## Roots of polynomials with identical magnitudes of all coefficients but alternating signs [closed]

Consider the following two polynomials: $Poly1: \lambda^{n}+a_{1} \lambda^{n-1}+a_{2} \lambda^{n-2}+ a_{3} \lambda^{n-3} + \cdots \quad poly2: \lambda^{n}-a_{1} \lambda^{n-1}+a_{2} \lambda^{n-2}-a_{3} \lambda^{n-3}+\cdots$

i.e., the magnitudes of the coefficients of each power of $\lambda$ are identical in both the polynomials but the signs alternate. For such polynomials, is there any relationship between the roots of these polynomials?

Thank you.

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## closed as too localized by Angelo, Andres Caicedo, Pietro Majer, Daniel Moskovich, Pete L. ClarkDec 26 2010 at 11:11

If I am not missing something in your question, this looks simple: $\lambda_{poly1} = - \lambda_{poly2}$...

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I am comparing the following two polynomials: s^6 + 2s^5 + 4s^4 + 5s^3 + 3s^2 + 6s + 20 and s^6 - 2s^5 + 4s^4 - 5s^3 + 3s^2 - 6s + 20 The alternating terms have negative coefficients. I would like to know if there is any relationship between their roots. – ndevarak Dec 26 2010 at 7:22
Hey, no homework questions please! :) – Alexey Lvov Dec 26 2010 at 7:26
I need this for my research. It is the basis for a conjecture which I feel is true but don't have a proper mathematical proof. A solution to this problem must have already been published but I am unable to find it. Which is why I need help to clarify if my assumption is right or wrong. – ndevarak Dec 26 2010 at 7:47
@Alexey Lvov: Those numbers for just for example. – ndevarak Dec 26 2010 at 7:52
Let $P(\lambda)$ be your first polynomial and $Q(\lambda)$ be your second polynomial. Then, if $n$ is even, $P(-\lambda) = Q(\lambda)$, and if $n$ is odd, $P(-\lambda) = -Q(\lambda)$. It's just a simple exercise in substitution! Having proven this it's obvious that the roots of Q are just the negatives of the roots of P. – Zhen Lin Dec 26 2010 at 9:51
Well, if n is even, then it seems like $$p_1(x) = p_2(-x)$$ and hence $x$ is a root of $p_1$ iff $-x$ is a root of $p_2.$
if $p_1(x)=\sum_k b_k x^k=0$, then $$\pm p_2(-x)=\sum_k (-1)^k b_k (-x)^k=\sum_k b_k x^k=0$$
Therefore this is indeed most probably homework $\smile$...