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Fourier series: x²

Short answer

x^2 is even, b_n=0. a_0=2pi^2/3, a_n=4(-1)^n/n^2, giving x^2 = pi^2/3 + 4 sum_{n=1}^infinity (-1)^n cos(nx)/n^2. Setting x=pi yields sum_{n=1}^infinity 1/n^2 = pi^2/6.

Obtain the Fourier series for f(x) = x² in the interval −π < x < π and hence prove that Σ 1/n² = 1/1² + 1/2² + 1/3² + ... = π²/6.

f(x) = x² on (−π, π) is even, so bₙ = 0.

a₀ = (2/π) ∫₀^π x² dx = 2π²/3
aₙ = (2/π) ∫₀^π x² cos nx dx = 4(−1)ⁿ/n²

(by parts: ∫₀^π x² cos nx dx = [x² sin nx/n + 2x cos nx/n² − 2 sin nx/n³]₀^π = 2π(−1)ⁿ/n².)

Fourier series

x² = π²/3 + 4 Σ_{n=1}^∞ (−1)ⁿ cos nx / n²

Deduction

x² is continuous at x = π (f(π) = f(−π) = π²). Put x = π, cos nπ = (−1)ⁿ, so (−1)ⁿ(−1)ⁿ = 1:

π² = π²/3 + 4 Σ 1/n²
4 Σ 1/n² = 2π²/3

Hence Σ 1/n² = 1/1² + 1/2² + 1/3² + ... = π²/6. ∎

✓ Answer checked by Claude Opus 5.5 (high effort)

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