Unformatted text preview: 1/52 + ...) − 2 (1/22 + 1/42 + 1/62 + ...) n2 π 2 π π = 4 1 42 (π /8) − 2 (π 2 /24) = 2 π π 3 = 1, i.e. so that f (0) = 2 3 + ∞ −4(−1)n n=1 n2 π 2 f (0) = 1 for the cosine series. The cosine series approximates an even (periodic) extension of (1−x2 /L2 ) to (−L, 0). Thus, from the left the value is 1 and from the right it is still 1 hence giving the average value of 1. This is the correct value because a continuous extension of (1−x2 /L2 ) to (−L, 0) is naturally even. For the fully periodic series, at x = 0, we have that (using Riemann Zeta with para2 mater 2) f (0) = 3 − ∞ n21π2 = 2/3 − 1/π 2 (π 2 /6) = 1/2, i.e. n=1 f (0) = 1/2 for the fully periodic series. The full series makes (1 − x2 /L2 ) become periodic with a period of length L (e.g. a sawtooth shape). Thus from the left the value is zero and from the right +1 – we get the average value of 1/2. Problem 2 Compute the Fourier sine series of the function f (x) = cos x − 1 + 2x π 0<x<π How fast for the Fourier coefﬁcients decrease with increasing...
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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.
 Winter '09
 NilesA.Pierce

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