Re: Integration of function containing if-test

[Søren Hauberg]

tor, 05 02 2009 kl. 18:17 +0100, skrev Torquil Macdonald Sørensen:
> Hi!
> 
> I am having some problems with integration of the following function 
> (this is really a simplification of my function, but the problem I 
> illustrate is the same).
> 
> I have a function nu(x) that is defined using an if-test. It has a 
> special value (0) at x = 0, and is defined in terms of an ordinary 
> function f(x) everywhere else. Therefore I have:
> 
> function y = nu(x)
>       if (x == 0)
>               y = 0;
>       else
>               y = f(x);
>       endif
> endfunction
> 
> When I pass it a 0 it returns a zero. But when I pass it a vector [0 1] 
> it doesn't work, because the if-test is apparently not executed on each 
> element of the vector individually, but on the vector x as a whole. 
> Since x = [0 1] is not equal to 0, the first part of the if-test is 
> never entered, and therefore f(x) is evaluated at both 0 and 1, thereby 
> returning the wrong result. In my case it returns [ NaN 1], since f(0) = 
> Nan and f(1) = 1.

I'm not quite I understand your problem, but:

1) Can you just do

   function y = nu(x)
     if any (x == 0)
       y = 0;
     else
       y = f(x);
     endif
   endfunction

? Notice the 'any'.

2) Since you only have a finite number of points where 'x' is zero, then
I would have thought you could just ignore them when doing integration.
Can't you just integrate 'f (x)' directly?

Søren