[GiNaC-devel] New in GiNaC: step function

[Chris Dams]

Dear Richy and Vladimir,

On Thu, 2006-03-09 at 22:44 +0100, Richard B. Kreckel wrote:
> Hi!
> 
> Vladimir Kisil wrote:
> 
> >    CD> That function already more or less exists as csgn(x).
> >
> >	I missed it. Should we add its power property like that: 
> >
> >static ex csgn_power(const ex & arg, const ex & exp)
> >{
> >	if (is_a<numeric>(exp) && exp.info(info_flags::positive)) {
> >		if (ex_to<numeric>(exp).is_odd())
> >			return csgn(arg);
> >		else
> >			return power(csgn(arg), _ex2).hold();
> >	} else
> >		return power(csgn(arg), exp).hold();
> >}
> >  
> >
> 
> But that appears to be missing the possibiliby of fractional powers.
> 
> >	
> >    CD>  The reason that it is useful to have both functions
> >
> >	Yes, I agree. So far I used home-brewed definition in
> >  http://arxiv.org/abs/cs.MS/0512073, now I can switch to the "official" one. ;-)
> >
> >    CD> (csgn(x)+1)/2 is kept together even when expanding and so on if
> >    CD> one writes it as step(x). When looking at output it is more easy
> >
> >	I agree again. My comment was intended only for the case, when the
> >  intermediate value is important. 
> >  
> >
> 
> I'm not totally convinced.  Just one experience: I once was collecting 
> lots of such theta functions in computations where they were falling out 
> as constraints from complex residue integrations: Is the contour around 
> the pole (parametrized as f(x)) or isn't it? The Laurent series of the 
> integrand at the poles was known to have only odd negative powers which 
> allows one to elegantly take care of the situation where the contour 
> crosses the pole: just add Pi*I times the residue (instead of 2*Pi*I).  
> Here, it was clearly an advantage to lump them all together with the 
> weight theta(f(x))=(1+csgn(f(x)))/2. Some of the term-rewriting rules 
> wouldn't work without the intermediate value.

Yes, one would, for instance, have
integral(x,-infty,infty, exp(I*k*x)/(x+I*epsilon)) = -2*Pi*I*step(k).
This would give precisely the same right result also at k=0 if
step(0)=1/2. For cases like this it seems that 1/2 is the best value. So
maybe the best thing to do would be to have step(0)=1/2 and not to do a
simplification step(x)^2 -> step(x). After all, if desired, such a
simplification could also be done by the user. (S)he could do
.subs(step(wild())*step(wild())==step(wild()), subs_options::algebraic).

Can everybody live with that?

Best,
Chris