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anonymous
 4 years ago
Why does 11/3+1/51/7... converge to Pi/4? In Latex,
\[\sum_{k=0}^{\infty}\frac{(1)^k}{2k+1}=\frac{\pi}{4}\]
Don't give me a copy/paste of what Wikipedia says though, it skips a couple steps and I don't understand!
anonymous
 4 years ago
Why does 11/3+1/51/7... converge to Pi/4? In Latex, \[\sum_{k=0}^{\infty}\frac{(1)^k}{2k+1}=\frac{\pi}{4}\] Don't give me a copy/paste of what Wikipedia says though, it skips a couple steps and I don't understand!

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0sure u know that\[\arctan x=\sum_{n=0}^{\infty} (1)^n \frac{x^{2n+1}}{2n+1}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\frac{1}{1+t^2}=1t^2+t^4t^6+t^8...=\sum_{n=0}^{\infty} (1)^n t^{2n} \ \ \ \ \ \ \ t<1\]integrate both sides from \(0\) to \(x\) for \(x<1\)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Wow should have caught that more easily. It seems so clear when you see it. But go on with the proof....

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0after integrating the geometric series u will get\[\arctan x=\sum_{n=0}^{\infty} (1)^n \frac{x^{2n+1}}{2n+1}\]let \(x=1\) u will have\[\arctan 1=\frac{\pi}{4}=\sum_{n=0}^{\infty} \frac{(1)^n }{2n+1}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0but this is not a complete proof

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Yeah but you can't just let x=1 cause that series only holds true for \[x<1\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0So really the question is how Wikipedia got that first substitution of arctanx with the series that had a larger radius of convergence. I get the rest.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0here we must use Abel's theorem that im not comfortable with

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Well can we try to reason this out? Clearly the introduction of \[\frac{(1)^{n+1}x^{2n+2}}{1+x^2}\]in the summation \[\frac{1}{1+x^2}=\sum_{k=0}^n(1)^nx^{2k}+\frac{(1)^{n+1}x^{2n+2}}{1+x^2}\]Had something to do with it...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0My guess is that Wikipedia made that sum finite (with n on top instead of infy), and so that messy term at the end is just a correction factor for when x=1 in the mclauren series 1/(1+x^2). Then, taking the limit n>infty later guarentees that the new series models arctan x, and, after integration, the correction factor approaches 0 at infity.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0So integration actually expanded our radius of convergence (as I understand it)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0If I'm not mistaken, then it looks something like this: \[\frac{1}{1+x^2}=\sum_{k=0}^n (1)^kx^{2k}+R\]where\[R=\{_{0,x<1}^{f(x, k), x=1}\]We know that there must be some existing f(x, k) for finite series. We also know, because of the radius of convergencIe for a geometric series, that \[\lim_{n\rightarrow\infty}\sum_{k=0}^nf(x, k)=0\]If and only if x<1, and for x=1, the same sum diverges.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0However, integrating both sides of my first equation in the previous reply gets us: \[arctan(x)=\sum_{k=0}^n\int(1)^kx^{2k}dx+\int Rdx\] Taking the limit n>infty of both sides, we find that, due to the properties of f(x, k), \[\lim_{n\rightarrow\infty}\int R dx=0\] For 1 < x <= 1. Then your proof works.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0EDIT: Last limit should be: \[\lim_{n\rightarrow\infty}\sum_{k=0}^n\int R dx=0\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0The whole thing is interesting, though. It leads me to another question, that I'll post separately (about whether or not such a function f(x, k) can exist with those properties.
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