Given some power series $f(x) = \sum_{n=1}^{\infty} a_n x^n$, is it generally true that $a_n = \frac{f^{(n)}}{n!}$? If so, why? We get this form when we develop a taylor series, but why is this case for every power series? Can't the $a_n$ terms be arbitrary? I'm a bit confused.
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You have to show that power series reps are unique. Since the Taylor form is one, it must be the one. – Randall Jun 18 '20 at 17:52
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See Matt E's answer on the question I've linked. In a nutshell, I would phrase the difference as: power series are from the perspective of the series, which has a corresponding function but Taylor series are from the perspective of the function, which has a corresponding series. If the series is convergent, they're the same thing. If the series isn't necessarily convergent (i.e., it's just a list of coefficients), it's no longer a Taylor series. – Jam Jun 18 '20 at 18:05
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Since$$f(x)=a_0+a_1x+a_2x^2+\cdots,$$you have $f(0)=a_0$ and $$f'(x)=a_1+2a_2x+3a_3x^2+\cdots\tag1$$and therefore $f'(0)=a_1$. On the other hand, since you have $(1)$, you have$$f''(x)=2a_2+6a_3x+12a_4x^2+\cdots$$and, in particular, $f''(0)=2a_2$ and so on…
José Carlos Santos
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If the series $$ f(x)=\sum_na_nx^n $$ converges in some neighborhood of zero, then $a_n=f^{(n)}(0)/n!$. This follows from term-by-term differentiation (which in turn follows form uniform convergence of power series inside the radius of convergence).
It is a fact that any sequence of real numbers $a_n$ are the Taylor coefficients of some function, i.e. there exists some $f$ with $f^{(n)}(0)=n!a_n$; however, the Taylor series doesn't necessarily converge (or converge to $f$). This goes by the name "Borel's lemma" in various places. See this post (and the monthly article giving Peano's earlier constructive proof) Every power series is the Taylor series of some $C^{\infty}$ function.
yoyo
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