Taylor Series expansion for a composition of functions The Taylor Series expansion for $\frac{1}{1-x}$ is convergent for every real number $-1 < x < 1$.
\begin{equation*}
\frac{1}{1 - x} = \sum_{n=0}^{\infty} x^{n} .
\end{equation*}
Since  $0 \leq x^{2} < 1$ if, and only if  $-1 < x < 1$, the Taylor Series expansion for $\frac{1}{1-x^{2}}$ is convergent for every real number $-1 < x < 1$.
\begin{equation*}
\frac{1}{1 - x^{2}} = \sum_{n=0}^{\infty} x^{2n} .
\end{equation*}
This is a special case of a theorem about the composition of certain functions. For example, the Taylor Series expansion for $e^{x}$ implies that
\begin{equation*}
e^{x^{2}} = \sum_{n=0}^{\infty} \frac{1}{n!} \, x^{2n} 
\end{equation*}
and that
\begin{equation*}
e^{(x-1)^2}
= \sum_{n=0}^{\infty} \frac{1}{n!} \, \left((x - 1)^2\right)^{n}
= \sum_{n=0}^{\infty} \frac{1}{n!} \, (x - 1)^{2n} .
\end{equation*}
I am looking for the precise statement for such a theorem and a rigorous demonstration of it (or a citation for a rigorous demonstration of it).
 A: You can substitute in another function but that might completely change the disk of convergence.

Let $f(z) = \sum_{j=0}^\infty a_j (z - z_0)^j$ be a Taylor series and $D$ its disk of convergence. If $g:S \to \mathbb{C}$, then $f \circ g$ is convergent in $g^{-1}[D]$ to $\sum_{j=0}^\infty a_j (g(z) - z_0)^j$.

Proof: Suppose $w = g(z) \in D$, then $f \circ g(z) = f(w) = \sum_{j=0}^\infty a_j (w - z_0)^j = \sum_{j=0}^\infty a_j (g(z) - z_0)^j$.
A: I don't think this theorem has a name, but see for instance Theorem 3.4 of Lang's Complex Analysis, which explains the convergence of formal power series under composition.
A: Complementing the other answers with an example in the context of reals:
In answer to one of your comments, the radius of convergence is not always the same:
$\sum_{n=0}^{\infty }\left ( \frac{1}{2} \right )^{n}x^{n}=\sum_{n=0}^{\infty }\left ( \frac{x}{2} \right )^{n}=\frac{1}{1-\frac{x}{2}}$ has radius of convergence $R=2$ which is not the same as the radius of convergence of $\sum_{n=0}^{\infty} x^{n}$.
Since the radius of convergence is not the same, the result you are aiming at, which could be stated as follows: 
"The limit of the composition is the same as the composition of the limit" is not true.
Taking the example above again to clarify why the statement is wrong:
The limit of $\sum_{n=0}^{\infty }\left ( \frac{x}{2} \right )^{n}$ is $\frac{1}{1-\frac{x}{2}}$ for $\left | x \right |< 2$.
So if we were to pick $x=1.5$, it is not true that $\sum_{n=0}^{\infty} x^{n}=\frac{1}{1 - (1.5)}$ as the series diverges to $\infty $ for $\left | x \right |\geq 1$
