Timeline for Solving for an operator by minimization
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
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| May 6, 2012 at 19:08 | comment | added | Brian Borchers | The operator $O$ here is computed by a numerical simulation that may or may not be noisy. If the simulation results are noisy, then using finite difference derivatives may be a poor choice. An alternative to consider in that case is the use of response surface methods. However, if the simulation results are reasonably good then finite difference derivatives are likely to work just fine. | |
| May 5, 2012 at 19:12 | comment | added | Pait | Gradient descent is problematic if you only have measured values of the function to be optimized, because measurements are always noisy. Taking derivatives of data, even if the it is obtained via numerical simulations, is almost always problematic. | |
| May 4, 2012 at 20:15 | comment | added | Brian Borchers | Gradient descent converges at only a linear rate, whereas Newton's method (and LM is essentially Newton's method) converges at a quadratic rate. LM can be much faster in practice. | |
| May 4, 2012 at 20:05 | comment | added | Ben Sprott | Will gradient descent be fine in the case I have mentioned? | |
| May 4, 2012 at 20:04 | comment | added | Ben Sprott | I should mention that I do not need to minimize for more than one pair at a time. I will want to take a single pair and find x,y that form a solution, and then a new pair will come along and new x,y values will do. Also, $\psi$ is fixed. | |
| May 4, 2012 at 19:13 | comment | added | Brian Borchers | One important issue with the LM method is that it a local search method. In some circumstances, if your least squares problem has local minima it may return a solution that is locally but not globally optimal. It's possible in that case that you'd miss a solution that really did satisfy $O(x,y)\psi=\phi$. A simple approach that can help with this problem is to start LM running from many different initial guesses and then take the best solution found. | |
| May 4, 2012 at 19:08 | comment | added | Ben Sprott | Hi, Yes $\phi$ and $\psi$ are 2 element complex vectors. Elements of a 2-d Hilbert space. | |
| May 4, 2012 at 19:06 | history | answered | Brian Borchers | CC BY-SA 3.0 |