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Unformatted text preview: AMATH 351 Homework 8 Due August 19, 2009 Section5.1 3, 9, 13 Section5.2 4, 12 Section5.4 6 Section5.5 10 Section5.6 6 Section5.7 7 Section 5.1 3. Determine the radius of convergence. X n =0 x 2 n n ! For the series to be convergent, lim n x 2( n +1) ( n +1)! x 2 n n ! = lim n x 2 n + 1 = 0 As indicated above, the ratio goes to for all the nite values of x , so the radius of convergence is . 1 9. sin x = x x 3 3! + x 5 5! x 7 7! + = X n =0 ( 1) n x 2 n +1 (2 n + 1)! lim n ( 1) n +1 x 2 n +3 (2 n + 3)! (2 n + 1)! ( 1) n x 2 n +1 = lim n  x 2 (2 n + 3) (2 n + 2) = 0 for all the nite values of x , the radius of convergence is . 13. ln x = ln (1 + x 1) = ( x 1) ( x 1) 2 2 + ( x 1) 3 3 ( x 1) 4 4 + = X n =1 ( 1) n +1 ( x 1) n n By ratio test lim n ( 1) n +2 ( x 1) n +1 n + 1 ( 1) n +1 n ( x 1) n = lim n  n n + 1 ( x 1) =  x 1  < 1 So radius of convergence is 1 . Section 5.2 4. y 00 + k 2 x 2 y = 0 , x = 0 , k a constant Since x = 0 is a regular point of the di erential equation, we assume y = X n =0 a n x n y = X n =1 na n x n 1 y 00 = X n =2 n ( n 1) a n x n 2 2 Substitue into the DE, we get X n =2 n ( n 1) a n x n 2 + X n =0 k 2 a n x n +2 = 0 Shift the index for the both of the series, we get...
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This note was uploaded on 01/10/2010 for the course MATH 124 taught by Professor Walker during the Spring '08 term at University of Washington.
 Spring '08
 WAlker
 Math, Calculus

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