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Fourier series: |x|

Short answer

Since |x| is even, b_n=0. a_0 = pi, a_n = -4/(pi n^2) for odd n, 0 for even n. This gives |x| = pi/2 - (4/pi)[cos x/1^2 + cos 3x/3^2 + ...]. Setting x=0 yields pi^2/8 = 1/1^2 + 1/3^2 + 1/5^2 + ...

Find the Fourier series of the function f(x) = |x| for −π ≤ x ≤ π and hence deduce that π²/8 = 1/1² + 1/3² + 1/5² + ...

f(x) = |x| on (−π, π) is an even function, so bₙ = 0.

Coefficients

a₀ = (2/π) ∫₀^π x dx = (2/π)(π²/2) = π
aₙ = (2/π) ∫₀^π x cos nx dx
   = (2/π) [ x sin nx/n + cos nx/n² ]₀^π
   = (2/(πn²)) [ (−1)ⁿ − 1 ]

So aₙ = 0 for even n, and aₙ = −4/(πn²) for odd n.

Series

|x| = π/2 − (4/π) [ cos x/1² + cos 3x/3² + cos 5x/5² + ... ]

Deduction

|x| is continuous at x = 0 with f(0) = 0:

0 = π/2 − (4/π) [ 1/1² + 1/3² + 1/5² + ... ]

Hence π²/8 = 1/1² + 1/3² + 1/5² + ... ∎

✓ Answer checked by Claude Opus 5.5 (high effort)

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