$$f_{i}\left ( x_{1} x_{2},...,x_{n}\right )=x_{i}\left [ \left ( \sum^{n} _{j = 1} x_{j}\right ) -x_{i}\right ]-k_{i}m_{i}=0$$
Where I have a set of non linear functions as above where n is the number of equations and k and m are measured constants (and therefore have some error). Is it possible to analytically or numerically solve this system of equations?
This is part of an electronics project working on reconstructing capacitor networks so although an analytical solution would be nice, a numerical solution would still be useful.
Considering that you have exactly $n$ equations $$f_{i}=x_{i}\left [ \left ( \sum^{n} _{j = 1} x_{j}\right ) -x_{i}\right ]-k_{i}m_{i}=0$$ let us define an extra variable $S$ $$S=\sum^{n} _{j = 1} x_{j}\implies x_i(S-x_i)=k_im_i\implies x_i=\frac{1}{2} \left(S\pm\sqrt{S^2-4 k_i m_i}\right)$$I suppose that we need to use the $-$. Then,summing over $i$, $$S=\frac n 2S-\frac 1 2\sum^{n} _{i = 1}\sqrt{S^2-4 k_i m_i}$$ let you with one equation for one unknown $S$ $$g(S)=(n-2)S-\sum^{n} _{i = 1}\sqrt{S^2-4 k_i m_i} =0$$ to be solved by a numerical method (such as Newton).
You can bracket the solution $$\frac{ n}{\sqrt{n-1}}\sqrt{(k_im_i)_{min}} < S <\frac{ n}{\sqrt{n-1}}\sqrt{(k_im_i)_{max}}$$
When you have the solution for $S$, you have all the $x_i$'s.
Edit
Just for illustration purposes, I used $n=10$ and $k_im_i=i^{\pi/2}$. The zero fo $g(S)=0$ corresponds to $S=13.9954$ from which are deduced the $x_i$ $$\left( \begin{array}{cc} i & x_i \\ 1 & 0.07182 \\ 2 & 0.21558 \\ 3 & 0.41353 \\ 4 & 0.66187 \\ 5 & 0.96130 \\ 6 & 1.31588 \\ 7 & 1.73349 \\ 8 & 2.22785 \\ 9 & 2.82357 \\ 10 & 3.57047 \end{array} \right)$$