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Proving Gregory's formula 
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#1
Dec803, 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 


#2
Dec803, 05:03 PM

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



#3
Dec803, 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}{1k} [/tex] k is the ratio between [tex] a_n [/tex] and [tex] a_{n1} [/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. 


#4
Dec903, 11:40 AM

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P: 2,538

Proving Gregory's formula
Do you know about Taylor / McLaurin series?



#5
Dec903, 04:58 PM

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P: 6,027

arctan(x)=xx^{3}/3+x^{5}/5x^{7}/7+...
Fill in the details. 


#6
Dec1003, 01:41 AM

P: 265

On the interval [tex](\pi,\pi][/tex] the function
[tex]f(x)=x[/tex] has the Fourierexpansion [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] 


#7
Dec1003, 10:54 AM

P: 41

Anyway, thanks everybody. 


#8
Dec1003, 04:57 PM

Sci Advisor
P: 6,027

To get the power series for arctan(x), use the derivative 1/(1+x^{2}). Expand the latter into a power series (binomial) and get
1/(1+x^{2})=1x^{2}+x^{4}x^{6}... Term by term integration gives you the desired result (using arctan(0)=0 for the constant of integration). 


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