Re: Re: Hyperbolic function identity
- To: mathgroup at smc.vnet.net
- Subject: [mg50964] Re: [mg50945] Re: [mg50932] Hyperbolic function identity
- From: "Wolf, Hartmut" <Hartmut.Wolf at t-systems.com>
- Date: Thu, 30 Sep 2004 04:52:05 -0400 (EDT)
- Sender: owner-wri-mathgroup at wolfram.com
No course to blame Mathematica. We do have TrigToExp and ExpToTrig, but cannot keep them both at the same time (no confluent ruleset). So without meta-rules, a choice must be made by the user. -- Hartmut >-----Original Message----- >From: Maxim A. Dubinnyi [mailto:maxim at nmr.ru] To: mathgroup at smc.vnet.net >Sent: Wednesday, September 29, 2004 9:15 AM >To: mathgroup at smc.vnet.net >Subject: [mg50964] [mg50945] Re: [mg50932] Hyperbolic function identity > > >This works correctly: > >FullSimplify[TrigToExp[ArcCosh[1+(z^2)/2]-2*ArcSinh[z/2]], z > 0] > >It looks like that mathematica works better with logarithms and >exponents then with trigonometric functions. > >Carlos Felippa wrote: > >>Why >> >> FullSimplify[ArcCosh[1+z^2/2]-2*ArcSinh[z/2],z>0]; >> >>does not evaluate to 0? >> >> >> > > >