pset_5sol

L solution b similarly we are looking for a series of

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Unformatted text preview: that ∞ f (x) = n=1 2 n3 π 3 [2 + n2 π 2 − 2(−1)n ] sin nπx . L SOLUTION (b) : Similarly, we are looking for a series of the form ∞ f (x) = n=0 Bn cos nπx . L Multiplying both sides by cos L 0 mπx L and integrating over the interval gives ∞ L mπx dx = f (x) cos L Bn n=0 0 cos mπx nπx cos dx. L L But by the orthogonality of the sine series, B0 = L 1 L L L f (x)dx 0 f (x) cos 0 mπx dx = Bm L L cos 0 L mπx nπx cos dx = Bm . L L 2 And so for n = 0,1,2,... 1 B0 = L (1 − x2 /L2 )dx = 0 2 3 Bn = 2L nπx dx f (x) cos L0 L 2L nπx = dx (1 − x2 /L2 ) cos L0 L 2 −2Lnπ cos(nπ ) = L n3 π 3 −4(−1)n = n2 π 2 ∞ so that nπx −4(−1)n 2 . cos f (x) = + 2π2 3 n=1 n L SOLUTION (c) : 2 To develop the fully periodic series, we want these functions to be periodic on 0 < x < L. That is, we look for ∞ f (x) = n=0 Bn cos 2nπx L ∞ + n=1 An sin 2nπx L . Multiplying both sides by cos orthogonality again yields B0 = 2mπx L and integrating over 0 < x < L and using 2 (1...
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