# Partial Derivative of Newton's Divided Difference

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

I have the base case, namely

$n=1$: \begin{align*} \frac{\partial}{\partial_{x_0}}f[x_0,x_1] &=\frac{\partial}{\partial_{x_0}}\left[\frac{f(x_1)-f(x_0)}{x_1-x_0}\right]\\[1.5ex] &=\frac{f(x_1)-f(x_0)-f'(x_0)(x_1-x_0)}{(x_1-x_0)^2}\\[1.5ex] &=\frac{\frac{f(x_1)-f(x_0)}{x_1-x_0}-f'(x_0)}{x_1-x_0}\\[1.5ex] &=\frac{f[x_0,x_1]-f[x_0-x_0]}{x_1,x_0}\\[1.5ex] &=f[x_0,x_0,x_1] \end{align*} $n=k$:

Now assume that $$\frac{\partial}{\partial_{x_0}} f[x_0,x_1,...,x_k]=f[x_0,x_1,...,x_i,x_i ,x_{i+1},...,x_k]$$

$n=k+1$: $$\frac{\partial}{\partial_{x_0}} f[x_0,x_1,...,x_{k+1}]=f[x_0,x_1,...,x_i,x_i ,x_{i+1},...,x_{k+1}]$$

This is where I'm stuck. Thank you so much for any help.

The definition of the forward divided difference says that: $$f[x_0,x_1,\ldots,x_{n-1},x_n] \equiv \frac{f[x_1,\ldots,x_{n-1},x_n]-f[x_0,x_1,\ldots,x_{n-1}]}{x_{n}-x_0}$$

Now if $0<i<n$ then

$$\frac{\partial f[x_0,x_1,\ldots,x_{n-1},x_n]}{\partial x_i} = \frac{\frac{\partial f[x_1,\ldots,x_{n-1},x_n]}{\partial x_i}-\frac{\partial f[x_0,x_1,\ldots,x_{n-1}]}{\partial x_i}}{x_{n}-x_0}$$

By the induction hypotesis (for $n-1$) we get

$$\frac{\partial f[x_0,x_1,\ldots,x_{n-1},x_n]}{\partial x_i} = \frac{f[x_1,\ldots,x_i,x_i,x_{i+1},\ldots, x_n] - f[x_0,\ldots,x_i,x_i,x_{i+1},\ldots, x_{n-1}] }{x_{n}-x_0}$$

and by the definition of the forward divided difference the right hand side is

$$f[x_0,x_1,\ldots,x_i,x_i,x_{i+1},\ldots, x_n]$$

which is the same as the induction hypotesis (for $n$).

• Thank you so much for your reply. I have two questions for you: 1) I see that you can "distribute" the partial derivative over the numerator b/c $x_n$ and $x_0$ are both considered as constants. However, does this eliminate the case where $x_i=x_0$ or $x_i=x_n$? Or does it not matter b/c you can simply move the terms around (invariance) and just say $x_i \neq$ to the two terms in the denominator? 2) I'm a little unclear on the terms $n-1$ vs $n$ for the assumption step vs induction step. Thanks in advance – hungryformath Nov 17 '15 at 2:19
• @hungryformath I will go through this again and add more details when I get time (hopefully later today). – Kibble Nov 17 '15 at 16:47
• Thanks for your response. I was able to clear up my question #2 by spending a little more time with the problem (it was just late last night). However, my question #1 still stands. Thanks in advance – hungryformath Nov 17 '15 at 20:01