Re: Re: A FullSimplify Problem
- To: mathgroup at smc.vnet.net
- Subject: [mg41105] Re: [mg41087] Re: [mg41050] A FullSimplify Problem
- From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
- Date: Sat, 3 May 2003 03:27:27 -0400 (EDT)
- Sender: owner-wri-mathgroup at wolfram.com
I feel compelled to apologize to the programer(s) of FullSimplify (and Simplify ) for even entertaining the possibility that it might have been the culprit. In fact these examples show clearly that FullSimplify would have produced the wanted answer had Integrate not made its typical wrong choice of branch error. The reason why I briefly thought Simplify might have been the culprit was that I tried substituting one randomly chosen value for a and the answer came up right. Of course this can easily happen in this sort of situation, and I should have really known better than to doubt the generally very reliable Simplify. Andrzej Kozlowski Yokohama, Japan http://www.mimuw.edu.pl/~akoz/ http://platon.c.u-tokyo.ac.jp/andrzej/ On Thursday, May 1, 2003, at 06:00 pm, Andrzej Kozlowski wrote: > This seems unlikely since Mathematica 4.2 returns > > > FullSimplify[Integrate[Log[1-2a Cos[x]+a^2],{x,0,Pi}],{a>1}] > > > 4 ¥ð Log[a] > > which is double the right answer. With the assumption a<-1 the answer > is even worse (if returning a complex number when a real one is > expected is "worse" than returning twice the right answer): > > FullSimplify[Integrate[Log[1 - 2*a*Cos[x] + a^2], > {x, 0, Pi}], {a < -1}] > > Out[7]= > Pi*(2*I*Pi + Log[1/a^2]) > > Of course this does not prove that the unsimplified answer returned by > Mathematica is wrong, in fact numerical tests suggest that it is > probably right and the problem may be with FullSimplify. But I would > not count on it. > > > > Andrzej Kozlowski > Yokohama, Japan > http://www.mimuw.edu.pl/~akoz/ > http://platon.c.u-tokyo.ac.jp/andrzej/ > > > On Wednesday, April 30, 2003, at 05:22 pm, Ersek, Ted R wrote: > >> At http://mathworld.wolfram.com/LeibnizIntegralRule.html >> I learned that >> Integrate[Log[1-2a Cos[x]+a^2],{x,0,Pi}] >> = 2*Pi*Log[Abs[a]] >> >> Mathematica knows how to do this integral, but gives a much more >> complicated >> result. Can anyone explain how to use FullSimplify and other >> transformations to show that the complicated result Mathematica gives >> is >> equivalent to the answer above? >> >> Thanks, >> Ted Ersek >> >> >> >> > > > >