1
$\begingroup$

I'm given a polynomial 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 $\sum r_1^2r_2$, that is, the sum of all different monomials one can form with permutations of the variables on the expression $r_1^2r_2$.

My approach

I know the relation between a polynomial an its roots. So, I know what is the value of the elementary symmetric polynomials $s_i$ and I just need to express all the polynomial $\sum r_1^2r_2$ in terms of $s_i$. There is a generic procedure to do so which I take from Cox's book "Galois theory".

My problem

However, I see that computing the value by hand apparently takes a lot of time. So for the first summand I should take the symmetric polynomial $f-s_1s_2$. Then I should repeat the process until I get to a zero polynomial.

I think there is a way to speed up the computations based in a form of symbolic computation. For instance in page 33, Cox reasons in the following way:

Given $f = \sum_4 x_1^3x_2^2x_3$ the leading term of $f$ is $x_1^3x_2^2x_3$ and therefore we should use polynomial $s_1s_2s_3$. Then $f_1 = f - s_1s_2s_3$ and here comes the magic:

$\sum s_1s_2s_3 = \sum_4 x_1^3x_2^2x_3+3\sum_4x_1^3x_2x_3x_4+3\sum x_1^2x_2^2x_3^2+ 8 \sum_4x_1^2x_2^2x_3x_4$

It is true that Cox claims that maybe a computer is helpful for this case. But how could I calculate this expression in order to avoid to directly calculate all the monomials and doing all the calculations?

$\endgroup$
2
$\begingroup$

It's not that hard. So, what you have is (I shall use $a$, $b$, $c$, and $d$, instead of $r_1$, $r_2$, $r_3$, and $r_4$)$$a^2b+a^2c+a^2d+b^2a+b^2c+b^2d+c^2a+c^2b+c^2d+d^2a+d^2b+d^2c.\tag{1}$$The term in which $a$ has the greatest exponent is $a^2b$ (and $a^2c$ and $a^2d$). So, from $(1)$ you subtract $(ab+ac+ad+bc+bd+cd)(a+b+c+d)$, thus getting$$-3(abc+abd+acd+bcd)$$and you're done!

$\endgroup$
2
$\begingroup$

$$\sum_{sym}r_1^2r_2=2\sum_{cyc}r_1^2(r_2+r_3+r_4)=\frac{1}{2}\sum_{cyc}r_1\sum_{sym}r_1r_2-6\sum_{cyc}r_1r_2r_3.$$

Actually, let $r_1=a$, $r_2=b$, $r_3=c$ and $r_4=d$.

Thus, $$\sum_{sym}a^2b=2\sum_{cyc}a^2(b+c+d)=$$ $$=2(a^2(b+c+d)+b^2(a+c+d)+c^2(a+b+c)+d^2(a+b+c));$$ $$\sum_{cyc}a=a+b+c+d;$$ $$\sum_{sym}ab=4(ab+ac+bc+ad+bd+cd);$$ $$\sum_{cyc}abc=abc+abd+acd+bcd.$$ In our case $$\sum_{sym}a^2b=2\sum_{cyc}a^2(b+c+d)=$$ $$=2(a+b+c+d)(ab+ac+ad+bc+bd+cd)-6(abc+abd+acd+bcd).$$

$\endgroup$
2
  • $\begingroup$ can i ask what does $\sum_{cyc}$ means exactly? you sum only on cycle permutation is it? $\endgroup$ – Rodrigo Sep 21 '17 at 21:57
  • $\begingroup$ @Rodrigo $\sum\limits_{cyc}$ it's a cyclic sum. $\endgroup$ – Michael Rozenberg Sep 22 '17 at 1:08

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.