Maclaurin series for $\cos(2x^3)$ I need some help here.

Find Maclaurin series representation for the function $f(x)=\cos(2x^3).$

I guess the easiest thing to do is using that $\displaystyle\cos x = \sum_{n=0}^{+\infty} \frac{(-1)^n}{(2n)!}{x^{2n}}.$
Using that identity I end up with: $\displaystyle \sum_{n=0}^{+\infty} \frac{(-1)^n}{(2n)!}{2x^{6n}},$
 which is wrong. Could anyone help?
 A: You were close, but you replaced $x$ in $\cos x$ with $(x^3)$ to get $(x^3)^{2n}$ and then  multiplied this by one factor of $2$ to get $2(x^3)^{2n}$. 
However, we need to replace $x$ with all of $(2x^3)$ to get $(2x^3)^{2n} = 2^{2n}\cdot (x^3)^{2n} = 4^nx^{6n}.$
Doing this gives us:
$$\cos(2x^3) = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!}(2x^3)^{2n} = \sum_{n=0}^\infty \frac{(-1)^n}{(2n)!} 4^nx^{6n}$$
A: I'm going to expound on this, because I feel that you shouldn't proceed by direct substitution unless you really have a solid foundation. I'll leave the derivation of the Taylor expansion for $cos(x)$ as an exercise, but will explain why the substitution can take place in this case. Now, here's a more structured argument:
Let $f(x) = cos(x)$ and $g(x) = 2x^{3}$; let $h(x) = (f\circ g)(x)$ ;let $y=2x^{3}$. 
Since $$0 = \text{lim}_{x \rightarrow 0} \frac{f(x)-P_{n,0,f}(x)}{(x-0)^{n}} \\
= \text{lim}_{y \rightarrow 0} \frac{f(y)-P_{n,0,f}(y)}{(y-0)^{n}} \\
= \text{lim}_{x \rightarrow 0} \frac{h(x)-P_{n,0,f}(g(x))}{(2x^{3}-0)^{n}} \\ = \text{lim}_{x \rightarrow 0} \frac{h(x)-P_{n,0,f}(g(x))}{(x-0)^{3n}}.$$
So, $P_{3n,0,h}(x) = P_{n,0,f}(g(x))$ and the substitution can take place.
In fact, we can generalise this substitution process to the composition of two analytic functions. I've seen it before, but I can't recall it. Maybe a proof is in order?
A: The formula with $4^n$ and $x^{6n}$ doesnt work, it can be checked with online calculators of Taylor series. I believe the reason is that using the substition is incorrect, because just plugging the $2x^3$ as x into the solution for cos x omitts the changes in the derivatives.
