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Re: Re: Re: Hyperbolic function identity

  • To: mathgroup at
  • Subject: [mg50987] Re: [mg50964] Re: [mg50945] Re: [mg50932] Hyperbolic function identity
  • From: Andrzej Kozlowski <akoz at>
  • Date: Fri, 1 Oct 2004 04:48:01 -0400 (EDT)
  • References: <>
  • Sender: owner-wri-mathgroup at

It seems to me that the problem lies elsewhere. Note that

FullSimplify[ArcCosh[1 + z^2/2] - 2*ArcSinh[z/2], z > 0,
    TransformationFunctions -> {Automatic, TrigToExp}]

does not work and you have to use:


It looks to me more likely that TrigToExp is actually used as a 
transformation function by FullSimplify while searching for the 
"simplest" form, but unless the condition z>0 is also used at the same 
time the expressions obtained in this way do not actually become 
"simpler" and are discarded. I don't think this can be helped unless 
FullSimplify with the condition z>0 is itself included among the 
transformation functions (which of course seems rather pointless in 
this case as you can just apply it to the entire expression itself).


On 30 Sep 2004, at 17:52, Wolf, Hartmut wrote:

> 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]
To: mathgroup at
>> Sent: Wednesday, September 29, 2004 9:15 AM
>> To: mathgroup at
>> Subject: [mg50987] [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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