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Hence the series in 5a and the series in 5c which can

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Unformatted text preview: ∞ (5c) −1 m=2 ∞ ζm −1 m m−1 ∞ Solution for (5a) and (5c) : m=2 . Recall the result of (2a): ∞ −f x + g x (5d) , , ζm −1 m · 2m m=2 ζm −1 m · 2m m−1 ζm −1 m (5b) , = −1 m=2 ζm −1 m m−1 · xm . Hence the series in (5a), and the series in (5c), which can be rewritten as ∞ m=2 −1 ∞ m=2 −1 m−1 ζm −1 m m−1 ζm −1 m · 1m , 1 2 · and m , respectively, are nothing else but −f 1 + g 1 , and respectively. Accordingly, substitute −f x + g x and obtain 1 − γ − ln 2, = and 1 2 −f x=1 1−γ 1−γ 2 13 and x− − ln +g 1 2 x= 1 2 , into ln Γ 2 + x 3√ π, 4 respectively. Solution for (5b) and (5d) : Recall the result of (2b): ∞ f x +g x = m=2 ζm −1 m · xm . Hence the series in (5b), and the series in (5d), which can be rewritten as ∞ ζm −1 m ζm −1 m m=2 · 1m , · 1 2 f 1 2 ∞ m=2 and m , respectively, are nothing else but f 1 +g 1, and x=1 respectively. Accordingly, substitute f x +g x and obtain 1 − γ, and = 1−γ 1−γ 2 x+ + ln 14 1 2 +g and x= , 1 2 into ln Γ 2 − x 1√ π, 2 respectively. Answers for (5a), (5b), (5c) and (5d) : ∞ (5a) m=2 ∞ (5b) m=2 −1 m=2 ∞ (5d) m=2 ζm −1 m ζm −1 m ∞ (5c) m−1 −1 m−1 = 1 − γ − ln 2, = 1 − γ, ζm −1 m · 2m 15 1−γ 2 − ln 3√ π, 4 = ζm −1 m · 2m = 1−γ 2 + ln 1√ π. 2...
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This note was uploaded on 01/12/2014 for the course MATH 116 taught by Professor Scholle,minho during the Fall '08 term at Kansas.

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