Mathematic Methods HW Solutions 33

# Mathematic Methods HW Solutions 33 - Chapter 7 33 8.11(a...

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Unformatted text preview: Chapter 7 33 8.11 (a) f (x) = (b) f (x) = ∞ ∞ π2 π2 (−1)n inx +2 e = +4 3 n2 3 −∞ n=0 ∞ 4π 2 +2 3 −∞ 1 iπ + n2 n (−1)n cos nx n2 1 4π 2 +4 3 einx = ∞ 1 n=0 8.12 (a) f (x) = = ∞ (−1)n (1 + in) inx e 1 + n2 −∞ sinh π π sinh π 2 sinh π + π π e2π − 1 (b) f (x) = 2π = ∞ (b) f (x) = 2 iπ −∞ n=0 ∞ −∞ n=0 ∞ 8 8.14 (a) f (x) = π ∞ 2 iπ 1 8.15 (a) f (x) = (b) f (x) = (c) f (x) = 8.16 f (x) = 1 − i π ∞ 1 (−1)n inπx/2 4 e =2+ n π ∞ ∞ 1 nπx (−1)n sin n 2 1 nπx sin n 2 1 ∞ 2 (−1) inπx e = n π 1 inπx 8 e =2 2 n π 1 n+1 sin nπx (−1) 1 odd n ∞ sin nπx 1 =1− n iπ 1 e2inπx −1 4 n2 −∞ ∞ 8 1 inπx e =3 n3 π −∞ odd n ∞ ∞ 2 cos 2nπx =− 2−1 4n π n −∞ odd n ∞ 1 1 1 (cos nx − n sin nx) 1 + n2 ∞ −4i π3 2 π ∞ n(−1)n+1 n(−1)n 2inπx 4i sin 2nπx = e 4 n2 − 1 π −∞ 4 n 2 − 1 −∞ n=0 ∞ 4 π2 1 (−1)n (cos nx − n sin nx) 1 + n2 4 1 inπx/2 e = n π 4 2 (b) f (x) = − π π ∞ ∞ 1 + in inx e 1 + n2 −∞ e2π − 1 1 + π 2 8.13 (a) f (x) = 2 + −∞ n=0 100 100 +2 3 π ∞ ∞ 1 100 = + 50 3 −∞ sin nπx . n3 1 inπx e n 1 nπx 100 cos − 2 n 5 π 1 1 − n2 π 2 inπ e . cos nπx . n2 1 odd n ∞ n 1 3 8.17 f (x) = 4 − π (cos πx − 1 cos 3πx + 1 cos 5πx · · · ) 2 3 2 5 2 1 + π (sin πx + 2 sin πx + 1 sin 3πx + 1 sin 5πx + 2 2 3 2 5 2 8.18 f (x) = cos nx − 4π n2 ∞ 1 2 6 sin 3πx · · · ) 1 nπx sin n 5 inπx/5 n=0 8.19 f (x) = 1 1 − 8 π2 ∞ 1 odd n 1 1 cos 2nπx + n2 2π ∞ 1 (−1)n+1 sin 2nπx n ∞ 1 sin nx n ...
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## This note was uploaded on 02/29/2012 for the course MHF 2312 taught by Professor Dr.chet during the Fall '11 term at UNF.

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