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MathGroup Archive 2007

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Re: "hard" simplification

  • To: mathgroup at smc.vnet.net
  • Subject: [mg75346] Re: [mg75315] "hard" simplification
  • From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
  • Date: Thu, 26 Apr 2007 03:28:30 -0400 (EDT)
  • References: <200704250931.FAA26921@smc.vnet.net>

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




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