My prof. gave this problem as a bonus in an exam, and I couldn't figure out a solution. Some hints or a general method of solving it would be very nice. Given the following polynomial: $$P(x)=a_{3}x^{3}+a_{5}x^{5}+a_{7}x^{7}+\dots+a_{199}x^{199}+a_{201}x^{201}$$

and knowing its solutions, provided in the following plot

(where the blue/red dots represent $\mathrm{Re[x]}/\mathrm{Im[x]}$) find the coefficients $a_{2k+1}$. The problem also stated that $a_{2k+1}\in\mathbb{R}_{>0}$ and $\lim_{k\rightarrow\infty}a_{2k+1}=1$ (in the exam we also had a very long list of all the numerical solutions for $P(x)$).

  • $\begingroup$ Use symmetric functions on the roots. $\endgroup$ – Vladhagen Jul 4 '14 at 1:09
  • 1
    $\begingroup$ I see $P(x) = x^3 Q(x)$ where $Q(x)$ is a degree-$99$ polynomial in $x^2$. Also, other than a multiple root at $0$, the roots in the plot seem almost (but not quite) like the $n$th roots of unity. $\endgroup$ – David K Jul 4 '14 at 2:35

Given the roots of a polynomial, call them $r_1, r_2, r_3, \ldots r_{201}$, then the coefficients are given by the elementary symmetric functions on the roots.

If you are not familiar with these, read this article on Wikipedia

So, $a_{2k+1} = \dfrac{e_{201-2k -1}(r_1, r_2, \ldots r_{201})}{a_{201}}$

This gives a "general" solution. Unless he/she gave you what $a_{201}$ is, I am not sure of a good method for getting it (since it actually can be whatever you want; the other coefficients will then depend on it, as we divide by $a_{201}$. This perhaps indicates that $a_{201} = 1$ could be a good choice!).

If you have the roots given to you, do some sort of thing like this: $$\prod_{i =1}^{201}(x-r_i)$$ on a computer. I am assuming that this is a numerical methods class and you have some computer ability. It should spit out what your polynomial is.

  • $\begingroup$ As you say, to get the roots right it doesn't matter what scaling factor you use, so you can start with $a_201=1$ to get a general relation. Then noticing that $\Re (P(100)) = 1$ and $\Im (P(100)) = 0$ according to the graph, you should be able to fix that scale appropriately. $\endgroup$ – ConMan Jul 4 '14 at 2:01

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