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Fourier series: x³

Short answer

Fourier series: f(x)=a0/2 + Σ(an cos nx + bn sin nx) with integrals for coefficients. For the odd function f(x)=x^3, a0=an=0, bn = 2(-1)^(n+1)(π^2/n - 6/n^3). The series is x^3 = 2 Σ (-1)^(n+1)(π^2/n - 6/n^3) sin nx for -π<x<π.

Define Fourier series for a function f(x). Obtain the Fourier series for f(x) = x³, −π ≤ x ≤ π.

Definition

A function f(x) satisfying the Dirichlet conditions in (−π, π) (single-valued, finitely many maxima/minima and finite discontinuities) has the Fourier series

f(x) = a₀/2 + Σ_{n=1}^∞ (aₙ cos nx + bₙ sin nx)
a₀ = (1/π) ∫_{−π}^{π} f(x) dx
aₙ = (1/π) ∫_{−π}^{π} f(x) cos nx dx
bₙ = (1/π) ∫_{−π}^{π} f(x) sin nx dx

f(x) = x³ on (−π, π)

x³ is an odd function, so a₀ = 0, aₙ = 0 and

bₙ = (2/π) ∫₀^π x³ sin nx dx

Integrate by parts repeatedly: ∫₀^π x³ sin nx dx = −π³(−1)ⁿ/n + 6π(−1)ⁿ/n³. So

bₙ = 2(−1)ⁿ⁺¹ [ π²/n − 6/n³ ]

Series

x³ = 2 Σ_{n=1}^∞ (−1)ⁿ⁺¹ (π²/n − 6/n³) sin nx
   = 2(π² − 6) sin x − 2(π²/2 − 6/8) sin 2x + 2(π²/3 − 6/27) sin 3x − ...

(The series converges to x³ for −π < x < π, and to 0 at x = ±π, the average of the end values.)

✓ Answer checked by Claude Opus 5.5 (high effort)

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