pset_5sol

# n putting it all together we nd that b1 bn 2 0 1 cos

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Unformatted text preview: s(x) sin(nx)dx cos(x) sin(nx)dx 0 π =⇒ cos(x) sin(nx)dx = =⇒ cos(x) sin(nx)dx = n(cos(nπ ) + 1) . n2 − 1 −n[cos(nπ ) + 1] (1 − n2 ) 0 π 0 We also have that for all n ≥ 1, π 0 1 sin(nx)dx = − cos(nx) n 5 π 0 = 1 − cos(nπ ) n and by a single integration by parts 2 π π 0 x sin(nx)dx = x cos(nx) π 1 π 2 + − cos(nx)dx 0 π n n0 π cos(nπ ) 1 2 π − + 2 sin(nx) 0 = π n n 2 cos(nπ ) . =− n Putting it all together we ﬁnd that b1 = bn 2 [0 − 1 + cos(π ) − 2 cos(π )] = 0 π 2 n(cos(nπ ) + 1) cos(nπ ) 1 2 cos(nπ ) = − −− π n2 − 1 n n n 2 1 + cos(nπ ) = π n(n2 − 1) for n ≥ 2. Thus, the sine series is given by 2 f (x) = π ∞ n=2 1 + cos(nπ ) n(n2 − 1) sin(nx)dx, x ∈ (0, π ). Clearly, the coefﬁcients decay like 1/n3 . This makes sense because the odd extension x of f (x) = cos(x) − 1 + 2π must have a discontinuous n − 1 = 2 derivative since integration by parts can was repeated at most twice. This is easily conﬁrmed by noting that at x = 0, x→0+ lim f (2) (x) = lim − cos(x) = −1, +...
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## This note was uploaded on 12/12/2009 for the course ACM 95b taught by Professor Nilesa.pierce during the Winter '09 term at Caltech.

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