Unformatted text preview: e Fourier coeﬃcients. For
simplicity, we will restrict our attention to the case where the period T = 2π ,
f (t) =
+ a1 cos t + a2 cos 2t + . . . + b1 sin t + b2 sin 2t + . . . .
The formulae for a general period T are only a little more complicated, and are
based upon exactly the same ideas.
The coeﬃcient a0 is particularly easy to evaluate. We simply integrate both
sides of (3.2) from −π to π :
π π f (t)dt =
−π −π π a0
2 −π π a1 cos tdt + π +
−π −π a2 cos 2tdt + . . . π b1 sin tdt + −π b2 sin 2tdt + . . . . Since the integral of cos kt or sin kt over the interval from −π to π vanishes, we
−π f (t)dt = πa0 ,
63 and we can solve for a0 , obtaining
a0 = 1
π π f (t)dt. (3.3) −π To ﬁnd the other Fourier coeﬃcients, we will need some integral formulae.
We claim that if m and n are positive integers,
π cos nt cos mtdt = π , for m = n,
0, for m = n, (3.4) sin nt sin mtdt = π , for m = n,
0, for m = n, (3.5) −π π
−π π sin nt cos mtdt = 0. (3.6) −π Let us verify the ﬁrst of these equations. We will use the trigonometric...
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- Winter '14