Instead each type of partial dierential equations

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Unformatted text preview: θ = . 2 2i Hence the Fourier expansion of f can be rewritten as cos θ = f (t) = a0 a1 a2 + (eit + e−it ) + (e2it + e−2it ) + . . . 2 2 2 + b1 it b2 (e − e−it ) + (e2it − e−2it ) + . . . , 2i 2i or f (t) = . . . + c−2 e−2it + c−1 e−it + c0 + c1 eit + c2 e2it + . . . , where the ck ’s are the complex numbers defined by the formulae c0 = a0 , 2 c1 = a1 − ib1 , 2 77 c2 = a2 − ib2 , 2 ... , (3.18) a1 + ib1 a2 + ib2 , c−2 = , 2 2 If k = 0, we can solve for ak and bk : c−1 = ... . bk = i(ck − c−k ). ak = ck + c−k , (3.19) It is not difficult to check the following integral formula via direct integration: π 2π, for m = n, 0 for m = n. eimt e−int dt = −π (3.20) If we multiply both sides of (3.18) by e−ikt , integrate from −π to π and apply (3.20), we obtain π −π f (t)e−ikt dt = 2πck , which yields the formula for coefficients of the complex form of the Fourier series: ck = 1 2π π f (t)e−ikt dt. (3.21) −π Example. Let us use formula (3.21) to find the comp...
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This document was uploaded on 01/12/2014.

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