An easily stated theorem sucient for many

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Unformatted text preview: e Fourier coeﬃcients. For simplicity, we will restrict our attention to the case where the period T = 2π , so that a0 f (t) = + a1 cos t + a2 cos 2t + . . . + b1 sin t + b2 sin 2t + . . . . (3.2) 2 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 dt + 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 conclude that π −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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