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Find by any method the first three terms of Taylor expansion about $x=0$ of of $\exp\{-\frac{1}{(x-a)^2}\}$

I found derivatives and plugged them into the general form for Taylor series and got

$e^-\frac{1}{a^2}[1-\frac{2}{a^3}x+\frac{4-6a^2}{2a^6}x^2]$

My question is, why is it that we can sometimes use the more direct method of simply using the exponential power series (which then turns out to be the same as the Taylor series) as for example we can with the function $e^{-x}$, whereas if we try to do that here we get a series that looks very different?

Thinking about this more I'm guessing the answer has to do with implicit expansions appearing within the first expansion.

Can the Taylor series expansion of certain functions ever be substantially different to other expansions of the same function, or just different in form? What would be most appropriate form in the given question?

James H
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    I believe you are asking something like this: suppose that you have a function $f(x) = \exp(g(x))$. One way to get the Taylor expansion of $f$ is using the definition $f(x) = f(0)+\frac{f'(0)}{1!}(x-0)+\frac{f''(0)}{2!}(x-0)^2 + \ldots$, and the other way is to note that $e^x = 1+x+\frac{x^2}{2!} + \frac{x^3}{3!} + ...$ and then $f(x) = e^{g(x)} = 1+g(x)+\frac {g(x)^2}{2!} + \ldots$, so you get two different expansions and you are asking why they are the same, is that right? The latter is probably the "direct" method you're talking about. – Sarvesh Ravichandran Iyer Jul 03 '23 at 11:05
  • Yes, that's exactly what I'm asking, thank you for clarifying. – James H Jul 03 '23 at 12:40
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    There is a theorem that guarantees equality of power series provided that the set of values on which they coincide has a limit point. In this case, the two infinite power series in question (one coming from the usual Taylor expansion and the other from substituting $g(x)$ in the Taylor expansion of $x$) can both be proved in different ways to equal $f(x)$ in a neighborhood of $x$ at least, therefore they're actually equal to each other : this is the uniqueness of the Taylor series, basically speaking. I used this here, for example. – Sarvesh Ravichandran Iyer Jul 03 '23 at 12:44
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    See Theorem 358 of this document, for example. The proof of the result is a little more involved, but you can use it quite freely. – Sarvesh Ravichandran Iyer Jul 03 '23 at 12:45

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