We will see that just as in the case of ordinary

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Unformatted text preview: lex Fourier coeﬃcients of the function f (t) = t for −π &lt; t ≤ π, extended to be periodic of period 2π . Then ck = 1 2π π te−ikt dt. −π We apply integration by parts with u = t, dv = e−ikt dt, du = dt and v = (i/k )e−ikt : π 1 ck = (i/k )e−ikt dt (it/k )e−ikt |π π − − 2π −π = 1 π i −iπk πi iπk (−1)k =i +e e . 2π k k k It follows from (3.19) that ak = (ck + c−k ) = 0, bk = i(ck − c−k ) = −2 (−1)k . k It is often the case that the complex form of the Fourier series is far simpler to calculate than the real form. One can then use (3.19) to ﬁnd the real form of the Fourier series. Exercise: 3.4.1. Prove the integral formula (3.20), presented in the text. 78 3.4.2. a. Find the complex Fourier coeﬃcients of the function for −π &lt; t ≤ π, f (t) = t2 extended to be periodic of period 2π . b. Use (3.19) to ﬁnd the real form of the Fourier series. 3.4.3. a. Find the complex Fourier coeﬃcients of the function f (t) = t(π − t) for −π &lt; t ≤...
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This document was uploaded on 01/12/2014.

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