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Let $x_0<x_1<\cdots<x_n$, and let $f$ be continuously differentiable. Show that $$\dfrac{\partial}{\partial x_i} f[x_0,x_1,\dots,x_n]=f[x_0,x_1,\dots,x_i,x_i,x_{i+1},\dots,x_n].$$

At first, I thought this question can be done by induction on $n$ but after attempting it, I got something weird.

Next, I tried using $$f[x_0,,x_1,\dots,x_n]=\sum_{i=0}^n f(x_i) \prod_{\substack{j=0 \\ j\neq i}}^n (x_i-x_j)^{-1}$$ But when you take the derivative of this equation. I will obtain a double sum because of the product rule.

Can someone help me get started on this question?

Username Unknown
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2 Answers2

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It is a matter of definitions. When arguments are distinct you have by the very definition of the divided differences, and using permutation symmetry of the objects: $$ f[x_0,...,x_{i-1},u,v,x_{i+1},..,x_n] = \frac{ f[x_0,...x_{i-1},u,x_{i+1}...,x_n] - f[x_0,...,x_{i-1},v,x_{i+1}...,x_n]} {u-v}$$

The case when arguments are equal are defined by taking limits (not described in the wiki page as far as I can see). Your function has to be differentiable in that case. As an example: $$ f'(x_0) = \lim_{x_1\rightarrow x_0} \frac{f(x_1)-f(x_0)}{x_1-x_0} = \lim_{x_1\rightarrow x_0} f[x_0,x_1]=:f[x_0,x_0]$$

In your case: $$ f[x_0,...,x_i,x_i,...,x_n] = \lim_{\epsilon\rightarrow 0^+} { f[x_0,...,x_i,x_i+\epsilon,...,x_n]} $$ Using the first part we may rewrite as: $$ \lim_{\epsilon\rightarrow 0^+} \frac{ f[x_0,...x_{i-1},x_i+\epsilon,x_{i+1}...,x_n] - f[x_0,...,x_{i-1},x_i,x_{i+1}...,x_n]} {(x_i+\epsilon) -x_i} $$ which equals the wanted partial derivative (assuming it exists).

H. H. Rugh
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There is something wrong. Some derivatives of $f$ should appear. $$f[x_0,x_1]=\frac{f(x_0)}{x_0-x_1}+\frac{f(x_1)}{x_1-x_0}$$ and so $$\frac{\partial f[x_0,x_1]}{\partial x_1}=\frac{f(x_0)}{(x_0-x_1)^2}+\frac{f'(x_1)}{x_1-x_0}-\frac{f(x_1)}{(x_0-x_1)^2}.$$ The term $f'(x_1)$ cannot disappear.

Gio67
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