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I want to taylor expand the following function (in the complex plane):

$$(1-\frac{z^2}{3!}+\frac{z^4}{5!})^{-1}$$

I am told that the result is:

$$1+\frac{z^2}{3!}+\frac{14}{6!}z^4+O(z^6)$$

I do not see how to obtain this. The only thing I have tried is using the taylor expansion of $(1-z)^{-1}$. But using this I can only obtain the second term of the expansion written above.

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HINT:

Recall from Taylor's Theoremthat $\frac{1}{1-x}=1+x+x^2+O(x^3)$

Now, set $x= z^2/3!-z^4/5!$ and don't forget the second order term $(z^2/3!-z^4/5!)^2$.

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  • $\begingroup$ Thank you. But isn't this only true for $|z|<1$? $\endgroup$ – john melon Apr 3 '17 at 20:23
  • $\begingroup$ OK. So the expression above is only valid in a domain that satisfies $\left|\frac{z^2}{3!}-\frac{z^4}{5!}\right|<1$? $\endgroup$ – john melon Apr 3 '17 at 20:26
  • $\begingroup$ No. The expansion is not a full series. Note from Taylor's Theorem that $\frac{1}{1-x}=1+x+x^2+o(x^2)$where we are using the little $o$ notation. $\endgroup$ – Mark Viola Apr 3 '17 at 20:26
  • $\begingroup$ What do you mean by full series? $\endgroup$ – john melon Apr 3 '17 at 20:28
  • $\begingroup$ We are truncating a series, not writing one. Taylor's Theorem is not Taylor's Series. If we write $\frac{1}{1-x}=\sum_{n=0}^\infty x^n$, then the series converges for $|x|<1$ and diverges elsewhere. But we can write $\frac{1}{1-x}=1+x+x^2+O(x^3)$ without restricting $x$. $\endgroup$ – Mark Viola Apr 3 '17 at 20:29

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