pset_5sol

# Clearly in this series a0 3 but n1 n2 a0 1 2l l x2

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Unformatted text preview: x→ 0 x→0− lim −f (2) (−x) = lim cos(x) = 1. − x→ 0 Problem 3 (a) For what values of x does the Fourier series π2 (−1)n +4 cos nx = x2 ? 3 n2 n=1 (b) What is the value of the Fourier series for all x? (c) From the relation above, show the identities ∞ ∞ n=1 ∞ 1 π2 = , n2 6 n=1 π2 (−1)n+1 = . n2 12 6 SOLUTION (a) : We have a cosine series which requires an even extension. Since x2 is already even, 2 we have that f (x) = x2 on some interval [−L, L] whose cosine series gives π3 + n 2 4 ∞ (−1) cos nx. Clearly in this series, a0 = π3 . But n=1 n2 a0 = 1 2L L x2 dx = −L 1 x3 2L 3 L −L = L3 − (−L)3 L2 = . 3 3 So by the uniqueness of the cosine series for an even f (x), we have that L = π , i.e. the series converges to x2 for −π ≤ x ≤ π. SOLUTION (b) : By the 2π -periodic nature of cos(nx), any x outside of the interval [−π, π ] will return a value equal to the value returned by an x0 = x − 2πm ∈ [−π, π ] for some m = 0, ±1, ±2, ... That is, the value...
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## This note was uploaded on 12/12/2009 for the course ACM 95b taught by Professor Nilesa.pierce during the Winter '09 term at Caltech.

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