Why can we just replace $z$ by $ z^2$ in this Taylor expansion? 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 $f(x^2) = \sum x^{2n} f^{(n)}(0)/n!$?
 A: Your confusion here stems from an abuse of notation.  In the case of differentiation, we are not merely making a substitution for the argument of a function - we are defining an entirely different function through function composition.  Strictly speaking, $\sin(z^2)$ has no derivative, because it is not a function - rather, it is a value in the codomain of $\sin$.
When we write $\frac{df(x^2)}{dx}$, we do not mean $\frac{df}{dx}(x^2)$.  The former can become quite messy; the latter is just a substitution, same as you've done with the power series.
A good way to avoid this type of confusion is to remember that differentiation acts on the function, not on its argument or on its value.  If you stick to explicitly defining your functions (e.g. avoid shortening $z \to \sin(z^2)$ to just $\sin(z^2)$), it should be much clearer what is actually going on.
A: It's better to go back to Taylor polynomials to convince yourself. If a polynomial $P$ of degree $n$ satisfies
$$\lim_{z\to 0} \frac{f(z)-P(z)}{z^n} = 0,$$
then it must be the $n$th degree Taylor polynomial of $f$. Using this, you can easily prove that if $P$ is the $n$th degree Taylor polynomial of $\sin$, then $Q(z)=P(z^2)$ is the $(2n)$th degree Taylor polynomial of $\sin(z^2)$. The rest follows immediately.
A: $f(\omega) = \sum \omega^n f^{(n)}(0)/n!$
Let $\omega=x$:  $f(x) = \sum x^n f^{(n)}(0)/n!$
Let $\omega=x^2$: $f(x^2) = \sum x^{2n} f^{(n)}(0)/n!$
A: Missing ingredients
Uniqueness of Taylor Series
If a function $f$ has a power series at $a$ that converges to $f$ on some open interval containing $a$, then that power series is the Taylor series for $f$ at $a$.
Uniqueness of Power Series
Let
$$\sum_{n=0}^{\infty}c_n(x−a)^n$$
and
$$\sum_{n=0}^{\infty}d_n(x−a)^n$$
be two convergent power series such that
$$\sum_{n=0}^{\infty}c^n(x−a)^n=\sum_{n=0}^{\infty}d_n(x−a)^n$$
for all $x$ in an open interval containing $a$. Then $c_n=d_n$ for all $n \geq 0$.
Bringing you to
Power replacement in Taylor series
If the functions $f(x),f(x^k)$ have power series at $a$ that converge to $f(x),f(x^k)$ respectively on some open interval containing $a$, then the formal substitution $(x-a) \to (u-a)^k$ in the Taylor series for $f(x)$ brings the Taylor series of $g(u)=f(u^k)$ at $a$ for $k \in \mathbb{N}$ that converges to $f(x^k)$ on the same open interval containing $a$.
A: 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.)
A: This is a good question! The extra ingredient that we need is the theorem that power series are unique. More precisely: if there exists some $\delta>0$ such that $\sum_{n=0}^\infty a_n (x-c)^n$ and $\sum_{n=0}^\infty b_n (x-c)^n$ both converge and are equal for all $|x-c|<\delta$, then $a_n=b_n$ for all $n\ge0$. From this theorem, and the fact that the power series you constructed definitely equals $f(x^2)$ which also equals the Taylor series of $f(x^2)$, you can indirectly conclude that the Taylor series of $f(x^2)$ is actually given by $\sum_{n=0}^\infty x^{2n} f^{(n)}(0)/n!$.
