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I understand that given that expanding a function onto polynomials is a valid thing, the equation for Taylor series follows, but why is expanding a function onto polynomials a valid thing to do? Why should we assume that something can be written as $a + a_1(x)^2+...$. It doesn't seem inherently obvious to me.

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Insofar as it is a valid thing to do, it's because the power series will "usually" converge to the function. $\hspace{.4 in}$ We shouldn't assume that. $\:$ – Ricky Demer Sep 10 '13 at 4:02
Your question is a little vague. Not all functions have a Taylor series expansion. A simple example is $x \mapsto |x|$ at $x = 0$. – copper.hat Sep 10 '13 at 4:04
Taylor series were created from crafting polynomials with derivatives that matched those of the original function. The series part comes in when observing the remainder term tends to $0$ for nice (read: analytic) functions. – oldrinb Sep 10 '13 at 4:09
Many useful functions have nice power series expansions, at least for values of $x$ in some interval. Quite a few don;t, but there are other tools, like Fourier series. – André Nicolas Sep 10 '13 at 4:41
I know that it's limited to some functions, and that many do converge, but was there really no initial formalism besides thinking that the functions derivatives should match the Taylor expansions derivatives? No arguments of the linear independence of the polynomials, like with Fourier series? – user82004 Sep 10 '13 at 14:37
up vote 1 down vote accepted

Taylor’s theorem applies to any differentiable function. So for a start, if the function isn't differentiable then Taylor's theorem doesn't apply.

Taylor's theorem says if a function $f(x)$ is differentiable n+1 times in an interval I containing x and c then there exists a point z between x and c such that

$f(x)=f(c)+f'(c).(x-c) + (f'' (c).(x-c)^2)/2! +⋯+(f^n (c).(x-c)^n)/n! +R_n (x) $ where $R_n (x)=(f^{n+1} (z).(x-c)^{n+1}$)/(n+1) !

So, I think that the answer to your question becomes simply what is the proof of Taylor's Theorem, and perhaps also, when the function $R_n $ tends to zero with n so that the series expansion converges for a function which is infinitely differentiable.

You can look this up

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P.S. (1) to recover the original form of a power series in x you need an interval that includes c = 0. (2) the function $R_n$ can be easily seen to tend to zero if $f^n(z)$ is bounded: so the sine and cosine expansions are very simple (as -1 <= $f^n(z)$ <= +1 for all n and all z). – Tom Collinge Mar 22 '14 at 12:34

I think you need to take a look at Stone–Weierstrass theorem. See and any standard analysis book (Rudin will be good enough)

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This is loosely related to the OPs question. – Pedro Tamaroff Sep 10 '13 at 4:37

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