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MathGroup Archive 2007

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Re: Showing that ArcSinh[2]/ArcCsch[2] is 3?

  • To: mathgroup at smc.vnet.net
  • Subject: [mg73660] Re: Showing that ArcSinh[2]/ArcCsch[2] is 3?
  • From: "dimitris" <dimmechan at yahoo.com>
  • Date: Sat, 24 Feb 2007 02:12:17 -0500 (EST)
  • References: <erjp6e$pv7$1@smc.vnet.net><ermcv4$hfo$1@smc.vnet.net>

Hello David and Dana.

Take a look at the following links:

http://mathematica6.wordpress.com/
http://gorithm.blogs.com/gorithm/2006/06/mathematica_6_o.html
http://www.gorithm.com/

Especially the last link contains very good news!

Regards
Dimitris


=CF/=C7 David W.Cantrell =DD=E3=F1=E1=F8=E5:
> "Dana DeLouis" <dana.del at gmail.com> wrote:
> > > TrigToExp[ArcSinh[2]/ArcCsch[2]]  yields
> > > Log[2 + Sqrt[5]]/Log[1/2 + Sqrt[5]/2].
> > > But then how should we transform that to 3?
> >
> > Hi.  Not ideal, but this worked with v 5.2
> >
> > equ = TrigToExp[ArcSinh[2]/ArcCsch[2]];
> >
> > avoid = Count[{#1}, _ArcSinh | _ArcCsch | _Log, Infinity] & ;
> >
> > FullSimplify[equ, ComplexityFunction -> avoid]
> >
> > 3
>
> Maybe not ideal, as you say, but, of all responses, it comes closest to
> doing what I wanted!
>
> Many thanks to all those who responded to my question. The title I used f=
or
> the thread was very badly worded; I apologize. Nonetheless, most
> respondents realized from the text of my message that I wanted to get
> Mathematica to _transform_ ArcSinh[2]/ArcCsch[2]] to 3, rather than to get
> it merely to verify that the value is 3.
>
> > Dana DeLouis
> > Windows XP & Mathematica 5.2
> > Ps.  Any guesses when v. 6.0 is released?
>
> Good question. Anyone?
>
> David
>
> > "David W.Cantrell" <DWCantrell at sigmaxi.net> wrote in message
> > news:erelp9$7mm$1 at smc.vnet.net...
> > >I hope I've just overlooked something very simple.
> > > I want to transform ArcSinh[2]/ArcCsch[2] to 3, using just "knowledge"
> > > already implemented in Mathematica. I tried FullSimplify first, and it
> > > doesn't help. I tried several other things. For example,
> > >
> > > TrigToExp[ArcSinh[2]/ArcCsch[2]]  yields
> > >
> > > Log[2 + Sqrt[5]]/Log[1/2 + Sqrt[5]/2].
> > >
> > > But then how should we transform that to 3?
> > >
> > > David
> > >



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