Re: Equality question?
- To: mathgroup at smc.vnet.net
- Subject: [mg59472] Re: Equality question?
- From: Jean-Marc Gulliet <jeanmarc.gulliet at gmail.com>
- Date: Wed, 10 Aug 2005 02:57:39 -0400 (EDT)
- Organization: The Open University, Milton Keynes, U.K.
- References: <dd9mq3$is1$1@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
ab at sd.com wrote:
> EQ1 = ((v*a*(x^2 - a^2))/(4*Pi))*Integrate[1, {\[Phi]p, 0, 2*Pi}]*
> Integrate[(-Sin[\[Theta]p])*((a^2 + x^2 - 2*a*x*Cos[\[Theta]p])^(-3/2) -
> (a^2 + x^2 + 2*a*x*Cos[\[Theta]p])^(-3/2)), {\[Theta]p, Pi/2, 0}]
>
> EQ2 = v*(1 - (z^2 - a^2)/(z*Sqrt[z^2 + a^2]))
>
> Is EQ1 equal to EQ2 ? or not?
>
Not in the general case, but sometimes. For instance:
In[1]:=
EQ1Bis = Simplify[Assuming[{v \[Element] Reals, a > 0, x > a},
((v*a*(x^2 - a^2))/(4*Pi))*
Integrate[1, {\[Phi]p, 0, 2*Pi}]*Integrate[(-Sin[\[Theta]p])*
((a^2 + x^2 - 2*a*x*Cos[\[Theta]p])^(-3/2) - (a^2 + x^2 +
2*a*x*Cos[\[Theta]p])^(-3/2)),
{\[Theta]p, Pi/2, 0}]]]
Out[1]=
(v*(a^2 + x*(-x + Sqrt[a^2 + x^2])))/(x*Sqrt[a^2 + x^2])
In[2]:=
EQ2Bis = v*(1 - (z^2 - a^2)/(z*Sqrt[z^2 + a^2])) /. z -> x
Out[2]=
v*(1 - (-a^2 + x^2)/(x*Sqrt[a^2 + x^2]))
In[3]:=
ComplexExpand[EQ1Bis - EQ2Bis]
Out[3]=
0
In[4]:=
$Version
Out[4]=
"5.2 for Microsoft Windows (June 20, 2005)"
Regards,
/J.M.