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vf321 Group Title

Why does 1-1/3+1/5-1/7... 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!

  • one year ago
  • one year ago

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  1. mukushla Group Title
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    sure u know that\[\arctan x=\sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+1}}{2n+1}\]

    • one year ago
  2. mukushla Group Title
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    for \(|x|<1\)

    • one year ago
  3. vf321 Group Title
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    how?

    • one year ago
  4. mukushla Group Title
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    \[\frac{1}{1+t^2}=1-t^2+t^4-t^6+t^8-...=\sum_{n=0}^{\infty} (-1)^n t^{2n} \ \ \ \ \ \ \ |t|<1\]integrate both sides from \(0\) to \(x\) for \(|x|<1\)

    • one year ago
  5. vf321 Group Title
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    Wow should have caught that more easily. It seems so clear when you see it. But go on with the proof....

    • one year ago
  6. mukushla Group Title
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    after 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\[\arctan 1=\frac{\pi}{4}=\sum_{n=0}^{\infty} \frac{(-1)^n }{2n+1}\]

    • one year ago
  7. mukushla Group Title
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    but this is not a complete proof

    • one year ago
  8. vf321 Group Title
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    Yeah but you can't just let x=1 cause that series only holds true for \[|x|<1\]

    • one year ago
  9. vf321 Group Title
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    So 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.

    • one year ago
  10. mukushla Group Title
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    here we must use Abel's theorem that im not comfortable with

    • one year ago
  11. vf321 Group Title
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    Well 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...

    • one year ago
  12. vf321 Group Title
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    My 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.

    • one year ago
  13. vf321 Group Title
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    So integration actually expanded our radius of convergence (as I understand it)

    • one year ago
  14. vf321 Group Title
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    If 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.

    • one year ago
  15. vf321 Group Title
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    However, 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.

    • one year ago
  16. vf321 Group Title
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    EDIT: Last limit should be: \[\lim_{n\rightarrow\infty}\sum_{k=0}^n\int R dx=0\]

    • one year ago
  17. vf321 Group Title
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    The 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.

    • one year ago
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