simplification

*To*: mathgroup at smc.vnet.net*Subject*: [mg75314] simplification*From*: dimitris <dimmechan at yahoo.com>*Date*: Wed, 25 Apr 2007 05:30:49 -0400 (EDT)

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 either by hand or by Mathematica. In fact, in Mathematica someone could try 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 order to show that these expressions are indeed equal? 2) Can someone show me a workaround that will simplify expression o1 to its equivalent (for a>1) o2? Thanks in advance! Dimitris