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Fourier Series and the Zeta Function

Date: 08/13/98 at 17:58:42
From: Patrick Perry
Subject: Evaluating the Riemann-Zeta Function

Dr. Math:

I was reading about the Zeta function in a complex analysis book, and 
I have seen the functional equation relating the Gamma function to the 
Zeta function. However, it is still not clear to me how to evaluate 
something like "zeta of 2." Could you please give me some idea of how 
it's done?

Date: 08/13/98 at 18:52:37
From: Doctor Anthony
Subject: Re: Evaluating the Riemann-Zeta Function


Z(2) is the series shown below:

   1 + 1/2^2 + 1/3^2 + 1/4^2 + 1/5^2 + . . . . . = pi^2/6

This can be proved by advanced trigonometric manipulations, but a more 
elegant method is with the aid of Fourier series.

A brief note on Fourier series is necessary if you are to understand 
the method.

For a great many functions within the range -pi < x < pi we can express 
f(x) as an infinite series of the form:

   f(x) = a(0)/2 + SUM(n=1 to inf.)[a(n)cos(nx) + b(n)sin(nx)]

where the a(i), b(i) are constants. So:

   f(x) = a(0)/2 + a(1)cos(x) + a(2)cos(2x) + a(3)cos(3x) + ....
                 + b(1)sin(x) + b(2)sin(2x) + b(3)sin(3x) + ....

This series is periodic with period 2pi and the convention is to take 
the interval -pi < x < pi.

The way to evaluate the constants a(i) is to multiply both sides by 
cos(nx) and then integrate between -pi and pi.

   INT(-pi to pi)[f(x)cos(nx)dx] = pi a(n)     (proof given below)

So a(n) = (1/pi)INT(-pi to pi)[f(x)cos(nx)dx]


   b(n) = (1/pi)INT(-pi to pi)[f(x)sin(nx)dx]

Also note:

   a(0) = (1/pi)INT(-pi to pi)[f(x) dx]

Proof of above results.

INT(-pi to pi)[cos(mx)cos(nx)dx]        m not equal n

   = (1/2)INT[cos(m+n)x + cos(m-n)x]dx  between -pi and pi 

   = 0    since sin(any multiple of pi) = 0


   INT(-pi to pi)[cos^2(nx)]dx = pi

So all the terms cos(mx)cos(nx) (m not equal to n) disappear and we 
get only the term in cos^2(nx)

This gives the result:

   INT[f(x)cos(nx)dx] = a(n) pi

and so:

   a(n) = (1/pi)INT[f(x) cos(nx) dx]

In the case n = 0 we get:

   INT[f(x) dx] = INT[a(0)/2 dx] + 0 + 0 + ... = a(0) pi


   a(0) = (1/pi)INT[f(x) dx]


   b(n) = (1/pi)INT[f(x)sin(nx)dx]

since all terms in sin(mx)sin(nx) (m not equal to n) disappear leaving 
only the term in sin^2(nx). Also all terms in sin(mx)cos(nx), whether 
or not m = n disappear on integration between -pi and pi.

This means that we can evaluate all the a(i), b(i) and express f(x) as 
an infinite series in sines and cosines of multiples of x. There are a 
great many functions that can be expressed in this way in the interval 
-pi to pi.

We can now turn our attention to finding the sum of the series:

   1 + 1/2^2 + 1/3^2 + 1/4^2 + ....

Let f(x) = x^2. This is an even function of x so we know that none of 
the sine terms of the Fourier series can be present. We need only look 
at the a(i) terms.

For all a(n) we get:

   pi a(n) = INT(-pi to pi)[x^2 cos(nx)dx]

When n = 0 this gives a(0) = (2/3)pi^2.

When n is greater than 0, integration by parts in 3 steps gives:

   a(n) = 4(-1)^n/n^2

The Fourier series is pi^2/3 + 4 SUM(1 to infinity)(-1)^n/n^2 cos(nx)

With x = pi, we get:

   pi^2 = pi^2/3 + 4 SUM(-1)^n/n^2 cos(n pi)

and since cos(n pi)= (-1)^n,  this produces +'s for each term:

   2 pi^2/3  =  4 SUM[1 + 1/2^2 + 1/3^2 + ..... to infinity]

   pi^2/6 =  1 + 1/2^2 + 1/3^2 + 1/4^2 + ..... to infinity

- Doctor Anthony, The Math Forum
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Associated Topics:
College Analysis
High School Analysis
High School Sequences, Series

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