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Given function is $f(x) = \frac{\cos(x)}{x}.$

$y = x - a , y = x - 1$.

$x = y+1 , f(y) = \frac{\cos(y+1)}{y+1}$

How to get Taylor series expansion about $1$ of this function?

If it was needed to determine Maclauren expansion it would be $f(x) = \frac{\cos(x)}{x} = \sum_{n=0}^\infty\frac{(-1)^nx^{2n-1}}{(2n)!}$

Should I multiply these two series:

(1.) $\frac{1}{y+1} = \sum_{n=0}^\infty(-1)^ny^n$
(2.) $\cos(y+1) = \sum_{n=0}^\infty\frac{(-1)^n(y+1)^{2n}}{(2n)!}$
(1.)$\times$(2.) = $\sum_{n=0}^\infty(\sum_{m=0}^n\frac{(-1)^{n-m}(y+1)^{2m}}{(2m)!})y^n = \sum_{n=0}^\infty(\sum_{m=0}^n\frac{(-1)^{n-m}(x)^{2m}}{(2m)!})(x-1)^n$

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  • $\begingroup$ Are you aware of the fact the $n$-th coefficient in a general series expansion $f(x)=\sum_n a_n (x-a)^n$ is given by $a_n=\frac{f^{(n)}(a)}{n!}$? $\endgroup$ Jun 24 '15 at 20:08
  • $\begingroup$ I think specifically in the case $a=1$, you could easily guess the general form of the $n$-th derivative at $1$ and then prove it by induction. Just evaluate the first few derivatives at $1$. $\endgroup$ Jun 24 '15 at 20:17
  • $\begingroup$ I'm not sure I managed to simplify the nth derivative: $f^{(n)}(1) = \sum_{k=0}^n(-1)^n(n-k)!\cos^{(n-k)}(1)$. How should I get $a_n=\frac{f^{(n)}(1)}{n!}$ then? $\endgroup$
    – purgerica
    Jun 24 '15 at 20:24
  • $\begingroup$ Yes, for $n=1,2,3, 4...$ you should spot the pattern by then. $\endgroup$ Jun 24 '15 at 20:25

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