# Taylor series of $f(x^2)$

If you know the taylor series for $f(x)$ can you find the taylor series for $f(x^2)$ by letting $x = x^2$? The taylor series in question is $\cos(x^2)$

I know the taylor series for $\cos(x)$ is $\displaystyle \sum_{n=0}^\infty \frac{(-1)^n}{(2n)!} x^{2n}$. Is the taylor series for $\cos(x^2)$ just $\displaystyle \sum_{n=0}^\infty \frac{(-1)^n}{(2n)!} x^{4n}$? If this works, does it work for all functions $f(x)$ with a closed form solution for the taylor series?

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## 2 Answers

Indeed you can, so long as $x^2$ is within the region of convergence for the taylor series.

It really is just like any other function. If $f(x)=\sum_n a_nx^n$, then $f(y)=\sum_n a_ny^n$ and $f(x^2)=\sum_n a_nx^{2n}$.

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Even if the radius of convergence is zero, still the "Taylor series" works this way. So there is no need to say $x$ is within the region of convergence. But even that is not the same as saying the Tayor series converges to the function. –  GEdgar Apr 17 '13 at 1:07
I suppose it depends on your perspective. I suppose I put it in there because if the region of convergence for $f(x)$ is $|x|\leq a$, then the region of convergence for $f(x^2)$ is $|x|^2\leq a$. –  Glen O Apr 17 '13 at 1:15
For example, if we know the Taylor series for the function $f(x) = \exp(-1/x^2)$ for $x\ne 0$ and $f(0)=0$, then we can find the Taylor series for the function $f(x^2)$ using this rule. Even though these Taylor series do not converge to the function except at $0$. –  GEdgar Apr 17 '13 at 14:11

You can calculate this directly from the Taylor series definition. Which is given here: http://en.wikipedia.org/wiki/Taylor_series (sorry I can't figure out how to type out LateX formulae here) You basically calculate the nth derivate and divide by n! at a. Normally Taylor series calculated at 0 are called MacLaurean series. I don't think you can directly plug it into the Taylor series for cos(x) because the derivative of cos(X^2) is different.

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