0
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ what have you done so far? $\endgroup$ – freak_warrior Nov 11 '13 at 14:49
0
$\begingroup$

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 $(x_i,y_i)\rightarrow (x_{\pi(i)},y_{\pi(i)})$.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.