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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 $x_{1,2} = \mu_1 \pm \sqrt{\frac{\mu_2}{2} - \mu_1^2}$

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  • $\begingroup$ Hem, your first equation is $\mu_0=\sum_{j=1}^n x_j^0=n.$ $\endgroup$
    – user65203
    Commented Jul 4, 2019 at 10:23
  • $\begingroup$ How is "good" convergence related to the existence of an analytical solution ??? $\endgroup$
    – user65203
    Commented Jul 4, 2019 at 10:26
  • $\begingroup$ 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. $\endgroup$
    – lrnv
    Commented Jul 4, 2019 at 13:53

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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.

enter image description here

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  • $\begingroup$ Thanks for the analysis. Do you think we can write the polynomial to be solved for any n ? $\endgroup$
    – lrnv
    Commented Jul 4, 2019 at 13:56
  • $\begingroup$ @lrnv: probably, using Grobner basis. I would expect the degree to explode. $\endgroup$
    – user65203
    Commented Jul 4, 2019 at 14:28
  • $\begingroup$ 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 ? $\endgroup$
    – lrnv
    Commented Jul 4, 2019 at 16:30
  • $\begingroup$ 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 ? $\endgroup$
    – lrnv
    Commented Jul 4, 2019 at 16:30
  • $\begingroup$ @lrnv: how could that help ? This is a more general problem, which includes the first. Or do you mean unknown weights ? $\endgroup$
    – user65203
    Commented Jul 5, 2019 at 6:58

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