Find the fourier series of this extension 314 the

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Unformatted text preview: 1 π π f (t) cos ktdt. (3.7) −π A very similar argument yields the formula bk = 1 π π f (t) sin ktdt. (3.8) −π Example. Let us use formulae (3.3), cients of the function −π, f (t) = π, 0, (3.7), and (3.8) to find the Fourier coeffifor −π < t < 0, for 0 < t < π , for t = 0, π , extended to be periodic of period 2π . Then a0 = average value of f = 0, 2 and am = 1 π π f (t) cos mtdt = −π = 1 π 0 −π −π cos mtdt + 1 π π π cos mtdt 0 1 −π 1π [sin mt]0 π + [sin mt]π = · · · = 0, − 0 πm πm while bm = 1 π π f (t) sin mtdt = −π 1 π 0 −π 65 −π sin mtdt + 1 π π π sin mtdt 0 3 2 1 -4 -2 4 2 -1 -2 -3 Figure 3.1: A graph of the Fourier approximation φ5 (t) = 4 sin t + (4/3) sin 3t + (4/5) sin 5t. = 1π 1 −π 2 2 [cos mt]0 π + [cos mt]π = − cos mπ − 0 πm πm mm = 2 (1 − (−1)m ) = m 4 m, 0, if m is odd, if m is even. Thus we find the Fourier series for f : f (t) = 4 sin t + 4 4 sin 3t + sin 5t + · · · . 3 5 The trigonometric polynomials φ1 (t) = 4 sin t, φ3 (t) = 4 sin t +...
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This document was uploaded on 01/12/2014.

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