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Re: with 5.2 Mac

  • To: mathgroup at smc.vnet.net
  • Subject: [mg59906] Re: with 5.2 Mac
  • From: sbjensen at midway.uchicago.edu (Steuard Jensen)
  • Date: Thu, 25 Aug 2005 06:33:40 -0400 (EDT)
  • Organization: The University of Chicago
  • References: <de45i8$qtf$1@smc.vnet.net> <200508230851.EAA03009@smc.vnet.net> <8B09B5D2-0003-4529-8281-677494C3F6D9@mimuw.edu.pl> <dehjl8$c6t$1@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

Quoth Andrzej Kozlowski <akoz at mimuw.edu.pl> in article
<dehjl8$c6t$1 at smc.vnet.net>:
> In fact using Complex[a_,b_]->Complex[a,-b] instead of Conjugate can  
> cause more serious problems and actually give wrong answers.

That's certainly what I'm worried about.  But...

> The reason is that if f is a function in the complex plane that is
> not holomorphic then it it not necessarily true that Conjugate[f[z]]
> = f [Conjugate[z]].
...
> As an example take
> u = Im[Sqrt[2 + 3*I]];

> N[Conjugate[u]]
> 0.895977

> but

> N[u/.Complex[a_,b_]->Complex[a,-b]]
> -0.895977


...but this example doesn't actually bother me at all (not yet, at
least!), nor do your earlier examples involving (-1)^(4/5):


> On 23 Aug 2005, at 15:27, Andrzej Kozlowski wrote:
> > w = Last[x /. Solve[x^5 == 1, x]]
> > (-1)^(4/5)

> > ComplexExpand[Conjugate[(-1)^(4/5)]]
> >
> > -(-1)^(1/5)
...
> > but Complex[a_,b_]->Complex[a,-b] will obviously have no effect.

As far as I can tell, the only difference between using, e.g.,
ComplexExpand[Conjugate[#]] and (#/.Complex[a_,b_]->Complex[a,-b]) in
these examples is which branch of the function you end up on.  (That
is, you'll still get a square root of (2-3 I) or a fifth root of
Conjugate[1]=1 using either method.)

So these changes will only be an issue if your work depends somehow on
the exact order in which Mathematica sorts roots or other branches of
functions.  (And for what it's worth, I think that both approaches
will give the same answer if the branch you've chosen gives an answer
on the real line.)

For my purposes, I'm not particular about what order the various
branches are listed in, unless that somehow leads to inconsistent
results.  (In fact, I'm not certain that my current work has any roots
that aren't of non-negative reals, but I'm still interested in a
solution that's reliable in general.)  So are there potential pitfalls
of the "simple substitution conjugation" approach that would lead to a
truly wrong answer (as opposed to simply a different one)?

						Steuard Jensen


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