Permutation of roots of a polynomial

econ101

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Hello!

new user to this cool website :)
I have a question on characterising the roots of n-degree polynomials. the polynomial comes from an eigenvalue problem, and the goal is to find relations between the roots of the polynomial (without finding the analytical solution).

to this regard, I am stucked with the following problem.

there is a 4-degree monic polynomial like that:
x^4 -a*x^3 + b*x^2 -a*(c^-1)*x + c^(-2),

where {a,b,c} are known parameters (real numbers).

in the article I am reading, the author rewrites the polynomial as such:

(x-x4)*(x-x3)*(x-x2)*(x-x1),

where {x4,x3,x2,x1} are the four roots of the polynomial.

and then he simply says "because of the symmetry of the polynomial, the four roots
satisfy: x1=(c*x2)^-1 and x3=(c*x4)^-1 "

What I understand: the 4-degree polynomial has 4 roots, so (1) can be rewritten as in (2). I understand that plugging the suggested permutation works out in the polynomial.

What I don’t understand: How did the author find the permutation? I cant find a way to come up with them.

I did a simple exercise with a second order monic polynomial and found a permutation for the roots (such as x1 is affine in x2), but I could only do so AFTER I had an analytical solution. I wonder how the guy found his permutations without explicitly solving?

thanks for any help!

truely yours
 
Your equation is \(\displaystyle x^4- ax^3 + bx^2 -ac^{-1}x + c^{-2}= 0\). Multiplying by \(\displaystyle c^2\), that is the same as \(\displaystyle c^2x^4- ac^2x^3+ bc^2x^2- acx+ 1= 0\). Let y= 1/cx. Then x= 1/cy so we have
\(\displaystyle \frac{1}{c^2y^4}- \frac{a}{cy^3}+ \frac{b}{y^2}- \frac{a}{y}+ 1= 0\). Now multiply through by \(\displaystyle y^4\): \(\displaystyle c^{-2}- ac^{-1}y+ by^2- ay^3+ y^4= 0\).

Exactly the original equation!
 
many thanks for your reply. I now better understand how the permutation works. Yet I still dont understand how you come up with the form of the permutation. Simply put: "Mmmmhh this polynomial looks like this. Let me try to use the permutation y=(c*x)^1 ". How do you think of such form of the permutation in the first place? (and not, say, y = x + b)
 
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