# 344 show that if a function f r r is smooth and

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Unformatted text preview: os(nπ/2) sin(nπ/2) 1 . − 2 [sin nt]|π 2 = + π/ 2n n 2n n2 Thus bn = 4 sin(nπ/2) , πn2 74 1.4 1.2 1 0.8 0.6 0.4 0.2 1 0.5 1.5 2 3 2.5 Figure 3.3: A graph of the Fourier sine approximations φ1 , φ3 , φ5 and φ7 . and the Fourier sine series of f (t) is f (t) = 4 4 4 4 sin t − sin 3t + sin 5t − sin 7t + . . . . π 9π 25π 49π (3.16) The trigonometric polynomials φ1 (t) = 4 sin t, π 4 4 sin t − sin 3t, . . . π 9π φ3 (t) = are better and better approximations to the function f (t). To ﬁnd the Fourier cosine series of t, for 0 ≤ t ≤ π/2, π − t, for π/2 ≤ t ≤ π , f (t) = we ﬁrst note that a0 = π /2 2 π π (π − t)dt = tdt + 0 π/2 21 π2 π 2 2 + 1 2 π 2 To ﬁnd the other an ’s, we use (3.14): an = 2 π π /2 π (π − t) cos ntdt . t cos ntdt + 0 π/2 This time, integration by parts yields π /2 t cos ntdt = 0 t sin nt n 75 π /2 π /2 − 0 0 1 sin ntdt n 2 = π . 2 = 1 π sin(nπ/2) π/2 + 2 [cos nt]|0 2n n π sin(nπ/2) cos(nπ/2) 1 + = −2 2n n2 n while π (π − t) cos ntdt = π/2 (π − t) sin(nt) n π π + π/2 π/2 1 sin ntdt n −π sin(nπ/2) 1 − 2 [cos...
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## This document was uploaded on 01/12/2014.

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