Re: Simplify problems for checking easy equalities...

*To*: mathgroup at smc.vnet.net*Subject*: [mg53866] Re: [mg53856] Simplify problems for checking easy equalities...*From*: DrBob <drbob at bigfoot.com>*Date*: Tue, 1 Feb 2005 04:08:16 -0500 (EST)*References*: <200501300818.DAA08843@smc.vnet.net>*Reply-to*: drbob at bigfoot.com*Sender*: owner-wri-mathgroup at wolfram.com

Here's a counterexample: Log[x^n] - n*Log[x] /. x -> I Log[I^n] - (I*n*Pi)/2 % /. n -> 3 -2*I*Pi and a formula that hints at how complicated the equation actually is: ComplexExpand[Log[x^n] - n*Log[x]] I*(ArcTan[Cos[n*Arg[x]], Sin[n*Arg[x]]] - n*Arg[x]) - (1/2)*n* Log[x^2] + Log[(x^2)^(n/2)] Some cases are easy, however: Simplify[Log[x^n]-n*Log[x],{x>0,n>0}] 0 Bobby On Sun, 30 Jan 2005 03:18:19 -0500 (EST), Cyrus Erik Eierud <cyruserik at tele2.se> wrote: > Please Help! > > Thanks for all great answers I've already found here! > My problem is that I can not simplify what to me seems as a very > simple equality task. This is what I want Mathematica to return zero > for: > > in[1]:= FullSimplify[Log[x^n] - n*Log[x]] > > out[1]= -n Log[x] + Log[x^n]) > > I have used Simplify to check equalities, but the one above (and many > other equations similar to the one above) just don't simplify. Am I > doing anything wrong or does anyone know of a better way to check > equalities? > > Appreciate any help, > Cyrus Eierud, Student > cyruserik at tele2.se > > > > -- DrBob at bigfoot.com www.eclecticdreams.net