Inverting a 2D numerical function?
csmallw
The best idea I've had for doing this so far is to extract contours from X_out and Y_out at a variety of z-positions and to figure out where the two contours cross (see my other post: http://www.igorexchange.com/node/4878, for a more specific question about this). Maybe there is a more efficient way of doing this, though?
Thanks,
Chris Smallwood
Specifically, If you can draw a contour line on the surface then, at least in principle you can have multiple pairs of (x,y) that match a given z value. If that is the case, how do you propose to create a unique inversion?
A.G.
WaveMetrics, Inc.
June 7, 2013 at 01:29 pm - Permalink
/Z=<desired value>,
. It solves AG's question of uniqueness by ignoring it :) That's a humorous way to say that FindRoots will, in most cases, find a root if one exists, but it won't find all the roots when more than one exists.Be sure to read the background material:
DisplayHelpTopic "Finding Function Roots"
John Weeks
WaveMetrics, Inc.
support@wavemetrics.com
June 7, 2013 at 04:00 pm - Permalink
Thanks!
Chris
June 10, 2013 at 01:01 pm - Permalink