obtain the Fourier Series expansion of the following function : f(x) = π+x for -π<x<0 & f(x) = π-x for 0<x<π?
obtain the Fourier Series expansion of the following function : f(x) = π+x for -π<x<0 & f(x) = π-x for 0<x<π
When i clculated then coefficients are as written below:
a ° = π
a^n = 0
and
b^n = 0
But, this the fourier series solution in expanded form becomes
f(x) = π + 0 +0
f(x) = π
but i think this solution may not be correct.
Can you people help me, how can we "estimate" that the solution i obtained is correct.
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I assume,
f(x) = π + x for -π ≤ x ≤ 0 & f(x) = π - x for 0 ≤ x ≤ π
f(x) = Ao + Σ(An sin nx + Bn cos nx)
for n = 1, 2, 3, . . .∞
. . . . . . . . . .0. . . . . . . . . .. . . . . π
Ao = 1/(2π) ∫ (π + x) dx + 1/(2π) ∫ (π - x) dx
. . . . . . . . -π. . . . .. . . . . . . . . . .0
Ao = ½ π
. . . . . . . . . .0. . . . . . . . . .. . . . . . . . .π
An = (1/π) ∫ (π + x) sin nx dx + (1/π) ∫ (π - x) sin nx dx
. . . . . . . . -π. . . . .. . . . . . . . . . .. . . 0
An = 0
. . . . . . . . . .0. . . . . . . . . .. . . . . . . . .π
Bn = (1/π) ∫ (π + x) cos nx dx + (1/π) ∫ (π - x) cos nx dx
. . . . . . . . -π. . . . .. . . . . . . . . . .. . . 0
Bn = (4/(n²π)) sin² (½ nπ)
f(x) = Ao + Σ(An sin nx + Bn cos nx)
f(x) = ½ π + Σ((4/(n²π)) sin² (½ nπ) cos nx)
for n = 1, 2, 3, . . .
This is an odd-numbered Fourier-series, with summations only building when n = [1, 3, 5, ..., etc.].
Series representation:
Ï/2 + â (4*cos(n*x) / n^2*Ï), from n = [1, â]
If you plot f, you'll see that it is even. So all of the sine coefficients will be zero. However, we certainly don't expect all of the a_n to be zero. We can use the symmetry
a_n = 1/pi int(x=-pi to pi) f(x) cos(nx) dx , n = 1,2,...
= 2/pi int(x=0 to pi) f(x) cos(nx) dx
Using the fact that f(x) cos(nx) is even for all n.
I agree with your finding of a_0 = pi.
For a_n, I get
a_n = 2/(pi n^2) [1 - (-1)^n]
I used integration by parts. That integral again is
a_n = 2/pi int(x=0 to pi) (pi-x) f(x) dx