Services & Resources / Wolfram Forums
-----
 /
MathGroup Archive
2004
*January
*February
*March
*April
*May
*June
*July
*August
*September
*October
*November
*December
*Archive Index
*Ask about this page
*Print this page
*Give us feedback
*Sign up for the Wolfram Insider

MathGroup Archive 2004

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

Search the Archive

Re: Re: Re: Hyperbolic function identity

  • To: mathgroup at smc.vnet.net
  • Subject: [mg50996] Re: [mg50964] Re: [mg50945] Re: [mg50932] Hyperbolic function identity
  • From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
  • Date: Fri, 1 Oct 2004 04:48:12 -0400 (EDT)
  • References: <200409300852.EAA26465@smc.vnet.net> <EC6856EA-12D4-11D9-8846-000A95B4967A@mimuw.edu.pl>
  • Sender: owner-wri-mathgroup at wolfram.com

Here is a puzzling observation.


FullSimplify[FullSimplify[ArcCosh[1 + z^2/2] -
     2*ArcSinh[z/2], ComplexityFunction ->
     (1/(1 + Count[#1, Log, Infinity, Heads ->
          True]) & )], z > 0]


0


I used a silly looking ComplexityFunction whose only purpse was to 
force FullSimplify to choose expressions involving logs in preference 
to those without them. FullSimplify obviously made use of TrigToExp or 
something like that to get the answer. However, the really curious 
thing is this:


FullSimplify[ArcCosh[1 + z^2/2] - 2*ArcSinh[z/2], z > 0,
   ComplexityFunction ->
    (1/(1 + Count[#1, Log, Infinity, Heads -> True]) & )]


2*(Log[2] - Log[z + Sqrt[z^2 + 4]]) +
   Log[z^2 + Sqrt[z^2 + 4]*z + 2] - Log[2]


FullSimplify[%, z > 0]

0

The fact that two applications of FullSimplify were needed seems 
strange to me and it might be a bug. One would have exected that having 
obtained the first expression above FullSimplify would make one more 
attempt at  FullSImplifying thus obtaining the answer 0. Somehow it 
fails to do so.

Andrzej



On 30 Sep 2004, at 20:36, Andrzej Kozlowski wrote:

> 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:
>
> Simplify[ArcCosh[1+z^2/2]-2*ArcSinh[z/
>          2],z>0,TransformationFunctions->{Automatic,
>      FullSimplify[TrigToExp[#],z>0]&}]
>
> 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).
>
> Andrzej
>
> 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 nmr.ru]
To: mathgroup at smc.vnet.net
>>> Sent: Wednesday, September 29, 2004 9:15 AM
>>> To: mathgroup at smc.vnet.net
>>> Subject: [mg50996] [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: Re: Fitting multiple data
  • Next by Date: Re: 3D data set
  • Previous by thread: Re: Re: Re: Hyperbolic function identity
  • Next by thread: Re: Re: Re: Hyperbolic function identity