Re: querries
- To: mathgroup at smc.vnet.net
- Subject: [mg119560] Re: querries
- From: "Dr. Wolfgang Hintze" <weh at snafu.de>
- Date: Fri, 10 Jun 2011 06:37:43 -0400 (EDT)
- References: <isq4o1$63f$1@smc.vnet.net>
You need to tell Mathematica something about mu and lambda in order to
make the expressions unambigious.
Then it works out fine.
Example
In[5]:=
Simplify[(mu*lambda)^(b) - (mu^b)*(lambda^b), {mu > 0, lambda > 0}]
Out[5]=
0
Regards,
Wolfgang
"Savits, Thomas H" <savits at pitt.edu> schrieb im Newsbeitrag
news:isq4o1$63f$1 at smc.vnet.net...
> 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