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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

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  • $\begingroup$ You need that $g(0)=0$. In that case it is true. $\endgroup$
    – logarithm
    May 27, 2019 at 14:29
  • $\begingroup$ @logarithm what is the problem with $0$? $\endgroup$
    – Anvit
    May 27, 2019 at 14:33
  • $\begingroup$ 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 $x$. $\endgroup$
    – logarithm
    May 27, 2019 at 14:47

1 Answer 1

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$ 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 $x->0$

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  • $\begingroup$ I was more looking for an anwser based upon the definition of taylor series applied on $h$ to conclude each term was equal $\endgroup$
    – Anvit
    May 27, 2019 at 14:32
  • $\begingroup$ whats the problem with $0$? $\endgroup$
    – Anvit
    May 27, 2019 at 14:33
  • $\begingroup$ A function can be expanded into taylor series only in an interval containing a zero of the function. $\endgroup$ May 27, 2019 at 14:35
  • $\begingroup$ As the taylor expansion gives you a good approximation of the function only in the interval close to the zero $\endgroup$ May 27, 2019 at 14:37

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