Re: simplification
- To: mathgroup at smc.vnet.net
- Subject: [mg78920] Re: [mg78852] simplification
- From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
- Date: Thu, 12 Jul 2007 05:31:08 -0400 (EDT)
- References: <200707111010.GAA05338@smc.vnet.net>
On 11 Jul 2007, at 19:10, dimitris wrote:
> Hello.
>
> (version 5.2)
>
> Let
>
> o = (-Log[-I + z] - Log[I + z] + Log[1 + z^2])*RootSum[1 + #1^2 + #1^3
> & , Log[z - #1]/(2*#1 + 3*#1^2) & ];
>
> The expression (seems that) is real in the whole real axis
>
> Plot[Chop@#@o, {z, -3, 3}] & /@ {Re, Im}
>
> I want to prove it (with Mathematica!)
>
> Interestingly,
>
> (FullSimplify[o, #1] & ) /@ {Element[z,Reals], z > 0}
>
> {(-(Log[1/(I + z)] + Log[I + z]))*RootSum[1 + #1^2 + #1^3 & , Log[z -
> #1]/(2*#1 + 3*#1^2) & ],
> (-(Log[1/(I + z)] + Log[I + z]))*RootSum[1 + #1^2 + #1^3 & , Log[z -
> #1]/(2*#1 + 3*#1^2) & ]}
>
> (If Mathematica 6 returns zero I find one more reason for
> upgrading...!)
>
> In Mathematica 5.2 I used the following method
>
> FullSimplify[o /. z -> 2]
> Simplify[FullSimplify[D[ff, z]] //. (a_.)*Log[b_] + (c_.)*Log[d_] ->
> Log[b^a*d^c]]
>
> 0
> 0
>
> (*the value for a given z, say 2 is zero and the expression has
> derivative equal
> to zero for any z*)
>
> Any other ideas?
>
> BTW,
> I am not very happy with the performance of FullSimplify
> below
>
> FullSimplify[-((Log[-I + z] + Log[I + z] - Log[1 + z^2])/(1 + z^2 +
> z^3)), z Reals]
> -((Log[1/(I + z)] + Log[I + z])/(1 + z^2 + z^3))
>
> (*the expression is actual zero*)
>
> Of course I talk about version 5.2.
>
> Dimitris
>
>
With Mathematica 6.01:
o = (-Log[-I + z] - Log[I + z] + Log[z^2 + 1])*
RootSum[#1^3 + #1^2 + 1 & ,
Log[z - #1]/(3*#1^2 + 2*#1) & ];
FullSimplify[o, Element[z, Reals]]
0
also:
FullSimplify[-((Log[-I + z] + Log[I + z] -
Log[z^2 + 1])/(z^3 + z^2 + 1)),
Element[z, Reals]]
0
Is that reason enough to upgrade?
Andrzej Kozlowski
- References:
- simplification
- From: dimitris <dimmechan@yahoo.com>
- simplification