Inverse of a particular function

Hi, I should get the inverse of this function, expressing v in dependence of x, y.

y = x * cos(norm(v)) + v * sin (norm(v))/norm(v);

Could you write here the Matlab code, if it's possible?

Answer 1

[Roger Stafford]

It is possible to solve your equation for v explicitly. However, it should be understood that if any solutions exist, there will always be infinitely many of them. The vectors x, y, and v can have any number of elements, though these should all three be the same.

 % The problem: Given vectors x and y, find vector v such that
 %  y = x*cos(*norm(v))+v/norm(v)*sin(norm(v))

 p = norm(x);
 i = x/p;         % Unit vector in vector x direction
 r = dot(x,y)/p;  % Component of y in x direction
 j = y-r*i;
 s = norm(j);     % Component of y orthogonal to x
 j = j/s;         % Unit vector in plane of x & y, orthogonal to vector x

 % v must be in the plane containing vectors x and y
 % and can therefore be expressed as:
 %  v = u*(cos(t)*i+sin(t)*j)
 % for some scalars, t and u.
 % The original vector equation can therefore be
 % expressed as the solution to two scalar equations
 % in two scalar unknowns, t and u, given the
 % values r, s, and p:
 %  r = p*cos(u) + cos(t)*sin(u)
 %  s = sin(t)*sin(u)
 %
 % For a solution to exist, r and s must lie within
 % the ellipse  r^2/(p^2+1)+s^2 < 1
 if r^2/(p^2+1)+s^2 >= 1, fprintf('No solutions exist\n'), else

 % It can be shown that we can find four solutions for t and u as follows:
 q = atan2(r^2-p^2-s^2,2*r*s);
 a = asin((r^2-p^2+s^2+2*p^2*s^2)/sqrt((r^2-p^2-s^2)^2+(2*r*s)^2));
 t = mod([(a-q)/2,(a-q)/2+pi,(pi-a-q)/2,(-pi-a-q)/2],2*pi);
 % The corresponding solutions for u are:
 u = mod(atan2(s./sin(t),(r*sin(t)-s*cos(t))./(p*sin(t))),2*pi);
 end

Each of these t,u pairs defines a solution vector, v:

v = u*(cos(t)*i+sin(t)*j)

Also infinitely many other solutions for v can be found by adding arbitrary positive multiples of 2*pi to u for each of the above four solutions.

[Roger Stafford]

Added note: The "elliptical" inequality, r^2/(p^2+1)+s^2 <= 1, follows from the necessary equality

 sin(2*t+q) = (r^2-p^2+s^2+2*p^2*s^2)/sqrt((r^2-p^2-s^2)^2+(2*r*s)^2)

by requiring that sin(2*t+q)^2 <= 1.

Answer 2

[John D'Errico]

Assuming that x and y are vectors, as is v? Note that your expression is ambiguous, as x could be a scalar.

I'll confidently state there is no analytical solution for v. You could probably write it as the solution of a nonlinear system of equations, so using fsolve.