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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 = ?