I have a function $g(x,y) = 1-(xy-3)^2$. Starting at $(x,y) = (0,0)$, is it possible to apply one step of the steepest ascent method to the problem $\underset{(x,y)\in R^2}{max}$ $g(x,y)$.
The gradient I've obtained is $\triangledown g$ = $((-2xy^2 +6y), (-2x^2y +6x))$. $\triangledown g(0,0) = (0,0)$ and $X_0 + t\triangledown g(X_0) = (0,0) $.
When I applied the step, I still got $(0,0)$ and is it what we should get when we start with the initial point of $(0,0)$?
If you're using the method to optimize (max or min) the function, then a zero derivative vector indicates you are at a local minimum / maximum. There is no progress possible beyond that.
By the way, are you looking for a global maximum of the function, given that you are using the steepest ascent method ?
In this case, I have a suggestion. Treat $xy=z$. The function becomes $1-(z-3)^2$. This is a concave function with $z=3$ as the global maximum. Now $xy = 3$ is a hyperbola and every point on it is a global max.
Ideally, the characteristic roots of n is dependent on x^(t+1)=x^(t). Use the characteristic root of the independent variables; that is {M^(t)}^-1 to find the reverse of X0 which assures the right hand side identity function.
