Taylor series of composition of function with polynomial

Say I have a taylor series around $$0$$ of some function $$f(x) = a_0 + a_1x + a_2x^2 + \cdots$$
Say $$g(x)$$ is a polynomial. Then is it true that taylor series of $$h(x) = f(g(x))$$ is term by term equal to $$a_0 + a_1g(x)+a_2g(x)^2+....$$?
The examples I've tried so far seem to work (on wolfram alpha)
Though I have no idea how to go about the proof

• You need that $g(0)=0$. In that case it is true. May 27, 2019 at 14:29
• @logarithm what is the problem with $0$? May 27, 2019 at 14:33
• The problem is that from the information that $f$ has a Taylor expansion at $0$ you don't know if it has it at other points. Your hypothesis is also imprecise. 'Having a Taylor expansion' at $x=0$ just means that $f$ has derivatives of all orders at $x=0$. Then you write the equation $f(x)=a_0+a_1x+a_2x^2+...$ but don't say for what values of $x$ it is assumed to be satisfied. The most general case would be that it is only satisfied for $x=0$. For function that are analytic at $x=0$ the equation is satisfied on a small neighborhood of $x=0$. For entire function it is satisfied for all $x$. May 27, 2019 at 14:47

$$h(x)=f(g(x))$$, since $$f(x)=a_0+a_1 x+a_2 x^2$$, $$h(x)=a_0+a_1 g(x)+a_2 g(x)^2$$,also since $$g(x)$$ is already polynomial, there is no need for further taylor expansion and the equality is prooven if the limit of $$g(x)=0$$ for $$x->0$$
• I was more looking for an anwser based upon the definition of taylor series applied on $h$ to conclude each term was equal May 27, 2019 at 14:32
• whats the problem with $0$? May 27, 2019 at 14:33