Re: "hard" simplification

*To*: mathgroup at smc.vnet.net*Subject*: [mg75356] Re: [mg75315] "hard" simplification*From*: Andrzej Kozlowski <akoz at mimuw.edu.pl>*Date*: Thu, 26 Apr 2007 03:33:50 -0400 (EDT)*References*: <200704250931.FAA26921@smc.vnet.net> <3D50FE62-46D7-4053-A78F-D7EBA1093C11@mimuw.edu.pl>

On 25 Apr 2007, at 21:28, Andrzej Kozlowski wrote: > > On 25 Apr 2007, at 18:31, dimitris wrote: > >> Hello. >> >> I have two expressions. (where a>1) >> >> In[56]:= >> o1 = (1/(16*(-1 + a^2)^(3/2)))*((3*(Sqrt[a - Sqrt[-1 + a^2]] - Sqrt[a >> + Sqrt[-1 + a^2]]) + >> a^2*(-Sqrt[a - Sqrt[-1 + a^2]] + Sqrt[a + Sqrt[-1 + a^2]]) + >> a*(Sqrt[(-(-1 + a^2))*(-a + Sqrt[-1 + a^2])] + >> Sqrt[(-1 + a^2)*(a + Sqrt[-1 + a^2])]))*Pi); >> >> In[58]:= >> o2 = ((2*a + 3)*Pi)/(2^(7/2)*(a + 1)^(3/2)); >> >> These expressions are equal. (They came from the evaluation of the >> same integral; first by hand, second >> by Mathematica). >> >> (If somebody is interested the integral is >> >> In[61]:= >> Integrate[1/(x^4 + 2*a*x^2 + 1)^2, {x, 0, Infinity}, Assumptions -> a >>> 1] >> >> Out[61]= >> ((3*(Sqrt[a - Sqrt[-1 + a^2]] - Sqrt[a + Sqrt[-1 + a^2]]) + a^2*(- >> Sqrt[a - Sqrt[-1 + a^2]] + Sqrt[a + Sqrt[-1 + a^2]]) + >> a*(Sqrt[(-(-1 + a^2))*(-a + Sqrt[-1 + a^2])] + Sqrt[(-1 + a^2)*(a >> + Sqrt[-1 + a^2])]))*Pi)/(16*(-1 + a^2)^(3/2)) >> >> and the simpler expression can be found by >> >> In[65]:= >> (1/(x^4 + 2*a*x^2 + 1)^2)*Dt[x] /. x -> Sqrt[y] >> Integrate[% /. Dt[y] -> 1, {y, 0, Infinity}, Assumptions -> a > 1] >> >> Out[65]= >> Dt[y]/(2*Sqrt[y]*(1 + 2*a*y + y^2)^2) >> >> Out[66]= >> ((3 + 2*a)*Pi)/(8*Sqrt[2]*(1 + a)^(3/2)) >> >> ) >> >> Indeed >> >> In[84]:= >> ({#1, RootReduce[Simplify[o /. a -> #1]]} & ) /@ Table[Random >> [Integer, >> {2, 100}], {20}] >> >> Out[84]= >> {{100, 0}, {54, 0}, {74, 0}, {63, 0}, {96, 0}, {44, 0}, {46, 0}, {26, >> 0}, {80, 0}, {14, 0}, {23, 0}, {46, 0}, {3, 0}, {64, 0}, >> {70, 0}, {59, 0}, {71, 0}, {17, 0}, {23, 0}, {56, 0}} >> >> 1) Can somebody show me one workaround (in a CAS!) in order to show >> that these expressions are indeed equal? >> >> 2) In the same vein with Vladimir's posts (and stealing his >> trademark!) can someone show me a workaround >> that will simplify expression o1 to its equivalent (for a>1) o2? >> >> Thanks in advance! >> >> Dimitris >> >> > > > I will only do (1) since I am in principle opposed to misusing CAS > for the purpose (2) (and perhaps also because I can't do it ;-)). > > First, we start with the definitions: > > > o1 = (1/(16*(-1 + a^2)^(3/2)))* > ((3*(Sqrt[a - Sqrt[-1 + a^2]] - > Sqrt[a + Sqrt[-1 + a^2]]) + > a^2*(-Sqrt[a - Sqrt[-1 + a^2]] + > Sqrt[a + Sqrt[-1 + a^2]]) + > a*(Sqrt[(-(-1 + a^2))*(-a + Sqrt[-1 + a^2])] + > Sqrt[(-1 + a^2)*(a + Sqrt[-1 + a^2])]))*Pi); > > o2 = ((2*a + 3)*Pi)/(2^(7/2)*(a + 1)^(3/2)); > > We now substitute 1+t^2 for a^2 (O.K. since a>1): > > p1 = Simplify[o1 /. a^2 -> 1 + t^2, t > 0]; > > Next, we substitute c^2 for a-t and d^2 for a+t, where c and d are >0: > > > p2 = Simplify[p1 /. {a - t -> c^2, a + t -> d^2}, > {c > 0, d > 0}] > > > (Pi*(c*(-t^2 + a*t + 2) + d*(t^2 + a*t - 2)))/(16*t^3) > > Now we use GroebnerBasis to eliminate t,c, and d: > > > sol = GroebnerBasis[{V - p2, a^2 - (1 + t^2), > a - t - c^2, a + t - d^2}, {V, a}, {t, c, d}] > > > {-16384*V^4*a^6 + 1024*Pi^2*V^2*a^5 + 49152*V^4*a^4 - > 16*Pi^4*a^4 - 3840*Pi^2*V^2*a^3 - 49152*V^4*a^2 + > 72*Pi^4*a^2 + 3840*Pi^2*V^2*a + 16384*V^4 - 81*Pi^4} > > Finally we solve for V and FullSimlify: > > FullSimplify[Solve[sol[[1]] == 0, V], a > 1] > > > {{V -> -(((2*a + 3)*Pi)/(8*Sqrt[2]*(a + 1)^(3/2)))}, > {V -> ((2*a + 3)*Pi)/(8*Sqrt[2]*(a + 1)^(3/2))}, > {V -> -((Pi*Abs[3 - 2*a])/(8*Sqrt[2]*(a - 1)^(3/2)))}, > {V -> (Pi*Abs[3 - 2*a])/(8*Sqrt[2]*(a - 1)^(3/2))}} > > > It's clear that only the second answer satisfies the conditions of > the problem. > > Of course this required some "hand steering" of Mathematica. I will > be surprised if someone shows that this coudl be done " on autmatic > pilot'. > > Andrzej Kozlowski > > I just noticed that the above argument does not constitute a strictly rigorous mathematical proof due to the substitution a-t->c^2, where c>0, which of course need not always hold. However, in the cases when it does not hold one can use the substitution a-t->-c^2, where c>0 and obtain the same solution for V. Andrzej Kozlowski

**References**:**"hard" simplification***From:*dimitris <dimmechan@yahoo.com>