solutions for quiz xxix

This is equivalent to 2 x 2 13 note the power series

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Unformatted text preview: s merely the outcome of the substitution of u = x2 in the series (∗) above with variable u . We have just proved that the series (∗) is convergent for an arbitrary real number u with −4 < u < 4. It is then clear that the original series is convergent for an arbitrary real number x such that x2 < 4. This is equivalent to −2 < x < 2. 13 Note : The power series ∞ ζ 2n ζ 2n x2 n −1 −1 n=1 is clearly x times the derivative of ∞ (#) x2 n . 2n n=1 This latter power series (#) is identiﬁed as fx 1 2 = ln Γ 2−x Γ 2+x − 2 < x < 2 , where Γ x is the Euler’s Gamma function, as usual, or, alternatively, fx − 2 < x < 2; = 1 2 ln 1 − x2 x = ± 1, 0 . · πx sin πx The reason why these two expressions of f x are equal is explained below. Accordingly, the original power series ∞ ζ 2n n=1 is identiﬁed as x f ′ x , that is, 14 −1 x2 n x 2 − Γ ′ Γ 2−x Γ ′ 2+x + 2−x Γ 2+x cos πx sin πx − 2 < x < 2 , or, alternatively, 1 − 3 x2 2 − 2 x2 − 2 < x < 2; ⋆ − πx 2 · x = ± 1, 0 . The reason why f x has two diﬀerent expressions Γ 1+x : Recall = x ·Γx the fundamental property of the Euler’s Gamma function . Hence Γ 2−x = 1−x Γ 1−x , Γ 2+x = 1+x and ·x ·Γ x. Accordingly, fx = the ﬁrst expression of f x 1 2 ln Γ 2−x Γ 2+x is rewritten as 15 1 2 ln = 1−x 1 2 ln 1+x x ·Γ x Γ 1−x 1 − x2 · x · Γ x Γ 1 − x . Here, use Γ x Γ 1−x π = sin πx the functional identity , and you may further rewrite the above as fx = 1 2 ln 1 − x2 the second expression of f x . 16 · πx sin πx...
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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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