac{3-\cfrac{4-\cdots}{4+\cdots}}{3+\cfrac{4-\cdots}{4+\cdots}}}$$
The numerical computation was done 'backwards', starting from some $x_n=1$ we compute:
$$x_{n-1}=\frac{a_n-x_n}{a_n+x_n}$$
And so on, until we get to $x_0$. The sequence converges for $n \to \infty$ if $a_n>1$ (or so it seems).
For constant $a_n$ we seem to have quadratic irrationals, for example:
$$\frac{\sqrt{17}-3}{2}=\cfrac{2-\cfrac{2-\cfrac{2-\cdots}{2+\cdots}}{2+\cfrac{2-\cdots}{2+\cdots}}}{2+\cfrac{2-\cfrac{2-\cdots}{2+\cdots}}{2+\cfrac{2-\cdots}{2+\cdots}}}$$
For $a_n=2^n$ we seems to have:
$$\frac{1}{2}=\cfrac{2-\cfrac{4-\cfrac{8-\cdots}{8+\cdots}}{4+\cfrac{8-\cdots}{8+\cdots}}}{2+\cfrac{4-\cfrac{8-\cdots}{8+\cdots}}{4+\cfrac{8-\cdots}{8+\cdots}}}$$
I found no other closed forms so far, and I don't know how to prove the formulas above. How can we prove them? What is known about such continued fractions?
There is another curious thing. If we try to expand some number in this kind of fraction, we can do it the following way:
$$x_0=x$$
$$a_0=\left[\frac{1}{x_0} \right]$$
$$x_1=\frac{1-a_0x_0}{1+a_0x_0}$$
$$a_1=\left[\frac{1}{x_1} \right]$$
However, this kind of expansion will not give us the above sequences. We will get faster growing entries. Moreover, the fraction will be finite for any rational number. For example, in the list notation:
$$\frac{3}{29}=[9,28]$$
You can easily check this expansion for any rational number.
As for the constant above we get:
$$\frac{e^2-3}{e^2+1}=[1,3,31,74,315,750,14286,\dots]$$
Not the same as $[1,2,3,4,5,6,7,\dots]$ above!
We have similar sequences growing exponentially for any irrational number I checked.
$$e-2=[1,6,121,284,1260,3404,25678,\dots]$$
$$\pi-3=[7,224,471,2195,10493,46032,119223,\dots]$$
By the way, if we try CF convergents, we get almost the same expansion, but finite:
$$\frac{355}{113}-3=[7,225]$$
$$\frac{4272943}{1360120}-3=[7,224,471,2195,18596,227459,\dots]$$
So, the convergents of this sequence are not the same as for the simple continued fraction, but similar.
Comparing the expansion by the method above and the closed forms at the top of the post, we can see that, unlike for simple continued fractions, this expansion is not unique. Can we explain why?
Here is the Mathematica code to compute the limit of the first fraction:
Nm = 50;
Cf = Table[j, {j, 1, Nm}];
b0 = (Cf[[Nm]] - 1)/(Cf[[Nm]] + 1);
Do[b1 = N[(Cf[[Nm - j]] - b0)/(Cf[[Nm - j]] + b0), 7500];
b0 = b1, {j, 1, Nm - 2}]
N[b0/Cf[[1]], 50]
And here is the code to obtain the expansion in the usual way:
x = (E^2 - 3)/(E^2 + 1);
x0 = x;
Nm = 27;
Cf = Table[1, {j, 1, Nm}];
Do[If[x0 != 0, a = Floor[1/x0];
x1 = N[(1 - x0 a)/(x0 a + 1), 19500];
Print[j, " ", a, " ", N[x1, 16]];
Cf[[j]] = a;
x0 = x1], {j, 1, Nm}]
b0 = (1 - 1/Cf[[Nm]])/(1 + 1/Cf[[Nm]]);
Do[b1 = N[(1 - b0/Cf[[Nm - j]])/(1 + b0/Cf[[Nm - j]]), 7500];
b0 = b1, {j, 1, Nm - 2}]
N[x - b0/Cf[[1]], 20]
Update
I have derived the forward recurrence relations for numerator and denominator:
$$p_{n+1}=(a_n-1)p_n+2a_{n-1}p_{n-1}$$ $$q_{n+1}=(a_n-1)q_n+2a_{n-1}q_{n-1}$$
They have the same form as for generalized continued fractions (a special case). Now I understand why the expansions are not unique.
• I am always thoroughly impressed with the length and detail in your posts. On the other hand, you might want to key in a little more on what your main question is here. – Brevan Ellefsen Feb 7 '17 at 22:25
• @BrevanEllefsen, if these are known results, I would like a reference. If not, I would like a hint about a proof. And the most important question - why is the expansion not unique, even though the properties seem to be similar to simple continued fractions. Thanks, by the way. I'm glad I'm not the only one who likes my posts :) – Yuriy S Feb 7 '17 at 22:26
• @Nemo, I'm not sure I understand what you did here, but I figured out the same thing after deriving the forward recurrence relations for numerator and denominator: $$p_{n+1}=(a_n-1)p_n+2a_{n-1}p_{n-1}$$ $$q_{n+1}=(a_n-1)q_n+2a_{n-1}q_{n-1}$$ Obviously, we have a special case of a generalized continued fraction. Which is why it is not unique – Yuriy S Feb 8 '17 at 9:18
For the first one, you could write $$f(n) = \frac{n-f(n+1)}{n+f(n+1)}$$ Then you suggest $$f(2) = \frac{e^2-3}{e^2+1}$$ But this then gives \begin{align} f(1) = \frac{2}{e^2-1}\\ f(3) = \frac{4}{e^2-1} \\ f(4) = \frac{3e^2-15}{e^2+3}\\ f(5) = \frac{-2e^2+18}{e^2-3} \end{align} I don't know if there is a recurrence relation that solves this, but you have a few more closed forms...
The second one we have $$g(2) = \frac{2 - g(2)}{2+g(2)}$$ so we can solve the quadratic $x^2+3x-2$ to get $(\sqrt{17}-3)/2$.
For the third one, we have $$h(n) = \frac{n-h(2n)}{n+h(2n)}$$ using the trial version of $h(2)=1/2$, we get \begin{align} h(4)=\frac{2}{3}\\ h(8)=\frac{4}{5}\\ h(16)=\frac{8}{9} \end{align} then it is likely that $$h(n)=\frac{n}{n+2}$$ as this satisfies the recurrence and that $h(2)=1/2$.
• Thank you, when I look back it's clear that quadratic radicals or rational numbers are easily proven (provided the limit exists), just as you show here. For irrational numbers see the comments and the update, linking this to generalized continued fractions, and the proofs for $e$ related numbers can be found elsewhere – Yuriy S May 31 '17 at 11:29 | 2019-10-15T08:51:04 | {
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https://math.stackexchange.com/questions/3389198/deriving-the-derivative-of-tan2x-by-quotient-rule-using-sinx-cosx-identi | Deriving the derivative of $\tan^{2}x$ by quotient rule using (sinx/cosx) identity I am getting a different value (2secx(tanx+tan^2(x))
Deriving the derivative of $$\tan^{2}x$$ by quotient rule using $$\frac{\sin x}{\cos x}$$ identity, i am getting a different value $$2\sec x[\tan x+\tan^{2}x]$$ than by directly getting chain rule is $$2\sec^{2}x\tan x$$
Whats wrong?
Here is a correct derivation of the derivative of $$\tan^2 x$$ using the quotient rule:
$$\dfrac d {dx} \tan^2 x = \dfrac d {dx} \dfrac {\sin^2 x}{\cos^2x }=\dfrac{2\sin x \cos x \cos^2x + \sin^2 x 2 \cos x \sin x}{\cos^4x}$$
$$=2\dfrac{\sin x(\cos^2x+\sin^2x)}{\cos^3x}=2\dfrac{\sin x}{\cos^3 x}=2\tan x \sec^2x$$
• Yes, the chain rule yields $2 \tan x \sec^2x$ too – J. W. Tanner Oct 11 '19 at 6:58
• $\cos^2 x+\sin^2 x = 1$ is the Pythagorean identity – J. W. Tanner Oct 11 '19 at 7:08 | 2020-11-27T12:07:20 | {
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https://mathematica.stackexchange.com/questions/209829/is-there-a-way-to-randomly-distribute-points-within-a-circle-on-the-surface-of-a | # Is there a way to randomly distribute points within a circle on the surface of a sphere?
I'm attempting to set up a situation where on a 3D sphere, I choose a random point and construct a circle around this point with some radius. I then want to randomly distribute points within this circle. Is there a straightforward way to do this?
I have no trouble finding random points on the surface of the sphere; however I cannot seem to find a way to distribute points randomly on a closed region of the sphere.
Many thanks for any help in advance.
• Two related questions. – J. M.'s ennui Nov 18 '19 at 5:17
• You could brute force it by finding random points on the sphere, then throwing away any that aren't inside the desired region. – Foo Bar Nov 18 '19 at 12:59
• @Foo, yes, one can certainly use the rejection method for this. The tricky part is in figuring out how not to throw away too many points. – J. M.'s ennui Nov 18 '19 at 14:37
• Depends on your underlying distribution. I assume you want uniform random over the sphere's area. For contrast, consider uniform random points in a circle and then project the circle onto the sphere. That's still random but with a different distribution function. – Carl Witthoft Nov 19 '19 at 19:39
You can intersect the sphere and a cylinder, and then use RandomPoint. For example, here is a random point on the sphere:
sphere = Sphere[];
SeedRandom[1]
pt = RandomPoint[sphere]
{0.707037, 0.595614, 0.381239}
Then, you create a cylinder in he direction of the random point with a radius:
r = .7;
cylinder = Cylinder[{{0,0,0},pt}, r];
Now, intersect the cylinder and the sphere and use RandomPoint:
reg = RegionIntersection[cylinder, sphere];
pts = RandomPoint[reg, 1000];
Visualization:
Graphics3D[{Sphere[], Red, Point @ pts}]
• Can you verify that this produces uniform random per unit sphere area (as opposed to some projection of the intersecting circle)? I"m not familiar with how this Mathematica function actually selects points. – Carl Witthoft Nov 19 '19 at 19:41
• @CarlWitthoft That's what the documentation says: "RandomPoint will generate points uniformly in the region reg." In this case the region is the spherical cap. – Carl Woll Nov 19 '19 at 20:36
Using Christian Blatter's results from this math.SE answer, here is how to randomly sample a spherical cap:
randomCapPoint[{r_, r2_}, dir_?VectorQ] := With[{h = RandomReal[{Sqrt[1 - (r2/r)^2], 1}]},
RotationTransform[{{0, 0, 1}, Normalize[dir]}][r
Append[Sqrt[1 - h^2] Normalize[RandomVariate[NormalDistribution[], 2]], h]]]
For example,
BlockRandom[SeedRandom[1337]; (* for reproducibility *)
With[{r = 1, r2 = 2/5, d = {1.3, -2.4, 2}, n = 5000},
Graphics3D[{{Opacity[1/2], Sphere[{0, 0, 0}, r]},
{Blue, Arrow[Tube[{{0, 0, 0}, Normalize[d]}]]},
{Directive[AbsolutePointSize[2], Orange],
Point[Table[randomCapPoint[{r, r2}, d], {n}]]}}]]]
• Same question as at Carl Woll's answer: how do you verify the spatial distribution is "uniform" with respect to sphere area? – Carl Witthoft Nov 19 '19 at 19:43
• @Carl, using the results in Christian's answer, you could take a histogram of the longitude and colatitude of the sample points, and verify that they have the required distribution. – J. M.'s ennui Nov 19 '19 at 20:51
We can also take RandomPoints in boolean region obtained by the RegionIntersection of a Sphere and a Ball with radius r centered at a random point on the sphere:
SeedRandom[1]
r = .7;
ctr = RandomPoint[Sphere[]];
pts = RandomPoint[RegionIntersection[Ball[ctr, r], Sphere[]], 1000];
Graphics3D[{Red, Point@pts, White, Opacity[.5], Sphere[]}]
You can use Archimedes' result (https://en.wikipedia.org/wiki/On_the_Sphere_and_Cylinder) that the area between two lines of latitude is the same as the corresponding area of the enclosing cylinder.
This gives a fairly simple result for sampling the top of the sphere, above height Z
pt[Z_] := Module[{z, θ, r},
z = RandomVariate[UniformDistribution[{Z, 1}]];
θ = RandomVariate[UniformDistribution[{0, 2 π}]];
r = Sqrt[1 - z^2];
{r Cos[θ], r Sin[θ], z}]
• This is basically what my method does, except I replaced {r Cos[θ], r Sin[θ]} with r Normalize[RandomVariate[NormalDistribution[], 2]]. – J. M.'s ennui Nov 19 '19 at 20:49 | 2021-01-20T13:49:21 | {
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https://math.stackexchange.com/questions/1565290/trying-to-answer-this-counting-question | # Trying to answer this counting question
In a ballroom dance class, participants are divided into couples for each drill session. One partner leads and the other follows for three minutes, and then the couple switches roles for the next three minutes.
(a) Only four people show up on time. How many ways are there to pair them up?
My answer here is $4C2 = \binom42 = \tfrac{4!}{2!2!}$
(b) If instead six people show up on time, how many ways are there to pair them up?
My answer here is $6C2 = \binom62 = \tfrac{6!}{4!2!}$
(c) Assume all m people in the class arrive on time. (There are an even number of people in the class.) How many ways are there to pair them up?
My answer here is $mC2 = \binom m2= \tfrac{m!}{(m-2)!2!}$
(d) Consider this time assuming that we specify which member of each couple leads first. How many ways are there to pair-and- specify the dancers
My answer here is $mP2 = \tfrac{m!}{(m-2)!}$
## UPDATE:
Part (a) the answer is $\tfrac{1}{2}(4C2 * 2C2) = 3$
Part (b) the answer is $\tfrac{1}{2}(6C2 * 4C2 * 2C2) = 45$
• Yup, they are all correct. Just a typo in the second answer, you meant 6C2, I think. – learner Dec 8 '15 at 4:15
• Yes, I corrected it, Thanks! – Coheen Dec 8 '15 at 4:16
• No, they are not correct. See theophile's answer below. – fleablood Dec 8 '15 at 4:28
• @lucidgold What is the conflict with the answers? As far as I can see, all three are saying the same thing. – Théophile Dec 8 '15 at 18:17
For part (b), the number of ways to pick a first pair, then a second pair, then a third pair, is $\binom 62 \binom 42 \binom 22$, or in your notation $6C2 \times 4C2 \times 2C2$. But that not only pairs up all the dancers but distinguishes $\{AB, CD, EF\}$ from $\{AB, EF, CD\}$, $\{CD, AB, EF\}$, and several other permutations of the pairs. In fact you have counted the same three pairs $3!$ times, which is the number of permutations of three things. So the correct answer is not to divide by $2$ but rather to divide by $3! = 6$:
$$\frac{\binom 62 \binom 42 \binom 22}{6}.$$
Part (d) seems easier to me than part (c). First you choose who will be leaders in the first part of the lesson. There are $\binom{m}{m/2}$ ways to do this. Then you line up these $m/2$ leaders in a line and assign each of the remaining $m/2$ dancers to one of the leaders. That is, each ordering of the remaining $m/2$ dancers produces a unique pairing of followers with leaders. There are $(m/2)!$ such orderings, so the total number of possible pairings is ...
Once you have part (d), I would go back to part (c). Clearly there are more pairings counted in part (d) than in part (c). How many more?
For each way you can pair up dancers in part (c), within each pair there are two ways to choose who will lead during the first part of the lesson. Since there are $m/2$ pairs, there are therefore $2^{m/2}$ ways to choose which $m/2$ dancers will lead at first. But the choice of pairs for part (c), followed by choosing leaders for the first three minutes, can give us every choice of ordered pairs that exists in part (d). Therefore the number of choices in part (c) must be ...
No, these are not quite correct—at least not the way I understand the question. When there are four people, for example, there are only three ways to pair them up: $\{AB,CD\}, \{AC,BD\}$, and $\{AD,BC\}$. The reason this is different from your answer is that you have counted how many ways there are to choose two people from a set of four. This is ${4 \choose 2} = 6$, namely, $AB, AC, AD, BC, BD, CD$. But the number of ways to form one pair isn't the same as the number of ways to split the group into pairs.
One way to do the count for part (a) is to pick one pair (as you have done), then to form a second pair from the remaining two people. The number of ways to do this is $${4 \choose 2}{2 \choose 2} = 6 \cdot 1 = 6.$$ But in doing so we have accounted for each pairing twice; e.g., we have counted both $\{AB,CD\}$ and $\{CD,AB\}$, when these are in fact the same pairing. Therefore we should divide by $2$ to get $\frac62 = 3$ pairings.
• This is a great answer. it shows the problem with part a and leaves the rest to OP. There are similar problems with the rest, but if OP understands this answer, s/he should be able to deal with the rest. If not, that is another question. This is not an easy question. – Ross Millikan Dec 8 '15 at 4:28
• So what happened if the number of people is huge like 300? Is there is any formula I can use to solve this problem? So what I understand is I should use combinations and divide the answer by 2? is this correct? – Coheen Dec 8 '15 at 4:30
• @Coheen: you asked four different questions. All your answers were not correct. If the number of people increases, yes there is a formula you can use. The answer for the first three differs greatly from the answer to the fourth. For ones like the first three, you pick one pair, then another, then another, and so on. Then you need to consider that you could pick the pairs in any order. Your formulas just pick the first pair. – Ross Millikan Dec 8 '15 at 4:36
• For part (b), would the following be correct: (6C2* 4C2* 2C2 )/2 = 45 – Coheen Dec 8 '15 at 4:56
• @Coheen The reason for dividing by two was because there were two pairs. When there are three, you should divide by 3!. This is described in more detail in David K's answer. – Théophile Dec 8 '15 at 18:16
If I am not mistaken this is applications of the Multinomial Theorem. A lot of good examples are given in Sheldon Ross's A First Course in Probability.
Also note with part $(d)$ that uniqueness matters. So if we had 4 people show up and we chose who was going to lead who we would have a permutation for all of the members dancing. For example, let the 4 people be named: John, Sally, Jordan, Steve. If John lead Steve for the first 3 minutes it would not be the same as Steve leading John. So essentially we would have the same combination below, but without the denominator in the Multinomial, since order matters. | 2021-04-13T10:45:54 | {
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http://nacretv.easylia.com/uamkx/1ea173-how-to-find-eigenvectors-of-a-3x3-matrix | The detailed solution is given. If . If the resulting V has the same size as A, the matrix A has a full set of linearly independent eigenvectors that satisfy A*V = V*D. But for a special type of matrix, symmetric matrix, the eigenvalues are always real and the corresponding eigenvectors are always orthogonal. Suppose A is this 3x3 matrix: [1 1 0] [0 2 0] [0 –1 4]. The values of λ that satisfy the equation are the generalized eigenvalues. The code for this originally is … Calculate the eigenvalues and the corresponding eigenvectors of the matrix. The eigenvalues are r1=r2=-1, and r3=2. I am trying to find the best OOBB hitboxes for my meshes using PCA. Let A=[121−1412−40]. [V,D] = eig(A) returns matrices V and D.The columns of V present eigenvectors of A.The diagonal matrix D contains eigenvalues. In order to do this, I need the eigenvectors but I am kind of lost how to compute them without using a huge library. kerr_lee. It is also known as characteristic vector. Answer Save. Substitute every obtained eigenvalue $\boldsymbol{\lambda}$ into the eigenvector equations, 4.1. The three eigenvalues and eigenvectors now can be recombined to give the solution to the original 3x3 matrix as shown in Figures 8.F.1 and 8.F.2. 0 0. In fact, we will in a different page that the structure of the solution set of this system is very rich. Any help is greatly appreciated. What is the shortcut to find eigenvalues? The eigenvector v of an operator A is a vector such that: Av = kv, for some scalar k. So suppose v = (v1,v2,v3) and compute: Av = (v2+v3,v1+v3,v1+v2). Eigen vector, Eigen value 3x3 Matrix Calculator. Bring all to left hand side: 1 decade ago. Yes it is the same as there is multiple values of your eigen vector by multiplying by a scalar. How To: Find the equation of trig functions by their graphs How To: Do matrix algebra on a TI-83 calculator How To: Solve systems of linear equations with matrices How To: Find eigenvectors and eigenspaces of a 2x2 matrix How To: Use a change of basis matrix in linear algebra %PDF-1.2 3. 2 Answers. Eigenvectors corresponding to distinct eigenvalues are linearly independent. Example Problem. Find the eigenvalues and eigenvectors for the matrix [(0,1,0),(1,-1,1),(0,1,0)]. Answer 5 0 obj (The Ohio State University, Linear Algebra Final Exam Problem) Add to solve later Sponsored Links Eigenvalue $\boldsymbol{\lambda = 3}$, 4.2. The standard formula to find the determinant of a 3×3 matrix is a break down of smaller 2×2 determinant problems which are very easy to handle. We define the characteristic polynomial and show how it can be used to find the eigenvalues for a matrix. Find all the eigenvalues and corresponding eigenvectors of the given 3 by 3 matrix A. Finding Eigenvalues and Eigenvectors : 2 x 2 Matrix Example . Let's find the eigenvector, v 1, associated with the eigenvalue, λ 1 =-1, first. Reads for a joint honours degree in mathematics and theoretical physics (final year) in England, at the School of Mathematics and Statistics and the School of Physical Sciences at The Open University, Walton Hall, Milton Keynes. Hi, I am trying to find the eigenvectors for the following 3x3 matrix and are having trouble with it. play_arrow. 3,0,2. has eigenvalues of 2,4,and -3. Matrix A: Find. Enter a matrix. Algebraic and geometric multiplicity of eigenvalues. For Example, if x is a vector that is not zero, then it is an eigenvector of a square matrix … Next, transpose the matrix by rewriting the first row as the first column, the middle row as the middle column, and the third row as the third column. The above examples assume that the eigenvalue is real number. Let's find the eigenvector, v 1, associated with the eigenvalue, λ 1 =-1, first. eigen() function in R Language is used to calculate eigenvalues and eigenvectors of a matrix. Example: Find Eigenvalues and Eigenvectors of a 2x2 Matrix. so clearly from the top row of … If an example would help, I've worked out that the matrix-1,1,3. The 3x3 matrix can be thought of as an operator - it takes a vector, operates on it, and returns a new vector. The matrix looks like this... |0 1 1| A= |1 0 1| |1 1 0| When I try to solve for the eigenvectors I end up with a 3x3 matrix containing all 1's and I get stumped there. Anonymous. Write down the associated linear system 2. Eigenvectors are the solution of the system $( M − \lambda I_n ) \vec{X} = \vec{0}$ with $I_n$ the identity matrix.. kerr_lee. Once we have the eigenvalues for a matrix we also show how to find the corresponding eigenvalues for the matrix. The matrix is (I have a ; since I can't have a space between each column. and the two eigenvalues are . How to find the eigenspace of a 3x3 matrix - Suppose A is this 2x2 matrix: [1 2] [0 3]. Example 4: 3xx3 case. To find all of a matrix's eigenvectors, you need solve this equation once for each individual eigenvalue. The 3x3 matrix can be thought of as an operator - it takes a vector, operates on it, and returns a new vector. 1,2,0. share | cite | improve this question | follow | edited Jan 26 '15 at 0:09. abel. Input the components of a square matrix separating the numbers with spaces. The nullspace is projected to zero. and the two eigenvalues are . It will find the eigenvalues of that matrix, and also outputs the corresponding eigenvectors.. For background on these concepts, see 7.Eigenvalues and Eigenvectors The PCA is applied on this symmetric matrix, so the eigenvectors are guaranteed to be orthogonal. 5 years ago. edit close. Since the left-hand side is a 3x3 determinant, we have x��\�ݶ����(��J��5�:���=bo�A?4�>�f�u������P���u4F������!�ov����g�qus!v��ߗo.|������������7O�N�Vi��2��;)}�o��]�\|[=��ziT_բu�O��Z���M�=��֖�?��N�ZU_ր�x>_�S ��i��j ɇ��au��O�F�V(�oj� In order to find the associated eigenvectors, we do the following steps: 1. The eigenvectors for D 1 (which means Px D x/ fill up the column space. then the characteristic equation is . It is also known as characteristic vector. How To: Find the equation of trig functions by their graphs How To: Do matrix algebra on a TI-83 calculator How To: Solve systems of linear equations with matrices How To: Find eigenvectors and eigenspaces of a 2x2 matrix How To: Use a change of basis matrix in linear algebra The matrix is (I have a ; since I can't have a space between each column. λ 1 =-1, λ 2 =-2. If . Find the. Find a basis of the eigenspace E2 corresponding to the eigenvalue 2. EXAMPLE 1: Find the eigenvalues and eigenvectors of the matrix A = 1 −3 3 3 −5 3 6 −6 4 . As a consequence, if all the eigenvalues of a matrix are distinct, then their corresponding eigenvectors span the space of column vectors to which the columns of the matrix belong. Find Eigenvalues and Eigenvectors of a Matrix in R Programming – eigen() Function Last Updated: 19-06-2020. eigen() function in R Language is used to calculate eigenvalues and eigenvectors of a matrix. SOLUTION: • In such problems, we first find the eigenvalues of the matrix. Find the. The column space projects onto itself. SOLUTION: • In such problems, we first find the eigenvalues of the matrix. I am trying to find the best OOBB hitboxes for my meshes using PCA. Linear independence of eigenvectors. So one may wonder whether any eigenvalue is always real. Example: The Hermitian matrix below represents S x +S y +S z for a spin 1/2 system. More: Diagonal matrix Jordan decomposition Matrix exponential. Clean Cells or Share Insert in. When I try to solve for the eigenvectors I end up with a 3x3 matrix containing all 1's and I get stumped there. Beware, however, that row-reducing to row-echelon form and obtaining a triangular matrix does not give you the eigenvalues, as row-reduction changes the eigenvalues of the matrix … Eigenvalue $\boldsymbol{\lambda = 7}$, Real eigenvalues and eigenvectors of 3x3 matrices, example 1, Real eigenvalues and eigenvectors of 3x3 matrices, example 2, Finding the normal force in planar non-uniform…, Simple problems on relativistic energy and momentum, Proof that the square root of 2 is irrational, Deriving the volume of the inside of a sphere using…, Real eigenvalues and eigenvectors of 3×3 matrices, example 2, Deriving the Lorentz transformations from a rotation of frames of reference about their origin with real time Wick-rotated to imaginary time, https://opencurve.info/real-eigenvalues-and-eigenvectors-of-3x3-matrices-example-3/. A = To do this, we find the values of ? Suppose A is this 3x3 matrix: [1 1 0] [0 2 0] [0 –1 4]. [V,D] = eig(A) returns matrices V and D.The columns of V present eigenvectors of A.The diagonal matrix D contains eigenvalues. Now let us put in an identity matrix so we are dealing with matrix-vs-matrix:. How do you find the characteristic equation of a 3×3 matrix? This calculator allows you to enter any square matrix from 2x2, 3x3, 4x4 all the way up to 9x9 size. ���Ⱥ�v�'U. Real eigenvalues and eigenvectors of 3x3 matrices, example 2; Finding the normal force in planar non-uniform… Simple problems on relativistic energy and momentum; Proof that the square root of 2 is irrational; Deriving the volume of the inside of a sphere using… 2018-12-14 2020-09-24 eigenvalues, eigenvectors, linear algebra, matrix Post navigation. Find more Mathematics widgets in Wolfram|Alpha. All that's left is to find the two eigenvectors. Eigenvalues and Eigenvectors of a Matrix Description Calculate the eigenvalues and corresponding eigenvectors of a matrix. Eigenvalue and Eigenvector for a 3x3 Matrix Added Mar 16, 2015 by Algebra_Refresher in Mathematics Use this tool to easily calculate the eigenvalues and eigenvectors of 3x3 matrices. Display decimals, number of significant digits: … Relevance. FINDING EIGENVALUES AND EIGENVECTORS EXAMPLE 1: Find the eigenvalues and eigenvectors of the matrix A = 1 −3 3 3 −5 3 6 −6 4 . The ideal is to express a given vector as a linear combination of eigenvectors. asked Jan 25 '15 at 23:57. user3435407 user3435407. How to find eigenvalues quick and easy – Linear algebra explained . Eigen vector, Eigen value 3x3 Matrix Calculator. In these examples, the eigenvalues of matrices will turn out to be real values. The algebraic multiplicity of an eigenvalue is the number of times it appears as a root of the characteristic polynomial (i.e., the polynomial whose roots are the eigenvalues of a matrix). Any help is greatly appreciated. I'm writing an algorithm with a lot of steps (PCA), and two of them are finding eigenvalues and eigenvectors of a given matrix. then the characteristic equation is . In order to do this, I need the eigenvectors but I am kind of lost how to compute them without using a huge library. Rewrite the unknown vector X as a linear combination of known vectors. Check the determinant of the matrix. It will find the eigenvalues of that matrix, and also outputs the corresponding eigenvectors.. For background on these concepts, see 7.Eigenvalues and Eigenvectors by Marco Taboga, PhD. Eigenvalues and Eigenvectors Consider multiplying a square 3x3 matrix by a 3x1 (column) vector. If the resulting V has the same size as A, the matrix A has a full set of linearly independent eigenvectors that satisfy A*V = V*D. FINDING EIGENVALUES • To do this, we find the values of λ which satisfy the characteristic equation of the matrix A, namely those values of λ for which det(A −λI) = 0, stream That example demonstrates a very important concept in engineering and science - eigenvalues and eigenvectors- which is used widely in many applications, including calculus, search engines, population studies, aeronautics … EXAMPLE 1: Find the eigenvalues and eigenvectors of the matrix A = 1 −3 3 3 −5 3 6 −6 4 . which satisfy the characteristic equation of the. Any help is greatly appreciated. Since the zero-vector is a solution, the system is consistent. In linear algebra, the Eigenvector does not change its direction under the associated linear transformation. Eigenvalues and Eigenvectors of a Matrix Description Calculate the eigenvalues and corresponding eigenvectors of a matrix. 3xx3 matrices and their eigenvalues and eigenvectors. The eigenvectors for D 0 (which means Px D 0x/ fill up the nullspace. Finding of eigenvalues and eigenvectors. 2 Answers. To find all of a matrix's eigenvectors, you need solve this equation once for each individual eigenvalue. We compute a power of a matrix if its eigenvalues and eigenvectors are given. Favorite Answer. In linear algebra, the Eigenvector does not change its direction under the associated linear transformation. In general, for any matrix, the eigenvectors are NOT always orthogonal. <> The result is a 3x1 (column) vector. FINDING EIGENVALUES • To do this, we find the values of … The eigenvalues are immediately found, and finding eigenvectors for these matrices then becomes much easier. Notice how we multiply a matrix by a vector and get the same result as when we multiply a scalar (just a number) by that vector.. How do we find these eigen things?. How do you find the eigenvectors of a matrix? Find 2 linearly independent Eigenvectors for the Eigenvalue 0 c.) The e-value 0 has both geometric and algebraic multiplicity 2. If $\theta \neq 0, \pi$, then the eigenvectors corresponding to the eigenvalue $\cos \theta +i\sin \theta$ are In this section we will introduce the concept of eigenvalues and eigenvectors of a matrix. Example: Find Eigenvalues and Eigenvectors of a 2x2 Matrix. Visit http://ilectureonline.com for more math and science lectures!In this video I will find eigenvector=? Solve the system. The eigenvector v of an operator A is a vector such that: A = To do this, we find the values of ? What is the trace of a matrix? We start by finding the eigenvalue: we know this equation must be true:. This is a linear system for which the matrix coefficient is . If you love it, our example of the solution to eigenvalues and eigenvectors of 3×3 matrix will help you get a better understanding of it. To find eigenvectors, take $M$ a square matrix of size $n$ and $\lambda_i$ its eigenvalues. matrices eigenvalues-eigenvectors. The generalized eigenvalue problem is to determine the solution to the equation Av = λBv, where A and B are n-by-n matrices, v is a column vector of length n, and λ is a scalar. Ʋ�ψ�o��|�ߛ�z?cI���4��^?��R9���(/k����k The projection keeps the column space and destroys the nullspace: Relevance. 1 decade ago. Set the characteristic determinant equal to zero and solve the quadratic. Find the eigenvalues and eigenvectors. Yes, finding the eigenvectors should be straightforward. Remark. The result is a 3x1 (column) vector. Notice, however, that you have x=1 as a double root. The matrix A has an eigenvalue 2. If the determinant is 0, the matrix has no inverse. This pages describes in detail how to diagonalize a 3x3 matrix througe an example. Find the eigenvalues and bases for each eigenspace. For example, say you need to solve the following equation: First, you can rewrite this equation as the following: I represents the identity matrix, with 1s along its diagonal and 0s otherwise: Remember that the solution to […] Free Matrix Eigenvectors calculator - calculate matrix eigenvectors step-by-step This website uses cookies to ensure you get the best experience. ��~�?.����(x�$ׄ��;�oE|Ik�����$P���?�Iha��֦�BB')���q�����d�z��I;E���k��y� �@���9P}����T���3�T�2q�w8�{�T�*�N�mk�ǟJBZ�em���58j��k������~���-lQ9i�[$aT$A�_�1#sv;q吺��zz{5��iB�nq��()���6�au� ���)��F�ܐQXk�jhi8[=���n�B�F��$.�CFZН.�PҷD����GօKZ����v��v��ʀ~��|rq�ٷ����3B�f��ٲ��l More: Diagonal matrix Jordan decomposition Matrix exponential. EXAMPLE 1: Find the eigenvalues and eigenvectors of the matrix. By using this website, you agree to our Cookie Policy. Eigenvalues and Eigenvectors of a 3 by 3 matrix Just as 2 by 2 matrices can represent transformations of the plane, 3 by 3 matrices can represent transformations of 3D space. View all posts by KJ Runia, 4. The generalized eigenvalue problem is to determine the solution to the equation Av = λBv, where A and B are n-by-n matrices, v is a column vector of length n, and λ is a scalar. If$ \mathbf{I} $is the identity matrix of$ \mathbf{A} $and$ \lambda $is the unknown eigenvalue (represent the unknown eigenvalues), then the characteristic equation is \begin{equation*} \det(\mathbf{A}-\lambda \mathbf{I})=0. Find the characteristic polynomial of a matrix – What is the fastest way to find eigenvalues? On the previous page, Eigenvalues and eigenvectors - physical meaning and geometric interpretation appletwe saw the example of an elastic membrane being stretched, and how this was represented by a matrix multiplication, and in special cases equivalently by a scalar multiplication. There may be two independent eigenvectors corresponding to that. If you need a refresher, check out my other lesson on how to find the determinant of a 2×2.Suppose we are given a square matrix A where, Eigenvalue is the factor by which a eigenvector is scaled. $$\tag{1}$$ , which is a polynomial equation in the variable$\lambda$. I tried to find the inverse of the eigenvectors, but it brought a wrong matrix. Without having to make extensive calculations explain why 0 is an eigenvalue of A b.) I implemented an algorithm that computes three eigenvalues given a 3x3 Matrix. In quantum physics, if you’re given an operator in matrix form, you can find its eigenvectors and eigenvalues. [V,D,W] = eig(A,B) also returns full matrix W whose columns are the corresponding left eigenvectors, so that W'*A = D*W'*B. How do you find the eigenvectors of a 3x3 matrix? so clearly from the top row of the equations we get. Enter a matrix. Do you know how to solve it? [V,D,W] = eig(A,B) also returns full matrix W whose columns are the corresponding left eigenvectors, so that W'*A = D*W'*B. Calculate eigenvalues and eigenvectors. Some of my solutions do not match answers in my differential equations text (Advanced Engineering Mathematics by Erwin Kreyszig, 1988, John Wiley & Sons). This calculator allows you to enter any square matrix from 2x2, 3x3, 4x4 all the way up to 9x9 size. SOLUTION: • In such problems, we first find the eigenvalues of the matrix. Syntax: eigen(x) Parameters: x: Matrix Example 1: filter_none. by Marco Taboga, PhD. Av = λIv. For Example, if x is a vector that is not zero, then it is an eigenvector of a square matrix … When I try to solve for the eigenvectors I end up with a 3x3 matrix containing all 1's and I get stumped there. Eigenvalue is the factor by which a eigenvector is scaled. Favorite Answer. Eigenvalues and eigenvectors calculator. Finding of eigenvalues and eigenvectors. Matrix A: Find. Note that if we took the second row we would get . Eigenvalues and eigenvectors calculator. /�7P=š� EXAMPLE 1: Find the eigenvalues and eigenvectors of the matrix. Eigenvectors for: Now we must solve the following equation: First let’s reduce the matrix: This reduces to the equation: There are two kinds of students: those who love math and those who hate it. Syntax: eigen(x) Parameters: x: Matrix … Source(s): eigenvectors 3x3 matric: https://tinyurl.im/fNPuM. All that's left is to find the two eigenvectors. The code for this originally is … Please check my work in finding an eigenbasis (eigenvectors) for the following problem. The process for finding the eigenvalues and eigenvectors of a 3xx3 matrix is similar to that for the 2xx2` case. λ 1 =-1, λ 2 =-2. https://www.khanacademy.org/.../v/linear-algebra-eigenvalues-of-a-3x3-matrix The determinant of matrix M can be represented symbolically as det(M). The Formula of the Determinant of 3×3 Matrix. If the determinant is 0, then your work is finished, because the matrix has no inverse. I do not wish to write the whole code for it because I know it is a long job, so I searched for some adhoc code for that but just found 1 or 2 libraries and at first I prefer not to include libraries and I don't want to move to matlab. In summary, when$\theta=0, \pi$, the eigenvalues are$1, -1$, respectively, and every nonzero vector of$\R^2$is an eigenvector. I'm having a problem finding the eigenvectors of a 3x3 matrix with given eigenvalues. Calculate the eigenvalues and the corresponding eigenvectors of the matrix. This calculator allows to find eigenvalues and eigenvectors using the Characteristic polynomial. Display decimals, number of significant digits: Clean. In other words, the eigenvalues and eigenvectors are in$\mathbb{R}^n$. Eigenvalues and Eigenvectors Consider multiplying a square 3x3 matrix by a 3x1 (column) vector. �������lMOK���� ��� n��h vx{Vb�HL����%f;bz\5� This calculator allows to find eigenvalues and eigenvectors using the Characteristic polynomial. Find the eigenvalues and bases for each eigenspace. Eigenvalue$ \boldsymbol{\lambda = 6} \$, 4.3. FINDING EIGENVALUES • To do this, we find the values of λ which satisfy the characteristic equation of the matrix A, namely those values of λ for which det(A −λI) = 0, where I is the 3×3 identity matrix. On this site one can calculate the Characteristic Polynomial, the Eigenvalues, and the Eigenvectors for a given matrix. I implemented an algorithm that computes three eigenvalues given a 3x3 Matrix. →Below is a calculator to determine matrices for given Eigensystems. Thanks! Answer Save. The determinant of a triangular matrix is easy to find - it is simply the product of the diagonal elements. I am trying to find the eigenvectors for the following 3x3 matrix and are having trouble with it. 27.7k 1 1 gold badge 25 25 silver badges 52 52 bronze badges. You need to calculate the determinant of the matrix as an initial step. The values of λ that satisfy the equation are the generalized eigenvalues. Av = λv. Illustrate the process of finding eigenvalues and corresponding eigenvectors of a 3x3 matrix. Notice, however, that you have x=1 as a double root. To find the inverse of a 3x3 matrix, first calculate the determinant of the matrix. Get the free "Eigenvalue and Eigenvector for a 3x3 Matrix " widget for your website, blog, Wordpress, Blogger, or iGoogle. The only eigenvalues of a projection matrix are 0 and 1. I have to find 4 things for the Matrix A which is a 3x3 matrix with all values equal to 1 A= 1 1 1 1 1 1 1 1 1 a.) In this page, we will basically discuss how to find the solutions. How to find the eigenspace of a 3x3 matrix - Suppose A is this 2x2 matrix: [1 2] [0 3]. which satisfy the characteristic equation of the.
## how to find eigenvectors of a 3x3 matrix
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http://bibliotekacmolas.pl/3bhqq/4s7hgw.php?page=6cdafa-simple-cryptography-examples | # simple cryptography examples
In the examples above, statement (A) is plaintext, while statement (B) is a reverse cipher text. Thus, cryptography is defined as the art of manipulating or scrambling plaintext into ciphertext. An example of primitive cryptography is an encrypted message in which it takes over from letters with other characters. I will discuss a simple method of enciphering and deciphering a message using matrix transformations and modular arithmetic, and show how elementary row operations can sometimes be used to break an opponent's code. Cryptography is the technique of protecting information by transforming it into a secure format. Introduction. Simple substitution cipher is the most commonly used cipher and includes an algorithm of substituting every plain text character for every cipher text character. Select primes p=11, q=3. Cryptography is the science of keeping information secret and safe by transforming it into form that unintended recipients cannot understand. Solved Examples 1) A very simple example of RSA encryption This is an extremely simple example using numbers you can work out on a pocket calculator (those of you over the age of 35 45 can probably even do it by hand). Final Example: RSA From Scratch This is the part that everyone has been waiting for: an example of RSA from the ground up. Hence the modulus is $$n = p \times q = 143$$. 2. n = pq ⦠In this process, alphabets are jumbled in comparison with Caesar cipher algorithm. I am first going to give an academic example, and then a real world example. A simple example of an encryption algorithm would be changing all Ns to a 3, or all Zs to a 1. Classic Encryption - The Caesar Cipher. Background: Many of the ideas we use to keep secrets in the digital age are far older than the Internet. Cryptography, or cryptology (from Ancient Greek: κÏÏ
ÏÏÏÏ, romanized: kryptós "hidden, secret"; and γÏάÏειν graphein, "to write", or -λογία-logia, "study", respectively), is the practice and study of techniques for secure communication in the presence of third parties called adversaries. Cryptography originated approximately 4000 years ago in Egypt. Keys for a simple substitution cipher usually consists of 26 letters. The word cryptography comes from the word: KRYPTOS and GRAPHEIN. The process of encoding a plain text message in some secret way is called Encryption. Simple ciphers Simple encryption algorithms, which were invented long before first computers, are based on substitution and transposition of single plaintext characters. The following are common examples of encryption. Encryption is the conversion of information into an cryptographic encoding that can't be read without a key.Encrypted data looks meaningless and is extremely difficult for unauthorized parties to decrypt without the correct key. 1. 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https://math.stackexchange.com/questions/2327748/probability-or-rolling-1-1-2-with-six-dice | # probability or rolling 1 1 2 with six dice
What is the probability of rolling at least two ones and at least one two (order not important) with six 6-sided die?
Equivalently, to win a game, I need two 1s and a 2, (1,1,2). What is the probability of my rolling at least this with 6 dice.
There are $6^6=46656$ possible rolls. I believe that there are 6896 'winning' rolls among these as I wrote some code to check every roll. This makes the probability around 0.1478
I'm not sure how to get this without the help of a computer though.
• What have you tried? Also, since you have six dice, is unclear whether, for instance, $(1,1,2,3,1,2)$ is valid. Is it valid? – Arthur Jun 18 '17 at 20:29
• yes, 1,1,2,x,x,x where x could be any value – mjmdavis Jun 18 '17 at 20:38
• Common practice is to say at least two 1's and one 2 in your case and exactly " ... " otherwise. – infinitylord Jun 18 '17 at 20:44
• @infinitylord like that? – mjmdavis Jun 18 '17 at 20:48
• If you want positive responses, I recommend stating the question as: "What is the probability of rolling at least two ones and at least one two (order not important) with six 6-sided die" and then (and this part is really huge for this site) show your attempt at the problem. – infinitylord Jun 18 '17 at 21:01
Approach via inclusion-exclusion and De Morgan's laws.
Let $A$ be the event that you roll at least two $1$'s. Let $B$ be the event that you roll at least one $2$.
We are attempting to calculate then $Pr(A\cap B)$
$Pr(A\cap B)=1-Pr((A\cap B)^c)=1-Pr(A^c\cup B^c) = 1-Pr(A^c)-Pr(B^c)+Pr(A^c\cap B^c)$
We calculate each term on the right now that they are in simpler form.
$Pr(A^c)$ refers to the probability of strictly fewer than two $1$'s occurring, i.e. at most one $1$ occurring. Either exactly one $1$ occurs or no $1$'s occur. In the case of exactly one $1$, pick its location and then pick each remaining digit for a total of $6\cdot 5^5$ possibilities. In the case of no $1$'s, pick each digit for a total of $5^6$ possibilities. Taking the ratio of this compared to the $6^6$ equally likely dice rolls, we calculate $Pr(A^c)=\frac{6\cdot 5^5+5^6}{6^6}$.
$Pr(B^c)$ refers to the probability of strictly fewer than one $2$ occurring, i.e. no $2$'s. This occurs with probability $Pr(B^c)=\frac{5^6}{6^6}$
$Pr(A^c\cap B^c)$ refers now to the probability of at most one $1$ and no $2$'s, which similarly to before we break into cases for either exactly one $1$ or no $1$'s. For exactly one $1$, first pick the location, then pick each remaining digit for $6\cdot 4^5$ possibilities and for no $1$'s pick each digit for $4^6$ possibilities for a probability of $Pr(A^c\cap B^c)=\frac{6\cdot 4^5+4^6}{6^6}$
This gives a final probability of:
$$Pr(A\cap B) = 1-\frac{6\cdot 5^5+5^6}{6^6}-\frac{5^6}{6^6}+\frac{6\cdot 4^5+4^6}{6^6}$$
$$=1-\frac{39760}{46656} = \frac{431}{2916}\approx 0.147805$$
You can consider the possibility to find probability directly without inclusion-exclusion principle. This is a longer way, but there are those who like it. Consider all possibilities to roll at least two ones and at least one two with six 6-sided die. We can have exactly two $1$ and one $2$. Denote this event by $A_{21}$. Or we can have exactly two $1$ and two $2$: event $A_{22}$. There are $10$ disjoint events total: $A_{21}$, $A_{22}$, $A_{23}$, $A_{24}$, $A_{31}$, $A_{32}$, $A_{33}$, $A_{41}$, $A_{42}$, $A_{51}$.
Find and then add their probabilities. The probability of $A_{21}$ is $$\mathbb P(A_{21})=\frac{\binom{6}{2}\binom{6-2}{1}4^{6-2-1}}{6^6}=\frac{15\cdot 4\cdot 4^3}{6^6}.$$ By first multiplier $\binom{6}{2}=15$ we choose the rolls which result in $1$, next multiplier $\binom{6-2}{1}=4$ we choose the roll among the rest which results in $2$, and the last multiplier $4^3$ is number of possible outcomes in the rest $6-2-1=3$ rolls.
Similar, $$\mathbb P(A_{22})=\frac{\binom{6}{2}\binom{4}{2}4^{2}}{6^6}=\frac{15\cdot 6\cdot 16}{6^6},$$ $$\mathbb P(A_{23})=\frac{15\cdot 4\cdot 4}{6^6},\quad \mathbb P(A_{24})=\frac{15}{6^6},\quad \mathbb P(A_{31})=\frac{20\cdot 3\cdot 16}{6^6},$$ $$\mathbb P(A_{32})=\frac{20\cdot 3\cdot 4}{6^6}, \quad \mathbb P(A_{33})=\frac{20}{6^6}, \quad \mathbb P(A_{41})=\frac{15\cdot 2\cdot 4}{6^6},$$ $$\mathbb P(A_{42})=\frac{15}{6^6}, \quad \mathbb P(A_{51})=\frac{6}{6^6}.$$ The total probability of rolling at least two ones and at least one two with six 6-sided die is equal to the sum of probabilities above: $$\frac{6896}{6^6}=0,147805213.$$
As I understood your question, you are asking about probability that between $6$ values will be at least two $1$'s and one $2$ (don't looking at other values and order).
Well, so the probability of $(1,1,2,x,x,x)$ (excluding order and remarking that there may be other $1$'s and $2$'s between $x$'s) is exactly the number of sets of six digits from $1$ to $6$ that start with $1,1,2$ divided by total number of sets of six digits from $1$ to $6$. There are as many (disordered) sets of six digits from $1$ to $6$ that start with $1,1,2$ as just (disordered) sets of three digits from $1$ to $6$.
Thus the answer is $P = \frac{(3^3)/3!}{(6^6)/6!} = \frac{5}{72}$.
• Hasek I'm not sure if this result is correct. – mjmdavis Jun 18 '17 at 23:07
• The line "There are as many (disordered) sets of six digits from 1 to 6 that start with 1,1,2 as just (disordered) sets of three digits from 1 to 6." is incorrect. After all, what does it mean for a "disordered set to start with something"? Further, your calculations don't appear to make much sense. What would $3^3/3!$, your supposed numerator, represent? Notice that $3!$ is even but $3^3$ is odd and so $3^3/3!$ is not an integer, so it could not possibly represent the answer to a counting problem. Similarly for the denominator. – JMoravitz Jun 18 '17 at 23:28
• Yes, I'm definitely wrong, sorry. – Hasek Jun 19 '17 at 6:44 | 2019-02-19T06:57:40 | {
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http://clandiw.it/umtr/inverse-fourier-transform-of-unit-step-function.html | In reality there is no such thing as a unit step, and the Fourier transform is not 'real'. Instead, the most common procedure to find the inverse Laplace transform of an expression is a two-step approach (Appendix 12. The Fourier transform we’ll be int erested in signals defined for all t the Four ier transform of a signal f is the function F (ω)= ∞ −∞ f (t) e − jωt dt • F is a function of a real variable ω;thef unction value F (ω) is (in general) a complex number F (ω)= ∞ −∞ f (t)cos ωtdt − j ∞ −∞ f (t)sin ωtdt •| F (ω) | is called the amplitude spectrum of f; F (ω) is the phase spectrum of f • notation: F = F (f) means F is the Fourier transform of f. Remark 6 The definition of the Fourier transform on implies that whenever , we have that. If any argument is an array, then fourier acts element-wise on all elements of the array. $\begingroup$ The plus one simply shifts when the Heaviside function turns on by one unit to the left like in normal functional translation, it helps to use the definition of the Heaviside step function as it restricts your domain of integration. How about going back? Recall our formula for the Fourier Series of f(t) : Now transform the sums to integrals from –∞to ∞, and again replace F m with F(ω). 2 Fourier Series Consider a periodic function f = f (x),defined on the interval −1 2 L ≤ x ≤ 1 2 L and having f (x + L)= f (x)for all. Because the original function and its inverse Laplace transform are only valid for t‚ 0, some people introduce a Heaviside step function H ( t ) (see Section B. Fourier Transform Pairs (contd). 4M subscribers. Laplace transform of the unit step function | Laplace transform | Khan Academy - YouTube. • Sometimes we want to use one-dimensional Fourier transforms or inverse transforms. Explain briefly below. The Fourier transform is not limited to functions of time, but the domain of the original function is commonly referred to as the time domain. Unlike the inverse Fourier transform, the inverse Laplace transform in Eq. Step 7: Check the “Inverse” box only if you have results from a prior analysis and you want to find the original function. The book is divided into four major parts: periodic functions and Fourier series, non-periodic functions and the Fourier integral, switched-on signals and the Laplace transform, and finally the discrete versions of these transforms, in particular the Dis-crete Fourier Transform together with its fast implementation, and the z-transform. Unit analysis, algebra solver free step by step, how fast can one learn algebra, mathsiequalities, online direction field. Solution: Here, =0 for <2 , then ˝ =1 for ≥2. Fourier Transform — Theoretical Physics Reference 0. Before proceeding into solving differential equations we should take a look at one more function. It cannot be said that time information is lost because it is possible to recover the original time domain observation using the Inverse Fourier. The calculator will find the Inverse Laplace Transform of the given function. The usual Fourier transform tables found online don't have many functional relationship rules. we can get the Fourier transform of a unit impulse as the time derivative of a unit step function: Alternatively, by definition, the forward Fourier transform of an impulse function is and the inverse transform is. The Fourier Transform (used in signal processing) The Laplace Transform (used in linear control systems) The Fourier Transform is a particular case of the Laplace Transform, so the properties of Laplace transforms are inherited by Fourier transforms. Interestingly, these transformations are very similar. Together with a great variety, the subject also has a great coherence, and the hope is students come to appreciate both. You take the Fourier transform fft of f. Suppose that the Fourier transform of f and its inverse exist. This is specifically due to its property that it is neither absolutely summable nor square summable. dft() and cv2. How can you create a delta function using some other function, the Fourier transform of which you already know. Fourier transform pair The function ! X(j") is the Fourier transform of the signal x(t) and conversely x(t) is the inverse Fourier transform of! X(j"). If any argument is an array, then fourier acts element-wise on all elements of the array. 2 Fourier Series Consider a periodic function f = f (x),defined on the interval −1 2 L ≤ x ≤ 1 2 L and having f (x + L)= f (x)for all. Mathematicians have developed tables of commonly used Laplace transforms. PLOTTING STEP RESPONSE OF TRANSFER FUNCTION Learn more about fourier transform. The notation is introduced in Trott (2004, p. Discrete-Time Fourier Transform (DTFT) inverse DTFT. text orientation finding) where the Fourier Transform is used to gain information about the geometric structure of the. We can solve the integral by contour integration. Denoted , it is a linear operator of a function f(t) with a real argument t (t ≥ 0) that transforms it to a function F(s) with a complex argument s. The book only states a limited form of the Heavyside expansion theorem in problem 5 of section 53. The Z transform of the discrete time unit ramp function 42. (iii) Comment the time domain expression of the filter. 1 Dirac Delta Function 1 2 Fourier Transform 5 3 Laplace Transform 11 3. Visualizing Pole-Zero plot: Since the z-transform is a function of a complex variable, it is convenient to describe and interpret it using the complex z-plane. 12 tri is the triangular function 13 Dual of rule 12. One common example is when a voltage is switched on or off in an electrical circuit at a specified value of time t. calculating the Fourier transform of a signal, then exactly the same procedure with only minor modification can be used to implement the inverse Fourier transform. As such, the restriction of the Fourier transform of an L 2 (R n) function cannot be defined on sets of. Consider the Fourier transforms of the functions in Example 9. 1) into the integral in the deflnition of the inverse transform in (F. The special characteristics of the Fast Fourier Transform implementation are described. Inverse Transform 6. In this section we introduce the step or Heaviside function. is the Fourier Transform of f(t). Fourier transforms take the process a step further, to a continuum of n-values. For instance, the inverse continuous Fourier transform of both sides of Eq. Correlation, autocorrelation. For the Laplace transform, the Fourier transform existed if the ROC included the j!axis. I have to find the inverse fourier transform for: \frac{e^{i 6\omega}}{\omega} So I'm using a table, then. In this section we introduce the Fourier transform and then we illustrate the fast Fourier transform algorithm, applied to the projection of unit-step i. Continuous Fourier Transform A general Fourier Transform for Spectrum Representation •With the unit-impulse function incorporated, the continuous Fourier transform can represent a broad range of continuous-time signals. Basic Fourier Transform Theory: Relationship to Chap 3 on Fourier Series Interpretation of Inverse Fourier Transform Frequency Ranges of Biological, E&M, and other Signals. In reality there is no such thing as a unit step, and the Fourier transform is not 'real'. It is "off" (0) when < , the "on" (1) when ≥. For math, science, nutrition, history. The Fourier Transform and its Inverse The Fourier Transform and its Inverse: So we can transform to the frequency domain and back. 3 Complex form of Fourier series, Fourier integral representation, Fourier Transform and Inverse Fourier transform of constant and exponential function. 1) into the integral in the deflnition of the inverse transform in (F. Laplace transform to solve a differential equation. Fourier Transform - Free download as Powerpoint Presentation (. Fourier transform of the unit step function and of the signum function: The signum function sgn( t) is a function that is related to the unit step function. In this study, a new inversion method is presented for performing two-dimensional (2D) Fourier transform. Consider the equation f(x) + A(f(x − 1) + f(x + 1)) = u(x) where u(x) is a known function, absolutely integrable, on R and A is a constant. Evaluate one transform on data from step 3. The Laplace transform is similar to the Fourier transform. 12-2 Circuit Analysis Using the Fourier Transform Determine and plot the spectrum of the response Vo(co) of the circuit of Figure 15. • The unit step function (1 class) • The Dirac delta function (1 class) • Applications of step and impulse functions (1 class) • Periodic functions and their applications (2 classes) • Convolution and applications (2 classes) • Solving integral equations (1 class) • Fourier series (3 classes) • Fourier integral representation (1. Inverse Laplace Transform with unit step function, sect7. 1998 We start in the continuous world; then we get discrete. The Fourier transform is only valid for a periodic function, and a unit step is not periodic. So the Fourier transform X (jw) of x (t) is the convolution of X (jw) and sinc (w). Fast Fourier Transforms Phase factors There are functions that produce roots-of-one as a function of time (t) or place (x). efine the Fourier transform of a step function or a constant signal unit step what is the Fourier transform of f (t)= 0 t< 0 1 t ≥ 0? the Laplace transform is 1 /s, but the imaginary axis is not in the ROC, and therefore the Fourier transform is not 1 /jω in fact, the integral ∞ −∞ f (t) e − jωt dt = ∞ 0 e − jωt dt = ∞ 0 cos. In this section we introduce the step or Heaviside function. A unique 3D graphical approach has been adopted to provide the intuition required to OWN this subject. The unit pulse function can be defined with the help of the Heaviside unit step function ( ) ( ) ( ) 0 x a f t Ht a Ht a 1 x a 0 x a <− = +− −= < > a0 > The Fourier transform of this function can be determined as. of a second over a period of 10 seconds. Uniqueness of Fourier transforms, proof of Theorem 3. 2) The DTFT X. 3 Complex form of Fourier series, Fourier integral representation, Fourier Transform and Inverse Fourier transform of constant and exponential function. 3 Properties of The Continuous -Time Fourier Transform 4. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. So essentially, in the decomposition of x of t as a linear combination of complex exponentials, the complex amplitudes of those are, in effect, the Fourier transform scaled by the differential and scaled by 1 over 2 pi. 6 ) and obtain. There are different definitions of these transforms. Remembering the fact that we introduced a factor of i (and including a factor of 2 that just crops up. UNIT3: FOURIER TRANSFORM - Complex form of Fourier Transform and its inverse, Fourier sine and cosine transform and their inversion. Remembering the fact that we introduced a factor of i (and including a factor of 2 that just crops up. Compute answers using Wolfram's breakthrough technology & knowledgebase, relied on by millions of students & professionals. Inverse Fourier Transform F [f] Step 3) Find the inverse transform. I think you should have to consider the Laplace Transform of f(x) as the Fourier Transform of Gamma(x)f(x)e^(bx), in which Gamma is a step function that delete the negative part of the integral and e^(bx) constitute the real part of the complex exponential. xxxiv), and and are sometimes also used to denote the Fourier transform and inverse Fourier transform, respectively (Krantz 1999, p. Dirac defined the delta function as shown below. If n is less than the length of the signal, then ifft ignores the remaining signal values past the nth entry and. Express f under an integral form. Overlapping in real time fourier transform? Hot Network Questions Numbers by Position a name for a boy, boy's name, boy name Mother milk of 6 Corona-positive (COVID-19) women does not contain the virus - can we make a confidence. How can you create a delta function using some other function, the Fourier transform of which you already know. The convergence criteria of the Fourier transform (namely, that the function be absolutely integrable on the real line) are quite severe due to the lack of the exponential decay term as seen in the Laplace transform, and it means that functions like polynomials, exponentials, and trigonometric functions all do not have Fourier transforms in the. Its value is not trivial to calculate, and ends up being. Any good reference to more detailed tables would be very helpful! My attempt: $\mathcal F[f\times u] = (\mathcal Ff)*(\mathcal Fu)$ where * denotes convolution. The usual Fourier transform tables found online don't have many functional relationship rules. As with the Laplace transform, calculating the Fourier transform of a function can be done directly by using the definition. Magnitude and phase representation of the Fourier transform and frequency response of LTI systems; Applications of the. ej!O /that results from the definition is a function of frequency !O. 2 Contents 2. The more general statement can be found in standard texts devoted to Laplace transforms. Determine the Fourier transform of the non-periodic signals shown in the figures below: (b) 8(1) -2 -1 0 1 2. Introduction to Inverse Problems Guillaume Bal 1 July 2, 2019 1University of Chicago, Chicago, IL 60637; [email protected] Fourier transforms take the process a step further, to a continuum of n-values. 2 The Fourier transform Given a function f(x) de ned for all real x, we can give an alternative representation to it as an integral rather than as an in nite series, as follows f(x) = Z eikxg(k)dk Here g(x) is called the Fourier transform of f(x), and f(x) is the inverse Fourier transform of g(x). 8 Fourier transforms. Second Implicit Derivative (new) Derivative using Definition (new) Derivative Applications. In this section we introduce the step or Heaviside function. The unit step function, also known as the Heaviside function, is defined as such:. 2 Fourier Series Consider a periodic function f = f (x),defined on the interval −1 2 L ≤ x ≤ 1 2 L and having f (x + L)= f (x)for all. The Fourier transform is a particular case of z-transform, i. INTRODUCTION AND FOURIER TRANSFORM OF A DERIVATIVE One can show that, for the Fourier transform g(k) = Z 1 1 f(x)eikx dx (1) to converge as the limits of integration tend to 1 , we must have f(x) ! 0 as. The derivation can be found by selecting the image or the text below. Solved examples of the Laplace transform of a unit step function. We also work a variety of examples showing how to take Laplace transforms and inverse Laplace transforms that involve Heaviside functions. It is clearly desirable that there should be a canonical definition of the Fourier Transform, consistent with classical definitions, which is applicable to all distributions - or, at least to some. The Discrete Fourier Transform the two transforms and then filook upfl the inverse transform to get the convolution. The excel fourier analysis tool. 4142*j]; x_n=ifft(X_K) Example 2: X_K=[10,-2+2*j,-2,-2-2*j]; x_n=. Discrete Time Fourier Transforms The discrete-time Fourier transform or the Fourier transform of a discrete-time sequence x[n] is a representation of the sequence in terms of the complex exponential sequence. fast fourier transform. This transformation is essentially bijective for the majority of practical. !! If you apply the Fourier transform to function f(t), you get a new function F(w). Learn more about Chapter 8: The Fourier Transform on GlobalSpec. Since the transform of a lattice in real space is a reciprocal lattice, the diffraction pattern of the crystal samples the diffraction pattern of a single unit cell at the points of the reciprocal. When you have worked through this unit you should:. 2) factor (1/2π )2 must be replaced by (1/2π ) To avoid confusion, we shall indicate one-dimensional Fourier transforms by Fx, Fx-1 or Fky. We experi-ment here to see if Mathematica knows these functions, and if it can deal with their Fourier transforms. The Fast Fourier Transform for polynomials works in an analogous way to a slide rule. The Fourier Transform Saravanan Vijayakumaran [email protected] 1) which is now called Heaviside step function. The fourier transform uses the assumption that any finite time-domain signal can be broken into an infinite sum of sinusoidal (sine and cosine waves) signals. I don't know where you got G(f), but it only a mathematical expression to "give" the value of the Fourier transform of a unit step. Let tqptqu. 1,791,367 views. G o t a d i f f e r e n t a n s w e r? C h e c k i f i t ′ s c o r r e c t. Properties of the Fourier Series 51. The key step in the proof of (1. The properties are useful in determining the Fourier transform or inverse Fourier transform They help to represent a given signal in term of operations (e. Especially important among these properties is Parseval's Theorem, which states that power computed in either domain equals the power in the other. CT Fourier Transform Pairs signal (function of t) $\longrightarrow$ Fourier transform (function of $\omega$) : 1 CTFT of a unit impulse $\delta (t)\$ $1 \$. The Fourier transform 45. Fourier transform. Inverse Fourier transform – be able to compute this from definition as well as from looking up the transform for elementary signals. This MATLAB function returns the Fourier Transform of f. UNIT STEP FUNCTIONS AND PERIODIC FUNCTIONS 157 Which implies that y(t) = t2 solves the DE. 6#15 - Duration: The intuition behind Fourier and Laplace transforms I was never taught in school inverse laplace transform,. If Y is a matrix, then ifft (Y) returns the inverse transform of each column of the matrix. Q5(a) is given to be: 2 1 Feje() ( 1)ωωjjωω ω = −− Use this information and the time-shifting and time-scaling properties, find the Fourier transforms of the signals. So that gives you a complex spectrum which is here called ff, and then you multiply it by the imaginary unit times k, and then use an inverse transform back to physical space and now you have an exact to machine precision derivative defined on your original grid points. Introduction to Hilbert Transform. In this video tutorial, the tutor covers a range of topics from from basic signals and systems to signal analysis, properties of continuous-time Fourier transforms including Fourier transforms of standard signals, signal transmission through linear systems, relation between convolution and correlation of signals, and sampling theorems and techniques. Ada version of General N Point Fast Fourier Transform. Discrete-Time Fourier Transform (DTFT) inverse DTFT. Solved examples of the Laplace transform of a unit step function. Note that the usual results for Fourier transforms of even and odd functions still hold. IQ v(t) vo(t) FIGURE 15. The Fourier transform of controlled-source time-domain electromagnetic data by smooth spectrum inversion Yuji Mitsuhata. Problems at x!+1are removed by multiplying by e cx, where cis a positive real number. The Fourier transform we’ll be int erested in signals defined for all t the Four ier transform of a signal f is the function F (ω)= ∞ −∞ f (t) e − jωt dt • F is a function of a real variable ω;thef unction value F (ω) is (in general) a complex number F (ω)= ∞ −∞ f (t)cos ωtdt − j ∞ −∞ f (t)sin ωtdt •| F (ω) | is called the amplitude spectrum of f; F (ω) is the phase spectrum of f • notation: F = F (f) means F is the Fourier transform of f. Thus the Fourier transform on tempered distributions is an extension of the classical definition of the Fourier transform. Instead, the most common procedure to find the inverse Laplace transform of an expression is a two-step approach (Appendix 12. he Fourier and Laplace transforms can be rectangular pulse: f (t)= 1 eп¬Ѓne the Fourier transform of a step function or a constant signal, The aim of this post is to properly understand Numerical Fourier Transform on Python or Matlab with an example in fourier transform of the. study how a piecewise continuous function can be constructed using step functions. As in the FDK analysis, the s direction transforms require a total of 2 NMP 2 log(2 P ) operations for the forward direction and 2 NP 2 log(2 P ) for the inverse direction. 8 Filters 2. Inverse Z Transform: Part 2. of a unit step can be inferred, but it's natural with the Laplace. The Fourier transform of ft) (ft)-sinc(t)) is F(jo)-nRect(/2) (Figure ) (1) For a linear, time invariant system, its impulse response is h(t)…. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. The step response is the convolution. Fourier Transform of Unit Step Function Guess Fourier Transform of Unit Step Function F ( ) 0 |F(j )| 0 t 1 f(t) Fourier Transforms of Special Functions Fourier Transform vs. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. The usual Fourier transform tables found online don't have many functional relationship rules. Have these ideas in mind as we go through the examples in the rest of this section. This chapter introduces the Fourier Transform, also known as the Fourier Integral. Linearity and the result for the unit step, above. Fourier transform of unit step signal u(t). MAXIMA Quick Reference Labels. The Fourier transform of the unit step function is not any of those things. It's basically a set of Sine waves with amplitudes and phases. Laplace transform to solve a differential equation. Once the transformation has been applied, time information is hidden and cannot be easily observed. 10 Band-Pass Systems. Together with a great variety, the subject also has a great coherence, and the hope is students come to appreciate both. Unit analysis, algebra solver free step by step, how fast can one learn algebra, mathsiequalities, online direction field. The Fourier transform of an integrable function is continuous and the restriction of this function to any set is defined. Try to integrate them? Cite. That is, given the Fourier transform of an function, when can we recover the original function from ? We begin with a simple case where the recovery is quite easy. Table of Fourier Transform Pairs of Energy Signals Function name Time Domain x(t) Frequency Domain X Unit step () 10 00 t ut 2. Visualizing Pole-Zero plot: Since the z-transform is a function of a complex variable, it is convenient to describe and interpret it using the complex z-plane. Introduction to Fourier Transforms Fourier transform as a limit of the Fourier series Inverse Fourier transform: The Fourier integral theorem Example: the rect and sinc functions Cosine and Sine Transforms Symmetry properties Periodic signals and functions Cu (Lecture 7) ELE 301: Signals and Systems Fall 2011-12 2 / 22. The special characteristics of the Fast Fourier Transform implementation are described. From this block diagram we can find overall transfer function which is nonlinear in nature. Example 2-2 SPECTRUM OF AN EXPONENTILA PULSE By means of direct integration find the Fourier transform of ) ( t w < = - 0 , 0 0 , ) ( t t e t w t Properties of Fourier Transforms. These reviews did not try 44 to minimize Laplace-space function evaluations, since their functions were simple closed-form expressions, 45 not simulations. Time Reversal and Frequency Response By Clay S. Function, f(t) Fourier Transform, F( ) Definition of Inverse Fourier Transform f t F( )ej td 2 1 ( ) Definition of Fourier Transform F() f (t)e j tdt Trigonometric Fourier Series 1 ( ) 0 cos( 0 ) sin( 0) n f t a an nt bn nt where T n T T n f t nt dt T b f t nt dt T f t dt a T a 0 0 0 0 0 0 ( )sin() 2. The Fourier transform of ft) (ft)-sinc(t)) is F(jo)-nRect(/2) (Figure ) (1) For a linear, time invariant system, its impulse response is h(t)…. Fourier Transform We will use the convention that a time function, g(t), and the Fourier Transform (FT) of that function, g(!), are in the time or frequency domain as indicated by the argument list rather than some variation on the function symbol. transforms on pairs from step 1. 7 Transmission of Signals Through Linear Systems 2. Tags: EMML, inner product, probability density functions, likelihood function, linear functional, orthonormal basis, linear transformation, vector, Linear Algebra. Fourier Transforms and the Dirac Delta Function A. The usual Fourier transform tables found online don't have many functional relationship rules. If any argument is an array, then fourier acts element-wise on all elements of the array. 16) Several important transforms are listed in the following table: f(t) F( ) a. Last time, we saw the equations that calculate the Fourier Transform and its inverse. Since sinc (w) has infinite duration in freqency domain, X (jw) convolved with sinc (w) also has infinite horizon in freqency domain. If you really want to understand the Fourier and Laplace transforms , how they work and why they work then this is the course for you. Characteristics of the Continuous Fourier Transform The plots in Figures 1-1 and 1-2 demonstrate two characteristics of the Fourier transforms of real time history functions: 1. 1 Dirac delta function The delta function –(x) studied in this section is a function that takes on zero values at all x 6= 0, and is inflnite at x = 0, so that its integral +R1 ¡1 –(x)dx = 1. So that gives you a complex spectrum which is here called ff, and then you multiply it by the imaginary unit times k, and then use an inverse transform back to physical space and now you have an exact to machine precision derivative defined on your original grid points. How to solve a basic math equation, foerester's algebra 1 suggested timelien, how to answe algebra problems, free algebra word problem solver, algebra1 answer keys g. 3 If f (x) is a good function with its Fourier transform g( y), then the Fourier transform of f (x) is 2πiyg( y), and the Fourier transform of f (ax + b) is |a|−1 e2πiby/a g( y/a). Function, f(t) Fourier Transform, F( ) Definition of Inverse Fourier Transform f t F( )ej td 2 1 ( ) Definition of Fourier Transform F() f (t)e j tdt Trigonometric Fourier Series 1 ( ) 0 cos( 0 ) sin( 0) n f t a an nt bn nt where T n T T n f t nt dt T b f t nt dt T f t dt a T a 0 0 0 0 0 0 ( )sin() 2. Since sinc (w) has infinite duration in freqency domain, X (jw) convolved with sinc (w) also has infinite horizon in freqency domain. The Fourier transform of ft) (ft)-sinc(t)) is F(jo)-nRect(/2) (Figure ) (1) For a linear, time invariant system, its impulse response is h(t)…. So we can write S2+S as S(S+1) now we can rewrite the equation as (S+2. Fessler,May27,2004,13:11(studentversion) Subtleties in dening the ROC (optional reading!) We elaborate here on why the two possible denitions of the ROC are not equivalent, contrary to to the book's claim on p. As such, it transforms one function into another, which is called the frequency domain representation of the original function (where the original function is often a function in the time-domain). " The full name of the function is "sine cardinal," but it is commonly referred to by its abbreviation, "sinc. Notice the minus sign! Usually, to get rid of that, the inverse transform is written with a minus sign inside the exponential. computation of the Z transform with contour integration 43. Trigonometric Polynomials 58. 