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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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