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Unformatted text preview: θ =
Hence the Fourier expansion of f can be rewritten as
cos θ = f (t) = a0
+ (eit + e−it ) + (e2it + e−2it ) + . . .
+ b1 it
(e − e−it ) + (e2it − e−2it ) + . . . ,
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
77 c2 = a2 − ib2
2 ... , (3.18) a1 + ib1
a2 + ib2
, c−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,
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
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.
- Winter '14