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Unformatted text preview: 7. (a) Given f ( x ) = ∞ summationdisplay n =∞ C n e inx , we want to mimic the approach suggested in class for finding the “ a k ”. To do this we multiply both sides by e ikx giving f ( x ) e ikx = ∞ summationdisplay n =∞ C n e inx e ikx and integrate from π to π giving integraldisplay π π f ( x ) e ikx dx = integraldisplay π π parenleftBigg ∞ summationdisplay n =∞ C n e inx e ikx parenrightBigg dx term by term = theorem ∞ summationdisplay n =∞ integraldisplay π π C n e i ( n k ) x dx . If n = k , we have integraldisplay π π C k e dx = 2 π C k . If n negationslash = k , we have integraldisplay π π C n e i ( n k ) x dx = C n bracketleftbigg i e i ( n k ) x n k bracketrightbigg π π = i C n n k ( e i ( n k ) π e i ( n k ) π ) = 0. Hence integraldisplay π π f ( x ) e ikx dx = 2 π C k = ⇒ C k = 1 2 π integraldisplay π π f ( x ) e ikx dx . (b) Since sin kx = e ikx e ikx 2 i , we have b k = 1 π integraldisplay π π f ( x ) sin kx dx = 1 π integraldisplay π π f ( x ) parenleftbigg e ikx e ikx 2 i parenrightbigg dx = 1 i parenleftbigg 1 2 π integraldisplay π π f ( x ) e ikx dx 1 2 π integraldisplay π π f ( x ) e ikx dx parenrightbigg = C k C k i = i ( C k C k ), k ≥ 1. In a similar way, we use cos kx = e ikx + e ikx 2 to get a k = 1 π integraldisplay π π f ( x ) parenleftbigg e ikx + e ikx 2 parenrightbigg dx = C k + C k , k ≥ 1. MATB42H Solutions # 2 page 6 Also, a = 1 π integraldisplay π π f ( x ) dx = 2 C . (c) From part (b) we have a = 2 C , so C = 1 2 a . Also, from part (b), we have a k = C k + C k 1 i b k = C k C k . Adding gives 2 C k = a k + 1 i b k = ⇒ C k = 1 2 ( a k i b k ), k ≥ 1. Subtracting the second from the first gives 2 C k = a k 1 i b k = ⇒ C k = 1 2 ( a k + i b k ), k ≥ 1....
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 Winter '10
 EricMoore
 Math, Multivariable Calculus

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