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 > > >