Re: Simplification shortcomings?
- To: mathgroup at smc.vnet.net
- Subject: [mg24644] Re: [mg24621] Simplification shortcomings?
- From: Andrzej Kozlowski <andrzej at tuins.ac.jp>
- Date: Mon, 31 Jul 2000 09:23:17 -0400 (EDT)
- Sender: owner-wri-mathgroup at wolfram.com
on 7/28/00 11:23 PM, Bob Harris at nitlion at mindspring.com wrote: > 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 > > > > > Yes. In[3]:= FullSimplify[(1 + Sqrt[5])/2 - Sqrt[(3 + Sqrt[5])/2]] Out[3]= 0 or In[5]:= RootReduce[(1 + Sqrt[5])/2 - Sqrt[(3 + Sqrt[5])/2]] Out[5]= 0 (Actually FullSimplify uses RootReduce so it is the latter that really does the job here). Andrzej -- Andrzej Kozlowski Toyama International University, JAPAN For Mathematica related links and resources try: <http://www.sstreams.com/Mathematica/>