Mathematica 9 is now available
Services & Resources / Wolfram Forums
-----
 /
MathGroup Archive
2006
*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 2006

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

Search the Archive

Re: Erf[InverseErf] -- making a function and its inverse cancel

  • To: mathgroup at smc.vnet.net
  • Subject: [mg69023] Re: Erf[InverseErf] -- making a function and its inverse cancel
  • From: "MKT" <mktippett at gmail.com>
  • Date: Sun, 27 Aug 2006 01:23:43 -0400 (EDT)
  • References: <ecoq4t$35n$1@smc.vnet.net><ecp1hs$803$1@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

Jean-Marc Gulliet wrote:
> mktippett at gmail.com wrote:
> > How can I get an expression like Erf[InverseErf[0,1/3]] to simplify to
> > 1/3?
> >
> > Cos[ArcCos[1/3]] gives 1/3. Is there some assumption that I'm missing?
> >
> > thanks,
> > Mike
> >
>
> For reason of efficiency, Mathematica does not call systematically the
> simplification rules. You must nudge Mathematica to force the
> simplification.
>
> In[1]:=
> FullSimplify[Erf[InverseErf[0, 1/3]]]
>
> Out[1]=
> 1
> -
> 3
>
> In[2]:=
> N[Erf[InverseErf[0, 1/3]]]
>
> Out[2]=
> 0.333333
>
> HTH,
> Jean-Marc

Thanks! I guess I need a newer version of mathematica :-). In version
4.1

Mathematica 4.1 for Linux
Copyright 1988-2000 Wolfram Research, Inc.
 -- Motif graphics initialized --

In[1]:= FullSimplify[Erf[InverseErf[0, 1/3]]]

                          1
Out[1]= Erf[InverseErf[0, -]]
                          3


  • Prev by Date: Re: General--Transformation of coordinates(from carteseian to polar)
  • Next by Date: Re: Graphics--If condition in Plot3D?????????
  • Previous by thread: Re: Erf[InverseErf] -- making a function and its inverse cancel
  • Next by thread: this expression doesn't do what I expected - what's wrong?