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I have a general and a specific question about the composition of Taylor series.

Let's say we have $f(x)$ and $g(x)$. We know that the normal composition of functions is something like this: $g \circ f = g(f(x))$. But if we now consider the Taylor approximation of each function $T(f)$ and $T(g)$ how would the composition of $T(g\circ f)$ look like?

Searching I found this: $T(g\circ f)(x;a)= T(Tf(Tg(x;a))(x;q))$. But it's pretty much confusing me and I'm not eve sure if it's right, so I'm asking your help.

To get a better understanding of the whole concept I show you the exercise I have to solve.

First of all I had to find the Taylor approximation of $f(x)=\cos(x)$ and then $g(x)=e^{x^2}$ at $x_0=0$ and till here it's okay.

Now it's given $h(x)=\cos(e^{x^2}-1)$ and I have to find its Taylor approximation at $x_0=0$.

I tried to follow the expression above, but I didn't get it.

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I do not know if ths is what you are expecting. So, forgive me if I am off-topic.

Consider $$h=\cos(u)=1-\frac{u^2}{2}+\frac{u^4}{24}-\frac{u^6}{720}+O\left(u^7\right)$$ Replace $u$ by $(e^{x^2}-1)$ which makes $$h=1-\frac{(e^{x^2}-1)^2}{2}+\frac{(e^{x^2}-1)^4}{24}-\frac{(e^{x^2}-1)^6}{720}+\cdots$$ Now, use $$e^{x^2}=1+x^2+\frac{x^4}{2}+\frac{x^6}{6}+O\left(x^7\right)$$ Replace and expand.

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Giving a formula the the Taylor coefficients of a composition can be a difficult problem. But you can manage to find a finite number of them. In your specific problem we have $$ \cos u=1-\frac{u^2}{2}+\frac{u^4}{4!}-\frac{u^6}{6!}+\dots\tag1 $$ and $$ u=e^{x^2}-1=x^2+\frac{x^4}{2!}+\frac{x^6}{3!}+\dots\tag2 $$ Sbstituting (2) in (1) we get $$\begin{align} \cos\bigl(e^{x^2}-1\bigr)&=1-\frac12\Bigl(x^2+\frac{x^4}{2!}+\frac{x^6}{3!}+\dots\Bigr)^2+\frac{1}{24}\Bigl(x^2+\frac{x^4}{2!}+\frac{x^6}{3!}+\dots\Bigr)^4+\dots\\ &=1-\frac{x^4}{2}-\frac{x^6}{2}-\frac{x^8}{4}+\dots \end{align}$$

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  • $\begingroup$ Same idea, same time ! Cheers (sabes que somos vecinos ?) $\endgroup$ – Claude Leibovici Dec 11 '15 at 18:17
  • $\begingroup$ Thanks to both, I understand how to deal with this kind of exercises, but I have a doubt now. @Julián if I had to calculate the Taylor approximation of the same function but at $x_0=2$ (for example) how would it be? $\endgroup$ – Ergo Dec 12 '15 at 11:24
  • $\begingroup$ To find the Taylor expansion of $g(f(x))$ around $x_0$, develop $g$ around $f(x_0)$. $\endgroup$ – Julián Aguirre Dec 12 '15 at 20:14

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