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MathGroup Archive 2007

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"hard" simplification

  • To: mathgroup at smc.vnet.net
  • Subject: [mg75315] "hard" simplification
  • From: dimitris <dimmechan at yahoo.com>
  • Date: Wed, 25 Apr 2007 05:31:21 -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 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



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