       Re: querries

• To: mathgroup at smc.vnet.net
• Subject: [mg119577] Re: querries
• From: Heike Gramberg <heike.gramberg at gmail.com>
• Date: Fri, 10 Jun 2011 06:40:47 -0400 (EDT)
• References: <201106090945.FAA06257@smc.vnet.net>

```1) Probably because it's not alway equal to 0. Consider for example

(mu*lambda)^(b) - (mu^b)*(lambda^b) /. {mu -> -1, lambda -> -1,  b -> 1/2}

which returns 2. This is because (-1*-1)^(1/2)=(1)^(1/2)=1 but (-1)^(1/2)*(-1)^(1/2)=I*I=-1.

If mu and lambda are both positive this expression is equal to 0, but since Mathematica doesn't
know is the case it leaves the expression as it is.

2) You could try PowerExpand[]

Heike.

On 9 Jun 2011, at 10:45, Savits, Thomas H wrote:

> 1)       Why does Mathematica not give the value 0 to the expression
> (mu*lambda)^(b) - (mu^b)*(lambda^b)? It seems to treat the expression (mu*lambda) as a new variable. I used the greek letters from the palette in the actual expression.
>
> 2)       Many times I cannot get Mathematica to simplify expressions involving the exponential functions, e.g., it won't combine the arguments of the Exp functions. Is there a way to force this?
>
> Thanks,
> Tom Savits
> Statistics Department
> University of Pittsburgh

```

• References:
• querries
• From: "Savits, Thomas H" <savits@pitt.edu>
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