I'm given a polynomial $p$ of degree $4$ and its roots, let's call them $r_1,$ $r_2$, $r_3$ and $r_4$. I'm asked to show what is the value of the expression of $\sum r_1^4$, that is, $r_1^4+r_2^4+r_3^4+r_4^4$.

My approach

If I try to use symmetric polynomials in the way that the fundamental theorem of symmetric polynomials does I have to substract $\sum r_1^4 - e_1^4$ where $e_1 = r_1 + r_2 + r_3+ r_4$ this gives a very complicated expression to compute by hand because I would have terms of the form $\sum r_1^3r_2r_3, \sum r_1^2r_2r_3, \cdots$

Instead I read a method which says that I can express my polynomial expression as a given combination of all symmetric polynomials of degree four that can be formed with symmetric polynomials of degree 4. So according to this I should write:

$p = a_1e_1^4+a_2e_2^2+a_3e_1e_3+a_4e_4$

The problem

The problem comes when evaluating this polynomial to obtain the $a_i$. I obtain different coefficients with different evaluations. So for the first step I could choose $r_1 = 0$ and $r_i = 0$ for the rest. Then I get $a_1 = 1$ and the equation becomes:

$p = e_1^4+a_2e_2^2+a_3e_1e_3+a_4e_4$

However, for the second coefficient I could put $r_1 = 1, r_2 = -1$ and the rest $0$. In this case I get $a_2 = 2$. Choosing $r_1 = r_2 = 1$ and the rest $0$, leads to $a_2 = -14$. Since the expression of a polynomial inas a function of symmetric polynomials should be unique this has to be wrong. Furthermore, the sum of the final result obtained by using Cardano-Vieta formulas does not correspond with the true roots of the polynomial I'm given.

What am I doing wrong?

Solution by GAVD

The method given by GAVD illustrates that the error in my developement was that I was missing some elementary symmetric polynomial, namely, $s_1^2s_2$ just by expressing in a canonical way the polynomial that results to him.

  • $\begingroup$ Can you tell us what $p$ is? $\endgroup$ – Mr. Chip Sep 22 '17 at 7:53
  • $\begingroup$ So when you say you're given it, you're actually only given it abstractly? $\endgroup$ – Mr. Chip Sep 22 '17 at 7:57
  • $\begingroup$ sorry for being naive, but what are the summations running on? $\endgroup$ – Math-fun Sep 22 '17 at 7:57
  • $\begingroup$ It is a notation meaning all expressions you can generate from that one by permuting variables.in this case we permute the four variables r $\endgroup$ – Rodrigo Sep 22 '17 at 8:13

Let $s_1 = \sum r_i$, $s_2 = \sum r_ir_j$, $s_3 = \sum r_ir_jr_k$, $s_4=r_1r_2r_3r_4$.

So, you have $$\sum r_i^2 = s_1^2 - 2s_2.$$

Then, $$\sum r_i^4 = (\sum r_i^2)^2 - 2\sum r_i^2r_j^2$$

but $$s_2^2 = (\sum r_ỉr_j)^2 = \sum r_i^2r_j^2 + 2(\sum r_i)(\sum r_ỉr_jr_k) - 2r_1r_2r_3r_4.$$ Thus, $$\sum r_i^4 = (s_1^2-2s_2)^2 - 2(s_2^2-2s_1s_3 + 2s_4).$$


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