✔ 最佳答案
Let f(x) = x^2 for –π< x <= π and f(x + 2π) = f(x)
By means of Fourier series, since f is an even function, we have
∞
f(x) = a0 / 2 + Σ an cos nx
1
π π π
where a0 = 1/π∫ x^2 dx = 2/π∫ x^2 dx = 2/π x^3/3│= 2π^2/3
-π 0 0
π
and an = 1/π∫ x^2 cos nx dx
-π
π
= 2/π∫ x^2 cos nx dx
0
π
= 2/nπ∫ x^2 d sin nx
0
π π
= 2/nπ(x^2 sin nx│ - ∫ sin nx dx^2 )
０ ０
π
= -4/nπ∫ x sin nx dx
0
π
= 4/n^2π∫ x d cos nx
0
π π
= 4/n^2π(x cos nx│ - ∫ cos nx dx )
0 0
π
= 4/n^2π(πcos nπ - sin nx/n│ )
0
= 4cos nπ/n^2
= 4(-1)^ n / n^2
Therefore
∞
f(x) = π^2 / 3 + Σ4(-1)^ n / n^2 cos nx
1
Put x = π,
∞
f(π) = π^2 / 3 + Σ4(-1)^ n / n^2 cos nπ
1
∞
π^2 = π^2 / 3 + Σ4 / n^2
1
∞
Σ1 / n^2 = π^2 / 6
1
