Re: Single Problem
- To: mathgroup at smc.vnet.net
- Subject: [mg5332] Re: [mg5311] Single Problem
- From: BobHanlon at aol.com
- Date: Wed, 27 Nov 1996 01:47:47 -0500
- Sender: owner-wri-mathgroup at wolfram.com
X = Sqrt[6 + Sqrt[6 + Sqrt[6 + Sqrt[6 + ...]]]] X = Sqrt[6 + X] X^2 = 6 + x X^2 - x - 6 = 0 (X - 3) (X + 2) = 0 X = 3 and X = -2 The latter is an extraneous root. X = 3 In Mathematica this can be solved using either Solve[X == Sqrt[6 + X], X] or FixedPoint[Sqrt[6. + #]&, Sqrt[6.]] Bob Hanlon FORWARDED MESSAGE: Subj: [mg5311] Single Problem From: yelride at ibm.net To: mathgroup at smc.vnet.net X-From: yelride at ibm.net (Redirley Matheus Santos) To: mathgroup at smc.vnet.net Hi everybody, Who would like solve this problem ? X is a Integer number X = Sqr(6+(Sqr(6+(Sqr(6+Sqr(6+Sqr(...))))))) infinite Sqr of 6 inside X = ?