pset_5sol

# L solution b similarly we are looking for a series of

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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 &lt; x &lt; 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 &lt; x &lt; L and using 2 (1...
View Full Document

Ask a homework question - tutors are online