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Prove... pi

Date: 7/5/96 at 20:2:1
From: Anonymous
Subject: Prove... pi

I have tried to prove:

Sum of 1/n^2 with n-> {1 to infinity} is <=> to pi^2/6, but I haven't 
made a correct proof of it yet. Could you help ...? 

Date: 7/6/96 at 11:47:4
From: Doctor Anthony
Subject: Prove... pi

This can be proved by some fairly unpleasant manipulation of trig 
series and Bernoulli numbers, but you can also use Fourier series to 
get the result.

In the interval -pi to +pi ANY function (provided it is continuous and 
differentiable save at a finite number of points in the interval) can 
be expressed in the form 

f(x) = (1/2)a0 + a1*cos(x) + a2*cos(2x) + a3*cos(3x) + infin.
               + b1*sin(x) + b2*sin(2x) + b3*sin(3x) + infin..

and it is easy to show that the a's and b's can be determined as 

an = (1/pi)INT(-pi tp pi)[f(x)cos(nx)dx]

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

If we use the half-interval 0<x<pi we can fill up the whole interval -
pi < x < pi by making f(x) have values of our own choosing for the 
interval -pi < x < o.  The series so obtained will be correct for the 
given interval 0 < x < pi. For example the function is made up to an 
even function by means of the definition f(-x) = f(x) and the Fourier 
series is then a cosine series.

This means that an = (2/pi)INT(0 to pi)[f(x)cos(nx)dx]

If we apply these ideas to the function x(a-x) in the range 0 < x < a 
in the series SUM(0 to infin.)cn*cos(n*pi*x/a)  where the cn are the 
coefficients, we obtain:

x(a-x) = (1/6)a^2 - (a/pi)^2{cos(2*pi*x/a) + (1/2^2)cos(4*pi*x/a) +         
                            (1/3^2)cos(6*pi*x/a) + infin.}

putting x=0, the left hand side is zero and putting the right hand 
side also equal to zero we get:

          (a/pi)^2{1 + 1/2^2 + 1/3^2 + ....} = (1/6)a^2

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

-Doctor Anthony,  The Math Forum
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Associated Topics:
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