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MathGroup Archive 1996

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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?

If I read your problem correctly, you want something like

   In[17]:=
      NestList[(6+#)^2&, x, 3]

   Out[17]=
                 2              2 2                   2 2 2
      {x, (6 + x) , (6 + (6 + x) ) , (6 + (6 + (6 + x) ) ) }

... and that grows to infinity really fast:

In[20]:=
NestList[(6+#)^2&, 1, 3]
Out[20]=
{1, 49, 3025, 9186961}
In[21]:=
NestList[(6+#)^2&, .1, 3]
Out[21]=
                               6
{0.1, 37.21, 1867.1, 3.50852 10 }
Out[22]=
      -16                       6
{1. 10   , 36., 1764., 3.1329 10 }
In[24]:=
$MachineEpsilon
Out[24]=
         -19
1.0842 10
In[25]:=
NestList[(6+#)^2&, $MachineEpsilon, 3]
Out[25]=
          -19                       6
{1.0842 10   , 36., 1764., 3.1329 10 }


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

In[31]:=
ListPlot[
   NestList[Sqrt[6+#]&, 1., 30],
   PlotRange -> All
];

To find that level off "limit", apply

In[32]:=
FixedPoint[Sqrt[6+#]&, 1.]
Out[32]=
3.
In[33]:=
FixedPoint[Sqrt[6+#]&, $MachineEpsilon]
Out[33]=
3.

Best,

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


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