a system of nonlinear equations (power sum)

[mmzaj]

greetings . i cam across this problem in a research paper i'm writing .

x_1+x_2+.....+x_m=k_1
x^2_1+x^2_2+.....+x^2_m=k_2
.......................
.......................
x^m_1+x^m_2+.....+x^m_m=k_m
k_j are constants

in compact form :
\sum^{m}_{i=1}x^{n}_{i}=k_{n}, n=1,2...m


is there a general solution(or algorithm to solve) for x_i !?!?

[Stephen Tashi]

I don't know a general solution, but it's an interesing problem. The thread at least deserves a bump. It's the kind of problem that I suspect has been the subject of some mathematical publications, but I don't know how to formulate good search terms for it.

If this arises in solving some applied math problem, can't you find papers related to that problem that deal with it?

A more general problem than yours is the problem of solving simultaneous polynomial equations in several variables. There is a method, which I have not studied in detail, called "Wu's elimination method" that supposedly does that.


If we tackle this as a math research problem, one approach would be to examine the transformation properties of the vector [k_i] when various transformations are applied to the vector [x_i] .

For example, let [y_i] be some initial guess for a solution and let [\alpha_i] be the vector [\sum_j y_j^i ] . If we multiply [y_i] by the constant \lambda then we change [\alpha_i] to [ \lambda^i \alpha_i] .

The thing to look for would be some more complicated type of transformations. A interesting daydream would be to find some that leave \alpha_i fixed for i = 1,2,..p and change \alpha_p to k_p and do whatever they want to the remaining \alpha_i . You could use a series of such transformations to change the results of an initial guess to the desired result.

If we tackle this problem as a problem in numerical methods, there are probably many ways to do it, but I'm not sure whether that's the sort of approach you are looking for. For example, I don't know whether the [k_i] that you have are the results of physical measurements or whether they might be integers or some other form of exact theoretical data.

[mmzaj]

one way to solve it is the following .

the elementary symmetric polynomials can be defined in terms of k_j :

e_{j}=\frac{1}{j!} \begin{vmatrix}
k_{1} & 1 & 0 & ... & & & & \\
k_{2} & k_{1} & 2 & 0 & ... & & & \\
. & ... & k_{1} & ... & & & & \\
. & & & & . & & & \\
. & & & & & . & & \\
. & & & & & & & \\
k_{j-1}&k_{j-2} & ... & & & & k_{1} & j-1 \\
k_{j} &k_{j-1} & ... & & & & k_{2} & k_{1}
\end{vmatrix}


define the polynomial :
f(x) = \sum_{j=0}^{m}(-1)^je_{j}x^{m-j}

( x_1,x_2,....x_m ) are the solutions for the equation f(x)=0

but i was hoping for some solution method that doesn't include polynomials and their roots !!

[Stephen Tashi]

At least we can confine our attention to solutions on a m-sphere.


Instead of the problem:

\sum^{m}_{i=1}x^{n}_{i}=k_{n}, n=1,2...m

Let \lambda = \sqrt{k_2} and consider the problem:

\sum^{m}_{i=1}y^{n}_{i}=k_{n}/ \lambda^n = \alpha_{n}, n=1,2...m

For n = 2 the equation requires that the \sum^{m}_{i=1} y^2_{i} = 1

If we can solve the second problem then the solution to the first problem is given by
x_i = \lambda y_i .

We should be able to put bounds on the possible values of the \alpha_n that allow a solution. For even n, I'd guess 1 \le \alpha_n \le m (\frac{1}{\sqrt{m}})^n .