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Re: [mg5311] Single Problem
Posted:
Nov 26, 1996 3:36 AM


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 Date: Sat, Nov 23, 1996 4:58 AM EDT From: yelride@ibm.net XFrom: yelride@ibm.net (Redirley Matheus Santos) To: mathgroup@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 = ?



