Re: querries
- To: mathgroup at smc.vnet.net
- Subject: [mg119576] Re: querries
- From: Bob Hanlon <hanlonr at cox.net>
- Date: Fri, 10 Jun 2011 06:40:36 -0400 (EDT)
- Reply-to: hanlonr at cox.net
Because it isn't always true. For example, ((-2)*(-8))^(1/2) - (-2)^(1/2)*(-8)^(1/2) 8 Read help for PowerExpand http://reference.wolfram.com/mathematica/ref/PowerExpand.html (mu*lambda)^b - (mu^b)*(lambda^b) // PowerExpand 0 (mu*lambda)^b - (mu^b)*(lambda^b) // Simplify[#, {mu > 0, lambda > 0}] & 0 Bob Hanlon ---- "Savits 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