       Re: Distributing square-root (1/2) power through

• To: mathgroup at smc.vnet.net
• Subject: [mg101332] Re: [mg101296] Distributing square-root (1/2) power through
• From: Leonid Shifrin <lshifr at gmail.com>
• Date: Wed, 1 Jul 2009 06:36:43 -0400 (EDT)
• References: <200906301035.GAA10672@smc.vnet.net>

```Hi Steven,

Mathematica can not simplify your expression further since it does not know
what your parameters Mu, Sy , Y1 ,Y2  are: if they are general complex
numbers, square root can not be taken unambiguously without indicating which
branch you want.

This will  do what you probably want:

In =

result = Assuming[Mu > 0 && Sy > 0 && Y1 > 0 && Y2 > 0, Simplify[Dist]];
result // InputForm

Out =

Abs[E^((Mu - Y1)^2/(2*Sy^2)) - E^((Mu - Y2)^2/(2*Sy^2))]/
(E^(((Mu - Y1)^2 + (Mu - Y2)^2)/(2*Sy^2))*Sqrt[2*Pi]*Sy)

Regards,
Leonid

On Tue, Jun 30, 2009 at 3:35 AM, Steven Matthew Anderson

> I'm playing with normal distributions, Two random points 1 and 2 with x and
> y coordinates given by:
>
> px1=PDF[NormalDistribution[Mu,Sx],X1]
> px2=PDF[NormalDistribution[Mu,Sx],X1]
> py1=PDF[NormalDistribution[Mu,Sy],Y1]
> py2=PDF[NormalDistribution[Mu,Sy],Y2]
>
> The square of the Euclidean Distance between them is
>
> SqD = (px2-px1)^2+(py2-py1)^2
>
> Take the square root and expand of that to get
>
> Dist = Sqrt[Expand[SqD]]
>
> Now the question:
>
> How do I get the square root to act just like another power so I can
> simplify this mess?  I have tried PowerExpand, FullSimplify, Expand,
> Simplify, and various combinations.  Not sure what I'm missing here.
>
>

```

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