Re: Re: Symbolic complex conjugation?

*To*: mathgroup at smc.vnet.net*Subject*: [mg89988] Re: [mg89947] Re: [mg89919] Symbolic complex conjugation?*From*: Andrzej Kozlowski <akoz at mimuw.edu.pl>*Date*: Thu, 26 Jun 2008 04:39:16 -0400 (EDT)*References*: <200806240731.DAA11098@smc.vnet.net> <200806251024.GAA18614@smc.vnet.net>

It might be useful to add that an expression like -a I does contain I *before* evaluation but not *after* evaluation. So: Unevaluated[-a I] /. HoldPattern[I] -> -I I a There is an analogous phenomenon involving expressions like 2/3 and Rational[2,3]: the first is evaluated into the second. Hence 2/3 /. (a_)/(b_) -> b/a 2/3 but Unevaluated[2/3] /. (a_)/(b_) -> b/a 3/2 Andrzej Kozlowski On 25 Jun 2008, at 19:24, Murray Eisenberg wrote: > I'll just type I in output to denote what you called Isymbol, and I > show > fractions in output as in-line, not 2D. > > The equations and Solve expressions are distractions here. so are the > complicated names and extra structure in the solution of the second > Solve result. > > You are asking, then, why the result of the second of the following is > not what you expect: > > test=Ia; test/.I -> -I > -I a > > actual=top/(-I something + else); actual/.I -> -I > top/(else - I something) > > The explanation is in what you see if you look at the FullForms: > > FullForm[test] > Times[Complex[0,1],a] > > FullForm[actual] > Times[Power[Plus[else,Times[Complex[0,-1],something]],-1],top] > > Or, look at this: > > FreeQ[#, I] & /@ {test, actual} > {False,True} > > What is I, really? > > FullForm[I] > Complex[0,1] > > In short, there is no I in the expression actual! > > To accomplish what you want, you could do this: > > actual/. -I -> I > top/(else + I something) > > AES wrote: >> I'm sorry, but I just don't understand why the following test case >> works >> just fine: >> >> [In copying it, I've substituted "Isymbol" for the \[ImaginaryI] that >> actually appears in the Out[] cells.] >> >> In[202]:= eqna={a+I b==0}; >> solna=Solve[eqna,b]; >> b=b/.solna[[1]]; >> bStar=b/.{I->-I}; >> {b, Star} >> >> Out[205]={ -Isymbol a, Isymbol a } >> >> but the actual calculation that prompted the test case doesn't: >> >> In[206]:= eqnp={((dwa/2)+I(w-wa))p-I (kappa/(2 w))dN e==0}; >> solnp=Solve[eqnp,p]; >> p=p/.solnp[[1]]; >> pStar=p/.{I->-I} >> >> Out[208]= { (dN e kappa)/(w (-Isymbol dwa+2 w-2 wa)), >> (dN e kappa)/(w (-Isymbol dwa+2 w-2 wa)) } >> >> And actually, I guess my real concern is not understanding "how it >> happens" -- but more "how it can happen" that Mathematica can do >> something this potentially damaging to some innocent user. >> > > -- > Murray Eisenberg murray at math.umass.edu > Mathematics & Statistics Dept. > Lederle Graduate Research Tower phone 413 549-1020 (H) > University of Massachusetts 413 545-2859 (W) > 710 North Pleasant Street fax 413 545-1801 > Amherst, MA 01003-9305 >

**References**:**Symbolic complex conjugation?***From:*AES <siegman@stanford.edu>

**Re: Symbolic complex conjugation?***From:*Murray Eisenberg <murray@math.umass.edu>