TITLE: Why can we just replace $z$ by $ z^2$ in this Taylor expansion?
QUESTION [9 upvotes]: When we expand $\sin(z^2)$, we can apparently just replace $z$ by $z^2$ in the Taylor series obtained by expanding $\sin(z)$. I can't see why. I mean, we can see that the derivatives of $\sin(z^2)$ will be a messy, with a lot of cos and sin, what make us conclude that just to replace $z²$ is enough to get the right series? The same thing goes to any Taylor series? I mean, if $f(x) = \sum x^n f^{(n)}(0)/n!$, when can we conclude that $f(x^2) = \sum x^{2n} f^{(n)}(0)/n!$?

REPLY [2 votes]: Let's start with an easier example. Suppose that a function $g$ is given by the equation
$$g(x) = x + 6.$$
Can we conclude that
$$g(8) = 8 + 6?$$
Yes, absolutely. When we say that $g(x) = x + 6$, what we actually mean to say is that for all real numbers $x$, $g(x) = x + 6$. We're saying that this equation is true for every possible real number that $x$ could possibly be. And the number $8$ is a real number, so the equation is true when we put $8$ in for $x$.
The same goes for Taylor series. Suppose we know that for all $x$,
$$f(x) = \sum x^n \frac{f^{(n)}(0)}{n!}.$$
Can we conclude that
$$f(8) = \sum 8^n \frac{f^{(n)}(0)}{n!}?$$
Yes, absolutely. The number $8$ is a real number, so the equation is true when we put $8$ in for $x$.
We have other options, though. We don't have to use a literal number; we can put anything in there, as long as that thing we're putting in represents a real number.
So, if $y^2$ is a real number, can we conclude that
$$f(y^2) = \sum (y^2)^n \frac{f^{(n)}(0)}{n!}?$$
Yes, absolutely. The number $y^2$ is a real number, so the equation is true when we put $y^2$ in for $x$.
(Technically, this process is called universal instantiation.)