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Proving Gregory's formula

  1. Dec 8, 2003 #1
    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
     
    Last edited: Dec 8, 2003
  2. jcsd
  3. Dec 8, 2003 #2

    Hurkyl

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    Can you make a modification to the sum to turn it into a power series?
     
  4. Dec 8, 2003 #3
    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.
     
  5. Dec 9, 2003 #4

    NateTG

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    Do you know about Taylor / McLaurin series?
     
  6. Dec 9, 2003 #5

    mathman

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

    Fill in the details.
     
  7. Dec 10, 2003 #6
    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]
     
  8. Dec 10, 2003 #7
    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.
     
    Last edited: Dec 10, 2003
  9. Dec 10, 2003 #8

    mathman

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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).
     
  10. Dec 11, 2003 #9
    I feel think that the Fourier-expansion that I showed earlier is much simpler than the arctan argument. :wink:
     
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