I don't know that exp(x)=%e^x is a theorem.
There are 3 roots in x^(1/3) unless x=0. and therefore also 3 roots in
%e^(1/3). But I think of exp(1/3) in terms of the power series for exp, and
it has only one value.
Internal to radcan, the conversion of x^y to e^(y*log(x)) and then back,
if possible, may be implicit in the processing.
solve(x^p-%e,x) gives what appears to be one solution, but should have some
multiplicity, at least if p is an integer. so %e^(1/p) is multiple valued.
solve(x^10-%e,x) has 10 solutions.. one of them %e^(1/10), presumably just
one of them...
I'm not proposing a solution here, just pointing out issues.
> -----Original Message-----
> From: maxima-bounces at math.utexas.edu
> [mailto:maxima-bounces at math.utexas.edu] On Behalf Of Robert Dodier
> Sent: Saturday, March 15, 2008 8:30 AM
> To: Stavros Macrakis
> Cc: fateman at EECS.Berkeley.EDU; maxima at math.utexas.edu
> Subject: Re: [Maxima] constants, simplification, numerical evaluation
>
> On 3/15/08, Stavros Macrakis <macrakis at alum.mit.edu> wrote:
>
> > exp(x)-%e^x => 0
> > log(exp(x)) => x
> > exp(x+y)/exp(x) => exp(y)
> > factor(exp(2*x)-1)
> > integrate
> > radcan
> > limit
> > etc. etc.
>
> Whatever simplifications are already implemented for %e^x,
> can also be applied to exp(x), with sufficient programming effort.
> ("Sufficient" in this case is much less than, say, the effort needed
> to repair or rewrite the definite integration code.)
>
> I am interested in finding a representation for the exponential
> function which is consistent with the representation of other
> mathematical functions. It is an interesting and useful theorem
> that exp(x) = %e^x but, I claim, that need not and should not
> be the basis of its representation in Maxima.
>
> > I suppose you could uniformly represent a^b as
> exp(b*log(a)) but then
> > you'd have to change everything in Maxima that handled "^"
> to handle
> > exp instead. Do you really want to represent x^2 as exp(2*log(x))?
>
> No.
>
> best
>
> Robert
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