# hws_104.2 - . Thus the interval of convergence is ( 1 / 3 ,...

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Math 104: Introduction to Analysis SOLUTIONS Alexander Givental HOMEWORK 12 31.2 Find the Taylor series for sinh x = ( e x e x ) / 2 and cosh x = ( e x + e x ) / 2. Solution. The result sinh x = s n 1 x 2 n 1 (2 n 1)! , cosh x = s n 0 x 2 n (2 n )! follows easily from either the expansion e x = x n /n !, or from (sinh x ) = cosh x, (cosh x ) = sinh x , togehter with sinh(0) = 0, cosh(0) = 1. Convergence of each series for any x follows from the ratio test: x 2 / 2 n (2 n + 1) 0 < 1 , and x 2 / (2 n + 1)(2 n + 2) 0 < 1 as n → ∞ . 23.2cd Determine the radius and exact interval of convergence for the series: (c) x n ! and (d) 3 n x 2 n +1 / n . Solution. (c) When | x | ≥ 1, the series diverges since its general term x n ! does not tend to 0 as n 0. For | x | < 1 the series converges absolutely by comparison with | x | m (note that x n ! = a m x m with a m = 1 when m = n !, and a m = 0 when m n = n !). Thus the convergence radius is 1, and the exact interval of convergence is ( 1 , 1). (d) The convergence radius is R = 1 / lim(3 n / n ) 1 / (2 n +1) = 1 3 lim(3 n ) 1 / (4 n +2) = 1 3 . When x = ± 1 / 3, the series turns into ± 1 / 3 n , which is known to diverge to
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Unformatted text preview: . Thus the interval of convergence is ( 1 / 3 , 1 / 3). 23.6b. Give an example of a series whose interval of convergence is exactly ( 1 , 1]. Solution. The series n&gt; ( x ) n /n converges (to ln(1+ x )) when | x | &lt; 1 (by the root test), diverges at x = 1 (since 1 /n = , and converges at x = +1 as an alternating series whose terms 1 /n monotonically tend to 0 in the absolute value. 23.8. Show that f n ( x ) := n 1 sin nx are dierentiable, tend to 0 for all x R , but lim f n ( x ) need not exist (at x = for instance). Solution. Indeed, f n ( x ) tends to 0 since | n 1 sin nx | n = 1 /n as n . However the sequence f n ( x ) = cos nx turns into ( 1) n at x = which has no limit as n . 1...
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