# Higher order divided differences and derivatives

Let $f:R\rightarrow R$. If there exists the finite limit $$\lim_{(x,y) \rightarrow a \atop x\neq y} \frac{f((y)-f(x)}{y-x}$$ then obviously there is a finite derivative $f'(a)$ and is equal this limit.

What about similar problem for higher order divided differences?

May is it true that existence of finite $$\lim_{(x_0,...,x_n)\rightarrow (a,...,a) \atop x_i \neq a} [x_0,...,x_n;f]$$ implies existence of $f^{(n)}(a)$?

If not is there connetion between high order divided differences and derivatives?

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Seems too easy for this forum. Anyway, what exactly is the limit you are looking for? Check this limit for simple functions not twice differentiable, like $x\sqrt{x^2}$, or $x^2\sin(1/x)$... –  Feldmann Denis Aug 24 '12 at 14:27

Assume we have a $\delta$ such that for $|x_i-a|<\delta$, we have $|[x_0,...,x_n;f]-\lim| < \epsilon$. Assume WLOG that $\lim = 0$ (by subtracting off a polynomial of degree $n$). Then if $y_0, ..., y_{n-1}$ and $z_0, ..., z_{n-1}$ are in the $\delta$-ball around $a$, we can show

$|[y_0,...,y_{n-1};f]-[z_0,...,z_{n-1};f]| < \epsilon\sum_{i=0}^{n-1}|y_i-z_i|$

by inducting on the number of $i$s such that $y_i\ne z_i$. Thus $\lim_{y_0,...,y_{n-1}\rightarrow y}[y_0,...,y_{n-1};f]$ exists, and by induction on $n$ is equal to $f^{(n-1)}(y)$. Letting the $y_i$s in the above approach $y$, and letting the $z_i$s approach $z$, we see

$|f^{(n-1)}(y)-f^{(n-1)}(z)| \le n\epsilon|y-z|$,

so $\left|\frac{f^{(n-1)}(y)-f^{(n-1)}(z)}{y-z}\right| \le n\epsilon$ for any $y,z$ in a $\delta$-ball around $a$. Taking $\epsilon$ to $0$, we see that $f^{(n)}(a)$ exists and is equal to $0$.

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In general the answer is no. For instance, if f is any odd function, then $$\lim_{h\to 0}\frac{f(h)+f(-h)-2f(0)}{h^2}=0,$$ without any assumptions on differentiability of f. So it certainly does not follow that f''(0) exists.
Thanks, but condition which is assumed (case $n=2$) is that there exists a finite: $$\lim_{(x_0,x_1,x_2)\rightarrow a \atop x_i \neq x_j} \frac{ \frac{f(x_2)-f(x_1)}{x_2-x_1}-\frac{f(x_1)-f(x_0)}{x_1-x_0}}{x_2-x_0}$$ |It seems be stronger limits $\lim_{h\rightarrow 0} \frac{f(x+h)+f(x-h)-2f(x)}{h^2}$. –  Jan Aug 24 '12 at 18:49
The document linked in Igor's answer seems to tacitly assume analyticity of the function. For example, equation (2) says that $F'(x)$ differs from $(F(x+h)-F(x))/h$ by $O(h)$, which fails at $x=0$ for the function $F(x)=x^{5/3}$. (It's possible that the OP also had analytic functions in mind.) –  Andreas Blass Aug 24 '12 at 16:15