Hi Andrei.
Sorry for replying so late, after so many threads
have been posted.
Mine is not properly an answer, I just wonder like
you what happen.
I don't know so much of Maxima, so I'll write what I
know about the exponential function.
First of all, the exponential function with base a
negative number (and *real* exponent x) is not
defined.
This is because the best way to approach the
exponential function that the mathematicians
have found is:
STEP 1 - Define the natural logarithm as the only
differentiable and injective solution of the
functional equation L(xy) = L(x) + L(y) with
L'(1) = 1. The answer is
L(x) = integral(1/t, t, 1, x), with x a positive
real number.
STEP 2 - Define the number "e" as L(e) = 1
STEP 3 - Define the exponential function with base e
as the inverse of L, so EXP(x) = y <=> L(y) = x
STEP 4 - Define the exponential function with
any POSITIVE base a as a^x = EXP(x * L(a))
You can see that if L, the logarithm, cannot have
negative argument, you cannot have exponential with
negative bases.
>From your experiments, I can guess that the first
thing the Maxima Semplifier does is to split like
(a/b)^x ----> a^x / b^x, and if a or b is negative,
Maxima doesn't nothing, althought x is integer
> (%i2) limit((-2/3)^x,x,inf);
> x
> (- 2)
> (%o2) limit ------
> x -> inf x
> 3
> (%i4) declare(x,integer);
> (%o4) done
> (%i6) limit((-2/%pi)^x,x,inf);
> x
> (- 2)
> (%o6) limit ------
> x -> inf x
> %pi
I thing this is an incorrect implementation: what I
said was about real exponential function; for
the descrete case, there is no problem at all (but
Maxima has).
It seems that the case (-1)^x is handled apart,
because Maxima seems to work with it with any kind
of x, real or integer:
> (%i1) limit((-1/2)^x,x,inf);
> (%o1) 0
> (%i4) declare(x,integer);
> (%o4) done
> (%i5) limit((-1/%pi)^x,x,inf);
> (%o5) 0
The only test that doesn't fit with my interpretation
is
> (%i3) limit((-1/%pi)^x,x,inf);
> (%o3) und
where x is a real number: according to what I
wrote, I expected 0 (although on my calculus books I
find that this expression has no sense, and the
same is for (-1/2)^x with real x). But, actually, I
dunno what "und" means.
Regards
Giovanni Gherdovich
Chiacchiera con i tuoi amici in tempo reale!
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