2 Transforms of Derivatives and Integrals 6. This Demonstration illustrates the relationship between a rectangular pulse signal and its Fourier transform. 11-9) give sketches of possible Fourier transform magni- tudes. The second channel for the imaginary part of the result. To see how the Fourier transform works, we will begin with a one-dimensional signal and consider a simple step function. We experi-ment here to see if Mathematica knows these functions, and if it can deal with their Fourier transforms. The sinc function sinc(x) is a function that arises frequently in signal processing and the theory of Fourier transforms. I used overlapping windows of 1024 points. Learn more about Chapter 8: The Fourier Transform on GlobalSpec. FTIR spectrometers (Fourier Transform Infrared Spectrometer) are widely used in organic synthesis, polymer science, petrochemical engineering, pharmaceutical industry and food analysis. Analyzing the frequency components of a signal with a Fast Fourier Transform. Third Derivative. Example 21 Find the Fourier transform of the function where represents unit step function Solution: Fourier transform of is given by = = or Result: Note: If Fourier transform of is taken as , then Example 22 Find the inverse transform of the following functions: i. The expression in (7), called the Fourier Integral, is the analogy for a non-periodic f (t) to the Fourier series for a periodic f (t). Let tqptqu. Modular graph functions associate to a graph an SL(2,Z)-invariant function on the upper half plane. How to do a fast fourier transform fft in microsoft excel 1. Any good reference to more detailed tables would be very helpful! My attempt: $\mathcal F[f\times u] = (\mathcal Ff)*(\mathcal Fu)$ where * denotes convolution. We denote by Sn,k the set of all k-SIIRVs of order n. Usually, to find the Inverse Laplace Transform of a function, we use the property of linearity of the Laplace Transform. The Fourier transform is a particular case of z-transform, i. Fourier transform of unit step signal u(t). To see how the Fourier transform works, we will begin with a one-dimensional signal and consider a simple step function. Fourier Transform We will use the convention that a time function, g(t), and the Fourier Transform (FT) of that function, g(!), are in the time or frequency domain as indicated by the argument list rather than some variation on the function symbol. Which condition then A should satisfy. 1,791,367 views. 1 and Table 4. For now we will use (5) to obtain the Fourier transforms of some important functions. To know Laplace transform of integral and derivatives (first and high orders derivatives. Online FFT calculator helps to calculate the transformation from the given original function to the Fourier series function. If the Fourier transform of In(t) is. 5 Application of the Fourier Transform. 4 Vector Algebra & Vector Differentiation. 7 Transmission of Signals Through Linear Systems 2. There are several variants of the discrete Fourier transform, with various normalization conventions, which are specified by the parameter DftNormalization. , This requirement can be stated as , meaning that belongs to the set of all signals having a finite norm ( ). 5 for t<0 and 0. DFT needs N2 multiplications. Singular Fourier transforms andthe Integral Representation of the Dirac Delta Function Peter Young (Dated: November 10, 2013) I. This analytic function, is called the Fourier-Laplace transform of. edu the inverse Fourier transform 11–1. 2 Fourier Transform of Impulse Function; 3. 5, 1 over 2, when t equals 0. Table of Fourier Transform Pairs of Energy Signals Function name Time Domain x(t) Frequency Domain X Unit step () 10 00 t ut 2. The circuit can be represented as a linear time. 2nd/12/10 (ee2maft. Fourier transform of the unit step function and of the signum function: The signum function sgn( t) is a function that is related to the unit step function. Plugging this equation into the Fourier transform, we get:. One common example is when a voltage is switched on or off in an electrical circuit at a specified value of time t. Instead, the most common procedure to find the inverse Laplace transform of an expression is a two-step approach (Appendix 12. 9 Low-Pass and Band-Pass Signals 2. Inverse Fourier transform – be able to compute this from definition as well as from looking up the transform for elementary signals. Signal and System: Fourier Transform of Basic Signals (Step Signal) Topics Discussed: 1. When you apply both of these rules, the Fourier Transform of the ramp is (1/jw)^2. where z = (x + iy) is a complex number. Fast Fourier Transform with APL. Which condition then A should satisfy. Impulse, rectangle, triangle, Heaviside unit step, sign functions. x/e−i!x dx and the inverse Fourier transform is. due to an initial unit impulse of heat at x = ξ. Below is a summary table with a few of the entries that will be most common for analysis of linear differential equations in this course. The accurate ISAL echo signal model is established for a space maneuvering target that quickly approximates the uniform acceleration motion. 24 Applications of Fourier Transforms to Generalized Functions Theorem 2. Fourier Transform of the Unit Step Function We have already pointed out that although L{u(t)} = 1 s we cannot simply replace s by iω to obtain the Fourier Transform of the unit step. The Fourier transfer of the signum function, sgn(t) is 2/(iω), where ω is the angular frequency (2Ï€f), and i is the imaginary number. That is, given the Fourier transform of an function, when can we recover the original function from ? We begin with a simple case where the recovery is quite easy. The Fourier Transform Saravanan Vijayakumaran [email protected] k{1 - e-t/T} 4. Solution for 3. function and the Fourier transformation C. Laplace Transforms of the Unit Step Function. Join 100 million happy users! Sign Up free of charge:. Time Displacement Theorem: [You can see what the left hand side of this expression means in the section Products Involving Unit Step Functions. ** The Fourier transform of the triangular pulse f(t) shown in Fig. both are piecewise continuous functions for all t>=0 , then fourier transform of. This set of Signals & Systems Multiple Choice Questions & Answers (MCQs) focuses on “Fourier Transforms”. Note that some authors (especially physicists) prefer to write the transform in terms of angular frequency instead of the oscillation frequency. The Fourier transform of ft) (ft)-sinc(t)) is F(jo)-nRect(/2) (Figure ) (1) For a linear, time invariant system, its impulse response is h(t)…. To obtain Laplace transform of simple functions (step, impulse, ramp, pulse, sin, cos, 7 ) 11. 4 Sampling Continuous-Time Signals. Odd functions have imaginary (and odd) Fourier transforms. The discrete-time Fourier transform or DTFT of a sequence xŒn is defined as Discrete-Time Fourier Transform X. Example 2-2 SPECTRUM OF AN EXPONENTILA PULSE By means of direct integration find the Fourier transform of ) ( t w < = - 0 , 0 0 , ) ( t t e t w t Properties of Fourier Transforms. Fourier Transform for Periodic Signal, Sampling Function. Real part of X(ω) is even, imaginary part is odd. , This requirement can be stated as , meaning that belongs to the set of all signals having a finite norm ( ). The Laplace transform is usually restricted to transformation of functions of with. Laplace transforms convert a function f (t) in the time domain into function in the Laplace domain F (s). Fessler,May27,2004,13:11(studentversion) Subtleties in dening the ROC (optional reading!) We elaborate here on why the two possible denitions of the ROC are not equivalent, contrary to to the book's claim on p. So that gives you a complex spectrum which is here called ff, and then you multiply it by the imaginary unit times k, and then use an inverse transform back to physical space and now you have an exact to machine precision derivative defined on your original grid points. Ada version of General N Point Fast Fourier Transform. If playback doesn't begin shortly, try restarting your device. 16) Several important transforms are listed in the following table: f(t) F( ) a. When faced with the task of finding the Fourier Transform (or Inverse) it can always be done using the synthesis and analysis equations. Another description for these analogies is to say that the Fourier Transform is a continuous representation (ω being a continuous variable), whereas the. As an example of the Laplace transform, consider a constant c. Find the Fourier transform of 3. Trigonometric Polynomials 58. 1 Introduction There are three definitions of the Fourier Transform (FT) of a functionf(t) - see Appendix A. Because the convolution of two tempered distributions isn't always defined, neither is their product in the above sense. 6#15 - Duration: The intuition behind Fourier and Laplace transforms I was never taught in school inverse laplace transform,. Materials include course notes, lecture video clips, practice problems with solutions, a problem solving video, and problem sets with solutions. 5): e( s): The Fourier transform of the odd part (of a real function) is imaginary (Theorem 5. The most. com – tashuhka Oct 14 '14 at 12:36. This is in fact very heavily exploited in discrete-time signal analy-sis and processing, where explicit computation of the Fourier transform and its inverse play an important role. A unique 3D graphical approach has been adopted to provide the intuition required to OWN this subject. (-0)-u(0) | 4. 4142*j,0,1-j*2. Fourier Transform — Theoretical Physics Reference 0. Proposition 8 Let be. Fourier Transforms and the Fast Fourier Transform (FFT) Algorithm Paul Heckbert Feb. Third Derivative. All real c. (8) below] is due to Zakharov and Shabat [10]. This remarkable result derives from the work of Jean-Baptiste Joseph Fourier (1768-1830), a French mathematician and physicist. The signum function is also known as the "sign" function, because if t is positive, the signum function is +1; if t is negative, the signum function is -1. 42 While these published numerical inverse Laplace transform algorithm reviews are thorough and useful, 43 they focus on computing a single time-domain solution as accurately as possible. $\begingroup$ The plus one simply shifts when the Heaviside function turns on by one unit to the left like in normal functional translation, it helps to use the definition of the Heaviside step function as it restricts your domain of integration. Fourier series of even and odd functions, Gibbs phenomenon, Fourier half-range series, Parseval's identity, Complex form of Fourier series. of a unit step can be inferred, but it's natural with the Laplace. The accurate ISAL echo signal model is established for a space maneuvering target that quickly approximates the uniform acceleration motion. So here is the first example. Join the initiative for modernizing math education. Express f under an integral form. Find the Fourier transform of re(r), where e(r) is the Heaviside function. MATLAB Program for Dicrete Unit Impulse Function; / MATLAB Videos / Discrete Fourier Transform in for image conversion step by step Why 2D to 3D image. Step Functions - In this section we introduce the step or Heaviside function. Sinosoidal Function 5. The forward FT is de ned as usual g(!) = Z 1 1 g(t) ei!t dt ; (1) where scaling constants have. A unique 3D graphical approach has been adopted to provide the intuition required to OWN this subject. has three possible solutions for its Fourier domain representation depending on the type of approach. 2) become single integrals, integrated over the appropriate variable. When dealing with Fourier cosine and sine series, you are actually extending a non-periodic function onto a periodic even or odd domain. 4 Properties of fourier transforms There are several properties of fourier transforms that can be used as tools for solving PDEs. sinc(f˝)has Fourier inverse 1 ˝ rect ˝(t). The explicit solution of dual Sturm-Liouville matrix problem serves as a kernel for an inverse integral Fourier matrix transform. , This requirement can be stated as , meaning that belongs to the set of all signals having a finite norm ( ). Laplace transform, Existence theorem, Laplace transforms of derivatives and integrals, Initial and final value theorems, Unit step function, Dirac- delta function, Laplace transform of periodic function, Inverse Laplace transform, Convolution theorem, Application to solve simple linear and simultaneous differential equations. The first is a function of location (x), the latter of time (t). UNIT V LAPLACE TRANSFORM: Definition-ROC-Properties-Inverse Laplace transforms-the S-plane and BIBO stability-Transfer functions-System Response to standard signals-Solution of. Inverse Laplace Transform with unit step function, sect7. PYKC - 11 Feb 08 2 5. Topics Covered: Partial differential equations, Orthogonal functions, Fourier Series, Fourier Integrals, Separation of Variables, Boundary Value Problems, Laplace Transform, Fourier Transforms, Finite Transforms, Green's Functions and Special Functions. Similarly if an absolutely integrable function gon R, has Fourier transform ˆgidentically equal to 0, then g= 0. The Fourier transform of ft) (ft)-sinc(t)) is F(jo)-nRect(/2) (Figure ) (1) For a linear, time invariant system, its impulse response is h(t)…. 2 Fourier Series of Functions: Exponential, trigonometric functions of any period =2L, even and odd functions, half range sine and cosine series. Laplace transform of the unit step function | Laplace transform | Khan Academy - YouTube. It can be thought of as a function of the real line (x-axis) which is zero everywhere except at the origin (x=0) where the. Both functions are constant except for a step discontinuity, and have closely related fourier transforms. CT Fourier Transform Pairs signal (function of t) $\longrightarrow$ Fourier transform (function of $\omega$) : 1 CTFT of a unit impulse $\delta (t)\$ $1 \$. Implicit Derivative. Any good reference to more detailed tables would be very helpful! My attempt: $\mathcal F[f\times u] = (\mathcal Ff)*(\mathcal Fu)$ where * denotes convolution. The inverse Z-transform can be derived using Cauchy's integral theorem. The Fourier transform of a periodic impulse train in the time domain with period T is a periodic impulse train in the frequency domain with period 2p /T, as sketched din the figure below. Fourier transform. Since sinc (w) has infinite duration in freqency domain, X (jw) convolved with sinc (w) also has infinite horizon in freqency domain. Apply partial fraction expansion to separate the expression into a sum of basic components. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Current time: 0:00 Total duration: 24:15. The motivation for this work is to develop a deeper understanding of the origin of the algebraic identities between modular graph functions which have been discovered recently, and of the relation. Its value is not trivial to calculate, and ends up being. Disclaimer: None of these examples are mine. Examples Fast Fourier Transform Applications Signal processing I Filtering: a polluted signal 0 200 400 600 800 1000 1200 f1. , This requirement can be stated as , meaning that belongs to the set of all signals having a finite norm ( ). other The first of each pair is usually called the direct Fourier transform and the other one is the matching inverse Fourier transform, The unit Dirac comb (shah function) is its own Fourier transform. But i could not find the fourier transform of x in such frequencies. [email protected] com - id: 73fc3d-YTM3O. 5) is called a Fourier series. Z transform, Convergence. (20 marks) Using Laplace transform methods, solve for t = 0 the following differential equation, d 2x dt2 - 3 dx dt + 2x = 1, subject to x = 0 and dx dt = 0 at t = 0. Solution for 3. 1 Occasionally the question arises as to how a signal's frequency content is affected when the signal is time reversed. 6#15 - Duration: The intuition behind Fourier and Laplace transforms I was never taught in school inverse laplace transform,. Solution: Here, =0 for <2 , then ˝ =1 for ≥2. The Fourier transform we’ll be int erested in signals defined for all t the Four Fourier tra nsform of f G efine the Fourier transform of a step function or a constant signal unit step. Complex exponential The spectrum of a complex exponential can be found from the above due to the frequency shift property: Sinusoids. Consider the Fourier transforms of the functions in Example 9. transforms of functions multiplied by tn, scale change property, transforms of functions divided by t, transforms of integral of functions, transforms of derivatives ; Evaluation of integrals by using Laplace transform ; Transforms of some special functions- periodic function, Heaviside-unit step function, Dirac delta function. We can solve this integral by considering. Applying the inverse Fourier Transform to the complex image yields According to the distributivity law, this image is the same as the direct sum of the two original spatial domain images. If n is less than the length of the signal, then ifft ignores the remaining signal values past the nth entry and. Mathematicians have developed tables of commonly used Laplace transforms. These are the sample pages from the textbook. A special case is the expression of a musical chord in terms of the volumes and frequencies of its constituent notes. The signal x(t) can be obtained back from Fourier transform X(t) by using the inverse Fourier transform. Consider a sinusoidal signal x that is a function of time t with frequency components of 15 Hz and 20 Hz. Fourier Transform - Free download as Powerpoint Presentation (. 5) ¶ The expansion (3. Instead, the most common procedure to find the inverse Laplace transform of an expression is a two-step approach (Appendix 12. But for a square-integrable function the Fourier transform could be a general class of square integrable functions. The Fourier transform is ) 2 (2 ( ) T 0 k T X j k p d w p w ∑ ∞ =−∞ = −. In this lecture, we will look at one way of describing discrete-time signals through their frequency content: the discrete-time Fourier transform (DTFT). FFT is a powerful signal analysis tool, applicable to a wide variety of fields including spectral analysis, digital filtering, applied mechanics, acoustics, medical imaging, modal analysis, numerical analysis, seismography, instrumentation, and communications. text orientation finding) where the Fourier Transform is used to gain information about the geometric structure of the. That is, we present several functions and there corresponding Fourier Transforms. Fourier series •Periodic function (𝑡)of period 1: 𝑡= 0 2 + =1 ∞ cos(2𝜋𝑛𝑡)+ 𝑘=1 ∞ sin(2𝜋𝑛𝑡) •Fourier coefficients: =2න. $\begingroup$ The plus one simply shifts when the Heaviside function turns on by one unit to the left like in normal functional translation, it helps to use the definition of the Heaviside step function as it restricts your domain of integration. The range of functions for which the Fourier transform may be used can be greatly extended by using general-ized functions -- that is, the Dirac delta function and its close relatives (sign function, step function, etc. Mathematicians have developed tables of commonly used Laplace transforms. com – tashuhka Oct 14 '14 at 12:36. In particular we shall obtain, intuitively rather than rigorously, various Fourier transforms of functions such as the unit step function which actually violate the basic conditions which guarantee the existence of Fourier transforms! Prerequisites. The unit pulse function is simply one time shifted step function, minus another shifted step function. X(jω) is called the Fourier transform of the time function () x t , whereas ( x t ) is the inverse Fourier transform of X () jω. Properties of Fourier Transforms. First Derivative. For instance, the inverse continuous Fourier transform of both sides of Eq. DEU, Electrical and Electronics Eng. 4M subscribers. We denote by Sn,k the set of all k-SIIRVs of order n. Fast Fourier Transform (FFT) Calculator. Inverse Fourier Transforms 59. Is my last statement correct, about rewriting my original integral in terms of the step function? If I take the (inverse) Fourier transform of the step function $u(\omega)$ and I end up with two terms (i. Express f under an integral form. function and the Fourier transformation C. Fourier transform is, by modern convention, 2 C ( ! ). Instead, the most common procedure to find the inverse Laplace transform of an expression is a two-step approach (Appendix 12. Half range series, Change of intervals, Harmonic analysis. If we weren't using the involutive definition of the Fourier transform, we would have to replace one of the occurences of "Fourier transform" in the above definition by "inverse Fourier transform". Which condition then A should satisfy. Fourier Transform of the Lorentzian. Fourier Transform The Fourier transform (FT) is the extension of the Fourier series to nonperiodic signals. DC Level, Unit Step Function, Switched Cosine, Pulsed Cosine, Exponential Pulse, Fourier Transforms of Periodic Functions, Summary, 5. The Fourier transform we’ll be int erested in signals defined for all t the Four Fourier tra nsform of f G efine the Fourier transform of a step function or a constant signal unit step. 6#15 - Duration: The intuition behind Fourier and Laplace transforms I was never taught in school inverse laplace transform,. The Inverse Fourier Transform The Fourier Transform takes us from f(t) to F(ω). Fourier series of even and odd functions, Gibbs phenomenon, Fourier half-range series, Parseval's identity, Complex form of Fourier series. Solution for 3. According to Stroud and Booth (2011. Lecture X Discrete-time Fourier transform. Follow Neso Academy on Instagram: @nesoacademy(https://bit. We also work a variety of examples showing how to take Laplace transforms and inverse Laplace transforms that involve Heaviside functions. So we can write S2+S as S(S+1) now we can rewrite the equation as (S+2. 4 Vector Algebra & Vector Differentiation. If you really want to understand the Fourier and Laplace transforms , how they work and why they work then this is the course for you. 2 Fourier Series of Functions: Exponential, trigonometric functions of any period =2L, even and odd functions, half range sine and cosine series. In this problem we will evaluate the Fourier transform of the given function. The response time is defined with respect to a unit step function, as the time it takes for the "smoothed" step function to rise from 10% to 90% of its original value. That tells us that the inverse Laplace transform, if we take the inverse Laplace transform-- and let's ignore the 2. You can also check “New Worksheet,” but having the Fourier Analysis results right next to your data will be more useful. k{1 - e-t/T} 4. , This requirement can be stated as , meaning that belongs to the set of all signals having a finite norm ( ). The #1 tool for creating Demonstrations and anything technical. The inverse Z-transform can be derived using Cauchy’s integral theorem. It includes Laplace transform of special functions, properties, operations and using Laplace transforms to solve ordinary and partial differential equations. If the first argument contains a symbolic function, then the second argument must be a scalar. Magnitude and phase spectrum. To make one more analogy to linear algebra, the Fourier Transform of a function is just the list of components of the. Basic Properties of Fourier Transform (1) (Linearity) If the Fourier transform of f1 and f2 exist, then (2. There are different definitions of these transforms. This is interesting because if we extract a section of a signal to analyse, and obtain its spectrum (via Fourier Transform), we are effectively multiplying the signal with a rectangular function (rect()). •It is the most general F. (Dirac & Heaviside) The Dirac unit impuls function will be denoted by (t). Compute the power and energy of 2 times t squared. !/, where: F. • is called the magnitude function • is called the phase function • Both quantities are again real functions of ω • In many applications, the DTFT is called the Fourier spectrum • Likewise, and are called the magnitude and phase spectra X(ejω) θ(ω) X(ejω) θ(ω). The Fourier Transform (used in signal processing) The Laplace Transform (used in linear control systems) The Fourier Transform is a particular case of the Laplace Transform, so the properties of Laplace transforms are inherited by Fourier transforms. The Fourier transform G(w) is a continuous function of frequency with real and imaginary parts. Fourier inverse step. 3 Fourier Transforms of Time Functions. Lecture X Discrete-time Fourier transform. We proceed via the Fourier Transform of the signum function sgn t which. Inverse Z Transform: Part 2. Solution: Here, =0 for <2 , then ˝ =1 for ≥2. In Chapter 6 we were able to derive some ad hoc extensions of the classical Fourier transform which applied to the unit step function, delta functions, end even to infinite series of delta functions. Another description for these analogies is to say that the Fourier Transform is a continuous representation (ω being a continuous variable), whereas the. 2 Transform or Series. The DTFT sequence x[n] is given by Here, X is a complex function of real frequency variable ω and it can be written as Where Xre. 1 Foreshortening 1. We can solve the integral by contour integration. An alias of itself By subtracting the constant 0. 9) to emphasize. 2) become single integrals, integrated over the appropriate variable. The Fourier transform of ft) (ft)-sinc(t)) is F(jo)-nRect(/2) (Figure ) (1) For a linear, time invariant system, its impulse response is h(t)…. Implicit Derivative. both are piecewise continuous functions for all t>=0 , then fourier transform of. in Department of Electrical Engineering Indian Institute of Technology Bombay July 20, 2012. Suppose that the Fourier transform of f and its inverse exist. DEU, Electrical and Electronics Eng. Translation (that is, delay) in the time domain goes over to complex phase shifts in the frequency domain. In particular we shall obtain, intuitively rather than rigorously, various Fourier transforms of functions such as the unit step function which actually violate the basic conditions which guarantee the existence of Fourier transforms! Prerequisites. Be able to use partial fraction expansions to compute the Inverse Fourier transform. MA 382 - Laplace and Fourier Transforms This course introduces the theoretical concepts and uses of the Laplace and Fourier transforms. • The unit step function (1 class) • The Dirac delta function (1 class) • Applications of step and impulse functions (1 class) • Periodic functions and their applications (2 classes) • Convolution and applications (2 classes) • Solving integral equations (1 class) • Fourier series (3 classes) • Fourier integral representation (1. Recall, that $$\mathcal{L}^{-1}\left(F(s)\right)$$$is such a function f(t) that $$\mathcal{L}\left(f(t)\right)=F(s)$$$. The ifft function allows you to control the size of the transform. Topics Covered: Partial differential equations, Orthogonal functions, Fourier Series, Fourier Integrals, Separation of Variables, Boundary Value Problems, Laplace Transform, Fourier Transforms, Finite Transforms, Green's Functions and Special Functions. Whereas its Fourier transform, or the magnitude of its Fourier transform, has the inverse property that as a gets smaller, in fact, this scales down in frequency. The Fourier-series expansions which we have discussed are valid for functions either defined over a finite range ( T t T/2 /2, for instance) or extended to all values of time as a periodic function. As I indicated last time, the Fourier transform is a complex function of frequency. If playback doesn't begin shortly, try restarting your device. Frequency Response Function For a 1storder system The FRF can be obtained from the Fourier Transform of Input-Output Time Response (and is commonly done so in practice) The FRF can also be obtained from the evaluation of the system transfer function at s=jω. This transform can be obtained via the integration property of the fourier transform. Fourier transforms take the process a step further, to a continuum of n-values. Of practical importance is the conjugate symmetry property: When s (t) is real-valued, the spectrum at negative. The forward Z-transform helped us express samples in time as an analytic function on which we can use our algebra tools. Fourier transform of typical signals. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Introduction to continuous time signals and systems: Basic continuous time signals, unit step , unit ramp, unit impulse and periodic signals with their mathematical representation and characteristics. The Heaviside function is a unit step at x = 0 and is shown below Differentiating the Heaviside function results in the Dirac /Delta function. Often the unit step function u. The Z transform of the geometric sequence 39. 2 Fourier Series Consider a periodic function f = f (x),defined on the interval −1 2 L ≤ x ≤ 1 2 L and having f (x + L)= f (x)for all. 4 Fourier Transform of One-Sided Exponential Function; 3. Example 1 Find the Fourier transform of the one-sided exponential function f(t) = ˆ 0 t < 0 e−αt t > 0 where α is a positive constant, shown below: f (t) t Figure 1 Solution. Unlike the inverse Fourier transform, the inverse Laplace transform in Eq. 5 Signals & Linear Systems Lecture 10 Slide 12 Fourier Transform of a unit impulse train XConsider an impulse train. 3 Fourier Transforms of Time Functions. Eventually, we have to return to the time domain using the Inverse Z-transform. fast fourier transform. 2 Transforms of Derivatives and Integrals 6. Solution for 3. Odd functions have imaginary (and odd) Fourier transforms. Z transform of step and related functions. This immediately tells us that the situation for compactly supported functions is very different from the situation for Schwartz functions — the Fourier transform of a compactly supported function is analytic, so it cannot be compactly supported or it would vanish identically. Notice the the Fourier Transform and its inverse look a lot alike—in fact, they're the same except for the complex. As I indicated last time, the Fourier transform is a complex function of frequency. The unit pulse function can be defined with the help of the Heaviside unit step function ( ) ( ) ( ) 0 x a f t Ht a Ht a 1 x a 0 x a <− = +− −= < > a0 > The Fourier transform of this function can be determined as. The unit step function "steps" up from 0 to 1 at t=0. Let us now substitute this result into Eq. is arbitrarily selected. The accurate ISAL echo signal model is established for a space maneuvering target that quickly approximates the uniform acceleration motion. Laplace Transform Calculator. For the Z-transform the DTFT exists if the ROC includes the unit circle. Rectangular Pulse Signal Some Examples of Fourier Transform. The first is a function of location (x), the latter of time (t). Unit-111: Fourier series: Trigonometric Fourier series and its convergence. Second Derivative. 11) is rarely used explicitly. For example: (lg is log base 2) Primary space: 4 * 8 = 32 Dual space: lg(4) + lg(8) = 2 + 3 = 5 = lg(32). 1 The Fourier transform. Especially important among these properties is Parseval's Theorem, which states that power computed in either domain equals the power in the other. Proven the the Heaviside function is a tempered distribution I must evaluate: \langle F Stack Exchange Network Stack Exchange network consists of 176 Q&A communities including Stack Overflow , the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Proposition 8 Let be. In that case the integrals in (4. This website uses cookies to ensure you get the best experience. 5D electrical modelling Shi-zhe Xu,1 Ben-chun Duan2 and Da-hai Zhang1 Abstract An optimization method is used to select the wavenumbers k for the inverse Fourier transform in 2. Lecture X Discrete-time Fourier transform. Determine the Fourier transform of the non-periodic signals shown in the figures below: (b) 8(1) -2 -1 0 1 2. Inverse Transform 6. The second channel for the imaginary part of the result. An alias of itself By subtracting the constant 0. Apply partial fraction expansion to separate the expression into a sum of basic components. The Fourier Transform is used to transform a process from the time domain to the frequency domain. MATLAB Program for Dicrete Unit Impulse Function; / MATLAB Videos / Discrete Fourier Transform in for image conversion step by step Why 2D to 3D image. Unit Impulse Response : We have Laplace transform of the unit impulse. The discrete Fourier transform (DFT) of the discrete signal is n =0, 1, … , N-1 Similarly, an inverse discrete Fourier transform is of this form: Note that the number of data points in x(n) and X(m) are always the same The frequency in the Fourier domain is related to the sampling frequency f s. The Fourier transform is one of the most useful mathematical tools for many fields of science and engineering. Inverse Fourier Transform. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. If Y is a matrix, then ifft (Y) returns the inverse transform of each column of the matrix. A similar analysis can be done in the frequency domain. To obtain Laplace transform of functions expressed in graphical form. Fourier Transform Symmetry (contd. 1) into the integral in the deflnition of the inverse transform in (F. Laplace transform with a Heaviside function by Nathan Grigg The formula To compute the Laplace transform of a Heaviside function times any other function, use L n u c(t)f(t) o = e csL n f(t+ c) o: Think of it as a formula to get rid of the Heaviside function so that you can just compute the Laplace transform of f(t+ c), which is doable. The equations describing the Fourier transform and its inverse are shown opposite. The Fourier transform of ft) (ft)-sinc(t)) is F(jo)-nRect(/2) (Figure ) (1) For a linear, time invariant system, its impulse response is h(t)…. Recently I came across finite Fourier transforms, which can be used for solving certain type of boundary value problem (BVP) of linear partial differential equation (PDE) with constant coefficient. : and, inverse, And we can reverse this, too. has three possible solutions for its Fourier domain representation depending on the type of approach. I don't know where you got G(f), but it only a mathematical expression to "give" the value of the Fourier transform of a unit step. The list given in Fourier [list] can be nested to represent an array of data in any number of dimensions. com – tashuhka Oct 14 '14 at 12:36.
exkds2ijwbfm8, scu0emspadvc8, 6jx7myor6dcbhqe, r76yr3voc6ebxht, hytmbmksppc0, k4lp9hjqws89, eqv1czdh0zmd1, mcjverfdlgenkw, lzszur4c6gtmz30, 436mnpveqe, i7xgkf69ctsuys, qnpao69epc298z, uv2pp8oaz6ka8a, 7kwals3rv1, n7p6yh671r, gtvjn9fwetc, 0kioxwtn1zo, pk13ifb40vhuxfd, 0e4czxbyadkf8, bue2ge4zxs77fm, zohxk1p95epev, 5s0aerhq8ps7oo, nq2oyhzvjns00m, 529efdfqv8p, 8pv8mktybs6zn, 546bvb64xeo | 2020-06-04T17:31:33 | {
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http://mathhelpforum.com/calculus/18409-papus-theorem.html | # Thread: Papus theorem
1. ## Papus theorem
Question 1
Use the Theorem of Pappus to find the volume of the solid generated when the region bounded by the x-axis and the semicircle y= sqrt(a^2 -x^2) is revolved about
a) the line y=-a
b) the line y=x-a
Question 2
Use the Theorem of Pappus to find the centroid of the triangular region with vertices (0,0), (a,0), and (0,b) , where a> 0 and b>0 .
Hint Revolve the region about the x-axis to obtain y bar and about the y-axis to obtain x bar.
I don't quite understand the Pappus theorem. Please do help you know how to solve them. Thank you very much.
2. #1:
$\displaystyle \overline{x}=0$ from the symmetry of the region, $\displaystyle \frac{{\pi}a^{2}}{2}$ is the area of the semicircle, $\displaystyle 2{\pi}\overline{y}$ is the distnace traveled by the centroid to generate the sphere so $\displaystyle \frac{4}{3}{\pi}a^{3}=(\frac{{\pi}a^{2}}{2})(2{\pi }\overline{y}), \;\ \overline{y}=\frac{4a}{3{\pi}}$
Now for part a, $\displaystyle V=\left[\frac{1}{2}{\pi}a^{2}\right]\left[2{\pi}\left(a+\frac{4a}{3{\pi}}\right)\right]=\frac{1}{3}{\pi}(3\pi+4)a^{3}$
See if you can tackle part b.
#2:
Revolve the region about the x-axis to get $\displaystyle \overline{y}$ and about the y-axis to get $\displaystyle \overline{x}$.
The region generates a cone of volume $\displaystyle \frac{1}{3}{\pi}ab^{2}$
when revolved about the x-axis, the area of the region is
$\displaystyle \frac{1}{2}ab$, so $\displaystyle \frac{1}{3}{\pi}ab^{2}$$\displaystyle =\left(\frac{1}{2}ab\right)(2{\pi}\overline{y})$$\displaystyle , \;\ \overline{y}=\frac{b}{3}$.
A cone of volume $\displaystyle \frac{1}{3}a^{2}b$ is generated when the
region is revolved about the y-axis so $\displaystyle \frac{1}{3}{\pi}a^{2}b$$\displaystyle =\left(\frac{1}{2}ab\right)(2{\pi}\overline{x}), \;\ \overline{x}=\frac{a}{3}$.
The centroid is.........?
3. Hi galatus,
Thank you very much for your reply.
I don't fully understand this question.
I know the Pappus theorem is
V = area R * distance traveled by the centroid.
You have showed me
So the second part is the distance traveled by the centroid. Why equals 2pi ( a + (4a)/(3pi))? UM... just don't get it!
4. Hello again, Kittycat. As before, you are giving up too soon. There is nothing here that should be a surprise to you.
1. Your axis of revolution is y = -a.
2. Your Centroid is at $\displaystyle (0,\frac{4a}{3\pi})$
3. The distance from the Centroid to the Axis of Rotation is the Radius of the circle travelled. $\displaystyle \left(\frac{4a}{3\pi} - (-a)\right)$
4. $\displaystyle 2\pi\;r$ is the formula for the circumference of a circle, given its radius.
P.S. Pep talk - pep talk - pep talk. More confidence. There is more stuff in your brain than you think. Reach in a bring it out.
5. hi TKHunny,
I am not that great - don't have so much good stuff in my brain as you thought.
I think you are great - with brilliant intellects!
What about for part b - the line y=x-a?
Can you explaining to me how to get the distance from the centroid to the line y=x-a?
Thank you very much.
6. Originally Posted by kittycat
I am not that great - don't have so much good stuff in my brain as you thought.
Hogwash!
Can you explaining to me how to get the distance from the centroid to the line y=x-a?
Several ways. One of the most useful formulas ever, the Distance from a point (a,b) to a line cx + dy + e = 0 is:
$\displaystyle \frac{|c*a + d*b + e|}{\sqrt{c^{2}+d^{2}}}$
It is a beautiful thing. | 2018-05-27T10:58:08 | {
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http://mathhelpforum.com/algebra/124720-peak-graph.html | # Math Help - Peak of graph
1. ## [Solved] Peak of graph
I read somewhere that you can get the coordinates for the peak of a quadratic equation by just rearranging the equation. I forgot how it was done though. Something like
$y=(x+p)^2+q$
is this correct? and how would you do for
$y=2x^2-4x+a^2-a$
2. Originally Posted by davidman
I read somewhere that you can get the coordinates for the peak of a quadratic equation by just rearranging the equation. I forgot how it was done though. Something like
$y=(x+p)^2+q$
is this correct? and how would you do for
$y=2x^2-4x+a^2-a$
If you write the quadratic equation as
$y = a(x - h)^2 + k$
then the turning point is $(x, y) = (h, k)$.
So for $y = 2x^2 - 4x + a^2 - a$, you complete the square
$y = 2\left(x^2 - 2x + \frac{1}{2}a^2 - \frac{1}{2}a\right)$
$y = 2\left[x^2 - 2x + (-1)^2 - (-1)^2 + \frac{1}{2}a^2 - \frac{1}{2}a\right]$
$y = 2\left[(x - 1)^2 + \frac{1}{2}a^2 - \frac{1}{2}a - 1\right]$
$y = 2(x - 1)^2 + a^2 - a - 2$.
So the turning point is:
$(x, y) = (1, a^2 - a - 2)$.
3. wooah, you made that so easy to understand.
thanks!
4. Originally Posted by davidman
I read somewhere that you can get the coordinates for the peak of a quadratic equation by just rearranging the equation. I forgot how it was done though. Something like
$y=(x+p)^2+q$
is this correct? and how would you do for
$y=2x^2-4x+a^2-a$
If the vertex of the parabola is V(p,q) then the equation of the parabola whose axis is parallel to the y-axis is written as:
$y = a(x-p)^2+q$
$y = 2x^2-4x+a^2-a = 2(x^2-2x+\tfrac12(a^2-a))$
$y=2(x^2-2x\bold{\color{red}+1-1}+\tfrac12(a^2-a)) = 2(x-1)^2+a^2-a-2$
So the coordinates of the vertex are $V\left(1, a^2-a-2\right)$
5. Thanks so much for the help!
I've been trying my hand at the second part of this question, but I can't seem to get to the answer shown in my book...
the second part is to get the minima of $f(a)=a^2-a-2$
and I thought the answer would reveal itself if I factorised and solved for $a$, but the answer is apparently $-\frac{9}{4}$... I was not even close... what exactly are you supposed to do to get to that answer?
6. Originally Posted by davidman
Thanks so much for the help!
I've been trying my hand at the second part of this question, but I can't seem to get to the answer shown in my book...
the second part is to get the minima of $f(a)=a^2-a-2$
and I thought the answer would reveal itself if I factorised and solved for $a$, but the answer is apparently $-\frac{9}{4}$... I was not even close... what exactly are you supposed to do to get to that answer?
Probably you are trapped by the text of the question(?) It reads " ... get the minima of f(a) ..."
I assume that you have calculated:
$f(a)=a^2-a-2$
$f(a)=a^2-a\bold{\color{blue}+\frac14-\frac14}-2$
$f(a)=\left(a-\frac12 \right)^2-\frac14-2$
$f(a)=\left(a-\frac12 \right)^2-\frac94$
Therefore the vertex has the coordinates $V\left(\tfrac12 ,~ -\tfrac94 \right)$
With a parabola opening up the f(a)-value of the vertex determines the minimum. The vertex is the lowest point of all points of the graph of f. Therefore
$\min(f(a)) = -\frac94$
7. Thanks a lot guys. earboth's post brings that question to perfection! | 2014-08-01T06:49:31 | {
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https://math.stackexchange.com/questions/2567104/roots-of-a-function-between-two-roots-of-another-function | # Roots of a function between two roots of another function
I've been asked the following problem:
Let $f$ and $g$ be two functions differatiable in the interval $I$. Proove that if $$f(x)g'(x) - f'(x)g(x) \ne 0 \quad \forall x \in I$$ there is one root of $g$ between every two roots of $f$
Now, the problem is, this question was asked in German, and the word for 'one' and the word for 'a' are both the same in German. In other words, I do not know, if the sentence says exactly one root of $g$ or at least one root of $g$. I've already spent a few hours on this problem and managed to show that there is at least one root of $g$ between two roots of $f$:
$$f(x)g'(x) - f'(x)g(x) = c, \quad c \ne 0$$ $$\Leftrightarrow g(x) = \frac{f(x)g'(x) - c}{f'(x)}$$ $$x_1, x2 \in I, \quad f(x_1) = 0, \quad f(x_2) = 0$$ $$g(x_1) = \frac{f(x_1)g'(x_1) - c}{f'(x_1)} = -\frac{c}{f'(x_1)}$$ $$g(x_2) = \frac{f(x_2)g'(x_2) - c}{f'(x_2)} = -\frac{c}{f'(x_2)}$$ Next, we can say If $g(x_1)$ and $g(x_2)$ have different signs, there exists a $\xi$ for which $g(\xi) = 0$ (IVT)
$g(x_1)$ and $g(x_2)$ have different signs if and only if $f'(x_1)$ and $f'(x_2)$ have different signs. Using Rolle's Theorem we can say:
There is an $\eta \in (x_1, x_2) : f'(\eta) = 0$
As $x_1 < \eta < x_2$ we can say, that either $f'(x_1)$ is negative and $f'(x_2)$ is positive or vice versa. Therefore the signs of $g(x_1)$ and $g(x_2)$ are different as well, and we know that there must be at least one root in the interval.
However, I just do not manage to create a proof that shows that there can't be more than 1 root in the interval. My guess would have been using Rolle's Theorem for a proof by contradiction, however there might perfectly be an extremum in the intervals, since $g(x)$ might have other roots outside the bounds of the interval. (Or did I get something wrong here)
So my question is:
Is the initial statement true, and if it is, how can I proove it?
• I believe the German statement asked for "a root". I'm a Spanish native speaker and the same ambiguity would have happened in Spanish; in this language the convention in maths is that if there's no precision "un/una" means "a/an", that is "at least one". – Alejandro Nasif Salum Dec 14 '17 at 22:54
• Also, the uniqueness comes from the fact that the statement is symmetric in $f$ and $g$, so under those conditions you proved that there is "at least one" root of $g$ between two given roots of $f$, but you've also proved that there is "at least one" root of $f$ between two given roots of $g$. If you add the precision "consecutive" roots, then this implies uniqueness. On the other hand, between non-consecutive roots of, say, $f$ there can be more than one root of $g$; precisely one more than the number of roots $f$ you have left in between. – Alejandro Nasif Salum Dec 14 '17 at 22:58
• This does not address your main concern, but the proof itself might be simplified by observing that under the assumption that $g$ has no roots, $f'(x)g(x)-f(x)g'(x)$ is just $g^2(x)$ times the derivative of $\frac{f(x)}{g(x)}$, and Rolle says that this must be zero somewhere in-between. – Hagen von Eitzen Dec 14 '17 at 23:01
• Just to be sure: Does this mean i can say: Assume $g(x)$ has no roots. Let $h(x) = \frac{f(x)}{g(x)}$. Then $h(x_1) = h(x_2) = 0$ because $f(x_1) = f(x_2) = 0$. Because of Rolle this means that $h'(\xi) = 0$, therefore $f'(\xi)g(\xi) - f(\xi)g'(\xi)$ must be $0$ as well, which contradicts the statement. Because of that, $g(x)$ must have roots. – zockDoc Dec 14 '17 at 23:21
The uniqueness comes from the fact that the statement is symmetric in $f$ and $g$, so under those conditions you proved that there is "at least one" root of $g$ between two given roots of $f$, but you've also proved that there is "at least one" root of $f$ between two given roots of $g$. If you add the precision "consecutive" roots, then this implies uniqueness.
On the other hand, between non-consecutive roots of, say, $f$ there can be more than one root of $g$; precisely one more than the number of roots $f$ you have left in between.
Anyway, I would say your proof works fine only under the hypothesis of consecutive roots. If you allow $f$ to be $0$ in the interval $(x_1,x_2)$, then I do not see how you could conclude that $f'(x_1)$ and $f'(x_2)$ have different signs. Think for instance of $f(x)=x^3-x$ in the interval $[-1,1]$, where Rolle's conditions are met, but $f'(-1)=f'(1)=2$. | 2020-07-09T11:12:17 | {
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https://math.stackexchange.com/questions/2957789/proof-of-f-1a-cap-f-1b-f-1a-cap-b | # Proof of $f^{-1}(A) \cap f^{-1}(B)= f^{-1}(A \cap B)$
First let me say I am aware of the other threads on this result. The reason for me making this thread is to find out whether or not my proof/proof attempt is correct.
The problem stated in full detail is given below.
Let $$X,Y$$ be sets and $$f:X \rightarrow Y$$, let $$C,D \subseteq X$$ and let $$A,B \subseteq Y$$. Prove that $$f^{-1}(A) \cap f^{-1}(B)= f^{-1}(A \cap B)$$.
Here is my attempt:
$$f^{-1}(A) \cap f^{-1}(B)= { \{x\in X \mid f(x) \in A\}}\cap { \{x\in X \mid f(x) \in B\}}= \{x\in X \mid f(x) \in A\wedge f(x) \in B\}= \{x\in X \mid f(x) \in A\cap B \}=f^{-1}(A \cap B)$$
If anyone would be kind enough to explain to me where I've gone wrong or made an unjustified assumption I would be very grateful!
• you have only proved $f^{-1}(A) \cap f^{-1}(B)\subseteq f^{-1}(A \cap B)$. You will need to prove that $f^{-1}(A \cap B)\subseteq f^{-1}(A) \cap f^{-1}(B)$ – Neo Oct 16 '18 at 11:32
• @Neo Really? As far as I can tell I have only used equality ad not implication rules... – krtgdl Oct 16 '18 at 11:33
• @Neo Nope. Every equality OP wrote is true, they proved equality, not just inclusion. – 5xum Oct 16 '18 at 11:35
• Learnt something new, I was marked wrong in an assignment for doing that once :/ – Neo Oct 16 '18 at 11:37
• @Neo If you did it the wrong way, it might have been marked correctly. It's possible to use the correct tool incorrectly. Or maybe you were cheated. I'd have to see the concrete example. – 5xum Oct 16 '18 at 11:56
\begin{align}x\in f^{-1}(A)\cap f^{-1}(B)\iff &(x\in f^{-1}(A)\land x\in f^{-1}(B)\\\iff& f(x)\in A\land f(x)\in B\\\iff& f(x)\in A\cap B\\\iff& x\in f^{-1}(A\cap B)\end{align} | 2019-09-24T08:35:13 | {
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https://www.qalaxia.com/questions/How-do-I-solve-this-geometry-question-involving-probability | D
How do I solve this geometry question involving probability?
59 viewed last edited 1 year ago
Anonymous
0
I saw this problem online and I am having trouble figuring how to answer this.
Here is what I have done so far: Let A, B and C be the randomly chosen points.
Let AA' be the diameter drawn from A; and BB' be the diameter drawn from B. Is the following claim correct?
The 3 points will not have a common semi-circle if and only if:
1) B and C lie on different sides of AA'; and
2) A and C lie on different sides of BB'
Prem Kumar
2
Here's another way to see the result and also to generalise to N points chosen randomly on the circle.
Pick one of these N points and draw the diameter through it. The probability that any other point lies on one side of the diameter is 1/2. So the probability that the remaining N-1 points all lie on one side of that diameter is \frac{1}{2^{N-1}}.
We have N points which are undistinguished, so the net probability that there is one point amongst them to which the above condition applies = \frac{N}{2^{N-1}}
For N=2 this gives 1 as you might expect: two points chosen at random lie on a semicircle trivially.
For N=3 it yields \frac{3}{4} as the previous arguments did too.
Mahesh Godavarti
1
What you have is reasonable. The order of the points does not matter. So, what you are saying is: pick A, B and C in that order.
The probability that A and B, do not share a semicircle is zero (obviously).
So, we have to wait for C. For A, B and C to not share a semicircle, C has to be such that neither B nor C lie on the same side of the diameter AA' and neither A nor C lie on the same side of the diameter BB'.
This makes sense, however, we can't go anywhere from here.
I would do the following:
Let x be the smaller of the distance between A and B. Then 0 \leq x \leq \pi r .
Then the probability density function of x , f(x) = \frac{1}{\pi r} .
Then, the probability that C will not be in the same semicircle as A and B, given the distance between A and B is x , g(\text{not semi-C}|x)=\frac{x}{2\pi r}.
Therefore, the total probability density function that A, B and C will not share a semicircle is given by \frac{x}{2 \pi^2 r^2} . I.e. C has to fall in the unbolded portion of the circle whose length is also x .
Therefore, total probability that A, B and C do not lie on the same semicircle is given by \int_0^{\pi r} \frac{x}{2 \pi^2 r^2} dx = \frac{1}{4} .
I am sure there is a much more elegant argument given that the answer has turned out to be this simple.
Anonymous
1
If C and B have to be opposite sides of AA', the probability of this is 1/2. If A and C have to be opposite sides of BB', the probability of this is 1/2. If C and B have to be on opposite sides of AA'; and A and C have to be on opposite sides of BB'; intuitively, the average probability of this appears to be 1/2 * 1/2 = 1/4.
Mahesh Godavarti
0
Let's formalize it.
Mahesh Godavarti
1
I think this is the elegant answer. Let A-SC be the semicircle defined by A and B-SC be the semicircle defined by B (A is at the middle of A-SC and B is at the middle of B-SC). Then C should not lie in the semicircle defined by both A and B. The probability that C does not lie in the semicircle defined by A = 1/2. The probability that C does not lie in the semicircle defined by B = 1/2. Since A-SC and B-SC are independent (note A and B are independent), then the probability that C does not lie in either semicircle = 1/2 X 1/2 = 1/4. I think this is the answer.
Vivekanand Vellanki
1
Continuing with Mahesh's response, for a given x, the probability that A, B and C dont lie on a semi-circle is g(\text{not semi-C}|x)=\frac{x}{2\pi r}.
Given x, this determines accurately the probability that the 3rd point is not on the same semi-circle.
Since x is uniformly distributed from 0 to \pi r, we need to find the expected probability.
For every x, there is an x' such that x+x'\ =\ \pi r; and
g(not semi C | x) + g(not semi C | x') = 1/2.
Hence, the expected value of g(not semi C | x) = 1/4 over 0\le x\le\pi r | 2020-01-24T04:57:18 | {
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https://math.stackexchange.com/questions/1990936/is-this-inequality-true-x-y-alpha-x-alpha-y-alpha-for-pos | # Is this inequality true? $(x + y)^{\alpha} < x^{\alpha} + y^{\alpha}$, for positive $x$ & $y$, and for $0 < \alpha < 1$
If $0 < \alpha < 1$, then
$$(x + y)^{\alpha} < x^{\alpha} + y^{\alpha}$$
for $x$, $y$ positive.
Is this inequality true in general?
I tried using Young's Inequality: For $z,t > 0$, and for $n$, $m$ such that $n+m=1$, then
$$z^n + t^m \leq nz + mt$$
So, using this we have
$$(x+y)^{\alpha} \cdot 1^{1 - \alpha} \leq \alpha(x+y) + (1-\alpha) = \alpha x + \alpha y + (1-\alpha)$$
which is not as tight as I want.
Hint: try rewriting the inequality as $$x^\alpha \left( 1 + t \right)^\alpha \le x^\alpha \left( 1 + t^\alpha \right)$$ where $t = y/x$.
Let $x\geq y$.
Hence, $(x+y,0)\succ(x,y)$ and since $f(x)=x^{\alpha}$ is a concave function, by Karamata we obtain: $$(x+y)^{\alpha}=(x+y)^{\alpha}+0^{\alpha}\leq x^{\alpha}+y^{\alpha}$$ and we are done!
LEt $f(t) = 1 + t^{\alpha} - (1+t)^{\alpha}$, $t > 0$ and $0 < \alpha < 1$. We have
$$f'(t) = \alpha t^{\alpha - 1} - \alpha(1+t)^{\alpha-1} = \alpha t^{\alpha-1} \left( 1 - \left( \frac{ 1 + t }{t} \right)^{\alpha-1} \right)$$
Since $\frac{1+t}{t} > 1$ and $\alpha - 1 <0$, then $1 - \left( \frac{ 1 + t}{t} \right)^{\alpha - 1} > 0$. Therefore,
$$f'(t) > 0 \; \text{all} \; t>0 \implies f(t) > f(0) \implies 1 + t^{\alpha} > (1+t)^{\alpha}$$
Assume $x,y > 0$, then $t = \frac{y}{x} > 0$ and thus
$$1 + (y/x)^{\alpha} > ( 1 + y/x)^{\alpha} \implies x^{\alpha} + y^{\alpha} > (x+y)^{\alpha}$$
$$(x+y)^a\lt x^a+y^a\iff x^a\left(1+\dfrac yx\right)^a=x^a\left(1+(\dfrac yx)^a\right)$$ Hence it is enough to prove $$(1+x)^a\lt 1+x^a$$
The function $$f(x)=1+x^a-(1+x)^a$$ has the derivative $$f'(x)=a(x^{a-1}-(1+x)^{a-1})$$ which is positive for $0\lt a\lt 1$ and $x$ positive because $$a(x^{a-1}-(1+x)^{a-1})\gt 0\iff \frac{1}{x^{1-a}}-\frac{1}{(1+x)^{1-a}}\gt 0$$ Consequently $f(x)$ is increasing for $x\gt 0$ so because $f(0)=0$ we have always $$f(x)\gt 0\iff(1+x)^a\lt 1+x^a$$
A hint that I like for problem solving or maybe a different way to conceptualize these problems. Consider these two expressions as functions of 3 variables, $x,y,\alpha$. In one variable calculus you can use derivative to find maximum points or minimums. You can use generalized methods here. Consider term one subtracted from term two. If it has a global minimum that says something related to this inequality.
Of course the case $\alpha=0$ or $\alpha=1$ is trivial, so suppose $0<\alpha<1$. Dividing both sides by $(x+y)^\alpha$ reduces the proof to the case $x+y=1$ (to see this, just do the substitution $x'=\frac{x}{x+y}, y'=\frac{y}{x+y}$). So suppose $x+y=1$. Then $0<x<1$, $0<y<1$, implying $x<x^\alpha$ and $y<y^\alpha$. Hence $$(x+y)^\alpha=1=x+y<x^\alpha+y^\alpha.$$ | 2019-06-25T09:34:17 | {
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https://stats.stackexchange.com/questions/325764/using-dnn-to-find-out-the-pdf-of-a-regression-problem | # using DNN to find out the pdf of a regression problem
When we use deep neural networks (DNNs) to solve a 1-dimention regression problem, we can approximate data distribution with the output of a DNN like the picture below.
My question is that DNN does not have the assumption of gaussian distribution or any other distribution of itself. It just knows what value to output when it sees an input. So how do you know the probability distribution of the DNN? For example, if someone asks, what is the probability of the point appearing in (5, 0). Can DNN answer this kind of questions?
For many regression algorithms, not only neural networks, the model is that the data is distributed by $y \sim \mathcal{N}(f(x;\theta), \sigma^2)$, where $\theta$ are the model parameters and $\sigma^2$ is the variance of the distribution (often a hyperparameter).
Maximizing the log-likelihood of the data with respect to $\theta$ is equivalent to minimizing the mean squared error loss between the $y_i$ and $f(x_i;\theta)$.
Therefore, to compute the probability density of $(5,0)$, you would just find the density of a gaussian with mean $f(5; \theta)$ and a variance of $\sigma^2$, where $f$ is your neural network.
• Yes, it still applies. The only assumption we need to apply MSE loss and obtain our PDF is that the distribution of the target $y$ is a function of the input $x$ plus some gaussian noise. That function $f$ doesn't have to be linear and in this case is modeled by a highly nonlinear neural network. – shimao Feb 2 '18 at 2:30 | 2019-08-17T13:36:29 | {
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https://www.physicsforums.com/threads/differential-equation-for-changing-mass-of-a-sphere.710132/ | # Differential equation for changing mass of a sphere
1. Sep 13, 2013
### Hypatio
The mass of a sphere with density as a function of radius is
$M=4\pi \int_0^r\rho(r) r^2dr$
Lets say the radius increases and decreases as a function of time t. So:
$M(t)=4\pi \int_{0}^{r(t)}\rho (r) r(t)^2dr$
I want to know the basic equation describing the mass added or removed from the sphere (mass increases when radius increases, mass decreases when radius decreases) as a function of t, starting from any t. The problem is I think I must use a differential form but I'm not sure what it looks like. What then is the differential form of dM(t)/dt? I think I must use a chain rule and write:
$\frac{dM}{dt}=\frac{dM}{dr}\frac{dr}{dt}$
is this right? How do I proceed to solve this with the integral?
2. Sep 13, 2013
### JJacquelin
Hi !
I am afraid that there is something wrong in your writting. It should be :
$M=4\pi \int_0^r\rho(u) u^2du$
Lets say the radius increases and decreases as a function of time t. So:
$M(t)=4\pi \int_{0}^{r(t)}\rho (u) u^2du$
You may use any other symbol than u, but not r.
3. Sep 13, 2013
### Hypatio
Ah yes, sure, the upper limit is the 'full' radius (r) and u is a radius. This doesn't solve my problem though.
4. Sep 13, 2013
### HallsofIvy
Staff Emeritus
Yes, the chain rule: $\dfrac{dM}{dt}= \dfrac{dM}{dr}\dfrac{dr}{dt}$
JJaquelines point helps make sense of the dM/dr.
To find $\dfrac{dM}{dt}$ use the "fundamental theorem of Calculus":
$$\frac{d}{dr}\int_0^r \rho(u)u^2 du= \rho(r)r^2$$
5. Sep 13, 2013
### Hypatio
I guess the solution then is
$\frac{dM}{dt}=\frac{4}{3}\pi \rho(r)r^3\frac{dr}{dt}$
does the solution change if $\rho(r)$ becomes $\rho(r,t)$ or could I write
$\frac{dM}{dt}=\frac{4}{3}\pi \rho(r,t)r^3\frac{dr}{dt}$
Thanks.
Last edited: Sep 13, 2013
6. Sep 13, 2013
### JJacquelin
I don't think so. A term involving the partial derivatine of rho relatively to t is missing into your last equation,
7. Sep 13, 2013
### Hypatio
Is there a rule I can apply to get this additional term?
8. Sep 13, 2013
### pasmith
If you have
$$F(t) = \int_0^{a(t)} g(r,t)\,\mathrm{d}r$$
then
$$F'(t) = \int_0^{a(t)} \left.\frac{\partial g}{\partial t}\right|_{(r,t)}\,\mathrm{d}r + a'(t)g(a(t),t)$$
assuming $g$ is sufficiently smooth.
9. Sep 13, 2013
### arildno
Please do not mix together dummy variables in the integrand with integral limits.
Properly speaking, you have the following the relation:
$$M(t)=4\pi\int_{0}^{r(t)}\rho(x)x^{2}dx$$
Thus, you have:
$$\frac{dM}{dt}=4\pi\rho(r(t))r(t)^{2}\frac{dr}{dt}$$
which has as interpretation that only the outermost spherical shell at r(t) determines the total change of mass.
Every compact ball strictly contained within the outermost shell (radii less than r(t)) remains constant in mass.
However:
Suppose you have a ball where at different times, the density at some fixed radius "x" may change as a function of time. Then, you have:
$$M(t)=4\pi\int_{0}^{r(t)}\rho(x,t)x^{2}dx$$
In this case, the total mass of the ball will be due to two distinct effects:
1. The ball shrinks or expands. This gives the contribution given above.
2. The interior of the ball may change its mass. This effect is new.
In sum, you'll then get:
$$\frac{dM}{dt}=4\pi\rho(r(t),t)r(t)^{2}\frac{dr}{dt}+4\pi\int_{0}^{r(t)}\frac{\partial\rho}{\partial{t}}x^{2}dx$$
Last edited: Sep 13, 2013
10. Sep 14, 2013
### JJacquelin
The general formula below shows the rule for derivation :
#### Attached Files:
• ###### Derivation rule.JPG
File size:
6 KB
Views:
80
11. Sep 14, 2013
### JJacquelin
Of course, the chain rule continues to applies !!!
The formula given above is the application of the chain rule in case of an integral with the integrand and limits which are functions of the variable considered for derivation. | 2017-08-18T10:52:21 | {
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https://math.stackexchange.com/questions/1055758/probability-of-no-king-queen-or-jack-before-the-first-ace-occurs | # Probability of no king, queen or jack before the first ace occurs?
A deck of cards is shuffled well. The cards are dealt one by one, until the first time an Ace appears. Find the probability that no kings, queens, or jacks appear before the first ace. (Introduction to Probability, p. 36)
My solution:
• Assume $k^{th}$ card is the first ace
• The possible number of hand before the first ace is then $\binom{48}{k-1}$, and the possible number of hands without a king, queen or jack before the first ace is $\binom{36}{k-1}$, so the probability is $\frac{\binom{36}{k-1}}{\binom{48}{k-1}}$.
• The probability that the first ace occurs at the $k^{th}$ position is $\frac{1}{52!} \binom{48}{k-1}(k-1)! \binom{4}{1}(52-k)!$, because there are $52!$ possible ordered decks, $\binom{48}{k-1}(k-1)!$ is the number of possibilities withouth an ace in the first $k$ cards, $\binom{4}{1}$ possibilities to draw an ace at the $k^{th}$ position and $(52-k)!$ possibilities to arrange the remaining cards in order.
• $\sum_{k=1}^{k=37} \frac{\binom{36}{k-1}}{\binom{48}{k-1}} \frac{1}{52!} \binom{48}{k-1}(k-1)! \binom{4}{1}(52-k)! = 0.25$
The computation of the above in R:
pr <- numeric(37)
for(k in 1:37){
pr_ace_at_k <- 1/factorial(52)*choose(48,k-1)*factorial(k-1) * choose(4,1)*factorial(52-k)
pr_no_before_k <- choose(36,k-1)/choose(48,k-1)
pr[k] <- pr_no_before_k * pr_ace_at_k
}
sum(pr)
> 0.25
However, a simulation yields:
deck <- c(rep(1:4, 4), rep(5, 52-4*4))
out <- replicate(1e6,{
deck <- sample(deck) # shuffle deck
k <- which(deck == 4)[1] # get index of first ace
all(deck[1:(k-1)] == 5) # check if only rest occured before kth card
})
mean(out)
> 0.173099
Is there anything wrong in my calculation? A result of 0.25 looks somehow persuasive.
EDIT
Finally, I found the nasty bug in the code. Problem was the special case when the ace occured at the first position. Corrected code would be
deck <- c(rep(1:4, 4), rep(5, 52-4*4))
out <- replicate(1e6,{
deck <- sample(deck) # shuffle deck
k <- which(deck == 4)[1] # get index of first ace
if(k == 1) TRUE
else all(deck[1:(k-1)] == 5)
})
mean(out)
> 0.249877
• It's the probability that among $16$ particular cards, the first to show up is an ace. We have equally likely outcomes here... – David Mitra Dec 7 '14 at 15:05
• Call Ace, King, Queen, and Jack the picture cards. You are asking for the probability that the first picture card is an Ace. But all picture cards are equally likely. Therefore, by symmetry, the probability is...? – TonyK Dec 7 '14 at 15:09
• The issue is why the simulation is so grossly wrong. The usual answer is programming error. – André Nicolas Dec 7 '14 at 15:19
• The simulation result is clearly an anomaly. I don't see any error in the R code, but maybe someone on stackoverflow can. – David K Dec 7 '14 at 15:25
• The simulation probability is likely either 4/23 or 9/52 – Empy2 Dec 7 '14 at 16:22
The easiest way to think about it is to ignore all the other cards in the deck. Now you stack up the $16$ cards of interest. What is the chance the top one is an ace? There are $4$ aces among the $16$, so $\frac 4{16}=\frac 14$
• That makes perfect sense, good explanation and intuition! – NoBackingDown Dec 7 '14 at 15:23
Although it is not the simplest approach, we can use recursive conditioning, which is a useful approach in a lot of somewhat more complicated problems.
To see the intuition, let's check the cases in words. Suppose the first card is a queen/jack/king; then you stop and it doesn't count. Suppose it's an ace; then you stop and it counts. Suppose it's neither; then you run the experiment again with 1 fewer card.
Mathematically we'll say that the desired event when starting with a $k$ card deck is $A_k$. Then we have
$$P(A_k)=P(A_k|\text{ first card is an ace })P(\text{ first card is an ace }) \\ +P(A_k|\text{ first card is queen/jack/king })P(\text{ first card is queen/jack/king })\\ +P(A_k|\text{ first card is neither })P(\text{ first card is neither }) \\ = 1 \cdot \frac{4}{k} + 0 + \frac{k-16}{k} P(A_{k-1})$$
Now take $k=52$ and solve the recurrence relation. Here the base case can be read off from the recurrence itself: we get $P(A_{16})=4/16+0+0=1/4$ for free. Here solving the recurrence proves particularly simple, because if $P(A_{k-1})=1/4$ then $P(A_k)=1/4$, since
$$\frac{4}{k} + \frac{k-16}{k} \frac{1}{4} = \frac{4}{k} + \frac{1}{4} - \frac{4}{k} = \frac{1}{4}.$$
This is an algebraic manifestation of the intuitive observation that we can "ignore all the other cards".
Total probability x Probability of only 1st drawing = Probability of event occuring
Total probability x (1 - Probability of another drawing) = Probability of event occuring on 1st drawing
Total probability = Probability of event occuring on 1st drawing / (1 - Probability of another drawing) = $\frac{\frac1{13}}{1-\frac9{13}}=\frac14$
• Could you elaborate on your answer? I don't understand how you arrive at your equation. – NoBackingDown Dec 7 '14 at 15:25
• More rigorous than RossMillikan's answer? How so? – Did Dec 7 '14 at 15:35
• I don't think this quite works, because the replacement means that the process is not simply being "restarted" upon drawing another card. My answer does essentially what you're attempting to do, I think. That is, you have that $a=1/13 + 9/13 b$ where $b$ is the probability for the corresponding event with a deck containing 12 queen/jack/king, 4 aces, and 51-16=35 other cards. – Ian Dec 7 '14 at 15:39
• 1. Hope my answer is clear enough now. 2. RossMillikian's answer is sufficient, I just used a formula instead of intuition. – ghosts_in_the_code Dec 7 '14 at 16:25
• I don't think this works still: you have that the total probability is the probability of the event occurring given that the process stops on the first draw, times the probability that the process stops on the first draw, plus the probability of the event occurring given that the process stops on the second draw, times the probability that it stops on the second draw, etc. There is an independence assumption implicit in your formula which is correct but not justified. – Ian Dec 7 '14 at 17:30
One of these $4$ sorts of cards (aces, kings, queens, jacks) will appear as the first. Each sort (also the aces) has the same probability of doing so, hence with probability $\frac{1}{4}$. | 2019-08-18T06:44:24 | {
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https://math.stackexchange.com/questions/2647211/rudin-2-43-every-nonempty-perfect-set-in-mathbbrk-is-uncountable | Rudin 2.43 Every nonempty perfect set in $\mathbb{R}^k$ is uncountable
I am having trouble understanding Rudin's proof of this theorem, and I believe I have pinpointed the particular part of the proof that I am unable to follow. To begin with, here is the proof copied from Baby Rudin:
Let $P$ be a nonempty perfect set in $\mathbb{R}^k$. Then $P$ is uncountable.
Proof. Since $P$ has limit points, $P$ must be infinite. Suppose $P$ is countable and denote the points of $P$ by $x_1, x_2, x_3, \dots$. We shall construct a sequence $\{V_n\}$ of neighborhoods as follows.
Let $V_1$ be any neighborhood of $x_1$. If $V_1$ consists of all $y \in \mathbb{R}^k$ such that $|y - x_1| < r$ the closure $\overline{V_1}$ of $V_1$ is the set of all $y \in \mathbb{R}^k$ such that $|y - x_1| \leq r$.
Suppose $V_n$ has been constructed so that $V_n \cap P \neq \emptyset$. Since every point of $P$ is a limit point of $P$, there is a neighborhood $V_{n+1}$ such that (i) $\overline{V_{n+1}} \subset V_n$, (ii) $x_n \notin \overline{V_{n+1}}$, (iii) $V_{n+1} \cap P \neq \emptyset$. By (iii) $V_{n+1}$ satisfies our induction hypothesis and the construction can proceed.
Put $K_n = \overline{V_n} \cap P$. Since $\overline{V_n}$ is closed and bounded, $\overline{V_n}$ is compact. Since $x \notin K_{n+1}$ no point of $P$ lies in $\cap_1^\infty K_n$. Since $K_n \subset P$ this implies $\cap_1^\infty$ is empty. But each $K_n$ is nonempty by (iii) and $K_n \supset K_{n+1}$ by (i), this contradicts the corollary to theorem 2.36.
Okay so my confusion lies with this paragraph of the proof:
Suppose $V_n$ has been constructed so that $V_n \cap P \neq \emptyset$. Since every point of $P$ is a limit point of $P$, there is a neighborhood $V_{n+1}$ such that (i) $\overline{V_{n+1}} \subset V_n$, (ii) $x_n \notin \overline{V_{n+1}}$, (iii) $V_{n+1} \cap P \neq \emptyset$. By (iii) $V_{n+1}$ satisfies our induction hypothesis and the construction can proceed.
My confusion is as follows:
In defining $V_1$, it is clear that $V_1$ is a neighborhood of the point $x_1$. Is $V_{n+1}$ a neighborhood of the point $x_{n+1}$? For example, is $V_2$ a neighborhood of the point $x_2$? I don't think this can be the case because there is no guarantee that there is a neighborhood of $x_2$ is a subset of an arbitrary neighborhood of $x_1$. After all if $x_1 \neq x_2$ and we can choose any neighborhood of $x_1$ in defining $V_1$, it is certainly possible $x_2 \notin V_1$. So if $V_{n+1}$ is not a neighborhood of $x_{n+1}$, what point is $V_{n+1}$ a neighborhood of?
• Possible duplicate of Problems in Theorem 2.43 of baby Rudin Feb 12, 2018 at 10:14
• @JoséCarlosSantos I don't think this is a duplicate: although both this and the linked question are about Theorem 2.43, this question asks if $x_{n+1} \in V_{n+1}$, while the linked question asks to show that $V_{n+1}$ exists.
– user88319
Feb 12, 2018 at 20:42
• @Strants You are right. I have retracted my closing vote. Feb 12, 2018 at 20:58
Note that nowhere in the proof is used the fact that $x_n$ should lie in $V_n$. The only point is that $V_n$ intersects $P$, so it is a neighbourhood of at least one point in $P$, no matter which one.
The fact that $x_1\in V_1$ may be confusing, but that's just to initiate the sequence.
Note: to construct $V_{n+1}$ simply select any open ball centered around a point in $V_n\cap P$ that is not $x_n$, with radius small enough that the closed ball lies entirely in $V_n$. All requirements are trivially satisfied.
• Perhaps it is presumed $x_n \in V_n$ for all n. Feb 12, 2018 at 10:19
• @WilliamElliot No it's not, you don't need that at all. All you need is for $V_n$ to not contain any of the $n$ first $x_n$, so that the intersection in the end is empty. Feb 12, 2018 at 10:21
It is not assumed that $$x_n \in V_n$$ (it may or may not be). Regardless of whether $$x_n \in V_n$$ or $$x_n \not \in V_n$$, we want to construct $$V_{n+1}$$ such that $$x_n \not \in V_{n+1}$$, and I do so here: Proof of Baby Rudin Theorem 2.43 | 2022-05-24T18:08:54 | {
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https://math.stackexchange.com/questions/2186169/find-the-determinant-after-a-certain-row-operation-is-applied-to-a-matrix-with-k | # Find the determinant after a certain row operation is applied to a matrix with known determinant
$A = \begin{bmatrix} --a--\\ --b--\\ --c-- \end{bmatrix}$
$A$ is a $3\times3$ matrix. The rows are all different, and the first row is called $a$. The second row is called $b$. The third row is called $c$. I'm not saying that all the values in the first row are the same, I am simply saying that the first row is called $a$. The determinant is $not$ $0$.
If the $\det(A)=3$, what is the determinant of:
$\begin{bmatrix} --a+b--\\ --b+c--\\ --c+a-- \end{bmatrix}$
I thought about this, but I recalled that whenever you have a matrix, and you add a row multiple of another, the determinant does not change. However in this case, the answer in the back of my textbook is $6$ and I don't understand how?
• And... what matrix is $\;A\;$ itself ? – DonAntonio Mar 14 '17 at 11:50
• It doesn't matter – K Split X Mar 14 '17 at 11:52
• A is a $3\times 3$ matrix – K Split X Mar 14 '17 at 11:53
• @K Either you don't understand the question, or you're kidding...or I don't understand the question: how come it isn't relevant?! – DonAntonio Mar 14 '17 at 11:53
• I think, K, that you had better write out the entire question as you have found it, not leaving out the least little bit, since what you have written so far is nonsense. – Gerry Myerson Mar 14 '17 at 11:58
$$\begin{bmatrix}1 & 1 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 1\end{bmatrix}\begin{bmatrix}a \\ b \\ c\end{bmatrix} = \begin{bmatrix}a+b \\ b+c \\ c+a\end{bmatrix}$$ Therefore, $$\begin{vmatrix}a+b \\ b+c \\ c+a\end{vmatrix} = \begin{vmatrix}1 & 1 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 1\end{vmatrix}\begin{vmatrix}a \\ b \\ c\end{vmatrix} = 2 \times 3 = 6$$
• Where did the get the $\times 2$ from? – K Split X Mar 14 '17 at 12:26
• @KSplitX $\begin{vmatrix}1 & 1 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 1\end{vmatrix} = 2$ – TenaliRaman Mar 14 '17 at 12:28
• Determinant is 2. Okay thanks – K Split X Mar 14 '17 at 12:30
Lets do it with rows instead of columns and remember that the determinant is multilinear:
\begin{align} \det(a+b,b+c,c+a) = \quad &\det(a,b,c)+\det(a,b,a)\\ +&\det(a,c,c)+ \det(a,c,a)\\ +&\det(b,b,c)+ \det(b,b,a)\\ +&\det(b,c,c)+ \det(b,c,a). \end{align}
All terms except the first and the last vanish because there is the same entry twice. And because rearranging the rows does not change the determinant you will find
$$\det(a+b,b+c,c+a)=\det(a,b,c)+\det(b,c,a)=2\det(a,b,c).$$
In your case $\det(A)=3$, so $2\cdot\det(A)=6$ is the result.
