# [Maxima] Operations on inf

Andrey G. Grozin A.G.Grozin at inp.nsk.su
Fri Mar 9 11:40:30 CST 2007

```On Fri, 9 Mar 2007, Jay Belanger wrote:
> "Stavros Macrakis" <macrakis at alum.mit.edu> writes:
> ...
>> Agreed that when you type in 1, you mean 1.  The problem comes when we start enabling
>> simplifications like 1/inf => 0.  Then as far as Maxima is concerned, it can't tell
>> the difference between the 1 you typed in and the 1 that comes from 1+1/inf.  Now I
>> believe we agree that (1+1/inf)^inf must be UND, since after all (1+1/x^a)^x = inf if
>> a<1, %e if a=1, and 1 if a>1, so how do we avoid the incorrect simplification (1+1/
>> inf)^inf => 1?
>>
>> As I say, there are sophisticated ways to avoid it (keep track of infinitesimals --
>> though of course that doesn't cover all cases) and there are simple ways (just
>> failsafe to UND).
>>
>> What would you propose?
>
> I see what you're saying now (finally, you say); I had been thinking
> of the results of explicitly using limits, but what happens when you
> enter (1+1/inf)^inf directly?
> I don't have any good ideas; I guess I'd have 1+1/inf evaluate to 1,
> have 1^inf evaluate to 1, (which would make (1+1/inf)^inf = 1, I
> realize) and expect the user to understand how Maxima and limits
> work.
In some other mail, there was an even more convincing proof that 1^inf
should be und:

1^inf = exp(inf*log(1)) = exp(inf*0) = exp(und) = und

I think simplifying this to 1 is incorrect.

Andrey
```