In particular we can use the error estimation in the

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Unformatted text preview: . The coefficient is then (−1)2 ( 1 1 )= (4(2) + 1)(2(2))! 9(4!) Therefore, a9 = 1 . 9(4!) Finally, this tells us that F (9) (0) = 9!a9 9! = 9(4!) = 5·6·7·8 = 1680 Next, to find F (16) (0) we look for the coefficient of x16 in the Taylor series for F (x). However, this time there is no n such that x4n+1 = x16 . This means that a16 = 0 and hence that F (16) (0) = 0. Suppose that we wanted to know the value of 1 2 0 1 dx. 1 + x9 1 Since 1+x9 looks like a rather simple rational function, we might be tempted to use partial fractions to try and calculate the integral exactly. At least theoretically, this should work. However, in p...
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This document was uploaded on 03/23/2013.

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