# Can anyone help me solve the following system of equations numerically?

I have the following 4 equations:

$$\begin{eqnarray} z &=& c_0 + c_1x + c_2y + c_3xy + c_4x^2 + c_5y^2 \\ 2(x-x_0) &=& \lambda(c_1 + c_3y + 2c_4x) \\ 2(y-y_0) &=& \lambda(c_2 + c_3x + 2c_5y) \\ 2(z-z_0) &=& -\lambda \end{eqnarray}$$

where $$x, y, z$$, and $$\lambda$$ are unknown, $$c_i$$ are known coefficients, and $$\mathbf{x}_0\!=\!(x_0, y_0, z_0)$$ is also known. The system is what results when trying to derive the nearest point to $$\mathbf{x}_0$$ on the surface defined by the first equation. Note that $$\mathbf{x}_0$$ does not lie on the surface, and therefore does not satisfy the first equation.

I originally posted here, hoping to find an analytical approach to solving this non-linear system, but to no avail. Can anyone advise me on solving this numerically? I have no experience with solving non-linear systems of equations.

• Why not symbolically? It's all polynomial. Dec 8, 2019 at 10:17
• Compute the residual $r$ with $z_0+r=c_0+c_1x_0+c_2y_0+c_3x_0y_0+c_4x_0^2+c_5y_0^2$ and then use homotopy continuation where the first equation is $z+(1-t)r=c_0+c_1x+c_2y+c_3xy+c_4x^2+c_5y^2$. This should work when ${\bf x}_0$ is close to the surface, if it is farther apart, the homotopy path could curve back or move to infinity. Dec 8, 2019 at 11:01

Eliminating the variables from the second, third and fourth equations $$x=\frac{-c_2 c_3 \lambda ^2-2 (c_1+c_3y_0) \lambda -4 x_0}{c_3(c_3+2 c_5) \lambda ^2+4 c_4 \lambda -4}$$
$$y=\frac{2 (x-x_0)}{c_3 \lambda }-\frac{2 c_4 x+c_1}{c_3}$$ $$z=z_0-\frac \lambda 2$$
Replace $$x,y,z$$ in the first equation to get an ugly polynomial in $$\lambda$$ of degree $$5$$.
• Thanks for the prompt response! Did you manipulate the 2nd and 3rd equations in my post to eliminate $y$? Dec 8, 2019 at 7:32
• @niran90. I used the second, third and fourth equations to solve for $(x,y,z)$. I kept the solution for $x$. Now, I solved the third to get $y$ and the fourth to get $z$. At this point, everything is function of $\lambda$. Dec 8, 2019 at 7:35