Understanding Gregory's Formula: A Mathematical Challenge

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In summary, the conversation was about trying to solve Gregory's formula, which involves finding the value of pi in a specific way. The speaker tried to break down the problem into a single sum and integrate it, but was having trouble. They also mentioned learning about imaginary numbers and calculus. Another person suggested using Taylor/McLaurin series to find the power series for arctan(x), which would help solve the problem. However, the original speaker later found a simpler solution involving Leibniz's arctan formula and integrating a new function. The conversation ends with the suggestion to use the Fourier-expansion instead, as it is simpler.
  • #1
Gunni
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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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  • #2
Can you make a modification to the sum to turn it into a power series?
 
  • #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.
 
  • #4
Do you know about Taylor / McLaurin series?
 
  • #5
arctan(x)=x-x3/3+x5/5-x7/7+...

Fill in the details.
 
  • #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]
 
  • #7
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.
 
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  • #8
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).
 
  • #9
I feel think that the Fourier-expansion that I showed earlier is much simpler than the arctan argument. :wink:
 

1. What is Gregory's formula?

Gregory's formula, also known as the Gregory-Leibniz series, is a mathematical formula used to approximate the value of pi. It was discovered by Scottish mathematician James Gregory in the 17th century.

2. How does Gregory's formula work?

Gregory's formula uses an infinite series of terms to approximate the value of pi. The formula is: pi = 4 - 4/3 + 4/5 - 4/7 + 4/9 - 4/11 + ... As more terms are added, the approximation becomes more accurate.

3. Is Gregory's formula accurate?

Gregory's formula is a good approximation of pi, but it is not exact. The more terms that are used in the series, the closer the approximation will be to the actual value of pi.

4. Why is Gregory's formula important?

Gregory's formula is important because it was one of the first methods for approximating the value of pi. It also paved the way for other mathematical techniques and formulas used in calculus and other areas of mathematics.

5. How can Gregory's formula be proven?

Gregory's formula can be proven using mathematical induction. This involves showing that the formula holds true for the first term, and then proving that if it holds true for the nth term, it also holds true for the n+1th term. This process is repeated infinitely to prove that the formula holds true for all values.

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