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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 $g'(x)=f[x_0,x_1,\ldots,x_n,x,x]$?

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    $\begingroup$ 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 $n$. $\endgroup$ – Jack D'Aurizio Dec 30 '16 at 18:12
  • $\begingroup$ @ Jack D'Aurizio, Thanks for your answer, however I can't understand why $\lim_{\epsilon\rightarrow 0}h[x,x+\epsilon]=h[x,x]$? $\endgroup$ – like_math Dec 30 '16 at 20:29
  • $\begingroup$ @ 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? $\endgroup$ – like_math Dec 30 '16 at 20:36
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    $\begingroup$ 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 $\frac{0}{0}$. $\endgroup$ – Jack D'Aurizio Dec 30 '16 at 20:49
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    $\begingroup$ 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 $h[x,x]$? $\endgroup$ – Jack D'Aurizio Dec 30 '16 at 20:55
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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?

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  • $\begingroup$ @Jack D'Aurizio, Can you tell me your comments about the above solution? $\endgroup$ – like_math Dec 31 '16 at 16:27
  • $\begingroup$ can you please find a formula for $$g''(x)$$? $\endgroup$ – Aman Gupta Nov 25 '19 at 12:29

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