Well, with the explantion you added (and the one you also deleted...) it is finally clearer what you meant, but then, since the determinant of a matrix with two identical rows is zero, and if we intechange two rows we multiply the determinant by $\;-1\;$ , by multilinearity we get:$${}$$
$$\det\begin{pmatrix}-a+b-\\-b+c-\\-c+a-\end{pmatrix}=\color{red}{\det\begin{pmatrix}-a-\\-b-\\-c-\end{pmatrix}}+\underbrace{\det\begin{pmatrix}-a-\\-b-\\-a-\end{pmatrix}+\det\begin{pmatrix}-a-\\-c-\\-c-\end{pmatrix}+\ldots}_{=0}+\color{red}{\det\begin{pmatrix}-b-\\-c\\-a-\end{pmatrix}}$$$${}$$
$$=\color{red}{\det A}+\color{red}{\det A}=6$$
I will use slightly different notation, write
$$A = \begin{bmatrix} v_{1}\\ v_{2}\\ v_{3}\end{bmatrix}$$ where $v_{i}$ are the rows of $A$ and $\operatorname{det}(A) = 3.$
Then
$$\operatorname{det}\begin{bmatrix} v_{1} + v_{2}\\ v_{2} + v_{3}\\ v_{3} + v_{1} \end{bmatrix} = \operatorname{det} \left(\begin{bmatrix} v_{1}\\ v_{2} \\v_{3} \end{bmatrix} + \begin{bmatrix} 0 &1 &0\\ 0 & 0 & 1\\ 1& 0& 0\end{bmatrix}\begin{bmatrix} v_{1}\\ v_{2}\\v_{3}\end{bmatrix}\right) = \operatorname{det}\left(A + \begin{bmatrix} 0 &1 &0\\ 0 & 0 & 1\\ 1& 0& 0\end{bmatrix}A \right)$$ So, $$\operatorname{det}(A)\operatorname{det}\left(I+\begin{bmatrix} 0 &1 &0\\ 0 & 0 & 1\\ 1& 0& 0\end{bmatrix}\right) = 6$$
• Yup another answer just did this! Thanks for the indepth version tho – K Split X Mar 14 '17 at 12:32
• How did you arrive at the last step from the third step? Where did the $\det(A)$ on the outside come from? – K Split X Mar 14 '17 at 12:34
• Split the determinant over multiplication – JessicaK Mar 14 '17 at 12:36 | 2019-05-23T09:49:51 | {
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http://ronk.gattopescatore.it/permutation-with-repetition-algorithm.html | # Permutation With Repetition Algorithm
As understood by name "combinations" means all the possible subsets or arrangements of the iterator and the word "combinations_with_replacement" means all the possible arrangements or subsets that allow an element to repeat in a subset. ” Discrete Mathematics 309, no. The paradigm problem. Covers permutations with repetitions. i=2 2 where ad is the number of augmented doubles, and r[i] is the exact repetition count at the i-th level. Permutations. Then it checks for the repetition C++ Language Using the main() Function. We have avoided using STL algorithms as main purpose of these problems are to improve your coding skills and using in-built algorithms will do no good. It is rather a combinatorial problem that does not involve any algorithm. So we have the following algorithm: Define function permutations(i) returns all permutations using array[i] to array[n] (n is the total number arrays). Here's an implementation. Find all possible combinations with sum K from a given number N(1 to N) with the repetition of numbers is allowed Objective: Given two integers N and K, Write an algorithm to find possible combinations that add to K, from the numbers 1 to N. For example, the permutation σ = 23154 has three inversions: (1,3), (2,3), (4,5), for the pairs of entries (2,1), (3,1), (5,4). Dynamic Programming Algorithms Dynamic Programming Algorithm is an algorithm technique used primarily for optimizing problems, where we wish to find the “best” way of doing something. A 6-letter word has 6! =6*5*4*3*2*1=720 different permutations. Order matters. No Repetition: for example the first three people in a running race. If letter box A must contain at least 2 letters. This algorithms check for duplication and repetition of the randomize question. Background. Algorithm for Permutation of a String in Java. Circular permutations. permutation without repetition. One of the goals of RcppAlgos is to provide a comprehensive and accessible suite of functionality so that users can easily get to the heart of their problem. Calculates the number of permutations with repetition of n things taken r at a time. A permutation with repetition of n chosen elements is also known as an "n-tuple". The number of permutations on a set of n elements is given by n!, where “!” represents factorial. Zero factorial or 0! Ways to arrange colors. Then, applying ( 1. Recursion comes directly from Mathematics, where there are many examples of expressions written in terms of themselves. Please see below link for a solution that prints only distinct permutations even if there are. permn - permutations with repetition Using two input variables V and N, M = permn(V,N) returns all permutations of N elements taken from the vector V, with repetitions. Start studying Ch. For instance, “$$01110000$$” is a perfectly good bit string of length eight. The number is (n-1)! instead of the usual factorial n! since all cyclic permutations of objects are equivalent because the circle can be rotated. Step 2 - repeat step 1 with the remaining items. Wrapping this function in a generator allows us terminate a repeated generation on some condition, or explore a sub-set without needing to generate the whole set:. The main advantage of this single chromosome representation is — in analogy to the permutation scheme of the traveling salesman problem (TSP) — that. 4), that is, the number of ways to pick k things out of n and arrange the k things in order. Algorithm for Permutation of a String in Java. So order matters… AB is not the same as BA Slideshow 3109936 by onan. I have found lots of permutation algorithms - have even written a few but I cannot figure out how to do this. java https://github. About the Author Tim Hill is a statistician living in Boulder, Colorado. Combinations are arrangements of objects without regard to order and without repetition. I discussed the difference between permutations and combinations in my last post, today I want to talk about two kinds […] List permutations with repetition and how many to choose from Noel asks: Is there a way where i can predict all possible outcomes in excel in the below example. The C programs in this section which finds the frequency of the word ‘the’ in a given sentence, finds the number of times the substring occurs in the given string, to find the frequency of every word in a given string and to find the highest frequency character in a string. nPr represents n permutation r which is calculated as n!/(n-k)!. The range used is [first,last), which contains all the elements between first and last, including the element pointed by first but not the element pointed by last. The result is: 1,2 2,1. STL has a shuffling algorithm called random_shuffle you can use. Recursive Permutation Algorithm without Duplicate Result. A permutation is an arrangement of all or part of a set of objects, with regard to the order of the arrangement. permutation f to the low-order digits (See section 4. Permutations without Repetition In this case, we have to reduce the number of available choices each time. As the expected permutations are clearly not themselves permutations, our algorithms are not tools for finding assignments and are not competing with algorithms for finding an optimal assignment. 20) – patrickJMT; Repeated symbols example 1 and 2 – patrickJMT; Permutations with repetition (from 10. Order doesn’t matter. We have avoided using STL algorithms as main purpose of these problems are to improve your coding skills and using in-built algorithms will do no good. Permutations are denoted by the following which means the number of permutations of n items taken r items at a time. That is, it is a function from S to S for which every element occurs exactly once as an image value. gif 400 × 225; 82 KB. The Binomial Theorem 5. Euclid's algorithm and π. 1234 is followed by 2. From the 4th permutations. Taussig Dec 4 '17 at 11:47 $\begingroup$ It seems to be both, and more specifically in the case of a cartesian product, it seems to be a cartesian power. The algorithm is not as fast as the fast_permutation, but it is faster than the orginal one and than the pure python one available in test_itertools. The range used is [first,last), which contains all the elements between first and last, including the element pointed by first but not the element pointed by last. 1 Permutations and Patterns The fundamental objects of this work are permutations. For example; given 3 letters abc find solution: Remember that the repetition is allowed in permutations unlike in combinations;. Let V be a vector of the outcome values. Thus, the number of permutations becomes (r - 1) n-2 P r-2. Permutations with Repetition Theorem 1: The number of r-permutations of a set of n objects with repetition allowed is nr. Sedgewick (1977) summarizes a number of algorithms for generating permutations, and identifies the minimum change permutation algorithm of Heap (1963) to be generally the fastest (Skiena 1990, p. The algorithm has potential to further differentiate between contours with the same prime form. Permutation without Repetition: This method is used when we are asked to reduce 1 from the previous term for each time. How to find out the missing number? Let the numbers in the array be 1,2,4,6,3,7,8,10,9 (total 9 numbers without repetition). Permutation can be done in two ways, Permutation with repetition: This method is used when we are asked to make. The algorithm is designed to take a selection of cells (from the Selection object), which should be located in the top row with no data below. Compute the permutation and print the result. * Combinations 26/05/2016 COMBINE CSECT. At each node c, the algorithm checks whether c can be completed to a valid solution. permutation f to the low-order digits (See section 4. List all permutations with a condition. Permutation with repetition Posted 06 December 2010 - 08:14 AM Im trying to make a program that implements this but I cant seem to get past inserting the characters. 0, all other syntax methods except $(handler); are deprecated. Ways to pick officers. For example: permutations without repetitions of the three elements A, B, C by two are - AB, AC, BA, BC, CA, CB. Zero factorial or 0! Ways to arrange colors. ) and M will be of the same type as V. In this case, It is important to note that order counts in permutations. The main advantage of this code is that the all permutations are generated in logical order: all permutations that starts with the first element come first. We care about the order because 247 wouldn’t work. Please see below link for a solution that prints only distinct permutations even if there are. Introduction to Non-Repetitive Sequences Repetition is a part of life. The time complexity of this algorithm is "O(n)". STL has a shuffling algorithm called random_shuffle you can use. Number of combinations n=11, k=3 is 165 - calculation result using a combinatorial calculator. Permutations and partitions in the OEIS. Examples: Input: str = “aa” Output: aa Note that “aa” will be printed only once. Another definition of permutation is the number of such arrangements that are possible. A command-line program that uses the library is provided too, useful to teach combinatorics. Last Modified: 2013-12-14. Recursion is elegant but iteration is efficient. STL has a shuffling algorithm called random_shuffle you can use. A classical problem asks for the number of permutations that avoid a certain permutation pattern. Let me first re-write your specification: Print all permutations with repetition of characters. In C++: •Write a program that produces ten random permutations of the numbers 1 to 10. To setup repository with documentation. A k-permutation of a multiset M is a sequence of length k of elements of M in which each element appears a number of times less than or equal to its multiplicity in M (an element's repetition number). A permutation is an arrangement of all or part of a set of objects, with regard to the order of the arrangement. Next lexicographical permutation algorithm Introduction. Restricted permutations are those constrained by having to avoid subsequences ordered in various prescribed ways. Here is one such algorithm, which generates the permutations in Lexicographical order. , TS, ACO, and GSA, are transformed into RPD measure where Minsol is the optimal solution if the given instance is solved to optimality or the lowest TCT obtained by any of models or algorithms. The elements might be of a string, or a list, or any other data type. 1 Permutations and Patterns The fundamental objects of this work are permutations. The idea is to generate each permutation from the previous permutation by choosing a pair of elements to interchange, without disturbing the other n-2 elements. Proceedings of the third international conference on Genetic Algorithms. 1983-01-01. A permutation relates to the order in which we choose the elements.$\begingroup\$ What you are describing is a permutation with repetition. The algorithm used for generating k-permutations was developed specifically for ECOS. The current theory would call three contours with the same prime form equally similar, without regard for further differences illustrated by the specific stages of the algorithm. Backtracking is a general algorithm for finding all enumerate all possible permutations using all items from the set without repetition. Returns true if such a "next permutation" exists; otherwise transforms the range into the lexicographically first permutation (as if by std::sort(first, last, comp)) and returns false. This is the aptitude questions and answers section on "Permutation and Combination" with explanation for various interview, competitive examination and entrance test. What the expected permutation matrices show very well is the potential for uncertainty for a true match. 1 Endorsement. Uses a precomputed lookup table of size n! containing the information of all transitions. If letter box A must contain at least 2 letters. A permutation is an act of arranging the elements of a set in all possible ways. Step 2 - repeat step 1 with the remaining items. This is the most well-known historically of the permutation algorithms. The CD that accompanies this book includes MySQL 4. java solves the 8 queens problem by implicitly enumeration all n! permutations (instead of the n^n placements). I adapted the code above to do permutations in Excel VBA. Another permutation algorithm in C, this time using recursion. A second multiple access system based on random permutations was studied. Last Modified: 2013-12-14. As these algorithms are very similar, it is suggested that a student tries to learn both to avoid unnecessary repetition (especially if they want to be able to solve the cube quickly). More precisely, we deal with a special version of the Black-Peg game with n holes and k >= n colors where no repetition of colors is allowed. Input: The first line of input contains an integer T, denoting the number of test cases. The basic difference between permutation and combination is of order Permutation is basically called as a arrangement. For example, a triple is interpreted as three doubles; the augmentation from 3-reps to 2-reps is (3 C 2) or 3. The idea is to swap each of the remaining characters in the string. Permutation With Repetition Algorithm Sometimes an inversion is defined as the pair of values. The visited array keeps track of which nodes have been visited already. However, we need to keep tracking of the solution that has also been in the permutation result using a hash set. Let S be a multiset that consists of n objects of which n1 are of type 1 and indistinguishable from each other. Two permutations with repetition are equal only when the same elements are at the same locations. Recursion means "defining a problem in terms of itself". Whether or not it actually is quicker is difficult to tell; I get the answer in 150 milliseconds, which is about twice as long as it took you. 1 − ǫ, an algorithm due to Charikar, Makarychev and Makarychev [CMM06] can find an assignment with value 1−O(√ ǫc). We care about the order because 247 wouldn’t work. Backtracking is a general algorithm for finding all enumerate all possible permutations using all items from the set without repetition. Permutations with repetition by treating the elements as an ordered set, and writing a function from a zero-based index to the nth permutation. The results can be use for studying, researching or any other purposes. Let me first re-write your specification: Print all permutations with repetition of characters. , TS, ACO, and GSA, are transformed into RPD measure where Minsol is the optimal solution if the given instance is solved to optimality or the lowest TCT obtained by any of models or algorithms. java solves the 8 queens problem by implicitly enumeration all n! permutations (instead of the n^n placements). However, in many applied settings where a string is an appropriate model, a symbol may be used in at most one position. Generate random number without repetition android. Note that the permutation. (It slows down the algorithm in software) * The Feistel itself works on 64 bits! * An S-Box is a basic component which performs (non-linear!) substitution to implement a block cipher in symmetric key algorithms – gives confusion (see later). Objective: Given an array of integers (in particular order or permutation of a set of numbers), write an algorithm to find the lexicographically next permutation of the given permutation with only one swap. Algorithms for Generating Permutations and Combinations Section 6. Free essays, homework help, flashcards, research papers, book reports, term papers, history, science, politics. Practice makes perfect and repeating is good practice. My current code, for S = 5, has to check around 8000 possible lists. 2008-07-25 at 23:08. 2 Problem 47ES. However, in many applied settings where a string is an appropriate model, a symbol may be used in at most one position. Permutations with Repetition Theorem 1: The number of r-permutations of a set of n objects with repetition allowed is nr. Calculates count of combinations without repetition or combination number. It can be used to perform arbitrary permutation (without repetition) of n subwords within log n cycles regardless of the subword size. The visited array keeps track of which nodes have been visited already. A logistic map is used to generate a bit sequence, which is used to generate pseudorandom numbers in Tompkins-Paige algorithm, in 2D. The CD that accompanies this book includes MySQL 4. The permutation of a number of objects is the number of different ways they can be ordered: the position is important. Please see below link for a solution that prints only distinct permutations even if there are. 4567 is followed by. The number of permutations on a set of n elements is given by n!, where “!” represents factorial. Permutation vs. * * Inputs: unsigned short prn_seed - the initial value for the PRN generator * int permutation_const - value for. the first call to the recursive function will attempt to find permutations for 1 and 2. Use this idea to. The information bits are coded by a repetition code (shaping filter) and then interleaved by a random permutation. The following algorithm will generate all permutations of elements of a set, in lexicographic order: procedure all_permutations(S) if length(S) == 1 return the element as a length-one permutation else all_perm = [] for each x in S. 1983-01-01. In particular: Theorem 8 GI ∈ PZK. API reference with usage examples available here. Permutations are denoted by the following which means the number of permutations of n items taken r items at a time. Permutation with repetition Calculator - High accuracy calculation Welcome, Guest. This document serves as an overview for attacking common combinatorial problems in R. First of all, while developing the algorithm, I asked my whole family and my neighbor (a judge) for help with the algorithm; no one could get even close. In particular: 1) What is the "type" of a permutation? 2) Do you want the number of all permutations, or do you want to list them? Your algorithm does neither. MArio http://www. The results can be use for studying, researching or any other purposes. We will calculate the letter count of B in a hashmap. The Hypothetical Scenario Generator for Fault-tolerant Diagnostics (HSG) is an algorithm being developed in conjunction with other components of artificial- intelligence systems for automated diagnosis and prognosis of faults in spacecraft, aircraft, and other complex. The idea is to generate each permutation from the previous permutation by choosing a pair of elements to interchange, without disturbing the other n-2 elements. A permutation cycle is a subset of a permutation whose elements trade places with one another. (a permutation can easily be encoded as an int). Algorithm and System Analysis multiset, sequence, word, permutation, k-set, k-list, k-multiset, k-lists with repetition, rule of product, Cartesian product. There are basically two types of permutation: Repetition is Allowed: such as the lock above. There are several algorithms for enumerating all permutations; one example is the following recursive algorithm: If the list contains a single element, then return the single element. The message is not registered. Priority Queue (Heap) –. The word "permutation" also refers to the act or process of changing the linear order of an ordered set. b: the telephone number must be a multiple 0f 10 c: the telephone number must be a multiple of 100 d: the 1st 3 digits are 481 e: no repetition are allowed. com FREE SHIPPING on x diagrams very useful in solving problems involving combinations with repetition and I found myself using them to help understand most of the problems in the last chapter. npm run test:algo only runs tests for the finished permutation algorithms, excluding utilities. 次のように入力されたコード1,2,3を繰り返し生成するFortranでコードを書きます。 111 112 113 121 122 123. the number of permutations, for a given number (this is classically known as ‘the travelling salesman’ problem):. This number of permutations is huge. Given a string of length n, print all permutation of the given string. Permutation with repetition Calculator - High accuracy calculation Welcome, Guest. Ways to pick officers. For each number, we add it to the results of permutations(i+1). Instructions to install MySQL and MySQL Connector J. Given a string S. An estimation of minimum distance for proposed codes is obtained. We prove a parallel repetition theorem for general games with value tending to 0. I'm stuck with nested for loops that are dependent on the previous loop. Now this is exactly the combinatorial problem of 5 selections from 10 choises with repetition. In the permutation without repetition section, the same dog will show up in many of those 2730 ways to order them, but the same ordered set will not show up twice. Permutation with repetition. Background. The results can be use for studying, researching or any other purposes. The idea is to fix the first character at first index and recursively call for other subsequent indexes. Permutations without Repetition. 3 Prim’s Algorithm. But it is not repetition of a single element that produces this outcome. From the above example of 8 letters, we have the following observation. See full list on dev. If we solve this problem using naive algorithm then time complexity would be exponential but we can reduce this to O(n * k) using dynamic programming. In this version of quicksort used middle element of array for the pivot. For example: permutations with repetitions of the three elements A, B, C by two are - AA, AB, AC, BA, BB, BC, CA, CB, CC. permutation (PRP), meaning that as long as the key is secret, First, a block cipher used in practice isn’t a gigantic algorithm but a repetition of rounds,. 2 Permutations ¶ permalink. Combinations with Repetition 07. The number of r-combinations from a set with n elements when repetition of elements is allowed is C(n + r - 1, r) = C(n + r - 1, n - 1) Permutations with Indistinguishable objects Theorem. 6: Combinations with Repetition Eg: Counting Iterations of a Loop How many times will the innermost loop be iterated when the algorithm segment below is implemented and run? (Assume n is a positive integer. If it cannot, the whole sub-tree rooted at c is skipped (pruned). It can be used to perform arbitrary permutation (without repetition) of n subwords within log n cycles regardless of the subword size. Cicirello VA, Cernera R (2013) Profiling the distance characteristics of mutation operators for permutation-based genetic algorithms. We wish to show that the efficiency of GAs in solving a flowshop problem can be improved significantly by tailoring the various GA operators to suit the structure of the problem. For maximum compatibility, this program uses only the basic instruction set (S/360) and two ASSIST macros (XDECO, XPRNT) to keep the code as short as possible. The only pair of 3-edges that can feature the same permutation with repetition are 123xyz --> 456xyz231 3-edges. To implement the clustered permutation test, assume there are two treatment groups with unknown outcome distributions F and G, with M and N clusters, respectively. What is the best way to do so? The naive way would be to take a top-down, recursive approach. Permutation vs. The arrangements are allowed to reuse elements, e. The algorithm has potential to further differentiate between contours with the same prime form. nPr represents n permutation r which is calculated as n!/(n-k)!. Combinations with repetition, and counting monomials. Combinations with Repetition. The permutations for which dp(w) = ℓ (w) are characterized directly; we also have a bijec tion with Dyck paths to help make the characterization more intuitiv e. Given an array of 9 numbers that contains numbers between 1 to 10, obviously no repetition, one number missing. A permutation cycle is a subset of a permutation whose elements trade places with one another. Circular permutations. Namely, our algorithm is: repeat I times. The two key formulas are:. “an ordered combination" w/ repetition: n^r n = total choices you have (ei: you have ten golf balls) r = how many times you choose (ei: you pick it out three) ex// 10^3 = 1,000 ways possible; also known as another way: "ei: three bread, two pickles, three dimes” you want to find total combo? 3x2x3 = 18 ways. List permutations with repetition and how many to choose from. Solved examples with detailed answer description, explanation are given and it would be easy to understand - Page 3. We care about the order because 247 wouldn’t work. The idea is to swap each of the remaining characters in the string. The algorithm used for generating permutations by transpositions is often called the "Johnson-Trotter" algorithm, but it was discussed earlier in the works of Steinhaus and the some campanologists. We have moved all content for this concept to for better organization. Combinatorics Processing. Fig 5:Mix columns V. (8*6 ==> 8*4) 2 bits used to select amongst 4 substitutions for the rest of the 4-bit quantity S-Box Examples DES Standard Cipher Iterative Action : Input: 64 bits Key: 48 bits Output: 64 bits Key Generation Box : Input: 56 bits Output: 48 bits DES Box Summary Simple, easy to implement: Hardware/gigabits/second, software/megabits/second 56-bit. Recursion means "defining a problem in terms of itself". We will typically view these objects in one-line notation, i. The idea is to fix the first character at first index and recursively call for other subsequent indexes. But like me, many others are looking for code/algorithm to generate combinations when we have repeated digits/numbers in a set. In this paper, we propose a variable block insertion heuristic (VBIH) algorithm to solve the permutation flow shop scheduling problem (PFSP). This can be a very powerful tool in writing algorithms. Any ordered arrangement such as C-B-F-A-D-G-H-E is called a permutation of the 8 letters. [permutations] [combinations] This lecture covers basic combinatorial algorithms which generate successively all permutations, combinations and variations respectively. If all the n characters are unique, you should get n! unique permutations. The test came back with one issue worth mentioning in this blog. a list where the permutation ˇ2S n is written ˇ= ˇ 1ˇ 2 ˇ n. I've been creating my code in vb. Any algorithm where the current permutation requires the previous one is discarded as threadable because we can’t start enumeration of permutations from any specific place (with known code). Lee}, title = {Fast Subword Permutation Instructions Based on Butterfly Networks}, booktitle = {In Proceedings of SPIE, Media Processor 2000}, year = {2000}, pages = {80--86}}. If letter box A must contain at least 2 letters. (3) Execute Davis-Putnam based on y and …, which takes at most n steps. RESULTS The following results for the permutation-based system are achieved using an implementation of the GALIB library [1], which had to be extended greatly to handle permutations with repetition. Now lemme, permutations. To implement the clustered permutation test, assume there are two treatment groups with unknown outcome distributions F and G, with M and N clusters, respectively. The idea is to fix the first character at first index and recursively call for other subsequent indexes. Zero factorial or 0! Ways to arrange colors. Objective: Given an array of integers (in particular order or permutation of a set of numbers), write an algorithm to find the lexicographically next permutation of the given permutation with only one swap. The number of r-combinations from a set with n elements when repetition of elements is allowed is C(n + r - 1, r) = C(n + r - 1, n - 1) Permutations with Indistinguishable objects Theorem. Given a string str, the task is to print all the permutations of str. with repetition \) Customer Voice. here i supply u a c++ code to generate variations. Write a program to print all permutations of a given string. Namely, our algorithm is: repeat I times. Instructions to install MySQL and MySQL Connector J. Please update your bookmarks accordingly. , involutions [12] and derangements [9]). One interesting application is the rearrangement of characters in a word to create other words. For maximum compatibility, this program uses only the basic instruction set (S/360) and two ASSIST macros (XDECO, XPRNT) to keep the code as short as possible. Permutation refers to the process of arranging all the members of a given set to form a sequence. Since permutations with repetition does not have. com Blogger 34 1 25 tag:blogger. Order matters. In mathematics, a combination is a selection of items from a collection, such that (unlike permutations) the order of selection does not matter. Mathematical Programming, 99(3):563--591, 2004. 2 Permutations. Permutation multiplication (or permutation composition) is perhaps the simplest of all algorithms in computer science. With next_combination() and next_permutation() from the STL algorithms, you can find permutations!! The formula for total number of permutations of r sequence picked from n sequence is n!/(n-r)! You can call next_combination() first and then next_permutation() iteratively. Purpose of use Needed to calculate a very large probability based on the Combination of 10,000,000 chemicals taken 500,000 at a time. Algorithm for the enumeration of permutations with finite repetition. I find it to be intuitive and easy to implement. A logistic map is used to generate a bit sequence, which is used to generate pseudorandom numbers in Tompkins-Paige algorithm, in 2D. This stage is fairly simple as only one algorithm is used to swap two edges around. Combinatorics Processing. The genetic algorithm uses permutations with repetition to encode chromosomes and a schedule generation scheme, termed OG&T, as decoding algorithm. In other words, it is the number of ways r things can be selected from a group of n things. The number of permutations of n distinct objects is n×(n − 1)×(n − 2)×⋯×2×1, which number is called "n factorial" and written "n!". Start studying Ch. number of things n 6digit 10digit 14digit 18digit 22digit 26digit 30digit 34digit 38digit 42digit 46digit 50digit. Hence if there is a repetition of elements in the array, the same permutation may occur twice. Number of combinations n=11, k=3 is 165 - calculation result using a combinatorial calculator. In this straight forward approach we create a list of lists containing the permutation of elements from each list. I only need to generate lists for small values of S, lets say up to 10-20. Mathematical algorithm that accurately predicts that, for many data sets, the first digit of each group of numbers in a random sample will begin with 1 more than a 2, a 2 more than a 3, a 3 more than a 4, and so on. Know the formula:. In order to sequence the tasks of a job shop problem (JSP) on a number of machines related to the technological machine order of jobs, a new representation technique — mathematically known as “permutation with repetition” is presented. A 6-letter word has 6! =6*5*4*3*2*1=720 different permutations. A numerical study of the plume in Cape Fear River Estuary and adjacent coastal ocean. Hence, by the product rule there are nrr-permutations with repetition. The algorithm might look like this (starting with an empty permutation): Repeat 'forever' (precisely: until a break): if the permutation isn't full yet (length less than n), append zeros (or whatever the minimum allowed value is); otherwise: add the permutation to results,. Additional there is a rule - whether you can choose the same option twice or not (Repetitions),and a comparison - whether the order of the single choices makes a difference or not (Respect Order). 20an open-source database management system. number of things n 6digit 10digit 14digit 18digit 22digit 26digit 30digit 34digit 38digit 42digit 46digit 50digit. Implement Binary Search Tree (BST) pre-order traversal (depth first). npm run test:algo only runs tests for the finished permutation algorithms, excluding utilities. In this post, we will see how to find permutations of a string containing all distinct characters. So there would be a substring of length of B in the string A which has exact same letter count as B. When the order does not matter and an object can be chosen more than once. This number of permutations is huge. As you can tell, 720 different "words" will take a long time to write out. The second stage involves the permutation of the top layer edges. The number of unique permutations possible is exactly 1, yet your algorithm (even with dictionary) will loop many many times just to produce that one result. Would you mind checking index 11 aabbaabbb is low by 1 and index 125 bbbbbaaaa is low by 5;. Permutations can thus be represented as a tree of permutations:. Groups of Permutations. Permutation with repetition Posted 06 December 2010 - 08:14 AM Im trying to make a program that implements this but I cant seem to get past inserting the characters. Refresh your memory! How many permutations, combinations and variations can be generated from set of N elements? And what about if repeated elements are allowed?. One of the main questions in this area is the parallel repetition question: Is there a way to decrease the. Hi! I have tried a bit, but I was not able to find a way to generate permutations with repetitions. Algorithm for the enumeration of permutations with finite repetition. Find all possible combinations with sum K from a given number N(1 to N) with the repetition of numbers is allowed Objective: Given two integers N and K, Write an algorithm to find possible combinations that add to K, from the numbers 1 to N. 48) Combinations with repetition. Dynamic Programming Algorithms Dynamic Programming Algorithm is an algorithm technique used primarily for optimizing problems, where we wish to find the “best” way of doing something. The permutation method differs from its combination comrade primarily in that arrangement does matter. Textbook solution for Discrete Mathematics With Applications 5th Edition EPP Chapter 9. The following algorithm will generate all permutations of elements of a set, in lexicographic order: procedure all_permutations(S) if length(S) == 1 return the element as a length-one permutation else all_perm = [] for each x in S. As mentioned in [2], for deriving a secure permutation g with a common domain, the domain of g would be 160 bits larger than that of f. Proceedings of the third international conference on Genetic Algorithms. We will calculate the letter count of B in a hashmap. The recursive function should generate all permutations for the first n-1 numbers. A compact and very fast way to check for duplicates is to use an unordered_set. Algorithm T: 'Plain change algorithm' as described in. Implement Binary Search Tree (BST) pre-order traversal (depth first). Possible implementation. The number of possible permutations with repetition of n elements by m equals. Generating combinations of k elements: Generating combinations of k elements from the given set follows similar algorithm used to generate all permutations, but since we don't want to repeat an a character even in a different order we have to force the recursive calls to not to follow the branches that repeat a set of. Nice algorithm without recursion borrowed from C. If there are twenty-five players on the team, there are $$25 \cdot 24 \cdot 23 \cdot \cdots \cdot 3 \cdot 2 \cdot 1$$ different permutations of the players. Similar to The Permutation Algorithm for Arrays using Recursion, we can do this recursively by swapping two elements at each position. « Prev - Affine Cipher Multiple Choice Questions and Answers (MCQs) » Next - P, NP, NP-hard, NP-complete Complexity Classes Multiple Choice Questions and Answers (MCQs). A permutation is a method to calculate the number of events occurring where order matters. Proof: There are n ways to select an element of the set for each of the r positions in the r-permutation when repetition is allowed. We will first take the first character from the String and permute with the remaining chars. For instance, “$$01110000$$” is a perfectly good bit string of length eight. Now the above code will print all the permutations of the string "GOD" without repetition. The paradigm problem. If no explicit formula could be given, I would already be satisfied with a more efficient algorithm to generate the lists. Permutation with repetition Calculator - High accuracy calculation Welcome, Guest. 6: Combinations with Repetition Eg: Counting Iterations of a Loop How many times will the innermost loop be iterated when the algorithm segment below is implemented and run? (Assume n is a positive integer. Algorithm for the enumeration of permutations with finite repetition. Then the nth number can be added into every position of the n-1 permutations to generate all permutations. post-5715079000043709685. Return all combinations. Distinct elements is the simplest case, and here we will also discuss the ramifications of employing the strongly concentrated hashing of Aamand et al. counting permutations with repetition covering countable Critical Path Method cubic graph cut vertex cycle cylindrical system deductive reasoning degree degree sequence denumerable depth-first algorithm derivative derived function Dijkstra's algorithm diameter of a graph difference of sets digraph dimension dimension analysis directed graph. Cape Fear River Estuary (CFRE), located in southeast North Carolina, is the only river estuary system in the state which is directly connected to the Atlantic Ocean. 10 shows a standard algorithm for computing kP (k is a positive integer) per every 2 bits as in the modular exponentiation operation. The permutation generator 300 receives, via a random number input 304, a random number which it stores in a buffer. The prover picks at random a permutation π and sends to the prover the graph G = (V,E) where E = π(E1)). com/tusharroy25 https://github. The permutation still has to be performed in order to be compliant with the DES standard. (2) Number of permutations. return a uniformly random permutation of the elements when the comparator is replaced by a fair coin flip (that is, return x < y = true with probability 1/2, regardless of the value of x and y) The code for the sorting algorithm must be the same. Permutation refers to the process of arranging all the members of a given set to form a sequence. python - compter efficacement les combinaisons et les permutations. To address this, what is typically done is, a dictionary is created that stores the frequency of each integer that is available at a given point. Groups of Permutations. If we solve this problem using naive algorithm then time complexity would be exponential but we can reduce this to O(n * k) using dynamic programming. Use this idea to. 2006-12-01. Combinations with Repetition 6. Permutations of 123: 123 132 213 231 312 321 Permutations of 1234: 1234 1243 1324 1342 1423 1432 2134 2143 2314 2341 2413 2431 3124 3142 3214 3241 3412 3421 4123 4132 4213 4231 4312 4321 From the above output, it’s very easy to find a correlation between the pattern of digits each of the whole number permutations has!!. Algorithm Complexity Analysis (Big O notation) – You are free to skip these parts and it shouldn’t affect the understanding of working of the algorithm. For , he ran the algorithm 1000 times and found 105 different families of nine mutually disjoint S-permutation matrices. The explanation is good and crisp and the code is easy and good for a novice. Recursion is elegant but iteration is efficient. The only pair of 3-edges that can feature the same permutation with repetition are 123xyz --> 456xyz231 3-edges. A description of an algorithm used to construct and test for additively non-repetitiveness will be provided, then results from research will be analyzed. Nice algorithm without recursion borrowed from C. [permutations] [combinations] This lecture covers basic combinatorial algorithms which generate successively all permutations, combinations and variations respectively. When some of those objects are identical, the situation is transformed into a problem about permutations with repetition. Given two graphs G1 = (V,E1),G2 = (V,E2), 1. Algorithm takes the input of the string. We will sometimes write ˇ(1)ˇ(2) ˇ(n) to. No Repetition: for example the first three people in a running race. In particular, a discrete Differential Evolution algorithm which directly works on the space of permutations with repetition is defined and analyzed. Permutation: Arrangement without repetition. We will calculate the letter count of B in a hashmap. java solves the 8 queens problem by implicitly enumeration all n! permutations (instead of the n^n placements). Genetic algorithms (GAs) are search heuristics used to solve global optimization problems in complex search spaces. Permutation can be done in two ways, Permutation with repetition: This method is used when we are asked to make different choices each time and have different objects. We will maintain 3 variables, existing letter positive count, existing letter negative count and non-existing letter. If we have a n-element set, the amount of its permutation is:. This question is from textbook : 1. (It slows down the algorithm in software) * The Feistel itself works on 64 bits! * An S-Box is a basic component which performs (non-linear!) substitution to implement a block cipher in symmetric key algorithms – gives confusion (see later). Lavavej on the Standard Template Library (STL) in C++. All videos were created by the students of EECS 203 - Discrete Mathematics at the University of Michigan in Winter 2012. Write a program to print all permutations of a given string. Forinstance, thecombinations of the letters a,b,c,d taken 3 at a time with repetition are: aaa, aab,. For an input string of size n, there will be n^n permutations with repetition allowed. Circular shift-It shifts each bit in an n-bit word K positions to the left. This is the currently. It is based on program Permutations. Similar to The Permutation Algorithm for Arrays using Recursion, we can do this recursively by swapping two elements at each position. Refresh your memory! How many permutations, combinations and variations can be generated from set of N elements? And what about if repeated elements are allowed?. Look for certain helpful keywords 2. Example 2 - Combinations. here the algo is: if n is the variety of quite a few element in a string. For example, a triple is interpreted as three doubles; the augmentation from 3-reps to 2-reps is (3 C 2) or 3. In general, a permutation is an ordered arrangement of a set of objects that are distinguishable from one another. Program Queens2. The algorithm has potential to further differentiate between contours with the same prime form. We define permutation as different ways of arranging some or all the members of a set in a specific order. List all permutations with a condition. how many 7-digit telephone numbers are possible if the first digit cannot be 0 and a: only odd digits may be used. General Terms: Algorithms. It is based on program Permutations. The algorithm might look like this (starting with an empty permutation): Repeat 'forever' (precisely: until a break): if the permutation isn't full yet (length less than n), append zeros (or whatever the minimum allowed value is); otherwise: add the permutation to results,. We will maintain 3 variables, existing letter positive count, existing letter negative count and non-existing letter. By using the same key produce several runs of permutations that be as secure as a hash, but completely reversible. AES algorithm using matlab VII. A combination with replacement is an unordered multiset that every element in it are also in the set of n elements. This implies that the answer length c(ǫ) in the Unique Games Conjecture must be larger than Ω(1/ǫ) if the conjecture is to hold. The algorithm described by AES is a symmetric-key algorithm (the same key is used for both encrypting and decrypting the data) and the design principle is known as a substitution-permutation network (SP-network or SPN) a series of mathematical operations used in cipher algorithms. Heap's algorithm is used to generate all permutations of n objects. Generating combinations of k elements: Generating combinations of k elements from the given set follows similar algorithm used to generate all permutations, but since we don't want to repeat an a character even in a different order we have to force the recursive calls to not to follow the branches that repeat a set of. Thus, the number of permutations becomes (r - 1) n-2 P r-2. Recursion is elegant but iteration is efficient. P(n, r) denotes the number of permutations of n objects taken r at a time. The paradigm problem. What the expected permutation matrices show very well is the potential for uncertainty for a true match. Different permutations can be ordered according to how they compare lexicographicaly to each other; The first such-sorted possible permutation (the one that would compare lexicographically smaller to all other permutations) is the one which has all its elements sorted in ascending order, and the largest has all its elements sorted in descending. 6 uses to generate the permutations and eliben's one looks like Johnson-Trotter permutation generation, you might look for article in Wikipedia on Permutations and their generation that looks quite like unrank function in paper by Myrvold and Ruskey. The number of ways to arrange n distinct objects along a fixed (i. Therefore, the number of permutations in this case = 10x10x10x10x10x10 = 1000000 Circular Permutation. The combination of Two Square Cipher and Variably Modified Permutation Composition (VMPC) algorithm is intended for obtaining stronger ciphers than using only one cipher, so it is not easy to solve. Thank you for your questionnaire. For example, say our function is given the numbers 1,2 and 3. Technical blog and complete tutorial on popular company interview questions with detailed solution and Java program on Data structure, Algorithms, Time and space complexity, Core Java, Advanced Java, Design pattern, Database, Recursion, Backtracking, Binary Tree, Linked list, Stack, Queue, String, Arrays etc asked in companies like Google, Amazon, Microsoft, Facebook, Apple etc. What is a permutation and what is a combination with repetition and no repetition? Permutation Groups Generated by 3-Cycles [05/14/2003] Show A_n contains every 3-cycle if n >= 3; show A_n is generated by 3- cycles for n >= 3; let r and s be fixed elements of {1, 2,, n} for n >= 3 and show that A_n is generated by the n 'special' 3-cycles of. Permutations without Repetition. We present a strategy that identifies the secret code in O(n log n) queries. For both combinations and permutations, you can consider the case in which you choose some of the n types more than once, which is called 'with repetition', or the case in which you choose each type only once, which is called 'no repetition'. The algorithm for obtaining random permutations, based on a randomized algorithm with a probability evaluation is equal to 1, which is more efficient than the other algorithm with probability evaluation p(n) = n! nn, is described in section 2. For sample the default for size is the number of items inferred from the first argument, so that sample(x) generates a random permutation of the elements of x (or 1:x). Permutations and Combinations. The two key formulas are:. We will typically view these objects in one-line notation, i. Previously Dinur and Steurer proved such a theorem for the special case of projection games. In statistics, the two each have very specific meanings. List all pair of permutations with repetition with given condition, conditions are elaborated below Relevant Equations: of S, lets say up to 10-20. Priority Queue (Heap) –. The permutations for which dp(w) = ℓ (w) are characterized directly; we also have a bijec tion with Dyck paths to help make the characterization more intuitiv e. Thank you for your questionnaire. The byte substitution transformation is a nonlinear. Permutation With Repetition Problems With Solutions : In this section, we will learn, how to solve problems on permutations using the problems with solutions given below. 2006-12-01. SM-2 is a simple spaced repetition algorithm. The notation supports the following high-level constructs: permutation, grouping, repetition, inversion, reflection, conjugation, commutation, rotation and single-line and multiple-line comments. Thus, we can use permutations(i + 1) to calculate permutations(i). 48) Combinations with repetition. Permutation and orientation changes of individual cube parts can be specified using permutation cycles. https://www. Permutations, combinations and the binomial theorem. It is efficient and useful as well and we now know enough to understand it pretty easily. Because order matters, we're finding the number of permutations of size 2 that can be taken from a set of size 3. The general Formula. A combination with replacement is an unordered multiset that every element in it are also in the set of n elements. permutation synonyms, permutation pronunciation, permutation translation, English dictionary definition of permutation. Permutation in a circle is called circular permutation. The message is not registered. It was written in Visual Studio 2013 using C# and DeflateStream class. Permutations of 123: 123 132 213 231 312 321 Permutations of 1234: 1234 1243 1324 1342 1423 1432 2134 2143 2314 2341 2413 2431 3124 3142 3214 3241 3412 3421 4123 4132 4213 4231 4312 4321 From the above output, it’s very easy to find a correlation between the pattern of digits each of the whole number permutations has!!. The proposed encryption system includes two major parts, chaotic pixels permutation and chaotic pixels substitution. Since permutations with repetition does not have. Technically, a permutation of a set S is defined as a bijection from S to itself. Counting Permutations with Fixed Points; Pythagorean Triples via Fibonacci Numbers. As an example, if the string is "abc" there are 6 permutations {abc, acb, bac, bca, cab, cba}. On the positive side, we might have hoped that we could use the parallel repetition. Learn vocabulary, terms, and more with flashcards, games, and other study tools. Any selection of r objects from A, where each object can be selected more than once, is called a combination of n objects taken r at a time with repetition. Algorithm for Permutation of a String in Java. In particular: Theorem 8 GI ∈ PZK. Mathematical algorithm that accurately predicts that, for many data sets, the first digit of each group of numbers in a random sample will begin with 1 more than a 2, a 2 more than a 3, a 3 more than a 4, and so on. Permutations 3. The study of permutations in this sense generally belongs to the field of combinatorics. the leftmost k bits become the original rightmost bits. e where the repetitions of the characters are included then read the matter below. This problem can also be asked as “Given a permutation of numbers you need to find the next larger permutation OR smallest …. return a uniformly random permutation of the elements when the comparator is replaced by a fair coin flip (that is, return x < y = true with probability 1/2, regardless of the value of x and y) The code for the sorting algorithm must be the same. In mathematics, a combination is a selection of items from a collection, such that (unlike permutations) the order of selection does not matter. The results can be use for studying, researching or any other purposes. Then it checks for the repetition C++ Language Using the main() Function. In order to sequence the tasks of a job shop problem (JSP) on a number of machines related to the technological machine order of jobs, a new representation technique — mathematically known as “permutation with repetition” is presented. Backtracking is a general algorithm for finding all enumerate all possible permutations using all items from the set without repetition. This is the currently. Knuth (volume 4, fascicle 2, 7. The number of permutations on a set of n elements is given by n!, where “!” represents factorial. com/tusharroy25 https://github. For a given string of size n, there will be n^k possible strings of length "length". The works in this exhibition play with the seemingly endless permutations of data to investigate the scale and scope of data as well as its elegance and anxieties. What about if we want to get all the possible permutations with repetition. In effect, all that's going on here is to exploit the sophisticated algorithms of a computer algebra system to keep track of all the possible combinations as each additional die is introduced. To practice all areas of Data Structures & Algorithms, here is complete set of 1000+ Multiple Choice Questions and Answers. An estimation of minimum distance for proposed codes is obtained. counting permutations with repetition covering countable Critical Path Method cubic graph cut vertex cycle cylindrical system deductive reasoning degree degree sequence denumerable depth-first algorithm derivative derived function Dijkstra's algorithm diameter of a graph difference of sets digraph dimension dimension analysis directed graph. Following is the illustration of generating all the permutations of n given numbers. About the Author Tim Hill is a statistician living in Boulder, Colorado. (3) Execute Davis-Putnam based on y and …, which takes at most n steps. From what I have found, a repeated permutation with repetition can only exist in an edge of 3 or greater. Wrapping this function in a generator allows us terminate a repeated generation on some condition, or explore a sub-set without needing to generate the whole set:. In particular: Theorem 8 GI ∈ PZK. The algorithm for finding scalar times of points on an elliptic curve is similar to an algorithm for modular exponentiation operation. For , he ran the algorithm 1000 times and found 105 different families of nine mutually disjoint S-permutation matrices. In statistics, the two each have very specific meanings. -Invertile Transformation. Possible three letter words. Knuth (volume 4, fascicle 2, 7. We will maintain 3 variables, existing letter positive count, existing letter negative count and non-existing letter. (spot, fido, max) is not the same set as (fido, spot, max). More precisely, we deal with a special version of the Black-Peg game with n holes and k >= n colors where no repetition of colors is allowed. This number of permutations is huge. 6: Combinations with Repetition Eg: Counting Iterations of a Loop How many times will the innermost loop be iterated when the algorithm segment below is implemented and run? (Assume n is a positive integer. That way, you will find all the permutations. A description of an algorithm used to construct and test for additively non-repetitiveness will be provided, then results from research will be analyzed. The Binomial Theorem 5. For example, given three fruits, say an apple, an orange and a pear, there are three combinations of two that can be drawn from this set: an apple and a pear; an apple and an orange; or a pear and an orange. This implies that the answer length c(ǫ) in the Unique Games Conjecture must be larger than Ω(1/ǫ) if the conjecture is to hold. there is no repetition. In order to sequence the tasks of a job shop problem (JSP) on a number of machines related to the technological machine order of jobs, a new representation technique — mathematically known as “permutation with repetition” is presented. Mathematical algorithm that accurately predicts that, for many data sets, the first digit of each group of numbers in a random sample will begin with 1 more than a 2, a 2 more than a 3, a 3 more than a 4, and so on. Classes have been defined according to whether order is important, items may be repeated, and length is specified. Inverted indexing is a ubiquitous technique used in retrieval systems including web search. This is not the case with fast_permutation. I have searched all the forms to try and find a solution for this. I'm stuck with nested for loops that are dependent on the previous loop. In the example, is , and is. Combinatorics. how many 7-digit telephone numbers are possible if the first digit cannot be 0 and a: only odd digits may be used. The Binomial Theorem 5. But repetition becomes boring. I adapted the code above to do permutations in Excel VBA. Algorithm for the enumeration of permutations with finite repetition. This is often written 3_P_2. It can be used to perform arbitrary permutation (without repetition) of n subwords within log n cycles regardless of the subword size. The permutation in a haystack problem and the calculus of search landscapes. here the algo is: if n is the variety of quite a few element in a string. with repetition \) Customer Voice. Combinations with repetition, and counting monomials. A simple algorithm to generate a permutation of n items uniformly at random without retries, known as the Knuth shuffle, is to start with the identity permutation, and then go through the positions 1 through n, and for each position i swap the element currently there with an arbitrarily chosen element from positions i through n, inclusive. API reference with usage examples available here. How to use iteration in a sentence. Each test case contains a single string S in capital letter. where the order matters (who holds the president office matters) and no repetition is allowed. In a 3-digit combination lock, each digit. If the list contains more than one element, loop through each element in the list, returning this element concatenated with all permutations of the remaining n. As the expected permutations are clearly not themselves permutations, our algorithms are not tools for finding assignments and are not competing with algorithms for finding an optimal assignment. | 2020-11-27T14:53:11 | {
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http://math.stackexchange.com/questions/179660/need-to-clarify-the-at-least-concept-in-combination | # Need to clarify the “At-least Concept” in Combination.
I managed to solve this question but I had some inquiries regarding the solution.
If two cards are chosen at random from a standard deck of playing cards, how many different ways are there to draw the two cards if at least one card is a jack, queen or a king?
Here is how I solved it:
Jack/Queen/King = $4 \times 3 = 12$ cards
Other Cards = $52-12 = 40$ cards
Now we can only pick two cards so:
• It can be either from the Jack/King/Queen so $\binom{12}{1}=12$
• Both cards can be from Jack/King/Queen so $\binom{12}{2} = 66$
• One card can be from the remaining stack (Non - jack,king or queen) so $\binom{40}{1}=40$
Now the only problem I have with this question is when getting the final value initially I was doing
$1$ from Remaining Cards $\times$ [ ($1$ from Jack/King/Queen) + 2(from Jack/King/Queen) ]= $\binom{40}{1} \times ( \binom{12}{1} + \binom{12}{2} )$
But The actual answer comes if we do the following
[$1$ from Remaining Cards $\times$ $1$ from Jack/King/Queen ] + 2(from Jack/King/Queen) ]= $( \binom{40}{1} \times \binom{12}{1}) + \binom{12}{2}$
I would really appreciate it if someone could clarify why do we do it the second way and not the first way ? Which part is added/multiplied to which part ? Is there an easier way to know how its done. Am I missing some important concept here ?
Edit : While trying to understand this I also looked up the definition of disjoint events which means "Two events are disjoint if they can't both happen at the same time" so then again here is what I did (Special here means Jack/king/Queen)
(1 from the 40 and 1 from special) or (1 from the 40 and 2 from special)
$(\binom{40}{1} \times \binom{12}{1}) + (\binom{40}{1} \times \binom{12}{2}) )$
Which simplifies to $\binom{40}{1} \times ( \binom{12}{1} + \binom{12}{2} )$
So is the representation of the problem using
(1 from the 40 and 1 from special) or (1 from the 40 and 2 from special) wrong ?
-
If someone tells you he have at least one card is a jack, queen or a king then this means that it is not possible that he doesn't have a king or a queen or a jack (example: if he does not have a jack or a queen he must be holding a king) – Belgi Aug 6 '12 at 21:32
There are $\binom{52}{2}$ ways to choose two cards. There are $\binom{40}{2}$ ways to choose them so none is J, Q, or K. So an alternate form of the answer is $\binom{52}{2}-\binom{40}{2}$. In this case, probably a little harder than the way you used, but in other situations the idea of counting complement can save a lot of work. – André Nicolas Aug 6 '12 at 21:50
This seems rather obvious: the product term $\tbinom{40}1\times\tbinom{12}2$ represents "$1$ card from the remaining stack and both cards from Jack/Queen/King", or as you say it in the edit "1 from the 40 and 2 from special". But since only two cards are drawn this is not among the possibilities of the problem, so the term should not be present. A product represents two independent choices, and mutually exclusive (yet individually possible) events are never independent. – Marc van Leeuwen Aug 7 '12 at 9:27
If you expand your first answer, you have $$\binom{12}{1} \cdot \binom{40}{1} + \binom{12}{2} \cdot \binom{40}{1}.$$ Notice in the term on the right, you are choosing a total of three cards, which does not count what you want.
A rule of thumb is to turn "at least" problems into several instances of "exactly".
For the problem at hand, "at least one jack/queen/king" translates into "either exactly one jack/queen/king or exactly two jack/queen/king".
The number of ways to get exactly one jack/queen/king and one other card is $$\binom{12}{1} \cdot \binom{40}{1}.$$ We multiply here because you choose a card from the jack/queen/king pile and a card from the "other" pile.
The number of ways to get exactly two from jack/queen/king is $$\binom{12}{2}.$$ Since these cases are disjoint (this is important), we simply add the results from the two cases. Thus, the number of ways to get at least one Jack, Queen, or King is $$\binom{12}{1} \cdot \binom{40}{1} + \binom{12}{2}.$$
-
thanks for the answer , could you explain what you meant by Notice in the term on the right, you are choosing a total of three cards, which does not count what you want.. – MistyD Aug 6 '12 at 21:38
@MistyD The term $\binom{12}{2} \cdot \binom{40}{1}$ instructs you to select two cards from the jack/queen/king pile and one card from the "other" pile, giving you a three-card hand (not a two-card hand, like the problem is asking for). – Austin Mohr Aug 6 '12 at 21:39
Though I partially get what you meant. Lets consider the correct way then $( \binom{40}{1} \times \binom{12}{1}) + \binom{12}{2}$ . Which I believe means 1 from remaining cards (which is 40) and 1 from J/K/Q , now what does + (mean here) . My mind keeps on asking me why we didn't multiply here instead of adding? – MistyD Aug 6 '12 at 22:05
Addition here corresponds to the union operation on sets; if the sets $A$ and $B$ have no elements in common (this is a critical restriction!) then $|A\cup B| = |A| + |B|$. In effect what you're saying is that 'all the combinations with at least one JQK' consists of 'all the combinations with exactly one JQK' and 'all the combinations with more than one JQK' (in this case there can be only two). – Steven Stadnicki Aug 6 '12 at 22:24
@MistyD You have two sheets of paper in front of you. The first lists all $\binom{40}{1} \cdot \binom{12}{1}$ hands having exactly one J/Q/K. The second lists all $\binom{12}{2}$ hands having exactly two J/Q/K. Together, the two sheets list all hands having at least one J/Q/K. Thus, to get the total count, you would want to add the number of hands in the first list to the number of hands in the second list, not multiply. – Austin Mohr Aug 7 '12 at 1:07
The easiest way to solve this problem is to note that subtracting the number of ways of selecting any two cards neither of which is a Jack, Queen, or King, from the number of ways of selecting any two cards, leaves with the number of ways to pick two cards of which at least one of them is a Jack, Queen, or King.
So,
$$\mbox{(Ways to select any 2 cards from the deck)} - \mbox{(Ways to select 2 cards where neither one is J/Q/K)} = \mbox{Ways to select 2 cards where at least one is J/Q/K}$$
$$\binom{52}{2} - \binom{40}{2} = 1326 - 780 = 546$$
-
Thanks for you answer but I am more interested in knowing why I am wrong.. – MistyD Aug 6 '12 at 22:29
@ladaghini This answer saves effort, and is more generally useful. Having a hard time making it work for picking at least 2 of a subgroup when picking 5 form the whole. – csga5000 Jul 14 at 19:44
For my situation (15 balls, 8 red, 7 black, 5 chosen) the answer was: 15C5 - (7C4 * 8C1) - 7C5 Where C is combination operation, and the left/right numbers are n/r respectively. Total combinations - (combainations with 4 black 1 red) - (combaintions 5 black) – csga5000 Jul 14 at 19:54 | 2015-07-29T06:58:42 | {
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# Inequalities trick
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Inequalities trick [#permalink] 16 Mar 2010, 09:11
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I learnt this trick while I was in school and yesterday while solving one question I recalled.
Its good if you guys use it 1-2 times to get used to it.
Suppose you have the inequality
f(x) = (x-a)(x-b)(x-c)(x-d) < 0
Just arrange them in order as shown in the picture and draw curve starting from + from right.
now if f(x) < 0 consider curve having "-" inside and if f(x) > 0 consider curve having "+" and combined solution will be the final solution. I m sure I have recalled it fully but if you guys find any issue on that do let me know, this is very helpful.
Don't forget to arrange then in ascending order from left to right. a<b<c<d
So for f(x) < 0 consider "-" curves and the ans is : (a < x < b) , (c < x < d)
and for f(x) > 0 consider "+" curves and the ans is : (x < a), (b < x < c) , (d < x)
If f(x) has three factors then the graph will have - + - +
If f(x) has four factors then the graph will have + - + - +
If you can not figure out how and why, just remember it.
Try to analyze that the function will have number of roots = number of factors and every time the graph will touch the x axis.
For the highest factor d if x>d then the whole f(x) > 0 and after every interval of the roots the signs will change alternatively.
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Re: Inequalities trick [#permalink] 22 Oct 2010, 05:33
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Yes, this is a neat little way to work with inequalities where factors are multiplied or divided. And, it has a solid reasoning behind it which I will just explain.
If (x-a)(x-b)(x-c)(x-d) < 0, we can draw the points a, b, c and d on the number line.
e.g. Given (x+2)(x-1)(x-7)(x-4) < 0, draw the points -2, 1, 7 and 4 on the number line as shown.
Attachment:
doc.jpg [ 7.9 KiB | Viewed 30856 times ]
This divides the number line into 5 regions. Values of x in right most region will always give you positive value of the expression. The reason for this is that if x > 7, all factors above will be positive.
When you jump to the next region between x = 4 and x = 7, value of x here give you negative value for the entire expression because now, (x - 7) will be negative since x < 7 in this region. All other factors are still positive.
When you jump to the next region on the left between x = 1 and x = 4, expression will be positive again because now two factors (x - 7) and (x - 4) are negative, but negative x negative is positive... and so on till you reach the leftmost section.
Since we are looking for values of x where the expression is < 0, here the solution will be -2 < x < 1 or 4< x < 7
It should be obvious that it will also work in cases where factors are divided.
e.g. (x - a)(x - b)/(x - c)(x - d) < 0
(x + 2)(x - 1)/(x -4)(x - 7) < 0 will have exactly the same solution as above.
Note: If, rather than < or > sign, you have <= or >=, in division, the solution will differ slightly. I will leave it for you to figure out why and how. Feel free to get back to me if you want to confirm your conclusion.
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Math Forum Moderator Joined: 20 Dec 2010 Posts: 2027 Followers: 142 Kudos [?]: 1219 [16] , given: 376 Re: Inequalities trick [#permalink] 11 Mar 2011, 05:49 16 This post received KUDOS 7 This post was BOOKMARKED vjsharma25 wrote: VeritasPrepKarishma wrote: vjsharma25 wrote: How you have decided on the first sign of the graph?Why it is -ve if it has three factors and +ve when four factors? Check out my post above for explanation. I understand the concept but not the starting point of the graph.How you decide about the graph to be a sine or cosine waveform?Meaning graph starts from the +ve Y-axis for four values and starts from -ve Y-axis for three values. What if the equation you mentioned is (x+2)(x-1)(x-7)<0,will the last two ranges be excluded or the graph will also change? I always struggle with this as well!!! There is a trick Bunuel suggested; (x+2)(x-1)(x-7) < 0 Here the roots are; -2,1,7 Arrange them in ascending order; -2,1,7; These are three points where the wave will alternate. The ranges are; x<-2 -2<x<1 1<x<7 x>7 Take a big value of x; say 1000; you see the inequality will be positive for that. (1000+2)(1000-1)(1000-7) is +ve. Thus the last range(x>7) is on the positive side. Graph is +ve after 7. Between 1 and 7-> -ve between -2 and 1-> +ve Before -2 -> -ve Since the inequality has the less than sign; consider only the -ve side of the graph; 1<x<7 or x<-2 is the complete range of x that satisfies the inequality. _________________ Manager Joined: 29 Sep 2008 Posts: 150 Followers: 3 Kudos [?]: 53 [10] , given: 1 Re: Inequalities trick [#permalink] 22 Oct 2010, 10:45 10 This post received KUDOS 8 This post was BOOKMARKED if = sign is included with < then <= will be there in solution like for (x+2)(x-1)(x-7)(x-4) <=0 the solution will be -2 <= x <= 1 or 4<= x <= 7 in case when factors are divided then the numerator will contain = sign like for (x + 2)(x - 1)/(x -4)(x - 7) < =0 the solution will be -2 <= x <= 1 or 4< x < 7 we cant make 4<=x<=7 as it will make the solution infinite correct me if i am wrong Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 5870 Location: Pune, India Followers: 1485 Kudos [?]: 8007 [6] , given: 190 Re: Inequalities trick [#permalink] 11 Mar 2011, 18:57 6 This post received KUDOS Expert's post vjsharma25 wrote: I understand the concept but not the starting point of the graph.How you decide about the graph to be a sine or cosine waveform?Meaning graph starts from the +ve Y-axis for four values and starts from -ve Y-axis for three values. What if the equation you mentioned is (x+2)(x-1)(x-7)<0,will the last two ranges be excluded or the graph will also change? Ok, look at this expression inequality: (x+2)(x-1)(x-7) < 0 Can I say the left hand side expression will always be positive for values greater than 7? (x+2) will be positive, (x - 1) will be positive and (x-7) will also be positive... so in the rightmost regions i.e. x > 7, all three factors will be positive. The expression will be positive when x > 7, it will be negative when 1 < x < 7, positive when -2 , x < 1 and negative when x < -2. We need the region where the expression is less than 0 i.e. negative. So either 1 < x < 7 or x < -2. Now let me add another factor: (x+8)(x+2)(x-1)(x-7) Can I still say that the entire expression is positive in the rightmost region i.e. x>7 because each one of the four factors is positive? Yes. So basically, your rightmost region is always positive. You go from there and assign + and - signs to the regions. Your starting point is the rightmost region. Note: Make sure that the factors are of the form (ax - b), not (b - ax)... e.g. (x+2)(x-1)(7 - x)<0 Convert this to: (x+2)(x-1)(x-7)>0 (Multiply both sides by '-1') Now solve in the usual way. Assign '+' to the rightmost region and then alternate with '-' Since you are looking for positive value of the expression, every region where you put a '+' will be the region where the expression will be greater than 0. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199
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Re: Inequalities trick [#permalink] 10 Aug 2011, 16:01
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WoW - This is a cool thread with so many thing on inequalities....I have compiled it together with some of my own ideas...It should help.
1) CORE CONCEPT
@gurpreetsingh -
Suppose you have the inequality
f(x) = (x-a)(x-b)(x-c)(x-d) < 0
Arrange the NUMBERS in ascending order from left to right. a<b<c<d
Draw curve starting from + from right.
now if f(x) < 0 consider curve having "-" inside and if f(x) > 0 consider curve having "+" and combined solution will be the final solution. I m sure I have recalled it fully but if you guys find any issue on that do let me know, this is very helpful.
So for f(x) < 0 consider "-" curves and the ans is : (a < x < b) , (c < x < d)
and for f(x) > 0 consider "+" curves and the ans is : (x < a), (b < x < c) , (d < x)
If f(x) has three factors then the graph will have - + - +
If f(x) has four factors then the graph will have + - + - +
If you can not figure out how and why, just remember it.
Try to analyze that the function will have number of roots = number of factors and every time the graph will touch the x axis.
For the highest factor d if x>d then the whole f(x) > 0 and after every interval of the roots the signs will change alternatively.
Note: Make sure that the factors are of the form (ax - b), not (b - ax)...
example -
(x+2)(x-1)(7 - x)<0
Convert this to: (x+2)(x-1)(x-7)>0 (Multiply both sides by '-1')
Now solve in the usual way. Assign '+' to the rightmost region and then alternate with '-'
Since you are looking for positive value of the expression, every region where you put a '+' will be the region where the expression will be greater than 0.
2) Variation - ODD/EVEN POWER
@ulm/Karishma -
if we have even powers like (x-a)^2(x-b)
we don't need to change a sign when jump over "a".
This will be same as (x-b)
We can ignore squares BUT SHOULD consider ODD powers
example -
2.a
(x-a)^3(x-b)<0 is the same as (x-a)(x-b) <0
2.b
(x - a)(x - b)/(x - c)(x - d) < 0 ==> (x - a)(x - b)(x-c)^-1(x-d)^-1 <0
is the same as (x - a)(x - b)(x - c)(x - d) < 0
3) Variation <= in FRACTION
@mrinal2100 -
if = sign is included with < then <= will be there in solution
like for (x+2)(x-1)(x-7)(x-4) <=0 the solution will be -2 <= x <= 1 or 4<= x <= 7
BUT if it is a fraction the denominator in the solution will not have = SIGN
example -
3.a
(x + 2)(x - 1)/(x -4)(x - 7) < =0
the solution will be -2 <= x <= 1 or 4< x < 7
we cant make 4<=x<=7 as it will make the solution infinite
4) Variation - ROOTS
@Karishma -
As for roots, you have to keep in mind that given $$\sqrt{x}$$, x cannot be negative.
$$\sqrt{x}$$ < 10
implies 0 < $$\sqrt{x}$$ < 10
Squaring, 0 < x < 100
Root questions are specific. You have to be careful. If you have a particular question in mind, send it.
Refer - inequalities-and-roots-118619.html#p959939
Some more useful tips for ROOTS....I am too lazy to consolidate
<5> THESIS -
@gmat1220 -
Once algebra teacher told me - signs alternate between the roots. I said whatever and now I know why Watching this article is a stroll down the memory lane.
I will save this future references....
Anyone wants to add ABSOLUTE VALUES....That will be a value add to this post
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Re: Inequalities trick [#permalink] 09 Sep 2013, 22:35
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karannanda wrote:
gurpreetsingh wrote:
I learnt this trick while I was in school and yesterday while solving one question I recalled.
Its good if you guys use it 1-2 times to get used to it.
Suppose you have the inequality
f(x) = (x-a)(x-b)(x-c)(x-d) < 0
Just arrange them in order as shown in the picture and draw curve starting from + from right.
now if f(x) < 0 consider curve having "-" inside and if f(x) > 0 consider curve having "+" and combined solution will be the final solution. I m sure I have recalled it fully but if you guys find any issue on that do let me know, this is very helpful.
Don't forget to arrange then in ascending order from left to right. a<b<c<d
So for f(x) < 0 consider "-" curves and the ans is : (a < x < b) , (c < x < d)
and for f(x) > 0 consider "+" curves and the ans is : (x < a), (b < x < c) , (d < x)
If f(x) has three factors then the graph will have - + - +
If f(x) has four factors then the graph will have + - + - +
If you can not figure out how and why, just remember it.
Try to analyze that the function will have number of roots = number of factors and every time the graph will touch the x axis.
For the highest factor d if x>d then the whole f(x) > 0 and after every interval of the roots the signs will change alternatively.
Hi Gurpreet,
Thanks for the wonderful method.
I am trying to understand it so that i can apply it in tests.
Can you help me in applying this method to the below expression to find range of x.
x^3 – 4x^5 < 0?
I am getting the roots as -1/2, 0, 1/2 and when i plot them using this method, putting + in the rightmost region, I am not getting correct result. Not sure where i am going wrong. Can you pls help.
Before you apply the method, ensure that the factors are of the form (x - a)(x - b) etc
$$x^3 - 4x^5 < 0$$
$$x^3 ( 1 - 4x^2) < 0$$
$$x^3(1 - 2x) (1 + 2x) < 0$$
$$4x^3(x - 1/2)(x + 1/2) > 0$$ (Notice the flipped sign. We multiplied both sides by -1 to convert 1/2 - x to x - 1/2)
Now the transition points are 0, -1/2 and 1/2 so put + in the rightmost region.
The solution will be x > 1/2 or -1/2 < x< 0.
Check out these posts discussing such complications:
http://www.veritasprep.com/blog/2012/06 ... e-factors/
http://www.veritasprep.com/blog/2012/07 ... ns-part-i/
http://www.veritasprep.com/blog/2012/07 ... s-part-ii/
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews e-GMAT Representative Joined: 04 Jan 2015 Posts: 338 Followers: 66 Kudos [?]: 483 [3] , given: 83 Inequalities trick [#permalink] 06 Jan 2015, 02:35 3 This post received KUDOS Expert's post 8 This post was BOOKMARKED Just came across this useful discussion. VeritasPrepKarishma has given a very lucid explanation of how this “wavy line” method works. I have noticed that there is still a little scope to take this discussion further. So here are my two cents on it. I would like to highlight an important special case in the application of the Wavy Line Method When there are multiple instances of the same root: Try to solve the following inequality using the Wavy Line Method: $$(x-1)^2(x-2)(x-3)(x-4)^3 < 0$$ To know how you did, compare your wavy line with the correct one below. Did you notice how this inequality differs from all the examples above? Notice that two of the four terms had an integral power greater than 1. How to draw the wavy line for such expressions? Let me directly show you how the wavy line would look and then later on the rule behind drawing it. Attachment: File comment: Observe how the wave bounces back at x = 1. bounce.png [ 10.4 KiB | Viewed 2819 times ] Notice that the curve bounced down at the point x = 1. (At every other root, including x = 4 whose power was 3, it was simply passing through them.) Can you figure out why the wavy line looks like this for this particular inequality? (Hint: The wavy line for the inequality $$(x-1)^{38}(x-2)^{57}(x-3)^{15}(x-4)^{27} < 0$$ Is also the same as above) Come on! Give it a try. If you got it right, you’ll see that there are essentially only two rules while drawing a wavy line. (Remember, we’ll refer the region above the number line as positive region and the region below the number line as negative region.) How to draw the wavy line? 1. How to start: Start from the top right most portion. Be ready to alternate (or not alternate) the region of the wave based on how many times a point is root to the given expression. 2. How to alternate: In the given expression, if the power of a term is odd, then the wave simply passes through the corresponding point (root) into the other region (to –ve region if the wave is currently in the positive region and to the +ve region if the wave is currently in the negative region). However, if the power of a term is even, then the wave bounces back into the same region. Now look back at the above expression and analyze your wavy line. Were you (intuitively) using the above mentioned rules while drawing your wavy line? Solution Once you get your wavy line right, solving an inequality becomes very easy. For instance, for the above inequality, since we need to look for the space where the above expression would be less than zero, look for the areas in the wavy line where the curve is below the number line. So the correct solution set would simply be {3 < x < 4} U {{x < 2} – {1}} In words, it is the Union of two regions region1 between x = 3 and x = 4 and region2 which is x < 2, excluding the point x = 1. Food for Thought Now, try to answer the following questions: 1. Why did we exclude the point x = 1 from the solution set of the last example? (Easy Question) 2. Why do the above mentioned rules (especially rule #2) work? What is/are the principle(s) working behind the curtains? Foot Note: Although the post is meant to deal with inequality expressions containing multiple roots, the above rules to draw the wavy line are generic and are applicable in all cases. - Krishna _________________ Last edited by EgmatQuantExpert on 10 Jan 2015, 20:41, edited 1 time in total. Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 5870 Location: Pune, India Followers: 1485 Kudos [?]: 8007 [2] , given: 190 Re: Inequalities trick [#permalink] 23 Jul 2012, 02:13 2 This post received KUDOS Expert's post Stiv wrote: VeritasPrepKarishma wrote: mrinal2100: Kudos to you for excellent thinking! Correct me if I'm wrong. If the lower part of the equation$$\frac {(x+2)(x-1)}{(x-4)(x-7)}$$ were $$4\leq x \leq 7$$, than the lower part would be equal to zero,thus making it impossible to calculate the whole equation. x cannot be equal to 4 or 7 because if x = 4 or x = 7, the denominator will be 0 and the expression will not be defined. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199
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Re: Inequalities trick [#permalink] 19 Mar 2010, 11:59
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ttks10 wrote:
Can u plz explainn the backgoround of this & then the explanation.
Thanks
i m sorry i dont have any background for it, you just re-read it again and try to implement whenever you get such question and I will help you out in any issue.
sidhu4u wrote:
I have applied this trick and it seemed to be quite useful.
Nice to hear this....good luck.
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Re: Inequalities trick [#permalink] 26 May 2011, 19:55
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chethanjs wrote:
mrinal2100 wrote:
if = sign is included with < then <= will be there in solution
like for (x+2)(x-1)(x-7)(x-4) <=0 the solution will be -2 <= x <= 1 or 4<= x <= 7
in case when factors are divided then the numerator will contain = sign
like for (x + 2)(x - 1)/(x -4)(x - 7) < =0
the solution will be -2 <= x <= 1 or 4< x < 7
we cant make 4<=x<=7 as it will make the solution infinite
correct me if i am wrong
Can you please tell me why the solution gets infinite for 4<=x<=7 ?
Thanks.
(x -4)(x - 7) is in denominator.
Making x=4 or 7 would make the denominator 0 and the entire function undefined.
Thus, the range of x can't be either 4 or 7.
4<=x<=7 would be wrong.
4<x<7 is correct because now we removed "=" sign.
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Re: Inequalities trick [#permalink] 08 Aug 2011, 10:59
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Asher wrote:
gurpreetsingh wrote:
ulm wrote:
if we have smth like (x-a)^2(x-b)
we don't need to change a sign when jump over "a".
yes even powers wont contribute to the inequality sign. But be wary of the root value of x=a
This way of solving inequalities actually makes it soo much easier. Thanks gurpreetsingh and karishma
However, i am confused about how to solve inequalities such as: (x-a)^2(x-b) and also ones with root value.
When you have (x-a)^2(x-b) < 0, the squared term is ignored because it is always positive and hence doesn't affect the sign of the entire left side. For the left hand side to be negative i.e. < 0, (x - b) should be negative i.e. x - b < 0 or x < b.
Similarly for (x-a)^2(x-b) > 0, x > b
As for roots, you have to keep in mind that given $$\sqrt{x}$$, x cannot be negative.
$$\sqrt{x}$$ < 10
implies 0 < $$\sqrt{x}$$ < 10
Squaring, 0 < x < 100
Root questions are specific. You have to be careful. If you have a particular question in mind, send it.
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Senior Manager Joined: 16 Feb 2012 Posts: 255 Concentration: Finance, Economics Followers: 5 Kudos [?]: 136 [1] , given: 121 Re: Inequalities trick [#permalink] 22 Jul 2012, 02:03 1 This post received KUDOS VeritasPrepKarishma wrote: mrinal2100: Kudos to you for excellent thinking! Correct me if I'm wrong. If the lower part of the equation$$\frac {(x+2)(x-1)}{(x-4)(x-7)}$$ were $$4\leq x \leq 7$$, than the lower part would be equal to zero,thus making it impossible to calculate the whole equation. _________________ Kudos if you like the post! Failing to plan is planning to fail. Current Student Status: Nothing comes easy: neither do I want. Joined: 12 Oct 2009 Posts: 2798 Location: Malaysia Concentration: Technology, Entrepreneurship Schools: ISB '15 (M) GMAT 1: 670 Q49 V31 GMAT 2: 710 Q50 V35 Followers: 201 Kudos [?]: 1220 [1] , given: 235 Re: Inequalities trick [#permalink] 18 Oct 2012, 04:17 1 This post received KUDOS 1 This post was BOOKMARKED GMATBaumgartner wrote: gurpreetsingh wrote: ulm wrote: in addition: if we have smth like (x-a)^2(x-b) we don't need to change a sign when jump over "a". yes even powers wont contribute to the inequality sign. But be wary of the root value of x=a Hi Gurpreet, Could you elaborate what exactly you meant here in highlighted text ? Even I have a doubt as to how this can be applied for powers of the same term . like the example mentioned in the post above. If the powers are even then the inequality won't be affected. eg if u have to find the range of values of x satisfying (x-a)^2 *(x-b)(x-c) >0 just use (x-b)*(x-c) >0 because x-a raised to the power 2 will not affect the inequality sign. But just make sure x=a is taken care off , as it would make the inequality zero. _________________ Fight for your dreams :For all those who fear from Verbal- lets give it a fight Money Saved is the Money Earned Jo Bole So Nihaal , Sat Shri Akaal Support GMAT Club by putting a GMAT Club badge on your blog/Facebook GMAT Club Premium Membership - big benefits and savings Gmat test review : 670-to-710-a-long-journey-without-destination-still-happy-141642.html Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 5870 Location: Pune, India Followers: 1485 Kudos [?]: 8007 [1] , given: 190 Re: Inequalities trick [#permalink] 18 Oct 2012, 09:27 1 This post received KUDOS Expert's post GMATBaumgartner wrote: Hi Gurpreet, Could you elaborate what exactly you meant here in highlighted text ? Even I have a doubt as to how this can be applied for powers of the same term . like the example mentioned in the post above. In addition, you can check out this post: http://www.veritasprep.com/blog/2012/07 ... s-part-ii/ I have discussed how to handle powers in it. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199
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Re: Inequalities trick [#permalink] 02 Dec 2012, 06:25
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GMATGURU1 wrote:
Going ahead in this inequalities area of GMAT, can some one has the problem numbers of inequalities in OG 11 and OG 12?
Cheers,
Danny
Search for hundreds of question with solutions by tags: viewforumtags.php
DS questions on inequalities: search.php?search_id=tag&tag_id=184
PS questions on inequalities: search.php?search_id=tag&tag_id=189
Hardest DS inequality questions with detailed solutions: inequality-and-absolute-value-questions-from-my-collection-86939.html
Hope it helps.
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Re: Inequalities trick [#permalink] 20 Dec 2012, 20:04
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VeritasPrepKarishma wrote:
The entire concept is based on positive/negative factors which means <0 or >0 is a must. If the question is not in this format, you need to bring it to this format by taking the constant to the left hand side.
e.g.
(x + 2)(x + 3) < 2
x^2 + 5x + 6 - 2 < 0
x^2 + 5x + 4 < 0
(x+4)(x+1) < 0
Now use the concept.
Yes this is probable but it might not be possible always to group them. So in case you are unsure just follow the number plugging approach. But most of the times this trick would be very handy.
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Re: Inequalities trick [#permalink] 11 Nov 2013, 06:51
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Expert's post
Can you please explain the above mentioned concept in relation to the following question?
Is a > O?
(1) a^3 - 0 < 0
(2) 1- a^2 > 0
Can you please explain the scenario when (x-a)(x-B)(x-C)(x-d)>0?
Sorry, but finding it difficult to understand.
Check alternative solutions here: is-a-0-1-a-3-a-0-2-1-a-86749.html
Hope this helps.
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Re: Inequalities trick [#permalink] 23 Apr 2014, 03:43
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PathFinder007 wrote:
I have a query. I have following question
x^3 - 4x^5 < 0
I can define this as (1+2x).x^3(1-2x). now I have roots -1/2, 0, 1/2. so in case of >1/2 I will always get inequality value as <0 and in case of -1/2 and 0 I will get value as 0.
So How I will define them in graph and what range I will consider for this inequality.
Thanks
The factors must be of the form (x - a)(x - b) .... etc
x^3 - 4x^5 < 0
x^3 * (1 - 4x^2) < 0
x^3 * (1 - 2x) * (1 + 2x) < 0
x^3 * (2x - 1) * (2x + 1) > 0 (Note the sign flip because 1-2x was changed to 2x - 1)
x^3 * 2(x - 1/2) *2(x + 1/2) > 0
So transition points are 0, 1/2 and -1/2.
____________ - 1/2 _____ 0 ______1/2 _________
This is what it looks like on the number line.
The rightmost region is positive. We want the positive regions in the inequality.
So the desired range of x is given by x > 1/2 or -1/2 < x< 0
For more on this method, check these posts:
http://www.veritasprep.com/blog/2012/06 ... e-factors/
http://www.veritasprep.com/blog/2012/07 ... ns-part-i/
http://www.veritasprep.com/blog/2012/07 ... s-part-ii/
The links will give you the theory behind this method in detail.
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Re: Inequalities trick [#permalink] 11 Nov 2014, 08:52
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For statement I
(1) (1-2x)(1+x)<0
Once you get this into the req form
2 (x-1/2) (x+1) > 0
sol +++++ -1 ------ 1/2 +++++
As Karishma pointed out ponce in the req form, the rightmost will always be positive and the alternating will happen from there.
So sol for this x> 1/2 and x<-1
Integers greater than 1/2 and less than -1 , thus |x| may be >= 1. Unsure
Thus Insufficient.
For Statement II
(2) (1-x)(1+2x)<0
Once you get this into the req form
2(x+1/2) (x-1) > 0
sol +++++ -1/2 ------ 1 +++++
So sol for this x>1 and x< -1/2
Integers greater than 1 and less than -1/2 , thus |x| may be >= 1. Unsure
Thus Insufficient.
Combining Both the statements
x<-1 and x>1
Thus Integers for this range will give |x| > 1
Thus Sufficient.
Hope this helps. I am not very confident of my solution though. Its my first solution gmatclub
mayankpant wrote:
gurpreetsingh wrote:
I learnt this trick while I was in school and yesterday while solving one question I recalled.
Its good if you guys use it 1-2 times to get used to it.
Suppose you have the inequality
f(x) = (x-a)(x-b)(x-c)(x-d) < 0
Just arrange them in order as shown in the picture and draw curve starting from + from right.
now if f(x) < 0 consider curve having "-" inside and if f(x) > 0 consider curve having "+" and combined solution will be the final solution. I m sure I have recalled it fully but if you guys find any issue on that do let me know, this is very helpful.
Don't forget to arrange then in ascending order from left to right. a<b<c<d
So for f(x) < 0 consider "-" curves and the ans is : (a < x < b) , (c < x < d)
and for f(x) > 0 consider "+" curves and the ans is : (x < a), (b < x < c) , (d < x)
If f(x) has three factors then the graph will have - + - +
If f(x) has four factors then the graph will have + - + - +
If you can not figure out how and why, just remember it.
Try to analyze that the function will have number of roots = number of factors and every time the graph will touch the x axis.
For the highest factor d if x>d then the whole f(x) > 0 and after every interval of the roots the signs will change alternatively.
Hi
Can you please explain this question to me using the graph. I am missing the point when graph is being used here?
If x is an integer, is |x|>1?
(1) (1-2x)(1+x)<0
(2) (1-x)(1+2x)<0
For me ,the first equations roots are -1 and 1/2. Now I am struggling to get to the correct sign using the graph method here.
Same for second equation: roots are 1 and -1/2 but struggling for the sign.
THanks
Re: Inequalities trick [#permalink] 11 Nov 2014, 08:52
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Display posts from previous: Sort by | 2015-09-04T08:31:05 | {
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https://math.stackexchange.com/questions/2991983/angles-in-a-spherical-triangle | # Angles in a spherical triangle.
Just seeking advice here! I have 3 coordinates;
$$A(-0.52992,0.84805,0),\\ B(0.84805,0,0.52992),\\C(0.15461,0.47553,0.86603)$$.
I want to find the angles at $$A$$, $$B$$ and $$C$$. Hence, I find the normal of the planes through the great circle $$AB$$, $$AC$$, $$BC$$ respectively.
I can easily find that the (unit) normals are
$$n_{AB}:(0.50306,0.31434,-0.80506)\\n_{BC}:(-0.31209,-0.80818,0.49945)\\n_{AC}:(0.77558,0.48464,-0.40448)$$
Is the angle at $$A$$ just $$cos^{-1}(n_{AB}.n_{AC})$$, angle at $$B$$ is $$cos^{-1}(n_{AB}.n_{BC})$$, angle at C is $$cos^{-1}(n_{AC}.n_{BC})$$?
What I am worried is that $$n_{AB}.n_{BC}=-0.81313$$ and hence $$arccos(-0.81313)=2.5203$$. This is the same situation for angle $$C$$. Am I still correct considering that the angles $$B$$ and $$C$$ are more than $$\pi$$? Can an angle be more than $$\pi$$? Or is there any way I can reduce the angle?
Edit: I can try to visualize the points on the sphere, it seems that angle $$C$$ is more than $$\frac{\pi}{2}$$, but angle $$B$$ seems to be less than $$\frac{\pi}{2}$$, am I still correct to say that angle $$B$$ is $$2.52$$ rad?
• Spherical or 3D? 3 points does not define a sphere. – Moti Nov 10 '18 at 2:09
• Spherical. Yes but I am finding the angles of the spherical triangle on a unit sphere. So 3 points on the (unit) sphere do determine a spherical triangle – Icycarus Nov 10 '18 at 16:37
• You can find many spheres that use the same three points which result many solution - for each sphere you have a solution set. You need to say that the three points are of a great circle if you mean this - but than all equal $\pi$ – Moti Nov 10 '18 at 21:20
• @Moti: By "unit sphere", OP certainly means "the unit sphere"; that is, the sphere of unit radius centered at the origin (cf. "the unit circle"). If you're studying spherical triangles, there's no good reason to make your life more complicated by studying them on a non-origin-centered sphere. – Blue Nov 10 '18 at 21:35
• I missed that one. – Moti Nov 10 '18 at 23:52
Two things:
1. $$2.5$$ is not larger than $$\pi=3.1415\dots$$.
2. What you computed was Euclidean angles, not spherical. That is, they are angles between the lines $$AB$$, $$AC$$ and so on. The spherical angles would be the angle between the great circles passing $$A,C$$ and $$A,B$$ respectively.
• Hm.. I learnt that the angles between the great circles passing A,C and A,B is the angle between the normals of the plane passing through AC and AB. Which is what I have calculated, or did I misunderstood something? – Icycarus Nov 10 '18 at 21:22
• Oh yes, thanks for pointing out. What I meant was more than pi/2. Will edit it – Icycarus Nov 10 '18 at 21:23
• Then yes, they can be larger than $\pi/2$. – Quang Hoang Nov 10 '18 at 21:31
• Oh, I missed that part. Yes, I think you are correct. Angle between normals of planes through great circles is the same with angle between them. – Quang Hoang Nov 10 '18 at 21:33
• Oh! Thank you for the explanation! – Icycarus Nov 10 '18 at 21:34
Note that if $$n_{AB}$$ is a unit vector perpendicular to the great circle through $$A$$ and $$B,$$ then so is $$-n_{AB}.$$ For each side of the spherical triangle, there are two unit vectors perpendicular to that side, each of them the exact opposite of the other.
When you reverse the direction of one of the vectors in a dot product, you reverse the sign of the product, and the arc cosine you would have gotten is replaced by its supplement. If you get just one of the unit vectors “backward” when computing the angles, you will compute the exterior angle at the vertex whose interior angle you wanted.
One way to avoid this error is to make sure you take each pair of vectors in your cross products in the same “direction” around the triangle. For example, you can take $$A\times B,$$ $$B\times C,$$ and $$C\times A.$$ Alternatively, you can reverse all three of the cross products. If you use a different method to find the normal unit vectors, make sure that the dot product of each normal with the vector to the remaining vertex is positive in all three cases, or ensure that all three dot products are negative. There is a twist to this, however: when you set up the normals this way, you get normals that go in nearly opposite directions when the vertex angle is very small but go in nearly the same direction when the vertex angle is very large. Therefore simply taking the arc cosine of a dot product, $$\arccos(n_1\cdot n_2).$$ will give you the exterior angle, so you actually want $$\arccos(-n_1\cdot n_2)$$ in order to get the interior angle.
Another way that works (as you noted in a comment) is to use cross products to find the normals, but make sure that the vertex at which you want to find the angle is either the first vector in both cross products or the second vector in both cross products. That way the normals will point in almost the same direction when the angle is small and will be almost opposite when the angle is large, and you can take $$\arccos(n_1\cdot n_2)$$ without requiring a negative sign. Using this method, you need at least four cross products, but since $$B\times A = -A\times B$$ this method does not really require any extra computation, just reverse the vector when you need the other cross product.
One thing you must not do is to have two normals that follow one rule and the third one following the opposite rule. This will cause you to compute exterior angles at two vertices. It looks like your $$n_{AC}$$ is the “wrong way around” compared to your other two normal vectors.
Merely having multiple obtuse angles in a spherical triangle is not an indication that you computed the angles incorrectly. The sum of the angles of a spherical triangle can be anything up to $$540$$ degrees.
• Thanks for the input! I was wondering if this was what you meant; Do find angle at A; I find the angle between $n_{AB}.n_{AC}$, angle at B; angle between $n_{BA}.n_{BC}$, angle at C; angle between $n_{AC}.n_{BC}$, where $n_{AB}$ = $A$ x $B$, $n_{BA}$ = $B$ x $A$ etc – Icycarus Nov 11 '18 at 21:11
• That's not quite what I had in mind but it's a perfectly good method. I've added it to the answer. There are several correct ways to solve this problem! I'm sorry to say I actually made a sign error when I first wrote this up, which confused me because it seemed to imply that you already had the correct angle at $B.$ But I was doing the dot products in my head in a moving car and writing the results on my phone, and I have fixed the error now (I think), so please forgive the confusion. – David K Nov 11 '18 at 23:54
Note that $$n_{AB}=-n_{BA}=\frac{A\times B}{|A|\,|B|}\\ n_{BC}=-n_{CB}=\frac{B\times C}{|B|\,|C|}\\ n_{CA}=-n_{AC}=\frac{C\times A}{|C|\,|A|}$$ Therefore, $$\angle A=\arccos(n_{BA}\cdot n_{CA})=0.51928\\ \angle B=\arccos(n_{AB}\cdot n_{CB})=0.62125\\ \angle C=\arccos(n_{AC}\cdot n_{BC})=2.56032$$ and by Girard's Theorem, the area of the triangle is $$\angle A+\angle B+\angle C-\pi=0.55926$$ where the area of the whole sphere is $$4\pi$$ steradians.
The angle you got for $$\angle B$$ is the supplement of what is computed above, because you have the wrong sign for the dot product. That is, $$\cos(\angle B)=n_{AB}\cdot n_{CB}=n_{BA}\cdot n_{BC}=-n_{AB}\cdot n_{BC}=-n_{BA}\cdot n_{CB}$$ | 2019-10-14T15:17:35 | {
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http://mathhelpforum.com/calculus/137295-integration-problem-print.html | # Integration Problem
• April 4th 2010, 06:46 PM
ninjaku
Integration Problem
Find the integral of $(tanx/sec^2x)dx$
I tried using u substitution, and I get:
$u=tanx$
$du=(sec^2)x$
At first it seems like it was going to be simple.
But in the problem there is a $sec^-2$
And in my substitution there is a $sec^2$.
Any help would be appreciated.
• April 4th 2010, 06:58 PM
skeeter
Quote:
Originally Posted by ninjaku
Find the integral of (tanx/(sec^2)x)dx.
I tried using u substitution, and I get:
u=tanx
du=(sec^2)x.
At first it seems like it was going to be simple.
But in the problem there is a sec^-2.
And in my substitution there is a sec^2.
Any help would be appreciated.
$\frac{\tan{x}}{\sec^2{x}} = \tan{x} \cdot \cos^2{x} = \sin{x} \cdot \cos{x}$
now integrate
• April 5th 2010, 12:41 AM
ninjaku
Quote:
Originally Posted by skeeter
$\frac{\tan{x}}{\sec^2{x}} = \tan{x} \cdot \cos^2{x} = \sin{x} \cdot \cos{x}$
now integrate
Ok after your suggestion. I did the problem two ways.
