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Unformatted text preview: = lim n ( n + 1) 2  x + 3  2 n 2 =  x + 3  2 . This gives convergence for  x + 3  < 2 and divergence for  x + 3  > 2. Thus R = 2. (Alternatively, you can do what I mentioned in class: Replace x + 3 with R and set L = 1: 1 = L = lim = R/ 2 and so R = 2.) The endpoints of the interval are given by x + 3 = 2 and x + 3 =2. The former gives the series n 2 and the latter gives n 2 (1) n . Both diverge since the terms do not go to zero. Thus the interval of convergence is5 < x <1. 3. Since e x = x n /n !, replacing x by2 x 2 gives e2 x 2 = X n =0 (2 x 2 ) n n ! = X n =0 (2) n x 2 n n ! . Multiply by 1 + x : (1 + x ) e2 x 2 = X n =0 (2) n ( x 2 n + x 2 n +1 ) n ! . The coecient of x 10 comes from the n = 5 term in the sum. The coecient is (2) 5 / 5! =4 / 15. This is also the coecient of x 11 ....
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 Fall '06
 Mohanty
 Math, Differential Equations, Equations

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