simplification
- To: mathgroup at smc.vnet.net
- Subject: [mg78852] simplification
- From: dimitris <dimmechan at yahoo.com>
- Date: Wed, 11 Jul 2007 06:10:21 -0400 (EDT)
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
- Follow-Ups:
- Re: simplification
- From: Andrzej Kozlowski <akoz@mimuw.edu.pl>
- Re: simplification