The first way I did using:
$u=sinx$
$du=cosx$
and got the answer to be $\frac{sin^2x}{2} + c.$
However when I checked it in my calculator, it said the answer was instead, $-\frac{cos^2x}{2} +c.$
Which I found to be the answer when I set :
$u=cosx$
$du=-sinx$
I would just like to know if both of those are correct, or only one.
And if only one of them is correct, how would I know which term I should pick to use for the substitution?
• April 5th 2010, 12:50 AM
harish21
Quote:
Originally Posted by ninjaku
Ok after your suggestion. I did the problem two ways.
The first way I did using:
$u=sinx$
$du=cosx$
and got the answer to be $\frac{sin^2x}{2} + c.$
However when I checked it in my calculator, it said the answer was instead, $-\frac{cos^2x}{2} +c.$
Which I found to be the answer when I set :
$u=cosx$
$du=-sinx$
I would just like to know if both of those are correct, or only one.
And if only one of them is correct, how would I know which term I should pick to use for the substitution?
$\frac{sin^2x}{2} + c = \frac{1-cos^2x}{2} + c= \frac{1}{2}-\frac{cos^2x}{2} +c = -\frac{cos^2x}{2} +c'$
Note: $c' = c + \frac{1}{2}$
these two terms $\frac{sin^2x}{2} + c$ and $-\frac{cos^2x}{2} +c'$ differ only by a constant. Either answer works in this case.
• April 5th 2010, 01:15 AM
ninjaku
That makes a lot of sense, thanks. | 2015-08-02T01:36:11 | {
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https://math.stackexchange.com/questions/1568288/volumes-of-solid-of-revolution-sinx-2-fake-proof | # Volumes of solid of revolution: sin(x) + 2 fake proof
I just recently learned about volumes of solids of revolution in my AP Calculus class and tried to create a problem to connect it to related rates. In this process, I found an error that neither my teacher or I seem to be able to figure out. It goes as follows:
The region contained by the function $y=\sin(x) + 2$, the line $x=2$, and the $y$-axis as shown below is revolved around the $x$-axis.
Example Image
We find the volume of the solid created by the disk method as follows:
$$V=\pi r^2 * \text{thickness}$$
$$r=\sin(x) + 2$$
$$V=\int_{0}^{2\pi} (\sin(x)+2)^2dx$$
$$V=9\pi ^ 2$$
We also know that $\int_{0}^{2\pi} \sin(x)dx=0$ therefore $\int_{0}^{2\pi}2dx=\int_{0}^{2\pi}(\sin(x)+2)dx$ so if we revolve a new region with the line $y=2$ in place of $y=\sin(x) + 2$ : The volume of the two solids should be the same.
Example Image
But we know that revolving the above solid around the $x$-axis would create a cylinder with volume $A=\pi r^2h$. In this problem, $r=2$, and $h=2$, so $V=8\pi^2$. Why are the volumes not equivalent?
• Why should the integral between 0 and 2 of the sine be zero? Should it be $2 \pi$? – Lonidard Dec 9 '15 at 22:03
• There are several pieces of your post that aren't rendering correctly, so it's difficult to say what your error may be. – Cameron Buie Dec 9 '15 at 22:06
• I fixed the formulas, that should be better, sorry, this is the first time I'm posting – ozay34 Dec 9 '15 at 22:10
• Since when is $$\int_{0}^{2 \pi} 2 dx = 0$$ ? – Mattos Dec 9 '15 at 22:12
• sorry, I fixed it – ozay34 Dec 9 '15 at 22:13
Hint:
The volume generated by the first half of the sine function (over $y=2$) is not the same as the volume generated by the second half (below $y=2$) because the radii of rotation are not the same.
In other words the same area generate different volumes if it is rotated with respect an axis with different radii of rotation.
One possible source of error is this: you are quite right that $$\int_0^{2\pi}2\,dx=\int_0^{2\pi}(2+\sin x)\,dx,$$ but this is not what you're dealing with! Rather, your integrand is $$(2+\sin x)^2=4+4\sin x+(\sin x)^2,$$ and so your volume is $$\pi\int_0^{2\pi}(2+\sin x)^2\,dx=4\pi\int_0^{2\pi}\,dx+\pi\int_0^{2\pi}(\sin x)^2\,dx.$$
Can you take it from there?
Another error is the assumption that the area of the solid of rotation $y=2$ is the same as the solid of rotation of $y=2+\sin x.$ It's true that the average radius is the same. However, the average cross-sectional area is not the same. (In general, the average of squared is not the square of the average.)
• but even if the average cross sectional area isn't the same for $y=2+sinx$, the added part for the first $\pi$ is compensated for by the second part from $\pi$ to $2\pi$, or so I thought. But as Emilio said, that isn't equivalent. I don't understand how that statement is true though because $\int_{0}^{2\pi}sinxdx=0$ – ozay34 Dec 9 '15 at 22:28
• Ok, thinking about it, I understand what Emilio is talking about, but I guess because I cant visualize it, it seems like it should work – ozay34 Dec 9 '15 at 22:31
• The kicker is that if $0<t\le1,$ then $$(2+t)^2-2^2=4+4t+t^2-4=4t+t^2$$ and $$2^2-(2-t)^2=4-(4-4t+t^2)=4-4+4t-t^2=4t-t^2.$$ Hence, $$(2-t)^2-2^2>2^2-(2-t)^2,$$ even though $$2+t-2=2-(2-t).$$ – Cameron Buie Dec 9 '15 at 22:39
• Translated into terms of volumes, the section between $x=0$ and $x=\pi$ has more extra volume (compared to the cylinder) than the section between $x=\pi$ and $x=2\pi$ loses (compared to the cylinder). – Cameron Buie Dec 9 '15 at 22:42 | 2019-07-18T11:49:09 | {
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http://math.stackexchange.com/questions/327159/matrix-inverse-question | # Matrix Inverse Question
Let $C$ be an invertible 2x2 matrix such that:
$$C^{-1} \cdot \begin{bmatrix}1 \\ 2\end{bmatrix} = \begin{bmatrix}3 \\ 4\end{bmatrix}$$
$$C^{-2} \cdot \begin{bmatrix}9 \\ 5\end{bmatrix} = \begin{bmatrix}3 \\ 4\end{bmatrix}$$
Find $2\times2$ matrices $A$ and $B$ so that $CA=B$ and solve for $C$.
-
Can you please format your question using LaTex / MathJax? I am having a tough time reading it. Regards – Amzoti Mar 11 '13 at 4:27
I am not familiar with putting in matrices :/ each matrix one column and 2 rowes; hope that helps – IndividualThinker Mar 11 '13 at 4:30
Is $C^{-1}, C^{-2}$ supposed to represent column 1 and column 2? Also, did I capture what you were trying to write? You can find guidance on Latex / MathJax in the FAQ (see link on right of top-of-page). Regards – Amzoti Mar 11 '13 at 4:35
$$\pmatrix{1\cr2\cr}=C\pmatrix{3\cr4\cr}$$
$$\pmatrix{9\cr5\cr}=CC\pmatrix{3\cr4\cr}=C\pmatrix{1\cr2\cr}$$
Now do you see what to use for $A$ and $B$?
-
Got it ! thanks! – IndividualThinker Mar 11 '13 at 15:17
Hint $\rm\ C\,(C^{-1} u = v)\:\Rightarrow\: u\, =\, \color{#C00}{C v}$
And $\rm\ C^2\, (C^{-2}w = v)\:\Rightarrow\: w = C(\color{#C00}{Cv}) = Cu$
Thus $\rm\ C\, (u,v) = (w,u)$
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https://math.stackexchange.com/questions/2420091/find-expectation-of-fracx-1-cdots-x-mx-1-cdots-x-n-when-x-1-l | # Find expectation of $\frac{X_1 + \cdots + X_m}{X_1 + \cdots + X_n}$ when $X_1,\ldots,X_n$ are i.i.d
Let $$X_1, \ldots, X_n$$ be i.i.d. random variables with expectation $$a$$ and variance $$\sigma^2$$, taking only positive values. Let $$m < n$$. Find the expectiation of $$\displaystyle\frac{X_1 + \cdots + X_m}{X_1 + \cdots + X_n}$$.
My attemps to solve this probles are rather straightforward. Denote $$X = X_1 + \cdots + X_m$$ and $$Y = X_{m+1} + \dots + X_n$$. So, $$X$$ has the expectation $$ma$$ and the variance $$m\sigma^2$$. And $$Y$$ has the expectation $$(n-m)a$$ and variance $$(n-m)\sigma^2$$. And also $$X$$ and $$Y$$ are independent. So we can compute the expectation by the definition $$\mathbb{E}\displaystyle\frac{X}{X+Y} = \int\limits_{\Omega^2}\frac{X(\omega_1)}{X(\omega_1) + Y(\omega_2)}\mathbb{P}(d\omega_1)\mathbb{P}(d\omega_2)$$. But we do not know the distribution, so we do not have chance to calculate it.
I would be glad to any help or ideas!
• Hint. You have $$\mathbb{E}\left[ \frac{X_1}{X_1+\cdots+X_n} \right] = \cdots = \mathbb{E}\left[ \frac{X_n}{X_1+\cdots+X_n} \right]$$ and $$\frac{X_1+\cdots+X_n}{X_1+\cdots+X_n} = 1.$$ – Sangchul Lee Sep 7 '17 at 10:29
• @SangchulLee how do you prove $\mathbb{E}\left[ \frac{X_1}{X_1+\cdots+X_n} \right] = \mathbb{E}\left[ \frac{X_n}{X_1+\cdots+X_n} \right]$ ? It doesn't look obvious to me. – Gabriel Romon Sep 7 '17 at 12:30
• @LeGrandDODOM, Since $X_1, \cdots, X_n$ are i.i.d., they are exchangable: for any permutation $\sigma$ on $\{1,\cdots,n\}$ we have the following equality in distribution: $$(X_1, \cdots, X_n) \stackrel{d}{=} (X_{\sigma(1)}, \cdots, X_{\sigma(n)} ).$$ And since the expectation depends only on the distribution, we have $$\mathbb{E}\left[\frac{X_1}{X_1+\cdots+X_n}\right] = \mathbb{E}\left[\frac{X_{\sigma(n)}}{X_{\sigma(1)}+\cdots+X_{\sigma(n)}}\right] = \mathbb{E}\left[\frac{X_{\sigma(n)}}{X_1+\cdots+X_n}\right].$$ – Sangchul Lee Sep 7 '17 at 12:33
• @SangchulLee if say $s_{i} = \frac{X_{i}}{\sum_{j=1}^{n}}X_{j}$, then while calculating expectation (in continuous case) we will have $s_{i}$ under the integrsl sign too,that is while calculating $E(s_{i})$ the term inside integral that is $s_{i}$ will be varing ,so how can we show that expectation of each $s_{i}$ are the same? or if $X_{i}$ are identically distributed then does that mean $s_{i}$ are identically distributed? I too get theintuition that they are same but how do i prove it? – BAYMAX Sep 16 '17 at 1:06
• @BAYMAX, If $X_i$ has common p.d.f. $f$, then $$\mathbb{E}\left[\frac{X_i}{X_1+\cdots+X_n}\right]=\int_{(0,\infty)^n}\frac{x_i}{x_1+\cdots+x_n}f(x_1)\cdots f(x_n)\,dx_1\cdots dx_n.$$ Now you can interchange the role of $x_1$ and $x_i$ to find that this expectation does not depend on $i$. This line of reasoning can be extended to arbitrary distribution on $(0,\infty)$. – Sangchul Lee Sep 16 '17 at 2:03
Suppose $$S_m=\sum\limits_{i=1}^{m} X_i$$ and $$S_n=\sum\limits_{i=1}^n X_i$$.
Now, $$\frac{X_1+X_2+\cdots+X_n}{S_n}=1\,, \text{ a.e. }$$
Therefore,
$$\mathbb E\left(\frac{X_1+X_2+\cdots+X_n}{S_n}\right)=1$$
Since $$X_1,\ldots,X_n$$ are i.i.d (see @SangchulLee's comments on main), we have for each $$i$$,
$$\mathbb E\left(\frac{X_i}{S_n}\right)=\frac{1}{n}$$
So for $$m\le n$$, $$\mathbb E\left(\frac{S_m}{S_n}\right)=\sum_{i=1}^m \mathbb E\left(\frac{X_i}{S_n}\right)=\frac{m}{n}$$
• How to prove that $\frac{X_i}{S_n}$ and $\frac{X_j}{S_n}$ are independent? – Alex Grey Sep 7 '17 at 11:10
• @AlexGrey That isn't really used anywhere. As the $X_i$'s are identically distributed, $E\left(\frac{X_1}{S_n}\right)=E\left(\frac{X_2}{S_n}\right)=...=E\left(\frac{X_n}{S_n}\right)$ and we simultaneously have $\sum_{i=1}^nE\left(\frac{X_i}{S_n}\right)=1$. – StubbornAtom Sep 7 '17 at 11:18
• @AlexGrey, And in general they are not independent. – Sangchul Lee Sep 7 '17 at 11:21
• @StubbornAtom How do you prove $E\left(\frac{X_1}{S_n}\right)=E\left(\frac{X_2}{S_n}\right)$ ? It doesn't look obvious to me. – Gabriel Romon Sep 7 '17 at 12:30
• @LeGrandDODOM I feel we don't need anything more after Sangchul's comment. – StubbornAtom Sep 7 '17 at 13:10 | 2021-05-07T09:34:35 | {
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https://math.stackexchange.com/questions/1964296/is-there-a-general-method-to-find-integer-x-y-solutions-to-ax-by-1 | # Is there a general method to find integer x,y solutions to $A^x=B^y-1$?
## Problem
For the equation:
$$12^x=5^y-1$$
I want to prove it has no solutions with x,y being positive integers.
## Question
Is there a general method for solving this type of equation? (It looks vaguely like a Pell equation, but not close enough that I can see how to solve it with standard methods)
If not, is there an elegant method to prove it for the particular case here (with $$A=12$$ and $$B=5$$)?
## What I've tried
Thinking modulo 12, the LHS = 0, and the RHS is 0 if and only if y is even. Writing $$y=2z$$, I can then factorize the RHS into $$(5^z+1)(5^z-1)$$
Both factors are even and by thinking modulo 3, only one of these factors can be divisible by 3. So I conclude that I need something like $$5^z+1=2^?3^x$$ and $$5^z-1=2^?$$ or vice versa.
Subtracting these equations I need $$2=2^?3^x-2^?$$.
If I now think in binary, these equations look like $$10_2 = (11_2)^x100..00_2 - 100...00_2$$.
It seems to make sense (but I don't see how to mathematically express this idea) that the only way this equation will work is as $$5^1+1=2.3$$ and $$5^1-1=2.2$$ but this solution results in a LHS of 24, which is not a power of 12.
However, I feel there must be a less convoluted proof!
• what is $\gcd(5^z - 1, 5^z + 1)?$ – Will Jagy Oct 11 '16 at 20:15
• Two, so this certainly cuts down the options for the factors a great deal (as one factor must have just a single 2). Is there more I should conclude? – Peter de Rivaz Oct 11 '16 at 20:18
• Mihăilescu's theorem kills all of these problems. – Fan Zheng Oct 11 '16 at 20:20
• that should be enough. Meanwhile, over the past few days I have been fiddling with questions such as $7^x - 3^y = 100,$ there seems to be a procedure but it is not easy math.stackexchange.com/questions/1946621/… – Will Jagy Oct 11 '16 at 20:21
As an illustration, let us solve $3^A - 2^B = 1,$ which will finish the other answer. We suspect the largest solution is $9-8=1.$ Take $3^A = 2^B + 1$ and subtract $9$ from both sides, for $3^A - 9 = 2^B - 8.$ Divide out both factors and introduce new variables, for $$9 (3^x - 1) = 8(2^y - 1).$$ We will show that this is impossible with $x,y \geq 1.$
Little explanation: given $m,n \geq 2$ with $\gcd(m,n) = 1,$ we know that $m^{\varphi(n)} \equiv 1 \pmod n.$ However, there may be a smaller $k$ with $m^{k} \equiv 1 \pmod n.$ If so, we take the smallest such $k$ and call it the order, sometimes multiplicative order, of $m \pmod n.$
We proceed under the assumption that $x \geq 1$ and $y \geq 1.$
Now, $2^y \equiv 1 \pmod 9.$ This means that $$6 | y.$$
jagy@phobeusjunior:~$./order 2 9 9 6 = 2 * 3 Furthermore,$2^6 - 1 | 2^y - 1.$$$2^6 - 1 = 63 = 3^3 \cdot 7.$$ Therefore$7 | (3^x - 1),$$$3^x \equiv 1 \pmod 7.$$ Therefore $$6 | x,$$ and$3^6 - 1$divides$3^x - 1.$jagy@phobeusjunior:~$ ./order 3 7
7 6 = 2 * 3
$$3^6 - 1 = 8 \cdot 7 \cdot 13.$$
Therefore $$2^y \equiv 1 \pmod {13},$$ $$12 | y.$$
jagy@phobeusjunior:~$./order 2 13 13 12 = 2^2 * 3 In particular $$4 | y,$$ and$2^y - 1$is divisible by$15,$especially divisible by$5.$$$3^x \equiv 1 \pmod 5,$$ so $$4 | x.$$ jagy@phobeusjunior:~$ ./order 3 5
5 4 = 2^2
However, $$3^4 - 1 = 80 = 5 \cdot 16.$$ This means that $8 (2^y - 1)$ is divisible by $16,$ a contradiction of $$9 (3^x - 1) = 8(2^y - 1)$$ with $x,y \geq 1.$
The powers of $5$ mod $11$ are $5,3,4,9$, and $1$. Thus $5^y-1\in\{4,2,3,8,0\}$ mod $11$. But $12^x\equiv1^x=1$ mod $11$. So $12^x=5^y-1$ can have no solutions in positive integers.
Remark (added later, on reading the question's request for a general method): A key feature of the equation $12^x=5^y-1$ that makes a simple congruence-based approach possible is the fact that we're proving the equation has no solutions. For equations like $2^x=3^y-1$ (see Will Jagy's excellent answer), where you're trying to prove it has just one solution, a simple congruence-based approach doesn't have a chance.
If $y$ is odd then $5^y-1=1 \pmod 3$ while $12^x=0 \pmod 3$. So there are no solutions whit $y$ odd.
Let $y=2z$. Then the equation is $12^x=25^z-1$. Thinking$\pmod {13}$, the left hand side is $1$ or $-1$ while the right hand side is $0$ or $2$. Then there are no solutions with $y$ even, and therefore no solutions at all.
That one is easy, clearly $y=1$ has no solutions.
working $\bmod 3$ we get $y$ is even, so $y=2k$ with $k\geq 1$.
From here $12^x=5^{2y}-1=(5^y+1)(5^y-1)$.
Clearly one factor must be $2\times 3^x$ and the other must be $4^{x-1}\times 2$.
Therefore we have $2\times 3^x=4^{x-1}\times 2+2$ or $2\times 3^x=4^{x-1}\times 2 -2$
The first is equivalent to $3^x=4^{x-1}+1$ and the second is equivalent to $3^x=4^{x-1}-1$.
It is easy to see neither have solutions by looking at the following table:
$$\begin{pmatrix} 1 & 1 \\ 3 & 4 \\ 9 & 16\\ 27 & 64\\ 81 & 256\\ 243 & 1024\\ \end{pmatrix}$$
• I do not understand how the table means there are no solutions, can you explain more? – Peter de Rivaz Oct 11 '16 at 20:46
• @PeterdeRivaz I put a proof about $3^A - 2^B = 1$ as an answer. – Will Jagy Oct 11 '16 at 22:48 | 2020-10-28T06:33:15 | {
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https://math.stackexchange.com/questions/1128552/poissons-equation-with-robin-boundary-conditions | # Poisson's equation with Robin boundary conditions
Explain how to define $u \in H^1(U)$ to be a weak solution of Poisson's equation with Robin boundary conditions: \begin{align} \begin{cases} \, \, \, \, -\Delta u = f & \text{in }U \\ u+\frac{\partial u}{\partial v}=0 & \text{on } \partial U. \end{cases} \end{align} Discuss the existence and uniqueness of a weak solution for a given $f \in L^2(U)$.
This is Exercise 5 in Chapter 6 of PDE Evans, 2nd edition.
I would like to define the bilinear form $B[u,v]$, for all $u,v \in H_0^1(U)$. But they did not really give that to the reader, unlike in Exercises 3 and 4 in the textbook.
Should I still define $B[u,v]$? If so, then I can try to satisfy the hypotheses of the Lax-Milgram Theorem, which would allow me to assert the existence of a weak solution to this problem.
It is not appropriate to work in the space $$H^1_0(U)$$ since nonzero boundary conditions are being considered. You will have to assume some regularity of the boundary of $$U$$.
One version of Green's theorem (see e.g. the appendix in Evans) is that $$- \int_U (\Delta u) v \, dx = \int_U Du \cdot Dv \, dx -\int_{\partial U} \frac{\partial u}{\partial \nu} v \, dS.$$
A weak solution to the problem at hand can be proposed by setting $$-\Delta u = f$$ in $$U$$ and $$\dfrac{\partial u}{\partial \nu} = -u$$ on $$\partial U$$ so that $$\int_U fv = \int_U Du \cdot Dv + \int_{\partial U} uv \, dS \quad \forall v \in H^1(U)$$ or a bit more precisely $$\int_U fv = \int_U Du \cdot Dv + \int_{\partial U} \newcommand{\tr}{\mathrm{Tr}\ \! }( \tr u )(\tr v) \, dS \quad \forall v \in H^1(U)$$ where $$\tr : H^1(U) \to L^2(\partial U)$$ is the trace operator.
An appropriate bilinear form is thus given by $$B[u,v] = \int_U Du \cdot Dv + \int_{\partial U} \newcommand{\tr}{\mathrm{Tr}\ \! }( \tr u )(\tr v) \, dS, \quad u,v \in H^1(U).$$ $$B$$ is clearly bounded. As far as coercivity goes, it may be helpful to use the Rellich-Kondrachov theorem. I can follow up with a hint if you like.
It remains to show that there is a constant $$\alpha > 0$$ with the property that $$\|u\|_{H^1}^2 \le \alpha B[u,u]$$ for all $$u \in H^1(U)$$. This can be proven by contradition. Otherwise, for every $$n \ge \mathbb N$$ there would exist $$u_n \in H^1(U)$$ with the property that $$\|u_n\|^2_{H^1} > n B[u_n,u_n]$$. For each $$n$$ define $$v_n = \dfrac{u_n}{\|u_n\|_{H^1}}$$. Then $$v_n \in H^1(U)$$, $$\|v_n\|_{H^1} = 1$$, and $$B[v_n,v_n] < \dfrac 1n$$ and all $$n$$.
Here we can invoke Rellich-Kondrachov. Since the family $$\{v_n\}$$ is bounded in the $$H^1$$ norm, there is a subsequence $$\{v_{n_k}\}$$ that converges to a limit $$v \in L^2(U)$$. However, since $$\|Dv_{n_k}\|_{L^2}^2 < \dfrac{1}{n_k}$$ it is also true that $$Dv_{n_k} \to 0$$ in $$L^2$$. Thus for any $$\phi \in C_0^\infty(U)$$ you have $$\int_U v D \phi \, dx = \lim_{k \to \infty} \int_U v_{n_k} D \phi \, dx = - \lim_{k \to \infty} \int_U D v_{n_k} \phi \, dx = 0.$$ This means $$v \in H^1(U)$$ and $$D v = 0$$, from which you can conclude $$v_{n_k} \to v$$ in $$H^1(U)$$. Since $$\|v_{n_k}\|_{H^1} = 1$$ for all $$k$$ it follows that $$\|v\|_{H^1} = 1$$ as well.
Next, since $$\|\tr v_{n_k}\|_{L^2(\partial U)}^2 < \dfrac{1}{n_k}$$ and the trace operator is bounded there is a constant $$C$$ for which $$\|\tr v\|_{L^2(\partial \Omega)} \le \|\tr v - \tr v_{n_k}\|_{L^2(\partial \Omega)} + \|\tr v_{n_k}\|_{L^2(\partial \Omega)} < \frac{1}{n_k} + C \|v - v_{n_k}\|_{H^1(U)}.$$ Let $$k \to \infty$$ to find that $$\tr v = 0$$ in $$L^2(\partial U)$$.
Can you prove that if $$v \in H^1(U)$$, $$Dv = 0$$, and $$\tr v = 0$$, then $$v = 0$$? Once you have established that fact you arrive at a contradiction, since $$v$$ also satisfies $$\|v\|_{H^1} = 1$$. It follows that $$B$$ is in fact coercive.
• I tried to edit your response to fix (what I thought) were typos, and tried to clarify minor things, to avoid ambiguity and make it easier for me to follow. Lastly, I changed the $\Omega$ to $U$, just to stay consistent with Evans' notation. Are all modifications changes correct? Because, for example, we never had $\Delta u = f$ on $\partial U$, when we were given $-\Delta u = f$ in $U$ by the problem. – Cookie Feb 2 '15 at 16:37
• Looks good. Thanks for pointing out the errors. – Umberto P. Feb 2 '15 at 17:03
• There's another one (I didn't modify), I think. Shouldn't the trace operator $T$ be mapped from $H^1(U) \to L^2(\color{red}{\partial}U)$? – Cookie Feb 2 '15 at 19:38
• Yeah, I just fixed it. – Umberto P. Feb 2 '15 at 19:46
• Fundamental question, I know, but does the $\frac{\partial u}{\partial \nu}$ in the "$u+\frac{\partial u}{\partial \nu}=0$ on $\partial U$" suggest that $\partial U$ is $C^1$? I'm asking since the problem doesn't explicitly state this, and the Trace Theorem requires this in its hypothesis. I want to be absolutely sure that we can conclude $Tu=u\vert_{\partial U}$ from the Trace theorem, as you are doing so in the line after writing "a bit more precisely" in your answer. – Cookie Feb 3 '15 at 20:12 | 2019-06-16T07:16:14 | {
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https://math.stackexchange.com/questions/720206/prove-binomnm-binommk-binomnk-binomn-km-k | # Prove $\binom{n}{m}\binom{m}{k} = \binom{n}{k}\binom{n-k}{m-k}$
I need to prove the following:
If $n,m,k\in \mathbb{N}$ and $k\leq m \leq n$, then
$$\binom{n}{m}\binom{m}{k} = \binom{n}{k}\binom{n-k}{m-k}$$.
I did the following steps:
\begin{align} \require{cancel} \binom{n}{m}\binom{m}{k} &= \binom{n}{k}\binom{n-k}{m-k} \\ \frac{n!}{m!(n-m)!}\cdot \frac{m!}{k!(m-k)!} &= \frac{n!}{k!(n-k)!}\cdot \frac{(n-k)!}{(m-k)!(n-k-m+k)!}\\ \frac{n!}{\cancel{m!}(n-m)!}\cdot \frac{\cancel{m!}}{k!(m-k)!} &= \frac{n!}{k!\cancel{(n-k)}!}\cdot \frac{\cancel{(n-k)}!}{(m-k)!(n-k-m+k)!} \\ \frac{n!}{k!(n-m)!(m-k)!} &= \frac{n!}{k!(n-m)!(m-k)!} \end{align}
The question is: is my proof correct? Are all my steps valid?
Thanks
• Maybe it is just me, but I am little confused of how you went in one step from $(m-k)!(n-k-m+k)!$ to $(n-m)!(m-k)!$ – imranfat Mar 20 '14 at 19:15
• @imranfat, $(n−k−m+k)!=(n-m)!$. – DKal Mar 20 '14 at 19:16
• Ok, I see it now, there has been a complete brainfart in my head going on... – imranfat Mar 20 '14 at 19:24
• Thanks for the edit Umberto! – Jeel Shah Mar 20 '14 at 19:32
It is correct!
An other way to prove this is the following:
$$\binom{n}{m}\binom{m}{k} = \frac{n!}{m!(n-m)!} \frac{m!}{k!(m-k)!}= \frac{n!}{(n-m)!} \frac{1}{k!(m-k)!}=\frac{n!}{k!} \frac{1}{(m-k)!(n-m)!}=\frac{n!}{k!(n-k)!} \frac{(n-k)!}{(m-k)!(n-m)!}=\binom{n}{k}\frac{(n-k)!}{(m-k)!((n-k)-(m-k))!}=\binom{n}{k} \binom{n-k}{m-k}$$
• The back of the book suggested that I use the following fact $\binom{n+1}{k+1} = \frac{n+1}{k+1}\binom{n}{k}$. How would I use that? Also, on the second last portion i.e. $(n-m)!$ going to $((n-k)-(m-k))!$ is that because we have $n = n-k$ and $m=m-k$? – Jeel Shah Mar 20 '14 at 19:31
• As regards the second question: $$(n-m)!=(n-m+0)!=(n-m+k-k)!=(n-k-m+k)!=((n-k)-(m-k))!$$ – Mary Star Mar 20 '14 at 19:34
• To use the relation $$\binom{n+1}{k+1}=\frac{n+1}{k+1} \binom{n}{k}$$ you could do the following: $$\binom{n}{m}=\frac{n}{m} \binom{n-1}{m-1}=...=\frac{n \cdot ... \cdot (n-k+1)}{m \cdot ... \cdot (m-k+1)} \binom{n-k}{m-k}=\frac{\frac{n!}{(n-k)!}}{\frac{m!}{(m-k)!}} \binom{n-k}{m-k}=\frac{\frac{n!}{k!(n-k)!}}{\frac{m!}{k!(m-k)!}} \binom{n-k}{m-k}= \frac{ \binom{n}{k} }{ \binom{m}{k} } \binom{n-k}{m-k}$$ So $$\binom{n}{m} \binom{m}{k}= \frac{ \binom{n}{k} }{ \binom{m}{k} } \binom{n-k}{m-k} \binom{m}{k}= \binom{n}{k} \binom{n-k}{m-k}$$ – Mary Star Mar 20 '14 at 22:41
You appear to do this in a non-standard way, but it looks alright (reading the equations from top to bottom instead of left to right).
Ever thought of a combinatoric proof?
Looks correct to me. Alternatively, you can give a combinatorial proof: both sides of the equation count the number of options to choose subsets $K\subseteq M\subseteq N$, where $N$ is a set of $n$ elements, $M\subseteq N$ is a subset of $m$ elements, and $K\subseteq M$ is a subset of $k$ elements. On the LHS you choose $M$ and then choose $K$ in $M$. On the RHS you choose $K$ in $N$ first, and then determine $M$ by choosing $m-k$ more elements from $N\setminus K$. | 2019-06-26T10:45:31 | {
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http://tenmg.com/15jcey/8c2ca1-multiply-a-diagonal-matrix | Effect of multiplying a matrix by a diagonal matrix. Each task will calculate a subblock of the resulting matrix C. If A and B are diagonal, then C = AB is diagonal. In this post, we explain how to diagonalize a matrix if it is diagonalizable. Matrix Multiplication. Next, we are going to calculate the sum of diagonal elements in this matrix using For Loop. But to multiply a matrix by another matrix we need to do the "dot product" of rows and columns ... what does that mean? Where do our outlooks, attitudes and values come from? Accelerating the pace of engineering and science. Explicitly: Way of enlightenment, wisdom, and understanding, America, a corrupt, depraved, shameless country, The test of a person's Christianity is what he is, Ninety five percent of the problems that most people Q. The mmult program will calculate C = AB, where C, A, and B are all square matrices. Diagonal matrices. What I actually need is a method to multiply each diagonal in A by some constant (i.e. Multiplying two matrices is only possible when the matrices have the right dimensions. Effect of multiplying a matrix by a diagonal matrix. if A is of size n*m then we have vector c of length (n+m-1)). Thanks Teja for that, I have updated my question to reflect a further requirement which I don't think your solution completes? De diagonale elementen kunnen al of niet gelijk zijn aan nul. The punishment for it is real. Self-imposed discipline and regimentation, Achieving happiness in life --- a matter of the right strategies, Self-control, self-restraint, self-discipline basic to so much in life. ; Step 3: Add the products. I will calculate these offline and store them in an 3-d array "J". Explicitly: Q. With the help of Numpy matrix.diagonal() method, we are able to find a diagonal element from a given matrix and gives output as one dimensional matrix.. Syntax : matrix.diagonal() Return : Return diagonal element of a matrix Example #1 : In this example we can see that with the help of matrix.diagonal() method we are able to find the elements in a diagonal of a matrix. Based on your location, we recommend that you select: . De ×-matrix = (,) is een diagonaalmatrix als voor alle , ∈ {,, …,}: , = ≠ Diagonaalmatrices worden volledig bepaald door de waarden van de elementen op de hoofddiagonaal. tl;dr Use loops. in good habits. Not all matrices are diagonalizable. Let us define the multiplication between a matrix A and a vector x in which the number of columns in A equals the number of rows in x . In mathematics, particularly in linear algebra, matrix multiplication is a binary operation that produces a matrix from two matrices. Thanks Teja Method 3 worked out to be faster. MathWorks is the leading developer of mathematical computing software for engineers and scientists. Yes, but first it is ONLY true for a matrix which is unitary that is a matrix A for which AA'=I. In order to multiply matrices, Step 1: Make sure that the the number of columns in the 1 st one equals the number of rows in the 2 nd one. Flip square matrices over the main diagonal. A. Unable to complete the action because of changes made to the page. In addition, m >> n, and M is constant throughout the course of the algorithm, with only the elements of D changing. In a previous post I discussed the general problem of multiplying block matrices (i.e., matrices partitioned into multiple submatrices). in .The mmult program can be found at the end of this section. Here it is for the 1st row and 2nd column: (1, 2, 3) • (8, 10, 12) = 1×8 + 2×10 + 3×12 = 64 We can do the same thing for the 2nd row and 1st column: (4, 5, 6) • (7, 9, 11) = 4×7 + 5×9 + 6×11 = 139 And for the 2nd row and 2nd column: (4, 5, 6) • (8, 10, 12) = 4×8 + 5×10 + 6×12 = 15… What is the effect of pre-multiplying a matrix. People are like radio tuners --- they pick out and For simplicity we assume that m x m tasks will be used to calculate the solution. for loop version Elapsed time is 0.000154 seconds. Poor Richard's Almanac. example. sparse matrix multiply Elapsed time is 0.000115 seconds. In a square matrix, transposition "flips" the matrix over the main diagonal. Multiplication of diagonal matrices is commutative: if A and B are diagonal, then C = AB = BA.. iii. I wish to find the most efficient way to implement the following equation, is a m*n dense rectangular matrix (with no specific structure), and, is a m*m diagonal matrix with all positive elements. Each other elements will move across the diagonal and end up at the same distance from the diagonal, on the opposite side. Choose a web site to get translated content where available and see local events and offers. Example1 Live Demo Other MathWorks country sites are not optimized for visits from your location. have come from personal foolishness, Liberalism, socialism and the modern welfare state, The desire to harm, a motivation for conduct, On Self-sufficient Country Living, Homesteading. The effect is that of multiplying the i-th column of matrix A by the factor ki i.e. i.e. '*B; toc; Again, depending on what m and n actually are, the fastest method may be different (for this choice of m and n, it seems method 3 is somewhat faster). Topically Arranged Proverbs, Precepts, Diagonalize the matrix A=[4−3−33−2−3−112]by finding a nonsingular matrix S and a diagonal matrix D such that S−1AS=D. There are two types of multiplication for matrices: scalar multiplication and matrix multiplication. Diagonal matrix. In our next example we program a matrix-multiply algorithm described by Fox et al. Reload the page to see its updated state. Multiplying a Vector by a Matrix To multiply a row vector by a column vector, the row vector must have as many columns as the column vector has rows. Common Sayings. Hell is real. Inverse matrix Let Mn(R) denote the set of all n×n matrices with real entries. My numbers indicate that ifort is smart enough to recognize the loop, forall, and do concurrent identically and achieves what I'd expect to be about 'peak' in each of those cases. A new example problem was added.) I reshape J to an [(n^2) x m] matrix since we want to take linear combinations of its columns by postmultiplying it with the elements in D. % Preallocate J for n*n*m elements of storage. https://www.mathworks.com/matlabcentral/answers/87629-efficiently-multiplying-diagonal-and-general-matrices#answer_97203, https://www.mathworks.com/matlabcentral/answers/87629-efficiently-multiplying-diagonal-and-general-matrices#comment_170160, https://www.mathworks.com/matlabcentral/answers/87629-efficiently-multiplying-diagonal-and-general-matrices#answer_97194, https://www.mathworks.com/matlabcentral/answers/87629-efficiently-multiplying-diagonal-and-general-matrices#comment_169818, https://www.mathworks.com/matlabcentral/answers/87629-efficiently-multiplying-diagonal-and-general-matrices#comment_170168. Find the treasures in MATLAB Central and discover how the community can help you! I then discussed block diagonal matrices (i.e., block matrices in which the off-diagonal submatrices are zero) and in a multipart series of posts showed that we can uniquely and maximally partition any square matrix into block… To multiply a matrix by a scalar, multiply each element by the scalar. Quotations. Wisdom, Reason and Virtue are closely related, Knowledge is one thing, wisdom is another, The most important thing in life is understanding, We are all examples --- for good or for bad, The Prime Mover that decides "What We Are". columns of the original matrix are simply multiplied by successive diagonal elements of the Let us see with an example: To work out the answer for the 1st row and 1st column: Want to see another example? A. This program allows the user to enter the number of rows and columns of a Matrix. D = diag(v,k) places the elements of vector v on the kth diagonal. (Update 10/15/2017. listen to one wavelength and ignore the rest, Cause of Character Traits --- According to Aristotle, We are what we eat --- living under the discipline of a diet, Personal attributes of the true Christian, Love of God and love of virtue are closely united, Intellectual disparities among people and the power Consider the matrix multiplication below For the product to be a diagonal matrix, a f + b h = 0 ⇒ a f = -b h and c e + d g = 0 ⇒ c e = -d g Consider the following sets of values The the matrix product becomes: Thus, as can be seen we can obtain a diagonal matrix that is a product of non diagonal matrices. But you can do something similar. Scalar Matrix Multiplication. As an example, we solve the following problem. In addition, m >> n, and, is constant throughout the course of the algorithm, with only the elements of, I know there are tricks for a related problem (D*M*D) to reduce the number of operations considerably, but is there one for this problem? Suppose there exists an n×n matrix B such that AB = BA = In. iii. If A is an m x n matrix and B is as n x p matrix The matrix multiplication algorithm that results of the definition requires, in the worst case, multiplications of scalars and (−) additions for computing the product of two square n×n matrices. This can only be done if the number of columns in the first matrix is equal to the number of rows in the second. Ideally is there a way to factorize / rearrange this so I can compute, offline (or something similar), and update. rows of the original matrix are simply multiplied by successive diagonal elements of the diagonal Now, I can use J to quickly calculate the answer for any D. We'll try all 3 methods. In de lineaire algebra is een diagonaalmatrix een vierkante matrix, waarvan alle elementen buiten de hoofddiagonaal (↘) gelijk aan nul zijn. Example in $\def\R{\Bbb R}\R^2$. Sometimes we need to find the sum of the Upper right, Upper left, Lower right, or lower left diagonal elements. tic; D = sparse(1:m,1:m,d); A = M'*D*M; toc; tic; B = bsxfun(@times,M,sqrt(d)); B = B. Here's an example of it in action - you can see that it far outperforms the standard dense multiply, sparse matrix multiply, and for loop versions: >> onesmatrixquestion dense matrix multiply Elapsed time is 0.000873 seconds. The main diagonal (or principal diagonal or diagonal) of a square matrix goes from the upper left to the lower right. Matrices where (number of rows) = (number of columns) For the matrices with whose number of rows and columns are unequal, we call them rectangular matrices. Let A be an n×n matrix. An m times n matrix has to be multiplied with an n times p matrix. diagonal matrix. In other words, the elements in a diagonal line from element a 11 to the bottom right corner will remain the same. Sin is serious business. The effect is that of multiplying the i-th row of matrix A by the factor kii.e. Left-multiplication be a diagonal matrix does not have any simple effect on eigenvalues, and given that eigenvalues are perturbed (or destroyed) what could one possibly want to say about "corresponding" eigenvectors? Notice how this expression is linear in the entries of D. You can express D as a sum of elementary basis functions. Definition. (The pre-requisite to be able to multiply) Step 2: Multiply the elements of each row of the first matrix by the elements of each column in the second matrix. Diagonal matrices have some properties that can be usefully exploited: i. C Program to find Sum of Diagonal Elements of a Matrix. For the following matrix A, find 2A and –1A. Further, C can be computed more efficiently than naively doing a full matrix multiplication: c ii = a ii b ii, and all other entries are 0. ii. Definition 3.9 An identity matrix is square and has with all entries zero except for ones in the main diagonal. Never multiply with a diagonal matrix. Numpy provides us the facility to compute the sum of different diagonals elements using numpy.trace() and numpy.diagonal() method.. One drawback, however, is that you need to be able to store a dense [n x n x m] array, and this may not be feasible if the n and m are too large. The time required to compute this matrix expression can be dramatically shortened by implementing the following improvements: W is a diagonal matrix. Diagonal Matrices, Upper and Lower Triangular Matrices Linear Algebra MATH 2010 Diagonal Matrices: { De nition: A diagonal matrix is a square matrix with zero entries except possibly on the main diagonal (extends from the upper left corner to the lower right corner). The reason for this is because when you multiply two matrices you have to take the inner product of every row of the first matrix with every column of the second. Then the matrix A is called invertible and B is called the inverse of A (denoted A−1). the successiverows of the original matrix are simply multiplied by … P.S. A diagonal matrix is a square matrix whose off-diagonal entries are all equal to zero. by Marco Taboga, PhD. gfortran, on the other hand, does a bad job (10x or more slower) with forall and do concurrent, especially as N gets large. Matrix diagonalization is the process of performing a similarity transformation on a matrix in order to recover a similar matrix that is diagonal (i.e., all its non-diagonal entries are zero). tensorized version Elapsed time is 0.000018 seconds. D = diag(v) returns a square diagonal matrix with the elements of vector v on the main diagonal. %Generate a new d (only the diagonal entries). What about division? The effect is that of multiplying the i-th row of matrix A by the factor ki i.e. In linear algebra, a diagonal matrix is a matrix in which the entries outside the main diagonal are all zero; the term usually refers to square matrices.An example of a 2-by-2 diagonal matrix is [], while an example of a 3-by-3 diagonal matrix is [].An identity matrix of any size, or any multiple of it (a scalar matrix), is a diagonal matrix. Q. Tools of Satan. Therefore computation sqrt (W) * B multiplies the i th row of B by the i th element of the diagonal of W 1/2. where M is a m*n dense rectangular matrix (with no specific structure), and D is a m*m diagonal matrix with all positive elements. by a diagonal matrix. OK, so how do we multiply two matrices? A. But each M'*ek*M is simply M(k,:)'*M(:,k). Scalar multiplication is easy. You may receive emails, depending on your. k=0 represents the main diagonal, k>0 is above the main diagonal, and k<0 is below the main diagonal. matrix. What is the effect of post-multiplying a matrix. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share … Add to solve later Sponsored Links Matrix Multiply . As such, it enjoys the properties enjoyed by triangular matrices, as well as other special properties. What is the effect of pre-multiplying a matrix. This implies that if you calculate all the M'*ek*M beforehand, then you just need to take a linear combination of them. In addition, I can exploit symmetry within M'*M and thus skip some of the rows in J*d, further reducing operations. example. where dk, a scalar, is the kth diagonal entry of D, and ek is a [m x m] matrix with all zeros except for a 1 in the kth position along the diagonal. Matrix Multiplication. Example. To understand the step-by-step multiplication, we can multiply each value in the vector with the row values in matrix and find out the sum of that multiplication. Tactics and Tricks used by the Devil. the successive You can also select a web site from the following list: Select the China site (in Chinese or English) for best site performance. A diagonal matrix is at the same time: upper triangular; lower triangular. Inverse matrix., Addition: two matrices of the same dimensions can be added by adding their corresponding entries. Once a matrix is diagonalized it becomes very easy to raise it to integer powers. We can add, subtract, and multiply elements of Mn(R). Scalar multiplication: to multiply a matrix A by a scalar r, one Multiplication of diagonal matrices is commutative: if A and B are diagonal, then C = AB = BA. I am almost certain you can't just find M'*M and somehow do something efficiently with only that. [PDF] Matrix multiplication. The best solution is going to depend on what your m and n actually are (if you know representative values of them, you should include those in your problem statement). Matrix a is of size n * M then we have vector C of length ( n+m-1 ) ):! Numpy.Diagonal ( ) method computing software for engineers and scientists multiplying block matrices ( i.e. matrices! Are going to calculate the solution changes made to the bottom right corner will remain the same can... Scalar multiplication and matrix multiplication is a binary operation that produces a matrix which unitary... Once a matrix is at the end of this section in mathematics, particularly in linear,... How this expression is linear in the first matrix is a method to each! For matrices: scalar multiplication and matrix multiplication above the main diagonal I the! Try all 3 methods k > 0 is above the main diagonal I actually need is a diagonal.... Solution completes discover how the community can help you and has with all entries zero except for ones in main. Recommend that you select: the scalar ) denote the set of all n×n with... We assume that M x M tasks will be used to calculate the.. Only be done if the number of columns in the second the end of this section compute the of. Any D. we 'll try all 3 methods used to calculate the solution need is a to! An n×n matrix B such that AB = BA.. iii enjoys properties! Solution completes scalar '' ) and numpy.diagonal ( ) and numpy.diagonal ). Is a matrix successive columns of a multiply a diagonal matrix matrix, transposition ''. = in integer powers the main diagonal, and B are diagonal, k ) the. Upper left, lower right, the elements of a square matrix, transposition flips '' the a... Solve the following problem changes made to the page I actually need is a operation... This matrix using for Loop enter the number of rows and columns of a ( A−1... B are diagonal, then C = AB = BA = in that of multiplying block (... The diagonal, k > 0 is above the main diagonal row of a! N×N matrices with real entries diagonal ( or principal diagonal or diagonal ) a! Compute this matrix expression can be added by adding their corresponding entries or something similar ), and k 0... We recommend that you select: number of rows in the second is a to... V on the opposite side successive rows of the upper right, or lower left elements... ), and multiply it on every entry in the entries of D. you can express as. To multiply each diagonal in a by the factor ki i.e optimized for visits from your location d... Of vector v on the main diagonal, on the main diagonal regular... Web site to get translated content where available and see local events and offers matrix such. P matrix not optimized for visits from your location is above multiply a diagonal matrix main.! There a way to factorize / rearrange this so I can use J quickly! Ab = BA = in d ( only the diagonal and end up at the.. M is simply M (:,k ),k ) following improvements: W is binary! Multiplication and matrix multiplication what I actually need is a square matrix, transposition flips the... 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http://cghb.bollola.it/triple-integral-pdf.html | # Triple Integral Pdf
(5 8 5) 4 5 60 3 3 3 x x x dx x x 3 2 9 5 9 2 2 1 1 2 1026 22 1001 2. View Math267-Triple-integrals_wNotes. But the real difficulty with triple integrals is-- and I think you'll see that your calculus teacher will often do this-- when you're doing triple integrals, unless you have a very easy figure like this, the evaluation-- if you actually wanted to analytically evaluate a triple integral that has more complicated boundaries or more complicated. Katz familiar to calculus students. Evaluating triple integrals A triple integral is an integral of the form Z b a Z q(x) p(x) Z s(x,y) r(x,y) f(x,y,z) dzdydx The evaluation can be split into an "inner integral" (the integral with respect to z between limits. 1 (Iterated Integrals). Then, parallel to the axis of walk,. 4 (approximate answer, depends on what you estimated the values at the midpoints to be). P 1 f(P1)=14kg=m3 f(P2)=7kg=m3 P 2 P 3 f(P3)=9kg=m3 P 4 f(P4)=21kg=m3 Suppose f 2C(R3) measures density (kg=m3) throughout W. INTEGRAL REVIEW July 2012 Vol. If it's a 2D surface, use a double integral. 1) is the signed volume bounded by the graph z f x y over the region; that is, the volume of the part of the solid below the xy-planeis taken to be negative. Triple integral is defined and explained through solved examples. Solutions to Midterm 1 Problem 1. Figure 1 In order for the double integral to exist, it is sufficient that, for example, the region D be a closed (Jordan) measurable region and that the function f(x, y ) be continuous throughout D. The inner limit is the easiest. Integration by Parts 21 1. • M yz = RRR S xδ(x,y,z)dV is the moment about the yz-plane. Cylindrical Coordinates, page 1040 In the cylindricalcoordinatesystem(柱坐標系), a point P in three-dimensional space is represented by the ordered triple (r,θ,z), where r and θ are polar coordinates of. Set up a triple integral of a function f(x,y,z) over a ball of radius 3 centered at (0,0,0) in R3. (So think of a wall around the perimeter of the floor area R, reaching up. above z = −√4x2 +4y2. 1 (Iterated Integrals). as an iterated integral (i. For any given θ, the angle φ that M makes with the z-axis runs from φ = φmin to φ = φmax. Sam Johnson (NIT Karnataka) Triple Integrals in Rectangular Coordinates October 24, 2019 17/62. patrickJMT 357,008 views. Then, parallel to the axis of walk,. Multiple Integrals 15. By symmetry, ¯x = 0 and ¯y = 0, so we only need ¯z. The sphere x2 +y2 +z2 = 4 is the same as ˆ= 2. Let's choose the order y-z-x. Multiple Integrals 1. 3 4 4 22 1 1 5 188 8 1. analysis of Legendre polynomials triple product integral. Line integrals Z C dr; Z C a ¢ dr; Z C a £ dr (1) ( is a scalar fleld and a is a vector fleld)We divide the path C joining the points A and B into N small line elements ¢rp, p = 1;:::;N. Suppose that E is a "Type 1" region between surfaces with equations z = h(x, y) and z = k(x, y) and has perpendicular projection D on the xy-plane. You may discuss the problems with other students. [Hint: The volume of a (solid) region DˆR3 is RRR D 1dxdydz] Solution: We integrate first with respect to z, keeping (x;y) fixed. ) We will turn triple integrals into (triple) iterated integrals. 7: Triple Integrals Outcome A: Evaluate a triple integral by iterated integration. 8 The Fubini’s Method to evaluate triple integrals is this: We first walk on a line parallel to one of the x-axis, y-axis, or z-axis. An orientable surface, roughly speaking, is one with two distinct sides. Calculadora gratuita de. Set up but do not evaluate the triple integral for the mass M of the solid bounded by the surface z = 1 x2 y2 and the xy plane, if the density is f(x;y;z) = x+1. Triple integration exercises 1. The volume is given by the. 1) is the signed volume bounded by the graph z f x y over the region; that is, the volume of the part of the solid below the xy-planeis taken to be negative. The dV in each of the integrals can be any of the 6 permutations of dx, dy, and dz. Proposition 17. ) Chapter 8 described the same idea for solids of revolution. The triple integral of a continuous function over a general solid region. See unit IV lesson 2 for a review. A good example to think about. The double integral of a nonnegative function f(x;y) deflned on a region in the plane is associated with the volume of the region under the graph of f(x;y). •Triple Integrals can also be used to represent a volume, in the same way that a double integral can be used to represent an area. Boise State Math 275 (Ultman) Worksheet 3. by evaluating the iterated integral using any of the six possible orders (Theorem 14. Cylindrical Coordinates, page 1040 In the cylindricalcoordinatesystem(柱坐標系), a point P in three-dimensional space is represented by the ordered triple (r,θ,z), where r and θ are polar coordinates of. Lady (December 21, 1998) Consider the following set of formulas from high-school geometry and physics: Area = Width Length Area of a Rectangle Distance = Velocity Time Distance Traveled by a Moving Object Volume = Base Area Height Volume of a Cylinder Work = Force Displacement Work Done by a Constant Force. (Hindi) Complete Engineering Mathematics for GATE 43 lessons • 5 h 53 m. Homework 6 Solutions Jarrod Pickens 1. For a triple integral in a region D ˆR3, it can be evaluated by using an iterated. The following concepts may or may not be seen on the exam and there may be concepts on the exam which are not covered on this sheet. I Project your region E onto one of the xy-, yz-, or xz-planes, and. It will come as no surprise that we can also do triple integrals---integrals over a three-dimensional region. 2 MATH11007 NOTES 18: TRIPLE INTEGRALS, SPHERICAL COORDINATES. : 0^ 2 ` E ³³³ yd 3. TRIPLE INTEGRALS IN SPHERICAL & CYLINDRICAL COORDINATES Triple Integrals in every Coordinate System feature a unique infinitesimal volume element. It's difficult to explain. Hence, the triple integral is given by Note that we can change the order of integration of r and theta so the integral can also be expressed Evaluating the iterated integral, we have find that the mass of the object is 1024*pi. Triple Integral Spherical Coordinates Pdf Download >>> DOWNLOAD (Mirror #1). This is an example of a triple or volume integral. That means we need to nd a function smaller than 1+e x. CALCULUS III DOUBLE & TRIPLE INTEGRALS STEP-BY-STEP A Manual For Self-Study prepared by Antony Foster Department of Mathematics (office: NAC 6/273) The City College of The City University of New York 160 Convent Avenue At 138th Street New York, NY 10031 [email protected] (b) Reverse the order of integration to dydzdx. Just as for double integrals, a region over which a triple integral is being taken may have easier representation in another coordinate system, say in uvw-space, than in xyz-space. Find materials for this course in the pages linked along the left. For example, the face of T in the xy-plane is given by x;y 0 and 2x + 3y 6. iosrjournals. Double and Triple Integrals. Fubini's theorem for triple integrals states that the value of a triple integral of a continuous function f over a region E in R 3 is a triple iterated integral. 3 Evaluate the integral RRR T. In the triple integral , , 0 If ( , , ) = 1 then this triple integral is the same as , which is simply the volume under the surface represented by z(x,y). Single Integral - the domain is the integral I (a line). (5), if ,, then the triple. 6: Triple Integrals Thursday, April 2, 2015 3:37 PM Section 15. P 1 f(P1)=14kg=m3 f(P2)=7kg=m3 P 2 P 3 f(P3)=9kg=m3 P 4 f(P4)=21kg=m3 Suppose f 2C(R3) measures density (kg=m3) throughout W. The double integral JSf(x, y)dy dx will now be reduced to single integrals in y and then x. A volume integral is a specific type of triple integral. Areas and Distances. To approximate a volume in three dimensions, we can divide the three-dimensional region into small rectangular boxes, each \$\Delta x\times\Delta y\times\Delta z. Compute volume, mass, and center of mass using triple integrals. (1) is deflned as Z C a ¢ dr = lim N!1 XN p=1 a(xp;yp;zp) ¢ rpwhere it is assumed that all j¢rpj ! 0. 1 (Iterated Integrals). Then for some continuous function f, f: \\3 →, the triple integral (),, R ∫∫∫f xyzdVcan be expressed 6 different ways in Cartesian (rectangular) coordinates. For example, nd out ∫ 1 0 1 (x+1) p x dx >## define the integrated function >integrand <- function(x) {1/((x+1)*sqrt(x))}. EXAMPLE 4 Find a vector field whose divergence is the given F function. Double and triple integrals 5 At least in the case where f(x,y) ≥ 0 always, we can imagine the graph as a roof over a floor area R. Using Iterated Integrals to find area. Triple integration of sum of two functions is explained. In this section we practice finding the integral of such functions. Today's Goals Today's Goals 1 Be able to set up and evaluate triple integrals in cartesian coordinates. Riemann Sums , Integral Representation for lengths, Areas, Volumes and Surface areas in Cartesian and polar coordinates multiple integrals – double and triple integrals – change of order of integration- change of variable. Both of the limits diverge, so the integral diverges. Thenthedefinite. That means we need to nd a function smaller than 1+e x. La integral triple de f sobre la caja B es ZZZ B f(x,y,z)dV = l´ım l,m,n→∞ Xl i=1 m j=1 Xn k=1 f(x∗ ijk,y ∗ ijk,z ∗ ijk)∆V si el l´ımite existe. R/ Nótese que la región de integración es la parte de la esfera de centro en el origen de coordenadas y radio 1 que está contenida en el primer octante, que se muestra en la siguiente figura:. ) Chapter 8 described the same idea for solids of revolution. Suppose that w= f(x,y,z) is a continuous function on the rectangular parallelipiped R: a≤ x≤ b, c≤ y≤ d, p≤ z≤ q. , 0 ≤ y ≤ 1. Triple Integrals Part 1: De–nition of the Triple Integral We can extend the concept of an integral into even higher dimensions. ANSWERS TO REVIEW PROBLEMS A. Spherical coordinates are pictured below: The volume of the \spherical wedge" pictured is approximately V = ˆ2 sin˚ ˆ ˚: The ˆ2. Set up the integral. Triple Integrals and Triple Iterated Integrals section 13. The solid below is enclosed by x= 0, x= 1, y= 0, z= 0, z= 1, and 2x+y+2z= 6. which is an integral of a function over a two-dimensional region. f Triple Integrals in Cylindrical and Spherical Coordinates We have already seen the advantage of changing to polar coordinates in some double integral problems. This is somewhat subtle in the physical interpretation but can be summarized as "generality". The infinite series forms of the two types of triple integrals can be obtained using binomial series and integration term by term theorem. For example, nd out ∫ 1 0 1 (x+1) p x dx >## define the integrated function >integrand <- function(x) {1/((x+1)*sqrt(x))}. (iii) Change the limits of the integral and include the "r" in the integral. Step 1: Add one to the exponent. Similarly, if f(x,y,z. At any point on an orientable surface, there exists two normal vectors, one pointing in the opposite direction of the other. Integral calculus that we are beginning to learn now is called integral calculus. 2 Sketch the solid whose volume is given by the integral R 1 0 R 1−x 0 R 2−2z 0 dydzdx. the triple integral of f over the solid and denote it by RRR S f(x,y,z)dV. (b) Reverse the order of integration to dydzdx. Use the Comparison Theorem to decide if the following integrals are convergent or divergent. Usually these integrals cannot be solved. via contour integration. (So think of a wall around the perimeter of the floor area R, reaching up. Just as with double integrals, the only trick is determining the limits on the iterated integrals. I'm trying to do the same with triple integrals using Simpson's rule in VBA. The solution is found in terms of a function which is determined by means of a Fredholm integral equation of the first kind. It will cover three major aspects of integral calculus: 1. To show this, let g and h be two functions having the same derivatives on an interval I. In this section we provide a quick discussion of one such system — polar coordinates — and then introduce and investigate their ramifications for double integrals. The analogy between single and. Show Step-by-step Solutions. Our first integral could equally well be ff(x, y)dx. 1 DOUBLE INTEGRALS OVER RECTANGLES TRANSPARENCIES AVAILABLE #48 (Figures 4 and 5), #49 (Figures 7 and 8), #50 (Figure 11), #51 (Figures 12 and 13) SUGGESTED TIME AND EMPHASIS 1 2 -1 class Essential Material POINTS TO STRESS 1. TRIPLE INTEGRALS IN SPHERICAL & CYLINDRICAL COORDINATES Triple Integrals in every Coordinate System feature a unique infinitesimal volume element. The paper also summarizes the results of the survey questions given to the students in two of the courses followed by the authors own critique of the enhancement project. Thank you! Part A: Triple. In particular, if then we have. 3 0 2ˇ 0 2 1 (r+ z)rdrd dz Region from Diagram 2 3 0 2ˇ 0 2 0 5zrdrd dz Region from Diagram 2 3 0 ˇ=2 0 2 1. BYJU’S online triple integral calculator tool makes the calculation faster, and it displays the integrated value in a fraction of seconds. line integrals, we used the tangent vector to encapsulate the information needed for our small chunks of curve. One should go to the original paper to admire the ingenuity displayed in finding (1. For example, nd out ∫ 1 0 1 (x+1) p x dx >## define the integrated function >integrand <- function(x) {1/((x+1)*sqrt(x))}. And a conversion can be made between 3 forms. We used a double integral to integrate over a two-dimensional region and so it shouldn’t be too surprising that we’ll use a triple integral to integrate over a three dimensional region. If we want to convert this triple integral to cylindrical coordinates we need to rewrite xand yusing the conversion formulas from above. projection of a function on i th and j th coordinates is calculated. Section 15. 1 DOUBLE INTEGRALS OVER RECTANGLES TRANSPARENCIES AVAILABLE #48 (Figures 4 and 5), #49 (Figures 7 and 8), #50 (Figure 11), #51 (Figures 12 and 13) SUGGESTED TIME AND EMPHASIS 1 2 -1 class Essential Material POINTS TO STRESS 1. All assigned readings and exercises are from the textbook Objectives: Make certain that you can define, and use in context, the terms, concepts and formulas listed below: • Evaluate triple integrals in Cartesian Coordinates. In Rectangular Coordinates, the volume element, " dV " is a parallelopiped with sides: " dx ", " dy ", and " dz ". Cylindrical and Spherical Coordinates General substitution for triple integrals. For the calculation of areas, we use majorly integrals formulas. fdV≡ Triple integral of f over R dV = volume element in coordinate system which describes R. Definición de integral triple Una integral triple es una generalización de una integral doble en el mismo sentido que una doble es una generalización de una integral sencilla. Double Integrals Definition (1. » 2 2 » ? 4 y2 0 » ? 4 x2 y2? 4 x2 y2 x2 a x2 y2 z2 dzdxdy:. Triple Integrals in Cylindrical Coordinates Useful for circle-symmetrical integration regions and integrand functions Switch to polar coordinates for 2 of the 3 coordinates, leave the third as is x r cos y r sin z z f ( x, y , z ) f (r , , z ) dx dy dz r dr d dz Equivalent to integrate first inz , then in polar coordinates. ! Evaluate a double integral as an iterated integral. Preface This book covers calculus in two and three variables. Multivariable Calculus Seongjai Kim Department of Mathematics and Statistics Mississippi State University Mississippi State, MS 39762 USA Email: [email protected] Compute ½ T. It is merely another tool to help you get started studying. 1 Find the centroid of the solid that is bounded by the xz-plane and the hemispheresy= √ 9−x2 −z2 andy= √ 16−x2 −z2 assumingthedensityisconstant. the integral calculus courses. Cylindrical and Spherical Coordinates General substitution for triple integrals. It will come as no surprise that we can also do triple integrals—integrals over a three-dimensional region. Each ordering leads to a di erent description of the region of integration in space, and to di erent limits of integration. The first input, fun, is a function handle. c 2019 MathTutorDVD. Triple Integrals 3 Figuring out the boundaries of integration. Review of Chapter 16: Multiple Integrals Note: This review sheet is NOT meant to be a comprehensive overview of what you need to know for the exam. z = 8 − x 2 − y 2. •Triple Integrals can also be used to represent a volume, in the same way that a double integral can be used to represent an area. It has been given many names - the Project Management Triangle, Iron Triangle and Project Triangle - which should give you an idea of how important the Triple Constraint. Question: 4 otalT Credit 2 2 GPA Credit Points Earned. TRIPLE INTEGRALS 3 5B-2 Place the solid hemisphere D so that its central axis lies along the positive z-axis and its base is in the xy-plane. Ejemplo: Calcular la integral triple de f(x,y,z) = xy en la región definida por D = {( x , y , z ) Î R 3 | x 2 + y 2 + z 2 £ 1, x ³ 0, y ³ 0, z ³ 0. The solid below is enclosed by x= 0, x= 1, y= 0, z= 0, z= 1, and 2x+y+2z= 6. This means we'll write the triple integral as a double integral on the outside and a single integral on the inside of the form We'll let the -axis be the vertical axis so that the cone is the bottom and the half-sphere is the top of the ice cream cone. ) We will turn triple integrals into (triple) iterated integrals. In this video, I start discussing how a particular order of integration for a given region and integral ' makes sense '! Then I go. Changes of variable can be made using Jacobians in much the same way as for double integrals. Application is made to the case of an electrified disc with a hole in it and numerical results for the capacity of the. wikiHow is a “wiki,” similar to Wikipedia, which means that many of our articles are co-written by multiple authors. I (x − 2)2 + y2 = 4 is a circle, since. 321 Example 53. Notes on Triple integrals: Wednesday, November 26 These are some notes for my lecture on triple integrals. Find the φ-limits of integration. ) Write ZZZ U xyzdV as an iterated integral in cylindrical coordinates. The two integrals that have dy as the innermost di erential are Z2 0 Zx 0 x 0 ex(y + 2z) dydzdx + 2 0 Z2x x x z x ex(y + 2z. We'll learn that integration and di erentiation are inverse operations of each other. I've uploaded my excel vba file. Integrals 6 1. Evaluating Triple Iterated Integrals. The following concepts may or may not be seen on the exam and there may be concepts on the exam which are not covered on this sheet. • M yz = RRR S xδ(x,y,z)dV is the moment about the yz-plane. Triple integration exercises 1. The two integrals that have dy as the innermost di erential are Z2 0 Zx 0 x 0 ex(y + 2z) dydzdx + 2 0 Z2x x x z x ex(y + 2z. We used a double integral to integrate over a two-dimensional region and so it shouldn't be too surprising that we'll use a triple integral to integrate over a three dimensional region. Here we study double integrals Z Z Ω f(x;y)dxdy (5. 6 to find their infinite series forms. The Evaluation Theorem 11 1. To begin with, suppose that ˚(x;y;z) is a piecewise continuous function. INTEGRAL REVIEW July 2012 Vol. Others come from using di erent coordinate systems. z = 8 − x 2 − y 2. Write down all the conditions (boundary surfaces). 7 Triple Integrals in Cylindrical and Spherical Coordinates Example: Find the second moment of inertia of a circular cylinder of radius a about its axis of symmetry. 7 Triple Integrals 4. Six of them can be obtained by permuting the order of the variables. 333 3 3 3 3 3 x dx x x x 4 32 1 5 5 5 5 75 4. We are given some solid region E in 3-space, and a function f(x,y,z), and we want to know 'how much of f is there in the region E'. Triple integral in cylindrical coordinates (Sect. Use cylindrical or spherical coordinates to evaluate the integral Z 1 1 Zp 1 2x 0 Zp 1 x2 y2 0 e (x 2+y2+z2)3=2dzdydx: Hint: Start by sketching the solid determined by the bounds of integration. Triple Integrals: A Hemisphere Example Let R be the region of three dimensional space bounded by z ≥0 and the surface of a sphere of radius a with a center at the origin. For any given θ, the angle φ that M makes with the z-axis runs from φ = φmin to φ = φmax. 6 Triple Integrals Comments. The two integrals that have dy as the innermost di erential are Z2 0 Zx 0 x 0 ex(y + 2z) dydzdx + 2 0 Z2x x x z x ex(y + 2z. Fubini's theorem for triple integrals states that the value of a triple integral of a continuous function f over a region E in R 3 is a triple iterated integral. Lady (December 21, 1998) Consider the following set of formulas from high-school geometry and physics: Area = Width Length Area of a Rectangle Distance = Velocity Time Distance Traveled by a Moving Object Volume = Base Area Height Volume of a Cylinder Work = Force Displacement Work Done by a Constant Force. Triple integrals are the analog of double integrals for three dimensions. 14 Vector Equation of a Plane ~n(~r ~r 0) = 0 where ~nis the vector orthogonal to every vector in the given plane and ~r ~r. With these substitutions, the paraboloid becomes z=16-r^2 and the region D is given by 0<=r<=4 and 0<=theta<=2*pi. Problem 1: Set up the limits of integration for a triple integral ∫ ∫ ∫E f(x,y,z) dV where E ={(x,y,z)| 0 b x b 2, 1 b y b 2-x, 0 b z b 4- x 2 -y 2 }. 3 Triple Integrals Triple integrals of functions f (x , y, Z) of flu-ee variables are a fairly straightforward gen- erahzation of double integrals. The paper also summarizes the results of the survey questions given to the students in two of the courses followed by the authors own critique of the enhancement project. Change of Variables in Multiple Integrals: Euler to Cartan Author(s): Victor J. Convert each of the following to an equivalent triple integ4al. 14 Vector Equation of a Plane ~n(~r ~r 0) = 0 where ~nis the vector orthogonal to every vector in the given plane and ~r ~r. Integrals 6 1. So let us give here a brief introduction on integrals based on the Mathematics subject to find areas under simple curves, areas bounded by a curve and a line and area between two. The rectangular. Asymptotics of integrals of n-fold products We determine precise asymptotics in spectral parameters for integrals of n-fold products of zonal spherical harmonics on SL2(C). We'll learn that integration and di erentiation are inverse operations of each other. The simplest application allows us to compute volumes in an alternate way. Triple integration of sum of two functions is explained. 99 USD for 2 months 4 months:. Integrals with Trigonometric Functions Z sinaxdx= 1 a cosax (63) Z sin2 axdx= x 2 sin2ax 4a (64) Z sinn axdx= 1 a cosax 2F 1 1 2; 1 n 2; 3 2;cos2 ax (65) Z sin3 axdx= 3cosax 4a + cos3ax 12a (66) Z cosaxdx=. In the field of FEM, triple integrals need to be evaluated while finding the stiffness matrix, mass matrix, body force vector, etc. Convert each of the following to an equivalent triple integ4al. V = ∭ U ρ 2 sin θ d ρ d φ d θ. Engineering Mathematics III: UNIT I: Linear systems of equations: Rank-Echelon form-Normal form – Solution of linear systems – Gauss elimination – Gauss Jordon- Gauss Jacobi and Gauss Seidel methods. The double integral gives us the volume under the surface z = f(x,y), just as a single integral gives the area under a curve. (iv) Evaluate. Divide the cube into LxMxN small rectangular elements, each having. In particular, the minimum x-value occurs on the plane z= x+ 2 and the maximum xp-value occurs on the cylinder x 2+4y = 4. Integrals over V can be found as iterated integrals with the following result: FIGURE 1. donde Si es un punto en el interior de uno de estos bloques, entonces el volumen del bloque puede ser aproximado por. Don't show me this again. Math 232 Calculus III Brian Veitch Fall 2015 Northern Illinois University 15. The triple integral of f over the box B is lim E E AV if this limit exists. 4 (approximate answer, depends on what you estimated the values at the midpoints to be). Faraday's Law :. Integrals 6 1. Solution Figure 15. Integration by Parts 21 1. The definition and properties of the double integral. We could express the result in the equiv-alent form ZZZ D f(x,y,z)dxdydz = Z b 3 a3 ˆZZ R f(x,y,z)dxdy ˙ dz with f ≡ 1. Solution: First sketch the integration region. Notation: double integral of f over R= I f x y dxdy( , ) ³³ Note: Area element = dA = dx dy. (b) Let’s guess that this integral is divergent. Because if your integration order takes care of Z first, i. Setting up a Triple Integral in Spherical Coordinates. James McKernan, Maths, 18. Then came a second integral to add up the slices. 'tiled' integral3 calls integral to integrate over xmin ≤ x ≤ xmax. LECTURE 3: TRIPLE INTEGRALS (II) 5 In that case, the bigger function is the function in front, and the smaller one is the one in the back, and Dis the shadow behind the surface. The divergence theorem can also be used to evaluate triple integrals by turning them into surface integrals. Triple Integrals in Spherical Coordinates If you are studying an object with spherical symmetry, it makes sense to use coordinates to re ect that. Sam Johnson (NIT Karnataka) Triple Integrals in Rectangular Coordinates October 24, 2019 17/62. The meaning of integration. TRIPLE INTEGRALS 385 4. LECTURE 3: TRIPLE INTEGRALS (II) 5 In that case, the bigger function is the function in front, and the smaller one is the one in the back, and Dis the shadow behind the surface. The usual “divide and conquer” approach for integrating f over B leads to the triple Riemann sum whose limit (if it exists) is the triple integral of f over B: ZZZ B f(x,y,z) dV = lim. c) Explain why any ordering starting with dz is not of Type I. Triple Integrals: A Hemisphere Example Let R be the region of three dimensional space bounded by z ≥0 and the surface of a sphere of radius a with a center at the origin. Triple Integrals over a General Bounded Region. Differential equations of first order and their. 7: Triple Integrals Outcome A: Evaluate a triple integral by iterated integration. patrickJMT 357,008 views. Math 232 Calculus III Brian Veitch Fall 2015 Northern Illinois University 15. Evaluating double integrals is similar to evaluating nested functions: You work from the inside out. the triple integral of f over the solid and denote it by RRR S f(x,y,z)dV. Try our award-winning software today. Partial Di erentiation and Multiple Integrals 6 lectures, 1MA Series Dr D W Murray Michaelmas 1994 Textbooks Most mathematics for engineering books cover the material in these lectures. First, lets describe the mass of a volume. For t2R, set F(t. Definición de integral triple Una integral triple es una generalización de una integral doble en el mismo sentido que una doble es una generalización de una integral sencilla. Cylindrical and Spherical Coordinates General substitution for triple integrals. Compute volume, mass, and center of mass using triple integrals. We partition a rectangular boxlike region containing D into. Triple Integrals in Cylindrical/Spherical Coordinates Multi-Variable Calculus. We shall use the following standard definitions for Laguerre polynomials (1) and Laguerre functions (2): (2) X„(x) = e"l/2L„(x) The Laguerre functions are known to constitute a complete orthonormal set in L2(0, a> ). pdf from MATH 267 at University of Calgary. Let us divide D into n subregions di whose areas are equal to si, choose a point (ξ i. 34 videos Play all MULTIPLE INTEGRALS (Complete Playlist) MKS TUTORIALS by Manoj Sir Triple Integrals, Changing the Order of Integration, Part 1 of 3 - Duration: 12:52. The limits for z arise from expressing the equation for the surface of the ellipsoid in the form z= c. 5: Triple Integrals. Triple integrals also arise in computation of Volume (if f(x,y,z)=1, then the triple integral equals the volume of R) Force on a 3D object Average of a Function over a 3D region Center of Mass and Moment of Inertia Triple Integrals in General Regions. Sam Johnson (NIT Karnataka) Triple Integrals in Rectangular Coordinates October 24, 2019 17/62. 1 The Double Integral over a Rectangle Let f = f(x, y) be continuous on the Rectangle R: a < x < b, c < y < d. The double integral of a nonnegative function f(x;y) deflned on a region in the plane is associated with the volume of the region under the graph of f(x;y). Just as for double integrals, a region over which a triple integral is being taken may have easier representation in another coordinate system, say in uvw-space, than in xyz-space. Triple integral of "height" w = f(x,y,z) times infinitesimal volume = total 4d hypervolume under 3d region. Z 1 1 1 + e x x dx Solution: (a) Improper because it is an in nite integral (called a Type I). Warning: This is only possible if all the limits of integration are numbers and the integrand is completely separable as a prod-uct of functions of a single variable. The corresponding multiple integrals are referred to as double integrals, triple integrals, and n-tuple integrals, respectively. 1 Structure and Process: Integral Philosophy and Triple Transformation Debashish Banerji1 Abstract: This paper looks at the ongoing debate between perennialism and pluralism in religious studies and considers the category of the integral, as described by Sri Aurobindo. The infinite series forms of the two types of triple integrals can be obtained using binomial series and integration term by term theorem. 7) Example Use cylindrical coordinates to find the volume in the z > 0 region of a curved wedge cut out from a cylinder (x − 2)2 + y2 = 4 by the planes z = 0 and z = −y. The simplest application allows us to compute volumes in an alternate way. Each ordering leads to a di erent description of the region of integration in space, and to di erent limits of integration. Now we define triple integrals for functions of three variables. • If δ(x,y,z) is the density of the solid at the point (x,y,z), then M = RRR S δ(x,y,z)dV gives the mass of the solid. 1, Introduction to Determinants In this section, we show how the determinant of a matrix is used to perform a change of variables in a double or triple integral. Multiple integrals possess a number of properties similar to those. boundary surface of E is equal to the triple integral of the divergence of F over E. (Note: The paraboloids intersect where z= 4. donde Si es un punto en el interior de uno de estos bloques, entonces el volumen del bloque puede ser aproximado por. Triple Integrals in Rectangular Form Note. pdf - Google Drive Sign in. In this section we convert triple integrals in rectangular coordinates into a triple integral in either cylindrical or spherical coordinates. We illustrate with some examples. TRIPLE INTEGRALS EXTRA CREDIT FOR MATH 222-01/02 DUE NOVEMBER 16, 2011 The answers to the problems below should be presented neatly, (either typed or written very neatly). We can compute R fdA on a region R in the following way. Because if your integration order takes care of Z first, i. The first variable given corresponds to the outermost integral and is done last. with respect to each spatial variable). Math symbols defined by LaTeX package «esint» No. 7 Triple Integrals in Cylindrical and Spherical Coordinates Example: Find the second moment of inertia of a circular cylinder of radius a about its axis of symmetry. Aplicaciones de de la integral Volumen de sólidos de revolución Definición Sea una función definida en el intervalo. Notice how the inequalities involve xand y. Triple integral is defined and explained through solved examples. above z = −√4x2 +4y2. Triple integral of infinitesimal volume = total volume of 3d region. I Project your region E onto one of the xy-, yz-, or xz-planes, and. Cylindrical and Spherical Coordinates General substitution for triple integrals. Z 1 1 1 + e x x dx Solution: (a) Improper because it is an in nite integral (called a Type I). The copyright holder makes no representation about the accuracy, correctness, or. wikiHow is a “wiki,” similar to Wikipedia, which means that many of our articles are co-written by multiple authors. Let x i = x i x i1 be the width of the i’th subinterval [x i1,x i] and let the norm of the partitionkPkbethelargestofthex i’s. For a triple integral in a region D ˆR3, it can be evaluated by using an iterated. The simplest application allows us to compute volumes in an alternate way. Look for a variable that has. Sometimes we can reduce a very difficult double integral to a simple one via a substitution. P 1 f(P1)=14kg=m3 f(P2)=7kg=m3 P 2 P 3 f(P3)=9kg=m3 P 4 f(P4)=21kg=m3 Suppose f 2C(R3) measures density (kg=m3) throughout W. If f is continuous, the triple integral exists and does not depend on the choice of (x∗ ijk,y ∗ ijk,y ∗ ijk) Same properties as double integrals Mth 254H – Fall 2006 3/10 Evaluating Fubini’s Theorem: If f(x,y,z) is continuous on B =[a,b]×[c,d]×[r,s], then ZZZ R f(x,y,z)dV = Z s r Z. This description is too narrow: it's like saying multiplication exists to find the area of rectangles. It can also evaluate integrals that involve exponential, logarithmic, trigonometric, and inverse trigonometric functions, so long as the. Sketch the solid whose volume is given by the iterated integral. 3 Find the volume of the region bounded by z = 50 10y, z = 10, y = 0, and y = 4 x2. Triple integral of infinitesimal volume = total volume of 3d region. 1 2x 2y 2= 1 2(x + y) = 1 r 2. Fubini's theorem for triple integrals states that the value of a triple integral of a continuous function f over a region E in R 3 is a triple iterated integral. Definición de integral triple Una integral triple es una generalización de una integral doble en el mismo sentido que una doble es una generalización de una integral sencilla. Express the big integral like that, and evaluate each single integral separately. It will be mostly about adding an incremental process to arrive at a \total". above z = −√4x2 +4y2. The assignment is due at the beginning of class, August 6th. We'll learn that integration and di erentiation are inverse operations of each other. The triple integral of a continuous function over a general solid region. Z 1 1 1 + e x x dx Solution: (a) Improper because it is an in nite integral (called a Type I). 4 EXERCISES Review Questions 1. Read Section 16. I've uploaded my excel vba file. History of the Integral from the 17 th Century. 34 videos Play all MULTIPLE INTEGRALS (Complete Playlist) MKS TUTORIALS by Manoj Sir Triple Integrals, Changing the Order of Integration, Part 1 of 3 - Duration: 12:52. Integrals of a function of two variables over a region in R 2 are called double integrals, and integrals of a function of three variables over a region of R 3 are called triple integrals. Challenge: 11,23 4. S8: Double integrals in polar co–ordinates. Double integrals in polar coordinates. Just as the definite integral of a positive function of one variable represents the area of the region between the. MULTIPLE INTEGRALS 154 15. \mathbf {F} = – Gm\,\mathbf {\text {grad}}\,u, where G is the gravitational constant. stackexchange [22], and in a slightly less elegant form it appeared much earlier in [18]. x y z x + y = 12 2 z = 1 - x - y 2 2 141. Engineering Mathematics 1 Notes Pdf – EM 1 Notes Pdf UNIT – V. Read Section 16. Triple Integrals in Spherical Coordinates If you are studying an object with spherical symmetry, it makes sense to use coordinates to re ect that. 1 Multiple-Integral Notation Previously ordinary integrals of the form Z J f(x)dx = Z b a f(x)dx (5. When we have to integrate a function of x,y,z over all space, we write a triple integral in this way: ∫ − ∞ + ∞ ∫ − ∞ + ∞ ∫ − ∞ + ∞ (,,). Approximating Integrals In each of these cases, the area approximation got better as the width of the intervals decreased. 2 Assignments 1. It's difficult to explain. Our first integral could equally well be ff(x, y)dx. Contents 1. Then using expan-sions of K and K he had obtained himself 30 years previously—Watson [6]—he was able to evaluate these last integrals. Multiple Integrals Double Integrals over Rectangles 26 min 3 Examples Double Integrals over Rectangles as it relates to Riemann Sums from Calc 1 Overview of how to approximate the volume Analytically and Geometrically using Riemann Sums Example of approximating volume over a square region using lower left sample points Example of approximating volume over a…. Sam Johnson (NIT Karnataka) Triple Integrals in Rectangular Coordinates October 24, 2019 17/62. (a) » 2 0 » 1 0 » 3 0 p xy z2q dzdydx (b) » 2 0 » z2 0 » y z 0 p 2x yq dxdydz (c) » ˇ{ 2 0 » y 0 » x 0 cosp x y zq dzdxdy 2. However each student is responsible. Fill in the limits of integration for the integral and put the dx, dy, dz in the correct order:. Therefore, the desired function is f(x)=1 4. Triple Integrals, Changing the Order of Integration, Part 1 of 3. All these methods are Numerical. The cylindrical coordinate system describes a point (x,y,z) in rectangular space in terms of the triple (r,θ,z) where r and θ are the polar coordinates of the projection. Welcome! This is one of over 2,200 courses on OCW. Both of the limits diverge, so the integral diverges. The analogy between single and. We used a double integral to integrate over a two-dimensional region and so it shouldn't be too surprising that we'll use a triple integral to integrate over a three dimensional region. Although the prerequisite for this. Triple Integrals: A Hemisphere Example Let R be the region of three dimensional space bounded by z ≥0 and the surface of a sphere of radius a with a center at the origin. In addition, some examples are used to demonstrate the calculations. INTEGRAL REVIEW July 2012 Vol. The corresponding multiple integrals are referred to as double integrals, triple integrals, and n-tuple integrals, respectively. and above the region in the xy. INTEGRALS 289 Thus, {F + C, C ∈ R} denotes a family of anti derivatives of f. Look for a variable that has. iosrjournals. Let Ube the solid inside both the cone z= p. Integrals are often described as finding the area under a curve. 3 0 2ˇ 0 2 1 (r+ z)rdrd dz Region from Diagram 2 3 0 2ˇ 0 2 0 5zrdrd dz Region from Diagram 2 3 0 ˇ=2 0 2 1. Convert integrals into another form For a double integral in a region D ˆR2, it can be evaluated by using an iterated integral in dxdy, or dydx, or using polar coordinate. If you want to calculate the area under the curve or some definite integral in the Symbolic (Analytical) way, then it is very hard to using C++ and not very useful. We'll learn that integration and di erentiation are inverse operations of each other. Z 1 1 1 + e x x dx Solution: (a) Improper because it is an in nite integral (called a Type I). Solution: ZZ D (x +y)dA = Z1 0 Z2y −y (x+y)dxdy = Z1 0 (x2 2 +xy) x=2y x=−y = Z1 0 9y2 2 dy = 3y3 2 y=1 y=0 = 3 2. The multiple integral is a type of definite integral extended to functions of more than one real variable—for example, $f(x, y)$ or $f(x, y, z)$. Imagine you have a cube that's gets denser as you move further out towards its corners. Triple Integrals over a General Bounded Region. Integral calculus that we are beginning to learn now is called integral calculus. Una integral triple es una generalización de una integral doble en el mismo sentido que una doble es una generalización de una integral sencilla. Elementary Approach to the Dirichlet Integral 1 2. Dirichlet integral, is often evaluated using complex-analytic methods, e. Integration Method Description 'auto' For most cases, integral3 uses the 'tiled' method. (We just add a third dimension. Free triple integrals calculator - solve triple integrals step-by-step. Evaluate the triple integral. This article is about the Euler–Poisson integral. Integrales Triples Hermes Pantoja Carhuavilca3 de 30. Double Integrals Definition (1. The line integral of a magnetic field around a closed path C is equal to the total current flowing through the area bounded by the contour C (Figure 2). patrickJMT 357,008 views. Homework 6 Solutions Jarrod Pickens 1. This sum has a nice interpretation. Engineering Mathematics III: UNIT I: Linear systems of equations: Rank-Echelon form-Normal form - Solution of linear systems - Gauss elimination - Gauss Jordon- Gauss Jacobi and Gauss Seidel methods. 8 The Fubini’s Method to evaluate triple integrals is this: We first walk on a line parallel to one of the x-axis, y-axis, or z-axis. Triple Integral Practice To Set Up A Triple Integral 1. Example Use cylindrical coordinates to find the volume of a curved wedge cut out from a cylinder (x − 2)2 + y2 = 4 by the planes z = 0 and. 7 We integrate the triple integral directly. Triple Integrals, Changing the Order of Integration, Part 1 of 3. and inside x2 +y2 = 4. Single Integral - the domain is the integral I (a line). set up the triple integral in terms of single integrals, but do not evaluate it). Z b a Z g 2(x) g 1(x) Z u 2(x;y) u 1(x;y) F(x;y;z)dzdydx : Now evaluate that iterated integral by go-ing from the inside to the outside. Example Use cylindrical coordinates to find the volume of a curved wedge cut out from a cylinder (x − 2)2 + y2 = 4 by the planes z = 0 and. Usually, one direction is considered to be positive, the other negative. To approximate a volume in three dimensions, we can divide the three-dimensional region into small rectangular boxes, each \(\Delta x\times\Delta y\times\Delta z. Problems: 5,7,9,13,17,33 3. Step 1: Draw a picture of E and project E onto a coordinate plane. Triple integral is an integral that only integrals a function which is bounded by 3D region with respect to infinitesimal volume. Triple Integrals in Cylindrical Coordinates A point in space can be located by using polar coordinates r,θ in the xy-plane and z in the vertical direction. Use the Comparison Theorem to decide if the following integrals are convergent or divergent. Triple integrals Triple integral examples 3c. Try our award-winning software today. To begin with, suppose that ˚(x;y;z) is a piecewise continuous function. Partial Di erentiation and Multiple Integrals 6 lectures, 1MA Series Dr D W Murray Michaelmas 1994 Textbooks Most mathematics for engineering books cover the material in these lectures. The two integrals that have dy as the innermost di erential are Z2 0 Zx 0 x 0 ex(y + 2z) dydzdx + 2 0 Z2x x x z x ex(y + 2z. The simplest application allows us to compute volumes in an alternate way. A solid region Eis said to be of type 1 if it lies between. Evaluating triple integrals A triple integral is an integral of the form Z b a Z q(x) p(x) Z s(x,y) r(x,y) f(x,y,z) dzdydx The evaluation can be split into an "inner integral" (the integral with respect to z between limits. MIT OpenCourseWare is a free & open publication of material from thousands of MIT courses, covering the entire MIT curriculum. (a)Set up but do not evaluate a single triple integral to nd the volume of Susing cylindrical coordinates. Math 21a Triple Integrals Fall, 2010 1 Evaluate the integral RRR E 2xdV, where E= {(x,y,z) : 0 ≤y≤2,0 ≤x≤ p 4 −y2,0 ≤z≤y}. b) Set up a triple integral over S in the dx dy dz ordering. The triple integral of a continuous function over a general solid region. We'll learn that integration and di erentiation are inverse operations of each other. Abraham de Moivre originally discovered this type of integral in 1733, while Gauss published the precise integral in 1809. This depends on finding a vector field whose divergence is equal to the given function. Accordingly, its volume is the product of its three sides, namely dV dx dy= ⋅ ⋅dz. Triple Integral Spherical Coordinates Pdf Download >>> DOWNLOAD (Mirror #1). Numerical Integration and Differentiation Quadratures, double and triple integrals, and multidimensional derivatives Numerical integration functions can approximate the value of an integral whether or not the functional expression is known:. 1 DOUBLE INTEGRALS OVER RECTANGLES TRANSPARENCIES AVAILABLE #48 (Figures 4 and 5), #49 (Figures 7 and 8), #50 (Figure 11), #51 (Figures 12 and 13) SUGGESTED TIME AND EMPHASIS 1 2 –1 class Essential Material POINTS TO STRESS 1. V = \iiint\limits_U {\rho d\rho d\varphi dz}. The cylindrical coordinate system describes a point (x,y,z) in rectangular space in terms of the triple (r,θ,z) where r and θ are the polar coordinates of the projection. TRIPLE INTEGRALS IN CYLINDRICAL AND SPHERICAL COORDINATES 5 3. Challenge: 11,23 4. Applications of Double/Triple Integrals. Such integrals arise whenever two functions are multiplied, with both the operands and the result represented in the Legendre polynomial basis. dimensional integrals. dimensional domain. Just as for double integrals, a region over which a triple integral is being taken may have easier representation in another coordinate system, say in uvw-space, than in xyz-space. Try to visualize the 3D shape if you can. P 1 f(P1)=14kg=m3 f(P2)=7kg=m3 P 2 P 3 f(P3)=9kg=m3 P 4 f(P4)=21kg=m3 Suppose f 2C(R3) measures density (kg=m3) throughout W. To obtain double/triple/multiple integrals and cyclic integrals you must use amsmath and esint (for cyclic integrals) packages. 5 36 Triple Integral Strategies The hard part is guring out the bounds of your integrals. Let D be the half-washer 1 x2 + y2 9, y 0, and let E be the solid region above D and below the graph z = 10 x2 y2. Integrals 6 1. where d is the radius of rotation. The divergence theorem can also be used to evaluate triple integrals by turning them into surface integrals. (5), if ,, then the triple. James McKernan, Maths, 18. Triple Integrals | x12. I Project your region E onto one of the xy-, yz-, or xz-planes, and use the boundary of this projection to nd bounds on domain D. We can interpret this result as the volume of the solid region because the integrand is 1. Aplicaciones de de la integral Volumen de sólidos de revolución Definición Sea una función definida en el intervalo. Triple integration of sum of two functions is explained. -plane defined by 0 ≤ x ≤ 2. This article considers two types of triple integrals and uses Maple for verification. z = 8 − x 2 − y 2. an integral of a function defined on some region in a plane and in three-dimensional or n-dimensional space. Notice how the inequalities involve xand y. Solution: We'll use the shadow method to set up the bounds on the integral. By using this website, you agree to our Cookie Policy. stackexchange [22], and in a slightly less elegant form it appeared much earlier in [18]. (1) is deflned as Z C a ¢ dr = lim N!1 XN p=1 a(xp;yp;zp) ¢ rpwhere it is assumed that all j¢rpj ! 0. Use a triple integral to determine the volume of the region below z = 4−xy. So then x2 +y2 = r2. Multiple integrals use a variant of the standard iterator notation. Challenge Problems. 25 3 4 3 12 4 tt t t dt 1. via contour integration. (Or vice versa. EXAMPLE 4 Find a vector field whose divergence is the given F function. MTH 254 LESSON 20. Triple Integral Calculator Added Mar 27, 2011 by scottynumbers in Mathematics Computes value of a triple integral and allows for changes in order of integration. The triple integral in this case is, Note that we integrated with respect to x first, then y, and finally z here, but in fact there is no reason to the integrals in this order. So let us give here a brief introduction on integrals based on the Mathematics subject to find areas under simple curves, areas bounded by a curve and a line and area between two. In Eastern Europe, it is known as Ostrogradsky’s. Notation: double integral of f over R= I f x y dxdy( , ) ³³ Note: Area element = dA = dx dy. 2, we showed how a function of two variables can be inte-grated over a region in 2-space and how integration over a region is equivalent to an iterated or double integral over two intervals. If we substitute back into the sum we get nX−1 i=0 G(yi)∆y. Recall that if you're looking for th volume then f. Hence h 1(y;z) = z 2 and h 2(y;z) = 4 4y2 = 2 p 1 y2. We would like to be able to integrate triple integrals for more general regions. pdf - Free download as PDF File (. 1 De nition of double integral Consider the function of two variables f(x,y) defined in the bounded region D. Hence the triple integral becomes Z Z Z D dV = Z 4 0 Z 1 1 Z 2 p 1 y2 z 2 1 dxdydz = Z 4. 6 to find their infinite series forms. fun(x,y,z) must accept a vector x and scalars y and z, and return a vector of values of the integrand. A solid region Eis said to be of type 1 if it lies between. Find the θ-limits of integration. However each student is responsible. We'll learn that integration and di erentiation are inverse operations of each other. Note, that integral expression may seems a little different in inline and display math mode - in inline mode the integral symbol and the limits are compressed. 5-8: Surface Area, Triple Integrals Friday, April 8 Surface Area Using the formula A(S) = ZZ D q 1 + f2 x + f y 2 dA, nd the surface area of a sphere of radius a. In this video, I start discussing how a particular order of integration for a given region and integral ' makes sense '! Then I go. V = ∭ U ρ 2 sin θ d ρ d φ d θ. Don't show me this again. By symmetry, ¯x = 0 and ¯y = 0, so we only need ¯z. Integrales Triples Hermes Pantoja Carhuavilca3 de 30. Triple integrals also arise in computation of Volume (if f(x,y,z)=1, then the triple integral equals the volume of R) Force on a 3D object Average of a Function over a 3D region Center of Mass and Moment of Inertia Triple Integrals in General Regions. Triple integrals are the analog of double integrals for three dimensions. Triple Integrals in Cylindrical/Spherical Coordinates Multi-Variable Calculus. Find materials for this course in the pages linked along the left. This website uses cookies to ensure you get the best experience. The meaning of integration. use the trapezoidal rule of integration to solve problems, 3. (iv) Evaluate. » Integrate can evaluate integrals of rational functions. In physics, triple integral arises in the computation of mass, volume, moment of inertia and force on a three dimensional object. If we want to convert this triple integral to cylindrical coordinates we need to rewrite xand yusing the conversion formulas from above. Triple Integrals Motivation Example An object conforms to the shape of a solid W in R3. Imagine you have a cube that's gets denser as you move further out towards its corners. (b) Let's guess that this integral is divergent. The solid below is enclosed by x= 0, x= 1, y= 0, z= 0, z= 1, and 2x+y+2z= 6. (Q9)Set up the triple integral W dV = W 1dV , using the order of integration dV = dz dy dx. 7 Triple Integrals in Cylindrical and Spherical Coordinates Example: Find the second moment of inertia of a circular cylinder of radius a about its axis of symmetry. 6 to find their infinite series forms. Compute the following integral by making a change in coordinates. Step 2: Determine the limits of integration. It is also useful in setting up triple integrals as iterated integrals to let Rbe the. Accordingly, its volume is the product of its three sides, namely dV dx dy= ⋅ ⋅dz. patrickJMT 357,008 views. The sphere x2 +y2 +z2 = 4 is the same as ˆ= 2. (Unfortunately, it's harder to draw in three dimensions. Solution: a) Sketch an arrow in the positive y direction: This arrow enters the solid at the xz-plane ( 1=0), passes through the interior (gray), and. 4 EXERCISES Review Questions 1. The Evaluation Theorem 11 1. This week’s review talks about Triple Integrals and Applications. Double Integrals Definition (1.
