Multi-dimensional root finding

[natski]

Hi all,

Consider that one has several functions, say 3, of the form f(x,y,z) and g(x,y,z) and h(x,y,z). You know the form of these equations and they are non-linear, long, messy equations. f', g' and h' are even longer and messier and therefore assume that they cannot be found.

Now consider you measure f, g and h and want to determine x, y and z.

What method is recommended for solving this in a numerical and computationally efficient (but simple to implement) manner?

Cheers,

Natski

 

[chaoseverlasting]

Well, its if 3 dimensions (x,y,z), and you can find the values of the functions, what you'll essentially be left with are equations of surfaces. Any solution you obtain will be the intersection of those surfaces. If you know the functions, I'm pretty sure you could use matrices to solve for them.

For example, you have three functions of the form

f(x,y,z)=a_{0}+a_{1}x+...+a_{n}x^n+b_0+b_1y+...+b_ny^n +c_0+c_1z+...+c_nz^n

along with all the xy, yz, zx terms of degree n, then you could create a matrix equation such that AX=B where A holds the coefficients, X holds the values x, x^2, etc and B holds the values of the 3 functions.

then X=A^-1B will give you a solution.

 

[natski]

Cramer's rule only applies for linear equations.
A matrix method would not be involve the Jacobian which cannot be computed in this case.

I think the solution lies within Broyden's method.

 

[chaoseverlasting]

Thats pretty messy though, perhaps another way to do so would be to assume that the function you have is of the form,

f(x,y,z)=(ax+by)^n + (cy+dz)^n +(ez+fx)^n

Since you know f, you could find out the values of the constants. Perhaps to simplify the above equation, you could apply the transformation

X=\frac{a}{\sqrt{a^2+b^2}}x+\frac{b}{\sqrt{a^2+b^2}}y

Y=\frac{c}{\sqrt{c^2+d^2}}y+\frac{d}{\sqrt{a^2+d^2}}z

Z=\frac{e}{\sqrt{e^2+f^2}}z+\frac{f}{\sqrt{e^2+f^2}}x

That would transform the above equation into

f(X,Y,Z)=k_1X^n +k_2Y^n+k_3Z^n

where

k_1=(a^2+b^2)^{\frac{n}{2}}

k_2=(c^2+d^2)^{\frac{n}{2}}

k_3=(e^2+f^2)^{\frac{n}{2}}

 

[chaoseverlasting]

I don't know much about Broyden's method, but from what wikipedia says, wouldn't you have to calculate the Jacobian there too?

If you use the transformation, I think you could use matrices to solve the equations and you can find the jacobian of the transformation. Then you could use cramer's rule to solve for Xⁿ, Yⁿ, Zⁿ.