Composite function $g(f(x))$ where $g$ is a power series Say we have
$$
g(x)=\sum_{k=0}^\infty a_kx^k\quad\text{for }\vert x\vert<C.
$$
I'm trying to see why
$$
g(f(x))=\sum_{k=0}^\infty a_kf(x)^k\quad\text{for }\vert f(x)\vert<C.
$$
In a sense, sure, it is just a composition of functions, but I am still a bit "worried" about the infinite series. How do we know we can replace all these infinitely many summands by $f(x)$? Could we use the definition perhaps, to ensure nothing goes wrong?;
$$
\lim_{n\to\infty}\sum_{k=0}^na_kx^k.
$$
I just feel like I'm missing something for this to feel completely safe to me. Any ideas? And please something more than: it's just a composite function, end of story.
 A: if $|x|<C $ then
$$g (x)=\sum_{k=0}^{+\infty}a_kx^k$$
if $|y|<C $ then
$$g (y)=\sum_{k=0}^{+\infty}a_ky^k $$
to define $g (z) $, we must have $|z|<C. $
So, to define $g (f (x)) $ you need that
$|f (x)|<C. $
in that case,
$$g (f (x))=\sum_{k=0}^{+\infty}a_k (f (x))^k. $$
For example
we can write $\sum_{k=0}^{+\infty}x^k $ if $|x|<1$
and
$$\sum_{k=0}^{+\infty}(x^2)^k $$ if $|x^2|<1$.
A: 
Given the series
  \begin{align*}
g(x)=\sum_{k=0}^\infty a_kx^k\quad\text{for }\vert x\vert<C.
\end{align*}
  we can take any $x_0$ with $|x_0|<C$ and evaluate $g$ at $x_0$
  \begin{align*}
g(x_0)=\sum_{k=0}^\infty a_kx_0^k
\end{align*}
  Since $|x_0|<C$ the series $g(x_0)$ converges.
Note, that with
  \begin{align*}
g(f(x))=\sum_{k=0}^\infty a_kf(x)^k\quad\text{for }\vert f(x)\vert<C.
\end{align*}
  we have the same situation, since we consider $g$ evaluated at $f(x)$ as we did before with $x_0$. The only aspect we have to assure in order to guarantee the convergence of the series $g(f(x))$ is $|f(x)|<C$. We do not have to consider anything more.

We note that $g\circ f$ is a composition of functions. But here we consider the composition of function evaluated at a specific point $f(x)$ which is, assuming real valued functions just an ordinary real.
