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Argument of a linear model

On Aug 30, 2012, at 12:08 AM, m4n14ccc wrote:

            
In a general sense such a problem is ill-defined mathematically  
because the inverse of a second order quadratic (my assumption  
regarding what you meant by "second order") will not necessarily be a  
function in the mathematical sense of being one-to-one. You could  
first plot and then see if it makes sense to fit a surface to the  
point set generated by:

   sapply( seq(-1, 1, by=0.01) , f)

(Which could then be limited in your more restricted domain.) You  
could then predict those elements in seq(-1, 1, 0.01) which had  
corresponding images in f(seq(.)).

It would be more productive if you assembled a test case in R code.