P4-15 - Solution to Problem 4.15 March 14, 2011 Here, R(x)...

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Solution to Problem 4.15 March 14, 2011 Here, R ( x ) = 1, P ( x ) = x 1, Q ( x ) = αx . We therefore have f ( s + k ) = ( s + k )( s + k 1) ( s + k ) = ( s + k )( s + k 2) (1) and g n ( s + k ) = ( s + k 1) α if n = 1 and 0 if n > 1. (2) The indicial equation is f ( s ) = ( s )( s 2) = 0, (3) so that s = 0 or 2. For k 1, we therefore have ( s + k )( s + k 2) A k = (( s + k 1) α ) A k - 1 . (4) Suppose s = 2. Then k ( k + 2) A k = ( k + 1 α ) A k - 1 . (5) Or, since k ( k + 2) is positive for all k , A k = α k 1 k ( k + 2) A k - 1 . (6) This uniquely determines all A k (with only A 0 arbitrary). If α is an integer 2, the series only has ±nitely many terms, and the solution is a polynomial of degree α 1
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Suppose s = 0. Then k ( k 2) A k = ( k 1 α ) A k - 1 . (7) For k = 1, A 1 = αA 0 . (8) For k = 2, 0 = (1 α ) A 1 . (9) For k 3, k ( k 2) is positive, and A k = α k + 1 k ( k 2) A k - 1 . (10) If α n = 0 or 1, we have that A 1 = 0, A 0 = 0, A 2 is arbitrary, and all subsequent A k . But this contradicts our assumption that A 0 n = 0. Therefore, there is no regular solution with
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P4-15 - Solution to Problem 4.15 March 14, 2011 Here, R(x)...

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