Proof that $f[x_{i0},…,x_{ik}] = f[x_0,…,x_k]$

i try to show, that $f[x_0,...x_k]$ is a symmetric function of $x_i$. What means, that for a permutation $x_{i0},...,x_{ik}$ of numbers $x_0, ...,x_k$ applies: $$f[x_{i0},...,x_{ik}] = f[x_0,...,x_k]$$

I got a hint: $f[x_0,...x_k]$ is the coefficient of the highest x-power of the interpolating polynomial $P_{0,...,k}(x)$ through the supporting points $x_0,...,x_k$

But until now, i didn't succeed. Thank you for your help.

• what have you done so far? – freak_warrior Nov 11 '13 at 14:49

1 Answer

The Lagrange Interpolation Polynomial for pairs $(x_1,y_1),\cdots,(x_n,y_n)$ is

$$P(x)=\sum_{j=1}^nP_j(x)$$

where

$$P_j(x)=y_j\prod_{k=1, k\neq j}^n\frac{x-x_k}{x_j-x_k}.$$

It follows that the coefficient of the highest power of $x$ is:

$$\sum_{j=1}^n\frac{y_j}{\prod_{k=1,k\neq j}(x_j-x_k)}$$

which is symmetric under permutations $(x_i,y_i)\rightarrow (x_{\pi(i)},y_{\pi(i)})$.

• How do you know the formular for the highest power of x? – mathNewbie Nov 11 '13 at 22:31
• @mathNewFag: Each polynomial $P_j(x)$ contributes to the sum. Look at the definition of $P_j$, if it helps write it as $P_j(x)=y_j\frac{1}{\prod_{k=1,j\neq j}(x_j-x_k)} \prod_{k=1,k\neq j}^n (x-x_k)$. – Alex R. Nov 12 '13 at 0:01
• Thanks! =) but why is this symmetric? under these permutations. – mathNewbie Nov 12 '13 at 23:15