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http://openstudy.com/updates/55c8a646e4b0016bb0157c45 | ## anonymous one year ago The question 1B-6?
1. phi
can you post the question?
2. phi
is it what is the locus of points for the equation $\vec{OP} \cdot \textbf{u}= c | \vec{OP}|$ ?
3. anonymous
yes, I can't understand: "Then the locus is the nappe of a right circular cone with axis in the direction u and vertex angle 2θ" How can I reach this conclusion
4. anonymous
1B-6 Let O be the origin, c a given number, and u a given direction (i.e., a unit vector). Describe geometrically the locus of all points P in space that satisfy the vector equation OP · u = c | OP | . In particular, tell for what value(s) of c the locus will be (Hint: divide through by | OP | ): a) a plane b) a ray (i.e., a half-line) c) empty
5. phi
if we divide both sides by |OP| we get $\frac{OP}{|OP|} \cdot u = c$ that is two unit vectors. the maximum value of v dot u is 1 when v and u are unit vectors
6. phi
now consider c=0 any vector OP perpendicular to u , when dotted with u will give 0 |dw:1439222905180:dw|
7. phi
any point P in the plane defined by u (as the normal) will form vector OP, and after dividing by |OP| will still be a vector (of unit length) in that plane. and dotted with u, will give 0 thus c=0 results in a plane
8. phi
if |c| = 1 (c=1 or c=-1) the locus will be a ray if c=1 then any point P where O to P is in the same direction as u will, after being divided by the length of OP will give us unit vector v = u and u dot v = 1
9. anonymous
If we applied the equation of Dot Product: $\left| A \right| \left| B \right| \cos \theta = A.B = \sum_{i }^{...} a _{i}.b_{i}$ W ell, u is a unit vecton, then $\left| u \right| = 1$ then, $\frac{ OP.u }{ \left| OP \right|}=\cos \theta = c$ Off course, this is the angle between these two vector, The points P need stay
10. anonymous
points P are at $\theta$ of in relation to vectior u. But and when the corretion says: Then the locus is the nappe of a right circular cone with axis in the direction u and vertex angle 2θ 0 . I can understand.
11. phi
for 0 < |c| < 1 we get a cone
12. anonymous
Ok Thanks Phi, I will see it carefully, thanks by help me!
13. phi
what part of the answer is not clear?
14. phi
***"Then the locus is the nappe of a right circular cone with axis in the direction u and vertex angle 2θ" How can I reach this conclusion**** |dw:1439223719035:dw|
15. phi
let c= 0.5 and OP/|OP| = v v dot u = 0.5 theta = 60 degrees any point P so that OP is in the direction 60 degrees away from u (in any direction) will be scaled to unit length. and then dotted with u = 0.5
16. anonymous
Ok, Itś clear now! Thaks... Now I will remake this exercise.
17. anonymous
Unfortunately I can´t get the concept on the last part of this question, c) empty response: Locus is the origin, if c > 1 or c < −1 (division by | OP | is illegal, notice).
18. anonymous
it is just because cos or sen modules can't be larger than 1. It is | sen(x) | <=1 and |cos(x)|<=1 ?
19. phi
yes, I was confused on that for a bit. if we want $\frac{OP}{|OP|} \cdot u = 2$ for all P not at the origin, the vector OP divided by its length will give us a unit vector, call it v; and v dot u =2 will never be true. I originally thought this meant "no solution" but if P is the origin, then OP is the zero length vector and $\vec{OP} \cdot \textbf{u} = 2| OP|$ becomes $\vec{OP} \cdot \textbf{u} = 2\cdot 0= 0$ or $<0,0,0> \cdot <a,b,c> = 0$ where I am assuming we are in 3D space.
20. phi
and <0,0,0> dot u gives 0 therefore the zero vector $$\vec{\textbf{0}}$$ is a solution for |c|>1
21. phi
v dot u =2 to elaborate v dot u = |u| |v| cos A because |u| and |v| are both 1 (both are unit vectors) and the max value of cos A is 1 the max value for v dot u is 1 we can never get a value larger than 1
22. anonymous
Ok, Let's a simplest way to show "a vector divided by its length" is a unit vector.... I used matlab... >> OP = [-4 -45 1] OP = -4 -45 1 >>mag=sqrt(4^2+45^2+1^2) mag = 45.1885 >>OP_unitVector=OP./mag OP_unitVector = -0.0885 -0.9958 0.0221 mag_unitVector=sqrt(0.0885^2 + 0.9958^2 + 0.0221^2 ) mag_unitVector = 1.0000
23. phi
we can do it algebraically say we have v= <a,b,c> then $$|v|= \sqrt{a^2+b^2 + c^2}$$ (which follows from pythagoras) and $|v|^2 = a^2 + b^2 + c^2$ $\frac{v}{|v|} \cdot \frac{v}{|v|}= \frac{v \cdot v}{|v|^2}$ $v \cdot v= a^2 + b^2 + c^2$ and therefore $\frac{v \cdot v}{|v|^2} = \frac{a^2+b^2+c^2}{a^2+b^2+c^2}=1$
24. phi
if we let $u = \frac{v}{|v|}$ we have shown $u \cdot u = |u|^2= 1 \\ \sqrt{|u|^2} = \sqrt{1}\\ |u|= 1$ and therefore $\Bigg|\frac{v}{|v|}\Bigg|= 1$ ie. is unit length | 2016-10-24T05:36:12 | {
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https://math.stackexchange.com/questions/3535654/prove-the-following-inequality-int-0-frac-pi-2-fracsinx-sqrt9 | # Prove the following inequality: $\int_{0}^{\frac{\pi }{2}}\frac{sin(x)}{\sqrt{9-sin^{4}(x)}}dx\geq \frac{1}{3}$
Prove the following inequality: $$\int_{0}^{\frac{\pi}{2}}\frac{\sin x}{\sqrt{9-\sin^{4}x}}\ \mathrm dx\geq\frac{1}{3}.$$
I am thinking of replacing the equation with $$\int_{0}^{\frac{\pi }{2}}\frac{\sin x}{9-\sin^{4} x }dx\geq \frac{1}{3}$$, however I am stuck at this point.
Do you have any suggestions?
• use the fact $\sqrt{9 - \sin^4x} \leq 3$ and $\int_0^{\pi/2} sin x dx = 1$ and $\sin x$ is nonegative on $[0,\pi/2]$. Feb 5 '20 at 18:07
• You can get the proper font and spacing for $\sin$ using \sin. For operators that don't have a command of their own, you can use \operatorname{name}. Feb 5 '20 at 18:19
\begin{align*}\int_{0}^{\pi/2} \frac{\sin{x}}{\sqrt{9-\sin^4{x}}}dx &\geq \int_{0}^{\pi/2}\frac{\sin{x}}{\sqrt{9}}dx\\& = \frac{1}{3}\int_{0}^{\pi/2} \sin{x}dx \\&=\frac{1}{3}[-\cos{\pi/2}-(-\cos{0})]\\&=\frac{1}{3}[0+1]\\&=\frac{1}{3} \end{align*}
The inequality is because $$\sqrt{9-\sin^4{x}}\leq \sqrt{9}$$ due to positivity of $$\sin^4{x}$$. So ... $$\frac{1}{\sqrt{9-\sin^4{x}}}\geq \frac{1}{\sqrt{9}}$$
• It's also $< 1/(2\sqrt{2})$. Feb 5 '20 at 20:23
The given inequality can be deduced from AM-GM. Indeed$$\begin{eqnarray*} I=\int_{0}^{\pi/2}\frac{\sin x}{\sqrt{9-\sin^4 x}}\,dx &=& \int_{0}^{1}\frac{z\,dz}{\sqrt{(9-z^4)(1-z^2)}}\\ &=& \frac{1}{2}\int_{0}^{1}\frac{du}{\sqrt{(3-u)(3+u)(1-u)}}=\int_{0}^{1}\frac{dv}{\sqrt{(2+v^2)(4-v^2)}}\end{eqnarray*}$$ and over $$[0,1]$$ we have $$\sqrt{(2+v^2)(4-v^2)}=\text{GM}(2-v^2,4+v^2)\leq\text{AM}(2-v^2,4+v^2)=3$$.
This also shows that $$I$$ is an incomplete elliptic integral of the first kind. | 2021-09-22T01:57:55 | {
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https://math.stackexchange.com/questions/175963/idempotents-in-mathbb-z-n | Idempotents in $\mathbb Z_n$
An element $$a$$ of the ring $$(P,+,\cdot)$$ is called idempotent if $$a^2=a$$. An idempotent $$a$$ is called nontrivial if $$a \neq 0$$ and $$a \neq 1$$.
My question concerns idempotents in rings $$\mathbb Z_n$$, with addition and multiplication modulo $$n$$, where $$n$$ is natural number. Obviously when $$n$$ is a prime number then there is no nontrivial idempotent. If $$n$$ is nonprime it may happen, for example $$n=4, n=9$$, that also there is no.
Is it known, in general, for what $$n$$ there are nontrivial idempotents and what is a form of such idempotents?
• They certainly do occur: $n=6,a=3.$ – Andrew Jul 27 '12 at 19:07
• There is a nontrivial idempotent in $\mathbb{Z}_n$ if and only if there is $1<a<n$ such that $n\, |\, a(a-1)$. This is because $a^2 \equiv a \pmod n \Leftrightarrow a^2-a \equiv 0 \pmod n$. I'm not entirely sure where you can go with this though! – Clive Newstead Jul 27 '12 at 19:14
• @CliveNewstead Since $a$ and $a-1$ are relatively prime, you get that such $a$ exists if and only if $n$ has two relatively prime nontrivial divisors, and this is equivalent to $n$ having two distinct primes... – N. S. Jul 27 '12 at 19:31
5 Answers
As often happens when dealing with $$\mathbf{Z}_n$$, the Chinese remainder theorem is your friend. If the prime factorization of $$n$$ is $$n=\prod_i p_i^{a_i},$$ then by CRT we have an isomorphism of rings $$\mathbf{Z}_n\cong\bigoplus_i \mathbf{Z}_{p_i^{a_i}}.$$ Observe that the isomorphism maps the residue class of an integer $$m$$ (modulo $$n$$) to a vector with all the components equal to the residue class of $$m$$ (this time modulo various prime powers): $$\overline{m}\mapsto(\overline{m},\overline{m},\ldots,\overline{m}).$$ So the residue class of $$m$$ is an idempotent if and only if it is an idempotent modulo all the prime powers $$p_i^{a_i}$$.
Let us look at the case of a prime power modulus $$p^t$$. The congruence $$x^2\equiv x\pmod{p^t}$$ holds, iff $$p^t$$ divides $$x^2-x=x(x-1)$$. Here only one of the factors of, $$x$$ or $$(x-1)$$, can be divisible by $$p$$, so for the product to be divisible by $$p^t$$ the said factor then has to be divisible by $$p^t$$. Thus we can conclude that $$x\equiv 0,1 \pmod{p^t}$$ are the only idempotents modulo $$p^t$$. Therefore we require that $$m\equiv 0,1\pmod{p_i^{a_i}}$$ for all $$i$$. By CRT these congruences are independent for different $$i$$, so the number of pairwise non-congruent idempotents is equal to $$2^\ell$$, where $$\ell$$ is the number of distinct prime factors $$p_i$$ of $$n$$.
• Can you please explain why $m\equiv 0,1\pmod{p_i^{a_i}}$ ? I understood everything up till that point – Belgi Jul 28 '12 at 0:55
• @Belgi, the notation was somewhat inconsistent, and the logic a bit of patchwork. Is it clearer now? – Jyrki Lahtonen Jul 28 '12 at 4:48
• What is "m"? ... – Stephen Mar 12 '15 at 19:32
• @Stephen: An integer. More precisely, an integer such that its residue class is an idempotent as explained on the sixth line. – Jyrki Lahtonen Mar 12 '15 at 19:35
• Sorry about the bump. I was reviewing some of my old answers and felt that this could be made a lot clearer (particular in view of Stephen's comment). Hopefully I succeeded. – Jyrki Lahtonen Aug 24 '18 at 9:51
Hint Idempotents in $$\:\Bbb Z_{ n}\:$$ correspond to factorizations of $$\:n\:$$ into two coprime factors. Namely, if $$\:e^2 = e\in\Bbb Z_n\:$$ then $$\:n\:|\:e(e\!-\!1)\:$$ thus $$\:n = jk,\ j\:|\:e,\ k\:|\:e\!-\!1,\:$$ so $$\:(j,k)= 1\:$$ by $$\:(e,e\!-\!1) = 1.\:$$ Conversely if $$\:n = jk\:$$ for $$\:(j,k)= 1,\:$$ then by CRT, $$\:\Bbb Z_n\cong \Bbb Z_j\times \Bbb Z_k\:$$ which has nontrivial idempotents $$\:(0,1),\,(1,0).\:$$ It is not that difficult to explicitly work out the details of the correspondence. Some integer factorization algorithms search for such nontrivial idempotents.
This correspondence between idempotents and factorizations holds more generally at the structural level - see the Peirce Decomposition.
• It's worth noting here that if $x$ is an idempotent corresponding to a given $j$ and $k$, then $(p+)1-x$ is the other one, so it suffices to compute the idempotent corresponding to either $(0,1)$ or $(1,0)$. – rogerl May 29 '15 at 15:17
• @rogerl Yes that is clear since $\,(0,1)+(1,0) = (1,1) = 1\ \$ – Bill Dubuque Jun 14 '15 at 20:54
By the Chinese remainder theorem $\mathbf Z/n\mathbf Z$ is a product of more than one (unital) ring if and only if $n$ has more than one prime factor, and in this case $\mathbf Z/n\mathbf Z$ certainly has nontrivial idempotents. If on the other hand $n=p^k$ is a prime power, then all elements are either invertible (if not divisible by $p$) or nilpotent (if divisible by $p$) and this excludes the possibility of nontrivial idempotents. The case $n=1$ is a special case of a prime power (but the unique element now is both invertible and nilpotent; there are still no nontrivial idempotents of course).
A start: You can figure it out! Let us start with a product $mn$ of relatively prime integers, neither equal to $1$. By the Chinese Remainder Theorem, there is an $x$ such that $x\equiv 0\pmod{m}$ and $x\equiv 1 \pmod{n}$. Then $x^2\equiv x \pmod{mn}$.
Generalize to a product of $k$ relatively prime integers. If you use the prime power factorization, you can get a complete list.
Let $m=p^{c_{1}}_{1}...p^{c_{n}}_{n}$ be a prime factorization of an integer $m$ with $c_{i}\geq1$ and $p_{i}$ are distinct prime numbers. Then the ring $\mathbb{Z}/m\mathbb{Z}$ has $2^{n}$ idempotents and each one is precisely of the form $\sum\limits_{k=1}^{n}h_{k}\epsilon_{k}+m\mathbb{Z}$ where $\epsilon_{k}\in\{0,1\}$ and $h_{k}\in(\prod\limits_{\substack{i=1,\\ i\neq k}}^{n}p^{c_{i}}_{i})\mathbb{Z}$ such that $h_{k}-1\in p^{c_{k}}_{k}\mathbb{Z}$. | 2020-02-18T16:46:01 | {
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https://www.physicsforums.com/threads/affine-subspace.917547/ | # I Affine Subspace
1. Jun 14, 2017
### Erik109
Hey,
I am struggling with developing an intuition behind 'Affine Subspaces'. So far I have read the theories concerning Affine Subspaces delivered by the course book and visited several websites, however none have made it 100% clear. I feel like I have some sort of intuition for it, but I fail to apply the intuition when it comes to solving problems.
At the moment, I am often required to show why a given set is an Affine Subspace of a certain space. Because I assume it is quite hard for you to convey the intuition behind it without writing a lot, I will try to convey the way I think by approaching a sample problem. I hope you can enlighten me on the errors/misconceptions I make or perhaps add something so I can learn more.
The sample problem:
Let X1 = {f : R → R : f(0) = 1 and f is continuous}. Prove that X1 is an affine subspace of C(R, R) (the space of all continuous function with domain R and mapping to R). Hint: You have to do something with the set X0 = {γ : R → R : f(0) = 0 and f is continuous}.
So, what I think they want me to do at first is two things:
- prove that the set X0 is a subspace of C(R, R) (Closed by addition/multiplication)
- define an arbitrary element β which is an element of C(R, R) but not an element of X0
Once I have done both, I feel like I have to define an arbitrary element of X1, for example, the function ƒ, and show that this function contains both an element of X0 (let's say γ) and the arbitrary element β. In other words:
ƒ(x) = γ(x) + β
If this holds then the set X1 is an Affine Subspace of C(R, R) because it contains the sum of a subspace of C(R, R) and an element of C(R, R)?
I feel like I don't fully understand how to construct this function ƒ(x) and which arbitrary β to define. I have scanned through the answer numerous times and I kinda get their reasoning why, but I can't develop my own reasoning. There must be a logical, stepwise approach to the problem.
Erik
2. Jun 14, 2017
### WWGD
As I understand it, an affine ( and a fine , haha) subspace is just a standard subspace followed by a translation. The subspace just does not go through the origin and there are issues with closure under operations, as in f(0)+g(0)=2 here. EDIT: Like you said, $X0$ is a subspace ( contains the origin) but $X1$ does not.
3. Jun 14, 2017
### Staff: Mentor
I'm not sure what the arbitrariness of $\beta$ has to do here. In general, an affine subspace is simply a linear subspace that got shifted from the origin $\vec{0}$ by a certain, fixed vector $\vec{f}_0\,$.
So you are right, the set $\{f \in C(\mathbb{R},\mathbb{R})\,\vert \,f(0)=0\}$ is the candidate for this subspace and you have to show that it is actually a subspace, i.e. closed under addition and scalar multiplication. (There are other multiplications on $C(\mathbb{R},\mathbb{R})$ so it's better to name them correctly for not to get confused later on.) You also have to make sure that $\vec{0}$ is in this set. This guarantees that this set isn't empty.
In the next step, you have to find this vector $\vec{f}_0\,$, which isn't an arbitrary $\beta$. But you are right as there is more than one possibility in this case. One $\beta = \vec{f}_0\,$ however, will do. What could be a candidate? Chose an easy one, this makes life easier if you want to do anything with this affine space.
4. Jun 14, 2017
### mathwonk
try translating by the constant function 1 or by the function cosine, or by any function at all with value 1 at 0.
5. Jun 14, 2017
### Erik109
Thanks a lot for the fast replies! They are all truly insightful and I surely understand the concept now.
@fresh_42 Special thanks for the extensive explanation.
The only thing is that I struggle a bit with proving a given affine subspace of $C(ℝ, ℝ)$ is indeed an affine subspace. I feel like I have to define 'arbitrary' elements to show that $X_1$ is an affine subspace in general. The answer I was provided started off with:
Fix an arbitrary $f ∈ X_1$. Let $γ: ℝ → ℝ$ be defined by $γ(x) = f(x) + h(x)$, where $h : ℝ → ℝ$ (the vector shifting from the origin, you mentioned) is given by $h ≡ −1$ Note that $γ$ is a continuous function with the property $γ(0) = 0$, i.e. $γ ∈ X_0$.
So, if I understand it correctly, the way they prove it is by showing that an arbitrary element from $X_1$ can be shifted back to $X_0$? So kind of 'deconstructing' an affine subspace to a linear subspace?
PS: Got the hang of the LaTeX structure on the forums, hope this helps with the layout.
6. Jun 14, 2017
### WWGD
Yes, essentially. You can " de-translate" the affine space into a vector space by finding the right amount to translate by.
7. Jun 14, 2017
### Staff: Mentor
Every affine subspace $A$ can be written $A=V +\vec{h}$ with a vector (linear) subspace $V$ and the shifting vector $\vec{h}$.
You can observe two things here:
• $\vec{h}$ is not unique. Every other vector $\vec{h}+\vec{v}$ with $\vec{v}\in V$ does the job as well.
• The difference between two elements $\vec{a}_1 = \vec{v}_1 + \vec{h} \, , \,\vec{a}_2 = \vec{v}_2 + \vec{h}$ in $A$ is $\vec{a}_1 -\vec{a}_2 = (\vec{v}_1 + \vec{h})- (\vec{v}_2 + \vec{h})= \vec{v}_1 - \vec{v}_2$ and always an element of $V$.
Now in your case, $V= \{f_V \in C(\mathbb{R},\mathbb{R})\,\vert \,f_V(0)=0\}$ and $A= \{f_A \in C(\mathbb{R},\mathbb{R})\,\vert \,f_A(0)=1\}$, so you need a function $h=\vec{h} \in C(\mathbb{R},\mathbb{R})$ that moves (shifts) all functions in such a way, that $1=\vec{f}_A(0) = \vec{f}_V(0) +\vec{h}(0) = 0 + h(0)$. @mathwonk mentioned a few candidates for $h$ in post #4. As said above, $\vec{h}$ isn't unique, only up to elements of $V$.
Affine subspaces are not closed under addition nor under scalar multiplication. However, all differences of two vectors are, as they are vectors in the corresponding linear subspace. This way you still have properties like flatness and can calculate similar to what you can do in vector spaces, but you don't have a zero and must be more careful with addition and scalar multiplication. E.g. all tangent spaces $T_p$ are linear spaces, but if you actually want to use them as tangents, you have to consider $\vec{p} + T_p$ which is affine. This is very important as usually it's spoken about the vector space $T_p\,$, e.g. a line $\mathbb{R} \cdot \vec{l}$. And as a line it is a vector space only if it contains the origin $\vec{0}$. The actual tangent are the points $\vec{p} + t\cdot \vec{l}$. I mention this, as it might be confusing if people talk about linearity of tangent spaces, e.g. when talking about a gradient $\nabla$, although it is strictly speaking the affine space shifted by $\vec{p}$. | 2017-11-20T14:31:10 | {
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http://math.stackexchange.com/questions/103569/use-pappus-theorem-to-find-the-moment-of-a-region-limited-by-a-semi-circunferen | # Use Pappus' theorem to find the moment of a region limited by a semi-circunference.
This is part of self-study; I found this question in the book "The Calculus with Analytic Geometry" (Leithold).
$R$ is the region limited by the semi-circumference $\sqrt{r^2 - x^2}$ and the x-axis. Use Pappus' theorem to find the moment of $R$ with respect to the line $y = -4$.
Pappus' theorem (also referred to as Guldinus theorem or Pappus-Guldinus theorem) is as follows:
If $R$ is the region limited by the functions $f(x)$ and $g(x)$, then, if $A$ is the area of $R$ and $\bar{y}$ is the y-coordinate of the centroid of $R$, the volume $V$ of the solid of revolution obtained by rotating $R$ around the x-axis is given by: $V = 2\pi\bar{y}A$.
Also, what I'm calling moment (I'm not sure if this is a common term) is the quantity that is divided by the area of the region in order to find a coordinate of the centroid of the region: for example, to find the x-coordinate of the centroid ($\bar{x}$), first one finds the moment around the y-axis ($M_y$), then divides it by the area of the region ($A$).
Since the region $R$ is symmetric with respect to the y-axis, the x-coordinate of the centroid ($\bar{x}$) is zero (therefore, the moment around the y-axis ($M_y$) is zero, because $M_y = \bar{x}A$, where $A$ is the area of the region). So, I only need to find the vertical coordinate of the centroid (with respect to the line $y = -4$) and the moment around the line $y = -4$.
The book's answer for the moment around the line $y = -4$ is: $\frac{1}{2}r^3\left (\pi+\frac{4}{3}\right )$. I included two attempts below; both arrive at a same result, which is different from the result of the book.
Attempt 1
First I tried to use Pappus' theorem to find the vertical coordinate of the centroid of the semi-circular region limited by $\sqrt{r^2 - x^2}$, with respect to the x-axis (not yet with respect to the line y = -4). I will call it $\bar{y}_x$.
Since the solid of revolution obtained by rotating this semi-circular region around the x-axis is a sphere, its volume is $V = \frac{4}{3}\pi r^3$. The area of the semi-circular region is $A=\frac{\pi r^2}{2}$. Substituting $V$ and $A$ into Pappus' theorem:
$\frac{4}{3}\pi r^3 = 2\pi\bar{y}_x\frac{\pi r^2}{2}$
$\bar{y}_x = \frac{4r}{3\pi}$.
This is the vertical coordinate of the centroid with respect to the x-axis. The vertical coordinate of the centroid with respect to the line $y = -4$ is:
$\bar{y} = \frac{4r}{3\pi} + 4$.
To find the moment around the line $y = -4$, I use the fact that $\bar{y} = \frac{M_x}{A}$:
$M_x = \bar{y}A = \left ( \frac{4r}{3\pi} + 4\right )\frac{\pi r^2}{2}$.
Attempt 2
The moment of a plane region with respect to the line $y = -4$ can be found by dividing this region into infinitesimal elements of area, then multiplying the area of each element of area by the vertical coordinate of the centroid.
So, if $f(x) = \sqrt{r^2 - x^2}$, then, if I divide the semi-circular region into several rectangles of length $dx$, the area of each rectangle is $f(x) dx$ and the vertical coordinate of the centroid of each rectangle with respect to the line $y = -4$ is $\frac{f(x)}{2} + 4$. So, the moment with respect to the line $y = -4$ is:
$M_x = \int_{-r}^r \left [ f(x)\times \left ( \frac{f(x)}{2} + 4 \right ) dx \right ]$.
Solving this integral gives the following result:
$M_x = \left ( \frac{4r}{3\pi} + 4\right )\frac{\pi r^2}{2}$,
which is the same result I found in the first attempt.
Both your attempts lead to the correct result, and the solution given in the book is wrong. To see that the book's solution cannot be correct do the following "Gedankenexperiment": Assume that we are told to compute the moment with respect to the line $y=-\eta$ for some given $\eta$ instead of $4$. For an $\eta \gg r$ this moment would be approximatively proportional to $r^2$ and to $\eta$, as is the case in your solution but not in the solution given by the book: The $O(r^3)$ dependency is unwarranted. | 2016-05-30T11:00:53 | {
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https://mathoverflow.net/questions/220440/non-negative-polynomials-on-0-1-with-small-integral | # Non-negative polynomials on $[0,1]$ with small integral
Let $P_n$ be the set of degree $n$ polynomials that pass through $(0,1)$ and $(1,1)$ and are non-negative on the interval $[0,1]$ (but may be negative elsewhere).
Let $a_n = \min_{p\in P_n} \int_0^1 p(x)\,\mathrm{d}x$ and let $p_n$ be the polynomial that attains this minimum.
Are $a_n$ or $p_n$ known sequences? Is there some clever way to rephrase this question in terms of linear algebra and a known basis of polynomials?
Following Robert Israel's answer, we also scale everything to $[-1,1]$ (thus multiplying the result by 2). As he mentions, the optimal polynomial is always a square of some other polynomial, $p_{2n}=p_{2n+1}=q_n^2$, and $q_n$ is either even or odd (see Lemma below). So we are left to find the minimal $L_2[-1,1]$-norm of an odd/even polynomial $q_n$ such that $\deg q_n\leq n$ and $q_n(\pm1)=(\pm1)^n$. In other words (recall the division by 2 in the first line!), $a_{2n}=a_{2n+1}=d_n^2/2$, where $d_n$ is the distance from $0$ to the hyperplane defined by $q(1)=1$ in the space of all odd/even polynomials of degree at most $n$ with $L_2[-1,1]$-norm.
Now, this hyperplane is the affine hull of the Legendre polynomials $P_i(x)=\frac1{2^ii!}\frac{d^i}{dx^i}(x^2-1)^i$, where $i$ has the same parity as $n$ (since we know that $P_i(1)=1$ and $P_i(-1)=(-1)^i$). Next, by $\| P_i\|^2=\frac{2}{2i+1}$, our distance is $$\left(\sum_{j\leq n/2}\| P_{n-2j}\|^{-2}\right)^{-1/2} =\left(\sum_{j\leq n/2}\frac{2(n-2j)+1}{2}\right)^{-1/2}=\sqrt{\frac4{(n+1)(n+2)}},$$ attained at $$q_n(x)=\left(\sum_{i\leq n/2}\frac{2}{2(n-2i)+1}\right)^{-1} \sum_{i\leq n/2}\frac{2P_{n-2i}(x)}{2(n-2i)+1}.$$ Thus the answer for the initial question is $a_{2n}=a_{2n+1}=\frac2{(n+1)(n+2)}$ and $p_{2n}=p_{2n+1}=q_n^2$.
Lemma. For every $n$, one of optimal polynomials on $[-1,1]$ is a square of an odd or even polynomial.
Proof. Let $r(x)$ be any polynomial.which is nonnegative on $[-1,1]$ with $r(\pm 1)=1$. We will replace it by some other polynomial which has the required form, has the same (or less) degree and the same values at $\pm 1$, is also nonnegative on $[-1,1]$, and is not worse in the integral sense. Due to the compactness argument, this yields the required result.
Firstly, replacing $r(x)$ by $\frac12(r(x)+r(-x))$, we may assume that $r$ is even (and thus has an even degree $2n$). Let $\pm c_1,\dots,\pm c_{n}$ be all complex roots of $r$ (regarding multiplicities); then $$r(x)=\prod_{j=1}^n\frac{x^2-c_j^2}{1-c_j^2},$$ due to $r(\pm 1)=1$.
Now, for all $c_j\notin[-1,1]$ we simultaneously perform the following procedure.
(a) If $c_j$ is real, then we replace $\pm c_j$ by $\pm x_j=0$. Notice that $$\frac{|x^2-c_j^2|}{|1-c_j^2|}\geq 1\geq \frac{|x^2-0^2|}{|1-0^2|}.$$ for all $x\in[-1,1]$.
(b) If $c_j$ is non-real, then we choose $x_j\in[-1,1]$ such that $\frac{|c_j-1|}{|c_j+1|}=\frac{|x_j-1|}{|x_j+1|}$. Notice that all complex $z$ with $\frac{|c_j-z|}{|x_j-z|}=\frac{|c_j-1|}{|x_j-1|}$ form a circle passing through $-1$ and $1$, and the segment $[-1,1]$ is inside this circle. Therefore, for every $x\in[-1,1]$ we have $$\frac{|c_j-x|}{|x_j-x|}\geq\frac{|c_j-1|}{|x_j-1|},$$ thus $$\frac{|x^2-c_j^2|}{|1-c_j^2|} =\frac{|c_j-x|}{|c_j-1|}\cdot\frac{|c_j+x|}{|c_j+1|} \geq \frac{|x_j-x|}{|x_j-1|}\cdot\frac{|x_j+x|}{|x_j+1|} =\frac{|x^2-x_j^2|}{|1-x_j^2|}.$$ So, we replace $\pm c_j$ and $\pm\bar c_j$ by $\pm x_j$ and $\pm x_j$ (or simply $\pm c_j$ by $\pm x_j$ if $c_j$ is purely imaginary).
After this procedure has been applied, we obtain a new polynomial whose roots are in $[-1,1]$ and have even multiplicities, and its values at $\pm1$ are equal to $1$. So it is a square of some polynomial which is even/odd (since the roots are still split into pairs of opposite numbers). On the other hand, its values at every $x\in[-1,1]$ do not exceed the values of $r$ at the same points, as was showed above. So the obtained polynomial is not worse in the integral sense, as required. The lemma is proved.
• Sorry to be dense, but please say why it has to be the square of another polynomial. I can see that any zeros in the critical interval have even multiplicity, but why can't the polynomial be negative elsewhere (as allowed by OP)? Oct 9 '15 at 22:54
• The last part of the answer (after `finally') is about it; I'll try to expand it. In short: If our polynomial has a non-real root, or if it has real roots outside $[-1,1]$, then we may replace these roots by some root on $[-1,1]$ so as to lower the values of the polynomial at all points in $(-1,1)$(and keep it nonnegative). Oct 10 '15 at 7:05
• Expanded (or, I would say, rewritten). Oct 10 '15 at 7:47
For better symmetry, let me change the boundary points to $(-1,1)$ and $(1,1)$. Then it is clear that we may assume $p_n$ is an even function (and thus you really want $P_n$ to be polynomials of degree at most $n$, otherwise you'll be likely to have an infimum rather than an actual minimum): $p_{2k+1} = p_{2k}$. I get \eqalign{p_0(x) = 1, & a_0 = 2\cr p_2(x) = x^2, & a_2 = \dfrac{2}{3}\cr p_4(x) = \dfrac{(5 x^2 - 1)^2}{16},& a_4 = \dfrac{1}{3}\cr p_6(x) = \dfrac{x^2}{16} (7 x^2 - 3)^2,& a_6 = \dfrac{1}{5}\cr p_8(x) = \dfrac{1}{64} (21 x^4 - 14 x^2 + 1)^2, & a_8 = \dfrac{2}{15}\cr p_{10}(x) = \dfrac{x^2}{64} (33 x^4 - 30 x^2 + 5)^2, & a_{10} = \dfrac{2}{21}\cr p_{12}(x) = \dfrac{1}{4096} (429 x^6 - 495 x^4 + 135 x^2 - 5)^2, & a_{12} = \dfrac{1}{14}} Well, so far it looks like $a_{2n} = \dfrac{4}{(n+1)(n+2)}$
• How were these computed? Oct 9 '15 at 0:03
• The leading coefficients look like Catalan numbers when the polynomials are normalized to $\sqrt{2^{2n} p_{2n}(x)}$. Oct 9 '15 at 1:51
• The next coefficients match $-{2n \choose n-2}$. The whole triangle looks like oeis.org/A234950 read skewed, with alternating signs. See oeis.org/A062991 for a signed version. Oct 9 '15 at 3:43
• For $n$ even, $$p_{2n} = \prod_{i=1}^{n/2} \dfrac{(x^2 -c_i^2)^2}{(1-c_i^2)^2}$$ For $n$ odd, there's also a factor of $x^2$. Differentiate $a_{2n} =\int_{-1}^1 p_{2n}(x)\; dx$ with respect to each $c_i$, solve, check which solution gives the least $a_{2n}$. Oct 9 '15 at 4:49 | 2021-11-28T23:38:04 | {
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https://byjus.com/question-answer/assertion-the-chord-of-contact-of-tangent-from-a-point-p-to-a-circle-passes/ | Question
# Assertion :The chord of contact of tangent from a point $$P$$ to a circle passes through $$Q$$. If $${ l }_{ 1 }$$ and $${ l }_{ 2 }$$ are the lengths of the tangents from $$P$$ and $$Q$$ to the circle, then $$PQ$$ is equal to $$\sqrt { { { l }_{ 1 } }^{ 2 }+{ { l }_{ 2 } }^{ 2 } }$$ Reason: The equation of chord of contact of tangents from the point $$P\left( { x }_{ 1 },{ y }_{ 1 } \right)$$ to the circle $${ x }^{ 2 }+{ y }^{ 2 }={ a }^{ 2 }$$ is $$x{ x }_{ 1 }+y{ y }_{ 1 }={ a }^{ 2 }$$
A
Assertion is true, Reason is true and reason is correct explanation for Statement-1.
B
Assertion is true, Reason is true and Reason is NOT correct explanation for Assertion.
C
Assertion is true, Reason is false
D
Assertion is false, Reason is true
Solution
## The correct option is A Assertion is true, Reason is true and reason is correct explanation for Statement-1.Let $$P\equiv \left( { x }_{ 1 },{ y }_{ 1 } \right)$$ and $$Q\equiv \left( { x }_{ 2 },{ y }_{ 2 } \right)$$Let the equation of the given circle be $${ x }^{ 2 }+{ y }^{ 2 }={ a }^{ 2 }$$The equation of chord of contact of tangents from the point $$P\left( { x }_{ 1 },{ y }_{ 1 } \right)$$ to the given circle is $$x{ x }_{ 1 }+y{ y }_{ 1 }={ a }^{ 2 }$$Since it passes through $$Q\left( { x }_{ 2 },{ y }_{ 2 } \right)$$$$\therefore { x }_{ 1 }{ x }_{ 2 }+{ y }_{ 1 }{ y }_{ 2 }={ a }^{ 2 }$$ ...(1)Now $${ l }_{ 1 }=\sqrt { { { x }_{ 1 } }^{ 2 }+{ { y }_{ 1 } }^{ 2 }-{ a }^{ 2 } } ,{ l }_{ 2 }=\sqrt { { { x }_{ 2 } }^{ 2 }+{ { y }_{ 2 } }^{ 2 }-{ a }^{ 2 } }$$and $$PQ=\sqrt { { \left( { x }_{ 2 }-{ x }_{ 1 } \right) }^{ 2 }+{ \left( { y }_{ 2 }-{ y }_{ 1 } \right) }^{ 2 } }$$$$=\sqrt { \left( { { x }_{ 1 } }^{ 2 }+{ { y }_{ 1 } }^{ 2 } \right) +\left( { { x }_{ 2 } }^{ 2 }+{ { y }_{ 2 } }^{ 2 } \right) -2\left( { x }_{ 1 }{ x }_{ 2 }+{ y }_{ 1 }{ y }_{ 2 } \right) } \\ =\sqrt { \left( { { x }_{ 1 } }^{ 2 }+{ { y }_{ 1 } }^{ 2 } \right) +\left( { { x }_{ 2 } }^{ 2 }+{ { y }_{ 2 } }^{ 2 } \right) -2{ a }^{ 2 } } \\ =\sqrt { \left( { { x }_{ 1 } }^{ 2 }+{ { y }_{ 1 } }^{ 2 }-{ a }^{ 2 } \right) +\left( { { x }_{ 2 } }^{ 2 }+{ { y }_{ 2 } }^{ 2 }-{ a }^{ 2 } \right) } =\sqrt { { { l }_{ 1 } }^{ 2 }+{ { l }_{ 2 } }^{ 2 } }$$Maths
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https://math.stackexchange.com/questions/3198820/find-the-smallest-n-in-mathbbn-such-that-the-group-is-isomorphic-to-the-di | # Find the smallest $n \in \mathbb{N}$ such that the group is isomorphic to the direct product of $n$ cyclic groups
Find the smallest $$n \in \mathbb{N}$$ such that the group $$\mathbb{Z}_{6} \times \mathbb{Z}_{20} \times \mathbb{Z}_{45}$$ is isomorphic to the direct product of $$n$$ cyclic groups.
I'm not sure but if I understand correctly,
$$\mathbb{Z}_{6} \times \mathbb{Z}_{20} \times \mathbb{Z}_{45}$$ is isomorphic to $$\mathbb{Z}_{2} \times \mathbb{Z}_{3} \times \mathbb{Z}_{4} \times \mathbb{Z}_{5} \times \mathbb{Z}_{5} \times \mathbb{Z}_{9}$$,
$$\mathbb{Z}_{2} \times \mathbb{Z}_{5} \times \mathbb{Z}_{9}$$ is isomorphic to $$\mathbb{Z}_{90}$$,
$$\mathbb{Z}_{3} \times \mathbb{Z}_{4} \times \mathbb{Z}_{5}$$ is isomorphic to $$\mathbb{Z}_{60}$$,
and therefore, $$\mathbb{Z}_{6} \times \mathbb{Z}_{20} \times \mathbb{Z}_{45}$$ is isomorphic to $$\mathbb{Z}_{60} \times \mathbb{Z}_{90}$$ and the answer is $$n = 2$$. Is this a correct solution?
• I agree that the answer is $n=2$. For the sake of being thorough, I would probably explain why $n=1$ is impossible (just because you've found one situation where $n=2$ doesn't necessarily mean it's the smallest $n$). – Theo C. Apr 23 at 21:21
• Yes, it is. You just have to find the smallest number of groups of moduli such that the moduli in each group are pairwise coprime. – Bernard Apr 23 at 21:24
• After you ask a question here, if you get an acceptable answer, you should "accept" the answer by clicking the check mark $\checkmark$ next to it. This scores points for you and for the person who answered your question. You can find out more about accepting answers here: How do I accept an answer?, Why should we accept answers?, What should I do if someone answers my question?. – Shaun May 24 at 14:27
You may find the invariant factors.
Actually $$\mathbb{Z}_6\times \mathbb{Z}_{20}\times \mathbb{Z}_{45}\cong (\mathbb{Z}_2\times\mathbb{Z}_4)\times(\mathbb{Z}_3\times\mathbb{Z}_9)\times(\mathbb{Z}_5\times\mathbb{Z}_5)$$. Now we can pick the largest ones in each bracket to form $$\mathbb{Z}_4\times\mathbb{Z}_9\times\mathbb{Z}_5\cong\mathbb{Z}_{180}$$. Then we pick the largest ones of the remaining, which is $$\mathbb{Z}_2\times\mathbb{Z}_3\times\mathbb{Z}_5\cong\mathbb{Z}_{30}$$, and nothing remain. Hence the group is isomorphic to $$\mathbb{Z}_{180}\times\mathbb{Z}_{30}$$ and so $$n\le 2$$. Since the group is obviously not cyclic, we have $$n = 2$$.
I think we can always get $$n$$ for any abelian group by finding the invariant factors.
According to GAP, the group is $$\Bbb Z_{180}\times \Bbb Z_{30}$$ and, of course, not cyclic.
gap> G:=DirectProduct(CyclicGroup(6) , DirectProduct(CyclicGroup(20), CyclicGroup(45)));
<pc group of size 5400 with 8 generators>
gap> StructureDescription(G);
"C180 x C30"
gap> IsCyclic(G);
false
gap> | 2019-06-27T12:31:00 | {
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# Quick tips for adding numbers x to y
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Quick tips for adding numbers x to y [#permalink]
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10 Jun 2009, 20:46
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Here are some examples on adding numbers from x to y:
1) Add the numbers from 40 to 70, inclusive.
2) Add the even numbers from 40 to 70, inclusive.
3) Add the odd numbers from 40 to 70, inclusive.
4) Add the numbers from 40 to 70.
5) Add the even numbers from 40 to 70.
6) Add the odd numbers from 40 to 70.
[Reveal] Spoiler:
1) 1,705
2) 880
3) 825
4) 1,595
5) 770
6) 825
Last edited by I3igDmsu on 05 Jul 2009, 19:59, edited 7 times in total.
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Re: Quick tips for adding numbers x to y [#permalink]
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10 Jun 2009, 22:58
1) Add the numbers from 40 to 70, inclusive.
Formula for sum of first N natural numbers is = N(N+1)/2
Sum from 1 to 70 Sa= 70*71/2
Sum from 1 to 40 Sb= 40*41/2
So sum from 40 to 70 = Sa-Sb
2) Add the even numbers from 40 to 70, inclusive.
3) Add the odd numbers from 40 to 70, inclusive.
========================
As a general solution to all these kind of problems learn the AP series.
google on Arithemic Progression series. It is kind of difficult to write the formula here but all these calculations are tooooooo simple using this series.
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Re: Quick tips for adding numbers x to y [#permalink]
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12 Jun 2009, 14:34
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If you don't mind remembering a formula or two then yes you can.
The examples that you have given can be grouped under Arithmetic Progressions...or a finite sequence of evenly spaced numbers.
There are two formulas:
$$1. S = \frac{n}{2} [2a + (n-1)d]$$
where
$$S=\text{sum of the all the numbers in the sequence}$$
$$n=\text{total number of numbers in the sequence}$$
$$a=\text{the first number of the sequence}$$
$$d=\text{the different between any two consecutive numbers in the sequence}$$
$$2. S = \frac{n}{2} [\text{First Term}+\text{Last Term}]$$
where
$$S=\text{sum of the all the numbers in the sequence}$$
$$n=\text{total number of numbers in the sequence}$$
You can use either equation based on what is provided in the question.
Lets take your examples and solve them. For all of these we know the first and the last number so we should be fine with equation 2.
1) Add the numbers from 40 to 70, inclusive.
Solution:
Here the sequence is 40,41, 42, ...., 69, 70.
To calculate n
$$n=\text{Last number}-\text{First number} + 1$$
$$n=70-40+1 = 31$$
Using equation 1.
$$S = \frac{n}{2} [\text{First Term}+\text{Last Term}]$$
$$S = \frac{n}{2} [\text{First Term}+\text{Last Term}]$$
$$S = \frac{31}{2} [40+70]$$
$$S = \frac{31}{2} [110]$$
$$S = 31\times55$$
$$S = 1705$$
2) Add the even numbers from 40 to 70, inclusive.
Here the sequence is 40,42, 44, ...., 68, 70.
Solve using 40 as the first term and 70 as the last term
To calculate n for even (or odd) number
$$n=\frac{\text{Last number}-\text{First number}}{2} + 1$$
3) Add the odd numbers from 40 to 70, inclusive.
Here is the sequence is 41,43,....67,69
Solve using 41 as the first term and 69 as the last term
and so on...
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Re: Quick tips for adding numbers x to y [#permalink]
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30 Jun 2009, 20:11
Could someone check my answers above please?
Thanks nookway, your help is appreciated!
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Re: Quick tips for adding numbers x to y [#permalink]
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03 Jul 2009, 07:51
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Expert's post
I3igDmsu wrote:
Could someone check my answers above please?
Thanks nookway, your help is appreciated!
Don't think you need our help - this can be easily done in Excel
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Re: Quick tips for adding numbers x to y [#permalink]
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05 Jul 2009, 20:00
Done. If anyone needs practice, the answers in the spoiler are correct.
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02 Aug 2009, 21:30
Great stuff! Thanks!
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Re: Quick tips for adding numbers x to y [#permalink]
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13 Oct 2011, 14:32
I did not get the logic of these.
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Re: Quick tips for adding numbers x to y [#permalink] 13 Sep 2017, 22:23
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# Quick tips for adding numbers x to y
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Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®. | 2018-03-20T12:04:25 | {
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http://mathhelpforum.com/statistics/208657-combinations-probability-question-please-help-print.html | • Nov 28th 2012, 06:08 PM
warElephant
Bill has a small deck of 12 playing cards made up of only 2 suits of 6 cards each. Each of the 6 cards within a suit has a different value from 1 to 6; thus, for each value from 1 to 6, there are two cards in the deck with that value. Bill likes to play a game in which he shuffles the deck, turns over 4 cards, and looks for pairs of cards that have the same value. What is the chance that Bill finds at least one pair of cards that have the same value?
8/33
62/165
17/33
103/165
25/33
this is my solution but it is wrong:
an Arbitrary deck of hands with at least one pair of similar cards = A, B, C, A
the number of patterns that A,B,C,A hand appears = 4C2
Therefore, number of ways A is chosen= 12
when the first A is chosen, it locks in the value for the second A
number of ways B is chosen = 10(because by choosing a value for A, two cards are taken off the choice list)
number of ways C is chosen = 9
P(at least two cards have the same value) = (12 x 10 x 9 x 4C2)/(12 x 11 x 10 x 9) = 18/33
I hope that your explaination you will elaborate:
1. why my method is wrong
2. the actual solution.
Thank you very much for even considering my post!!!:)
• Nov 28th 2012, 09:49 PM
Scopur
P(at least 1) = 1 - P(None). This is probably more useful here.
So there are $6C4$ possible ways to select 4 distinct cards. Then there are $2^4$ different suits for those cards. Finally there are $12C6$ different cards.
So you have
$P(at least one) = 1 - P(none) = 1 - \frac{6C4*2^4}{12C6} = 1 - \frac{16}{33} = \frac{17}{33}$
• Nov 28th 2012, 10:15 PM
Soroban
Hello, warElephant!
Another approach . . .
Quote:
Bill has a small deck of 12 playing cards made up of only 2 suits of 6 cards each.
Each of the 6 cards within a suit has a different value from 1 to 6.
Bill likes to play a game in which he shuffles the deck, turns over 4 cards,
and looks for pairs of cards that have the same value.
What is the probability that Bill finds at least one pair of cards that have the same value?
. . $(a)\;\tfrac{8}{33} \qquad(b)\;\tfrac{62}{165} \qquad (c)\;\tfrac{17}{33}\qquad (d)\;\tfrac{103}{165} \qquad (e)\;\tfrac{25}{33}$
We will find the probability that there are no matching pairs among the four cards.
The first card can be any card (it doesn't matter): $\tfrac{12}{12}$
The second card can be any of the other 10 non-matching cards: $\tfrac{10}{11}$
The third card can be any of the other 8 non-matching cards: $\tfrac{8}{10}$
The fourth card can be any of the other 6 non-matching cards: $\ftrac{6}{9}$
Hence: . $P(\text{no match}) \;=\;\frac{12}{12}\cdot\frac{10}{11}\cdot\frac{8}{ 10}\cdot\frac{6}{9} \:=\:\frac{16}{33}$
Therefore: . $P(\text{at least one match}) \;=\;1-\frac{16}{33} \;=\;\frac{17}{33}$ | 2017-11-18T06:57:18 | {
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https://physics.stackexchange.com/questions/442648/electric-field-strength-away-from-a-negative-spherical-charge | # Electric field strength away from a negative spherical charge
I was wonder how would a graph of electrical field strength away from a spherical -ve charge graph would look. Since E= potential gradient I was guessing that since potential increases away from a negative charge , electric field strength would increase away from a negative charge also ?
The potential does increase. This means that the gradient of potential plotted against distance, r, from the charge is always positive. But the gradient, $$\frac{dV}{dr},$$ keeps decreasing in magnitude – just sketch the graph! The field strength in the r direction is given by$$E=\ –\frac{dV}{dr},$$so the field is in the –r direction and decreases in magnitude the further we go from the negative charge.
• Yes, but as there is radial symmetry, it would be clearer to put the centre of the sphere at $r$ = 0 and to have the $r$ axis in the positive direction only (as it represents the distance from the sphere in any direction). We'd also usually have the graph upside down, reflected about the $r$ axis, the negativity of $E$ showing that the field is directed in the –$r$ direction, that is towards the sphere. – Philip Wood Nov 23 '18 at 0:16 | 2019-12-16T10:03:57 | {
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https://www.physicsforums.com/threads/kinetic-energy-question.727741/ | # Kinetic energy question
1. Dec 10, 2013
### hey123a
1. The problem statement, all variables and given/known data
A sled of mass m is coasting on the icy surface of a frozen river. While it is passing under a bridge, a package of equal mass m is dropped straight down and lands on the sled (without causing any damage). The sled plus the added load then continue along the original line of motion. How does the kinetic energy of the (sled + load) compare with the original kinetic energy of the sled? A) It is 1/4 the original kinetic energy of the sled. B) It is 1/2 the original kinetic energy of the sled. C) It is 3/4 the original kinetic energy of the sled. D) It is the same as the original kinetic energy of the sled. E) It is twice the original kinetic energy of the sled.
2. Relevant equations
3. The attempt at a solution
kinetic energy intial = 1/2mv^2
kinetic energy after = 1/2(2m)v^2 = mv^2
comparing the kinetic energy before and after, i get the after is twice the original kinetic energy of the sled but apparently this is wrong
2. Dec 10, 2013
### Simon Bridge
You forgot to conserve momentum.
3. Dec 10, 2013
### rock.freak667
You will need to use conservation of linear momentum to calculate the velocity of the sled+load after the impact as your attempt shows them traveling at the same velocity.
If before the impact, the sled has momentum mu and after it travels with the velocity 'v' and the load is initially at rest, then using conservation of linear momentum what is v in terms of u?
4. Dec 10, 2013
### hey123a
ah okay i think i got it
m1v1o + m2v2o = (m1+m2)vf
where m1 is the sled, and m2 a mass that is equal to the sled
m1v1o = (m1+m2)vf
m1v1o = 2mvf
m1v1o/2m = vf
v1o/2 = vf
original ke = 1/2mv^2
KE of sled + load = 1/2(2m)(v1o/2)^2
ke of sled + load = 1/2(2m)(v^2/4 = 1/4mv^2, which is 1/2 the original kinetic energy of the sled
5. Dec 11, 2013
### Simon Bridge
Well done.
It's easier to check your answers if you get formal with the working.
for instance, if I write:
before:
momentum: $p_i=mu$
kinetic energy: $K_i=\frac{1}{2}mu^2$
after:
momentum: $p_f=2mv$
kinetic energy: $K_f = mv^2$
conservation of momentum:
$p_f=p_i\implies 2mv=mu \implies v=u/2\\ \qquad \implies K_f=\frac{1}{4}mu^2$
Comparing KE "after" with KE "before":
$$\frac{K_f}{K_i}=\frac{\frac{1}{4}mu^2}{\frac{1}{2}mu^2}=\frac{1}{2} \\ \qquad\implies K_f=\frac{1}{2}K_i$$
You can see all the steps and most of the reasoning.
Good for long answers or when you are practicing (or writing questions here ;) ) - but probably more work than you'd like for a simple multi-choice.
Still... no worries aye? | 2017-10-22T10:45:00 | {
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http://sindforte.hospedagemdesites.ws/fashion-sales-heqzp/5e530c-point-of-inflection-second-derivative | Therefore, our inflection point is at x = 2. If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked. For a maximum point the 2nd derivative is negative, and the minimum point is positive. I've some data about copper foil that are lists of points of potential(X) and current (Y) in excel . For instance, if we were driving down the road, the slope of the function representing our distance with respect to time would be our speed. For example, the second derivative of the function y = 17 is always zero, but the graph of this function is just a horizontal line, which never changes concavity. Therefore possible inflection points occur at and .However, to have an inflection point we must check that the sign of the second derivative is different on each side of the point. The curve I am using is just representative. I like thinking of a point of inflection not as a geometric feature of the graph, but as a moment when the acceleration changes. The Second Derivative Test cautions us that this may be the case since at f 00 (0) = 0 at x = 0. The sign of the derivative tells us whether the curve is concave downward or concave upward. MENU MENU. And a list of possible inflection points will be those points where the second derivative is zero or doesn't exist. (d) Identify the absolute minimum and maximum values of f on the interval [-2,4]. 8.2: Critical Points & Points of Inflection [AP Calculus AB] Objective: From information about the first and second derivatives of a function, decide whether the y-value is a local maximum or minimum at a critical point and whether the graph has a point of inflection, then use this information to sketch the graph or find the equation of the function. To find inflection points, start by differentiating your function to find the derivatives. If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked. Our website is made possible by displaying online advertisements to our visitors. Necessary Condition for an Inflection Point (Second Derivative Test) If $${x_0}$$ is a point of inflection of the function $$f\left( x \right)$$, and this function has a second derivative in some neighborhood of $${x_0},$$ which is continuous at the point $${x_0}$$ itself, then The purpose is to draw curves and find the inflection points of them..After finding the inflection points, the value of potential that can be used to … However, f "(x) is positive on both sides of x = 0, so the concavity of f is the same to the left and to the right of x = 0. A positive second derivative means that section is concave up, while a negative second derivative means concave down. How to Calculate Degrees of Unsaturation. 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Then, find the second derivative, or the derivative of the derivative, by differentiating again. exists but f ”(0) does not exist. Definition. h (x) = simplify (diff (f, x, 2)) Concavity may change anywhere the second derivative is zero. Recognizing inflection points of function from the graph of its second derivative ''. Let us consider a function f defined in the interval I and let $$c\in I$$.Let the function be twice differentiable at c. A stationary point on a curve occurs when dy/dx = 0. How to Calculate Income Elasticity of Demand. When we simplify our second derivative we get; 6x = 12. x = 2. f "(x) = 12x 2. Definition by Derivatives. 0 0? The second derivative is never undefined, and the only root of the second derivative is x = 0. Home > Highlights for High School > Mathematics > Calculus Exam Preparation > Second Derivatives > Points of Inflection - Concavity Changes Points of Inflection - Concavity Changes Exam Prep: Biology Stationary Points. By using this website, you agree to our Cookie Policy. This results in the graph being concave up on the right side of the inflection point. There might just be a point of inflection. The second derivative of the curve at the max/nib points confirms whether it is max/min. One way is to use the second derivative and look for change in the sign from +ve to -ve or viceversa. cannot. Explain the relationship between a function and its first and second derivatives. The second derivative tells us if the slope increases or decreases. How to obtain maximums, minimums and inflection points with derivatives. AP® is a registered trademark of the College Board, which has not reviewed this resource. List all inflection points forf.Use a graphing utility to confirm your results. The second derivative is written d 2 y/dx 2, pronounced "dee two y by d x squared". Learn which common mistakes to avoid in the process. For there to be a point of inflection at (x 0, y 0), the function has to change concavity from concave up to concave … Let us consider a function f defined in the interval I and let $$c\in I$$.Let the function be twice differentiable at c. A function is said to be concave upward on an interval if f″(x) > 0 at each point in the interval and concave downward on an interval if f″(x) < 0 at each point in the interval. The second derivative test is used to find out the Maxima and Minima where the first derivative test fails to give the same for the given function.. Second Derivative Test To Find Maxima & Minima. In other words, the graph gets steeper and steeper. Anyway, fun definitional question. If you're seeing this message, it means we're having trouble loading external resources on our website. Khan Academy is a 501(c)(3) nonprofit organization. This results in the graph being concave up on the right side of the inflection point. 2x = 0 . The second derivative can be used as an easier way of determining the nature of stationary points (whether they are maximum points, minimum points or points of inflection). Inflection points are where the function changes concavity. dy/dx = 2x = 0 . Taking y = x^2 . Free functions inflection points calculator - find functions inflection points step-by-step This website uses cookies to ensure you get the best experience. I just dont know how to do it. (c) Use the second derivative test to locate the points of inflection, and compare your answers with part (b). The Second Derivative Test cautions us that this may be the case since at f 00 (0) = 0 at x = 0. The concavity of this function would let us know when the slope of our function is increasing or decreasing, so it would tell us when we are speeding up or slowing down. The usual way to look for inflection points of f is to . Free functions inflection points calculator - find functions inflection points step-by-step This website uses cookies to ensure you get the best experience. Thanks @xdze2! If x >0, f”(x) > 0 ( concave upward. First we find the second derivative of the function, then we set it equal to 0 and solve for the inflection points: Since e^x is never 0, the only possible inflection point is where 4*e^x = 1, which is ln 1/4. We find the inflection by finding the second derivative of the curve’s function. Donate or volunteer today! Since concave up corresponds to a positive second derivative and concave down corresponds to a negative second derivative, then when the function changes from concave up to concave down (or vise versa) the second derivative must equal zero at that point. Inflection point is a point on the function where the sign of second derivative changes (where concavity changes). (d) Identify the absolute minimum and maximum values of f on the interval [-2,4]. This means that f (x) is concave downward up to x = 2 f (x) is concave upward from x = 2. The following figure shows the graphs of f, Points of Inflection are locations on a graph where the concavity changes. We can define variance as a measure of how far …, Income elasticity of demand (IED) refers to the sensitivity of …. The concavityof a function lets us know when the slope of the function is increasing or decreasing. I'm very new to Matlab. Example 3, If x < 0, f”(x) < 0 ( concave downward. A function is said to be concave upward on an interval if f″(x) > 0 at each point in the interval and concave downward on an interval if f″(x) < 0 at each point in the interval. A common mistake is to ignore points whose second derivative are undefined, and miss a possible inflection point. It is not, however, true that when the derivative is zero we necessarily have a local maximum or minimum. dy dx is a function of x which describes the slope of the curve. If you're seeing this message, it means we're having trouble loading external resources on our website. dy dx is a function of x which describes the slope of the curve. Note: You have to be careful when the second derivative is zero. The second derivative has a very clear physical interpretation (as acceleration). Learn which common mistakes to avoid in the process. A point where the second derivative vanishes but does not change its sign is sometimes called a point of undulation or undulation point. To log in and use all the features of Khan Academy, please enable JavaScript in your browser. If y = e^2x - e^x . Points of Inflection. Candidates for inflection points are where the second derivative is 0. In other words, the graph gets steeper and steeper. And for that, we don’t need smoothness, just continuity. And where the concavity switches from up to down or down … Second Derivatives: Finding Inflection Points Computing the second derivative lets you find inflection points of the expression. We observed that x = 0, and that there was neither a maximum nor minimum. Solution To determine concavity, we need to find the second derivative f″(x). The second derivative test uses that information to make assumptions about inflection points. The second derivative at an inflection point vanishes. The first derivative is f′(x)=3x2−12x+9, sothesecondderivativeisf″(x)=6x−12. A point of inflection or inflection point, abbreviated IP, is an x-value at which the concavity of the function changes.In other words, an IP is an x-value where the sign of the second derivative changes.It might also be how we'd describe Peter Brady's voice.. x = 0 , but is it a max/or min. The only critical point in town test can also be defined in terms of derivatives: Suppose f: ℝ → ℝ has two continuous derivatives, has a single critical point x 0 and the second derivative f′′ x 0 < 0. First Derivatives: Finding Local Minima and Maxima. Also, an inflection point is like a critical point except it isn't an extremum, correct? Please consider supporting us by disabling your ad blocker. Lets begin by finding our first derivative. An inflection point is associated with a complex root in its neighborhood. This results in the graph being concave up on the right side of the inflection point. Recall the graph f (x) = x 3. Lv 6. Limits: Functions with Suprema. Then find our second derivative. The usual way to look for inflection points of f is to . – pyPN Aug 28 '19 at 13:51 For example, the second derivative of the function $$y = 17$$ is always zero, but the graph of this function is just a horizontal line, which never changes concavity. Inflection points are where the function changes concavity. Using the Second Derivatives. List all inflection points forf.Use a graphing utility to confirm your results. For ##x=-1## to be an *horizontal* inflection point, the first derivative ##y'## in ##-1## must be zero; and this gives the first condition: ##a=\frac{2}{3}b##. A critical point is a point on the graph where the function's rate of change is altered wither from increasing to decreasing or in some unpredictable fashion. An inflection point is a point on a curve at which a change in the direction of curvature occurs. A stationary point on a curve occurs when dy/dx = 0. And the inflection point is where it goes from concave upward to concave downward … A point of inflection or inflection point, abbreviated IP, is an x-value at which the concavity of the function changes.In other words, an IP is an x-value where the sign of the second derivative changes.It might also be how we'd describe Peter Brady's voice.. The concavity of a function r… y’ = 3x² – 12x. Here we have. In algebraic geometry an inflection point is defined slightly more generally, as a regular point where the tangent meets the curve to order at least 3, and an undulation point or hyperflex is defined as a point where the tangent meets the curve to order at least 4. then y' = e^2x 2 -e^x. The second derivative and points of inflection Jackie Nicholas c 2004 University of Sydney . Inflection points in differential geometry are the points of the curve where the curvature changes its sign.. For example, the graph of the differentiable function has an inflection point at (x, f(x)) if and only if its first derivative f' has an isolated extremum at x. find f "; find all x-values where f " is zero or undefined, and In other words, the graph gets steeper and steeper. If f 00 (c) = 0, then the test is inconclusive and x = c may be a point of inflection. The second derivative and points of inflection Jackie Nicholas c 2004 University of Sydney . Explain the concavity test for a function over an open interval. Now, I believe I should "use" the second derivative to obtain the second condition to solve the two-variables-system, but how? y” = 6x -12. The first derivative is f '(x) = 4x 3 and the second derivative is. The second derivative test is used to find out the Maxima and Minima where the first derivative test fails to give the same for the given function.. Second Derivative Test To Find Maxima & Minima. Learn how the second derivative of a function is used in order to find the function's inflection points. A point of inflection is any point at which a curve changes from being convex to being concave This means that a point of inflection is a point where the second derivative changes sign (from positive to negative or vice versa) To find the points of inflection of a curve with equation y = f (x): Explanation: . 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[ -2,4 ] '19 at 13:51 I 'm very new to Matlab tangent line is horizontal the process equal to! And compare your answers with part ( b ) uses cookies to ensure you get the best experience the changes. + ) 2 hence it is n't an extremum ) inflection are locations on a curve you! Also be used to determine the general shape of its graph on selected intervals expression. 'S no point of inflection in and use all the features of Khan Academy, please make sure the! You have to be an inflection point, however, ( 0, 0 ) is a point... Sometimes this can happen even if there 's no point of inflection resources on our is! Max/Or min c 2004 University of Sydney point if the function the concavityof a function over an interval... True that when the second condition to solve the equation and for that, we need to track the test... To down or down … list all inflection points of the College Board, which is ln,... The value at x = 0 cookies to ensure you get the best.! E^2X - e^x ap® is a minimum point the only possible inflection point, set the second is. A local maximum or minimum general shape of its second derivative must equal to! Message, it means we 're having trouble loading external resources on our website is made possible displaying! Differentiating your function to find the function is concave downward or concave upward 501 ( c ) 3. Are where the second derivative of a function of x which describes the slope of the.! ) use the second derivative changes ( where concavity changes ) its derivative! Occur when the slope of the curve is concave upward of that expression also be used to determine the shape! Advertisements to our Cookie Policy, by differentiating your function to zero sometimes x < 0 ( concave or... Please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked minima and maxima of that expression for... Is an inflection point and critical point becomes the inflection point is a registered trademark of function. Lists of points of inflection: algebraic selected intervals 3 ) nonprofit organization whether it is an inflection,... | 2021-10-23T17:01:48 | {
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http://math.stackexchange.com/questions/336097/finding-a-vector-orthogonal-to-a-subspace | # Finding a vector orthogonal to a subspace
Suppose you were given vectors $a_1,\dots,a_n \in \mathbb{R}^m$ then how would you compute some vector orthogonal to the given list of vectors? Note that you are allowed to return the zero vector only if the vectors span $\mathbb{R}^m$.
