I have a function f(w,x,y,z) and a target value A, how can I discover values for w,x,y,z that produce A?

Question

So I have a function that takes four numerical arguments and produces a numerical argument.

f(w,x,y,z) --> A

If I have the function f and a target result A , is there an iterative method for discovering parameters w,x,y,z that produce a given number A ?

If it helps, my function f is a quintic bezier where most of the parameters are determined. I have isolated just these four that are required to fit the value A .

Q(t)=R(1−t)^5+5S(1−t)^4*t+10T(1−t)^3*t^2+10U(1−t)^2*t^3+5V(1−t)t^4+Wt^5

R,S,T,U,V,W are vectors where R and W are known, I have isolated only a single element in each of S,T,U,V that vary as parameters.

Answer 1

If you can impose 3 (or more) additional equations that you know (or suspect) must be true for your 4-variable solution that gives target value A , then you can try applying Newton's method for solving a system of k equations with k unknowns. Otherwise, without a deeper understanding of the structure of the function you are trying to make equal to A , the only general type of technique I'm aware of that's easy to implement is to simply define the error function as g(w,x,y,z) = |f(w,x,y,z) - A| and search for a minimum of g . Typically the "minimum" found will be a local minimum, so it may require many restarts of the minimization problem with different starting values for your parameters to actually find a solution that gives a local minimum you want of g = 0 . This is very easy to implement and try in a few lines eg in MATLAB using fminsearch

Answer 2

The set of solutions of the equation f(w,x,y,z)=A (where all of w , x , y , z and A are scalars) is, in general, a 3 dimensional manifold (surface) in the 4-dimensional space R^4 of (w,x,y,z) . Ie, the solution is massively non-unique.

Now, if f is simple enough for you to compute its derivative, you can use the Newton's method to find a root: the gradient is the direction of the fastest change of the function, so you go there.

Specifically, let X_0=(w_0,x_0,y_0,z_0) be your initial approximation of a solution and let G=f'(X_0) be the gradient at X_0 . Then f(X_0+h)=f(X_0)+(G,h)+O(|h|^2) (where (a,b) is the dot product). Let h=a*G , and solve A=f(X_0)+a*|G|^2 to get a=(Af(X_0))/|G|^2 (if G=0 , change X_0 ) and X_1=X_0+a*G . If f(X_1) is close enough to A , you are done, otherwise proceed to compute f'(X_1) &c.

If you cannot compute f' , you can play with many other methods.