2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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 $\endgroup$ – Javier Sep 22 '17 at 8:13
2
$\begingroup$

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, $$\sum r_i^4 = (s_1^2-2s_2)^2 - 2(s_2^2-2s_1s_3 + 2s_4).$$

$\endgroup$

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.