[Maxima-bugs] [ maxima-Bugs-1079521 ] freeof (exp, exp(x)) yields true

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Bugs item #1079521, was opened at 2004-12-05 14:19
Message generated for change (Comment added) made by macrakis
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Category: Lisp Core
Group: None
Status: Open
Resolution: None
Priority: 2
Submitted By: Robert Dodier (robert_dodier)
Assigned to: Nobody/Anonymous (nobody)
Summary: freeof (exp, exp(x)) yields true

Initial Comment:
freeof (exp, exp(x))  =>  true
although
freeof (sin, sin(x))  => false 
(kill (f), freeof (f, f(x)))  => false

Exponentials are always stored as %e^something, so 
freeof (%e, exp(x))  =>  false

Dunno what is the right policy here.  Substituting %e
for exp in the argument list (which is consistent with
the otherwise universal policy of substituting %e^x for
exp(x)) won't work right:
freeof (exp, exp(x))  hypothetically =>  false  (OK)
freeof (exp, %e*2) hypothetically => false (OOPS)

Another answer is to not substitute %e^x for exp(x) but
that will have widespread effects (to put it mildly).


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>Comment By: Stavros Macrakis (macrakis)
Date: 2005-01-24 17:42

Message:
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user_id=588346

This is not a freeof issue.

It is not that exponentials are "stored as" %e^x, it is that
the expression exp(x) is simplified to %e^x.  The function
'exp' just never appears in a simplified expression.

Similarly sin(%i) simplifies to %i*sinh(1).  Surely you
don't expect freeof to "find" the sin in %i*sinh(1) or to
"find" the exp in 2^n (which is precisely equivalent to
exp(n*log(2)).

Of course, Maxima is not completely consistent about when it
canonicalizes and when it does not (even when this is
feasible).  For example, by default it does *not* simplify
exp(n*log(2)) to 2^n, but to %e^(log(2)*n); it does not
simplify n! to gamma(n+1); etc.

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