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Re: querries

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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@pitt.edu> schrieb im Newsbeitrag 
news:isq4o1$63f$1@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 

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Thread

querries "Savits, Thomas H" <savits@pitt.edu> - 2011-06-09 09:45 +0000
  Re: querries "Kevin J. McCann" <Kevin.McCann@umbc.edu> - 2011-06-10 10:40 +0000
  Re: querries "Dr. Wolfgang Hintze" <weh@snafu.de> - 2011-06-10 10:37 +0000

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