Why not define your own ' power' operator? Then you can write whatever
rules you want, including seeing what it simplifies to with the built-in
'^' operator. You could also change simpexpt to have a flag controlling
the 0^-n case for your special needs... The power of open source!
-s
On Apr 24, 2012 6:33 PM, "Richard Hennessy" <rich.hennessy at verizon.net>
wrote:
> Carrying around undefined forms like this has been proposed before, but
> first think about what semantics you'd want around them.
>
> Is 3/0 equal to 2/0? In that case, perhaps they should both simplify to
> 1/0.
>
> Is 1/0 > 2345?
>
> Is -1/0 < 1/0?
>
> Is -1/0 the same as minf?
>
> Is 1/0 - 1/0 equal to 0? What is 3/0 - 2/0 ?
>
> What is (3/0)*0? Should that simplify to 3? to 1? Or just remain
> unsimplified? If unsimplified, what use is it?
>
> Keeping terms like this unsimplified is of some use to programmers who
> write their own simplification rules taking the current problem context
> into account. That is why I would like to keep terms like this from blowing
> up Maxima before my program that I might have written can even get the
> input.
>
> Rich
>
>
>
> On Tue, Apr 24, 2012 at 17:27, Richard Hennessy <
> rich.hennessy at verizon.net> wrote:
>
>> If you type in :
>>
>> (%i5) kill(all);
>> (%o0) done
>> (%i1) 3/4;
>> (%o1) 3/4
>> (%i2) 3/2;
>> (%o2) 3/2
>> (%i3) 3/1;
>> (%o3) 3
>> (%i4) 3/0;
>> expt: undefined: 0 to a negative exponent.
>> -- an error. To debug this try: debugmode(true);
>>
>> Since Maxima is not trying to compute the answer, it just simplifies the
>> input in these cases, why the error? Would it be better if you could get
>> the unsimplified form back?
>>
>> (%i5) 3/0;
>> (%o5) 3/0
>>
>> Rich
>>
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>>
>
>