Mathematic Methods HW Solutions 31

# Mathematic Methods HW Solutions 31 - Chapter 7 7.5 31 2 an...

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Unformatted text preview: Chapter 7 7.5 31 2 an = − nπ sin nπ 2 bn = cn = a 0 / 2 = c0 = 0 nπ 1 nπ 2 cos 2 − 1 −inπ/2 2inπ 2e c− n = cn 4 1 − cos nπ = − nπ {0, 1, 0, 0, and repeat} − 1 − e−inπ = 1 nπ {−1, 2i, 1, 0, and repeat }, n > 0 1 f (x) = − π eix + e−ix − 7.6 7.7 2i 2ix − e−2ix − 1 (e3ix + e−3ix ) 2e 3 2i 6ix 1 5ix −5ix ) − 6 (e − e−6ix ) · · · + 5 (e + e 2 1 1 = − π cos x − 3 cos 3x + 5 cos 5x · · · 1 1 41 − π 2 sin 2x + 6 sin 6x + 10 sin 10x · · · 1 inx 2 1 (n = ±2, ± 6, ± 10, · · · ) f (x) = 2 + iπ ne 1 1 4 =2+π (n = 2, 6, 10, · · · ) n sin nx ∞ ∞ i inx 1 i π einx + + e f (x) = − 2π 4 n 2n 2n −∞ −∞ even n=0 odd n ∞ ∞ n = 7.8 π − 4 f (x) = 1 + 1 ∞ −∞ n=0 7.9 f (x) = 7.10 f (x) = 2 π − 2 π π 2 + 2 π (−1) 2 sin nx − n π (−1)n ∞ −∞ odd n ∞ −∞ odd n i inx e = 1+2 n einx π 4 =− n2 2 π einx π 4 =+ n2 2 π eix − e−ix 1 1 − 7.11 f (x) = + π 4i π = 1 odd n ∞ 1 2 1 + sin x − π 2 π ∞ −∞ even n=0 ∞ 2 even n 1 cos nx n2 (−1)n+1 1 ∞ 1 sin nx n cos nx n2 1 odd n ∞ cos nx n2 1 odd n einx n2 − 1 cos nx n2 − 1 1 7.13 an = 2 Re cn , bn = −2 Im cn , cn = 2 (an − ibn ), c−n = 1 (an + ibn ) 2 8.1 8.2 f (x) = 1 i + 2π ∞ −∞ odd n 2 1 inπx/l 1 e =− n 2π ∞ 1 odd n 1 nπx sin n l 1 nπ 1 nπ 1 nπ sin 2 , bn = nπ 1 − cos 2 , a0 /2 = c0 = 4 i −inπ/2 − 1) 2nπ (e 1 = 2nπ {1 − i, −2i, −(1 + i), 0, and repeat}, n > 0; c−n = cn 1 f (x) = 4 + 21 (1 − i)eiπx/l + (1 + i)e−iπx/l − 2i (e2iπx/l − e−2iπx/l ) π 2 − 1+i e3iπx/l − 1−i e−3iπx/l + 1−i e5iπx/l + 1+i e−5iπx/l · · · 3 3 5 5 1 1 = 1 + π cos πx − 3 cos 3πx + 1 cos 5πx · · · 4 l l 5 l 1 2 2 + π sin πx + 2 sin 2πx + 1 sin 3πx + 1 sin 5πx + 6 sin 6πx · · · l l 3 l 5 l l an = cn = ...
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