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Fourier series: x²

Short answer

x^2 is even, so b_n=0. a_0 = 2pi^2/3, a_n = 4(-1)^n/n^2. The Fourier series is x^2 = pi^2/3 + 4 sum_{n=1}^infinity (-1)^n cos(nx)/n^2. Setting x=0 gives pi^2/12 = 1/1^2 - 1/2^2 + 1/3^2 - ...

Find the Fourier series for the function f(x) = x² in the interval −π ≤ x ≤ π and deduce that π²/12 = 1/1² − 1/2² + 1/3² − ...

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

Coefficients

a₀ = (2/π) ∫₀^π x² dx = 2π²/3
aₙ = (2/π) ∫₀^π x² cos nx dx
   = (2/π) [ x² sin nx/n + 2x cos nx/n² − 2 sin nx/n³ ]₀^π
   = (2/π)(2π (−1)ⁿ/n²) = 4(−1)ⁿ/n²

Series

x² = π²/3 + 4 Σ_{n=1}^∞ (−1)ⁿ cos nx / n²
   = π²/3 − 4cos x + cos 2x − (4/9) cos 3x + ...

Deduction

Put x = 0 (x² is continuous there, value 0):

0 = π²/3 + 4 [ −1/1² + 1/2² − 1/3² + ... ]
4 [ 1/1² − 1/2² + 1/3² − ... ] = π²/3

Hence π²/12 = 1/1² − 1/2² + 1/3² − 1/4² + ... ∎

✓ Answer checked by Claude Opus 5.5 (high effort)

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