Fourier series

Suppose e^x = (1/2)a_0 + Σ(a_n)cos(nx) + Σ(b_n)sin(nx), (n runs from 1 to infinity), x in [-π, π].
Then
a_0 = (1/π)∫(e^x) dx,
a_n = (1/π)∫(e^x)cos(nx) dx,
b_n = (1/π)∫(e^x)sin(nx) dx,
where the integrals are integrated from -π to π.

Let A(x) = ∫(e^x)cos(nx) dx and B(x) = ∫(e^x)sin(nx) dx.
Then A(x) + iB(x)
= ∫(e^x)(e^nix) dx
= ∫[e^(1+ni)x] dx
= e^(1+ni)x / (1 + ni) + C, where C is an arbitrary complex constant
= (1 - ni)(e^x)[cos(nx) + isin(nx)]/(1+n^2) + C
= (e^x)[cos(nx) + nsin(nx)]/√(1 + n^2) + i(e^x)[sin(nx) - ncos(nx)]/(1+n^2) + C.

By comparing real and imaginary parts respectively,
∫(e^x)cos(nx) dx = (e^x)[cos(nx) + nsin(nx)]/√(1 + n^2) + k, and
∫(e^x)sin(nx) dx = (e^x)[sin(nx) - ncos(nx)]/(1+n^2) + k',
where k and k' are real constants.
[This result may also be obtained by integrating by parts.]

Taking limits from -π to π for both integrals, we get the a_n's and b_n's:
a_0 = [e^π - e^(-π)]/π = 2sinh π / π ,
a_n = (-1)^n * [e^π - e^(-π)]/(1+n^2)π = (-1)^n * 2sinh π / (1+n^2)π,
b_n = (-1)^(n-1) * n[e^π - e^(-π)]/(1+n^2)π = (-1)^(n-1) * 2nsinh π / (1+n^2)π.

Hence, e^x = sinh π /π + (2sinh π /π) * Σ{(-1)^n * [cos(nx) - sin(nx)]/(1+n^2)}, (n runs from 1 to infinity), x in [-π, π].

Put x = π. Then
e^π = sinh π /π + (2sinh π /π) * Σ[1/(1+n^2)].
Put x = -π. Then
e^(-π) = sinh π /π + (2sinh π /π) * Σ[1/(1+n^2)].
Summing up, get
e^π + e^(-π) = 2sinh π /π + (4sinh π /π) * Σ[1/(1+n^2)].
2cosh π = (2sinh π /π) [1 + 2Σ[1/(1+n^2)]}.
Hence Σ[1/(1+n^2)] = [(π / tanh π) - 1]/2.

Similarly, put x = 0. Then
1 = sinh π /π + (2sinh π /π) * Σ[(-1)^n /(1+n^2)].
Hence Σ[(-1)^n /(1+n^2)] = [(π/sinh π) - 1]/2.

Using the method for a generalized Fourier series, the usual Fourier series involving sines and cosines is obtained by taking
圖片參考：http://mathworld.wolfram.com/images/equations/FourierSeries/inline22.gif
is given by

圖片參考：http://mathworld.wolfram.com/images/equations/FourierSeries/equation1.gif
(6) where

圖片參考：http://mathworld.wolfram.com/images/equations/FourierSeries/inline35.gif
.
..........................................................................................

Referring to the defintion above, now, we have,
a0 = (eπ - e-π)/π = {2sinh(π) }/π
an = (-1)n(eπ + e-π) /π(1+n2) = (-1)n2sinh(π) /π(1+n2) .....using integration by parts
bn = - nan = (-1)n+12nsinh(π) /π(1+n2) .....using integration by parts

Hence,
ex » {sinh(π) /π} + Σ k=1 to ∞ { (-1)n2sinh(π) cos(nx) /π(1+n2) } + Σ k=1 to ∞ { (-1)n+12nsinh(π) sin(nx) /π(1+n2) }

Now,
1= e0 » {sinh(π) /π} + Σ k=1 to ∞ { (-1)n2sinh(π) cos(0) /π(1+n2) } +0
1 = {sinh(π) /π} + Σ k=1 to ∞ { (-1)n2sinh(π) /π(1+n2) }
π = sinh(π) + Σ k=1 to ∞ { (-1)n2sinh(π) /(1+n2) }
{π - sinh(π) }/2sinh(π) = Σ k=1 to ∞ { (-1)n /(1+n2) }
{π/sinh(π) -1 }/2 = Σ k=1 to ∞ { (-1)n /(1+n2) }

On the other hand, noticing that (-1)n2sinh(π) cos(nπ) /π(1+n2) = 2sinh(π) /π(1+n2) , we must have Dirichlet conditions in order to obtain the Full Fourier Approximation at the end points. We apply the method of shifting data.

Suppse F(x)= ex - (Ax+B) and F(π)=F(-π)= 0 , then
F(x) = ex - ( {sinh(π) /π}x + cosh(π) ) by solving the boundary conditions.
F(x)= a'0 /2 + Σ k=1 to ∞ { a'n cos(nx) } + Σ k=1 to ∞ { b'n sin(nx) }

Now,
a'0 = {2sinh(π) }/π - 2cosh(π)
a'n = an + {integration of odd function} + {integration of periodic function} = an +0 +0 = (-1)n2sinh(π) /π(1+n2)

Hence,
F(x)= {sinh(π) }/π - cosh(π) + Σ k=1 to ∞ { (-1)n2sinh(π) cos(nx) /π(1+n2) } + Σ k=1 to ∞ { b'n sin(nx) }
0= F(π)= {sinh(π) }/π - cosh(π) + Σ k=1 to ∞ { (-1)n2sinh(π) cos(nπ) /π(1+n2) } + Σ k=1 to ∞ { b'n sin(nπ) }
0= {sinh(π) }/π - cosh(π) + Σ k=1 to ∞ { 2sinh(π) /π(1+n2) } + 0
0= 1 - {π/tanh(π) } + Σ k=1 to ∞ { 2 /(1+n2) }
( {π/tanh(π) } - 1)/2 = Σ k=1 to ∞ { 1 /(1+n2) }



圖片參考：http://i175.photobucket.com/albums/w130/bjoechan2003/My%20Cat%2020070228/DSCN0687.jpg?t=1176005263