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?
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 $f(x^2)=\sum_n a_nx^{2n}$.
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.