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Re: querries

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Concerning your first question, try this:

  (mu*lambda)^(b) - (mu^b)*(lambda^b) // PowerExpand

0

Concerning the second one, it is not quite clear, what do you have in mind, since this:

Exp[a]*Exp[b]

E^(a + b)

works. You have probably a more complex example in mind.

Have fun, Alexei



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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Re: querries Alexei Boulbitch <alexei.boulbitch@iee.lu> - 2011-06-10 10:39 +0000

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