We will see that just as in the case of ordinary

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: lex Fourier coefficients of the function f (t) = t for −π < 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 < 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 < t ≤...
View Full Document

This document was uploaded on 01/12/2014.

Ask a homework question - tutors are online