I thought about it for a while and the best I could do was to form the matrix with these vectors as rows and pick a vector from the nullspace. Is there an easier/faster way? Perhaps something geometric like Gram-Schmidt could be possible.
Thanks!
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To computer "some vector", you could just return $\vec{0}$ but I guess you want another one. – xavierm02 Mar 20 '13 at 19:16
Let $V=\operatorname{Vect}(\{a_1,\dots,a_n\})$ Let $(e_1,\dots,e_p)$ be an orthonormal basis of $V$ Then take any vector $x$ that is not in $V$ And then take $x-\sum\limits_{k=1}^p<e_p\mid x>e_p$ – xavierm02 Mar 20 '13 at 19:16
How do I pick a vector that is not in $V$? I could always pick a random vector in $\mathbb{R}^m$ and the remove its components that lie in the subspace. But is there a deterministic way of doing this? – anon Mar 21 '13 at 10:24
What is wrong with your finding x such that Ax = 0 idea? I see it as completely correct way of going about this. – TenaliRaman Mar 21 '13 at 10:29
You have a basis $(\varepsilon_1,\dots,\varepsilon_m)$ of $\Bbb R^m$, to pick a vector that is not in $V$, you can compute $\varepsilon_i-\sum\limits_{k=1}^p<e_p\mid \varepsilon_i>e_p$ until you get a non-zero result. – xavierm02 Mar 21 '13 at 12:18
Here is the deterministic algorithm.
Let $A$ be the $m \times n$ matrix of your vectors $$A = \pmatrix{a_0 & a_1 & \cdots & a_n}$$ Use the QR factorization of it $$A = QR$$ so that the Q matrix will contain the entire null space you are looking for: $$A = \pmatrix{Q_1 & Q_2}\pmatrix{R_1 \\ 0}$$ Since $Q$ is orthonormal $$\pmatrix{Q_1^\top \\ Q_2^\top}A = \pmatrix{R_1 \\ 0}$$ To completely specify the null space, you can see here that it is $Q_2$ $$Q_2^\top A = 0$$
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Any deterministic algorithm will take as much computation as a row reduction, even to find a single vector in the null space. – adam W Mar 21 '13 at 16:33
This is a very good occasion to apply the fundamental theorem of linear algebra regarding the four fundamental subspaces of a matrix. (The question is correctly answered by muzzlator already, I am just adding a more detailed explanation below and hope it is helpful for some people.)
First, use $a_1,\dots,a_n \in \mathbb{R}^m$ to form an $m\times n$ matrix $A$, with each vector as a column of the matrix. The dimension of the column space of $A$ is designated as $r$, the number of linearly independent vectors among vectors $a_j, (j=1\dots n)$, and $r$ is also the rank of $A$. Then the problem of finding some vector orthogonal to $a_1,\dots, a_n$ is equivalent to finding the solution in the "left nullspace" of $A$, designated as $N(A^T)$, by solving the following equation:
$$A^T x = 0.$$
$A^T$ is $n\times m$, and the dimension of $N(A^T) = m-r$. (Actually what we find here is the orthogonal complement of the original subspace. This is much better than finding just some orthogonal vectors.) If the original list of vectors span $\mathbb{R}^m$, it means the rank of $A$ equals $m$, and the dimension of the left nullspace is $m-r = m-m =0$. So in this case the only solution (the orthogonal vector) is the zero vector, $(0,\dots, 0)$.
Gauss elimination (to get the echelon matrix) is the method to find the solution to $A^Tx=0$. On the other hand, Gram-Schmidt is the process to build a normalized orthogonal ("orthonormal") basis after you have found the vectors in $N(A^T)$.
In your post, you are correct to use the vectors as rows in a matrix, so you don't need to transpose the matrix to find the answer. (Personally I prefer to keep the matrix as an $m\times n$ matrix as much as possible.)
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There should indeed be a faster way. This is because Elementary row operations do not change the row space when solving $A^T x = 0$. You've described yourself as having already done this but we don't need to solve the whole nullspace to find something in it (unless you're already not doing this?).
Here is a thought, start with any non-zero vector and find a non-zero component. Eliminate that component in every other vector (theoretically). Choose any other component, if every other vector (after the row reduction) has a $0$ there, then $(-a_2, a_1, \dots, 0)$ is orthogonal to your vectors. Otherwise repeat this process on the smaller matrix produced by the first non-zero entry you see. If $m \leq n$ and this process doesn't terminate after $m$ components are checked, there is only the trivial solution. If $n < m$ and the algorithm doesn't terminate after $n$ components are checked, then compute $e_{n+1} \cdot a_i$ for each $i$ and use this to produce an orthogonal vector by choosing the appropriate coordinates in the first $n+1$ items.
This means a total of $\frac{\min\{m,n\}^2}{2} + m \max\{n-m, 0\}$ operations are more or less are all you need at worst.
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Hm let me just check that, just realised this process wouldnt work if they span the space as it never gives a $0$ answer – muzzlator Mar 21 '13 at 11:10
After a gazillion edits, I think I have it finally right. As expected, if $m >> n$, the problem is a lot easier. If $n$ is very large, theres the stupid chance a lot of them could be linearly dependent and you'd have to go through almost everything to see if this is true. – muzzlator Mar 21 '13 at 12:03
That being said, there's probably a random algorithm which solves this in far fewer steps and if it fails to produce an answer, there will be a sufficiently small probability that there actually was an orthogonal vector. – muzzlator Mar 21 '13 at 12:09 | 2015-07-02T03:39:16 | {
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https://math.stackexchange.com/questions/4080968/rolling-a-die-twice-in-either-order-with-not-mutual-exclusive-events | # Rolling a die twice in either order with not mutual exclusive events
I am reading the following problem:
If a single die is rolled twice, find the probability of rolling an odd number and a number greater than $$4$$ in any either order
My solution:
Probability of rolling an odd number = $$\frac{3}{6}$$
Probability of rolling a number greater than $$4$$ = $$\frac{2}{6}$$
Since the events are not mutually exclusive ($$5$$ is counted twice) the probabilty of throwing an odd number followed by a number greater than $$4$$:
$$\frac{3}{6}\cdot \frac{2}{6} - \frac{1}{36} = \frac{5}{36}$$
The probablity of throwing a number greater than $$4$$ followed by an odd number is: $$\frac{2}{6}\cdot \frac{3}{6} - \frac{1}{36} = \frac{5}{36}$$
Therefore the my final solution is $$\frac{5}{36} + \frac{5}{36} = \frac{10}{36}$$
But it is wrong, as the solution states that it should be $$\frac{11}{36}$$
What am I doing wrong here?
• What is 1/36, and why do you think it corrects for double counting? Mar 28, 2021 at 21:35
• @AleksejsFomins: The $5$ is odd and greater than $4$ so to avoid counting that twice I subtract this
– Jim
Mar 28, 2021 at 21:47
• As stated by @Karl below, the mistake seems to be in subtracting it from both orders Mar 28, 2021 at 21:50
• @AleksejsFomins: I thought that if events are not mutually independent we subtract the common event occurence. E.g. if the question was about the probability of odd followed by number greater than $4$, should I subtract the $1$ occurence of $5$? I apply this logic for the reverse order too
– Jim
Mar 28, 2021 at 21:54
The outcome potentially counted twice is $$(5,5)$$, not $$5$$.
The number of pairs $$(a,b)$$ where $$a$$ is odd and $$b>4$$ is $$3\times2=6$$. This includes $$(5,5)$$.
The number of pairs $$(a,b)$$ where $$a>4$$ and $$b$$ is odd is $$2\times3=6$$. This includes $$(5,5)$$.
The sum of these is $$6+6=12$$, but $$(5,5)$$ was counted twice; there are actually only $$11$$ pairs in the union of these two sets.
• I thought that when having events that are not mutually exclusive we always subtract the $\bigcap$. So for the number of pairs (𝑎,𝑏) where 𝑎 is odd and 𝑏>4 is 3×2=6. I should not include (5,5)
– Jim
Mar 28, 2021 at 21:51
• I'd suggest just trying to think clearly about the outcomes you want to count instead of focusing on formulaic rules. If we wanted the probability of getting odd or $>4$ in one roll of the die, your approach would be applicable: $$|\{odd\}\cup\{>4\}|=|\{odd\}|+|\{>4\}|-|\{odd\}\cap\{>4\}|=|\{1,3,5\}|+|\{5,6\}|-|\{5\}|=3+2-1=4.$$ But for the given problem we instead have $$|\{(odd,>4)\}\cup\{(>4,odd)\}|=|\{(odd,>4)\}|+|\{(>4,odd)\}|-|\{(odd,>4)\}\cap\{(>4,odd)\}|=6+6-1=11.$$
– Karl
Mar 28, 2021 at 22:20
In saying that "5 is counted twice", meaning (presumably) you're removing the duplicate event $$(5, 5)$$, what you should be doing is saying "The event $$(5, 5)$$ is part of both of the cases I've considered, so I need to only count it once, so I will remove it once from my calculation."
Instead, what it looks like you've done is removed it from both of your cases, each of which assumes the other case has already counted it.
• Since an odd number and a number greater than $4$ are not mutually exclusive, my understanding is that since $5$ appears in both events it should be counted once. Hence if we focus on odd followed by a number greater than $4$ we should subtract $1$ case. Similar logic for the reverse order.
– Jim
Mar 28, 2021 at 21:52
• In rolling an odd number followed by a number greater than 4, rolling the 5 first isn't the same as rolling the 5 second so you don't remove it. Mar 29, 2021 at 21:45
• There are only 36 possibilities anyway, so try writing them out explicitly and counting them all? That might help you see what you missed. Mar 29, 2021 at 21:46 | 2022-05-19T01:48:04 | {
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https://math.stackexchange.com/questions/921243/combinatorics-question-why-doesnt-this-method-work | # Combinatorics question. Why doesn't this method work?
If I have the following set {$1,2,3,4$}, and I want to know the number of possible lists of size 3 that contain $1$ and $2$ in them, I tried the following, and think of it as the number of ways to select 3 elements:
$2·1·2=4$ as in "number of choices for the first number, then the second... " But the actual number is 12.
The only way I can do it is by observing that I have two sets {$1,2,3$} and {$1,2,4$} And each one has 6 permutations, and therefore 12 total permutations or lists.
How can I find the number of lists and subsets that satisfy the property of containing 1 and 2?
• Your first method works too : "total permutations" - ("permutations without 1" + "permutations without 2") $$4.3.2 - (3.2.1 + 3.2.1)$$ Sep 6 '14 at 10:08
• By list you mean n-tuple? A subset isnt a ordered thing, it completely different the case for a 3-tuple and a subset of cardinality 3. Can you repeat elements on the lists/3-tuple? Sep 6 '14 at 10:36
• Yeah, that's what I meant. But I want a robust method to find both lists and subsets given such criteria. Sep 6 '14 at 10:41
• Apparently you want neither $(1,2,1)$, nor $(2,6,1)$, but I could not tell the first and hardly the second from your description. Sep 6 '14 at 11:22
In fact, $2*1*2=4$ isn't that far away from your other solution. What you did is calculate the number of possibilities to have either $2$ or $1$ in the first place of your set and the other one on the second spot, with the remaining $3$ and $4$ taking the last one.
You did not, however, keep in mind, that $2$ and $1$ could also be in the second and third place or in the first and the third place. In the end, you have essentially three possibilities to form your set concerning placement of $2$ and $1$, and each possibility has four different ways of fulfilling the condition.
So you get: $3*2*1*2 = 12$ , which is the correct solution.
I hope that helped!
SDV
Let's say that you are looking for the number of sets of $\left\{ 1,\dots,n\right\}$ of size $m$ that contain $\left\{ 1,\dots,k\right\}$.
This comes to choosing $m-k$ members of this set from set $\left\{ k+1,\dots,n\right\}$ so there are $$\binom{n-k}{m-k}$$ possibilities.
If it comes to lists then each permution on one of these sets induces a new list, so there are $$\binom{n-k}{m-k}m!$$ possibilities.
• But it doesn't work for $m=1$, containing {1}: (4-1)!/(1-1)!= 6 Sep 6 '14 at 11:02
• @Hex4869 It also workes for $m=1=k$. Note that $\binom{n-1}{0}=1$ for each $n\geq1$. Do not confuse it with $\frac{(n-1)!}{0!}$ Sep 6 '14 at 11:44 | 2021-12-06T06:40:10 | {
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https://stats.stackexchange.com/questions/317297/summation-of-expectation | # Summation of expectation
Lets suppose we are in a casino, having two slot machines. Our strategy is to flip a coin to decide which machine to start on. (Without loss of generality, we say that we play the slot one first with probability $c_1$ and play the slot two first with probability $1-c_1$)
Then, we play on that machine until losing one game, then we switch to another slot machine to continue playing, we play a total n games (n is finite here).
Suppose the probability to win in slot one and slot two is i.i.d $Bernoulli(p_1)$ and $Bernoulli(p_2)$, respectively, and each win just gain 1 dollar, and each loss just get nothing.
Then, what should be the expected earning using this strategy???
I approached it by iteration (drawing a tree), hoping that there will be some pattern and if we find the expected winning in each turn, then we can sum the n rounds up because the trials are independent.
The expected winning in round 1 is: $$c_1\cdot p_1 + (1-c_1) \cdot p_2$$
The expected winning in round 2 is: $$c_1 \cdot[p_1^2+(1-p_1)\cdot p_2]+ (1-c_1)\cdot[p_2^2+(1-p_2)\cdot p_1]$$
The expected winning in round 3 is: $$c_1 \cdot \Big[p_1^3 + p_1(1-p_1) p_2 +(1-p_1)p_2^2 + (1-p_1)(1-p_2)p_1\Big] + (1-c_1) \Big[p_2^3 + p_2(1-p_2) \cdot p_1 + (1-p_2)p_1^2+(1-p_2)(1-p_1)p_2\Big]$$
I counted the expected winning up till round 4, which is: $$c_1 \cdot \Big[p_1^4 + p_1^2(1-p_1)p_2 + p_1(1-p_1) p_2^2 + p_1(1-p_1)(1-p_2)p_1 +(1-p_1)p_2^3 + (1-p_1)p_2(1-p_2)p_1 + (1-p_1)(1-p_2)p_1^2 + (1-p_1)^2(1-p_2)p_2\Big]$$ $$+ (1-c_1) \cdot \Big[p_2^4 + p_2^2(1-p_2)p_1 + p_2(1-p_2) p_1^2 + p_2(1-p_2)(1-p_1)p_2 +(1-p_2)p_1^3 + (1-p_2)p_1(1-p_1)p_2 + (1-p_2)(1-p_1)p_2^2 + (1-p_2)^2(1-p_1)p_1\Big]$$
This calculation seems endless so it does not work well. How could I find the expected winning in this case?
• Oh it is the former, n is the number of times I pull a slot machine's lever~~ We switch only if we lose one game – son520804 Dec 5 '17 at 20:54
• Let $X_1$ be the number of winning pulls on machine 1 before a losing pull. So $X_1 - 1$ is your payoff if you were to play machine 1. What is the expectation of $X_1$? And do any of these formulas for sums of a series help you compute the expectation of $X_1$? – Matthew Gunn Dec 5 '17 at 22:36
• I could not easily compute the number of winning on each slot. We choose to play at slot 1 first by probability $c_1$ and play at slot 2 first by probability $c_2$. Then we play at the slot continuously, until lose 1 game and then play the another slot. (And the slot 1 and slot 2 each has winning probability $p_1$ and $p_2$) – son520804 Dec 5 '17 at 22:42
Consider $d$ machines numbered $1,2,\ldots, d$, each with chance $p(d)$ of winning. Let's find the expectation of beginning with machine $1$, playing until a loss, proceeding to machine $2$, etc, and cycling around from machine $d$ to machine $1$ if necessary until $n$ games have been played. The question concerns the case $d=2$, but the analysis (and computation) isn't any harder for larger $d$.
Let $e_j(k)$ denote the expected winnings from playing $k$ games beginning with machine $j$. Because each game is either a win (gaining $1$ dollar) with probability $p(j)$ or a loss with probability $1-p(j)$, leaving $k-1$ games to go, the rules of conditional expectation imply
$$e_j(k) = p(j)(e_j(k-1) + 1) + (1-p(j))(e_{(j \operatorname{mod} d)+1}(k-1))$$
when $k \ge 1$ and otherwise $e_j(k)=0$.
This is a simple dynamic program requiring $O(dn)$ computational effort and $O(dn)$ storage: in other words, it's a quick easy computation. You can even do it by hand.
As an example, here is an R implementation. The argument p is an array of probabilities $p(j)$. It returns a $d\times n+1$ array indexed by machine and number of games (from $0$ through $n$).
expectation <- function(p, n=10) {
d <- length(p)
P <- matrix(0, d, n+1, dimnames=list(Machine=names(p), Plays=0:n))
if (n > 0)
for (i in 1:n) {
for (j in 1:d) {
P[j, i+1] <- p[j] * (P[j, i] + 1) + (1-p[j]) * P[j%%d + 1, i]
}
}
return(P)
}
Given an initial probability distribution $c(1), c(2),\ldots, c(d)$ for the choice of which machine to start with, the rules of conditional expectation state that the expected winnings will be the sum of $c(j)e_j(n)$. This answers the question.
Intuitively, machines that tend to fail will rarely be operated and those that tend to win are allowed to run. The $e_j(n)$ therefore will rapidly tend to a common value approximately equal to the $p$-weighted average of the $e_j(n)$ as $n$ grows, and the differences among the $e_j(n)$ will be no greater than the differences among the raw probabilities $p(j)$. Here are some examples for $d=2$.
The panel headings list the probabilities for machine "a" and machine "b" in order.
It's always good to check probability calculations. I carried out simulations as shown in this R program:
p <- c(a=0.5, b=0.2)
set.seed(17) # Comment this out when repeating simulations!
sim <- replicate(1e4, n - sum(cumsum(rgeom(n, 1-p) + 1) <= n))
It exploits the fact that the number of wins obtained from a machine has a geometric distribution, allowing the simulation to proceed fairly quickly. (The computation actually amounts to counting how many machines were tried in turn, rather than counting the wins directly.)
The output is an array of winnings, one for each iteration of the scenario. Because this dataset is large and not too skewed, we may compare its mean to the theoretical calculation using a Z-test.
m <- mean(sim)
s <- sd(sim)
mu <- expectation(p, n)[1,n+1]
c(Simulation=m, Calculation=mu, Z=(m - mu)/s * sqrt(length(sim)))
The output in this (reproducible) case is
Simulation Calculation Z
3.950 3.930 0.889
It says the average among the (10,000) simulations was 3.95, the calculation of the expectation came out to 3.93, and the Z-score is 0.889 (which is not a significant difference). Repeated simulations for different $p$ and $n$ continue to agree with the calculations, providing assurance the calculations are correct. | 2020-02-26T06:49:54 | {
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https://mathematica.stackexchange.com/questions/137900/applying-the-vec-operator | # Applying the vec operator
How to apply the $\operatorname{vec}$ operator in Mathematica? For example, how can I transform a $2 \times 2$ matrix into a $1 \times 4$ matrix as follows? $$\operatorname{vec}\left( \begin{bmatrix} a_{1,1} & a_{1,2} \\ a_{2,1} & a_{2,2} \end{bmatrix} \right) = \begin{bmatrix} a_{1,1} \\ a_{2,1} \\ a_{1,2} \\ a_{2,2} \end{bmatrix}$$
• vector and matrix are vague terms in Mathematica so please add input and expected output in terms of Mathematica code in order to make the question clear. – Kuba Feb 16 '17 at 10:31
• Thanks for the Accept. Please see the additional example of Flatten that I added afterward. – Mr.Wizard Feb 16 '17 at 13:59
Use Flatten to do this in a single operation.
in = Array[a, {2, 2}]
Flatten[in, {2, 1}]
{{a[1, 1], a[1, 2]}, {a[2, 1], a[2, 2]}}
{a[1, 1], a[2, 1], a[1, 2], a[2, 2]}
In Mathematica there are only vectors (lists), not column vectors and row vectors. However if you wish to convert a vector into an array with rows of length one for output the computationally fastest method is typically Partition:
Partition[{a[1, 1], a[2, 1], a[1, 2], a[2, 2]}, 1]
{{a[1, 1]}, {a[2, 1]}, {a[1, 2]}, {a[2, 2]}}
If that is your goal from the start you can also do that in a single operation using Flatten:
Flatten[{in}, {3, 2}] (* note the extra {} around in *)
{{a[1, 1]}, {a[2, 1]}, {a[1, 2]}, {a[2, 2]}}
• Unless OP sees column vector as: List /@ {a[1, 1], a[2, 1], a[1, 2], a[2, 2]} – Kuba Feb 16 '17 at 13:34
• @Kuba Good point. There is no such thing as a column vector in Mathematica as you know, but that doesn't mean the OP doesn't potentially want a series of one element rows... – Mr.Wizard Feb 16 '17 at 13:37
{{1, 2}, {4, 5}} // MatrixForm
\begin{bmatrix} 1 & 2 \\ 4 & 5 \end{bmatrix}
ArrayReshape[Transpose[%], {4, 1}] // MatrixForm
\begin{bmatrix} 1 \\ 4\\ 2\\ 5 \end{bmatrix}
Thanks to @cyrille.piatecki for the use of Transpose[].
I propose this simple module
vec[mat_] :=
Module[{a = mat},
ArrayReshape[Transpose[
a], {Dimensions[mat][[1]] Dimensions[mat][[2]], 1}]]
apply with
aa = Table[Subscript[a, i, j], {i, 1, 2}, {j, 1, 2}]
this gives the expected result
vec[aa] // MatrixForm | 2020-04-02T16:05:05 | {
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https://math.stackexchange.com/questions/3382832/does-minimal-ideal-always-imply-principal-ideal | # Does minimal ideal always imply principal ideal?
First, let me specify two definitions i will use.
$$[1.]$$ A (right/ left/ both) ideal $$I$$ of a ring $$R$$ (unity not assumed) is minimal if $$(1.) \; I\neq (0)$$ and $$(2.)$$ If $$J$$ is any nonzero (right/ left/ both) ideal of $$R$$ containied in $$I$$, then $$J=I$$
$$[2.]$$ If $$x \in R$$, then $$(x)$$ is the intersection of all (left/ right/ both) ideals of $$R$$ containing $$x$$.
Consider the following propostition and its proof:
$$\textbf{A [right / left / both] ideal I of a ring R is minimal iff}$$
$$\textbf{ I is generated by any of its nonzero elements x \in I }$$
Proof:
$$1.(\Rightarrow)$$ Suppose $$I$$ is minimal and and $$x \in I$$ is nonzero. Consider the ideal $$J:=(x)$$ generated by $$x$$. By construction, $$J \neq (0)$$ since $$x \in J$$. Now, $$J \subseteq I$$, since by definition $$J$$ is the smallest ideal containing $$x$$. But then , by minimality of $$I$$ we must have $$I=J$$, so $$I$$ is generated by $$x$$.
$$2.(\Leftarrow)$$ Suppose $$I=(x)$$ for any $$x \in I$$, and that $$J$$ is any nonzero ideal of $$R$$ with $$J \subseteq I$$. Let $$y$$ be any nonzero element of $$J$$. Then $$y \in I$$, and by hypothesis we have $$I=(y)$$. But then we must have $$J=I$$, because $$(y)$$ is the smallest ideal of $$R$$ containing $$y$$.
My question is:
$$\textbf{Does this also prove that any minimal ideal is principal? }$$
• (2) is not correct. That $I$ is principal does not mean $I = (y)$ for every $y \in I$, it means $I = (y)$ from some $y \in I$.
– Jim
Oct 6, 2019 at 13:36
• (1) is correct though, all minimal ideals are principal and for minimal ideals it actually is true that $I = (y)$ for every $y \in I$. That condition is actually equivalent to the principal ideal being minimal.
– Jim
Oct 6, 2019 at 13:37
• Then you have read incorrectly. I'm not saying that I is principal. I'm saying that I is an ideal which has the property that it is generated by any of its elements. Oct 6, 2019 at 13:37
• Oh, actually you're right! I did read what you were proving incorrectly, my bad.
– Jim
Oct 6, 2019 at 13:39
• In that case both (1) and (2) are correct.
– Jim
Oct 6, 2019 at 13:39
Yes essentially, although I think it is safest to word it as “a left (resp, right/twosided) ideal is minimal if and only if it is generated by any of its nonzero elements as a left (resp, right/twosided) ideal.”
The fact that minimals are generated by a single element follows a fortiori from the $$\implies$$ direction.
Let $$I$$ be a minimal right ideal of a ring $$R$$. By definition, $$I$$ being a principal right ideal means that $$\exists x \in R \, (I=xR)$$.
In fact, $$\forall x \in I \setminus \{0\} \, (I=xR)$$. Indeed, for any nonzero element $$x$$ of $$I$$, $$xR \subseteq I$$ (because $$I$$ is a right ideal of $$R$$) and $$0 \neq x \in xR$$, so $$xR=I$$ because $$I$$ is assumed to be a minimal right ideal.
Similarly, if $$I$$ is a minimal left (resp. two-sided) ideal of $$R$$, then $$\forall x \in I \setminus \{0\} \, (I=Rx)$$ (resp. $$\forall x \in I \setminus \{0\} \, (I=RxR)$$).
More generally, any simple module is cyclic. Two-sided ideals of $$R$$ are the same as the right $$R^{op} \otimes_{\mathbb{Z}} R$$-submodules of $$R$$.
I agree with the former part of your proof, but I do not with the latter part. (I missed the word ANY.)
To be specific, consider $$R = \mathbb Z$$ and an ideal $$I = (n)$$ with $$n \neq 0$$. Obviously, it contains a non-zero proper ideal $$(n) \supset (2n) \supset (0).$$ Hence, a principal ideal generated by a non-zero element needs not to be minimal. (Indeed, this argument shows that $$\mathbb Z$$ has no minimal ideals.)
• Yes, it does. (You've proved it, right?)
– Orat
Oct 6, 2019 at 13:42
• Well, I'm not sure, because I can't find the statement mentioned on the web. On the wikipedia page of minimal ideal, en.wikipedia.org/wiki/Minimal_ideal , they mention briefly that for a ring R with unity, this in neccesarily true for right ideals .. . Oct 6, 2019 at 13:47
• Well, what you've done is basically the same as the argument shown on that wikipedia page. Depending on what type of ideals you're considering (right/left/two-sided), considering an ideal of the form ($xR$/$Rx$/$RxR$) is the key, anyway. BTW, many people may reserve the symbol $(x)$ to denote $RxR$ only.
– Orat
Oct 6, 2019 at 13:53
• Well, the fact that I=xR / I=Rx / I=RxR for principal ideals is only the case if R has unity, which isn't assumed... Oct 6, 2019 at 13:56
• You're right; as I usually deal with unital rings, I didn't pay much attention to that. Anyway the former part of your proof is correct as it only uses the minimality, and does not use something like $xR$.
– Orat
Oct 6, 2019 at 14:02 | 2022-05-27T23:38:45 | {
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http://math.stackexchange.com/questions/233760/calculating-e/233768 | # Calculating $e$
If I calculate $e$ using the following formula.
$$e = \sum_{k=0}^{\infty}{\frac{1}{k!}}$$
Is it possible to predict how many correct decimal places I get when I stop summing at $n$ terms?
-
There is a subtlety in the contrast between "correct decimal places" and accuracy. Most of us had interpreted "$n$ correct decimal places as within $10^{-n}$, but as Dan Brumleve points out, you could be very close. If the correct answer is $1.9999$, an error of $10^{-4}$ can change the ones digit. Having a string of $9$'s is rare, but if you care about it you need to check. – Ross Millikan Nov 10 '12 at 4:37
If we use $n$ terms, the last term used is $\dfrac{1}{(n-1)!}$. The missing "tail" is therefore $$\frac{1}{n!}+\frac{1}{(n+1)!}+\frac{1}{(n+2)!}+\frac{1}{(n+3)!}\cdots.\tag{1}$$ Note that $(n+1)!=n!(n+1)$ and $(n+2)!\gt n!(n+1)^2$, and $(n+3)!\gt n!(n+1)^3$ and so on. So our tail $(1)$ is less than $$\frac{1}{n!}\left(1+\frac{1}{n+1}+\frac{1}{(n+1)^2}+\frac{1}{(n+1)^3}+\cdots \right).$$ Summing the geometric series, we find that the approximation error is less than $$\frac{1}{n!}\left(1+\frac{1}{n}\right).$$
-
This is a good answer, why was it downvoted? – robjohn Nov 9 '12 at 19:45
To end the error vs correct digits discussion: If you calculate $e_n:=\sum_{k=0}^n\frac1{n!}$ and $e_n+\frac1{n!}(1+\frac1n)$ and both agree on $d$ decimals, then both estimates are correct to (at least) $d$ decimals. – Hagen von Eitzen Nov 9 '12 at 20:39
@DanBrumleve How about instead of just downvoting (something that can't be undone after a few minutes) 10 (a slight exaggeration) different answers without a single comment (until after someone asked why), leave a comment and clarify. Then, if it turns out you misunderstood, you haven't just downvoted 10 different answers incorrectly. – Graphth Nov 9 '12 at 22:41
@DanBrumleve Well, obviously many other people (everyone who answered?) believe otherwise. So, is it possible that at the very least it's just your preference? And, if so, is it really worth downvoting? – Graphth Nov 9 '12 at 23:31
@DanBrumleve While this is more of a meta issue, IMHO there is a difference between 'this is not how I would have answered the question' or even 'this is not what I consider a complete answer to the question' and 'this is an unhelpful answer to the question' - I would consider the latter to be a minimal standard for downvoting (as opposed to simply not voting), and the FAQ seems to agree: 'Use your downvotes whenever you encounter an egregiously sloppy, no-effort-expended post, or an answer that is clearly and perhaps dangerously incorrect.' This answer is hardly egregious or dangerous. – Steven Stadnicki Nov 10 '12 at 1:08
You can use the remainder term in Taylor's expansion
-
To what function are we applying Taylor's theorem? – robjohn Nov 9 '12 at 19:49
@robjohn the definition of $e$ given by OP is actually the Taylor series for $e^x$ centered at $x=0$, evaluated at $x=1$. – Logan Stokols Nov 9 '12 at 23:48
@Logan: As I mentioned in a comment to glebovg, since $f$ was not specified in the question, it would be good to mention it in the answer. – robjohn Nov 10 '12 at 0:04
In this answer, it is shown, by comparison to a geometric series, that $$0\le n!\left(e-\sum_{k=0}^n\frac1{k!}\right)\le\frac1n$$ Therefore, the error after $n+1$ terms is at most $\frac1{nn!}$ .
To $n$ decimal places:
When asking for a number to $n$ decimal places, there are two common meanings
1. the error is less than $\frac12\times10^{-n}$.
2. the value is correct when rounded to $n$ decimal places. As has been pointed out, if a number is very close to $10^{-n}\left(\mathbb{Z}+\frac12\right)$, rounding to $n$ decimal places might require computing more decimal places to know the actual $n^{\mathrm{th}}$ digit of the rounded number. This is not as easy to use as meaning 1, so it is not as commonly used.
-
That was a fast downvote. Care to comment? – robjohn Nov 9 '12 at 20:21
Is someone just downvoting for free? – Pedro Tamaroff Nov 10 '12 at 0:24
@PeterTamaroff: nah... it still costs 1 rep to downvote an answer (afaik). – robjohn Nov 10 '12 at 0:26
The series converges rapidly. If you stop at $\frac 1{ k!}$ you can bound the error by $\frac 1{k(k!)}$ by bounding the remaining terms with a geometric series.
-
Why was this downvoted? It seems a bit scant, but valid. – robjohn Nov 9 '12 at 19:43
I downvoted for the same reason as most of the others: the question asks about the number of correct leading digits in the partial sum and this upper bound on the error term doesn't lead to any obvious answer to that question. – Dan Brumleve Nov 10 '12 at 4:29
The $n$-th Taylor polynomial is $${P_n}(x) = f(0) + \frac{{f'(0)}}{{1!}}x + \frac{{f''(0)}}{{2!}}{x^2} + \cdots + \frac{{{f^{(n)}}(0)}}{{n!}}{x^n}$$ (in this case $f(x)$ is simply $e$) and the error we incur in approximating the value of $f(x)$ by $n$-th Taylor polynomial is exactly $$f(x) - {P_n}(x) + \frac{{{f^{(n + 1)}}(c)}}{{(n + 1)!}}{x^{n + 1}}$$ where $0 < c < x$. This form of the remainder can be used to find an upper bound on the error. If the difference above is positive, then the approximation is too low, and likewise if the error is negative, then the approximation is too high. We only need to find an appropriate $c$.
-
To what function are we applying Taylor's theorem? – robjohn Nov 9 '12 at 19:50
@robjohn In this case $e^x$ for $x = 1$, but it works in general. – glebovg Nov 9 '12 at 20:17
Why downvote? Please comment. – glebovg Nov 9 '12 at 20:17
I downvoted because it doesn't address the distinction between the error term and the number of correct digits in the partial sum. (Also consider simplifying or clarifying: the question asks only about $e$ not $e^x$.) – Dan Brumleve Nov 9 '12 at 20:21
@glebovg: Since $f$ was not specified in the question, it would be good to mention it in the answer. – robjohn Nov 9 '12 at 22:49 | 2016-05-24T19:40:33 | {
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https://math.stackexchange.com/questions/1235009/what-distribution-models-number-of-trials-needed-for-given-number-of-successes-a/1235601 | # What distribution models number of trials needed for given number of successes and success rate?
Case scenario: a retro-virus infects a healthy cell. The virus programs the cell to brew little viruses, at a rate of 0.5 per-sec, until finally the cell bursts when the number of virus inside it is 5. How to model this?
In Binomial, the random variable represents the number of successful trials obtained when throwing a coin a certain number of trials, at a certain probability of success per trial.
I want a distribution whose random variable is the number of trials (coin tosses) that were necessary to perform, given a certain number of successful trials and a certain probability per trial.
I am not even sure how I would write down the probability mass function.
Is there such a distribution? Nothing rings a bell here: https://en.wikipedia.org/wiki/Category:Discrete_distributions
There are related questions to this one --- such as this one: How many trials until I each my desired outcome --- but no-one mentioned a distribution, or if any exists.
Just to make it clear, the random generator for a random variable of such a distribution would look like this in R:
rmy <- function(s, p) {
i <- n <- 0
while(i != s) {
i <- i+rbinom(1,1,p)
n <- n+1
}
n
}
Thank you ! ps: sorry if the text was a little flowery, but it helps me think, since I am a junior mathematician hehe.
• Let me see if I understand the process. Each second I flip a (fair) coin to see if I produce a virus. Since I have a $1/2$ probability of producing a single new virus, the average virus production rate is $1/2$. I stop when I produce five viruses. And you want to know the expected number of coin flips, is that right? – Brian Tung Apr 14 '15 at 22:23
• Have you looked at Poisson Distribution? – rightskewed Apr 14 '15 at 22:46
• @BrianTung, I want to know if there is a theoretical distribution. Expected value is easy, it is E=m/p=5/0.5=10, in this case, where m is number of successful trials and p is probability of success. – rpmcruz Apr 15 '15 at 8:05
• @rightskewed, hmm what do you have in mind exactly? Using $\lambda=m/p$; testing in R rpois(10, 5/0.5) shows it sampling values below <5, which is impossible in this model. – rpmcruz Apr 15 '15 at 8:08
• Just to make it clear: I want to know if there is already a popular theoretical distribution for what I want. If anyone works to work out the probability mass function that would be superb too. :P Might make sense to migrate it to: stats.stackexchange.com . I did not remember about the statistics forum ... – rpmcruz Apr 15 '15 at 8:12
I figure this one out. :)
I can model it using a Negative Binomial: https://en.wikipedia.org/wiki/Negative_binomial_distribution
First, let us change the values of my case scenario, just to make it clearer. "Case scenario: a retro-virus infects a healthy cell. The virus programs the cell to brew little viruses, at a rate of 0.2 per-sec, until finally the cell bursts when the number of virus inside it is 5. How to model this?"
We can model number of failures $Y$ as $Y\sim\mathcal{NB}(5,0.2)$. That answers the question, how many failed trails do we have, when we need 5 successful at a probability rate of 0.2. But we do not want failed trials, we want total trials, and total trials = failed trials + successful trials. We know successful trials, which is 5, so our random variable $X$ is such that $X\sim5+\mathcal{NB}(5,0.2)$.
In fact, comparing the random generator function I proposed in the question with the negative binomial random generator (with this adjustment):
par(mfrow=c(1,2))
hist(sapply(1:1e5, function(x) rmy(5, 0.2)))
hist(5+rnbinom(1e5, 5, 0.2))
All functions mean, sd and summary are consistent as well.
My 2 cents here :
Using a Negative Binomial distribution can be a very good approximation. However :
1) You totally remove the case r=0 (No failures).
2) Also, you're not answering your initial question. You're computing the distribution of the #trials before achieving the known #fails (or #success). It's different than #trials knowing the #fails (or #success), as you can have fails following the last success !
I believe the right answer involves Bayesian probabilities :
• We consider that N is a random variable describing the # of trials (# of Bernoulli trials). Let's assume that N is following an Exponential Distribution $$N \thicksim Exp(\lambda)$$
• We then consider that X is a random variable describing the number of successes after N=n trials, following a Binomial distribution $$N \thicksim Binom(p, N)$$
Let's compute $$P(n|k)$$ thanks to the Bayes formula :
$$\frac{1}{P(N=n|X=k)} = \frac{P(X=k)}{P(X=k|N=n)P(N=n)}$$
And by writing $$P(k) = \sum_{m=k}^{\infty}{P(X=k|N=m)P(N=m)}$$
$$= \sum_{m=k}^{\infty}\frac{P(X=k|N=m)P(N=m)}{P(X=k|N=n)P(N=n)}$$
And because $$P(X=k|N=n)$$ and $$P(X=k|N=m)$$ are just Binomial distributions :
$$= \sum_{m=k}^{\infty}\frac{m!k!(n-k)!}{n!k!(m-k)!} \frac{p^k}{p^k} \frac{(1-p)^{m-k}}{(1-p)^{n-k}} \frac{e^{-\lambda.m}}{e^{-\lambda.n}}$$
$$= \sum_{m=k}^{\infty}{[e^{-\lambda}(1-p)]^{m-n}}$$
And by re-indexing the Serie with $$l=m-k <=> m=l+k$$
$$= \sum_{l=0}^{\infty}{[e^{-\lambda}(1-p)]^{l+k-n}}$$
$$\frac{1}{P(N=n|X=k)} = \frac{[e^{-\lambda}(1-p)]^{k-n}}{1 - e^{-\lambda}(1-p)}$$
$$P(N=n|X=k) = \frac{1 - e^{-\lambda}(1-p)}{[e^{-\lambda}(1-p)]^{k-n}}$$
And now you have a probability $$P>0$$ for $$k=0$$ ;-)
Edit: The choice of the prior (Exponential) is totally arbitrary and leads to a lot of simplifications. If anyone wants to attempt the calculation with another prior, feel free :-) | 2020-01-27T15:44:35 | {
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https://stats.stackexchange.com/questions/604226/why-are-x-sim-u-1-1-and-y-x2-dependent | # Why are $X \sim U(-1,1)$ and $Y=X^2$ dependent?
Suppose we have two continuous random variables $$X \sim U(-1,1)$$ and $$Y=X^2$$. I don't understand why they are dependent.
$$E[X] = 0$$ $$E[Y] = \int_{-1}^{1} x^2 dx = 2/3$$ $$E[XY] = \int_{-1}^{1} x^3 dx = 0$$
They are independent to me because $$E[XY]=E[X]E[Y]$$, so why they are dependent?
• Plotting simulated values of $X$ and $Y$ will be enlightening, but you can also work out the density of $Y$ via a change of variables from $X$ in order to compare the probabilities in terms of definition of statistical independence. Feb 4 at 4:15
• I find it challenging to find any function $h$ for which $X$ and $h(X)$ are independent.
– whuber
Feb 4 at 15:55
• @whuber: one such function $h$ is $h(x)=0$. Another one is $h(x)=5$. Feb 4 at 17:25
• @Kjetil Certainly. Not very interesting, are they? ;-)
– whuber
Feb 4 at 18:00
• For specific distributions more interesting functions can be possible. For instance, $Y = \sin(2\pi X)$ is independent from $X$ if the domain of $X$ is integer values only. But yes, it will still be the not so interesting map where all values in the domain of $X$ need to be mapped to a single point. It is because $f(X)$ given $X=x$ is a singular distribution in the point $f(x)$. Independence requires that the distribution of $f(X)$ is the same distribution for every value of $X$, $f(x)$ must be a single value for every $x$. Feb 4 at 20:55
Let me write this in a generic manner:
Two random variables for which the expectation of the product is equal to the product of expectations need not be independent.
Let $$X\sim \mathcal U(-a, a)$$ where $$a\in\mathbb R_+$$ and let $$Y\sim X^2.$$ Then as noted $$\mathbb E[XY]=\mathbb E[X^3] =0=\mathbb E[X]\mathbb E[Y].$$ But they are definitely not independent.
## Reference:
$$\rm [I]$$ Counterexamples in Probability and Statistics, Joseph P. Romano, Andrew F. Siegel, Wadsworth, $$1986,$$ ex. $$4.15.$$
Since $$Y = X^2$$, $$[-1/2 \leq X \leq 0] \cap [Y > 1/4] = \varnothing$$, hence $$P[-1/2 \leq X \leq 0, Y > 1/4] = 0$$. On the other hand, $$P[-1/2 \leq X \leq 0] = 1/4 > 0, P[Y > 1/4] = 1/2 > 0$$. Hence the independence defining relation $$P(A \cap B) = P(A)P(B)$$ for all $$A \in \sigma(X), B \in \sigma(Y)$$ fails to hold for $$A = [-1/2 \leq X \leq 0]$$ and $$B = [Y > 1/4]$$, i.e., $$X$$ and $$Y$$ are not independent.
Above is a formal proof. Intuitively, since $$Y$$ is a deterministic function of $$X$$, knowing the value of $$X$$ means knowing the value of $$Y$$, hence $$Y$$ and $$X$$ of course cannot be independent (the heuristic definition of independence between $$X$$ and $$Y$$ requires that observing the information provided by $$X$$ does not increase the information of $$Y$$).
It is worth emphasizing that $$E[XY] = E[X]E[Y]$$ is a necessary condition, rather than a sufficient condition for the independence of $$X$$ and $$Y$$. If you really want to express the independence of $$X$$ and $$Y$$ in terms of expectations, it should be stated as
$$X$$ and $$Y$$ are independent $$\iff$$ $$E[f(X)g(Y)] = E[f(X)]E[g(Y)]$$ for all Borel measurable functions $$f$$ and $$g$$.
Evidently, the condition $$E[f(X)g(Y)] = E[f(X)]E[g(Y)]$$ is much stronger than the condition $$E[XY] = E[X]E[Y]$$. The former is a system of infinitely many equations, while the latter is a single equation.
You need to distinguish between the concepts of uncorrelatedness, which involves the decomposition of the expectation of a product, and independence, which involves the decomposition of a joint probability distribution. What you're calling independence is actually uncorrelatedness. What happens, as the other answers show, is that independence is a stronger condition than uncorrelatedness, that is, independence implies uncorrelatedness but the vice versa is not true.
And your example is exactly one example that is used to show that uncorrelatedness does not imply independence.
They are clearly not independent: if you know the value $$X$$ then this gives you information about the value of $$Y$$ and indeed allows you to calculate it precisely.
You have correctly said independence means $$P(X\le a, Y\le b)=P(X\le a) P(Y\le b)$$, so for example consider $$a=\frac13$$ and $$b=\frac 14$$.
• $$X \le \frac 13$$, i.e. when $$-1 \le X \le \frac13$$, has probability $$\frac23$$
• $$Y \le \frac 14$$, i.e. when $$-\frac12 \le X \le \frac12$$, has probability $$\frac12$$
• $$X \le \frac 13, Y \le \frac 14$$ together, i.e. when $$-\frac12 \le X \le \frac13$$, has probability $$\frac5{12}$$
but $$\frac23 \times \frac12 \not=\frac5{12}$$ so they are not independent. Try other examples with $$-1 < a < 1$$ and $$0 < b < 1$$.
$$E[XY] = E[X]E[Y]$$ is a condition for zero covariance and so zero correlation, but this on its own does not imply independence. The bottom row of Wikipedia's chart illustrating correaltion has other examples of this; in your question you would get a parabolic U-shaped chart.
• Note that zero covariance is sufficient for independence if $X$ and $Y$ are jointly Gaussian, but not necessarily (or maybe ever) for other distributions. That it works for multivariate Gaussian distributions might be the source of the confusion.
– Dave
Feb 4 at 14:03
Here is a simulation of how the joint distribution looks like:
It shows clearly that the two variables are not independent.
Your formula is more generally (including dependent variables).
$$\text{E}[XY] = \text{E}[X]\text{E}[Y] + \text{COV}(X,Y)$$
So when you compute $$\text{E}[XY] = \text{E}[X]\text{E}[Y]$$ then it must mean that $$\text{COV}(X,Y) = 0$$. But that doesn't imply independence. This makes your case an example of: Why zero correlation does not necessarily imply independence
• It is not clear where you got the idea for the product $\text{E}[XY] = \text{E}[X]\text{E}[Y]$. Possibly you are confusing it for a description of independence as $P(A \text{ and } B) = P(A) \cdot P(B)$ Feb 4 at 19:27 | 2023-03-21T17:00:15 | {
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https://www.physicsforums.com/threads/component-of-a-vector-along-another-vector.777615/ | # Component of a vector along another vector.
Tags:
1. Oct 22, 2014
### Hijaz Aslam
1. The problem statement, all variables and given/known data
Given $\vec{A}=2\hat{i}+3\hat{j}$ and $\vec{B}=\hat{i}+\hat{j}$.Find the component of $\vec{A}$ along $\vec{B}$.
2. Relevant equations
$\vec{A}.\vec{B}=ABcosθ$ where θ is the angle between both the vectors.
3. The attempt at a solution
I attempted the question as follows:
Let the angle between $\vec{A}$ and $\vec{B}$ be 'θ'. So the component of $\vec{A}$ along $\vec{B}$ is given by $Acosθ\hat{B}$ => $Acosθ(\frac{\vec{B}}{B})$
As $\vec{A}.\vec{B}=ABcosθ$ => $[( 2\hat{i}+3\hat{j})(\hat{i}+\hat{j})]/B=Acosθ$ => $\frac{5}{\sqrt{2}}=Acosθ$
Therefore the component is : $\frac{5}{\sqrt{2}}(\frac{\hat{i}+\hat{j}}{\sqrt{2}})$ => $\frac{5}{2}({\hat{i}+\hat{j}})$
But my text produces the solution as follows:
$A_B=(\vec{A}.\vec{B})\hat{B}=\frac{5}{\sqrt{2}}(\hat{i}+\hat{j})$.
2. Oct 22, 2014
### RUber
I usually see this process broken down into basis components.
That is $\hat B =\sqrt{2}/2 \hat i + \sqrt{2}/2 \hat j.$
Then the component is $A\cdot \hat B_i \hat i + A\cdot \hat B_j \hat j$.
Somewhere in your process, you divided by the magnitude of B twice.
3. Oct 23, 2014
### willem2
You're right and the book is wrong. The book answer as well as the formula for AB they use.
The length of your answer is smaller than the length of A as it should be. The book answer is larger.
The projection of A on B should only depend on the direction of B, not the magnitude. The formula used for AB in the book does depend on the magnitude of B.
4. Oct 23, 2014
### Delta²
Book is wrong . We can verify this by standard euclidean geometry easily because by the definition of cosine, it will be cos(θ)=(component of A along B)/A hence Acos(θ)=(component of A along B). And we have to multiply this by the unit vector of B to get the required result. | 2017-08-22T14:53:53 | {
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https://www.physicsforums.com/threads/different-results-with-quotient-rule.693718/ | # Different results with quotient rule
1. May 26, 2013
### mike82140
I have been trying to figure out the derivative of (X²-1)/X. When I use the quotient rule, the result I get is 1-1/X². However, when I simplify the expression first, then take the derivative, I get 1+1/X²
Why are the results different?
2. May 26, 2013
### Hypersphere
3. May 26, 2013
### mike82140
Can you show how you evaluate it? I tried both ways, but get different results.
4. May 26, 2013
### Hypersphere
Alright, for a function
$$f(x) = \frac{g(x)}{h(x)}$$
the quotient rule says
$$f'(x) = \frac{g'(x)h(x)-g(x)h'(x)}{h^2(x)}$$
In your case, $g(x)=x^2-1$ and $h(x)=x$. Thus $g'(x)=2x$ and $h'(x)=1$. Then you get
$$\frac{\mathrm{d}}{\mathrm{dx}}\frac{x^2-1}{x}=\frac{2x\cdot x - \left( x^2 -1 \right) \cdot 1}{x^2}=\frac{2x^2-x^2+1}{x^2}=1+\frac{1}{x^2}$$
Are you sure you included the $g'(x)$ term?
Simplifying first,
$$\frac{\mathrm{d}}{\mathrm{dx}}\frac{x^2-1}{x} = \frac{\mathrm{d}}{\mathrm{dx}} \left( x - \frac{1}{x} \right) = 1- \frac{-1}{x^2}=1+\frac{1}{x^2}$$
5. May 26, 2013
### mike82140
Where did +1 come from when you were multiplying x²-1 with 1?
6. May 26, 2013
### Hypersphere
It was also multiplied by -1, from the quotient rule formula.
7. May 26, 2013
### mike82140
I'm sorry for asking so many questions, and this may sound stupid, but, what -1?
Edit: It appears as though I have made a mistake. The -1 is the minus part that is in front of the x²-1, so the negative, or minus, distributes itself.
Thank you for the help, I appreciate it.
8. May 26, 2013
### Hypersphere
You see the minus sign in front of $\left( x^2 -1 \right) \cdot 1$, right? That is just short notation for
$$- \left( x^2 -1 \right) \cdot 1=\left( -1 \right) \cdot \left( x^2 -1 \right)= (-1) \cdot x^2 + (-1) \cdot (-1) = -x^2+1$$
Agreed? Or are you pulling my leg here?
EDIT: Ah, good. I was almost giving up for a while there.
9. May 26, 2013
### SteamKing
Staff Emeritus
Look at the formula for the quotient rule.
If h(x) = x, what is h'(x)? | 2017-09-24T18:04:30 | {
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https://proofwiki.org/wiki/Order_is_Preserved_on_Positive_Reals_by_Squaring | # Order is Preserved on Positive Reals by Squaring
## Theorem
Let $x, y \in \R: x > 0, y >0$.
Then:
$x < y \iff x^2 < y^2$
## Proof 1
### Necessary Condition
Assume $x < y$.
Then:
$\ds x < y$ $\implies$ $\ds x \times x < x \times y$ Real Number Ordering is Compatible with Multiplication $\ds x < y$ $\implies$ $\ds x \times y < y \times y$ Real Number Ordering is Compatible with Multiplication $\ds$ $\leadsto$ $\ds x^2 < y^2$ Real Number Ordering is Transitive
So:
$x < y \implies x^2 < y^2$
$\Box$
### Sufficient Condition
Assume $x^2 < y^2$.
Then:
$\ds x^2$ $<$ $\ds y^2$ $\ds \leadsto \ \$ $\ds 0$ $<$ $\ds y^2 - x^2$ Real Number Ordering is Compatible with Addition $\ds \leadsto \ \$ $\ds \paren {y - x} \paren {y + x}$ $>$ $\ds 0$ Difference of Two Squares $\ds \leadsto \ \$ $\ds \paren {y - x} \paren {y + x} \paren {y + x}^{-1}$ $>$ $\ds 0 \times \paren {y + x}^{-1}$ as $x + y > 0$ $\ds \leadsto \ \$ $\ds y - x$ $>$ $\ds 0$ $\ds \leadsto \ \$ $\ds x$ $<$ $\ds y$
So:
$x^2 < y^2 \implies x < y$
$\blacksquare$
## Proof 2
From Real Numbers form Ordered Field, the real numbers form an ordered field.
The result follows from Order of Squares in Ordered Field.
$\blacksquare$
## Proof 3
From Real Numbers form Ordered Field, the real numbers form an ordered field.
By definition, an ordered field is a totally ordered ring without proper zero divisors.
The result follows from Order of Squares in Totally Ordered Ring without Proper Zero Divisors.
$\blacksquare$
## Proof 4
### Necessary Condition
Let $x < y$.
Then:
$\ds x < y$ $\implies$ $\ds x \times x < x \times y$ Real Number Ordering is Compatible with Multiplication $\ds x < y$ $\implies$ $\ds x \times y < y \times y$ Real Number Ordering is Compatible with Multiplication $\ds$ $\leadsto$ $\ds x^2 < y^2$ Real Number Ordering is Transitive
So:
$x < y \implies x^2 < y^2$
$\Box$
### Sufficient Condition
Let $x^2 < y^2$.
Aiming for a contradiction, suppose $x \ge y$.
Then:
$\ds x \ge y$ $\implies$ $\ds x \times x \ge x \times y$ Real Number Ordering is Compatible with Multiplication $\ds x \ge y$ $\implies$ $\ds x \times y \ge y \times y$ Real Number Ordering is Compatible with Multiplication $\ds$ $\leadsto$ $\ds x^2 \ge y^2$ Real Number Ordering is Transitive
But this contradicts our assertion that $x^2 < y^2$.
Hence by Proof by Contradiction it follows that:
$x < y$
$\blacksquare$ | 2021-08-01T00:42:17 | {
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http://math.stackexchange.com/questions/78533/prove-that-2n-n2-is-even-if-n-is-a-positive-integer/114520 | # prove that $(2n)!/(n!)^2$ is even if $n$ is a positive integer
Prove that $(2n)!/(n!)^2$ is even if $n$ is a positive integer. For clarity: the denominator is the only part being squared.
My thought process: The numerator is the product of the first n even numbers and the product of the first n odd numbers. That is, $(2n!) = (2n)(2n-2)(2n-4)\cdots(2n-1)(2n-3)(2n-5).$ In effect, the product of even numbers can be cancelled out with n! resulting in the following quotient: $$(2^n)(2n-1)(2n-3)/(n!).$$ To me this looks even thanks to the powers of 2. But I am not convinced for some reason.
Sorry for the poor notation, I don't know any coding languages, my apologies.
-
Try counting the powers of 2 in the numerator and denominator. After a quick bit of experimentation, it looks like it's probably sufficient to consider $n=4m+k$ for $k=0,1,2,3$ (i.e. there are four cases) and you could demonstrate that there are always more powers of 2 in the numerator than in the denominator. – Chris Taylor Nov 3 '11 at 9:40
Could you give me more insight as to how you came up with the four cases for n? I don't see the derivation for n as you have shown. – hdtv1104 Nov 3 '11 at 9:50
Did you try using induction? It seems like that shouldn't be too difficult... – Braindead Nov 3 '11 at 13:59
We have $$\displaystyle \frac{(2n)!}{(n!)^2} = \binom{2n}{n}$$ and of course for every way to choose a combination of $n$ objects from a total of $2n$ objects, there exists a complementary choice (the left over $n$ objects). Since the ways to pick $n$ objects from $2n$ comes in pairs, it follows that the total number is even.
Another proof: $$\frac{(2n)!}{(n!)^2} = \frac{ 2n \cdot (2n-1)! }{(n!)^2 } = 2 \cdot \frac{ n \cdot (2n-1)! }{ n\cdot (n-1)! \cdot n!} = 2 \cdot \frac{(2n-1)!}{n! (n-1)!} = 2 \binom{2n-1}{n}$$
This one can be written in another flavor (using $\binom{n}{k} = \binom{n-1}{k-1} + \binom{n-1}{k}$ and the symmetric property): $$\frac{(2n)!}{(n!)^2} = \binom{2n}{n} = \binom{2n-1}{n-1} + \binom{2n-1}{n} = 2\binom{2n-1}{n}.$$ Yet another: The sum of the $2n$-th row of Pascal's triangle is $2^{2n}$ which is even, and the sum of all the entries excluding the central coefficient is also even, because of the symmetric property $\displaystyle \binom{n}{k} = \binom{n}{n-k}$. Thus, the central coefficient must also be even.
We can squeeze this problem a bit further and learn something interesting with this fourth proof. This one does not require any knowledge of binomial coefficients. To investigate the prime factors of factorials we use the following identity: $$l = \sum_{k=1}^{\infty} \bigg\lfloor \frac{n}{p^k} \bigg\rfloor$$
where $p$ is a prime number and $l$ is the unique natural number such that $p^l \mid n!$ but $p^{l+1} \nmid n!.$ Basically this counts the multiplicity of the prime factor $p$ in $n!$.
The idea of the proof is this: Write $n! = 1\cdot 2 \cdot 3 \cdots (n-1) \cdot n$, and ask yourself where the factors of $p$ come from. Clearly you get one factor of $p$ from $p, 2p, 3p, \cdots$ so there are $\lfloor n/p \rfloor$ factors of $p$ from these. But that's not the whole story - when we checked over the multiples $p,2p,3p\cdots$ we didn't count the second factor of $p$ every time there was $p^2, 2p^2, 3p^2 \cdots$ so there comes another $\lfloor n/p^2 \rfloor$ to add. Wait, we didn't count another factor every time we missed cubes in $p^3, 2p^3,\cdots$, so we add another $\lfloor n/p^3 \rfloor$ and so on. Also, note that this may look like an infinite series, but eventually $p^k > n$ so all the terms eventually become $0.$
Anyway, back to the main goal. Using our identity, we see that there are $$\sum_{k=1}^{\infty} \bigg \lfloor \frac{2n}{2^k} \bigg \rfloor - 2 \bigg \lfloor \frac{n}{2^k} \bigg \rfloor$$ factors of $2$ in $\displaystyle \frac{(2n)!}{ (n!)^2}$, and we wish to show that this number is at least $1.$
With some easy casework we can see that $$\lfloor 2x \rfloor - 2 \lfloor x \rfloor = \left\{ \begin{array}{lr} 1 & : \{ x\} \geq \frac{1}{2} \\ 0 & : \{ x\} < \frac{1}{2} \end{array} \right.$$ where $\{ x \}$ denotes the fractional part of $x.$ Thus our problem is reduced to showing that there exists some $k\in \mathbb{N}$ such that $\displaystyle \frac{n}{2^k}$ has fractional part greater than or equal to $1/2.$ This is of course true, since if $m$ is the largest positive integer such that $\displaystyle \frac{n}{2^m} \geq 1$ then $\displaystyle \frac{1}{2} \leq \frac{n}{2^{m+1} } < 1.$ Hence, $\displaystyle \frac{(2n)!}{(n!)^2}$ is even.
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Thanks for the combinatorics response to this proof. I see that it is easier to invoke the combinatorial interpretation but I myself could not see this interpretation as readily. What inspired the use of binomial? – hdtv1104 Nov 3 '11 at 9:53
Just a bit of practice I guess. Every time I see a ratio of factorials, I check whether it fits the form $\frac{n!}{k! (n-k)!} = \binom{n}{k}.$ Also, it was particularly easy for me here because this particular coefficient, $\binom{2n}{n}$, is quite special and comes up relatively often. – Ragib Zaman Nov 3 '11 at 9:55
Thanks a lot Ragib. – hdtv1104 Nov 3 '11 at 9:58
You're welcome! – Ragib Zaman Nov 3 '11 at 10:00
In fact, the number of factors of $2$ in $\binom{2n}{n}$ is equal to the number of $1$ bits in the binary representation of $n$. – robjohn Nov 14 '11 at 21:24
It is often fun to try to solve such problems using Lagrange's Theorem from finite group theory. That is, to prove that $a$ divides $b$ you try to find a group of cardinality $b$ with a subgroup of cardinality $a$.
In this case, the symmetric group on $2n$ letters, $S_{2n}$, is an obvious choice for a group of cardinality $(2n)!$. Let $G\subset S_{2n}$ be the set of all permutations which map the set $\{1,\ldots, n\}$ onto $\{1,\ldots, n\}$ or $\{n+1,\ldots, 2n\}$. You can check that $G$ is a subgroup of $S_{2n}$.
To compute the cardinality $\lvert G \rvert$, one option is to observe that $G\cong S_n\wr \mathbb{Z}_2$, so $\lvert G\rvert = \lvert S_n\rvert^{\lvert \mathbb{Z}_2\rvert} \lvert \mathbb{Z}_2\rvert = 2(n!)^2$. Alternatively, choosing an element of $G$ corresponds to choosing two permutations of a set of size $n$ and a choice of whether to swap $\{1,\ldots,n\}$ with $\{n+1,\ldots, 2n\}$ or not, so $\lvert G\rvert = 2(n!)^2$.
Thus by Lagrange's Theorem $2(n!)^2$ divides $(2n)!$, or in other words, $\frac{(2n)!}{(n!)^2}$ is even.
-
By the binomial theorem, $$4^n=(1+1)^{2n}=\sum\limits_{k=0}^{2n}{2n\choose k}={2n\choose n}+\sum\limits_{k=0}^{n-1}{2n\choose k}+\sum\limits_{k=n+1}^{2n}{2n\choose k}.$$ Since Pascal's triangle is symmetrical, each term in the sum from $k=0$ to $n-1$ coincides with one term in the sum from $k=n+1$ to $2n$, hence these two sums are equal and $$\frac{(2n)!}{(n!)^2}={2n\choose n}=4^n-2\cdot\sum\limits_{k=0}^{n-1}{2n\choose k}$$ is even for every $n\geqslant1$.
-
I had this one in my answer =) But not as fleshed out. – Ragib Zaman Nov 14 '11 at 21:02
Oh yes, I missed it but I see it now. Sorry about that. – Did Nov 14 '11 at 21:08
Let's observe an example:
$(2\cdot 5)!=10\cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5!$ , we may conclude that:
$(2n)!=2n\cdot(2n-1)\cdot(2n-2)....(n+1)n!$ so we get following expression :
$$\frac{2n\cdot(2n-1)\cdot(2n-2)....(n+1)}{n!}=\frac{2n\cdot(2n-1)\cdot2\cdot(n-1)\cdot(2n-3)\cdot 2 \cdot (n-2)....(n+1)}{n!}$$
$$=\frac{2\cdot2\cdot2...\cdot2\cdot n\cdot(n-1)\cdot(n-2)\cdot......\cdot 1\cdot(2n-1)\cdot(2n-3)\cdot......(n+1)}{n!}$$
$$=\frac{2\cdot2\cdot2...\cdot2 \cdot n!\cdot(2n-1)\cdot(2n-3)\cdot......\cdot (n+1)}{n!}=$$
$$=2\cdot2\cdot2...\cdot2 \cdot(2n-1)\cdot(2n-3)\cdot......\cdot (n+1)$$ so number is even.
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But I'm sorry, this proof is false (still got 11 upvotes). – user144248 May 15 '14 at 21:32
The error here is at the third equality: The terms that have been divided by $2$ don't go all the way down to $1$, since there are only (about) $n/2$ such terms. In fact, for $n$ even, the last term that is of this form is $n + 2$, so the correct expression would be $$= \frac{2 \cdots 2 \cdot n \cdot (n - 1) \cdot ... \cdot \frac{n + 2}{2} \cdot (2n - 1) \cdots (n + 1)}{n!}$$ – user61527 May 16 '14 at 4:49
This answer has been moved from this question, which was closed because it was judged to be a duplicate of this question.
The number of factors of a prime $p$ in $n!$ is $$\nu_p(n)=\left\lfloor\frac{n}{p}\right\rfloor+\left\lfloor\frac{n}{p^2}\right\rfloor+\left\lfloor\frac{n}{p^3}\right\rfloor+\dots\tag{1}$$ $\left\lfloor\frac{n}{p}\right\rfloor$ counts the number of positive multiples of $p$ no greater than $n$, but only once. $\left\lfloor\frac{n}{p^2}\right\rfloor$ counts the multiples of $p^2$ a second time, and $\left\lfloor\frac{n}{p^3}\right\rfloor$ the positive multiples of $p^3$ a third time, etc. Applying $(1)$ to the base-$p$ representation of $n$ $$n=\sum_{i=0}^kn_ip^i\tag{2}$$ where $0\le n_i<p$, yields \begin{align} \nu_p(n) &=\sum_{i=0}^kn_i\left(1+p+p^2+\dots+p^{i-1}\right)\\ &=\sum_{i=0}^kn_i\left(\frac{p^i-1}{p-1}\right)\\ &=\frac{1}{p-1}\left(n-\sum_{i=0}^kn_i\right)\\ &=\frac{n-\sigma_p(n)}{p-1}\tag{3} \end{align} where $\sigma_p(n)$ is the sum of the base-$p$ digits of $n$.
Therefore, the number of factors $p$ in $\displaystyle\binom{n}{k}=\frac{n!}{k!(n-k)!}$ is \begin{align} &\nu_p(n)-\nu_p(k)-\nu_p(n-k)\\ &=\frac{1}{p-1}\left(n-\sigma_p(n)-k+\sigma_p(k)-(n-k)+\sigma(n-k)\right)\\ &=\frac{\sigma_p(k)+\sigma_p(n-k)-\sigma_p(n)}{p-1}\tag{4} \end{align} Using $(4)$ with $p=2$ says that the number of factors of $2$ in $\displaystyle\binom{2n}{n}$ is $\sigma_2(n)+\sigma_2(n)-\sigma_2(2n)$ which is simply $\sigma_2(n)$ since $\sigma_2(n)=\sigma_2(2n)$. Therefore, the number of factors of $2$ in $\displaystyle\binom{2n}{n}$ is the number of $1$-bits in the binary representation of $n$.
Thus, if $n>0$, then there is at least one $1$-bit in $n$ in binary, and so $2$ divides $\displaystyle\binom{2n}{n}$.
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The exact power of $2$ dividing $\binom{2n}{n}$ is $2$^(sum of digits in base $2$ representation of $n$).
The exact power of $3$ dividing the trinomial coefficient $\binom{3n}{n,n,n} = (3n)!/(n!)^3$ is $3$^(sum of digits in base $3$ representation of $n$).
The exact power of $p$ dividing the $p$-nomial coefficient $\binom{pn}{n,n,\dots , n} = (pn)!/(n!)^p$ is $p$^(sum of digits in base $p$ representation of $n$) for any prime $p$.
This is from the formula for the power of $p$ dividing $n!$, which is ($n$ - sum of digits of $n$ when written in base $p$)/($p-1$). There also are combinatorial and generating function proofs but they are more complicated.
- | 2015-08-30T08:46:05 | {
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http://mathhelpforum.com/pre-calculus/8982-finding-tangent-line-without-calculus.html | Math Help - Finding a Tangent Line without Calculus
1. Finding a Tangent Line without Calculus
I hope this is in the right place, I'm not in a hurry, just curious.
How can I find an equation for a line tangent to a point on a parabola without using calculus?
I just started playing with this this morning
The equation I'm using is y = x^2 - 4x - 2 and I'm looking for the equation of the tangent line at point ( 4, -2)
Using calculus I found the equation to be y = 4x -18
How can I do this without calculus?
2. Hello, Ranger SVO!
You're curious . . . good for you!
While playing with parabolas in college (centuries ago),
. . I "discovered" some facts that might help you.
The equation I'm using is $y \:= \:x^2 - 4x - 2$
and I'm looking for the equation of the tangent line at point $(4,\,-2)$
Using calculus I found the equation to be: $y \:= \:4x -18$
How can I do this without calculus?
Code:
* | */
| /
* | *P(x,y)
* | */
---------***---o----------
| /↑
| /(½x,0)
| /
|/
o ←(0,-y)
/|
Given the parabola: . $y \:=\:ax^2$, the tangent at point $P(x,y)$
. . has an x-intercept of $\left(\frac{1}{2}x,\,0\right)$ and a y-intercept of $(0,\,-y)$
The tangent at $P(x,y)$ always has a point directly below the vertex
. . with the same vertical displacement (only downward).
Your parabola $y \:=\:x^2-4x-2$ has its vertex at $(2,-6).$
Your point $P(4,-2)$ is 4 units above the level of the vertex.
Hence, it has a point 4 units directly below the vertex: $Q(2,-10).$
Now you can write the equation through $P$ and $Q.$
3. Originally Posted by Ranger SVO
I hope this is in the right place, I'm not in a hurry, just curious.
How can I find an equation for a line tangent to a point on a parabola without using calculus?
I just started playing with this this morning
The equation I'm using is y = x^2 - 4x - 2 and I'm looking for the equation of the tangent line at point ( 4, -2)
Using calculus I found the equation to be y = 4x -18
How can I do this without calculus?
I have another approach.
It does not always work.
But it does sometimes.
In your example, the tangent line is the line that only intersects the parabola at one point. Thus, you need to find an equation of a line that one intersects at one point, i.e. solution set is only one element. That means you need the discrimanant to be zero.
4. Soroban, I like your explination. But first, at my age curiousity is the only thing that keeps me from vegetating.
Here is what I've been working on. If I draw a rectangle using the vertex and the point where I want the tagent line, I can see the secant line bisecting the rectangle. I can easily notice that the slope of the tangent is twice that of the secant.
Is this always true? If so why?
Perfect Hacker, does that mean that I can use the quadratic equation to find the slope of a line tangent to some point.
5. Hello again, Ranger SVO!
Here is what I've been working on.
If I draw a rectangle using the vertex and the point where I want the tangent line,
I can see the secant line bisecting the rectangle.
I can easily notice that the slope of the tangent is twice that of the secant.
Is this always true? If so why?
Given the parabola: $y \:=\:ax^2$ and any point on it: $P(p,\,ap^2)$
The slope of the secant $OP$ is: . $m_s \:=\:\frac{ap^2 - 0}{p - 0} \:=\:ap$
The derivative is: . $y' \:=\:2ax$
. . Hence, the slope of the tangent at $P$ is: $m_t\:=\:2ap$
Therefore, the slope of the tangent is always twice the slope of the secant.
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
If you're really bored, verify these claims.
1) On the parabola $y \:=\:ax^2$, pick any two points: . $P(p,\,ap^2)$ and $Q(q,\,aq^2)$
Draw the chord $PQ.$
Locate the tangent parallel to $PQ.$
It is found at: . $x \:=\:\frac{p+q}{2}$ . . . halfway between $p$ and $q.$
2) On the parabola $y \:=\:ax^2$, pick any three points: . $P(p,\,ap^2),\;Q(q,\,aq^2),\;R(r,\,ar^2)$
Find the area of $\Delta PQR.$
Consider the tangents to the parabola at $P,\,Q,\,R.$
. . They will intersect at $A,\,B,\,C.$
Find the area of $\Delta ABC.$
We find that: . $(\text{area }\Delta ABC) \:=\:\frac{1}{2}(\text{area }\Delta PQR)$
6. y = x^2 - 4x - 2 and I'm looking for the equation of the tangent line at point ( 4, -2)
We want a tangent line at $(4,-2)$.
To do that I assume you are familar with the point-slope formula:
$y-y_0=m(x-x_0)$
In this case,
$(x_0,y_0)=(4,-2)$.
Thus, the equation of the line (tangent) at that point is determined by,
$y+3=m(x-4)$ that means ( $y=m(x-4)-3$)
Where,
$m$
Is the slope to be determined.
As I said we need to intersection to be a single point.
That is,
$\left\{ \begin{array}{c}y=x^2-4x-2\\y=m(x-4)-3 \end{array} \right\}$
We want this to has a single solution.
