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Proving Gregory's formula

by Gunni
Tags: formula, gregory, proving
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Gunni
#1
Dec8-03, 11:56 AM
P: 41
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
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Hurkyl
#2
Dec8-03, 05:03 PM
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Can you make a modification to the sum to turn it into a power series?
Gunni
#3
Dec8-03, 05:36 PM
P: 41
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
#4
Dec9-03, 11:40 AM
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Proving Gregory's formula

Do you know about Taylor / McLaurin series?
mathman
#5
Dec9-03, 04:58 PM
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arctan(x)=x-x3/3+x5/5-x7/7+...

Fill in the details.
suyver
#6
Dec10-03, 01:41 AM
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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
#7
Dec10-03, 10:54 AM
P: 41
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
#8
Dec10-03, 04:57 PM
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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
#9
Dec11-03, 01:34 AM
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P: 265
I feel think that the Fourier-expansion that I showed earlier is much simpler than the arctan argument.


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