Services & Resources / Wolfram Forums
-----
 /
MathGroup Archive
2000
*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 2000

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

Search the Archive

Re: Simplification shortcomings?

  • To: mathgroup at smc.vnet.net
  • Subject: [mg24649] Re: Simplification shortcomings?
  • From: "Allan Hayes" <hay at haystack.demon.co.uk>
  • Date: Mon, 31 Jul 2000 09:23:20 -0400 (EDT)
  • References: <8lsutf$27c@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

Bob,

FullSimplify uses more information and tries harder than Simplify:

FullSimplify[(1 + Sqrt[5])/2 - Sqrt[(3 + Sqrt[5])/2]]

        0

--
Allan
---------------------
Allan Hayes
Mathematica Training and Consulting
Leicester UK
www.haystack.demon.co.uk
hay at haystack.demon.co.uk
Voice: +44 (0)116 271 4198
Fax: +44 (0)870 164 0565

"Bob Harris" <nitlion at mindspring.com> wrote in message
news:8lsutf$27c at smc.vnet.net...
> Howdy,
>
> I'm a relative novice to Mathematica.  While working with it today, I had
> occasion to want to know if a result was equal to (1 + Sqrt[5])/2.  The
> result was shown as Sqrt[(3 + Sqrt[5])/2].  After some pancil and paper
> work, I figured out that these two are equal.  Or, I should say, that the
> former is one of the values that the latter can have.
>
> I was frustrated in my attempts to get Mathematica to answer that question
> for me.  Simplify[(1 + Sqrt[5])/2 - Sqrt[(3 + Sqrt[5])/2]] didn't provide
> any improvement.  Calculating this value to many decimal digits showed it
> was near zero (probably close enough that I could have applied the
> techniques shown in Scheinerman's recent article in American Mathematical
> Monthly).  The only way I got Mathematica to show the equality was to
square
> both numbers;  Simplify[((1 + Sqrt[5])/2)^2 - (3 + Sqrt[5])/2] is zero.
>
> Is there any better way to do this?  I have some other, more complicated
> numbers that I need to compare.
>
> Thanks,
> Bob Harris
>
>
>
>




  • Prev by Date: Re: Need help defining an Octahedron
  • Next by Date: Re: Simplification shortcomings?
  • Previous by thread: Re: Simplification shortcomings?
  • Next by thread: Re: Simplification shortcomings?