solutions for b

1 10 2 x10 2 x9 dierentiate this twice j

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Unformatted text preview: too is convergent for an arbitrary real number x. 5 Note ⋆ Jx = : Let’s duplicate (2b): 1 0!! 2 − 1 2!! 2 x2 + 2 x+ 1 4!! 1 2 x4 − 2 x3 − 6!! x4 + 8 ·7 6!! 2 x6 + 2 x5 + 1 2 x8 − 2 x7 − 8!! 1 10!! 2 x10 + ··· . 2 x9 + ··· , Differentiate this twice: J′ x = 2 − 2!! + 4 ·3 4 4!! 6 8 8!! 10 10!! and J ′′ x = − 2 ·1 2!! 2 4!! 2 x2 − 6 ·5 6!! 2 8!! 2 x6 − 10 ·9 10!! 2 x8 + 12 ·11 12!! 2 x10 − ···. J ′′ x can be written as follows: J ′′ x =− 1 2 0!! 2 + 3 4 2!! 2 x2 − 5 6 4!! 2 x4 + 7 8 6!! 2 x6 − 9 10 8!! 2 x8 + 11 12 10!! 2 x10 − ··· . 2 x10 + ··· . 2 x10 − ··· . Add up J x and J ′′ x : Jx J ′′ x = =− 1 0!! 2 1 2 0!! 2 − + 1 2!! 2 3 4 2!! x2 + 2 2 x− 1 4!! 2 5 6 4!! x4 − 4 2 x+ 1 6!! 2 7 8 6!! x6 + 6 2 x− 1 8!! 2 9 10 8!! x8 − 8 2 x+ 1 10!! 11 12 10!! + J x + J ′′ x = 1 2 0!! 2 − 1 4 2!! 2 x2 + 1 6 4!! 2 x4 − 6 1 8 6!! 2 x6 + 1 10 8!! 2 x8 − 1 12 10!! 2 x10 + ··· . Realize that x multipled to this exactly equals −J ′ x . Hence we obtain x J ′′ x + J′ x + xJ x =0 Bessel’s differential equation J x is called the Bessel function of the first kind . [II] (30pts) Let ε be an arbitrary positive real number. Then the integral +∞ ( ∗) −1 ζs s=1+ε ds is convergent . Proof: Let = ζs fs f s is well-defined over (i) f s > 0 over (ii) fs 1 + ε, 1 + ε, − 1. +∞ , +∞ , and the following (i) and (ii) hold: and is monotonically decreasing over 1 + ε, Hence the Integral Test is applicable. Now, we claim ∞ (iii) = 1. fn n=2 Indeed, ∞ fn...
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