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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.

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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.

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  • $\begingroup$ I see we can evaluate $g(f(x))$ at $x$ for which $\vert f(x)\vert<C$, but are you saying we can only evaluate, and we don't really have a composite function? Because that got me confused. $\endgroup$ – Sha Vuklia May 1 '17 at 16:32
  • $\begingroup$ @ShaVuklia: Note the difference between $f$ which is a function and $f(x)$ which is a real value, namely $f$ evaluated at $x$. Similarly we do have a composition of functions, namely $g\circ f$. But $(g\circ f)(x)=g(f(x))$ is the composite $g\circ f$ evaluated at $x$ or in other words, the function $g$ evaluated at $f(x)$. $\endgroup$ – Markus Scheuer May 1 '17 at 16:35
  • $\begingroup$ Right, I didn't realise we were making the distinction between $(g\circ f)$ and $(g\circ f)(x)$ here. I usually read those as the same, but in this case I shouldn't. $\endgroup$ – Sha Vuklia May 1 '17 at 16:43
  • $\begingroup$ @ShaVuklia: Good to see my answer was helpful! :-) $\endgroup$ – Markus Scheuer May 1 '17 at 16:45
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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 $|x^2|<1$.

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