Prove... piDate: 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) + ...to infin. + b1*sin(x) + b2*sin(2x) + b3*sin(3x) + ...to infin.. and it is easy to show that the a's and b's can be determined as follows: 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) + .....to 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 Check out our web site! http://mathforum.org/dr.math/ |
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