I want to try to understand a proof of the following statement. Every proof I found took an easier way with the highest coefficient. But I want to understand this proof.
the value $$f[x_j,...,x_{j+k}] is independent of the order of the points $x_j,...,x_{j+k}$.
The proof uses the equation:
$$f[x_j,...,x_{j+k}]=\sum_{i=j}^{j+k}f_i \prod_{\substack{i=j\\ m \neq i}}^{j+k}\frac{1}{x_i-x_m} \ \ \ \ \ (*)$$.
And then does a induction with $k$ For $k=0$ it is clear.
For showing the step from $k-1 \rightarrow k$ we assume that we already proofed $(*)$ for $k-1$ and all $j$. Therefore we can use $(*)$ for $f[x_{j+1},...,x_{j+k}]$ and for $f[x_j,...,x_{j+k-1}]$
Then the proof says to show is:
$$\sum_{i=j}^{j+k}f_i \prod_{m=j \\ m \neq i}^{j+k} \frac{1}{x_i-x_m}=\frac{1}{x_{j+k}-x_j}\left(\sum_{i=j+1}^{j+k}f_i \prod_{m=j+1\\ m\neq i}^{j+k} \frac{1}{x_i-x_m}-\sum_{i=j}^{j+k-1}f_i \prod_{m=j\\ m\neq i}^{j+k-1} \frac{1}{x_i-x_m}\right)$$
But why, does it help to proof that the divided differences are invarinat under permutation I know that I can rewrite the equation wiht $(*)$ to
$$f[x_j,...,x_{j+k}]:=\frac{f[x_{j+1},...,x_{j+k}]-f[x_{j},...,x_{j+k-1}]}{x_{j+k}-x_j}$$.
But I dont understand how I show the statement with that.
Thanks for helping me. And sorry for my horrible english.