Why is the Maclaurin series for $\cos(x^2)$ simply the Maclaurin series for $\cos(x)$ with $x^2$ substituted for $x$?

Question

So I'm trying to understand why we can use expansions of Maclaurin series in this form.

If I try to convert the following into a Maclaurin Series $$f(x) = x^3 \cos(x^2)$$ Using the following: $$\cos(x) = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!}$$ I get $$x^3 \cos(x^2) = \sum_{n=0}^{\infty} \frac{(-1)^n x^{4n+3}}{(2n)!}$$

Why can we plug in $(x^2)$ in in the place of $\cos(x)$? Isn't there a chain rule to consider, since the Maclaurin/Taylor Series have something to do with derivatives? Why is it that we can just plug in $x^2$ and it still works?

Answer 1

Since it is true that, for any real number $x$, you have$$\cos(x)=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\cdots,$$then, in particular, if $x\in\Bbb R$, then you also have $x^2\in\Bbb R$, and therefore\begin{align}\cos(x^2)&=1-\frac{(x^2)^2}{2!}+\frac{(x^2)^4}{4!}-\cdots\\&=1-\frac{x^4}{2!}+\frac{x^8}{4!}-\cdots\end{align}And now you multiply both sides of this equality by $x^3$, in order to get that$$x^3\cos(x^2)=x^3-\frac{x^7}{2!}+\frac{x^{11}}{4!}-\cdots.$$

Answer 2

It turns out that inside their respective radii of convergence you can manipulate power series "like polynomials". That isn't too hard to prove, makes a nice exercise. In your case the power series are $x^3$ (clearly everywhere convergent) and the Maclaurin series for $\cos x$, also everywhere convergent. Thus the resulting series is everywhere convergent, and as power series are unique, is the (Maclaurin) power series of the function in question.