# Solving a system of algebraic equations

I am trying to find complex solutions with positive real part $\{t_j \;|\;{\rm Re}\;t_j>0, j = 1, 2, 3, \dots, n\}$ of the system of equations $$0 = 1 + \sum_j \left(t_j^{2l+1} + {t_j^*}^{2l+1}\right),\; l = 1,2,3,\dots m.$$ Where for a given $n$ I would like to make $m$ as large as possible. Since, this system is non-analytic and thus for $n=m$ most likely under-constrained, I had the idea to just fix the magnitude of all solutions to 1: $t_j = e^{i\phi_j}$ with $-{\pi\over 2} < \phi_1 \le \phi_2 \le \dots \le \phi_n <{\pi\over 2}$. In terms of these the system becomes: $$0 = 1 + 2\sum_j \cos{\left[\phi_j(2l+1)\right]},\; l = 1,2,3,\dots n.$$ This definitely has solutions up to $n=m=2$, but already for $n=3$, my naive attempt at numerically solving this (Mathematica's NSolve) is taking quite long. Is there some better way to find or at least confirm the existence of such solutions?

Thanks, Nik

• What is $t_j^*$? Is that the complex conjugate? – Gerry Myerson Oct 11 '12 at 22:34
• So letting $\phi_0=1$ and $\phi_{-i}=-\phi_i$ you want that for $0 \le \ell \le m$ the values $(e^{i\phi_k})^{2\ell+1}$ for $-n \le k \le n$ to have average real part $0$. If they were, in each case, equally distributed around the unit circle, that would suffice. – Aaron Meyerowitz Oct 12 '12 at 0:42
• Gerry, yes that is correct, I should have said that! – nik Oct 12 '12 at 15:15
• Aaron, yes, but is it possible to use those considerations to construct solutions with ${-\pi\over2} < \phi_j < {\pi\over 2}$? I.e. they must lie in the positive half-plane. – nik Oct 12 '12 at 15:22

I would try to get rid of the trigonometric functions, and rather rewrite the system as a polynomial system. If $$x_j=\cos(\phi_j)$$, then $$\cos(\phi_j(2\ell+1))=T_{2\ell+1}(x_j)$$, where $$T_k$$ is the $$k$$-th Chebyshev polynomial of degree $$k$$. So your system of equations is $$\begin{equation} 0=1+2\sum_j T_{2\ell+1}(x_j),\;\;l=1,2,\dots,m, \end{equation}$$ with the requirement that $$0. Such a system can be discussed via Groebner bases, see here how to do that with Sage. For $$m=n\le4$$ there are only finitely many solutions. Among them pick those which fit your inequalities. For instance, for $$m=n=4$$, an approximation of a solution seems to be

\begin{align*} x_1 &= 0.963494595276259\\ x_2 &= 0.852773246361416\\ x_3 &= 0.600336170417163\\ x_4 &= 0.058262327046178 \end{align*}

Of course, you can use this technique also to handle the original case, where $$t_j$$ need not have length $$1$$. For instance, for $$n=2$$ one can show that $$m\le4$$, and and approximate solution for $$m=4$$ is $$t_1=0.466916296430820 + 0.717248344919154i$$, $$t_2=0.856453001234213 + 0.445264622297009i$$.

One cannot expect simple expressions for the $$t_j$$. For instance, the absolute values of the $$t_j$$ are roots of an irreducible (over the rationals) polynomial of degree $$60$$.

• So what does the final solution for $m=n=4$ come out to be? – Aaron Meyerowitz Oct 13 '12 at 4:57
• @Aaron: Not sure what you mean. We have $t_j=e^{i\phi_j}=\cos(\phi_j)+i\sin(\phi_j)=x_j\pm i\sqrt{1-x_j^2}$. Since nik adds $t_j^{2\ell+1}$ and its complex conjugate, it does not matter which sign you choose in $\pm i\sqrt{1-x_j^2}$ for each $j$. – Peter Mueller Oct 13 '12 at 9:10

Your problem may be understood as solving a system of equations and inequalities over $\mathbb{R}$. It is solvable in the sense that there exists an algorithm to find solutions (e.g. to find a point in every connected component specified by the system just mentioned).

Such algorithms are desribed, e.g., in this book: "Algorithms in Real Algebraic Geometry" by Saugata Basu, Richard Pollack, Marie-Françoise Roy.

How practical they are at present, is another question.

Consider the case $m=n=4.$ If we take $t_j=\cos(\frac{2j\pi}{9})+I\sin(\frac{2j\pi}{9})$ then $\sum_{j=0}^{8}t_j^q= 1 + \sum_{j=1}^{4} \left(t_j^{q} + {t_j^*}^{q}\right)=0$ for $1 \le q \le 8.$ This is because the set of values $t_j^q$ is just the nine $9$th roots of unity (or in two cases, the third roots of unity taken three times). Admittedly, this is not exactly what you wanted.

If you take just the real parts for $j=1,2,3,4$ you get

• $t_1=t_1^*=.766044443118979$
• $t_2=t_2^*=.173648177666934$
• $t_3=t_3^*=-.500000000000$
• $t_4=t_4^*=-0.939692620785905$

It can be seen that this makes $1 + \sum_{j=1}^4 \left(t_j^{q} + {t_j^*}^{q}\right)=0$ correct for $q=1,3,5,7.$

• Hmm, maybe I am missing something, but please see my response to your comment above. – nik Oct 12 '12 at 15:24