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MathGroup Archive 2005

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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


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