# 7(a given f x = ∞ summationdisplay n =-∞ c n e

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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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