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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: [mg73617] Re: Showing that ArcSinh[2]/ArcCsch[2] is 3?
  • From: "David W.Cantrell" <DWCantrell at sigmaxi.net>
  • Date: Fri, 23 Feb 2007 04:33:16 -0500 (EST)
  • References: <erjp6e$pv7$1@smc.vnet.net>

"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 for
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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