# Solve a simple multivariate polynomial

Suppose that, given constants $$\mu_0,...\mu_{n-1}$$, there exists a solution $$(x_1,...,x_n) \in \mathbb{R}_{+}^{n}$$ to the following system :

$$\mu_i = \sum\limits_{j=1}^{n} x_j^i$$

Can we find it for any n ?

In the over way around, this is related to Vandermonde matrices. I tried some optimisation algorithms on it and it converges quite well, so I'm pretty sure there exist an analytical solution.

E.g, for $$n=2$$, take $$x_{1,2} = \mu_1 \pm \sqrt{\frac{\mu_2}{2} - \mu_1^2}$$

• Hem, your first equation is $\mu_0=\sum_{j=1}^n x_j^0=n.$
– user65203
Jul 4 '19 at 10:23
• How is "good" convergence related to the existence of an analytical solution ???
– user65203
Jul 4 '19 at 10:26
• My first equation is indeed $\mu_0 = n$. When supposing that a solutin exist, i assume that $\mu_0 = n$. Furthermore you are right, good convergence that not mean there is an analytical solution.
– lrnv
Jul 4 '19 at 13:53

## 1 Answer

I am pretty sure that there is no analytical solution, as the system involves polynomials of degree up to $$n$$.

In the case $$n=3$$, the equations define a plane, a sphere and a cubic surface.

The intersection of the plane and the sphere is a circle. The intersection of the plane and the cubic surface can be three straight lines forming an equilateral triangle, or a set of curves asymptotic to these. There can be six intersections with the circle, and if I am right, you can obtain them by solving a sextic polynomial.

As $$n$$ grows, this quickly becomes intractable.

• Thanks for the analysis. Do you think we can write the polynomial to be solved for any n ?
– lrnv
Jul 4 '19 at 13:56
• @lrnv: probably, using Grobner basis. I would expect the degree to explode.
– user65203
Jul 4 '19 at 14:28
• This is a different question, but if i change my equations to include weights that depend only on j (and sum up to a finite number, say n or 1) $$\mu_i = \sum\limits_{j=1}^^{n} \rho_j x_j^i$$ would that help or make it worst ?
– lrnv
Jul 4 '19 at 16:30
• This is a different question, but if i change my equations to include weights that depend only on j (and sum up to a finite number, say n or 1) $$\mu_i = \sum\limits_{j=1}^^{n} \rho_j x_j^i$$ would that help or make it worst ?
– lrnv
Jul 4 '19 at 16:30
• @lrnv: how could that help ? This is a more general problem, which includes the first. Or do you mean unknown weights ?
– user65203
Jul 5 '19 at 6:58