MathGroup Archive 2004

[Date Index] [Thread Index] [Author Index]

Search the Archive

Re: Re: Hyperbolic function identity


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?
>>
>>  
>>
>
>
>


  • Prev by Date: Fitting multiple data
  • Next by Date: Re: Hyperbolic function identity
  • Previous by thread: Re: Hyperbolic function identity
  • Next by thread: Re: Hyperbolic function identity