Proving Gregory's formula

[Gunni]

Hello,

I was talking to a friend of mine that's studying math at the university here and he gave me this problem to solve: Prove Gregory's formula. I'm going nuts. I've broken it down into a single sum like this:

\frac{\pi}{4} = 1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7} ... = \sum_{n=0}^\infty \frac{1}{(1+2n)(-1)^n}

Now, from there I've tried integrating it with the upper limits at infinity and lower at 0, tried connecting it to a circle with a radius of 1/2 and pretty much everything I can think of. I'm not really asking for a complete proof of the formula as I'd like to try to do it myself, just a little help. Am I doing the totally wrong thing or would this approach work out if I did something different?

Thanks

[Hurkyl]

Can you make a modification to the sum to turn it into a power series?

[Gunni]

I don't think so, the ratios between the numbers are never constant, I only know how to calculate infinite power series of the form:

S_n = \frac{a_1}{1-k}
k is the ratio between a_n and a_{n-1}
Where the series only converges if -1 < k < 1.

I'm beginning to think that the solution might be to think of it as a function and calculate the integral from zero to infinity. I've been trying that and I can't get around integrating the function, it's slightly more complex than what I've been doing so far (I just finished the course on how to integrate). I think that might be it since pi is related to the area of a circle, so it might work if I calculate the area of the function. Something like this:

\int_{0}^{\infty} \frac{1}{(1+2n)(-1)^n} dn

I hit a brick wall in relation to that earlier when I tried to calculate a smoother graph than the one I'd done before, with my function I get an imaginary number whenever n isn't a whole number.

Right now might be a good time to mention that I start learning about imaginary numbers next semester and that I just finished the starter courses on calculus. This is a problem I got from a friend, it's from the final exam on mathematical analysis at the University of Iceland, I'm doing my final year in the equivalent of high school here.

[NateTG]

Do you know about Taylor / McLaurin series?

[mathman]

arctan(x)=x-x3/3+x5/5-x7/7+...

Fill in the details.

[suyver]

On the interval (-\pi,\pi] the function

f(x)=x

has the Fourier-expansion

x=\lim_{N\rightarrow\infty} \left(-\sum_{n=1}^N\frac{(-1)^ni}{n}e^{-inx}+\sum_{n=1}^N\frac{(-1)^ni}{n}e^{inx}\right)

=2\sum_{k=1}^\infty\frac{(-1)^{k+1}}{k}\sin(kx)

Just substitute x=\pi/2 to find

\pi=4\sum_{k=0}^\infty\frac{(-1)^{k}}{2k+1}=4\left(1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\ldots\right)

[Gunni]

Originally posted by mathman
arctan(x)=x-x3/3+x5/5-x7/7+...

Fill in the details.
I caved in last night and just asked him how it's done. The proof he had was based around making another function, integrating that, inserting t so that it looked somewhat like the equation I have above and inserting x=1 to attain arcan(1) = pi/4. Something I would never have thought of since I'd never seen Leibinz's arctan formula, the Taylor / McLaurin series or Fourier functions before. Oh, well, that's something to do during the christmas vacation, then.

Anyway, thanks everybody.

[mathman]

To get the power series for arctan(x), use the derivative 1/(1+x2). Expand the latter into a power series (binomial) and get
1/(1+x2)=1-x2+x4-x6...
Term by term integration gives you the desired result (using arctan(0)=0 for the constant of integration).

[suyver]

I feel think that the Fourier-expansion that I showed earlier is much simpler than the arctan argument. [;)]