Re: Re: Re: Simplifying algebraic expressions

*To*: mathgroup at smc.vnet.net*Subject*: [mg67008] Re: [mg66959] Re: [mg66881] Re: [mg66839] Simplifying algebraic expressions*From*: Andrzej Kozlowski <akoz at mimuw.edu.pl>*Date*: Tue, 6 Jun 2006 06:29:33 -0400 (EDT)*References*: <200606011054.GAA20566@smc.vnet.net> <200606020809.EAA18083@smc.vnet.net> <200606050748.DAA16346@smc.vnet.net> <4484EC71.7060108@wolfram.com>*Sender*: owner-wri-mathgroup at wolfram.com

On 6 Jun 2006, at 11:46, Carl K. Woll wrote: > Andrzej Kozlowski wrote: >> Here is, I think, an optimized version of the "simplification" I >> sent earlier: >> rule1 = {2x -> u, 3y -> v}; rule2 = Map[Reverse, rule1]; >> Simplify[TrigFactor[Simplify[ExpToTrig[ >> Simplify[ExpToTrig[(-1)^(2*x + 3*y) /. rule1], >> Mod[u, 2] == 0] /. rule2], y â?? Integers]], >> y â?? Integers] >> (-1)^y >> "Optimized" means that I can't see any obvious way to make this >> simpler. >> Unlike other answers to the original post, this works in >> Mathematica 5.1 and 5.2, and involves only reversible operations. >> In other words, it constitutes a proof. On the other hand, >> obviously, it would be ridiculous to use this approach in >> practice: there must be a simple transformation rule (or rules) >> missing from Simplify, which apparently has already been added in >> the development version. > > Andrzej, > > It seems to me that your simplification approach isn't as general > as the replacement method I advocated (a variant of which is given > below), as rule1 and rule2 are specific to the input. Also, my > replacement method is based on the fact that: > > In[30]:= Simplify[((-1)^a)^b == (-1)^(a b), Element[{a, b}, Integers]] > > Out[30]= True > > So, I think my approach is generally valid. You are right, I apologise. I did not look carefully at your code. But now that I see some challenge ;-) I also see that I can completley remove rule1 and rule 2 form my code and make it equally general: Simplify[TrigFactor[Simplify[ExpToTrig[ Simplify[ExpToTrig[(-1)^PolynomialMod[2*x + 3*y, 2]], Mod[u, 2] == 0]], y â?? Integers]], y â?? Integers] (-1)^y What's more, I can actually produce an even shorter code that will produce the required simplification: f[(-1)^(n_)] := (-1)^PolynomialMod[n, 2] Simplify[f[(-1)^(2*x + 3*y)], TransformationFunctions -> {Automatic, f}] (-1)^y Of course this ought to be modified so as to be performed only under the assumption that x and y are integers (not very hard to do). I suspect that this is close to what has been done in the development version. Andrzej Kozlowski > >> Note also the following problem, which I suspect is related: >> Simplify[(-1)^(u + v), (u | v) â?? Integers && >> Mod[u, 2] == 0 && Mod[v, 2] == 1] >> -1 >> Simplify[(-1)^(u + v), (u | v) â?? Integers && >> Mod[u, 2] == 0 && Mod[v, 2] == 0] >> 1 >> But >> Simplify[(-1)^(u + v), (u | v) â?? Integers && >> Mod[u, 2] == 0] >> (-1)^(u + v) >> I am sure this also gives (-1)^v in the development version. > > To obtain the desired simplification in 5.2, we can modify my > previous approach as follows: > > power[e_, a_Plus] := Times @@ (Simplify /@ (power[e, #1]&) /@ List > @@ a) > power[e_, a_Integer b_?(Refine[Element[#,Integers]]&)] := (e^a)^b > power[e_, a_] := e^a > > simp[expr_, assum_] := Block[{$Assumptions = assum}, > Simplify[expr /. Power -> power] > ] > > In the above I rely on the following two equivalences, verified by > Mathematica: > > In[50]:= > Simplify[e^a e^b == e^(a+b)] > > Out[50]= > True > > In[51]:= > Simplify[(e^a)^b == e^(a b), Element[{a,b},Integers]] > > Out[51]= > True > > The tricky part is that Mathematica automatically converts e^a e^b > --> e^(a+b), and as you point out above Simplify[(-1)^(u+v),...] > doesn't work. However, Simplify[(-1)^u, ...] does work, which is > why the Simplify appears in the definition of power[e_,a_Plus] > before Times is applied. > > Here are some examples: > > In[56]:= > simp[(-1)^(u+v),Mod[u,2]==0] > > Out[56]= > v > (-1) > > In[57]:= > simp[(-2)^(u+v),Mod[u,2]==0] > > Out[57]= > v u + v > (-1) 2 > > In[58]:= > simp[(-1)^(2x+3y),Element[{x,y},Integers]] > > Out[58]= > y > (-1) > > In[59]:= > simp[(-2)^(2x+3y),Element[{x,y},Integers]] > > Out[59]= > y 2 x + 3 y > (-1) 2 > > Carl Woll > Wolfram Research > >> Andrzej Kozlowski >> On 2 Jun 2006, at 17:09, Andrzej Kozlowski wrote: >>> On 1 Jun 2006, at 19:54, Amitabha Roy wrote: >>> >>> >>>> Hello: >>>> >>>> I would like Mathematica to be able to take an expression, say, >>>> >>>> (-1)^{2 x + 3 y} and be able to simplify to (-1)^y. >>>> >>>> Is there a way one can do this ? >>>> >>>> Thanks >>>> >>> >>> Note that you are using {} instead of (). Different kind of brackets >>> have completely different meaning in Mathematica. >>> I have found it amazingly hard to force Mathematica to perform this >>> simplification. The best I could do is this. We need two rules. >>> rule1 >>> will replace 2x by u and 3y by v. rule 2 does the opposite: it >>> replaces u by 2x and v by 2y. >>> >>> rule1 = {2x -> u, 3y -> v};rule2 = Map[Reverse, rule1]; >>> >>> Now: >>> >>> >>> Simplify[TrigFactor[ >>> FullSimplify[ExpToTrig[ >>> FullSimplify[ >>> ComplexExpand[ >>> (-1)^(2*x + 3*y) /. >>> rule1], Mod[u, 2] == >>> 0] /. rule2], >>> y â?? Integers]], >>> y â?? Integers] >>> >>> >>> (-1)^y >>> >>> Uff... Surely, this ought to be easier... >>> >>> Andrzej Kozlowski >>> Tokyo, Japan >>> >

**References**:**Simplifying algebraic expressions***From:*Amitabha Roy <aroy@cs.bc.edu>

**Re: Simplifying algebraic expressions***From:*Andrzej Kozlowski <akoz@mimuw.edu.pl>

**Re: Re: Simplifying algebraic expressions***From:*Andrzej Kozlowski <akoz@mimuw.edu.pl>