What can be deduced about the roots of this polynomial?

In summary, the polynomial has a characteristic polynomial of the form:$$\begin{align*} P(z) &= z^{m_1+m_2+m_3+m_4}-z^{m_2+m_3+m_4}+z^{m_1+m_3+m_4}+z^{m_1+m_2+m_4}-z^{m_1+m_2+m_3} \\&= z^{(m_1+m_2+m_3+m_4)^2}-z^{m_2+m_3+m_
  • #1
Dowland
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0
Hello everyone,

I'm currently doing some research about feedback systems in engineering and right now I'm playing around with special types of feedback matrices. In the process, I stumbled upon a potentially interesting polynomial, which is actually the characteristic polynomial of the system. It looks as follows:

$$\begin{align*} P(z) &= z^{m_1+m_2+m_3+m_4}
- z^{m_2+m_3+m_4} + z^{m_1+m_3+m_4} + z^{m_1+m_2+m_4} - z^{m_1+m_2+m_3} \\
&\quad - z^{m_3+m_4} - z^{m_2+m_4} - z^{m_1+m_3} - z^{m_1+m_2}
- z^{m_4} + z^{m_3} + z^{m_2} - z^{m_1} + 2\end{align*}$$

Where the m_i's are positive integers. Since my background in algebra is a little weak, I'm basically curious if anyone with more insight can see something interesting about this polynomial? Maybe someone could even give a hint about some theorem or similar that could be helpful? I would like to be able to say something about its roots over the complex numbers but so far I have not find any useful theorem in my basic abstract algebra textbook. It almost looks symmetric, except for the annoying switching of signs in front of the different terms, and unfortunately, it's not skew-symmetric either. The only thing I can say is that I've been trying different values for the m_i's in Matlab (like more or less random positive integers) and all the roots always end up on the unit circle, which is one of the reasons I think the issue is interesting.
 
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  • #2
This is not my field, but ##x_k=z^{m_k}## might work better.
 
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Likes anuttarasammyak
  • #3
Hi. I observe:

P(1)=1 so P(1)-1 has factor (z-1).

P(z)-2 has factor z^n where n=min{m1,m2,m3,m4}.

m1 and m4 are exchangeable. m2 and m3 are exchangeable. To show it explicitly
[tex]P(z)=abcd+ad(b+c)-bc(a+d)-(a+d-1)(b+c+1)+3[/tex]
[tex]=(1-a)(1-d)(1+b)(1+c)+3-ad-bc[/tex]
[tex]=(1-a)(1-d)(1+b)(1+c)-(1-a)(1-d)-(1+b)(1+c)+(1-a)+(1-d)+(1+b)+(1+c)+1[/tex]
[tex]=[1-(1-a)(1-d)][1-(1+b)(1+c)]+(1-a)+(1-d)+(1+b)+(1+c)[/tex]
where ##a=z^{m_1}## and so on. This shows (a and -b) and (d and -c) are exchangeable and so on.
 
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  • Informative
Likes Keith_McClary
  • #4
I find your question interesting; and especially, your empirical observation that all the roots are on the unit circle. I have not tried to think about your question. I wrote to you because of your question itself, and about your other post about G Chrystal's Textbook of Algebra. I have been browsing the book for a while now. I am looking for a mathematics friend with whom I can discuss it.
 

FAQ: What can be deduced about the roots of this polynomial?

How can I find the roots of a polynomial?

To find the roots of a polynomial, you can use various methods such as factoring, synthetic division, the Rational Root Theorem, or numerical methods like Newton's method or the bisection method.

Can all polynomials be factored to find their roots?

Not all polynomials can be factored easily to find their roots. Some polynomials may have complex roots or roots that cannot be expressed in terms of real numbers.

What is the significance of finding the roots of a polynomial?

Finding the roots of a polynomial can provide valuable information about the behavior and properties of the polynomial function, such as the x-intercepts, the number of real roots, and the nature of the roots (real or complex).

How do the roots of a polynomial relate to its graph?

The roots of a polynomial correspond to the x-intercepts of its graph. The graph of a polynomial will intersect the x-axis at the points where the polynomial equals zero, which are the roots of the polynomial.

Can the roots of a polynomial be used to factorize it?

Yes, the roots of a polynomial can be used to factorize it using the Factor Theorem. If you know the roots of a polynomial, you can use them to write the polynomial as a product of linear factors.

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