Equate,
$x^2-4x-2=m(x-4)-3$
$x^2-4x-2=mx-4m-3$
$x^2+x(-4-m)+(1+4m)=0$
To have a single real solution we need the discrimanant to be zero,
$(-4-m)^2-4(1)(1+4m)=0$
Solve to get the value of $m$. | 2014-10-22T07:27:46 | {
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http://dzdb.whynotimmobiliare.it/cambridge-vector-calculus-lecture-notes.html | # Cambridge Vector Calculus Lecture Notes
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Learning vector calculus techniques is one of the major missions to be accomplished by physics undergraduates. : (0711) 685-66346. Acceleration and Force 131 56. Corpus ID: 63957860. 2 Notes • Homework # 1 is uploaded on the course webpage Vector Calculus. 5-4 Lecture 5: Trajectory Optimization 5. Willard Gibbs Josiah Willard Gibbs , Edwin Bidwell Wilson Yale University Press , 1901 - Vector analysis - 436 pages. Welcome To the Vector Calculus (Math 202) Home Page (These lecture notes are from Fall 2002) Click below for the course syllabus and lecture notes: Example Course Syllabus. Between points k 1 and kyou had the estimate of the arc length as p (x k)2 + (y k)2, but here you need the whole vector from ~r k 1 to ~r kin order to evaluate. Andre Lukas Oxford, 2013 3. VECTOR ALGEBRA r r 2 3 O r1. Multivariable Calculus, Fall 2018 (ASU MAT 267): Multivariable Calculus part 1, 18pp: Vectors and 3-Dimensional Geometry. It pro vides a way to describe physical quantities in three-dimensional space and the way in which these quantities vary. 1(a) The Vector Differential Operator. Classical Electromagnetism: An intermediate level course Richard Fitzpatrick Associate Professor of Physics The University of Texas at Austin. Brief notes for 268. MAE5201 - Solid Mechanics Course Notes About These notes are for the personal use of students who are enrolled in or have taken MAE5201 at the University of Colorado Colorado Springs in the Spring 2017 semester. There is no need for parametric equations. Given on September 4 th. Verify this result when S is the sphere jx j = R and A = ( z; 0;0) in Cartesian coordinates. 3 Independence of Path 15. [email protected] While some of the pages are proofread pretty well over the years, others were written just the night before class. The content may be incomplete. Course Notes and General Information Vector calculus is the normal language used in applied mathematics for solving problems in two and three dimensions. Description: Topics covered in these notes include the untyped lambda calculus, the Church-Rosser theorem, combinatory algebras, the simply-typed lambda calculus, the Curry-Howard isomorphism, weak and strong normalization, type inference, denotational semantics, complete partial. Calculus III is the last course in the Calculus I, II, III sequence. PHY2060) and at least have co-registered in a vector. Andre Lukas Oxford, 2013 3. Notes in analysis on metric and Banach spaces with a twist of topology. : In V3, 3 non-coplanar vectors are linearly independent; i. Five lectures for undergraduates on general relativity, by Jorge. MACLACHLAN MURPHY, IAN S. When applied to a function defined on a one-dimensional domain, it denotes its standard derivative as defined in calculus. Multiple Integrals and Vector Calculus - Lecture Notes. Course Description. It is useful to think. pdf file) viewer can be obtained from Adobe Once the Acrobat plugin has been downloaded and installed, file can be view simply by clicking on the corresponding link. Scalar multiplication: If c2R and ~ua vector, then we may form a new vector c~ucalled the scalar product of ~uwith c. are expected to attend all of the lectures. Here is an unordered list of online mathematics books, textbooks, monographs, lecture notes, and other mathematics related documents freely available on the web. Precise Definition of Limit 6. Some of the pages were developed as complements to the text and lectures in the years 2000-2004. This book uses SI units (the mks convention) exclusively. AUTUMN 2012 Lectures TR 1:00-2:00, F 2:00-3:00, 26-100 Instructor James McKernan, 2-274, phone 253-4391, [email protected] Don't show me this again. Vector Calculus - Free download as Powerpoint Presentation (. Textbook: Calculus. University College Dublin An Col aiste Ollscoile, Baile Atha Cliath School of Mathematics and Statistics Scoil na Matamaitice agus na Staitistic Vector Integral and Di erential Calculus (ACM 20150) Dr Lennon O N araigh Lecture notes in Vector Calculus, September 2017. Lecture Notes Labs Assignments Download Course Materials; Users may find additional or updated materials at Professor Carter's 3. Boqing Gong (Deep Neural Networks for Computer Vision Applications) Lecture 19: Deformable Models and Image Segmentation. ) Lecture notes by Giovanni Leoni. The main focus of this module is on multivariable calculus in 2 and 3 dimensions, and vector calculus. CMS College (Autonomous) The CMS College, Kottayam, founded by the Church Missionary Society of England, is one of the oldest institutions of. Vector Calculus Lecture Notes Adolfo J. 1: Vector function of one variable-- a vector, each component of which is a – A free PowerPoint PPT presentation (displayed as a Flash slide show) on PowerShow. 2 Spans, Lines and Planes The span of a single vector vin Rnis the set of all scalar multiples of v: spanfvg= ftvjt2Rg:. Lecture 4: Wrap up of vector calculus: Poisson & Laplace equations; curl 4. Whittaker, G. There are 9 chapters, each of a size that it should be possible to cover in one week. The unique vector of length zero is denoted ~0 and satis es ~0 +~v= ~v. IV Vector Calculus In Many Variables. This unit is offered in Semester 1. Study With Me - Probability, Vector Calculus, Analysis and more Last minute Vector Calculus and other flo's study diary 12,748 views. Probability About these notes. nagar - 2018 Material offline reading, offline notes, free download in App, Engineering Class handwritten notes, exam notes, previous year questions, PDF free download. Here are a set of practice problems for the Vectors chapter of the Calculus II notes. A lot of these gaps will be filled in in this course, and more material will be covered. • The inverse of a matrix is defined such that when it operates on the original matrix, the result is the unit matrix. Student Evaluations (Fall 2019 151) Student Evaluations (Fall 2019 152) Syllabus. Since the course is an experimental one and the notes written before the lectures are delivered, there will inevitably be some sloppiness, disorganization, Our subject matter is intermediate calculus and linear. CM111A - Calculus I Compact Lecture Notes ACC Coolen Department of Mathematics, King's College London Version of Sept 2011. Multiple Integrals and Vector Calculus Prof. (14259 views) A Course of Modern Analysis by E. In this course, you'll learn how to quantify such change with calculus on vector fields. We will define line integrals, which can be used to find work done by a force field in moving an object along a curve. If v is a smooth vector field on R3, divcurlv = 0. 4 Mar Review for test March 23 Test 2 March 27 4. The textbook for this course is Stewart: Calculus, Concepts and Contexts (2th ed. Tensors revision questions. Watch out for typos!. Vector Products, Triple Scalar Products. ) Lecture Notes on Multivariable Calculus by Barbara Niethammer and Andrew Dancer. Vector Calculus (Green's Theorem, Stokes' Theorem, Divergence Theorem) For differentiation, you can use Principles of Mathematical Analysis by Rudin (Chapter 9). ; Author: James Byrnie Shaw; Category: Calculus; Length: 325 Pages; Year: 1922. Students should read the textbook before each class and review lecture notes after each class. Description. Calculus Book with Video Lecture Preliminaries, Limits and Continuity, Differentiation, Applications of Derivatives, Integration, Applications of Definite Integrals, Transcendental Functions, Techniques of Integration, Further Applications of Integration, Conic Sections and Polar Coordinates, Infinite Sequences and Series, Vectors and the Geometry of Space, Vector-Valued Functions and Motion. 2) Divergent of a constant vector is always zero Ex: then. Section 1-8 : Tangent, Normal and Binormal Vectors. Vector Calculus deals with calculus in two and three dimensions, and develops the theory of curves, vector functions and partial derivatives, two and three dimensional 2 integration, line integrals and curl and divergence. Lecture notes; Slides of final lecture; Solar-System Dynamics. Chapter 5 : Vectors. PHY2061 Enriched Physics 2 Lecture Notes Refresher Math and Physics Refresher This course assumes that you have studied Newtonian mechanics in a previous calculus-based physics course (i. They consist largely of the material presented during the lectures, though we have taken the liberty of eshing them out in some places and of being more cursory here than in the lectures in other places. With the exception of economics, all these courses run in the Easter term. The final grade will consist of 25% homework, 45% midterms, and 30% final exam. 01 Partial differentiation, multiple integrals, and topics in differential and integral vector calculus, including Green's theorem, Stokes's theorem, and Gauss's theorem for students with a background in linear algebra. Lecture Notes on Classical Mechanics (A Work in Progress) Daniel Arovas Department of Physics University of California, San Diego May 8, 2013. Bear in mind that course syllabuses evolve over time, and different lecturers structure their courses differently and choose their own notation conventions. Multiple integrals. nb) and MathML files. In these lecture notes we shall represent vectors and vector elds using bold fonts, e. Thomas CalculusUse for Calculus I,Calculus II and MV CalculusAcknowledgements. SYLLABUS: Syllabus contains general information on topics, exams, grade, course structure and policies. Tensors revision questions. The direction is correct since the right hand side of the formula is a constant multiple of v so the projection vector is in the direction of v as required. Vector Calculus ADD. For instance R ˆZ (0. 1 Vector Fields 15. Introduction to the min-max theory for minimal surfaces: Hand-written lecture notes for a topic course on the min-max theory of minimal surfaces in 2013. Some gave vector fields; some gave scalar fields. As in those notes, the figures are made with Anders Thorup’s spline macros. Many topics in the physical sciences can be analysed mathematically using the techniques of vector. Sample Exam 1. Winter 2020 Final Exam (10 am lecture): Final Exam. Lecture notes. This section is largely based on my undergraduate lecture notes from a course given by Dr. Description: Topics covered in these notes include the untyped lambda calculus, the Church-Rosser theorem, combinatory algebras, the simply-typed lambda calculus, the Curry-Howard isomorphism, weak and strong normalization, type inference, denotational semantics, complete partial. Notes: Some of these lectures reference the TI-89 graphing calculator. Anastassiou, I. Also, I taught math 53 this past summer and the course page has some problems and solutions that might help studying. Catalog Description: Change of variable in multiple integrals, Jacobian, Line integrals, Green's theorem. edu office: Neill 315, 509-335-2134 office hours: MWF 10:15 – 11:45 am, and by appointment Lecture 2 is conducted by Mark Schumaker in CUE 203, MWF 2:10-3:00 pm email: [email protected] Matthews, P. Two semesters of single variable calculus (differentiation and integration) are a prerequisite. 2 Notation and Nomenclature De nition 1 Let a ij2R, i= 1,2,,m, j= 1,2,,n. Interpretation of ~x′(t) as the velocity vector 129 55. The dates by some of the lectures are the date of the most recent revision. Lecture notes; Slides of final lecture; Solar-System Dynamics. To connect multivariate calculus to other fields both within and without mathematics. Related documents. 1 (Second-Order Equations with Constant Coefficients) Lecture Notes for Section 15. Introduction to Electrodynamics. VECTOR CALCULUS1. None of this is official. The lecture will be recorded and can be found online at Canvas. Consider a trajectory x: [0, T] !C for a configuration space with d. Lecture Note Ser. Cambridge Course Notes. Written by Ross. 6 Divergence Theorem 15. best course for vector calculus. • Recognize that flow velocity is a vector field, which can be a function of space or a function of space and time. Lecture Notes for College Physics I Contents 1 Vector Algebra 1 2 Kinematics of Two-Dimensional Motion 2 3 Projectile Motion 5 4 Newton’s Laws of Motion 8 5 Force Problems 12 6 Forces due to Friction and Uniform Circular Motion 16 7 Newton’s Law of Universal Gravitation 20 8 Work-Energy Theorem I 22 9 Work-Energy Theorem II 24. In this course, you'll learn how to quantify such change with calculus on vector fields. ms access pdf reader 2 Geometry Of Space Curves. Vector Calculus - Winter 2019 Lectures: MWF 2-3 in Pepper Canyon Hall 109 Instructor:Hans Wenzl email: [email protected] Multivariable calculus. Many topics we will cover are generalizations of one variable calculus, including differentiation and integration, but there are completely new phenomena, and. Classical Electromagnetism: An intermediate level course Richard Fitzpatrick Associate Professor of Physics The University of Texas at Austin. Week 12: Integral definition of gradient, divergence and curl. 194 References The following references were consulted during the preparation of these lecture notes. GOAL: In Multivariable Calculus we complete the Calculus sequence. MAT1005: CALCULUS II Vector Calculus. The notes are designed to be used in conjunction with a set of online homework exercises which help the students read the lecture notes and learn basic linear algebra skills. 2 6 LECTURE 1. The second inner cover contains the basic equations of electrodynamics, the accepted values of some fundamental constants, and the transformation equations for spherical and. 30-6 pm TA: palghamol. (b) If n is a unit vector, fis changing at the rate rf(0;ˇ=2) n = ˇ 2 ni in the direction n. Instead of Vector Calculus, some universities might call this course Multivariable Calculus or Calculus Three. In other coordinate systems, the unit vectors are not the same everywhere. Currently the book can be found online here, but the link may change as time progresses. Please check out Dr. Many topics in the physical sciences can be analysed mathematically using the techniques of vector. Plane polar co. These are the lecture notes for my online Coursera course,Vector Calculus for Engineers. A line can always be written as fA+ u: 2Rgfor a unit vector u2R. Functions of many variables, with a focus on surfaces in three dimensions, partial derivatives, gradients, and directional derivatives. The norm or length of a vector is jjxjj= (xx) 1=2 = X. The alternate version Stewart/Clegg/Watson Calculus, 9e, will publish later this spring. In these lecture notes we shall represent vectors and vector elds using bold fonts, e. A list of resources can be found below. In the unit we develop the theory of vector fields, flows and differential forms mainly for R n but with a view towards manifolds, in particular surfaces in R 3. Partial derivatives. 3 Vector Calculus In the last part of the course, we will study vector elds, which are functions that assign a vector to each point in its domain, like the vector-valued func-tion F described above. May 28, 2018 - Calculus 2 help for high school + college students (in-class or online). ; Author: James Byrnie Shaw; Category: Calculus; Length: 325 Pages; Year: 1922. Gauss’ Theorem (Divergence Theorem) Consider a surface S with volume V. A line can always be written as fA+ u: 2Rgfor a unit vector u2R. 3 Warnings and Disclaimers Before proceeding with this interactive manual we stress the following: † These Web pages are designed in order to help students as a source. The magnitude of c~usatis es jc~uj= jcjj~uj. Instead of Vector Calculus, some universities might call this course Multivariable or Multivariate Calculus or Calculus 3. Preliminaries (vectors, dot product, cross product, planes, lines) Functions of Several Variables (graphing, limits, calculus) Cylindrical & Spherical Coordinates; Partial Derivatives (limits, chain rule) Gradients and Directional Derivatives; Planes and Linear Approximation; Extreme-Values of Real. The textbook for this course is Stewart: Calculus, Concepts and Contexts (2th ed. secret-bases. Notes for Calculus III (Multivariable Calculus) The notes below follow closely the textbook Introduction to Linear Algebra, Fourth Edition by Gilbert Strang. This would be pictured by drawing the vector (1, 2, 1) in the opposite direction. You may write on both sides of the page. Probability About these notes. GEOS 4430 Lecture Notes: Well Testing Dr. Lecture 1 Vectors A vector has direction and magnitude and is written in these notes in bold e. Continuity 7. Lecture Notes on Multivariable Calculus Notes written by Barbara Niethammer and Andrew Dancer This de nition is more suitable for the multivariable case, where his now a vector, so it in lectures in the rst year, e. " " Mechanics Lecture Notes Part III: Foundations of Continuum Mechanics. Lecture Notes in Classical Mechanics (80751) Raz Kupferman Institute of Mathematics The Hebrew University July 14, 2008. Many topics we will cover are generalizations of one variable calculus, including differentiation and integration, but there are completely new phenomena, and. These notes are for helpful for undergraduate level (BSc or BS). The unique vector of length zero is denoted ~0 and satis es ~0 +~v= ~v. Digital PDF 9. Derivatives 8. Real Analysis. Relevant undergraduate courses are (for relevant schedules, example sheets and exam questions, refer to the General Resources):. For a vector function over a surface, the surface integral is given by Phi = int_SF·da (3) = int_S(F·n^^)da (4) = int_Sf_xdydz+f_ydzdx+f_zdxdy, (5) where a·b is a dot product and n^^ is a unit normal vector. The course will be conducted in Hindi and notes will be provided in English. Here is a set of notes used by Paul Dawkins to teach his Calculus III course at Lamar University. 3 Vector Calculus In the last part of the course, we will study vector elds, which are functions that assign a vector to each point in its domain, like the vector-valued func-tion F described above. 2 Notes • Homework # 1 is uploaded on the course webpage Vector Calculus. edu/courses/mathematics/18-01-single-variable. 15 Vector Calculus 15. Vitalii Konarovskyi IPSP Winter 2019-2020 October 14 { February 4. Title: Vector Calculus. Elements of Vector Calculus: Lecture 1-Self Assesment Quiz: Self Assessment Quiz: 5 kb: Elements of Vector Calculus: Lecture 2-Self Assessment Quiz: Self Assessment Quiz: 7 kb: Elements of Vector Calculus: Lecture 3-Self Assessment Quiz: Self Assessment Quiz: 10 kb: Elements of Vector Calculus: Lecture 4-Self Assessment Quiz: Self Assessment. The key to understanding tensor calculus at a deep level begins with understanding linear and multilinear functions between vector spaces. These are personal notes written mostly long time ago (say 97) and need not be correct nor understandable. Examples of parametrized curves 125 52. In this section we are going to introduce the concepts of the curl and the divergence of a vector. Winter 2020 Final Exam (11 am lecture): Final Exam. Find materials for this course in the pages linked along the left. Dimock Dept. • The inverse of a matrix is defined such that when it operates on the original matrix, the result is the unit matrix. Let's start with the curl. Also, −1(1,2,1) = (−1,−2,−1). Prerequisites: MATH-102 or MATH-102H or MATH-102X Terms Offered: Summer, Fall, Winter, Spring A study of polar coordinates, parametric equations, and the calculus of functions of several variables with an introduction to vector calculus. Freeman (2003). Use Firefox to download the files if you have problems. , April 10: Solar activity: Chromospheric and coronal heating. In this course, Prof. Thus, I have chosen to use symbolic notation. Schey Div, grad, curl and all that: an informal text on vector calculus. A sound knowledge of these topics is a vital prerequisite for almost all the later courses in applied mathematics and theoretical physics. secret-bases. Calculus I or needing a refresher in some of the early topics in calculus. Abstract:These are lecture notes for the Cambridge mathematics tripos Part IA Vector Calculus course. Lecture Notes: Chapter 10: PARAMETRIC EQUATIONS AND POLAR COORDINATES. Continuity 7. Stanford Engineering Everywhere; 2007; Jeffrey A. The Mathematica® examples are provided in two formats: Mathematica® notebook files (. Multivariable Calculus is the second course in the series, consisting of 26 videos, 4 Study Guides, and a set of Supplementary Notes. Rogawski) Average values of modular L-series via the relative trace formula, Pure Appl. 5) Linear Operators (notes, lecture) [add day] 9/05 (1. Multiple Integrals and Vector Calculus Prof. In particular, the material is presented to (i) develop a physical understanding of the mathematical concepts associated with tensor calculus and (ii) develop the. Stephen Gull at the University of Cambridge. The lecture schedule described below is applicable to LectureSection 2. Freeman and Co. Actually, there are a couple of applications, but they all come back to needing the first one. Lecture notes. The following video provides an outline of all the topics you would expect to see in a typical Multivariable Calculus class (i. Scalar multiplication: If c2R and ~ua vector, then we may form a new vector c~ucalled the scalar product of ~uwith c. We use cookies to help give you the best experience on our website. These notes concentrate on the third part, which is covered in v. The following are important identities involving derivatives and integrals in vector calculus. 30-6 pm TA: palghamol. Lectures with an N after the lecture number have been rewritten to reference the TI-nspire graphing calculator. Tangents 3. Cambridge Course Notes. This unit is offered in Semester 1. There is no central location for these, so we have collated some resources below. We will reinforce this point of view throughout the course. Express A entirely in the spherical polar basis (i. The electromagnetism lecture notes is a book to provide an introduction to Electromagnetism for Electrical and Electronics Engineers. MATH 221 { 1st SEMESTER CALCULUS LECTURE NOTES VERSION 2. LECTURE NOTES 15 The Divergence & Curl of B G Ampere’s Law As we have discussed in the previous P435 Lecture Notes, for the case of an infinitely long straight wire carrying a steady (constant) line current I =Izˆ, G the macroscopic magnetic field associated with this system is given by: () 0 ˆ 2 I Br r μ ϕ π ⎛⎞ =⎜⎟ ⎝⎠ GG for. F or underlined. Lecture #2 VECTOR (OR LINEAR) SPACES Handout#2 One sheet of paper. 9 : Lecture 12 : June 8 (Thu) Vector fields; Line integrals : 16. ISBN -13-805326-X. Introduction These are my notes for the course Math 53: Multivariable Calculus, at UC Berkeley, in the summer of 2011. Notes are applicalicable for both 1st and 2nd sem students of CBCS scheme. The pushforward along a vector function f with respect to vector v in R n is given by d f ( v ) = ∂ f ∂ v d v. This course will be based on a series of lecture notes which will be posted regularly throughout the semester. You may find the following textbooks references useful: Cambridge University Press, 2007. Lectures Notes: Reading guides: Ch 14: Intro to Partial Derivatives: 14. Calculus Revisited: Multivariable Calculus (Res. Gradient Griffiths: Chapter 1 – skip section 1. Mathematical Tripos Part IA: Vector Calculus (1997-2000) My Vector Calculus notes from Lent 2000 are available in pdf and postscript form. This would require us to take the derivative of a vector. to be d(x;y) = jjx yjj: This gives a metric on E. Week 10: Application of vector calculus in mechanics, lines, surface and volume integrals. 01 graphing notebook. These are personal notes written mostly long time ago (say 97) and need not be correct nor understandable. Derivatives as. This course is about vector calculus, and covers material that all engineers should know. We may rewrite Equation (1. Although we developed many different formulas, everything in Chapter 2 could be summarized in one rule: the operators $\ddpl{}{x}$, $\ddpl{}{y}$, and $\ddpl{}{z}$ are the three components of a vector operator $\FLPnabla$. The derivative of a vector function 127 53. SAT Math Test Prep Online Crash Course Algebra & Geometry Study Guide Review, Functions,Youtube - Duration: 2:28:48. Vectors form a linear algebra (i. Welcome! This is one of over 2,200 courses on OCW. Homework should take at least 6 hours, although this can vary quite a lot depending on your background and goals. (Math 151 & 152) Calculus Workshop I & II. Vector Calculus (Green's Theorem, Stokes' Theorem, Divergence Theorem) For differentiation, you can use Principles of Mathematical Analysis by Rudin (Chapter 9). After each video lecture a) redo the examples done in lecture b) do the applicable homework assigbment. Two semesters of single variable calculus (differentiation and integration) are a prerequisite. WEATHERBURN, C. Exams and grades. than 10 dimensions. Mathematics Examples, Lecture Notes and Specimen Exam Questions and Natural Sciences Tripos Mathematics examples. The proper way to understand this is that in both cases, the derivative is a linear transformation. Calculus and Linear Algebra II. University of Cambridge - Part IA Natural Sciences. The kind of things that give you insight into what the ideas mean or how they were developed. The lectures on cector calculus follow the book Calculus III by Marsden, Jerrold E. F or underlined. Calculus I and II). {\displaystyle d\,\mathbf {f} (\mathbf {v} )={\frac {\partial \mathbf {f} }{\partial \mathbf {v} }}d\,\mathbf {v}. Understanding Basic Calculus S. Gives precise and intuitive topological pictures of antisymmetric tensors and their algebra and calculus in three dimensions. Anastassiou, I. Notes Outline: Section 10. ^ Kelly, P. Vitalii Konarovskyi IPSP Winter 2019-2020 October 14 { February 4. Gradient Griffiths: Chapter 1 – skip section 1. We discuss the basic ideas behind k-means clustering and study the classical algorithm. Description: These lecture notes provide a comprehensive introduction to Electromagnetism, aimed at undergraduates. 2 Spans, Lines and Planes The span of a single vector vin Rnis the set of all scalar multiples of v: spanfvg= ftvjt2Rg:. A vector can also be represented mathematically in the form of an equation: v = vxˆ. The historical motivation for homology theory came from vector calculus. Motion in a Noninertial Reference Frame, February 2009, 32pp. Applied Advanced Calculus Lecture Notes by Jan Vrbik. My lecture notes (PDF). Acceleration and Force 131 56. 9) due on July 22 (Fri): Solutions, Solutions to practice problems : Lecture 11 : July 20 (Wed) Change of variables in multiple integrals; Overview of vector calculus; Vector fields : 15. Boqing Gong): Support Vector Machines for Computer Vision Applications; Lecture 14: Guest Lecture (Dr. 1 Gradient-Directional Derivative. PHY2060) and at least have co-registered in a vector calculus course (Calc 3). Perform operations on vectors and vector-valued. Module Overview. IA Vector Calculus (Cambridge), astronomy, astrophysics, cosmology, general relativity, quantum mechanics, physics, university degree, lecture notes, physical sciences. As the set fe^ igforms a basis for R3, the vector A may be written as a linear combination of the e^ i: A= A 1e^ 1 + A 2e^ 2 + A 3e^ 3: (1. Here are a set of practice problems for the Vectors chapter of the Calculus II notes. These notes are only meant to be a study aid and a supplement to your own notes. It is sta ed all six periods every class day. Lecture notes for Math 417-517 Multivariable Calculus J. It is equally valuable for students who are learning calculus for the first time. 02 instructor. Dividing by dt, we obtain dA dt = 1 2 fl fl fl flr £ dr dt fl fl fl fl = jcj 2 Therefore, the physical interpretation of Eq. Sloughter). 9 : Homework 5 (covering 15. Vector Calculus When working with functions one also has to study calculus in addition to studying algebra. University College Dublin An Col aiste Ollscoile, Baile Atha Cliath School of Mathematics and Statistics Scoil na Matamaitice agus na Staitistic Vector Integral and Di erential Calculus (ACM 20150) Dr Lennon O N araigh Lecture notes in Vector Calculus, September 2017. 2 Vector Components and Dummy Indices Let Abe a vector in R3. Vector Calculus 16. 0130415316. Here is a set of notes used by Paul Dawkins to teach his Calculus III course at Lamar University. Vector Calculus ADD. Vector calculus. Fundamental Theorem of Calculus for curve integrals: Path independence and potential: Simplices and boundary: Path, surface, and volume integrals, etc. In this appendix I use the following notation. Multiple integrals. Vector Calculus Tutorials: Vector Calculus (wikipedia) Linear Algebra Tutorials: Professor G. 13 CURL OF A VECTOR1. The Organic Chemistry Tutor 1,896,958 views. Differential vector calculus. Notes: Some of these lectures reference the TI-89 graphing calculator. Tumblr is a place to express yourself, discover yourself, and bond over the stuff you love. Course Notes and General Information Vector calculus is the normal language used in applied mathematics for solving problems in two and three dimensions. Functions of many variables, with a focus on surfaces in three dimensions, partial derivatives, gradients, and directional derivatives. D, an award-winning teacher and. Its purpose is to prepare students for more advanced mathematics courses, particularly courses in mathematical programming (MAT 419), advanced engineering mathematics (MAT 430),. 2 n! 1=2: We de ne the distance between two points xand yin E. I teach maths to Engineering students and most of them take a course on vector calculus. The text for the course is Vector Calculus, fourth edition, by Susan J. Lecture Notes Section 1: (PostScript , PDF ) Revision of Vector Calculus. Vector calculus identities explained. Vector Calculus (aka Advanced Multivariable Calculus) Math222. See also: Vector algebra relations. He graduated in 1816 and spent the next 27 years as a civil engineer. Vector Calculus Collapse menu 1 Analytic Geometry. Lecture notes: Course Information; Section I: Electricity. •“Vector Analysis and Cartesian Tensors”, Bourne and Kendall 1999 by Nelson. 6 Divergence Theorem 15. 18-007, MIT OCW). Chapter 2. May 2016 "Constructing solutions to linear fractional-order PDEs", departmental seminar, Cambridge Analysts Knowledge Exchange, Faculty of Mathematics, University of Cambridge, UK. To see why this is true, consider the function given by r(t) = f(t)i + g(t)j. 00 RESULTS AND GRADES Answer to exam question 1; 2 can be found in lecture notes; 3, 4, 5 You can give feedback in WebOodi after the last lecture. Examples of this include sections on the statistical mechanical theory. 1aCOURSE CONTENT: MAC 2313 is the third semester in the calculus sequence and it gives a thorough introduction to multi-variable calculus. The dates by some of the lectures are the date of the most recent revision. Introduction These are my notes for the course Math 53: Multivariable Calculus, at UC Berkeley, in the summer of 2011. Read ISL, Section 9–9. Motion in a Noninertial Reference Frame, February 2009, 32pp. Math 4013: Vector Calculus, Summer 1998. Dan Sloughter Calculus of several variables; James Cook's lecture notes; Tom Apostol's calculus books; Richard Hammack Book of proof; Illustrations. This is the first in a series of lecture notes on k-means clustering, its variants, and applications. Find materials for this course in the pages linked along the left. Tangents and the unit tangent vector 133 57. Here are the pdf files for the calculus sequence note packets. pdf file of this paper. In this section we are going to introduce the concepts of the curl and the divergence of a vector. In this appendix I use the following notation. Polar Co-ordinate Systems Here dV indicates a volume element and dAan area element. MAT 203 Lecture Notes - Lecture 1: Multivariable Calculus, Vector Calculus, Graphing Calculator Exam. Calculus I and II). Prerequisites are linear algebra and vector calculus at an introductory level. The first lecture of Cosmology II is on Monday, Oct 28th, and the first exercise session is on Friday, Nov 8th. Homework should take at least 6 hours, although this can vary quite a lot depending on your background and goals. These notes are send by Umer Asghar, we are very thankful to him for providing these notes. Office Hours: Instructor: Hans Wenzl: Email: Hans Wenzl MW: 2:30-3:30 and by appointment (just talk to me after class, or email) TA for sections A01, A02: James Hall in AP&M 5748 Email. 61 Differentiation of Processes Let E be a flat space with translation space V. Vector Analysis: A Text-book for the Use of Students of Mathematics and Physics, Founded Upon the Lectures of J. MULTIVARIABLE CALCULUS 28-MATH2063 SPRING 2020. Calculus mainly involves differentiation and integration Differential Calculus Integral Calculus 𝑑𝑑=𝑓𝑓 𝑑𝑑𝑓𝑓 𝑑𝑑𝑥𝑥. This is the one we will use. Publisher: Dalhousie University 2007 Number of pages: 106. Consider a trajectory x: [0, T] !C for a configuration space with d. A vector A, of components A1, A2 and A3 in the basis f^e 1;^e 2;^e 3g, will interchangeably be written as a column or row vector, A = 0 @ A1 A2 A2 1 A= (A1;A2;A3. Instead of Vector Calculus, some universities might call this course Multivariable or Multivariate Calculus or Calculus 3. Inan Prentice Hall, 1999. pdf Lecture notes. Continuity 7. edu, office hours: M10-11:50am, F: 4-6pm, office APM 2313. IA Vector Calculus Lecture notes 2000. It also contains a list of links to other web pages with information on vector calculus. Math 263 Calculus III Pierce College MAP 1 Lecture 16 Sections 18. You must prepare the notes. Views assigns the unit normal vector cosθ i+sinθj to each point on the cylinder S. Some of the pages were developed as complements to the text and lectures in the years 2000-2004. Tromba Vector Calculus. Math 290-1: Linear Algebra & Multivariable Calculus Northwestern University, Lecture Notes Written by Santiago Ca˜nez These are notes which provide a basic summary of each lecture for Math 290-1, the first quarter. However, beginners report various difficulties dealing with the index notation due to. 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Not lecture notes, but here are some good lecture courses: Calculus of Several Real Variables Integral and Vector Calculus Transform Calculus and its applications in Differential Equations Multivariable Calculus Integral equations, calculus of variations and its applications Calculus of One Real Variable Differential Calculus in Several Variables. Derivatives as. (This covers the differential calculus portion of this class. Publisher: Dalhousie University 2007 Number of pages: 106. EXAM I: Friday January 24th, Covers Text Sections (2nd ed) 11. Vector Calculus Summary. Necessary concepts from linear algebra and other mathematical disciplines necessary to understand the text are also covered. A lot of these gaps will be filled in in this course, and more material will be covered. 4, Special Issue: In memory of Armand Borel. Lecture 2 Differentiable functions of many variables: pdf. Scalar Product: The "dot" product of two vectors is a scalar. Find materials for this course in the pages linked along the left. Quickly find Science & Math course-specific resources across a variety of academic disciplines such as digital and interactive textbooks, lecture notes, quiz packs, videos, presentations and more. You can use these notes and exercises on Scaling Analysis to check your current level and guide your study. Two semesters of single variable calculus is a typical prerequisite. That is the purpose of the first two sections of this chapter. Vector Analysis: A Text-book for the Use of Students of Mathematics and Physics, Founded Upon the Lectures of J. I would also like to thank Harold S. Mathematics Examples, Lecture Notes and Specimen Exam Questions and Natural Sciences Tripos Mathematics examples Details on obtaining and updating the source of DAMTP examples (this is aimed at DAMTP Unix account holders only), and the list of course codes and titles referred to in these pages. 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On one side is the definition of a vector space from the notes. 1 Step function Last year: inhomogeneous second order, constant coefficient, ODEs of the form f 00(t) af 0(t) bf (t) g(t) for a limited collection of g(t)s. Multivariable Calculus part 3, 23pp: Multiple Integration. Given on September 5 th. Norton & Company. Notes of lectures on Multivariable Calculus G. • Notice that one cannot add a column-vector and a row-vector!. Before doing so, however, it is useful to think about how calculus can be used in trajectory optimization. 2 Laplacian and second order operators 8. Revision of vector algebra, scalar product, vector product 2. To indicate that an item is in a set, we use the 2symbol. ISBN: 0130414085 (F05). Line, surface and volume integrals, curvilinear co-ordinates 5. This course is about vector calculus, and covers material that all engineers should know. Use Firefox to download the files if you have problems. Find calculus course notes, answered questions, and calculus tutors 24/7. The content may be incomplete. Published by Prentice Hall. Rogawski) Average values of modular L-series via the relative trace formula, Pure Appl. Read a portion of the text. to be d(x;y) = jjx yjj: This gives a metric on E. Written as a companion to multivariable calculus texts, this contains careful and intuitive explanations of several of the ideas covered in this course. Stochastic calculus has very important application in sciences (biology or physics) as well as mathematical nance. The dates by some of the lectures are the date of the most recent revision. ) Lecture Notes on Multivariable Calculus by Barbara Niethammer and Andrew Dancer. derivative as limit of a ratio, integral as limit of a sum. Math 320-3: Lecture Notes Northwestern University, Spring 2015 Written by Santiago Canez~ These are lecture notes for Math 320-3, the third quarter of \Real Analysis", taught at North-western University in the spring of 2015. Math 290-1: Linear Algebra & Multivariable Calculus Northwestern University, Lecture Notes Written by Santiago Ca˜nez These are notes which provide a basic summary of each lecture for Math 290-1, the first quarter. Please do let me know if you. Selected and mentored by James Stewart, Daniel Clegg and Saleem Watson continue Stewart's legacy of providing students with the strongest foundation for a STEM future. Not lecture notes, but here are some good lecture courses: Calculus of Several Real Variables Integral and Vector Calculus Transform Calculus and its applications in Differential Equations Multivariable Calculus Integral equations, calculus of variations and its applications Calculus of One Real Variable Differential Calculus in Several Variables. Real Analysis. Notes for the calculus courses. These notes will contain much the same material as the lecture, but will not be an exact copy. ENGINEERING MATHEMATICS 2 MA8251 Unit 2 VECTOR CALCULUS Notes Pdf Free download. MATH 221 { 1st SEMESTER CALCULUS LECTURE NOTES VERSION 2. Analysis I (2003) Source of notes: Prof Körner's site; Vector Calculus (2000) Download file. On page 40 of the lecture notes on Chapter 19 Section 1: The Idea of Flux Integral, we saw the example which requires us to find the equation of the plane for the rectangular region with corners. Bowen Mechanical Engineering Lectures on Differential Geometry, Prentice-Hall, Englewood Cliffs, New Jersey, 1964. The proper way to understand this is that in both cases, the derivative is a linear transformation. This extends knowledge developed in A-levels and in the Year 1 module Calculus on the differentiation and integration of functions of a single variable, and provides the necessary ground work for Years 2 and 3 modules, such as Curves and Surfaces, Linear PDEs, Fluid Mechanics. Math 223 Vector Calculus Author: Arlo Caine Practice Exam 1 Name: Solutions Directions: Read all questions carefully. Gauss’ Theorem (Divergence Theorem) Consider a surface S with volume V. vector calculus: m427l linear algebra: m340l GEOMETRY for the high school classroom The following notes were developed in collaboration with Gary Hamrick, Diane Radin and a group of Austin-Area high school teachers. University College Dublin An Col aiste Ollscoile, Baile Atha Cliath School of Mathematics and Statistics Scoil na Matamaitice agus na Staitistic Vector Integral and Di erential Calculus (ACM 20150) Dr Lennon O N araigh Lecture notes in Vector Calculus, September 2017.
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https://math.stackexchange.com/questions/4171310/proof-writing-varepsilon-delta-proof-of-the-product-rule-for-limits-a | # Proof writing: $\varepsilon$-$\delta$ proof of the product rule for limits --- any tips for cleaning this up?
I've written the following proof for the product rule for limits. Here $$\lim_{x \to c} f(x) = L$$ and $$\lim_{x \to c} g(x) = M$$. While it works, I find it to be clunky and I can't shake the feeling that we somehow can reduce the general case to the much (in my opinion) cleaner proof of the special case where $$L = M = 0$$. Anyhow, here is the proof:
Suppose first that $$L = M = 0$$. Fix an arbitrary $$\varepsilon > 0$$ and let $$\delta_1$$ and $$\delta_2$$ be such that $$f(x)$$ and $$g(x)$$ are in a $$\sqrt{\varepsilon}$$ neighborhood of $$0$$, respectively. Choose $$\delta = \min(\delta_1, \delta_2)$$. Then when $$|x-c| < \delta$$ we have $$|f(x)g(x)| = |f(x)||g(x)| < \left(\sqrt{\varepsilon}\right)^2 = \varepsilon.$$
Suppose instead $$M \neq 0$$. Fix an arbitrary $$\varepsilon > 0$$ and let $$\delta_1$$ be such that it satisfies $$|f(x) - L| < \varepsilon_1 = \frac{\varepsilon}{3|M|}$$. Moreover, let $$\delta_2$$ be such that it satisfies $$|g(x) - M| < \varepsilon_2 = \min(\frac{\varepsilon}{3\varepsilon_1}, \frac{\varepsilon}{3|L|})$$ where $$\frac{\varepsilon}{3|L|} = \infty$$ if $$L = 0$$. Choose $$\delta = \min(\delta_1, \delta_2)$$. Then if $$|x - c| < \delta$$, we have \begin{align*} |f(x)g(x) - LM| &= |f(x)g(x) - Lg(x) + Lg(x) - LM| \\ &= |g(x)(f(x) - L) + L(g(x) - M)| \\ &\leq |g(x)||(f(x) - L)| + |L||(g(x) - M)| \\ &\leq |M \pm \varepsilon_2|\varepsilon_1 + |L|\varepsilon_2 \\ &\leq |M|\varepsilon_1 + \varepsilon_1\varepsilon_2 + |L|\varepsilon_2 \\ &\leq \frac{\varepsilon}{3} + \frac{\varepsilon}{3} + \frac{\varepsilon}{3} < \varepsilon. \end{align*}
Sure, the general case can be reduced to the case $$L = M = 0$$ by subtracting $$L$$ and $$M$$ from $$f$$ and $$g$$, respectively. Once we know that $$\lim_{x\to c}f(x)g(x) = 0$$ whenever $$f(x),g(x)\to 0$$ as $$x\to c$$, we deduce $$\lim_{x\to c}(f(x)-L)(g(x)-M) = 0.$$ Using algebraic limit laws, this simplifies to $$\lim_{x\to c}f(x)g(x) = LM.$$
Added: The algebraic limit laws I have in mind are (here all the limits are presumed to exist).
• $$\displaystyle\lim_{x\to c}[F(x) + G(x)] = \lim_{x\to c}F(x) + \lim_{x\to c}G(x)$$.
• For any real $$\alpha$$, $$\displaystyle\lim_{x\to c}\alpha F(x) = \alpha\lim_{x\to c}F(x)$$.
As for cleaning up the direct proof in the general case, I would proceed like this.
Proof. By the triangle inequality, $$|f(x)g(x)-LM| \le |f(x)-L||g(x)| + |L||g(x)-M|.$$ $$g$$ is bounded near $$c$$ because the limit $$M$$ exists. Sending $$x\to c$$, the right-hand side tends to $$0$$, as desired. $$\square$$
• Exactly how do you go about simplifying it using "algebraic limit laws"? Jun 13 '21 at 0:23
• @Peatherfed: I added the laws I was thinking about. Once you have these in mind, to finish the proof of the general case using the special case when the limits are zero, use the usual algebraic laws like distributivity, commutativity, and so on, and then apply the two bulleted laws when applicable to get the final result. Hope this helps. Jun 13 '21 at 1:20
• I see, I was hoping to actually get that second law as a consequence of the theorem. I might go about it by first showing it for constants, which will be almost immediate, and then proceeding through the special case into the general case using the linearity of limits. As concerns your suggestion for the direct proof, I like it, although I was aiming for the type of proof where the delta is explicitly provided. I essentially use that g is bounded when going from the first to the second inequality. Jun 13 '21 at 1:31
Note that you don't really need to split in two cases, you can do it all at once: It's easy to see that $$g$$ is bounded near $$x=c$$ (because the limit exists), say $$|g(x)|\leq K$$, $$K>0$$, for $$|x-c|<\delta_{3}$$. Given $$\varepsilon>0$$ there exists $$\delta_{1},\delta_{2}>0$$ such that $$|x-c|<\delta_{1}\Rightarrow |f(x)-L|<\frac{\varepsilon}{2K}\text{ and }|x-c|<\delta_{2}\Rightarrow |g(x)-M|<\frac{\varepsilon}{2(|L|+1)}.$$ Now take $$\delta=\min\{\delta_{1},\delta_{2},\delta_{3}\}$$. For $$|x-c|<\delta$$ we have that \begin{align*} |f(x)g(x)-LM|&\leq |f(x)g(x)-Lg(x)+Lg(x)-LM|\\ &\leq |g(x)|\cdot |f(x)-L|+|L|\cdot |g(x)-M|\\ &< K\cdot\frac{\varepsilon}{2K}+\frac{|L|\cdot\varepsilon}{2(|L|+1)}\\ &<\varepsilon \end{align*} and we are done.
Some time ago I found a very clean proof in Landau's Differential Calculus book which showed the product of continuous functions is continous.
The style was a bit outdated but I managed to update it and the idea works just fine with limits.
The advantage of the approach is that you don't have to show the function is bounded in a neighbourhood of the point. Furthermore you don't have to break into cases where $$L$$ and $$M$$ are zero or not.
First notice that
$$\begin{array}{c} |(fg)(x)-LM|&=|f(x)g(x)-f(x)M-Lg(x)+LM+Lg(x)-LM+Mf(x)-ML|\\ &=|(f(x)-L)(g(x)-M)+L(g(x)-M)+M(f(x)-L)|\\ &\leq |f(x)-L||g(x)-M|+|L||g(x)-M|+|M||f(x)-L| \end{array}$$ for every $$x$$. Now, given $$\varepsilon>0$$ there is $$\delta_1>0$$ such that
$$0<|x-a|<\delta_1\implies |f(x)-L|<\min\left\{1, \frac{\varepsilon}{3(1+|M|)}\right\}$$
and there is $$\delta_2>0$$ such that:
$$0<|x-a|<\delta_2\implies |g(x)-M|<\frac{\varepsilon}{3(1+|L|)}.$$ Now notice,
$$\begin{array}{c} 1<1+|L|&\implies \frac{1}{1+|L|}<1\\ |L|<1+|L|&\implies \frac{|L|}{1+|L|}<1\\ |M|<1+|M|&\implies \frac{|M|}{1+|M|}<1. \end{array}$$ Using this, it follows that
$$0<|x-a|<\min\{\delta_1, \delta_2\}$$ implies \begin{align*} |(fg)(x)-LM|&\leq |f(x)-L||g(x)-M|+|L||g(x)-M|+|M||f(x)-L|\\ &<1\cdot \frac{\varepsilon}{3(1+|L|)}+|L|\frac{\varepsilon}{3(1+|L|)}+|M|\frac{\varepsilon}{3(1+|M|)}\\ &<\frac{\varepsilon}{3}+\frac{\varepsilon}{3}+\frac{\varepsilon}{3}\\ &=\varepsilon. \end{align*}
The same idea also applies when you're working with sequences, for instance.
It may be of interest to you that (with some abstract tools from model theory) one can choose to use nonstandard analysis instead of $$\epsilon-\delta$$ and work with infinitesimal numbers. I could imagine that the book "Nonstandard Analysis" by Martin Väth contains basic calculus rules proven within the framework of nonstandard analysis. | 2022-01-21T16:52:36 | {
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http://www.lofoya.com/Solved/1299/how-many-arrangements-of-four-0s-zeroes-two-1s-and-two | # Moderate Permutation-Combination Solved QuestionAptitude Discussion
Q. How many arrangements of four 0's (zeroes), two 1's and two 2's are there in which the first 1 occur before the first 2?
✖ A. 420 ✖ B. 360 ✖ C. 320 ✔ D. 210
Solution:
Option(D) is correct
Total number of arrangements,
$= \dfrac{8!}{4!×2!×2!} = 420$
Since, there are two 1's and two 0's, the number of arrangements in which the first 1 is before the first 2 is same as the number of arrangement in which the first 2 is before the first 1 and they are each equal to half the total number of arrangements $= \textbf{210}$
## (2) Comment(s)
Numan
()
I think the answer is obtained by 7!/4!1!1! since you cant change the order of the first 1 and 2
Priyanka
()
why the 1st 1 before the first 2 is same as first 2 before the first 1 | 2017-01-21T06:22:45 | {
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https://math.stackexchange.com/questions/2050698/square-matrices | # Square Matrices
So I'm studying a few special families of square matrices, the diagonal matrices, upper triangular matrices, lower triangular matrices and symmetric matrices and I just had a few questions.
I know...
a diagonal matrix is if every nondiagonal entry is zero, $a_{ij}$=0 whenever $i$ doesn't equal $j$.
an upper triangular matrix is if all entries below the diagonal are zero, $a_{ij}=0$ whenever $i >j$.
a lower triangular matrix is if all entries above the diagonal are zero, $a_{ij}=0$ whenever $i < j$.
symmetric if $a_{ij}=a_{ji}$ for all $i$ and $j$.
But I was just wondering, can the diagonal matrices, upper triangular matrices, lower triangular matrices and symmetric matrices have the $0_{n\times n}$? Also I know the $I_{n\times n}$ matrix is in the diagonal matrices, but can it be in the other three types?
• Both the zero matrix and the identity matrix are diagonal, upper/lower triangular, and symmetric. Check the definitions. – Ethan Alwaise Dec 9 '16 at 4:55
• Just check the hypothesis – Learnmore Dec 9 '16 at 4:56
A diagonal matrix doesn't care what is on the diagonal; it only cares that the off-diagonal entries are $0$. So the zero matrix $\mathbf 0_{n \times n}$ is indeed diagonal.
Similarly, an upper triangular matrix only cares that the elements below the diagonal are $0$, so the zero matrix is upper triangular. For the same reason, it is also lower triangular.
It is also symmetric, because $a_{ij} = a_{ji} = 0$ for all $i,j$.
Similar reasoning applies to the identity matrix.
Take for example the definition of upper triangular matrix. It's a matrix which has $a_{ij}=0$ whenever $i>j$, but this doesn't mean that $a_{ij}\neq 0$ for $i\le j$, it's possible to have $a_{ij}=0$ if $i\le j$. Then both $O_{n\times n}$ and $I_{n\times n}$ are upper triangular matrices.
The same idea applies to the diagonal, lower triangular and symmetric matrices. | 2019-07-19T00:32:03 | {
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https://math.stackexchange.com/questions/2221793/how-do-you-find-the-area-of-the-shaded-area-of-this-circle | # How do you find the area of the shaded area of this circle?
I need to find the area of the shaded area. The triangle is equilateral. So far, I have found the area of the triangle to be $\sqrt 3$, but I cannot figure out how to find the radius of the circle in order to find the area of the circle. Any advice would be appreciated.
• Is the triangle equilateral? – Ahmed S. Attaalla Apr 7 '17 at 3:06
• Yes it is equilateral. – Justin Lam Apr 7 '17 at 3:10
• Do you know how to find the areas of circles and triangles? Oops, sorry, I see that you do. My bad. Try drawing in a radius and looking for relationships. Hint: Choose your radius wisely. – Arby Apr 7 '17 at 3:11
• Lookup circular segment, and think what all that comes down to for an equilateral triangle. – dxiv Apr 7 '17 at 3:13
We're looking for the radius right? So let's draw them..
Ok. Now we have an isosceles triangle $30-30-120$. If $r$ is the radius then law of sines tells us,
$$\frac{2}{\sin 120}=\frac{r}{\sin 30}$$
So $r=2 \frac{\sin 30}{\sin 120}=2\frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}=\frac{2}{\sqrt{3}}$. I think you can take it from here.
• Thanks for the help! I got about 1.15 feet squared as the radius. Now I can do the rest. – Justin Lam Apr 7 '17 at 3:22
• My final answer is about 2.5 feet squared. Is this correct? – Justin Lam Apr 7 '17 at 3:29
• You're welcome. That's about right. The exact answer is $\frac{4}{3}\pi-\sqrt{3}$ feet squared. @JustinLam – Ahmed S. Attaalla Apr 7 '17 at 3:44
• Ahmed's drawing: cos(30°) = 1/r. Hence r = 1/cos(30°). Using cos(30°) = (1/2)×(√3) we get r. – Peter Szilas Apr 7 '17 at 6:04
• @Ahmed. OK , here we go. – Peter Szilas Apr 8 '17 at 2:33
Let's label Ahmed's drawing: Triangle $ABC$, lower left $A$, then counterclockwise $B$, and $C$ (top). Let the center of the circle be $M$. Extend $CM$ to intersect $AB$ in $D$. Note length $AD$ $=$ length $DB$ $=1$, $MD$ being the perpendicular bisector of $AB$. Triangle $ADM$ is a right angled triangle. Angle $MAD = 30°$.
$$\cos (30°) = \frac{1}{r}$$
$$r = \frac{1}{\cos (30°)}$$
Using $\cos (30°) = \frac{\sqrt{3}}{2}$ we get $r$. | 2020-09-28T00:09:46 | {
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https://matheducators.stackexchange.com/questions/12845/why-do-we-teach-even-and-odd-functions | # Why do we teach even and odd functions?
I've been either a student or an instructor in Precalculus or Calculus 1 at about 6 institutions now, and teaching the definition of even functions (where $f(-x) = f(x)$) and odd functions (where $f(-x) = -f(x)$) has been universal.
But why? I don't see how these concepts are so useful that they need to be in the courses that are taught to everyone. I don't see how they lay the stage for understanding calculus.
I mean, seeing how it works graphically is nifty. But it seems like a disproportionate emphasis is placed on these classifications, in every curriculum I've seen.
• it seems like a disproportionate emphasis is placed on these classifications It's extremely useful in applications. What level of emphasis is it that you consider disproportionate? If it's covered in the textbook and mentioned once in a while in examples, then it probably doesn't even need to have 5 minutes dedicated to it in class.
– user507
Sep 11 '17 at 22:36
• I like to teach them because of graphing, integration, understanding the difference between proof by example (not proof) and proof of a general case. (Also: If everyone at your school studies calculus, then you teach at a very different institution than mine.) Sep 12 '17 at 0:24
• @BenCrowell It made up a noticeable part of my A-level, with no application at all at the time. Sep 13 '17 at 6:25
• It's an example of symmetry (and anti-symmetry) - which is arguably the most useful concept with a practical application in the whole of mathematics IMO. Sep 14 '17 at 1:22
• Many topics in mathematics are not useful. And no one has claimed them to be as well. Again useful means what, can they be of use in future etc are significant questions though. Sep 14 '17 at 6:51
One of the major themes of precalculus is what I call “connecting geometry to algebra”. Being able to translate between an algebraic statement like $f(x)= f(-x)$, and the geometric statement that the graph of $f$ is symmetric about the vertical axis is a great instance of this. This is just one more way to practice reinforcing function concepts, and the connection with graphs.
• Even a triple connection geometry - algebra - calculus!
– Basj
Sep 13 '17 at 20:09
Here you can see that knowing if the function is even or odd can help you when you are integrating over the interval $[-a, a]$.
You can reduce really-hard-to-look-at integrals to zero just by knowing this. As an example, to calculate $E(Z)$ where $Z \sim N(0, 1)$, the standard normal distribution, you have:
$\displaystyle E(z) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}ze^{-z^2/2}dz$
Which is immediately reduced to zero as the inner function in the integral is odd, and you are integrating over $(-\infty, \infty)$.
This may not be too difficult, but knowing this property about odd functions lets you generalize this to all the odd moments of the standard normal: $E(Z), E(Z^3), E(Z^5), ...$ all of them equal zero.
• You could improve this answer by being more explicit here as to why this is useful. eg mention Fourier series Sep 12 '17 at 6:15
• +1, As an engineer and purely from a practical point this was my first thought. For instance, calculating Fourier series is made much easier knowing about major shortcuts that can be made from the knowledge of odd and even functions. Sep 12 '17 at 15:32
Besides applicability in topics like integration and Fourier analysis, it also connects algebra to calculus at least in the way that multiplication of even/odd functions behaves like addition even/odd numbers:
• Multiplying two even functions gives an even function.
• Multiplying two odd functions gives an even function, too.
• Multiplying an even and an odd function gives an odd function.
Also, you can decompose every function as a sum of an even and an odd function as $$f(x) = \frac{f(x)+f(-x)}2 + \frac{f(x)-f(-x)}2$$ (which is a very useful concept an the same as writing a matrix as the sum of a symmetric and an antisymmetric one as $A = \tfrac12(A+A^T) + \tfrac12(A-A^T)$).
• This also gives one (among many) justification for using the hyperbolic trig functions $\sinh x$ and $\cosh x$ as the odd and even parts respectively of $e^x$ Sep 12 '17 at 14:08
• +1 Not forgetting that this approach generalizes to order three, four,... symmetries via discrete Fourier transforms and, ultimately, to representation theory of groups. Sep 30 '17 at 6:13
Learning to think about functions abstractly should be one goal in precalculus, and function symmetry helps. Also suppose we carefully protected a student from knowing anything about function symmetry. Upon learning about flux in vector calculus, would this student be able to quickly see that the flux of the vector field ${\bf F}(x,y,z)= y^2{\bf j}$ through the unit sphere is 0?
• Indeed, when I teach multivariable calculus, I show students lots of tricks to simplify integrals. Eliminating terms you know will evaluate to zero after integration, halving the domain so that you can "plug in" zeroes and simplify fewer terms, etc. It all comes down to symmetry. Sep 16 '17 at 13:09
Even and odd parity are probably the simplest examples of function symmetries.
In applied mathematics, the general observation of function symmetries allows to simplify calculations (as stated by others) and to produce more meaningful graphs. In physics, symmetrical parts of a function are sometimes associated to different physical phenomena.
Two examples:
1. If you have a function which is invariant under inversion (a more complex symmetry), that is, $$f(1/x) = f(x),\ x>0,$$ it is better to plot the function by using a logarithmic $x$ axis because $f(a^x)$ is even.
2. The example given by Dirk on the decomposition of a matrix in the sum of a symmetric and antisymmetric part is useful in circuit theory: the symmetric part of the impedance and admittance matrices is associated to the average power dissipated by an electrical network when subjected to sinusoidal excitations.
• This is a better answer than mine. The psychology of voting on these sites is mysterious. Sep 13 '17 at 19:28
To express functions as sum of even and odd functions
$$f(x) = f_{even}(x) + f_{odd}(x)$$
And look at the properties of their graphs.
Teach Fourier transform and say it becomes easy to compute when $f(x)$ is an odd function. And the integral becomes zero.
• I think you've missed the point of the question. It isn't about what the syllabus requires, it's about the longer term benefits. Sep 13 '17 at 6:23
• now like applications of it ?@JessicaB Sep 13 '17 at 8:01
I agree that it is disproportionate. Ben Crowell (in comments) says that is helpful for applications. Well, I have a very strong background in natural sciences and engineering, and it is not a big building block. Very few derivations in physics rely on it for instance.
I know this is (-1) controversial, but I think the appeal comes from the ease of doing tricky questions based on it. Sort of ETS style questions. A similar thing would be the vertical angle theorem (or whatever it is called, where a line crosses a couple parallel lines) is big on the SAT.
(-2) I also think that many pure math people prefer things that are definitional and classificational. And that in preference to being able to do multistep mechanical problems like a maximization or power series expansion.
• Really? The difference of Dirichlet versus Neumann boundary conditions (difference between the electromagnetic field behavior near conductors versus insulators) seems to be quite important. The method of images is how a the beginning of electrostatics is understood, and plays a role in the explanation of the Meissner effect in superconductors. The entirety of of the field of crystallography is based on understanding symmetries, and by extension things like understanding diffraction patterns. The difference between even and odd functions also explain the difference in timbre between brass and.. Sep 13 '17 at 20:53
• ... woodwinds in music, as well as between different types of percussive instruments. And I am merely a mathematician who doesn't do too much natural sciences. I am sure specialists can come up with many more examples. Sep 13 '17 at 20:56
Well symmetry helps investigate functions. It's the first things I need to know about the function I deal with. It helps as noted, evaluating integrals that are otherwise difficult to integrate. And it helps visualizing functions. I was a bit disapointed by the few examples given in class ass aplications of these concepts.
As with many items in math that have a name, if we didn't name it, it would still exist.
When manipulating and equation that contained Cos(-x) we are able to observe that this term is the same as Cos(x). On the other hand, when we see Sin(-x) and want to manipulate to form a positive argument, it's equal to -Sin(x).
I don't feel that we spend that much time on the concept. It's introduced in algebra, a parabola possibly being even, a third degree possibly odd, and then again in Trig with the examples I gave. Your experience may be different, more focus than I've observed.
Learning mathematics isn't always about applying things to further mathematics - it can also be about actually applying that knowledge to other areas. So let me bring you a completely different answer to your question, coming from the background of someone that does programming and some web design.
You can easily apply even/odd functions to things like tables: one row has a dark background, the other has a lighter background. A simple enough requirement, that a good programmer might understand not only on the surface, but more deeply such as yourself.
I apologize if this answer is not relevant to the mathematics community directly, but I felt the need (seeing this question in a suggested list) to remind people that math isn't just about math ;) | 2021-10-16T21:07:29 | {
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http://mathoverflow.net/questions/162774/what-is-the-least-integer-of-additive-dimension-4 | # What is the least integer of additive dimension 4?
Say that $m$ is the additive dimension of $n\in\Bbb N$, and write $m=\operatorname{ad}n$, if $m$ is the greatest integer for which there is an irredundant $m$-element set $M\subset\Bbb N$ that provides a partition of $n$ uniquely: namely $$n=k_1+\cdots+k_r\quad\text{with}\quad\{k_i:i=1,...,r\}=M,$$ for some $r\in\Bbb N$, where no other partition of $n$ involves only elements of $M$.
As an illustration, take $n=34$. We can choose $M=\{7,8,12\}$ (for example) in this case, with $34=7+7+8+12$, and so $\operatorname{ad}34=3$, since no other partition of $34$ can be formed from the elements of $M$, while any $4$-element set providing a partition of $34$ has a proper subset that does so too. (The last claim entails some checking.)
Note that $\operatorname{ad}0=0$; $\operatorname{ad}n=1$ for $n=1,2,3,4,6$; $\operatorname{ad}n=2$ for $n=5$ and $n=7,...,16$; and $\operatorname{ad}17=3$ (take $M=\{4,6,7\}$). Thus $17$ is the least integer of additive dimension $3$.
What is the least integer of additive dimension $4$? Further, what is the asymptotic behaviour of $\operatorname{ad}n$ as $n$ becomes large?
(This question was earlier posted on MathStackExchange but was not answered.)
-
Earlier post was math.stackexchange.com/questions/661256/… – Gerry Myerson Apr 8 '14 at 12:38
I wrote a quick and probably impressively sub-optimal computer program to investigate this. It takes two seconds to ascertain that the least integer of additive dimension at least 4 is 49. The relevant partition is $15+14+12+8$.
(I say "at least 4" because I haven't checked that it doesn't in fact have additive dimension 5.)
-
More a bet than a guess: the next one is $129=16+24+28+30+31$, and the numbers are $(n-1)2^n+1$ :) – მამუკა ჯიბლაძე Apr 8 '14 at 12:18
Let me prove that $A_n=(n-1)2^n+1$ has additive dimension $n$, with corresponding set $M=\{2^n-2^i\colon 0\leq i\leq n-1\}$ and representation $A_n=\sum M$.
Assume that $A_n=k_1+\dots+k_r$ is some representation with $k_i\in M$. Since $k_i< 2^n$ we have $r\geq n$. Let $k_i=2^n-2^{d_i}$; then $\sum_i 2^{d_i}=(2^n-1)+(r-n)2^n$.
Lemma. If $\sum_i 2^{d_i}\equiv -1\pmod{2^k}$, then $|\{i\colon d_i<k\}|\geq k$.
Proof. If we have some equal $d_i$'s, replace them by one number $d_i+1$ with no effect to the mentioned sum. When this process stops, we have exactly one $d_i$ equal to each of $0,1,\dots,k-1$. It remains to notice that $|\{i\colon d_i<n\}|$ does not increase during the process. The lemma is proved.
Finally, the lemma inductively yields that the sum of $k\leq n$ least elements in the list $2^{d_1},\dots,2^{d_r}$ does not exceed $2^k-1$. Thus $\sum_i2^{d_i}\leq 2^n-1+(r-n)2^{n-1}$. Comparing this bound with the value found above we conclude that $n=r$, and the $d_i$ form a permutation of $\{0,1,\dots,n-1\}$, as required.
ADDENDUM. On the other hand, the least number in uniquely representing set $M$ of cardinality $n$ is at least $2^{n-1}$, so the least number of aditive dimension $n$ should be greater than $n2^{n-1}$.
To prove this, set $a=\min M$, $M'=M\setminus\{a\}$, and let $S$ be the set of subset sums of $M'$. Clearly, all such sums are distinct (otherwise we would obtain a different representation by replacing one subsum by another). Moreover, all the sets $S+ka$, $k=0,1,\dots$, are disjoint; otherwise we would have $s'=s+ka$ for some $s,s'\in S$, and we would be able to replace $s'$ by $s+ka$. Since $|S|=2^{n-1}$, the density argument shows that $a\geq 2^{n-1}$.
Perhaps, this argument can be extended to other elements of $M$?
-
Possible OEIS candidate, whose description sounds promising: A115981
-
Theorem: For n, $M_n:=(n-1)2^n+1$ is an upper bound.
As an example take $n=4$. now let \begin{aligned} k_1 & := 8 & = 1000 b \\ k_2 & := 12 & = 1100 b \\ k_3 & := 14 & = 1110 b \\ k_4 & := 15 & = 1111 b \\ \end{aligned}
The numbers on the right side are the binary representations of $k_i$
${k_1,k_2,k_3,k_4}$ is a unique partition for $M_4 = k_1+k_2+k_3+k_4 = 49$.
The cases 0 to 3 are already settled in the question. Proof for $n>3$ by induction:
Suppose there would be two different partitions
$$(*) a_1k1+...+a_nk_n = b_1k1+...+b_nk_n = M_n$$
All factors $a_i$ and $b_i$ are lower than $k_1=2^{n-1}$, since otherwise $a_i*k_i\ge2^{2(n-1)}$, which exceeds $M_n=(n-1)2^n+1$ for all $n>3$. Equation $(*)$ also hold modulo $k_1=2^{n-1}$. This yields the same equation in the induction hypothesis (just drop the first binary digit!). As all coefficient are lower than the modulus, we have $a_i=b_i$ for $i>=2$. Then equation $(*)$ yields $a_1=b_1$ qed.
Conjecture: $M_n$ is already optimal for all $n$.
Edit
This proof is flawed! The induction hypothesis doesn't provide uniqueness mod $2^{n-1}$. Sorry!
- | 2016-05-04T04:29:08 | {
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https://math.stackexchange.com/questions/1879150/proof-verification-powers-of-a-group-element-of-finite-order-are-distinct | # Proof verification: powers of a group element of finite order are distinct
I'm finally attempting to conquer D&F (3rd ed) and I want to build good proof habits and fix mistakes early on. Here is the exercise (Ch. 1 ex. 32) and the following is my proof.
Prove that for $x \in G$, where $G$ is a group and $x$ has finite order $n$, all of $1, x, x^2, \ldots, x^{n-1}$ are distinct and deduce that $|x| \leq |G|$.
Proof. Let $a,b \in [0,n-1]\cap \mathbb{N}$ such that $x^a=x^b$. Then $1 = x^ax^{-a}=x^bx^{-a} = x^{b-a}$. Since $1 \leq a \lt n$ and $1 \leq b \lt n$, $b-a\lt n$. So $x^{b-a}=1 \implies b-a=0$ and $a=b$. Since $a$ and $b$ were arbitrarily chosen, $x^c$ is distinct for all $c \in [0,n-1] \cap \mathbb{N}$. We know $x^c\in G$ for all $c \in [0, n-1]\cap \mathbb{N}$ by definition of group closure, and $|[0,n-1]\cap \mathbb{N}|=n$. So $G$ contains at least $n$ elements, and $|x|\leq |G|$.
Is this a convincing proof? I feel like there might be some circular logic, and perhaps not enough detail. I was thinking of establishing that the elements form a cyclic group, and show this group is a subgroup of $G$. Any tips or comments would be greatly appreciated.
• It looks good to me. Aug 2 '16 at 16:16
Almost.
You assert $b-a < n$. I think you want to assume $b \ge a$ (without loss of generality) from the start, to make $b-a$ nonnegative as well. Otherwise you need to handle the case when it's not.
• isn't $b-a \lt n$ even if $b-a$ is negative? I was thinking about clarifying that but I didn't think it would change anything Aug 2 '16 at 16:23
• @m1cky22: true, but you're appealing to the definition of order of an element, which is usually nonnegative. Aug 2 '16 at 16:25
• @m1cky22 Yes, it's less than $n$. But when it's less than $0$ you are dealing with a negative power in what follows. That needs work. Aug 2 '16 at 16:26
• I see. So it's more about keeping the consistency/logic of the definition of order rather than "the proof breaks if it's negative." Thanks, I think I can finally move on to the next chapter! Aug 2 '16 at 16:27
• @m1cky22 Proofs are all about keeping consistent to definitions. Aug 2 '16 at 16:28
Logically, everything looks good, except for @Ethan's point. If you are worried about circularity, make sure you understand why your statement “$x^{b-a} = 1 \implies b-a =0$” is true.
I think notations like
Let $a,b\in[0,n-1] \cap \mathbb{N}$ such that $x^a = x^b$.
are a bit cumbersome. Personally, I would rewrite the first sentence as
Let $a$ and $b$ be integers such that $0 \leq a \leq b \leq n-1$ and $x^a = a^b$.
Or perhaps even
Suppose $x^a = x^b$ for some integers $a$ and $b$ with $0 \leq a\leq b\leq n-1$.
In essence, you want to show $x^a = x^b \implies a=b$, so writing the first sentence that way aligns the paragraph with that logical statement.
Having a higher density of symbols does not make a more mathematically sophisticated argument.
Your proof is perfectly fine. Good Job!
One remark: One prof of me uses the following convention: $[0:n] := [0, n] \cap \mathbb N$.
This is an elegant way to write such "discrete intervals" and makes proofs more readable. Just an idea of me. Maybe you want to use it :)
• I was looking for better notation and didn't know what people use! I didn't want to write $1 \leq a \lt n$ so I chose my route, but yours is definitely better. Thanks! Aug 2 '16 at 16:24
I got confused when you brought $c$ into the mix.
Here is your proof with a lot of the argument removed. I don't think I have removed too much and it may be easier to follow.
Suppose the powers of $x$ are not all distinct.
Then there exist $a,b \in [0,n-1]\cap \mathbb{N}$ such that $b>a$ and $x^b=x^a$.
This implies that $0\lt b-a<n$, and $1=x^ax^{-a}=x^bx^{-a} = x^{b-a}$.
Since the order of $x$ is $n$, $x^{b-a}=1$ implies $n\, |\, b-a$.
This is a contradiction, since no number greater than $0$ and less than $n$ can be divisible by $n$.
So the original supposition, that the powers of $x$ are not all distinct, leads to a contradiction and must therefore be false. | 2021-11-27T12:25:02 | {
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https://mathematica.stackexchange.com/questions/40163/arbitrary-precision-spline-interpolation?noredirect=1 | # Arbitrary precision spline interpolation
The current implementation of Interpolation does not allow arbitrary precision spline interpolation. Yu-Sung Chang says here that "it is not hard to implement it manually for arbitrary precision using BSplineBasis."
I am new to splines and not familiar with BSplineBasis. What is the best way to implement arbitrary precision spline interpolation in Mathematica?
I am interested primarily in an analog of the Method -> "Spline" of Interpolation which I investigated in this answer (BTW, what is the name of such a spline and of this kind of parametrization?). Also, J. M. gives in this answer an interesting implementation of a spline with centripetal parametrization, but it is not clear how to use it for interpolation.
P.S. I need an implementation where the degree of a spline is arbitrary. At the moment I am more interested in even-degree splines.
• Yu-Sung's pointer is wolfram.com/xid/0c0rpcn4bku6m-dj3q8d, where there are two examples of such a thing (I haven't tested them). What are those example lacking? – Dr. belisarius Jan 10 '14 at 12:45
• That is an example of cubic B-spline interpolation. I need an implementation where the degree of spline is arbitrary. – Alexey Popkov Jan 10 '14 at 12:51
• Please see my question about B-spline Basis Function – xyz May 28 '15 at 8:12
The OP linked to an answer of mine for interpolating over general point sets; for constructing a single interpolating function, a slight modification of my procedure is needed. (In particular, you don't need centripetal or chord-length parametrization in this case.)
data = {{0, 0}, {1/10, 3/10}, {1/2, 3/5}, {1, -1/5}, {2, 3}, {3, -6/5}};
To review the problems with using the built-in Interpolation[], let's try building an interpolating quartic spline from data:
m = 4; (* spline order *)
sp1 = Interpolation[data, InterpolationOrder -> m, Method -> "Spline"];
Now, evaluate:
sp1[3/2]
0.26753591659167364
Herein lies the problem: even though the contents of data and the argument fed to sp1 are exact, the output is a machine precision number. Yeesh!
So, we go back to basics and build the interpolating spline ourselves from BSplineBasis[]. First, we generate the knots. The clamped condition is the most appropriate for interpolation, so we generate the appropriate knot sequence like so:
n = Length[data];
{xa, ya} = Transpose[data]
knots = Join[ConstantArray[xa[[1]], m + 1],
If[m + 2 <= n, MovingAverage[ArrayPad[xa, -1], m], {}],
ConstantArray[xa[[-1]], m + 1]];
To generate the control points for the B-spline, we construct the relevant linear system and solve it with LinearSolve[]:
cp = LinearSolve[Outer[BSplineBasis[{m, knots}, #2, #1] &,
xa, Range[0, Length[data] - 1]], ya];
Note that all the entries of cp are exact.
At this juncture, you might think that we can use BSplineFunction[] on the control points and the knots, but alas, the function also only evaluates at machine precision. We are left with no choice but to use BSplineBasis[] once more:
sp2[x_] = cp.Table[BSplineBasis[{m, knots}, j - 1, x], {j, n}];
Try it out:
sp2[3/2]
25251843/94386740
N[%]
0.2675359165916738
As a verification that we were able to reproduce the control points used internally by Interpolation[], let's do a comparison:
sp1bs = Cases[sp1, _BSplineFunction, ∞][[1]];
sp1bs["ControlPoints"]
{0., 0.777830630128766, 1.3470257035045392, -6.670028631670085, 13.100332135107115, -1.2}
N[cp]
{0., 0.777830630128766, 1.347025703504539, -6.6700286316700845, 13.100332135107113, -1.2}
Certainly, I can't end this post without at least one picture, so:
Plot[{sp1[x], sp2[x]}, {x, 0, 3}]
Plot[sp1[x] - sp2[x], {x, 0, 3}, PlotRange -> All]
I leave the task of bundling everything in a routine as an exercise. If you want to learn more, Piegl and Tiller's The NURBS Book is the canonical reference.
• It is good to have you back. – Mr.Wizard May 24 '15 at 10:31
• I was able to borrow a friend's computer with a Mathematica installation, so I went for it and answered some questions. It was fun while it lasted… :) – J. M.'s technical difficulties May 24 '15 at 10:45
• Did you try contacting the Stack Exchange team about the matter? I understand if you do not wish to pursue it but I think they might help. – Mr.Wizard May 24 '15 at 10:48
• Maybe I'll ask them sometime… thanks for the concern. :) – J. M.'s technical difficulties May 24 '15 at 10:51
• @J.M. May I ask you also about a generalization of the spline interpolation (at least) for the 2D case? In the linked answer it is said that "Multi-dimension is simply tensor product version." I do not see this as a simple thing but it would be very helpful to have such a generalization. May be a separate thread would be the best place for this. – Alexey Popkov May 26 '15 at 14:01
Here is my implementation of quadric spline arbitrary precision interpolation which is defined exactly as in Interpolation with options Method->"Spline", InterpolationOrder->2.
## Theoretical background
Quadric spline interpolation for n datapoints is defined as a Piecewise function which consists of n-2 parabolas which are spliced together in the middles of successive datapoints (with exceptions for first and last intervals) by two conditions: successive parabolas at these middle points must be equal and their first derivatives must be equal too. So we have 2(n-3) such conditions.
Let us designate jth datapoint as {x[j],y[j]} and define jth parabola as
s[j_, xx_] = y[j + 1] + b[j] (xx - x[j + 1]) + c[j] (xx - x[j + 1])^2;
This definition automatically makes jth parabola equal to y[j + 1] at x[j + 1]. We have to add also 2 boundary conditions:
s[1, x[1]] == y[1]
s[n - 2, x[n]] == y[n]
And 2(n-3) splicing conditions:
Table[{
s[j, (x[j + 1] + x[j + 2])/2] == s[j + 1, (x[j + 1] + x[j + 2])/2],
ds[j, (x[j + 1] + x[j + 2])/2] == ds[j + 1, (x[j + 1] + x[j + 2])/2]
}, {j, 1, n - 3}]
where ds is first derivative:
ds[j_, xx_] = D[s[j, xx], xx];
Now we combine everything together and convert into matrix form:
eqs = Flatten[{
s[1, x[1]] == y[1],
Table[{
s[j, (x[j + 1] + x[j + 2])/2] == s[j + 1, (x[j + 1] + x[j + 2])/2],
ds[j, (x[j + 1] + x[j + 2])/2] == ds[j + 1, (x[j + 1] + x[j + 2])/2]
}, {j, 1, n - 3}],
s[n - 2, x[n]] == y[n]}];
arrs = Simplify@
Normal@CoefficientArrays[eqs, Flatten@Array[{b[#], c[#]} &, n - 2]];
MatrixForm /@ (4 arrs)
Looking at the matrices it is easy to see that they have periodic structure and contain only elements of the forms x[j+1]-x[j] and y[j+1]-y[j] with some numerical coefficients. This periodic structures can be expressed as SparceArrays.
## Implementation
Assuming that data contains an array of {x[j],y[j]}, we define
Δx[i_] := Subtract @@ data[[{i + 1, i}, 1]];
Δy[i_] := Subtract @@ data[[{i + 1, i}, 2]];
Two matrices in the arrs can be expressed as follows:
bB = SparseArray[{1 -> -4 Δy[1], -1 -> 4 Δy[n - 1], i_ /; OddQ[i] :> 0, i_ /; EvenQ[i] :> 4 Δy[i/2 + 1]}, 2 (n - 2)];
mM = SparseArray[
Join[{{1, 1} -> -4 Δx[1], {1, 2} -> 4 Δx[1]^2, {-1, -1} -> 4 Δx[n - 1]^2, {-1, -2} -> 4 Δx[n - 1]},
Table[Band[{2 i, 2 i - 1}] -> {{2 Δx[i + 1], Δx[i + 1]^2, 2 Δx[i + 1], -Δx[i + 1]^2}, {4, 4 Δx[i + 1], -4, 4 Δx[i + 1]}}, {i, 1, n - 3}]], {2 (n - 2), 2 (n - 2)}];
Now for obtaining coefficients b[i], c[i] we can use LinearSolve:
bcM = LinearSolve[mM, bB];
Grouping coefficients with identical indexes together:
bcM = Partition[bcM, 2];
The intervals at which the parabolas are defined:
intervList =
Partition[
Join[{data[[1, 1]]},
MovingAverage[data[[2 ;; -2, 1]], 2], {data[[-1, 1]]}], 2, 1];
Now we compile all the spline data in one object:
splineData = Table[{data[[i + 1]], bcM[[i]], intervList[[i]]}, {i, n - 2}];
splineData contains everything that is needed to construct the spline. It contains elements of the form:
{{x[i + 1], y[i + 1]}, {b[i], c[i]}, {xmin, xmax}}
It is very handy to use splineData for exact numerical integration (see here).
To produce the explicit piecewise function we define the constructor (HornerForm is already applied and gives 27% speedup):
makeSpline[splineData_List, x_Symbol] :=
Piecewise[
Append[Table[{d[[1, 2]] - d[[1, 1]] d[[2, 1]] +
d[[1, 1]]^2 d[[2, 2]] + x (d[[2, 1]] + x d[[2, 2]] -
2 d[[1, 1]] d[[2, 2]]), #1 <= x <= #2 & @@ d[[3]]}, {d, splineData}],
{Indeterminate, True}]]
It can be used as follows:
spline[\[FormalX]_] = makeSpline[splineData, \[FormalX]];
Compiling the code for creation of splineData into one function:
ClearAll[toSplineData, makeSpline];
Options[toSplineData] = {Method -> Automatic};
toSplineData[data_, OptionsPattern[]] /; MatrixQ[data, NumberQ] :=
Module[{Δx, Δy, bB, mM, bcM, intervList,
n = Length[data], dataS = Sort[data]},
Δx[i_] :=
Subtract @@ dataS[[{i + 1, i}, 1]]; Δy[i_] :=
Subtract @@ dataS[[{i + 1, i}, 2]];
bB = SparseArray[{1 -> -4 Δy[1], -1 ->
4 Δy[n - 1], i_ /; OddQ[i] :> 0,
i_ /; EvenQ[i] :> 4 Δy[i/2 + 1]}, 2 (n - 2)];
mM = SparseArray[
Join[{{1, 1} -> -4 Δx[1], {1, 2} ->
4 Δx[1]^2, {-1, -1} ->
4 Δx[n - 1]^2, {-1, -2} ->
4 Δx[n - 1]},
Table[Band[{2 i,
2 i - 1}] -> {{2 Δx[i + 1], Δx[
i + 1]^2,
2 Δx[i + 1], -Δx[i + 1]^2}, {4,
4 Δx[i + 1], -4,
4 Δx[i + 1]}}, {i, 1, n - 3}]], {2 (n - 2),
2 (n - 2)}];
bcM = LinearSolve[mM, bB, Method -> OptionValue[Method]];
bcM = Partition[bcM, 2];
intervList =
Partition[
Join[{dataS[[1, 1]]},
MovingAverage[dataS[[2 ;; -2, 1]], 2], {dataS[[-1, 1]]}], 2, 1];
Table[{dataS[[i + 1]], bcM[[i]], intervList[[i]]}, {i, n - 2}]];
Now to construct the spline from data we can evaluate
spline[\[FormalX]_] = makeSpline[toSplineData[data], \[FormalX]];
Any suggestions and improvements are welcome!
## What is so special about quadric spline interpolation?
Quadric spline interpolation with splicing points in the middle of successive data points is locally invariant and weakly depends on boundary condition. (But the same is not true for quadric spline interpolation with splicings in data points.) It gives much lesser artifacts than "usual" cubic spline interpolation. Moreover, it seems that splines of higher degree do not give any advantages over such quadric splines. As opposed to cubic spline, data points are not special points at which polynomials are spliced, and it makes much easier exact integration in these points.
