I came across this question in my textbook for Nonlinear Optimisation and I don't know what to do:
Consider the function: $$ f(x_1,x_2)=(r-1)^2-\frac{1}{2}(r-1)^2\cos \left( \frac{1}{r-1}-\phi \right) $$
where $0\cos(1/0)=0$ and where $r$ and $\phi$ are the polar coordinates of $(x_1,x_2)$, that is, $r=\sqrt{(x_1^2+x_2^2)}$ and $\phi=bgtan(x_1/x_2)$.
The minimum of this function is attained at every point of the circle $r = 1$.
Consider the following optimisation method: the gradient is determined in a point $x_k$.
Subsequently $x_{k+1}$ is chosen as the first local minimum along the half line $$ \{ x_k - \alpha \nabla f(x_k): \alpha \geq 0 \} $$ If $x_0$ is not optimal, that is, $||x_0||\neq r$, how does this method behave?
I don't need a full calculation but maybe a picture or some indication of what is special about this function's behaviour would be very helpful!
Any tips are greatly appreciated!