pset_5sol

n2 2 2mx l dx similarly multiplying both sides by sin

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Unformatted text preview: x2 /L2 )dx = , 3 1 L L 0 Bn = 2nπx 2L dx f (x) cos L0 L 2L 2nπx = (1 − x2 /L2 ) cos L0 L 2 −2nπ cos(2nπ ) = L L 4n3 π 3 −1 . = n2 π 2 2mπx L dx Similarly, multiplying both sides by sin using orthogonality again yields L and integrating over 0 < x < L and 2mπx L f (x) sin 0 2mπx L L = Am 0 sin2 dx = Am L/2. And so for all n = 1, 2, ..., An = 2L f (x) sin L0 1 = . nπ 2nπx L dx Thus, the fully periodic series is given by 2 2nπx (−1) f (x) = + cos 2π2 3 n=1 n L SOLUTION (d) : Clearly by the above results, at x = 0, f (0) = 0 for the sine series, 3 ∞ ∞ + n=1 1 sin nπ 2nπx L . The sine series approximates an odd (periodic) extension of (1 − x2 /L2 ) to (−L, 0). Thus, from the left the value is -1 and from the right it is +1, and the average of the two is 0 (as expected from the theorem on uniform convergence). For the cosine series, at x = 0, we have (recalling series approximating π , e.g. Riemann Zeta function), ∞ n=1 −4(−1)n 4 4 = 2 (1 + 1/32 +...
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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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