# How to expand undifferential function as power series?

If a function has infinite order derivative at $0$ and $\lim_{n\to \infty}(f(x)-\sum_{i=1}^{n} a_{n}x^n)=0$ for every $x \in (-r,r)$,then it can be expand as power series$\sum a_{n}x^n$,

My question is if this function is not differential at $0$,how to expand it as $\sum a_{n}x^n$ satisfied with $\lim_{n\to \infty}(f(x)-\sum_{i=1}^{n} a_{n}x^n)=0$ for every $x \in (-r,r)$?Is it unique ?

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If one of the derivatives is infinite, then how do you obtain $a_n$, seeing that it depends on the derivatives themselves... –  Guess who it is. Apr 15 '12 at 17:25
Re-derive Fourier series? –  Mr. F Apr 15 '12 at 17:38
Check out the basics linked here. Your question was something that spurred decades of math research and resulted in radically changing the notion of a function. Fourier series is one method to get some kinds of convergence properties for series expansions of non-differentiable functions, when the points causing non-differentiability are sufficiently well-behaved. –  Mr. F Apr 15 '12 at 17:53

If $\sum_{j=0}^\infty a_j x^j$ converges for every $x$ in an interval $(-r,r)$, then the radius of convergence of the series is at least $r$, and the sum is analytic in the disk $\{z \in {\mathbb C}: |z| < r\}$. So if $f(x)$ is not analytic in $(-r,r)$, in particular if it is not differentiable at $0$, there is no way to represent it as $\sum_n a_n x^n$ with $\sum_{j=0}^n a_j x^j \to f(x)$.

However, you can try a series $\sum_n a_n x^n$ such that some subsequence of partial sums $P_N(x) = \sum_{j=0}^N a_j x^j$ converges to $f(x)$. Suppose $f$ is continuous on $[-r,r]$ except possibly at $0$. I'll let $a_0 = f(0)$ and $N_0 = 0$. Given $a_j$ for $0 \le j \le N_k$, let $g_k(x) = (f(x) - P_{N_k}(x))/x^{N_k}$ for $x \ne 0$, $g_k(0) = 0$. Since $g_k$ is continuous on $E_k = [-r, -r/(k+1)] \cap \{0\} \cap [r/(k+1), r]$, Stone-Weierstrass says there is a polynomial $h_k(x)$ with $|g_k(x) - h_k(x)| < r^{-N_k}/(k+1)$ on $E_k$. Moreover we can assume $h_k(0) = g_k(0) = 0$. Let $N_{k+1} = N_k + \deg(h_k)$, and let $a_j$ be the coefficient of $x^j$ in $x^{N_k} h_k(x)$ for $N_k < j \le N_{k+1}$. Thus $P_{N_{k+1}}(x) = P_{N_k}(x) + x^{N_k} h_k(x)$ so that $|P_{N_{k+1}}(x) - f(x)| = |x|^{N_k} |g_k(x) - h_k(x)| < 1/(k+1)$ for $x \in E_k \backslash \{0\}$ (we already know $P_{N_{k+1}}(0) = f(0)$). Since the union of the $E_k$ is all of $[-r,r]$, the partial sums $P_{N_k}(x)$ converge to $f(x)$ pointwise on $[-r,r]$.

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If $f$ is continuous in a neighborhood, then you can use the Stone-Weierstrass theorem, and write is as a sum of $x^n$, but you loose the nice interpretation of a Taylor series.

Also if $f$ is only locally integrable, then you can to the same thing.

But note that you always will have to change the notion of the convergence ... e.g. uniform or in $L^p$-sense.

As pointed out in the comments, this does not really answer your question, since you require the coeffiecients not to change (I have overlooked that by first reading), but perhaps it will be sufficient for whatever application you have in mind.

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Stone-Weierstrass gives polynomial approximations, but not a series. That is, for every $\epsilon > 0$ you can approximate $f(x)$ within $\epsilon$ on a given interval by some polynomial, but the coefficients of the polynomials for different $\epsilon$ are not necessarily related. –  Robert Israel Apr 15 '12 at 18:14
@RobertIsrael yes you are right ,do you have any idea about this problem? –  noname1014 Apr 15 '12 at 19:06