# sol12 - Math 104, Solution to Homework 12 Instructor:...

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Unformatted text preview: Math 104, Solution to Homework 12 Instructor: Guoliang Wu August 6, 2009 Ross, K. A., Elementary Analysis: The theory of calculus : 26.3 (a) Since X n =1 nx n = x (1- x ) 2 for | x | &amp;lt; 1 , d dx X n =1 nx n = d dx x (1- x ) 2 X n =1 n 2 x n- 1 = 1 (1- x ) 2 + 2 x (1- x ) 3 = 1 + x (1- x ) 3 X n =1 n 2 x n = 1 (1- x ) 2 + 2 x (1- x ) 3 = x (1 + x ) (1- x ) 3 , | x | &amp;lt; 1 . (b) Plug in x = 1 2 and x = 1 3 , X n =1 n 2 2 n = 1 2 (1 + 1 2 ) (1- 1 2 ) 3 = 6 , X n =1 n 2 3 n 1 3 (1 + 1 3 ) (1- 1 3 ) 3 = 3 2 . 26.6 (a) s ( x ) = X n =0 (- 1) n x 2 n +1 (2 n + 1)! = X a m x m , where a m = , m is even (- 1) n (2 n +1)! , m = 2 n + 1 1 So = lim sup | a m | 1 /m = lim n (- 1) n (2 n + 1)! 1 / (2 n +1) = lim k 1 k ! 1 /k = lim k k ! ( k + 1)! = 0 . Thus R = 1 = + . So s is differentiable on R and can be differentiated term by term: s ( x ) = X n =0 (- 1) n (2 n + 1) x 2 n (2 n + 1)! = X n =0 (- 1) n x 2 n (2 n )! = c ( x ) . Moreover, the series for c ( x ) also has radius of convergence R = and can be differentiable term by term on R , so c ( x ) = X n =1 (- 1) n (2 n ) x 2 n- 1 (2 n )! = X n =1 (- 1) n x 2 n- 1 (2 n- 1)! =- X n =0 (- 1) n x 2 n +1 (2 n + 1)! =- s ( x ) . (b) ( s 2 + c 2 ) = 2 ss + 2 cc = 2 sc + 2 c (- s ) = 0 . (c) Since ( s 2 + c 2 ) = 0 , the function s 2 + c 2 is a constant. Be- cause s 2 (0) + c 2 (0) = 0 + 1 = 1 , s 2 + c 2 = 1 , x R ....
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## sol12 - Math 104, Solution to Homework 12 Instructor:...

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