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MathGroup Archive 2004

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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?
>>
>>  
>>
>
>
>


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