# A question about Newton's Divided Difference and continuity of function $f$

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 Newton's Divided Difference and continuity of function $f$. Let $f$ be continuous. Assume that $$X=\{x=(x_0,x_1,\ldots,x_n)\in\mathbb{R}^{n+1}\quad\text{s.t.}\quad x_0,x_1,\ldots,x_n\quad\text{are distinct}\}.$$ Show that $f(x)=f[x_0,x_1,\ldots,x_n]$ is continuous on $X$.

• I would use a different letter for the lastly defined function $f$. – ajotatxe Dec 30 '16 at 16:48

For $j=0,\ldots, n$, fix $x_k$ for $k\neq j$ and define $f_j:\Bbb R\setminus\{x_k:k=0,\ldots, n, k\neq j\}\to\Bbb R$ as $$f_j(t)=f(x_0,\ldots,x_{j-1},t,x_{j+1},\ldots,x_n)$$

You can easily prove (using induction on $n$) that every $f_j$ is continuous.

• Thank you for your answer. I can not show what you mentioned. Is it possible that you prove it. Thanks. – like_math Dec 30 '16 at 17:10
• Can you tell me your comments about this solution? Let $P_{n-1}$ be the polynomial which interpolates the function $f$ at $x_i$, $i=0,1,\ldots,n-1$. So we have $$f(x)-P_{n-1}(x)=(x-x_0)(x-x_1)\cdots(x-x_{n-1})f[x_0,x_1,\ldots,x_{n-1},x],$$ where $x\neq x_i$, $i=0,\ldots,n-1$. Suppose that $x=x_n$. Since $x_n\neq x_i$, $i=0,1,\ldots,n-1$, it follows that $$f[x_0,x_1,\ldots,x_n]=\frac{f(x_n)-P_{n-1}(x_n)}{\prod_{i=0}^{n-1}(x_n-x_i)}.$$ $f$ and $P_{n-1}$ are continuous and $\prod_{i=0}^{n-1}(x_n-x_i)\neq 0$, so $f[x_0,x_1,\ldots,x_n]$ is continuous. – like_math Dec 31 '16 at 16:14

We will prove by induction on $n$.

Base case: Assume that $x_0$ is fixed and $g(x)=f[x_0,x]$. But $$g(x)=\frac{f(x)-f(x_0)}{x-x_0}.$$ Since $x\neq x_0$ and $f$ is continuous, it is obvious $g(x)$ is continuous on
$$X=\{(x_0,x)\in\mathbb{R}^{2}\quad\text{s.t.}\quad x_0,x\quad\text{are distinct}\}.$$ Induction step: Let $n$ be given and suppose $f(x)=f[x_0,x_1,\ldots,x_n]$ is continuous on $$X=\{(x_0,x_1,\ldots,x_n)\in\mathbb{R}^{n+1}\quad\text{s.t.}\quad x_0,x_1,\ldots,x_n\quad\text{are distinct}\}.$$ Now, assume that $g(x)=f[x_0,x_1,\ldots,x_n,x]$ where $x_0,x_1,\ldots,x_n$ are fixed and distinct. But $$g(x)=\frac{f[x_1,\ldots,x_n,x]-f[x_0,x_1,\ldots,x_n]}{x-x_0}.$$ Since $x\neq x_0,x_1,\ldots,x_n$ and $f[x_1,\ldots,x_n,x]$ is continuous by induction hypothesis, $g(x)$ is continuous too. So $f(x)=f[x_0,x_1,\ldots,x_n,x_{n+1}]$ is continuous on $$X=\{(x_0,x_1,\ldots,x_n,x_{n+1})\in\mathbb{R}^{n+2}\quad\text{s.t.}\quad x_0,x_1,\ldots,x_n,x_{n+1}\quad\text{are distinct}\},$$ and the proof of the induction step is complete.

• @ajotatxe, Can you tell me your comments about the above solution? – like_math Dec 31 '16 at 16:17