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 $n$).

  • $\begingroup$ 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 $\endgroup$ – hungryformath Nov 17 '15 at 2:19
  • 1
    $\begingroup$ @hungryformath I will go through this again and add more details when I get time (hopefully later today). $\endgroup$ – Kibble Nov 17 '15 at 16:47
  • $\begingroup$ 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 $\endgroup$ – hungryformath Nov 17 '15 at 20:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.