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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?

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    $\begingroup$ The chain rule aspect is already taken care of here as well, because term $n$ is no longer from taking $n$ derivatives. Think of it this way: $\cos(z)$, regardless of what $z$ is, is $\sum_{n=0}^\infty \frac{(-1)^n z^{2n}}{(2n)!}$. Then you just take $z=x^2$. $\endgroup$
    – Ian
    Mar 6, 2021 at 18:00
  • $\begingroup$ As long as the thing you're plugging in lies in the domain of the function, and within the domain of convergence of the series, then you can plug it in. This is literally what the domain of a function means. Take a look at this answer for a much more lengthy and elaborate explanation $\endgroup$
    – peek-a-boo
    Mar 6, 2021 at 18:01
  • $\begingroup$ You can make the substitution because of the uniqueness of the Maclaurin series. $\endgroup$
    – Bernard
    Mar 6, 2021 at 18:55

2 Answers 2

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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$$x^3\cos(x^2)=x^3-\frac{x^7}{2!}+\frac{x^{11}}{4!}-\cdots.$$

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    $\begingroup$ +1 Worth noting: at the end we have a power series that converges to $x^3\cos(x^2)$. Such a series is unique, therefore this series must equal the Maclaurin series for $x^3\cos(x^2)$ (as defined using derivatives at $0$.) $\endgroup$
    – 2'5 9'2
    Mar 6, 2021 at 18:31
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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.

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