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MathGroup Archive 2007

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

  • To: mathgroup at smc.vnet.net
  • Subject: [mg73685] Re: [mg73623] RE: [mg73550] Re: Showing that ArcSinh[2]/ArcCsch[2] is 3?
  • From: János <janos.lobb at yale.edu>
  • Date: Sat, 24 Feb 2007 02:25:47 -0500 (EST)
  • References: <200702230936.EAA17628@smc.vnet.net>

Interestingly the front end can be quite misleading to the eye :)
In[18]:=
N[ArcSinh[2]/ArcCsch[2], {Infinity, 423}]

/423 is the maximum achievable numerical accuracy I have on my G5/

In the notebook I see:

3.0000000000000000000000000000000000000000000000000000000000000000000000=

000000\
000000000000000000000000000000000000000000000000000000000000000000000000=

000000\
000000000000000000000000000000000000000000000000000000000000000000000000=

000000\
000000000000000000000000000000000000000000000000000000000000000000000000=

000000\
000000000000000000000000000000000000000000000000000000000000000000000000=

000000\
00000000000000000000000000000000000

However,
In[23]:=
IntegerPart[N[ArcSinh[2]/
     ArcCsch[2], {Infinity, 423}]]
Out[23]=
2

Hehe...haha...

This is the situation when just by looking in the front end, the 
integer part of a  Real 3. becomes 2

J=E1nos
P.S.  By the way if I increase my accuracy to 424 in N[], the local 
kernel quits :)




On Feb 23, 2007, at 4:36 AM, Tony Harker wrote:

> Dear Oleksandr,
>
>               That's interesting: but looking at what is produced by
>   ArcSinh[2]/ArcCsch[2] == 3 // FullSimplify
>  makes me wonder just how Mathematica achieved this. In other 
> words, what
> does the message that Bob Hanlon so carefully suppressed mean? Similar
> messages sometimes occur when checking the results of simple 
> equations by
> back-substituting, especially when the solutions contain surds, and =

> appear
> to show that Mathematica is diving off into real numbers to prove 
> results
> involving integers and powers of integers.
>
>    Tony
>
>
> Dr A.H. Harker
> Department of Physics and Astronomy
> University College London
> Gower Street
> London
> WC1E 6BT
>
>
>
> ]->-----Original Message-----
> ]->From: sashap [mailto:pavlyk at gmail.com]
> ]->Sent: 21 February 2007 11:04
> ]->To: mathgroup at smc.vnet.net
> ]->Subject: [mg73550] Re: Showing that ArcSinh[2]/ArcCsch[2] is 3?
> ]->
> ]->Dear David,
> ]->
> ]->It is the case in the CAS and in the real life as well that
> ]->proving left hand side equal to right hand side is easier
> ]->than deriving what right hand side the left hand side equals to.
> ]->
> ]->I can not think of a direct way to use built-in 'knowledge'
> ]->to reduce ArcSinh[2]/ArcCsch[2] to 3, but want to point out
> ]->that proving it post-factum does not seem problematic:
> ]->
> ]->In[3]:=
> ]->ArcSinh[2]/ArcCsch[2]-3//FullSimplify
> ]->
> ]->Out[3]=
> ]->0
> ]->
> ]->Sincerely,
> ]->
> ]->Oleksandr Pavlyk
> ]->Special Functions Developer
> ]->Wolfram Research Inc
> ]->
> ]->On Feb 20, 5:24 am, "David W.Cantrell"
> ]-><DWCantr... at sigmaxi.net> wrote:
> ]->> 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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