Mathematica 9 is now available
Services & Resources / Wolfram Forums / MathGroup Archive
-----

MathGroup Archive 2011

[Date Index] [Thread Index] [Author Index]

Search the Archive

Re: querries

  • To: mathgroup at smc.vnet.net
  • Subject: [mg119571] Re: querries
  • From: Alexei Boulbitch <alexei.boulbitch at iee.lu>
  • Date: Fri, 10 Jun 2011 06:39:42 -0400 (EDT)

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


-- 
Alexei Boulbitch, Dr. habil.
Senior Scientist
Material Development

IEE S.A.
ZAE Weiergewan
11, rue Edmond Reuter
L-5326 CONTERN
Luxembourg

Tel: +352 2454 2566
Fax: +352 2454 3566
Mobile: +49 (0) 151 52 40 66 44

e-mail: alexei.boulbitch at iee.lu

www.iee.lu

--

This e-mail may contain trade secrets or privileged, undisclosed or
otherwise confidential information. If you are not the intended
recipient and have received this e-mail in error, you are hereby
notified that any review, copying or distribution of it is strictly
prohibited. Please inform us immediately and destroy the original
transmittal from your system. Thank you for your co-operation.



  • Prev by Date: Re: Using the Combinatorica Package in my own package
  • Next by Date: Re: Fast way to find neighbours of vertices in a Graph & possible performance bug in NeighborhoodGraph
  • Previous by thread: Re: querries
  • Next by thread: Re: querries