# A question about properties of Newton's Divided Difference

Recall that Newton's Divided Difference: $$f[x_0,x_1]=\frac{f(x_1)-f(x_0)}{x_1-x_0},$$ and $$f[x_0,x_1,\ldots,x_n]=\frac{f[x_1,\ldots,x_n]-f[x_0,\ldots,x_{n-1}]}{x_n-x_0},$$ where $x_0,x_1,\ldots,x_n$ are distinct.

Now I have a question about properties of Newton's Divided Difference. Let $f(x)$ be a function and $x_0,x_1,\ldots,x_n$ are distinct. Define $$g(x)=f[x_0,x_1,\ldots,x_n,x].$$ How can I show that $g'(x)=f[x_0,x_1,\ldots,x_n,x,x]$?

• Since $h'(x) = \lim_{\varepsilon\to 0}\frac{h(x+\varepsilon)-h(x)}{(x+\varepsilon)-x} = \lim_{\varepsilon\to 0} h[x,x+\varepsilon]$ it is enough to apply induction on $n$. – Jack D'Aurizio Dec 30 '16 at 18:12
• @ Jack D'Aurizio, Thanks for your answer, however I can't understand why $\lim_{\epsilon\rightarrow 0}h[x,x+\epsilon]=h[x,x]$? – like_math Dec 30 '16 at 20:29
• @ Jack D'Aurizio, We know that if $h$ is a differentiable function, then $h[x,x+\epsilon]=h'(\xi_x)$, where $\xi_x\in(x,x+\epsilon)$. So, when $\epsilon\rightarrow 0$, then $\lim_{\epsilon\rightarrow 0}h[x,x+\epsilon]=\lim_{\epsilon\rightarrow 0}h'(\xi_x)=h'(x)$. Is that correct? – like_math Dec 30 '16 at 20:36
• If $h$ has a continuous derivative, yes, that is correct. On the other hand, how it is possible to define $h[x,x]$ without a limit? According to the original definition, it should be $\frac{h(x)-h(x)}{x-x}$, but that is just $\frac{0}{0}$. – Jack D'Aurizio Dec 30 '16 at 20:49
• I simply took that as a reasonable definition of $h[x,x]$, $$h[x,x]\stackrel{\text{def}}{=}\lim_{\varepsilon\to 0}h[x,x+\varepsilon].$$ Otherwise, how would you define $h[x,x]$? – Jack D'Aurizio Dec 30 '16 at 20:55

If we define $$f[x_0,x_1,\ldots,x,x]\overset{\text{def}}{=}\lim_{\epsilon\rightarrow 0}f[x_0,x_1,\ldots,x_n,x,x+\epsilon],$$ then \begin{align*} g'(x)&=\lim_{\epsilon\rightarrow 0}\frac{g(x+\epsilon)-g(x)}{\epsilon}\\ &=\lim_{\epsilon\rightarrow 0}\frac{f[x_0,x_1,\ldots,x_n,x+\epsilon]-f[x_0,x_1,\ldots,x_n,x]}{\epsilon}\\ &=\lim_{\epsilon\rightarrow 0}\frac{f[x_0,x_1,\ldots,x_n,x+\epsilon]-f[x,x_0,x_1,\ldots,x_n]}{\epsilon}\\ &=\lim_{\epsilon\rightarrow 0}f[x,x_0,x_1,\ldots,x_n,x+\epsilon]\\ &=\lim_{\epsilon\rightarrow 0}f[x_0,x_1,\ldots,x_n,,x,x+\epsilon]\\ &=f[x_0,x_1,\ldots,,x_n,x,x]\qquad(\text{by definition}). \end{align*} Therefore, we do not need to use induction on $n$. Is that right?
• can you please find a formula for $$g''(x)$$? – Aman Gupta Nov 25 '19 at 12:29