On Thursday 03 August 2006 12:13, giovanni gherdovich wrote:
Hello, Giovanni.
I've got a reply actually before, though it missed the mailing list.
Indeed, with integer powers it seems to be a bug, so I posted it as a bug:
http://sourceforge.net/tracker/index.php?func=detail&aid=1528607&group_id=4933&atid=104933
As for real powers, the expression is defined once you allow for complex
numbers, and in that case the limit is still zero, i.e. limit a^x = 0, if x
-> +inf, abs(a) < 1. So it would be nice if Maxima could handle those cases
as well.
As for the definition, you may define an exponential function in several ways.
To add few: either as a solution to f'(x) = f(x), f(0) = 1, or as a sum of
series. I understand that the latter definition is used when you expand the
definition for complex numbers.
> 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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