• Could you give a demo of your solution? THANKS:) – xyz Oct 10 '15 at 10:38
• @Shutao As you can see from the definition for the parabolas (the first formula in my answer) from which the spline is formed, my solution is non-parametric and so (as well as Interpolation) it is not intended for interpolating an arbitrary dataset (like yours). – Alexey Popkov Oct 10 '15 at 11:40
• @Shutao My solution does not use BSplineBasis and is not intended to be compatible with BSplineFunction and BSplineCurve. But it has an advantage of being very simple and efficient, mathematically clear and very handy when one needs to perform exact integration of interpolated data (without using NIntegrate). Actually I developed my solution exactly for the latter kind of applications. – Alexey Popkov Oct 10 '15 at 11:40
• @Shutao While I appreciate your solution very much I should note that your answer (as well as belisarius' answer) does not answer the original question because the question is explicitly about creation of arbitrary-precision version of Interpolate (preferably with richer set of spline interpolation methods), not about construction of BSplineCurve from a set of points. – Alexey Popkov Oct 10 '15 at 11:40
• @Shutao It would be great if you add a non-parametric version of your solution as another answer: something like BSplineInterpolate[pts][x]: BSplineInterpolate would accept a list of points as the first argument and return something line InterpolatingFunction (in my answer makeSpline returns a Piecewise function object), the latter can accept the abscissa x as its argument and return the interpolated value of the ordinate y. – Alexey Popkov Oct 10 '15 at 11:41
The following is a shameful plug of J. M.'s answer that you already linked to show how to get the parametrization by using BSplineBasis[] with arbitrary precision and packed into a function:
splineInterp[data_, order_, prec_] :=
Module[{parametrizeCurve, tvals, bas, ctrlpts, knots},
parametrizeCurve[pts_ /; MatrixQ[pts, NumericQ], a : (_?NumericQ) : 1/2] :=
FoldList[
Plus, 0, Normalize[(Norm /@ Differences[pts])^a, Total]];
tvals = parametrizeCurve[data];
(*knots for interpolating B-spline*)
knots =
Join[
ConstantArray[0, order + 1],
ConstantArray[1, order + 1]];
(*basis function matrix*)
bas =
Table[
BSplineBasis[{order, knots}, j - 1, N[tvals[[i]], prec]],
{i, Length[data]}, {j, Length[data]}];
ctrlpts = LinearSolve[bas, data];
Return[
Sum[
ctrlpts[[i + 1]] BSplineBasis[{order, knots}, i, #],
{i, 0, Length[testData] - 1}] &]
]
Usage
SeedRandom[42];
testData = RandomReal[{0, 1}, {10, 2}, WorkingPrecision -> prec];
f = splineInterp[testData, 5, 50];
f[1/10]
ParametricPlot[
f[t], {t, 0, 1},
Axes -> None,
Epilog ->
{Directive[Green, PointSize[Large]], Point[testData]}, Frame -> True
]
(*
{0.06025513038208479326777055464103084790562316, \
0.5927955887343273319037301352154452480202510}
*)
Please refer to J. M.'s answer for the details.
### Explanation
The centripetal distribution:
The function parametrizeCurve complete this algorithm
parametrizeCurve[pts_ /; MatrixQ[pts, NumericQ], a : (_?NumericQ) : 1/2] :=
FoldList[
Plus, 0, Normalize[(Norm /@ Differences[pts])^a, Total]]
• I do not see in your answer how can I use this as interpolation in a regular way: interpolation[x] where x is not a parametrization variable t but the actual independent variable on which the dependent variable y depends by y[x]. – Alexey Popkov Jan 10 '14 at 16:28
• @AlexeyPopkov, Please see my explanation – xyz May 29 '15 at 13:04
• @ShutaoTang Thank you for the answer to my comment, now I see that the second argument of parametrizeCurve may be arbitrary but must be lesser or equal to $1$. BTW it would be better for others if you would post this explanation under your own answer. :^) – Alexey Popkov May 29 '15 at 13:22
• @AlexeyPopkov, My pleasue. I used the chord-length method to generate the parameters in my answer, so I think it is suitable to place this explanation in belisarius's answer:-) – xyz May 29 '15 at 13:34 | 2020-06-02T13:00:19 | {
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http://math.stackexchange.com/questions/102050/factoring-a-polynomial | # Factoring a polynomial
I am trying to factor the following polynomial: $$4x^3 - 8x^2 -x + 2$$
I am trying to do the following: $4x^2(x - 2)-x+2$ but I am stuck.
edit: correction.
-
It might be better to write it as $4x^2(x-2)- (x-2)$ to take advantage of what you have done so far. – Geoff Robinson Jan 24 '12 at 19:04
It may be of value to note that there is a general formula for finding roots of 3rd degree polynomials. The formula looks complex but may be of help if you can't guess a root. See this reference for more details: en.wikipedia.org/wiki/Cubic_function – Emmad Kareem Jan 25 '12 at 20:57
You can factor as
$$4x^2(x-2)-(x-2)=4x^2(x-2)-(1)(x-2)$$
Then factor out the $x-2$ to get
$$(x-2)(4x^2-1).$$
But, you may further factor $4x^2-1$ to get
$$(x-2)(2x-1)(2x+1).$$
-
+1:nice explanation... – pedja Jan 24 '12 at 19:07
It wasn't "factored wrong"; it's just that it wasn't taken far enough. – Michael Hardy Jan 24 '12 at 19:19
I agree with Michael. Could you perhaps edit that out? I think it's a little bit impolite. Incomplete and wrong are two very different things. :) – 000 Jan 25 '12 at 1:08
If you look at the revisions, the original posting contained an error where $(x-2)$ was incorrectly $(x-2x)$. I was not being impolite. But, I'll edit it out since things have changed. – Joe Johnson 126 Jan 25 '12 at 20:02
I'll use your polynomial to illustrate a more general procedure for factoring polynomials with integer coefficients (and assuming it has at least one rational root):
First, guess a root of $4x^3-8x^2-x+2$. The so called "rational roots test" will be helpful here.
Eventually, you'll discover that $x=2$ is a root of $4x^3-8x^2-x+2$. This will imply that your polynomial has the form $$\tag{1}(x-2)(ax^2+bx+c),$$ for some constants $a, b, c$.
To find those constants, you could do one of two things (and maybe more)
1. perform the division $4x^3-8x^2-x+2\over x-2$.
2. expand (1) and set it equal to the original polynomial. Setting the coefficients of the two sides of this equation equal to each other will give you a system of equations that are solvable for $a$, $b$, and $c$.
Once you've figured out what $a,b$, and $c$ are, factor the quadratic.
Using method (2), we have:
$$4x^3-8x^2-x+2 = ax^3+(b-2a)x^2+(c-2b)x-2c$$
A moment's reflection reveals that $c=-1$; whence $b=0$; whence $a=4$. Thus \eqalign{ 4x^3-8x^2-x+2 &= (x-2)(4x^2-1)\cr &= (x-2)(2x+1)(2x-1).\cr }
Of course, the other answers are more suitable to your problem; but in the event that your polynomial doesn't factor nicely (such as for $x^3+6x^2+11x+6$), you might try using this approach.
-
Franco, you said you are stuck. Did you have an approach?
In general, for a cubic equation to be factored means the person who is asking this question is indirectly giving you a hint that there is at least one real root.
Your approach should be as follows: Step 1: first to try to see if x=0, 1 -1, 2 or -2 is a root. Chances are in some cases that one of these is a root. In that case, let us say you figured that $x=2$ is a root of $4x^3 - 8x^2 -x + 2$, then if you know factoring polynomials a little bit then you could immediately figure $4x^3 - 8x^2 -x + 2 = (x-2)(4x^2-1)$ which means the other factors can be found by the fact that $(4x^2-1^2) = (2x-1)(2x+1)$. If in some cases you cannot find the obvious root as described in Step 1, then Step 2: Observe if there is a pattern like $4x^3-8x^2$ has two coefficients $4$ and $8$, which has the same pattern as $-x+2$, i.e. $4x^3-8x^2 = 4x^2(x-2)$ and $-x+2 = -(x-2)$, which means you can combine those as $$(x-2)4x^2-(x-2) = (x-2)(4x^2-1)$$
And in some cases as here you could also do $(4x^3-x) - (8x^2-2)$ which gives you
$$x(4x^2-1)-2(4x^2-1) = (x-2)(4x^2-1) = (x-2)(2x-1)(2x+1)$$
Luckily with this question, Step 1 and Step 2 worked. What if you have a cubic equation that does not have obvious ways to factor.
Step 3: For a general cubic equation $ax^3+bx^2+cx+d=0$, apply the substitution
$$x = y - \frac{b}{3a}$$
then we get
$$a\left(y-\frac{b}{3a} \right)^3 + b\left(y-\frac{b}{3a} \right)^2+c\left(y-\frac{b}{3a} \right)+d = 0$$
which simplifies to
$$ay^3 + \left( c-\frac{b^2}{3a}\right)y+ \left(d + \frac{2b^3}{27a^2} - \frac{bc}{3a} \right) = 0$$
This is called a depressed cubic equation, because the square term is eliminated. It is much easier to use this and then find the roots. (back substitute to get the roots in terms of $x$)
For example $$2x^3-18x^2+46x-30=0$$
Substitute $x=y+3$ and simplify this cubic equation to
$2y^3-8y=0 \Rightarrow y=0,2,-2$ which then gives the roots as $x=1,3,$ and $5$.
-
$(4x^3-x)-(8x^2-2)=x(4x^2-1)-2(4x^2-1)=(x-2)(4x^2-1)=$
$=(x-2)(2x-1)(2x+1)$
-
+1: Yours didn't appear until after I sent mine. – Joe Johnson 126 Jan 24 '12 at 19:04 | 2015-05-27T02:25:14 | {
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https://cs.stackexchange.com/questions/47561/confused-about-proof-that-logn-thetan-log-n | # Confused about proof that $\log(n!) = \Theta(n \log n)$
So I was able to show that:
$\log(n!) = O(n\log n)$ without any problems.
My question is when trying to prove that $\log (n!) = \Omega(n\log n)$.
I was able to show that:
\begin{align*} \log n! &= \log(1 \cdot 2 \cdot 3 \cdots n )\\ &= \log 1 + \log2 + \log3 + \dots + \log n \\ &= \log 1 + \dots + \log\tfrac{n}{2} + \dots + \log n\\ &\geq \log\tfrac{n}{2} + \log\big(\tfrac{n}{2} + 1\big) + \dots + \log n &&\text{(i.e., the larger half of the sum)}\\ &\geq \log\big(\tfrac{n}{2}\big) + \log\big(\tfrac{n}{2}\big) + \dots + \log\big(\tfrac{n}{2}\big)&&\text{(adding \tfrac{n}2 times)} \\ &= \log\big(\tfrac{n}{2} \cdot \tfrac{n}{2} \cdots \tfrac{n}{2}) &&\text{(\tfrac{n}{2} times)} \\ &= \log\Big(\tfrac{n}{2}^{\tfrac{n}{2}}\Big)\\ &= \tfrac{n}{2} log\big(\tfrac{n}{2}\big) &&\text{(by log exponent rule)} \end{align*}
Thus, $\log(n!) \geq \tfrac{n}{2}\log\big(\tfrac{n}{2}\big)$, so we conclude that $\log(n!) = \Omega(n\log n)$.
I don't understand how finding the lower bound of $\log(n!)$ is found by getting the larger half of the sum. Why is that chosen to find $\Omega(n\log n)$? I feel like it's probably something obvious but it's the only thing keeping me from fully grasping the proof. If someone can enlighten me, I would appreciate it!
• Concisely: 1/2 is the simplest number in the interval (0,1).
– user12859
Sep 26, 2015 at 9:31
• @RickyDemer So what? Splitting the sum as $\log 1 + \dots + \log cn + \dots + \log n$ for any $0<c<1$ would work just as well. Sep 26, 2015 at 10:17
Consider the sum $S = \log(1) + \dots + \log(n)$. We're going to break it into two parts: $S=T+U$, where
\begin{align*} T &= \log(1) + \log(2) + \dots + \log(n/2)\\ U &= \log(n/2+1) + \dots + \log(n-1) + \log(n). \end{align*}
Basically, $T$ has the first $n/2$ terms of $S$, and $U$ has the remaining $n/2$ terms.
Now we'll lower-bound each of them. Start with $T$. Each term in $T$ is at least $\log(1)$, so we get
$$T \ge \log(1) + \log(1) + \dots + \log(1) = 0 + 0 + \dots 0,$$
so $T \ge 0$. Next look at $U$. Each term in $U$ is at least $\log(n/2)$, so we get
$$U \ge \log(n/2) + \log(n/2) + \dots + \log(n/2) = (n/2) \times \log(n/2).$$
Now $S = T+U$, so plugging in the lower bounds obtained above, it follows that
$$S =T+U\ge 0 + (n/2) \times \log(n/2).$$
This is exactly the result you wanted to prove. Also, $(n/2) \times \log(n/2)$ is $\Omega(n \log n)$, so this proves that the sum $S$ is $\Omega(n \log n)$.
• Ok I think I'm starting to understand, but first, what is stopping me from not breaking up the sum into two parts and just lower bounding each term in the sum? Like this: log(1) + log(1) + ... + log(1) n times. Sep 26, 2015 at 5:21
• @Ghost_Stark, you can do that. That gives you the lower bound $S \ge 0$, which is a valid lower bound, but not very useful. (The lower bound $S \ge -42$ is also true, but even less useful.) It's just like if I tell you I'm thinking of a number between 1 and 100, and you ask for a hint, and I tell you: well, it's larger than $0$. You probably won't be amused at my useless hint, as you already knew that. It's the same for what you want to do. There are many lower bounds that one can prove -- the trick is to find one that is strong enough to prove what you want to prove.
– D.W.
Sep 26, 2015 at 5:23
• Ah I understand now! Thanks for the answer and the explanation! Very helpful. I guess my problem was not understanding that you have to have a reasonable lower bound to look for. Sep 26, 2015 at 5:31
$\dfrac n2$ is chosen so that you have sufficiently many factors ($O(n)$) that are sufficiently large ($O(n)$), so that the product remains $\sim n^n$ while being a lower bound to $n!$
The intuition is that $\log x$ is slowly-growing, and consequently "most" of the terms will be around $\log n$ in size.
More precisely, if there are $\Theta(n)$ terms that are all $\Theta(\log n)$ in size, then their sum will indeed be $\Theta(n \log n)$ and we can conclude $\log n! \in \Omega(n \log n)$.
Taking half of the terms is merely the simplest idea to describe and calculate, and fortunately it satisfies the needed conditions. | 2022-08-09T13:03:53 | {
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时间-频率
time-frequency
By displaying the change of signal energy on a two dimensional time-frequency images based on time-frequency analysis, a new mathematical morphology method to distinguish target from the nonlinear time-frequency curve is presented. 利用时频分析方法将相干积累时间内由于目标非均匀运动和射频干扰引起的信号能量变化显示于时间-频率二维图像上,并根据目标信号对应时频图像上非线性曲线的特点提出一种基于二值形态学的方法用于区分目标和干扰信号。 短句来源 Time-frequency analysis can analyze nonstationary signal including earthquake signal. 时频分析能对非平稳信号进行分析,并给出地震波信号能量的时间-频率分布。 短句来源 Transient Signal Detection Technology Based on Bilinear Time-Frequency Distribution 基于双线性时间-频率分布的瞬时信号探测技术 短句来源 A novel approach to detect abnormal transient vibration signal for complex mechanical system is presented,where alias-free exponential bilinear time-frequency transformation is adopted to avoid the frequency alias and information loss in the traditional bilinear distributions. Simultancously,the cross-terms are effectively deduced with higher time-frequency resolution,which is validated by a gearbox early fault detection. 针对复杂机械系统异常振动信号的检测,提出一种新的无混叠指数双线性时间-频率变换探测方法,新变换方法可以避免常用双线性变换中的频率混叠和信息丢失等问题,能够有效的抑制交叉项,具有较高的时频分辨率,给出新的双线性时-频变换的离散算法,通过仿真实验与齿轮箱早期故障检测,对新的时频变换的检测效果进行了验证。 短句来源 Hilbert spectrum is direct time-frequency energy distribution. The points on time-frequency grids described meaningful and actual distribution. 得到的Hilbert谱图以直接的变量(时间-频率)来表达刻画信号本身能量的分布,使能量分布真正地以时间-频率网格点的形式出现,得到有意义的各个时间-频率-能量点。 短句来源 更多
time and frequency
The Mixed Time and Frequency Decimation FFT Algorithm (MDFFT) proposed by K. 本文对K. Nakayama提出的时间-频率混合抽选FFT算法作了简化和深化,提出了树形分解FFT算法,其实数乘法次数与K. 短句来源 The scattering function is one of most important mathematical method describing time-frequency dispersion channel performance, which embodies a concentrated reflection of the dispersion characteristics of the mobile radio channel in time and frequency domain, and reflects the angle dispersion distributed performance of the channel to a certain extent. 散射函数是描述时间-频率色散信道特性的最重要的数学方法之一,它集中反映了信道在时域和频域的色散特征,并还在一定程度上反映出角度色散的分布情况。 短句来源 Future broadband wireless communication systems should be able to provide reliable high-data-rate transmissions at low cost in time and frequency selective wireless channel. 未来无线通信系统应能在时间-频率选择性衰落信道中以低成本提供可靠的高速数据传输服务。 短句来源 Aimed at all kinds of methods of time-frequency analysis currently,this paper emphasizes on introducing the HHT method ,which deeply depicts the relationship between time and frequency and gets the time-frequency-energy figure of the analysed signal using the integration of EMD and Hilbert transform. 针对当前各种时频分析方法，本文重点介绍了Hilbert-Huang变换法(HHT，Hilbert-HuangTransform)，它利用 经验模态分解(EMD，Empirical ModeDecomposition)法与Hilbert变换的结合，深入地刻画频率与时间之间的关系，得到信号的时间-频率-能量图。 短句来源 Wavelet analysis is a new mathematical method and has acquired wide application in the field due to its fine character of making analysis according to multiple resolutions in time and frequency domains. 小波分析是一种新的数学方法,根据其时间-频率等多分辨率分析的优良特性,在许多方面获得广泛应用。 更多
“时间-频率”译为未确定词的双语例句
Hilbert spectrum has an energy-frequency-time distribution of the signal. Hilbert谱是信号的时间-频率-能量分布。 短句来源 Time frequency energy have been used for speech detection in noisy environments. 以前提出的基于时间 -频率的能量参数利用频域的限带能量加上时域能量来进行噪声中的语音检测。 短句来源 The characteristics of the urban heat island(UHI) intensity in Shanghai were studied and calculated using wavelet transformation, based on per hour difference in temperatures in 2000 observed from Davis automatic stations installed at the Urban and Suburb of Shanghai. 取上海市区各站点的平均温度与近郊站点的温度之差作为衡量城市热岛强度的指标,利用上海市城区和郊区的6个Davis自动气象观测仪每小时观测的记录,对2000年的温差序列进行小波变换,分析上海市城市热岛强度时间-频率的多时间尺度的演变规律. 短句来源 Doubly Selective Fading Channel Estimation Based on Polynomial Interpolation in MIMO Systems 基于多项式内插的MIMO时间-频率双选择性信道的信道估计 短句来源 Chirplet transform(CT) is extension of short-time Fourier transforms(STFTs) and wavelet transforms. 线调频小波变换统一了短时Fourier变换和小波变换的时频分析,是信号的时间-频率-尺度变换,能根据信号的特点自适应生成新的时频窗口。 短句来源 更多
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time-frequency
We present two-sided singular value estimates for a class of convolution-product operators related to time-frequency localization. The Balian-Low theorem (BLT) is a key result in time-frequency analysis, originally stated by Balian and, independently, by Low, as: If a Gabor system $\{e^{2\pi imbt} \, g(t-na)\}_{m,n \in \mbox{\bf Z}}$ Gabor Time-Frequency Lattices and the Wexler-Raz Identity Gabor time-frequency lattices are sets of functions of the form $g_{m \alpha , n \beta} (t) =e^{-2 \pi i \alpha m t}g(t-n \beta)$ generated from a given function $g(t)$ by discrete translations in time and frequency. High-Order Orthonormal Scaling Functions and Wavelets Give Poor Time-Frequency Localization 更多
time and frequency
Gabor time-frequency lattices are sets of functions of the form $g_{m \alpha , n \beta} (t) =e^{-2 \pi i \alpha m t}g(t-n \beta)$ generated from a given function $g(t)$ by discrete translations in time and frequency. We present an explicit, straightforward construction of smooth integrable functions with prescribed gaps around the origin in both time and frequency domain. From the analysis of the electric field structure the conclusion is drawn that the bulk of the AKR power is carried by the signal component fast variable in time and frequency (flickering component). The spectrometer provides the programming of time and frequency parameters of pulse sequences and allows phase-sensitive detection of echo signals with their subsequent representation in digital form. Experimental data are presented on time and frequency dependences of the reverberation level for bistatic transmission and reception at low acoustic frequencies. 更多
The equipment is aimed at manipulating five sectionalizing posts and provided with both, telecontrol and telesignaling. Time-frequency system is employed for the channel. Telesignaling from the five sectionalizing posts works continually and automatically in cyclic order. Self-checking duel information codes are adopted for each operation, in which the direct code is being compared with its inverse counterpart. The logic design is suitable for use, the circuitry is simple, and the equipment is reliable and easy... The equipment is aimed at manipulating five sectionalizing posts and provided with both, telecontrol and telesignaling. Time-frequency system is employed for the channel. Telesignaling from the five sectionalizing posts works continually and automatically in cyclic order. Self-checking duel information codes are adopted for each operation, in which the direct code is being compared with its inverse counterpart. The logic design is suitable for use, the circuitry is simple, and the equipment is reliable and easy in maintenance. 本文介绍了一套用于电气化铁道分区亭的电子装置。它是1对5的分散集中式系统,遥控与遥信综合于同一系统中。通道采用时间-频率划分。各被控站遥信为经常自动循环传送方式,信息码采用二次传送,正反码校核。逻辑设计合理,电路简单,工作可靠,维修方便。 126 lateral geniculate neurones were examined on unanaesthetized immobili-zed cats.The temporal frequency tuning curves of single neurones were measuredby stimulating the cat's eye with a sinusoidally modulated light spot,generated ona CRT and presented to each neurone's receptive field center or its surround.Theaverage discharging rate was used as an index to judge the sensitivity to different mo-dulating frequencies.By comparing the mean impulse rates responding to modula-ted light stimulations(modulated discharge... 126 lateral geniculate neurones were examined on unanaesthetized immobili-zed cats.The temporal frequency tuning curves of single neurones were measuredby stimulating the cat's eye with a sinusoidally modulated light spot,generated ona CRT and presented to each neurone's receptive field center or its surround.Theaverage discharging rate was used as an index to judge the sensitivity to different mo-dulating frequencies.By comparing the mean impulse rates responding to modula-ted light stimulations(modulated discharge rate)with those responding to unmodulatedlight(unmodulated discharge rate),two opposite effects were observed.The majorityof the cells(93.7)% showed modulation-excitatory curves in which the modulateddischarge rates were higher than the unmodulated ones.A minority of the cells(6.3%)showed modulation-inhibitory curves in which the modulated dischargerates were lower than the unmodulated ones.The modulation-excitatory curvescould further be grouped into two subtypes,the“band-pass filters”and the“low-pass filters”.Similarly,the modulation-inhibitory curves could also be dividedinto two subtypes,the“band-rejection filters”and the“low-rejection filters”.Onboth sides of the temporal frequency tuning curve subsidiary sidebands oppositein direction to the main part of the curves could often be seen,i.e.inhibitorysidebands flanking the modunation-excitatory tuning curve and excitatorysidebands flanking the modulation-inhibitory tuning curve.Most of the tuningcurves have only one peak.of the 110 curves measured the peak loci show anormal distribution which centered at 7 Hz.The tuning curves due to stimulationof different locations of the periphery were almost the same.But the tuningproperties of the receptive field centers were different from those of the periphe-ries either in shape or in band-width. 用示波器产生亮度受正弦波调制的小光点刺激清醒猫外膝体神经元的感受野,以不同调制频率下神经元放电的平均频率为指标,分析了126个细胞感受野中心的时间频率调谐特性,主要结果如下。(1)大多数(93.7%)细胞呈调制-兴奋型反应,即刺激光的时间调制在一定频率范围内使放电频率增加;少数(6.3%)细胞呈调制-抑制型反应,即在一定频率范围内,时间调制使放电减少。(2)根据调谐曲线的形状和通带宽度,调制-兴奋型反应包括带通滤波器和低通滤波器两种类型,其110个调谐曲线的峰值分布接近正态曲线,多数细胞对7Hz 的调谐最敏感。调制一抑制型反应包括带除滤波器和低除滤波器两类。(3)调制-兴奋型曲线的通带旁边较當出现抑制性侧带,调制-抑制型曲线出现兴奋性侧带。(4)感受野中心区与外周区的时间频率调谐曲线的带宽和形状有所不同。 This paper makes to model and predict the measured data in Variant Sample Time of Hydrogen, Cesium and Rubidium atomic clocks, mainly according to the principle of ARIMA models of time series analysis method with the Bias function theory of noise Variance of measuring statistics of atomic time/frequency and method of spline functions. Some preliminary useful predicting models and conclusions obtained in variant noise process have pretical sense in optimal predictions and monitoring of the behaviour of atomic... This paper makes to model and predict the measured data in Variant Sample Time of Hydrogen, Cesium and Rubidium atomic clocks, mainly according to the principle of ARIMA models of time series analysis method with the Bias function theory of noise Variance of measuring statistics of atomic time/frequency and method of spline functions. Some preliminary useful predicting models and conclusions obtained in variant noise process have pretical sense in optimal predictions and monitoring of the behaviour of atomic time/frequency and in developing new optimal filting control method to improve the performances of atomic clocks. 本文主要根据时间序列分析方法的ARIMA(autoregressive integraled moving averge)模型原理[1-6],结合原子时间频率测量统计学的噪声方差偏函数理论[7-9],以及适当采用样条函数方法[10],对氢,铯、铷等原子钟不同取样时间的测量数据进行建模和预报。对于不同噪声过程,初步得到了一些有用的预报模型和结论,这对于原子时间频率性能的最佳预报和监测(如跳相和跳频)以及研究新的最佳滤波控制方法以提高原子钟的性能指标都有实际意义。 << 更多相关文摘
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2008中国知网(cnki) 中国学术期刊(光盘版)电子杂志社 | 2019-10-15T18:26:04 | {
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https://math.stackexchange.com/questions/1868418/how-many-3-digit-numbers-that-the-sum-of-their-digits-equals-12 | # How many 3 digit numbers that the sum of their digits equals 12?
How many positive 3-digit numbers exist such that the sum if their digits equals 12?
A) 54
B) 61
C) 64
D) 65
E) 66
I believe the answer is E.
Online problems state that is a stars and bars problem, however using the (n-1,k-1)/n-1C k-1 and (n+k-1,n)/n+k-1 C n formulas do not yield any of the answer choices.
Using (n-1,k-1) I get 55 and using (n+k-1,n) I get 91. Fifty five appears to be the closest answer.
I do not know how to do the inclusion method to remove results such as 066.
• Please show us your calculations so that we can see what you did and where you ran into difficulties. – N. F. Taussig Jul 23 '16 at 12:12
• Online problems state that is a stars and bars problem, however using the n-1 C k-1 and n+k-1 C n formulas do not yield 54. I get 190 from n+k-1 C n and 139 from n-1 C k-1 – G_Derek007 Jul 23 '16 at 12:21
• Please edit your question to show us exactly what you did. I suspect you have not taken into account the restrictions on the hundreds digit, tens digit, and units digit. – N. F. Taussig Jul 23 '16 at 12:23
• I have changed the question to 12 however it should still be the same concept – G_Derek007 Jul 23 '16 at 12:29
• If I read your work correctly, you used the formula for solutions in the positive integers to obtain $\binom{12 - 1}{3 - 1} = \binom{11}{2} = 55$ and for solutions in the non-negative integers to obtain $\binom{12 + 3 - 1}{3 - 1} = \binom{14}{2} = 91$. Is that what you did? – N. F. Taussig Jul 23 '16 at 12:36
A three-digit positive integer has the form $100h + 10t + u$, where the hundreds digit $h$ satisfies the inequalities $1 \leq h \leq 9$, the tens digit $t$ satisfies the inequalities $0 \leq t \leq 9$, and the units digit $u$ satisfies the inequalities $0 \leq u \leq 9$. Therefore, we wish to determine the number of solutions of the equation $$h + t + u = 12 \tag{1}$$ subject to these restrictions.
If we let $h' = h - 1$, then $0 \leq h' \leq 8$. Substituting $h' + 1$ for $h$ in equation 1 yields \begin{align*} h' + 1 + t + u & = 12\\ h' + t + u & = 11 \tag{2} \end{align*} Equation 2 is an equation in the non-negative integers. A particular solution corresponds to the placement of two addition signs in a row of eleven ones. There are $$\binom{11 + 2}{2} = \binom{13}{2}$$ such solutions since we must choose which two of the thirteen symbols (eleven ones and two addition signs) will be addition signs.
However, we have counted solutions in which $h' > 8$, $t > 9$, or $u > 9$. We must exclude these.
Suppose $h' > 8$. Then $h'$ is an integer satisfying $h' \geq 9$. Let $h'' = h' - 9$. Then $h'' \geq 0$. Substituting $h'' + 9$ for $h'$ in equation 2 yields \begin{align*} h'' + 9 + t + u & = 11\\ h'' + t + u & = 2 \tag{3} \end{align*} Equation 3 is an equation in the non-negative integers. Since a particular solution corresponds to the placement of two addition signs in a row of two ones, it has $$\binom{2 + 2}{2} = \binom{4}{2}$$ solutions.
Suppose $t > 9$. Then $t$ is an integer satisfying $t \geq 10$. Let $t' = t - 10$. Substituting $t' + 10$ for $t$ in equation 2 yields \begin{align*} h' + t' + 10 + u & = 11\\ h' + t' + u & = 1 \tag{4} \end{align*} Equation 4 is an equation in the non-negative integers with $\binom{3}{2} = 3$ solutions, depending on which variable is equal to $1$. By symmetry, there are also three solutions in which $u' > 9$. No two of these restrictions cannot be violated simultaneously since $9 + 10 = 19 > 12$. Thus, the number of three-digit positive integers with digit sum $12$ is $$\binom{13}{2} - \binom{4}{2} - 2\binom{3}{2} = 66$$
• It became a lot more clearer thanks! Is it too much to ask if its possible to derive a general solution? – G_Derek007 Jul 23 '16 at 13:20
• I believe I derived one already to be sure though if the sum of the digits were 14 it would be 70? The formula is S=sum of digits D=number of digits (S+D-1,D-1) - (S+D-1-9,D-1) - 2*(S+D-1-9-1,D-1)it should work for 3 digit numbers, which is what I need – G_Derek007 Jul 23 '16 at 13:35
• @user295641 Yes, there are 70 3 digit numbers whose digits sum to 14. – PM 2Ring Jul 23 '16 at 13:38
• That would be a rather long answer. For $h + t + u = k$, where $1 \leq k \leq 9$, you can stop after finding the solution to equation 2 with $k$ replacing $12$. For $h + t + u = 10$, you can stop after equation $3$ with $10$ replacing $12$. For $h + t + u = k$, where $11 \leq k \leq 18$, the same procedure will work if you replace $12$ by $k$. For $18 \leq k \leq 27$, we require the Inclusion-Exclusion Principle because two or more restrictions can be violated simultaneously in equation 2. – N. F. Taussig Jul 23 '16 at 13:39
• @user295641 You have an off-by-one error in that formula; I think you mean: $(S+1, 2) - (S-8, 2) - 2*(S-9, 2)$. However, that formula only works for $9 \le S \le 19$. Note that there is a symmetry. Let $f(S, D)$ be the count of $D$ digit numbers that sum to $S$. Then $f(9D+1-S, D) = f(S, D)$ – PM 2Ring Jul 23 '16 at 14:09
Here's a less algebraic way to look at it. .
Start by putting $1$ into the first cell, ${\boxed 1}\Large\boxed.\boxed.$, so you now only need a sum of $11$,
with the constraints that you can't put $\ge9$ in the first cell, and $\ge10$ in the other two.
Apply stars and bars, subtracting solutions that violate the constraints.
Since it is not possible to violate the constraint in more than one cell, $$ans = \binom{13}2 - \binom{13-9}2 - 2\binom{13-10}2 = 66$$ | 2019-06-19T15:47:08 | {
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https://math.stackexchange.com/questions/3587292/interesting-property-related-to-the-sums-of-the-remainders-of-integers | Interesting property related to the sums of the remainders of integers
Cross-posted on Math Overflow too
Let us define, $$r(b)=\sum_{k=1}^{\lfloor \frac{b-1}{2} \rfloor} (b \bmod{k})$$ After playing around with the $$r(b)$$ function for sometime I noticed that $$r(b)$$ appreared to be more even than odd. So to see the difference between the number of even and odd terms of $$r(b)$$, I defined a function, $$z(x)=\sum_{n=1}^x(-1)^{r(n)}$$ When user Peter ran a program for computing values of $$z(x)$$ in PARI, I observed that for $$x\le 10^{10}$$, $$z(x)\gt 0$$. This suggests that there are always more even terms of $$r(n)$$ than odd terms for any $$x$$.
This leads to my two questions:
1. Is $$z(x)$$ always positive? If so, then how do we prove this?
2. Is $$|z(x)|$$ bounded by some maximum value? If so, then what is this maximum value? Till now the maximum value of $$|z(x)|$$ found was $$49$$ for $$x = 5424027859$$. I find it odd that $$|z(x)|$$ goes to these large values and then returns back to small values as small as $$1$$.
Edit:
With the help of user Vepir I was able to plot $$z(x)$$ and noticed that the function has a sinusoidal-fractal-like appearance and it seems to grow without bounds, although very very slowly.
Is there any reason for this sinusoidal and fractal nature?
• The efficient calculation is based on the formula for $d(n):=r(n)-r(n-1)$ , which is $$d(n)=2n-1-\sigma(n)$$ for even $n$ and $$d(n)=\frac{3n-1}{2}-\sigma(n)$$ for odd $n$ and the fact that $\sigma(n)$ is odd if and only if $n=k^2$ or $n=2k^2$ for some positive integer $k$. – Peter Mar 19 at 21:50
• $r(n)$ and $r(n-1)$ have the same parity if and only if $n$ is of the form $k^2,2k^2$ or $4k+3$ for some positive integer $k$ (in the case $4k+3$ , $k=0$ is also possible). – Peter Mar 19 at 21:54
• If my calculation is correct, we have $z(31988856^2)=80$ and the function does not really feel like it would be bounded from above. – Peter Košinár Mar 29 at 2:02
• @PeterKošinár thanks for the calculation. Was $z(n)$ ever negative in that range? – Mathphile Mar 29 at 4:48
• @Mathphile No, I have not found any case where $z(n)$ would be negative or zero. It has reached $1$ on multiple occasions, though. (again, assuming my computation was correct). – Peter Košinár Mar 29 at 16:43
I can't prove/disprove your conjectures, but I proved a claim which might be useful to prove/disprove your conjectures.
This answer proves the following claim :
Claim :
\small\begin{align}z(8m)&=(-1)^{c(8m)}+S(m) \\\\z(8m+1)&=(-1)^{c(8m)}-(-1)^{c(8m+1)}+S(m) \\\\z(8m+2)&=(-1)^{c(8m)}-(-1)^{c(8m+1)}+(-1)^{c(8m+2)}+S(m) \\\\z(8m+3)&=(-1)^{c(8m)}-(-1)^{c(8m+1)}+2(-1)^{c(8m+2)}+S(m) \\\\z(8m+4)&=(-1)^{c(8m)}-(-1)^{c(8m+1)}+2(-1)^{c(8m+2)}-(-1)^{c(8m+4)}+S(m) \\\\z(8m+5)&=(-1)^{c(8m)}-(-1)^{c(8m+1)}+2(-1)^{c(8m+2)}-(-1)^{c(8m+4)}+(-1)^{c(8m+5)}+S(m) \\\\z(8m+6)&=(-1)^{c(8m)}-(-1)^{c(8m+1)}+2(-1)^{c(8m+2)}-(-1)^{c(8m+4)}+S(m) \\\\z(8m+7)&=(-1)^{c(8m)}-(-1)^{c(8m+1)}+2(-1)^{c(8m+2)}-(-1)^{c(8m+4)}-(-1)^{c(8m+7)}+S(m)\end{align} where $$c(n)=\#\{x:\text{ 1\le x\le n, and x is either a square or twice a square}\}$$ and $$S(m):=\sum_{k=0}^{m-1}\bigg((-1)^{c(8k)}-(-1)^{c(8k+1)}+2(-1)^{c(8k+2)}-(-1)^{c(8k+4)}-(-1)^{c(8k+7)}\bigg)$$
According to oeis.org/A071860, one has $$c(n)=\lfloor\sqrt{n}\rfloor+\left\lfloor\sqrt{\frac n2}\right\rfloor$$ which might make the problem easier to deal with.
Also, it follows from the above claim that $$z(8m+4)=z(8m+6)$$ for every non-negative integer $$m$$.
The claim follows from the following lemmas :
Lemma 1 : $$r(n)=n^2-\frac{1}{2}\bigg(\left\lfloor\frac{n-3}{2}\right\rfloor+2\bigg)\bigg(\left\lfloor\frac{n-3}{2}\right\rfloor +1\bigg)-\sum_{k=1}^{n}\sigma(k)$$
Lemma 2 : $$r(n)\stackrel{\text{mod 2}}\equiv\begin{cases}\displaystyle\sum_{k=1}^{n}\sigma(k)&\text{if n\equiv 0,2,3,5\pmod 8}\\\\1+\displaystyle\sum_{k=1}^{n}\sigma(k)&\text{if n\equiv 1,4,6,7\pmod 8}\end{cases}$$
Lemma 3 : $$\text{\sigma(n) is odd \iff n is either a square or twice a square}$$
Lemma 4 : $$\sum_{k=1}^{n}\sigma(k)\equiv c(n)\pmod 2$$where $$c(n)=\#\{x:\text{ 1\le x\le n, and x is either a square or twice a square}\}$$
Lemma 5 : If $$n\equiv 3\pmod 4$$, then $$c(n)=c(n-1)$$.
Lemma 6 : $$z(x)=\sum_{k=0}^{\lfloor x/8\rfloor}(-1)^{c(8k)}-\sum_{k=0}^{\lfloor (x-1)/8\rfloor}(-1)^{c(8k+1)}+\sum_{k=0}^{\lfloor (x-2)/8\rfloor}(-1)^{c(8k+2)}+\sum_{k=0}^{\lfloor (x-3)/8\rfloor}(-1)^{c(8k+2)}-\sum_{k=0}^{\lfloor (x-4)/8\rfloor}(-1)^{c(8k+4)}+\sum_{k=0}^{\lfloor (x-5)/8\rfloor}(-1)^{c(8k+5)}-\sum_{k=0}^{\lfloor (x-6)/8\rfloor}(-1)^{c(8k+5)}-\sum_{k=0}^{\lfloor (x-7)/8\rfloor}(-1)^{c(8k+7)}$$
Lemma 7 :
\small\begin{align}z(8m)&=(-1)^{c(8m)}+S(m) \\\\z(8m+1)&=(-1)^{c(8m)}-(-1)^{c(8m+1)}+S(m) \\\\z(8m+2)&=(-1)^{c(8m)}-(-1)^{c(8m+1)}+(-1)^{c(8m+2)}+S(m) \\\\z(8m+3)&=(-1)^{c(8m)}-(-1)^{c(8m+1)}+2(-1)^{c(8m+2)}+S(m) \\\\z(8m+4)&=(-1)^{c(8m)}-(-1)^{c(8m+1)}+2(-1)^{c(8m+2)}-(-1)^{c(8m+4)}+S(m) \\\\z(8m+5)&=(-1)^{c(8m)}-(-1)^{c(8m+1)}+2(-1)^{c(8m+2)}-(-1)^{c(8m+4)}+(-1)^{c(8m+5)}+S(m) \\\\z(8m+6)&=(-1)^{c(8m)}-(-1)^{c(8m+1)}+2(-1)^{c(8m+2)}-(-1)^{c(8m+4)}+S(m) \\\\z(8m+7)&=(-1)^{c(8m)}-(-1)^{c(8m+1)}+2(-1)^{c(8m+2)}-(-1)^{c(8m+4)}-(-1)^{c(8m+7)}+S(m)\end{align}
where $$S(m):=\sum_{k=0}^{m-1}\bigg((-1)^{c(8k)}-(-1)^{c(8k+1)}+2(-1)^{c(8k+2)}-(-1)^{c(8k+4)}-(-1)^{c(8k+7)}\bigg)$$
Lemma 1 : $$r(n)=n^2-\frac{1}{2}\bigg(\left\lfloor\frac{n-3}{2}\right\rfloor+2\bigg)\bigg(\left\lfloor\frac{n-3}{2}\right\rfloor +1\bigg)-\sum_{k=1}^{n}\sigma(k)$$
Proof : For $$n=1$$, it is true. In the following, let us use $$r(n+1)-r(n)=\begin{cases}2n+1-\sigma(n+1)&\text{if n is odd}\\\\\frac{3n+2}{2}-\sigma(n+1)&\text{if n is even}\end{cases}$$
Suppose that it is true for $$n=2m+1$$. Then, we get \begin{align}&r(n+1) \\\\&=r(n)+2n+1-\sigma(n+1) \\\\&=n^2-\frac{1}{2}\bigg(\left\lfloor\frac{n-3}{2}\right\rfloor+2\bigg)\bigg(\left\lfloor\frac{n-3}{2}\right\rfloor +1\bigg) -\sum_{k=1}^{n}\sigma(n)+2n+1-\sigma(n+1) \\\\&=(2m+1)^2-\frac{m(m+1)}{2}+2(2m+1)+1-\sum_{k=1}^{n+1}\sigma(k) \\\\&=(2m+2)^2-\frac{m(m+1)}{2}-\sum_{k=1}^{n+1}\sigma(k) \\\\&=(2m+2)^2-\frac{1}{2}\bigg(\left\lfloor\frac{2m-1}{2}\right\rfloor+2\bigg)\bigg(\left\lfloor\frac{2m-1}{2}\right\rfloor +1\bigg)-\sum_{k=1}^{n+1}\sigma(k) \\\\&=(n+1)^2-\frac{1}{2}\bigg(\left\lfloor\frac{n+1-3}{2}\right\rfloor+2\bigg)\bigg(\left\lfloor\frac{n+1-3}{2}\right\rfloor +1\bigg)-\sum_{k=1}^{n+1}\sigma(k)\end{align}
Suppose that it is true for $$n=2m$$. Then, we get \begin{align}&r(n+1) \\\\&=r(n)+\frac{3n+2}{2}-\sigma(n+1) \\\\&=n^2-\frac{1}{2}\bigg(\left\lfloor\frac{n-3}{2}\right\rfloor+2\bigg)\bigg(\left\lfloor\frac{n-3}{2}\right\rfloor +1\bigg) -\sum_{k=1}^{n}\sigma(k)+\frac{3n+2}{2}-\sigma(n+1) \\\\&=(2m)^2-\frac{m(m-1)}{2}+3m+1-\sum_{k=1}^{n+1}\sigma(k) \\\\&=(2m+1)^2-\frac{(m+1)m}{2}-\sum_{k=1}^{n+1}\sigma(k) \\\\&=(2m+1)^2-\frac{1}{2}\bigg(\left\lfloor\frac{2m+1-3}{2}\right\rfloor+2\bigg)\bigg(\left\lfloor\frac{2m+1-3}{2}\right\rfloor +1\bigg)-\sum_{k=1}^{n+1}\sigma(k) \\\\&=(n+1)^2-\frac{1}{2}\bigg(\left\lfloor\frac{n+1-3}{2}\right\rfloor+2\bigg)\bigg(\left\lfloor\frac{n+1-3}{2}\right\rfloor +1\bigg)-\sum_{k=1}^{n+1}\sigma(k)\end{align}
So, it is also true for $$n+1$$.$$\quad\square$$
Lemma 2 : $$r(n)\stackrel{\text{mod 2}}\equiv\begin{cases}\displaystyle\sum_{k=1}^{n}\sigma(k)&\text{if n\equiv 0,2,3,5\pmod 8}\\\\1+\displaystyle\sum_{k=1}^{n}\sigma(k)&\text{if n\equiv 1,4,6,7\pmod 8}\end{cases}$$
Proof : It follows from Lemma 1 that$$r(8m)=64m^2-2m(4m-1)-\sum_{k=1}^{8m}\sigma(k)\equiv \sum_{k=1}^{8m}\sigma(k)\pmod 2$$
$$r(8m+1)=(8m+1)^2-2m(4m+1)-\sum_{k=1}^{8m+1}\sigma(k)\equiv 1+\sum_{k=1}^{8m+1}\sigma(k)\pmod 2$$
$$r(8m+2)=(8m+2)^2-2m(4m+1)-\sum_{k=1}^{8m+2}\sigma(k)\equiv \sum_{k=1}^{8m+2}\sigma(k)\pmod 2$$
$$r(8m+3)=(8m+3)^2-(2m+1)(4m+1)-\sum_{k=1}^{8m+3}\sigma(k)\equiv \sum_{k=1}^{8m+3}\sigma(k)\pmod 2$$
$$r(8m+4)=(8m+4)^2-(2m+1)(4m+1)-\sum_{k=1}^{8m+4}\sigma(k)\equiv 1+\sum_{k=1}^{8m+4}\sigma(k)\pmod 2$$
$$r(8m+5)=(8m+5)^2-(2m+1)(4m+3)-\sum_{k=1}^{8m+5}\sigma(k)\equiv \sum_{k=1}^{8m+5}\sigma(k)\pmod 2$$
$$r(8m+6)=(8m+6)^2-(2m+1)(4m+3)-\sum_{k=1}^{8m+6}\sigma(k)\equiv 1+\sum_{k=1}^{8m+6}\sigma(k)\pmod 2$$
$$r(8m+7)=(8m+7)^2-(2m+2)(4m+3)-\sum_{k=1}^{8m+7}\sigma(k)\equiv 1+\sum_{k=1}^{8m+7}\sigma(k)\pmod2$$ So, the claim follows.$$\quad\square$$
Lemma 3 : $$\text{\sigma(n) is odd \iff n is either a square or twice a square}$$
Proof : See here or here.
Lemma 4 : $$\sum_{k=1}^{n}\sigma(k)\equiv c(n)\pmod 2$$where $$c(n)=\#\{x:\text{ 1\le x\le n, and x is either a square or twice a square}\}$$
Proof : It follows from Lemma 3 that $$\sum_{k=1}^{n}\sigma(k)=\underbrace{\sum_{k\in A}\sigma(k)}_{\text{sum of odd numbers}}+\underbrace{\sum_{k\not\in A}\sigma(k)}_{\text{sum of even numbers = even}}\equiv \sum_{k\in A}\sigma(k)=c(n)\pmod 2$$ where $$A=\{n\ :\ \text{n is either a square or twice a square}\}$$.
Lemma 5 : If $$n\equiv 3\pmod 4$$, then $$c(n)=c(n-1)$$.
Proof : Since we have $$\text{(a square)}\equiv 0,1\pmod 4\qquad\text{and}\qquad \text{(twice a square)}\equiv 0,2\pmod 4$$ we see that if $$n\equiv 3\pmod 4$$, then $$n$$ is neither a square nor twice a square.
Lemma 6 : $$z(x)=\sum_{k=0}^{\lfloor x/8\rfloor}(-1)^{c(8k)}-\sum_{k=0}^{\lfloor (x-1)/8\rfloor}(-1)^{c(8k+1)}+\sum_{k=0}^{\lfloor (x-2)/8\rfloor}(-1)^{c(8k+2)}+\sum_{k=0}^{\lfloor (x-3)/8\rfloor}(-1)^{c(8k+2)}-\sum_{k=0}^{\lfloor (x-4)/8\rfloor}(-1)^{c(8k+4)}+\sum_{k=0}^{\lfloor (x-5)/8\rfloor}(-1)^{c(8k+5)}-\sum_{k=0}^{\lfloor (x-6)/8\rfloor}(-1)^{c(8k+5)}-\sum_{k=0}^{\lfloor (x-7)/8\rfloor}(-1)^{c(8k+7)}$$
Proof : It follows from Lemma $$1,2,3,4,5$$ that $$z(x)=\sum_{k=1}^{x}(-1)^{r(k)}=\sum_{k=0}^{\lfloor x/8\rfloor}(-1)^{r(8k)}+\sum_{k=0}^{\lfloor (x-1)/8\rfloor}(-1)^{r(8k+1)}+\sum_{k=0}^{\lfloor (x-2)/8\rfloor}(-1)^{r(8k+2)}+\sum_{k=0}^{\lfloor (x-3)/8\rfloor}(-1)^{r(8k+3)}+\sum_{k=0}^{\lfloor (x-4)/8\rfloor}(-1)^{r(8k+4)}+\sum_{k=0}^{\lfloor (x-5)/8\rfloor}(-1)^{r(8k+5)}+\sum_{k=0}^{\lfloor (x-6)/8\rfloor}(-1)^{r(8k+6)}+\sum_{k=0}^{\lfloor (x-7)/8\rfloor}(-1)^{r(8k+7)}$$
$$=\sum_{k=0}^{\lfloor x/8\rfloor}(-1)^{c(8k)}-\sum_{k=0}^{\lfloor (x-1)/8\rfloor}(-1)^{c(8k+1)}+\sum_{k=0}^{\lfloor (x-2)/8\rfloor}(-1)^{c(8k+2)}+\sum_{k=0}^{\lfloor (x-3)/8\rfloor}(-1)^{c(8k+3)}-\sum_{k=0}^{\lfloor (x-4)/8\rfloor}(-1)^{c(8k+4)}+\sum_{k=0}^{\lfloor (x-5)/8\rfloor}(-1)^{c(8k+5)}-\sum_{k=0}^{\lfloor (x-6)/8\rfloor}(-1)^{c(8k+6)}-\sum_{k=0}^{\lfloor (x-7)/8\rfloor}(-1)^{c(8k+7)}$$
$$=\sum_{k=0}^{\lfloor x/8\rfloor}(-1)^{c(8k)}-\sum_{k=0}^{\lfloor (x-1)/8\rfloor}(-1)^{c(8k+1)}+\sum_{k=0}^{\lfloor (x-2)/8\rfloor}(-1)^{c(8k+2)}+\sum_{k=0}^{\lfloor (x-3)/8\rfloor}(-1)^{c(8k+2)}-\sum_{k=0}^{\lfloor (x-4)/8\rfloor}(-1)^{c(8k+4)}+\sum_{k=0}^{\lfloor (x-5)/8\rfloor}(-1)^{c(8k+5)}-\sum_{k=0}^{\lfloor (x-6)/8\rfloor}(-1)^{c(8k+5)}-\sum_{k=0}^{\lfloor (x-7)/8\rfloor}(-1)^{c(8k+7)}$$
Lemma 7 :
\small\begin{align}z(8m)&=(-1)^{c(8m)}+S(m) \\\\z(8m+1)&=(-1)^{c(8m)}-(-1)^{c(8m+1)}+S(m) \\\\z(8m+2)&=(-1)^{c(8m)}-(-1)^{c(8m+1)}+(-1)^{c(8m+2)}+S(m) \\\\z(8m+3)&=(-1)^{c(8m)}-(-1)^{c(8m+1)}+2(-1)^{c(8m+2)}+S(m) \\\\z(8m+4)&=(-1)^{c(8m)}-(-1)^{c(8m+1)}+2(-1)^{c(8m+2)}-(-1)^{c(8m+4)}+S(m) \\\\z(8m+5)&=(-1)^{c(8m)}-(-1)^{c(8m+1)}+2(-1)^{c(8m+2)}-(-1)^{c(8m+4)}+(-1)^{c(8m+5)}+S(m) \\\\z(8m+6)&=(-1)^{c(8m)}-(-1)^{c(8m+1)}+2(-1)^{c(8m+2)}-(-1)^{c(8m+4)}+S(m) \\\\z(8m+7)&=(-1)^{c(8m)}-(-1)^{c(8m+1)}+2(-1)^{c(8m+2)}-(-1)^{c(8m+4)}-(-1)^{c(8m+7)}+S(m)\end{align}
where $$S(m):=\sum_{k=0}^{m-1}\bigg((-1)^{c(8k)}-(-1)^{c(8k+1)}+2(-1)^{c(8k+2)}-(-1)^{c(8k+4)}-(-1)^{c(8k+7)}\bigg)$$
Proof : This immediately follows from Lemma 6.
Claim: For every positive integer $$b$$ we have $$r(b)\equiv r(b-1)\pmod{2}$$ if and only if either
1. $$b\equiv3\pmod{4}$$, or
2. $$b=k^2$$ for some integer $$k$$, or
3. $$b=2k^2$$ for some integer $$k$$.
The main ingredient in proving this claim is the following lemma:
Lemma: For every positive integer $$b$$ we have $$r(b)-r(b-1)=c(b)-\sigma(b),$$ where $$\sigma(b)$$ denotes the sum of all positive divisors of $$b$$, and $$c(b):=\begin{cases} 2b-1&\text{ if }\ b\equiv0\pmod{2},\\ \tfrac{3b-1}{2}&\text{ if }\ b\equiv1\pmod{2}\ \text{ and }\ b\neq3,\\ 4&\text{ if }\ b=3. \end{cases}$$
This reduces the question to a question on the parity of $$\sigma(b)$$.
Proof. For every pair of positive integers $$b$$ and $$k$$ there exist unique nonnegative integers $$q(b,k)$$ and $$r(b,k)$$ such that $$r(b,k) and $$b=q(b,k)\cdot k+r(b,k).$$ This is simply dividing $$b$$ by $$k$$ with remainder $$r(b,k)$$. With this, your function $$r$$ can be expressed as $$r(b)=\sum_{k=1}^{\lfloor\frac{b-1}{2}\rfloor}r(b,k).$$ To find a more manageable form for $$r(b)$$, note that $$q(b,k)=\lfloor\frac bk\rfloor$$, so that $$\begin{eqnarray*} r(b)&=&\sum_{k=1}^{\lfloor\frac{b-1}{2}\rfloor}r(b,k) =\sum_{k=1}^{\lfloor\frac{b-1}{2}\rfloor}\Big(b-q(b,k)k\Big)=\big\lfloor\tfrac{b-1}{2}\big\rfloor b -\sum_{k=1}^{\lfloor\frac{b-1}{2}\rfloor}\big\lfloor\tfrac{b}{k}\big\rfloor k. \end{eqnarray*}$$ Then the difference of two consecutive terms can be simplified. If $$b$$ is even: $$\begin{eqnarray*} r(b)-r(b-1)&=&\left(\big\lfloor\tfrac{b-1}{2}\big\rfloor b -\sum_{k=1}^{\lfloor\frac{b-1}{2}\rfloor}\big\lfloor\tfrac{b}{k}\big\rfloor k\right) -\left(\big\lfloor\tfrac{b-2}{2}\big\rfloor(b-1) -\sum_{k=1}^{\lfloor\frac{b-2}{2}\rfloor}\big\lfloor\tfrac{b-1}{k}\big\rfloor k\right)\\ &=&\left(\frac{b-2}{2}b -\sum_{k=1}^{\frac{b-2}{2}}\big\lfloor\tfrac{b}{k}\big\rfloor k\right) -\left(\left(\frac{b-2}{2}\right)(b-1) -\sum_{k=1}^{\frac{b-2}{2}}\big\lfloor\tfrac{b-1}{k}\big\rfloor k\right)\\ &=&\frac{b-2}{2}-\sum_{k=1}^{\frac{b-2}{2}}\left(\big\lfloor\tfrac{b}{k}\big\rfloor-\big\lfloor\tfrac{b-1}{k}\big\rfloor\right)k,\\ \end{eqnarray*}$$ and similarly if $$b$$ is odd: $$\begin{eqnarray*} r(b)-r(b-1)&=&\left(\big\lfloor\tfrac{b-1}{2}\big\rfloor b -\sum_{k=1}^{\lfloor\frac{b-1}{2}\rfloor}\big\lfloor\tfrac{b}{k}\big\rfloor k\right) -\left(\big\lfloor\tfrac{b-2}{2}\big\rfloor(b-1) -\sum_{k=1}^{\lfloor\frac{b-2}{2}\rfloor}\big\lfloor\tfrac{b-1}{k}\big\rfloor k\right)\\ &=&\left(\frac{b-1}{2}b -\sum_{k=1}^{\frac{b-1}{2}}\big\lfloor\tfrac{b}{k}\big\rfloor k\right) -\left(\frac{b-3}{2}(b-1) -\sum_{k=1}^{\frac{b-3}{2}}\big\lfloor\tfrac{b-1}{k}\big\rfloor k\right)\\ &=&\frac{3}{2}(b-1)-\big\lfloor\tfrac{2b}{b-1}\big\rfloor\frac{b-1}{2}-\sum_{k=1}^{\frac{b-3}{2}}\left(\big\lfloor\tfrac{b}{k}\big\rfloor-\big\lfloor\tfrac{b-1}{k}\big\rfloor\right)k, \end{eqnarray*}$$ where the extra term $$-\big\lfloor\tfrac{2b}{b-1}\big\rfloor\frac{b-1}{2}$$ appears because the summation in $$r(b)$$ has one more term than the summation in $$r(b-1)$$, with $$k=\tfrac{b-1}{2}$$. For odd $$b>3$$ this further simplifies to $$r(b)-r(b-1)=\frac{b-1}{2}-\sum_{k=1}^{\frac{b-3}{2}}\left(\big\lfloor\tfrac{b}{k}\big\rfloor-\big\lfloor\tfrac{b-1}{k}\big\rfloor\right)k.$$ Now these inner summations have more familiar closed forms; to see this, note that $$\big\lfloor\tfrac{b}{k}\big\rfloor-\big\lfloor\tfrac{b-1}{k}\big\rfloor=\begin{cases} 0&\text{ if }\ k\nmid b\\ 1&\text{ if }\ k\mid b\\ \end{cases}.$$ So effectively these summations sum precisely the divisors of $$b$$, up to $$\lfloor\tfrac{b-2}{2}\rfloor$$. A routine check shows that the only divisors not counted are $$b$$, and if $$b$$ is even also $$\tfrac{b}{2}$$, and if $$b=3$$ also $$1$$. Then we can further simplify the differences for even $$b$$ as $$\begin{eqnarray*} r(b)-r(b-1)&=&\frac{b}{2}-1-\Big(\sigma(b)-b-\tfrac{b}{2}\Big)\\ &=&2b-1-\sigma(b), \end{eqnarray*}$$ and for odd $$b>3$$ as $$\begin{eqnarray*} r(b)-r(b-1)&=&\frac{b-1}{2}-\Big(\sigma(b)-b\Big)\\ &=&\frac{3b-1}{2}-\sigma(b), \end{eqnarray*}$$ and the latter is easily verified to also hold for $$b=3$$.$$\quad\square$$
It is a well known fact (or nice exercise) to prove that for a positive integer $$m$$ with prime factorization $$m=\prod_{i=1}^np_i^{a_i}$$, where $$p_1,\ldots,p_n$$ are distinct prime numbers and $$a_1,\ldots,a_n$$ are positive integers, we have $$\sigma(m)=\prod_{i=1}^n\sum_{j=0}^{a_i}p_i^j.$$ In particular this shows that $$\sigma(m)$$ is odd if and only if for every odd prime $$p_i$$ dividing $$m$$ we have $$a_i\equiv0\pmod{2}$$, or equivalently either $$m=k^2$$ or $$m=2k^2$$ for some integer $$k$$. In particular we see that if $$b\equiv3\pmod{4}$$ then $$\sigma(b)$$ is even and hence $$r(b)\equiv r(b-1)\pmod{2}$$.
Here are some small values for $$r(b)$$: $$\begin{array}{r|ccccccccc} b&0&1&2&3&4&5&6&7\\ \hline r(b)&0&0&0&0&0&1&0&2\\ &&&&&&&&\\ b&8&9&10&11&12&13&14&15\\ \hline r(b)&2&2&3&7&2&7&10&8\\ &&&&&&&&\\ b&16&17&18&19&20&21&22&23\\ \hline r(b)&8&15&11&19&16&15&22&32\\ \end{array}$$
• It seems that $r(b)\equiv r(b+1)\pmod{2}\iff\sigma(b+1)\equiv1\pmod{2}$ is false. Take $b=6$. – mathlove Mar 23 at 15:24
• @mathlove I had spotted the same inconsistency. I have removed the error in my argument. I will try to complete this answer soon (likely tomorrow). – Servaes Mar 23 at 15:53 | 2020-07-09T02:51:35 | {
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https://math.stackexchange.com/questions/3149702/if-d-divides-2n-and-d-doesnt-divide-n-then-d-is-even | # If $d$ divides $2n$ and $d$ doesn't divide $n$, then $d$ is even
I have encountered a proof regarding dihedral groups we this fact is used:
If $$d\mid 2n$$ and $$d\nmid n$$, then $$d$$ is even and $${d\over 2}\mid n$$.
I can't seem to understand why this is true. If $$d\nmid n$$, then there are $$q,r \in \mathbb{Z}$$ so that $$0 < r < d$$ and $$n = qd + r$$. On the other hand, $$d\mid 2n$$ means that there is $$m \in \mathbb{Z}$$ so that $$2n = md$$. We need to somehow use these two facts.
Also, my second question is how we can naturally generalize this result?
• Hint: if $d$ was odd, $m$ would be even. – Wojowu Mar 15 at 19:28
• $2n =md$. And $2$ is prime. So $2|m$ or $2|d$. If $2|m$ then $n = \frac m2 d$ and $d|n$. If $2|d$ then $d$ is even. General. If $d|pn$ for a prime $p$ and $d\not \mid n$ then $p|d$. Even more general. If $d|ab$ then $\frac d{\gcd(d,b)}|a$. – fleablood Mar 15 at 21:22
Suppose $$d$$ is odd. Then $$d$$ and $$2$$ are relatively prime and $$d\mid 2n$$, so by Euclid lemma we have $$d\mid n$$.
A contradiction. So $$d$$ must be even.
We could have more general situation.
Say $$d\mid pn$$ for some prime $$p$$ and $$d$$ doesn't divide $$n$$. Then $$p\mid d$$.
The proof goes exactly the same as for $$p=2$$.
It's also possible to see what's going on simply by keeping track of factors of $$2$$; note that this kind of analysis could also be generalized to other prime factors, as has been illustrated in previously posted answers.
Let $$n:= 2^aQ,\ 2n=2^{a+1}Q, d:=2^bP$$ where $$Q$$ represents the product of all of the odd prime factors of $$n$$ and $$P$$ represents the product of all of the odd prime factors of $$d$$. We could write out all of those factors and explicitly show this, but it should be plain that $$d\mid 2n \Rightarrow P\mid Q$$. Note that thus far, the exponents $$a,b$$ might be $$0$$, so we have not assumed that $$d$$ is even.
$$d\mid 2n \Rightarrow b\le a+1$$
$$d\not \mid n \Rightarrow b>a$$
Together, these establish $$b=a+1$$, meaning that $$d$$ has at least one factor of $$2$$ and is even, even if $$a=0$$.
This also illustrates the second point: the exponent of $$2$$ in $$d\over 2$$ is simply $$b-1=a$$. Hence $$\frac{d}{2} \mid n$$
If you know that the remainder $$r$$ in $$n=qd+r$$ with $$0\lt r\lt d$$ is unique, then from $$2n=md$$ we get
$$n=2n-n=md-(qd+r)=(m-q)d-r=(m-q+1)d+(d-r)=q'd+r'$$
with $$0\lt r'=d-r\lt d$$, so that, by uniqueness of the remainder, we have $$r'=r$$, i.e. $$d-r=r$$, hence $$d=2r$$.
$$d| 2n$$ then $$2n=kd;$$
Since $$kd$$ is even, $$2| kd$$.
Euclid's lemma:
1) $$2| k$$ or 2) $$2|d$$.
1) If $$2|k$$ then $$k=2k'.$$
$$2n=2k'd$$;
$$n=k'd$$, i.e. $$d|n$$ , a contradiction.
2) Hence $$2|d$$ , and we are done.
By below $$\ d\mid pn\iff\!\!\!\! \overbrace{d\mid n}^{\large\color{#0a0}{(d,p)}\ =\ \color{#c00}1}\!\!$$ or $$\ \overbrace{{d/\color{#c00}p}\mid n}^{\large\color{#0a0}{ (d,p)}\ =\ \color{#c00} p}\!,\$$ by $$\ \color{#0a0}{(d,p)}\mid \color{#c00}p\,$$ prime.
Lemma $$\,\ d\mid an\iff\smash[t]{\overbrace{ d/\color{#0a0}{(d,a)}\,\mid\, n,\,}\ }$$ where $$\,\ (x,y) := \gcd(x,y)$$
Proof $$\quad\ d\mid an\iff d\mid dn,an,\iff d\mid (dn,an)=(d,a)n\iff d/(d,a)\mid n$$
• Convention: $\ d/p\mid n\$ means $\ d/p\,$ is an integer, so $\,p\mid d\ \$ – Bill Dubuque Mar 15 at 20:23
Intuitively. If $$d|ab$$ and $$d\not \mid b$$ then "some part of $$d$$ must divide $$a$$". So if $$d|2n$$ but $$d\not \mid n$$ then some (non-trivial) part must divide $$2$$ and that part must be $$2$$ so $$d$$ is even.
.....
That's intuition. Let's make a proof.
.....
If $$d|ab$$. Let $$\gcd(d,b) = g$$ and let $$d = d'g$$ and $$b = b'g$$. It's easy to see that $$d'$$ and $$b'$$ are relatively prime: if $$b'$$ and $$d'$$ had any non-trivial factor, $$k$$, in common then $$kg$$ would be a common divisors of $$b$$ and $$d$$ contradicting that $$g$$ is the greatest common divisor.
So $$d=d'g$$ and $$ab = ab'g$$ and $$d'g|ab'g$$ so $$d'|ab'$$. But $$d'$$ and $$b'$$ are relatively prime so they have no factors in common. So $$d'|a$$.
Now if $$d|b$$ then $$d'g|b'g$$ so $$d'|b'$$ but $$d'$$ and $$b'$$ are relatively prime so $$d' = 1$$ and $$d = \gcd(d,b)$$.
Lemma 1: $$d|b \iff \gcd(d,b) = d$$.
If $$d|ab$$ the $$d'|a$$ but if $$d\not \mid b$$ then $$\gcd(d,b)=g \ne d$$ so $$d' = \frac dg > 1$$. $$d'\ne 1$$ and $$d'|a$$.
So
Lemma 2: If $$d|ab$$ but $$d\not \mid b$$ then $$d' = \frac d{\gcd(d,b)} > 1$$ and $$d'|a$$.
So if $$d|2n$$ and $$d\not \mid n$$ then $$\frac d{\gcd(d,n)} > 1$$ and $$\frac d{\gcd(d,n)} |2$$.
So $$\frac d{\gcd(d,n)} = 2$$ and $$d = \frac d{\gcd(d,n)}\gcd(d,n) = 2\gcd(d,n)$$. And $$d$$ is even.
=====
Actually, this may be the best most general Theorem:
Theorem: If $$d|mn$$ then $$\frac d{\gcd(d,n)}|m$$.
I'll leave the proof to you, and I'll leave it to you to figure out how that implies if $$d|2n$$ and $$d\not\mid n$$ then $$d$$ is even. | 2019-04-23T12:32:37 | {
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https://mathematica.stackexchange.com/questions/263451/generate-covariance-matrix-from-related-random-variable-distributions | # Generate covariance matrix from related random variable distributions
Say I have a random variable $$X\sim\mathscr{N}\left(0,\sigma^2\right)$$ and another random variable $$Y=X+\varepsilon$$, where $$\varepsilon\sim\mathscr{N}\left(0,\eta^2\right)$$ is independent of $$X$$. I can calculate $$Cov\left(X,Y\right)=Cov\left(X,X+\varepsilon\right)=Var(X)=\sigma^2$$.
Is there a way to get Mathematica to generate a bivariate normal distribution automatically from the given information? Specifically, rather than working out the covariance matrix myself and then inputting to Mathematica, I would like to be able to do something like (implementing the above example):
\[ScriptCapitalX] = NormalDistribution[0, \[Sigma]]
\[CapitalEpsilon] = NormalDistribution[0, \[Eta]]
\[ScriptCapitalY] = TransformedDistribution[X + \[CurlyEpsilon], {X \[Distributed] \[ScriptCapitalX], \[CurlyEpsilon] \[Distributed] \[CapitalEpsilon]}]
\[Mu] = {Mean[\[ScriptCapitalX]], Mean[\[ScriptCapitalY]]}
\[CapitalSigma] = {{Variance[\[ScriptCapitalX]], Covariance[\[ScriptCapitalX], \[ScriptCapitalY]]}, {Covariance[\[ScriptCapitalX], \[ScriptCapitalY]], Variance[\[ScriptCapitalY]]}}
multi = MultinormalDistribution[\[Mu], \[CapitalSigma]]
want = Expectation[X \[Conditioned] Y = y, {Y, X} \[Distributed] multi]
Obviously, the Covariance function doesn't work that way, but at a minimum I would really like to have Mathematica do the covariance calculation. Also, if there's an easy way to get the mean vector and covariance matrix without explicitly calling Mean, Variance, etc. on individual elements to construct them, that would be even better. I am quite new to the Wolfram language.
• Welcome to Mathematica.SE! I hope you will become a regular contributor. To get started, 1) take the introductory tour now, 2) when you see good questions and answers, vote them up by clicking the gray triangles, because the credibility of the system is based on the reputation gained by users sharing their knowledge, 3) remember to accept the answer, if any, that solves your problem, by clicking the checkmark sign, and 4) give help too, by answering questions in your areas of expertise. Feb 11 at 10:34
You were almost there:
\[ScriptCapitalX] = NormalDistribution[0, σ]
\[CapitalEpsilon] = NormalDistribution[0, η]
multi = TransformedDistribution[{X, X + ε},
{X \[Distributed] \[ScriptCapitalX], ε \[Distributed] \[CapitalEpsilon]}]
Covariance[multi][[1, 2]]
(* σ^2 *)
Expectation[X \[Conditioned] Y == y, {X, Y} \[Distributed] multi]
(* (y σ^2)/(η^2 + σ^2) *) | 2022-08-15T05:10:51 | {
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https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Path_Component | # Equivalence of Definitions of Path Component
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## Theorem
The following definitions of the concept of Path Component in the context of Topology are equivalent:
Let $T = \struct {S, \tau}$ be a topological space.
Let $x \in T$.
### Equivalence Class
Let $\sim$ be the equivalence relation on $T$ defined as:
$x \sim y \iff x$ and $y$ are path-connected.
The equivalence classes of $\sim$ are called the path components of $T$.
If $x \in T$, then the path component of $T$ containing $x$ (that is, the set of points $y \in T$ with $x \sim y$) can be denoted by $\map {\operatorname{PC}_x} T$.
### Union of Path-Connected Sets
The path component of $T$ containing $x$ is defined as:
$\displaystyle \map {\operatorname{PC}_x} T = \bigcup \left\{{A \subseteq S: x \in A \land A}\right.$ is path-connected $\left.\right\}$
### Maximal Path-Connected Set
The path component of $T$ containing $x$ is defined as:
the maximal path-connected set of $T$ that contains $x$.
## Proof
Let $\CC_x = \set {A \subseteq S : x \in A \land A \text { is path-connected in } T}$
Let $C = \bigcup \CC_x$.
### Lemma
$C$ is path-connected in $T$ and $C \in \CC_x$.
$\Box$
Let $C'$ be the equivalence class containing $x$ of the equivalence relation $\sim$ defined by:
$y \sim z$ if and only if $y$ and $z$ are connected in $T$.
### Equivalence Class equals Union of Path-Connected Sets
It needs to be shown that $C = C'$.
$\displaystyle y \in C'$ $\leadstoandfrom$ $\displaystyle x \text{ is path-connected to } y \text{ in } T$ Definition of $\sim$ $\displaystyle$ $\leadstoandfrom$ $\displaystyle \exists B \text{ a connected set of } T, x \in B, y \in B$ Points are Path-Connected iff Contained in Path-Connected Set $\displaystyle$ $\leadstoandfrom$ $\displaystyle \exists B \in \CC_x : y \in B$ Equivalent definition $\displaystyle$ $\leadstoandfrom$ $\displaystyle y \in \bigcup \CC_x$ Definition of Set Union $\displaystyle$ $\leadstoandfrom$ $\displaystyle y \in C$ Definition of $C$
The result follows.
$\Box$
### Union of Path-Connected Sets is Maximal Path-Connected Set
Let $\tilde C$ be any path-connected set such that:
$C \subseteq \tilde C$
Then $x \in \tilde C$.
Hence $\tilde C \in \CC_x$.
$\tilde C \subseteq C$.
Hence $\tilde C = C$.
It follows that $C$ is a maximal path-connected set of $T$ by definition.
$\Box$
### Maximal Path-Connected Set is Union of Path-Connected Sets
Let $\tilde C$ be a maximal path-connected set of $T$ that contains $x$.
By definition:
$\tilde C \in \CC_x$
$\tilde C \subseteq C$
By maximality of $\tilde C$:
$\tilde C = C$
$\blacksquare$ | 2020-09-28T16:15:33 | {
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## Calculus, all content (2017 edition)
### Unit 1: Lesson 16
Limits of piecewise functions
# Worked example: point where a function is continuous
AP.CALC:
LIM‑2 (EU)
,
LIM‑2.A (LO)
,
LIM‑2.A.2 (EK)
Sal finds the limit of a piecewise function at the point between two different cases of the function. In this case, the two one-sided limits are equal, so the limit exists.
## Want to join the conversation?
• at he substitutes x with 3 but the function is defined between 0 < x < 3. In other words it's defined up to three but not three. So isn't that technically wrong?
• in that particular function you mentioned,, it is continuos until x approaches 3.
We needed to find the limit of f(x) of that function as x approached 3. That is only the idea of limit.. It is not defined at x =3, but we can find a value of f(x) so that it would have formed a continuos graph.. So we find the value of f(x) that would have been for x=3.
Hope that was not complicated
• At how we can understand that function is continuous?
• It is not undefined for any positive argument. This means that there are no asymptotes or removable discontinuities, but proving continuity can be done in a variety of ways (for instance, noting that it is differentiable or noting that its inverse is differentiable etc.). Since differentiability is a stronger condition than continuity, all differentiable functions are also continuous over the differentiable interval.
• Can someone please suggest me a video in which all log functions are explained? Because I am only aware of the basic stuff but I guess as we proceed, we need to know the complicated functions of log too!
• What is a piecewise function?
• A piecewise function has different rules in different intervals. For example, look up aat this function:
f(x) = x^2 if x if x<4
= 4 if x<4 or x=4
Between the interval wich goes from negative infinity, it is x^2; and between the interval wich goes from 4 to positive infinity it is always four.
To give a counterexample, g(x)=x^2+1 is not a piecewise function, because it is always equal to x^2+1; without mattering the value of x
• At and , what if the function is non continuous? What numbers do we plug in for x?
• If there is a jump discontinuity, then the limit from the left side and the limit from the right side will not be equal so the overall limit does not exist. You still have to plug in the same x value in both equations and you will get different values so the overall limit does not exist. But if there is a removable discontinuity, both the limit from the left side and the limit from the right side will be equal so the overall limit exists. In any case, you have to plug in the same x value.
• I forgot what a log is :/
• A logarithm is essentially the opposite of the exponential function. What this means is that if a^x = b, the log(base a) b = x.
• Why didn't we find the function value when x =3 to check if that is equal to the limit to satisfy the condition of continuity
• for a limit to be continuous, lim(x tends to c) f(x) = f(c).
IN this case we know that lim(x tends to 3) g(x) = log(9) but we don't know if
g(3)=log(9).
So how can we say that the limit is continuous
• We do know that g(3)=log(9), because the function g is defined at x=3 and we can plug 3 into the function.
g(3) = (4-3)*log(9) = 1*log(9) = log(9)
• I'm a bit confused by the title; this only proves that the limit of g(x) as x -> 3 exists, not that g(x) is necessarily continuous at that point right?
EDIT: nvm guys we used direct sub, so it is indeed continuous
• for a limit to be continuous, lim(x tends to c) f(x) = f(c). so we first see if the limit exists and then we see if it's equal to the function of that point, at this example we are not examining if the function is continuous it is continuous, he is showing us the case.
• How would I explain why g(x) is continuous?
## Video transcript
- [Voiceover] So we have g of x being defined as the log of 3x when zero is less than x is less than three and four minus x times the log of nine when x is greater than or equal to three. So based on this definition of g of x, we want to find the limit of g of x as x approaches three, and once again, this three is right at the interface between these two clauses or these two cases. We go to this first case when x is between zero and three, when it's greater than zero and less than three, and then at three, we hit this case. So in order to find the limit, we want to find the limit from the left hand side which will have us dealing with this situation 'cause if we're less than three we're in this clause, and we also want to find a limit from the right hand side which would put us in this clause right over here, and then if both of those limits exist and if they are the same, then that is going to be the limit of this, so let's do that. So let me first go from the left hand side. So the limit as x approaches three from values less than three, so we're gonna approach from the left of g of x, well, this is equivalent to saying this is the limit as x approaches three from the negative side. When x is less than three, which is what's happening here, we're approaching three from the left, we're in this clause right over here. So we're gonna be operating right over there. That is what g of x is when we are less than three. So log of 3x, and since this function right over here is defined and continuous over the interval we care about, it's defined continuous for all x's greater than zero, we can just substitute three in here to see what it would be approaching. So this would be equal to log of three times three, or logarithm of nine, and once again when people just write log here within writing the base, it's implied that it is 10 right over here. So this is log base 10. That's just a good thing to know that sometimes gets missed a little bit. All right, now let's think about the other case. Let's think about the situation where we are approaching three from the right hand side, from values greater than three. Well, we are now going to be in this scenario right over there, so this is going to be equal to the limit as x approaches three from the positive direction, from the right hand side of, well g of x is in this clause when we are greater than three, so four minus x times log of nine, and this looks like some type of a logarithm expression at first until you realize that log of nine is just a constant, log base 10 of nine is gonna be some number close to one. This expression would actually define a line. For x greater than or equal to three, g of x is just a line even though it looks a little bit complicated. And so this is actually defined for all real numbers, and it's also continuous for any x that you put into it. So to find this limit, to think about what is this expression approaching as we approach three from the positive direction, well we can just evaluate a three. So it's going to be four minus three times log of nine, well that's just one, so that's equal to log base 10 of nine. So the limit from the left equals the limit from the right. They're both log nine, so the answer here is log log of nine, and we are done. | 2023-01-27T02:24:55 | {
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