Mathematic Methods HW Solutions 30

Mathematic Methods HW Solutions 30 - Chapter 7 30 5.1 to...

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Unformatted text preview: Chapter 7 30 5.1 to 5.11 The answers for Problems 5.1 to 5.11 are the sine-cosine series in Problems 7.1 to 7.11. x→ −2π −π −π/2 0 π/2 π 2π 6.1 1/2 1/2 1 1/2 0 1/2 1/2 6.2 1/2 0 0 1/2 1/2 0 1/2 6.3 0 1/2 0 0 1/2 1/2 0 6.4 −1 0 −1 −1 0 0 −1 6.5 −1/2 1/2 0 −1/2 0 1/2 −1/2 6.6 1/2 1/2 1/2 1/2 1/2 1/2 1/2 6.7 0 π/2 0 0 π/2 π/2 0 6.8 1 1 1 1 6.9 0 π π/2 0 π/2 π 0 6.10 π 0 π/2 π π/2 0 π 6.11 0 0 0 0 1 0 0 6.13 and 6.14 7.1 7.2 7.3 π 2 1 1+ π 2 At x = π/2, same series as in the example. 1 i f (x) = + 2π ∞ −∞ odd n 1 inx 1 2 e =− n 2π ∞ 1 odd n 1 sin nx n 1 1 an = nπ sin nπ , bn = nπ 1 − cos nπ , a0 /2 = c0 = 1 , 2 2 4 i −inπ/2 cn = 2nπ (e − 1), n > 0; c−n = cn 1 f (x) = 4 + 21 (1 − i)eix + (1 + i)e−ix − 2i (e2ix − e−2ix ) π 2 − 1+i e3ix − 1−i e−3ix + 1−i e5ix + 1+i e−5ix · · · 3 3 5 5 1 1 1 1 = 4 + π cos x − 3 cos 3x + 5 cos 5x · · · 1 1 1 2 + π sin x + 2 sin 2x + 3 sin 3x + 5 sin 5x + 6 sin 6x · · · 2 1 an = − nπ sin nπ , a0 /2 = c0 = 2 1 bn = nπ i cn = 2nπ 1 f (x) = 4 7.4 1− cos nπ 2 −inπ e + 21 π 1 4 1 nπ {1, −2, 1, 0, − cos nπ = and repeat} −inπ/2 −e , n > 0; c−n = cn ix −(1 + i) e − (1 − i) e−ix + 2i (e2ix − e−2ix ) 2 + 1−i e3ix + 1+i e−3ix − 1+i e5ix − (1−i) e−5ix · · · 3 3 5 5 1 1 1 1 = 4 − π cos x − 3 cos 3x + 5 cos 5x · · · 1 1 1 + π sin x − 2 sin 2x + 3 sin 3x + 5 sin 5x · · · 2 c0 = a0 /2 = −1/2; for n = 0, coefficients are 2 times the coefficients in Problem 7.3. 1 1 f (x) = − 2 − π (1 + i)eix + (1 − i)e−ix − 2i (e2ix − e−2ix ) 2 1−i 3ix − 3 e − 1+i e−3ix + 1+i e5ix + 1−i e−5ix · · · 3 5 5 1 2 1 1 = − 2 − π cos x − 3 cos 3x + 5 cos 5x · · · 1 1 2 2 + π sin x − 2 sin 2x + 3 sin 3x + 5 sin 5x − 6 sin 6x · · · 2 ...
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