       Re: Single Problem

• To: mathgroup at smc.vnet.net
• Subject: [mg5322] Re: Single Problem
• From: haberndt at dnai.com (Harald Berndt)
• Date: Wed, 27 Nov 1996 01:47:38 -0500
• Organization: DNAI ( Direct Network Access )
• Sender: owner-wri-mathgroup at wolfram.com

```In article <5764ab\$28t at dragonfly.wolfram.com>, Redirley Matheus Santos
<yelride at ibm.net> wrote:

> 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 = ?

Do you really mean Sqr?

In:=
NestList[(6+#)^2&, x, 3]

Out=
2              2 2                   2 2 2
{x, (6 + x) , (6 + (6 + x) ) , (6 + (6 + (6 + x) ) ) }

... and that grows to infinity really fast:

In:=
NestList[(6+#)^2&, 1, 3]
Out=
{1, 49, 3025, 9186961}
In:=
NestList[(6+#)^2&, .1, 3]
Out=
6
{0.1, 37.21, 1867.1, 3.50852 10 }
Out=
-16                       6
{1. 10   , 36., 1764., 3.1329 10 }
In:=
\$MachineEpsilon
Out=
-19
1.0842 10
In:=
NestList[(6+#)^2&, \$MachineEpsilon, 3]
Out=
-19                       6
{1.0842 10   , 36., 1764., 3.1329 10 }

It's quite different for the square root, as one can demonstrate with

In:=
ListPlot[
NestList[Sqrt[6+#]&, 1., 30],
PlotRange -> All
];

To find that level off "limit", apply

In:=
FixedPoint[Sqrt[6+#]&, 1.]
Out=
3.
In:=
FixedPoint[Sqrt[6+#]&, \$MachineEpsilon]
Out=
3.

Best,

--
Harald Berndt, Ph.D.                            Research Specialist,
Voice: 510-652-5974                                      Consultant
FAX:   510-215-4299

```

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