extending limit

Thank you for the guidance--I'll refine simplim%signum. In general, maybe
the limit
code could call the simplim function sooner?

 (%i1) :lisp(trace simplim%signum);

simplim%signum isn't called (but it would be OK if it did)

 (%i1) limit(signum(x),x,inf);
 (%o1) 1

simplim%signum is called:

 (%i2) limit(signum(x + a),x,inf);
 0> Calling (SIMPLIM%SIGNUM ((%SIGNUM SIMP) ((MPLUS SIMP) $A $X)) $X $INF)
 <0 SIMPLIM%SIGNUM returned 1
 (%o2) 1

Barton

-----drdieterkaiser at web.de wrote: -----


>Am?Montag,?den?25.01.2010,?12:54?-0600?schrieb?Barton?Willis:
>>?I?know?little?to?nothing?about?the?limit?code;?nevertheless,?how?about:
>>
>>?(setf?(get?'%signum?'simplim%function)?'simplim%signum)
>>
>>?(defun?simplim%signum?(e?x?pt)
>>???(let?((sgn))
>>?????(setq?e?(limit?(cadr?e)?x?pt?'think))
>>?????(setq?sgn?($csign?e))
>>????(cond?((eq?sgn?'$zero)?'$ind)
>>???????????((memq?sgn?'($neg?$pn?$pos))
>>????????????(take?'(%signum)?e))?;;?OK,?this?does?sign(e)?twice,...
>>???????????(t??(throw?'limit?())))))
>>
>>?Examples:
>>
>>?(%i2)?limit(signum(x^2-3),x,5);
>>?(%o2)?1
>>
>>?(%i3)?limit(signum(x),x,a);
>>?(%o3)?'limit(signum(x+a),x,0)
>>
>>?(%i4)?assume(a?>?0);
>>?(%o4)?[a?>?0]
>>
>>?(%i5)?limit(signum(x),x,a);
>>?(%o5)?1
>>
>>?(%i6)?limit(signum(x^3-1),x,inf);
>>?(%o6)?1
>>
>>
>>?(%i7)?limit(signum(x^3-7),x,minf);
>>?(%o7)?-1
>>
>>?(%i8)?limit(x?*?signum(x),x,0);
>>?(%o8)?0
>
>Hello?Barton,
>
>I?think?a?simplim%function?is?the?right?way?to?implement?the?handling?of
>specific?arguments?of?a?function,?which?are?not?handled?by?the
>simplifier.?If?present?it?is?called?before?the?limit?routines?call?the
>simplifying?function?to?get?an?answer?for?an?argument.
>
>Therefore?the?simplim%function?should?be?as?complete?as?possible.?We
>have?a?lot?of?functions?which?might?benefit?from?a?simplim%function?too.
>
>To?be?consistent?I?think?the?default?of?a?simplim%function?should?be?to
>call?the?simplifier.?Only?known?specific?arguments?are?handled.
>
>This?is?another?short?example.?Only?an?argument?1?is?handled,?all?other
>arguments?are?passed?to?the?simplifier:
>
>(defun?simplim%elliptic_kc?(expr?var?val)
>??;;?Look?for?the?limit?of?the?argument
>??(let?((m?(limit?(cadr?expr)?var?val?'think)))
>????(cond?((onep1?m)
>???????????;;?For?an?argument?1?return?$infinity.
>???????????'$infinity)
>??????????(t
>????????????;;?All?other?cases?are?handled?by?the?simplifier?of?the
>function.
>????????????(simplify?(list?'(%elliptic_kc)?m))))))
>
>Dieter?Kaiser