Re: Simplifying equations for Mathematica

*To*: mathgroup at smc.vnet.net*Subject*: [mg65863] Re: Simplifying equations for Mathematica*From*: Maxim <m.r at inbox.ru>*Date*: Wed, 19 Apr 2006 04:54:22 -0400 (EDT)*References*: <200604160749.DAA11245@smc.vnet.net> <e22h8k$e56$1@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

On Tue, 18 Apr 2006 11:07:00 +0000 (UTC), Daniel Lichtblau <danl at wolfram.com> wrote: > Yaroslav Bulatov wrote: >> [This post has been delayed due to email problems - moderator] >> >> >> I'm trying to solve some likelihood equations, and Mathematica will not >> >> finish in reasonable time. I'm wondering if there is a way I can >> >> rewrite it so that Mathematica can do them >> >> >> >> The concrete example is >> >> f[t1,t2,t3,t4,t5]=Log[Exp[0]+Exp[t1]+Exp[t2]+Exp[t1+t2]+Exp[t5]+Exp[t5+t1 >> +t2]+Exp[t5+t2+t4]+Exp[t1+t2+t3+t4+t5]] >> >> >> >> The gradient of f defines a map R^5->R^5, and I need to invert that >> >> map. (This particular example can be solved by hand, but I'm wondering >> about other cases of the same form) >> >> >> >> Here's the command I use to solve it (works on Mathematica 5.2 only) >> >> Solve[Map[Apply[Equal,#]&,Thread[{D[Log[Exp[0]+Exp[t1]+Exp[t2]+Exp[t1+t2] >> +Exp[t5]+Exp[t5+t1+t2]+Exp[t5+t2+t4]+Exp[t1+t2+t3+t4+t5]], >> {{t1,t2,t3,t4,t5},1}],{m1,m2,m3,m4,m5}}]],{t1,t2,t3,t4,t5}] >> >> >> >> It's been running for several days on a Pentium 2Ghz. Are there simple >> >> transformations I can do to help Mathematica solve it? > > > One method is to make it explicitly polynomial by working, in effect, > with "variables" Exp[t1], etc. This can be done as below. We work with > expressions rather than equations. > > exprs = Map[Apply[Subtract,#]&, > Thread[{D[Log[Exp[0]+Exp[t1]+Exp[t2]+Exp[t1+t2]+Exp[t5]+ > Exp[t5+t1+t2]+Exp[t5+t2+t4]+Exp[t1+t2+t3+t4+t5]], > {{t1,t2,t3,t4,t5},1}],{m1,m2,m3,m4,m5}}]]; > > Below we make exponentials into variables, keeping the same names. This > will not work if the variables appear other than in powers of > exponentials. > > exprs2 = exprs /. {Exp[a_]:>a, Exp[Plus[a__]]:>Apply[Times,a]}; > vars = {t1,t2,t3,t4,t5}; > > This we can solve. > > In[75]:= Timing[soln = Solve[exprs2==0, vars];] > Out[75]= {0.344022 Second, Null} > > Now take logs of results since in effect we solved for Exp[tj]'s rather > than tj's. > > In[76]:= InputForm[Log[vars] /. First[soln]] > Out[76]//InputForm= > {Log[-(5*m1 - 3*m2 - 5*m3 + 4*m4 - m5)/(4*(-2 + m1 + m2 - m3 + m5))], > Log[-(3*m1 - 5*m2 - 3*m3 + 4*m4 + m5)/(4*(2 - m1 - m2 + m3 - m5))], > Log[-(-5*m1 + 3*m2 + 21*m3 - 12*m4 + m5)/(4*(-2 + m1 + m2 - m3 + m5))], > Log[-((m1 - m2 - 3*m3 + 4*m4 - m5)/(-2 + m1 + m2 - m3 + m5))], > Log[-(-m1 - m2 + m3 - 4*m4 + 5*m5)/(4*(-2 + m1 + m2 - m3 + m5))]} > > > Daniel Lichtblau > Wolfram Research > I have my doubts about this method, because the rule (Exp[a_] :> a) will be tried first and so it will replace E^(t1 + t2) with (t1 + t2). Numerical tests show that after we replace ti with Log[ti] and get rid of denominators one of the solutions is t1 == t2 == -1, t5 == 0. This suggests that the ordering of the variables can matter: In[1]:= f[t1_, t2_, t3_, t4_, t5_] := Log[Exp[0] + Exp[t1] + Exp[t2] + Exp[t1 + t2] + Exp[t5] + Exp[t5 + t1 + t2] + Exp[t5 + t2 + t4] + Exp[t1 + t2 + t3 + t4 + t5]] In[2]:= Lvar = {t1, t2, t3, t4, t5}; Lvar2 = {t4, t3, t2, t1, t5}; Lm = {m1, m2, m3, m4, m5}; Lexpr = D[f @@ Lvar, {Lvar}] - Lm; Lexpr2 = Lexpr /. Thread[Lvar -> Log[Lvar]] // Together // Numerator; In[7]:= SetOptions[Roots, Cubics -> False, Quartics -> False]; (Lsol = Solve[GroebnerBasis[Lexpr2, Lvar2] == 0, Lvar2] // Simplify) // ByteCount // Timing Out[8]= {10.922*Second, 204216} The last three seem to be generic solutions: In[9]:= Lexpr /. Thread[Lvar -> Log[Lvar /. #]]& /@ Rest@ Lsol /. Thread[Lm -> Array[Random[Real, {-10, 10}, 20]&, Length@ Lm]] // MatrixQ[#, # == 0&]& Out[9]= True Maxim Rytin m.r at inbox.ru

**Follow-Ups**:**Re: Re: Simplifying equations for Mathematica***From:*Daniel Lichtblau <danl@wolfram.com>

**References**:**Simplifying equations for Mathematica***From:*Yaroslav Bulatov <yaroslavvb@gmailnospa.com>