anonymous
  • anonymous
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!
Mathematics
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

anonymous
  • anonymous
sure u know that\[\arctan x=\sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+1}}{2n+1}\]
anonymous
  • anonymous
for \(|x|<1\)
anonymous
  • anonymous
how?

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
\[\frac{1}{1+t^2}=1-t^2+t^4-t^6+t^8-...=\sum_{n=0}^{\infty} (-1)^n t^{2n} \ \ \ \ \ \ \ |t|<1\]integrate both sides from \(0\) to \(x\) for \(|x|<1\)
anonymous
  • anonymous
Wow should have caught that more easily. It seems so clear when you see it. But go on with the proof....
anonymous
  • anonymous
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\[\arctan 1=\frac{\pi}{4}=\sum_{n=0}^{\infty} \frac{(-1)^n }{2n+1}\]
anonymous
  • anonymous
but this is not a complete proof
anonymous
  • anonymous
Yeah but you can't just let x=1 cause that series only holds true for \[|x|<1\]
anonymous
  • anonymous
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.
anonymous
  • anonymous
here we must use Abel's theorem that im not comfortable with
anonymous
  • anonymous
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...
anonymous
  • anonymous
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.
anonymous
  • anonymous
So integration actually expanded our radius of convergence (as I understand it)
anonymous
  • anonymous
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.
anonymous
  • anonymous
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.
anonymous
  • anonymous
EDIT: Last limit should be: \[\lim_{n\rightarrow\infty}\sum_{k=0}^n\int R dx=0\]
anonymous
  • anonymous
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.

Looking for something else?

Not the answer you are looking for? Search for more explanations.