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The examples given here for example, show that once you know the form of a taylor polynomial as a function of $x$, you can replace the $x$ with another function. It works when you work out the problems but I don't understand why. It would seem that the chain rule would make the taylor polynomial something else.

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Could you please be more specific about what in that link seems unclear? It may be easier to explain with an example in mind. –  Jonas Meyer Jan 27 '11 at 2:27
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If $\displaystyle{f(x)=\sum_{n=0}^\infty a_n(x-c)^n}$ is a Taylor series that converges in the interval $(c-R,c+R)$, and if $g$ is a function, then $\displaystyle{f(g(x))=\sum_{n=0}^\infty a_n(g(x)-c)^n}$ is valid as long as $g(x)$ is in the interval $(c-R,c+R)$. $f(y)=\sum\limits_{n=0}^\infty a_n(y-c)^n$ and $f(t)=\sum\limits_{n=0}^\infty a_n(t-c)^n$ converge if $y$ and $t$ are in $(c-R,c+R)$. In the last statement, let $t=g(x)$. The point is that whatever you plug in in place of the original "$x$", the series converges as long as that thing is in $(c-R,c+R)$. I don't know if that helps. –  Jonas Meyer Dec 13 '11 at 5:59
    
The new series isn't necessarily a Taylor series. You still get a series, but the terms aren't (necessarily) of the form $c_n (x-x_0)^n$ (e.g. the substitution $x\mapsto x^2-1$). The only cases where you do get a Taylor series back out are linear substitutions $x\mapsto ax-b$, in which case it's straightforward to check that the resulting series is consistent with the true Taylor series. –  Riley E Dec 13 '11 at 8:36
    
(My above comment is only in regards to the "It would seem that the chain rule would make the taylor polynomial something else." part of the question) –  Riley E Dec 13 '11 at 8:46
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closed as not a real question by Jonas Meyer, t.b., Jonas Teuwen, Srivatsan, J. M. Dec 13 '11 at 10:35

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1 Answer

See this (theorem concerning substitution in Taylor series).

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