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:

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

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:

[tex] S_n = \frac{a_1}{1-k} [/tex]
k is the ratio between [tex] a_n [/tex] and [tex] a_{n-1} [/tex]
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:

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

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 [tex](-\pi,\pi][/tex] the function

[tex]f(x)=x[/tex]

has the Fourier-expansion

[tex]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)[/tex]

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

Just substitute [tex]x=\pi/2[/tex] to find

[tex]\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)[/tex]

Gunni

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. [;)]