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Topic: Single Problem
Replies: 2   Last Post: Nov 26, 1996 3:42 AM

 Messages: [ Previous | Next ]
 Harald Berndt Posts: 22 Registered: 12/7/04
Re: Single Problem
Posted: Nov 26, 1996 3:30 AM

In article <5764ab\$28t@dragonfly.wolfram.com>, Redirley Matheus Santos
<yelride@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[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

Date Subject Author
11/23/96 Redirley Matheus Santos
11/26/96 Harald Berndt
11/26/96 Brian J. Albright