Well, log(%i) and log(-%i) have all the same issues as log(-1) (except for
(2)). If you convert log(%i*x)+log(%i) to log(%i*x)+%i*%pi/2, you have
the same contraction problem as before. That is why I'm suggesting that
solving this reasonable may involve adding some functionality to
logcontract as well as changing the default simplifications for log.
-s
On Tue, Feb 21, 2012 at 11:43, Henry Baker <hbaker1 at pipeline.com> wrote:
> But I asked about log(-%i), which should be pretty non-controversial.
>
> If there is a switch for log, it should indicate what your favorite branch
> is: e.g., [-%pi,+%pi), [0,2*%pi), etc.
>
> At 08:30 AM 2/21/2012, Stavros Macrakis wrote:
> >Good question. For that matter, log(-1) only simplifies to %i*%pi if
> lognegint = true (default is false).
> >
> >Here are the possible arguments I can think of for this behavior (I don't
> guarantee that they correspond to the original rationale...):
> >
> >1) log is multivalued, and the user may want to control whether log(-1)
> => %pi or -%pi etc.
> >2) Maxima assumes the user is working in the real domain, so shouldn't
> introduce %i unless explicitly asked to. But domain:complex doesn't change
> the behavior.
> >3) log(-1) may appear as an intermediate result of a real calculation
> which goes away with simplification, e.g. logcontract(log(-1)+log(-x)), so
> why confuse matters?
> >4) Once you have an expression like log(x)+%i*%pi, converting back to
> log(-x) is slightly tricky -- actually, simply log(exp(EXPR)) will do it,
> but there is no easy way to apply that globally, like logcontract.
> >
> >I suspect that 3 and 4 are the most important.
> >
> >So perhaps the default is OK, but log(-C) and log(%i*C) should expand out
> when domain=complex or logexpand=ultra. Conversely, perhaps there should
> be a logconmode=all which contracts things like log(x)+C so that arguments
> 3 and 4 go away....
> >
> >Thoughts?
> >
> > -s
> >
> >On Mon, Feb 20, 2012 at 23:38, Henry Baker <hbaker1 at pipeline.com> wrote:
> >How come log(-%i) doesn't already simplify by itself, instead of
> requiring rectform to do it?
> >
> >Is there some switch that Maxima uses to force log(constant) to simplify
> if it can?
> >
> >At 01:12 PM 2/20/2012, Stavros Macrakis wrote:
> >>Henry,
> >>
> >>Yes, this derivation is essentially the same as the bottomup rectform.
> In fact:
> >>
> >> expr: logcontract(logarc(4*atan(1/5)-atan(1/239)));
> >> rectform(substpart(e1:rectform(piece),expr,1,2,1)) => %pi/4
> >>
> >>You'll see that e1 here is -%i.
> >>
> >>What rectform does internally by default looks more like
> >>
> >> rectform(substpart(e2:polarform(piece),expr,1,2,1));
> >>
> >>where e2 isn't at all helpful.
> >>
> >>I agree that most mathematical functions (sin, etc.) and form-changing
> routines (factor etc.) should distribute over "=".
> >>
> >> -s
> >>
> >>On Mon, Feb 20, 2012 at 15:45, Henry Baker <hbaker1 at pipeline.com> wrote:
> >>result=4*atan(1/5)-atan(1/239);
> >> 1 1
> >>(%o1) result = 4 atan(-) - atan(---)
> >> 5 239
> >>(%i2) lhs(%)=logcontract(logarc(rhs(%)));
> >> 4
> >> (%i - 5) (%i + 239)
> >> %i log(- --------------------)
> >> 4
> >> (%i - 239) (%i + 5)
> >>(%o2) result = ------------------------------
> >> 2
> >>(%i3) %*2/%i;
> >> 4
> >> (%i - 5) (%i + 239)
> >>(%o3) - 2 %i result = log(- --------------------)
> >> 4
> >> (%i - 239) (%i + 5)
> >>(%i4) exp(lhs(%))=exp(rhs(%));
> >> 4
> >> - 2 %i result (%i - 5) (%i + 239)
> >>(%o4) %e = - --------------------
> >> 4
> >> (%i - 239) (%i + 5)
> >>(%i5) lhs(%)=rectform(rhs(%));
> >> - 2 %i result
> >>(%o5) %e = - %i
> >>(%i6) log(lhs(%))=log(rhs(%));
> >>(%o6) - 2 %i result = log(- %i)
> >>(%i7) rectform(lhs(%))=rectform(rhs(%));
> >> %i %pi
> >>(%o7) - 2 %i result = - ------
> >> 2
> >>(%i8) solve(%,result);
> >> %pi
> >>(%o8) [result = ---]
> >> 4
> >>
> >>This example shows why rectform, log, exp, logcontract, logarc, etc.,
> should distribute over "=".
> >>
> >>At 11:20 AM 2/20/2012, Stavros Macrakis wrote:
> >>>Consider the expression 4*atan(1/5)-atan(1/239) (Machin's formula).
> >>>
> >>>How can we use Maxima to simplify it to %pi/4?
> >>>
> >>>Well, here's one approach:
> >>>
> >>>atan(trigexpand(trigexpand(tan(ex))))
> >>>
> >>>And here's another way that doesn't depend on trig identities:
> >>>
> >>>(%i1) ex:4*atan(1/5)-atan(1/239);
> >>>(%o1) 4*atan(1/5)-atan(1/239)
> >>>(%i2) llex: logcontract(logarc(ex));
> >>>(%o2) %i*log(-(%i-5)^4*(%i+239)/((%i-239)*(%i+5)^4))/2
> >>>(%i3) rectform(llex);
> >>>(%o3)
> (atan(sin(4*atan(1/5))/cos(4*atan(1/5)))-atan(sin(4*(%pi-atan(1/5)))/cos(4*(%pi-atan(1/5))))-2*atan(1/239))/2
> >>> <<< oops, not so helpful
> >>>(%i4) scanmap(rectform,llex,bottomup);
> >>> <<< but reorganizing the rectform calculation gives a better
> result
> >>> <<< makes me think that rectform could be improved...
> >>>(%o4) %pi/4
> >>>
> >>>Are any of these techniques packaged into some analog of trigsimp?
> >>>
> >>> -s
>
>