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From: scavo@cie.uoregon.edu (Tom Scavo)
Subject: Re: Challenging limit problem (involving x^2 + c for c=-3/2)
Summary: FALSE, the limit does not exist
References: <5071@network.ucsd.edu>
Reply-To: scavo@cie.uoregon.edu (Tom Scavo)
Followup-To: comp.theory.dynamic-sys
Organization: University of Oregon Campus Information Exchange
Keywords: recurrence sequence, iterative mappings, limit point
Message-ID: <1991Mar29.222557.20208@ariel.unm.edu>
Date: Fri, 29 Mar 91 22:25:57 GMT
Lines: 54

In article <5071@network.ucsd.edu> revans@euclid.UUCP (Ron Evans) writes:
>
>
>Consider the sequence x  , x  ,...  defined by  x  = 0, 
>                       0    1                    0              
>         2
>x    = x   -  3/2 .  TRUE OR FALSE:  lim inf x   =  - 3/2 ?
> n+1    n                                     n
>

In article <1991Mar26.215525.27585@ariel.unm.edu> I gave a reference
and wrote:
>
>(iii) For  -3/4 < c < 1/4 ,  Q  has a pair of fixed points  x1  and  x2  
>given by
>
>                   1 + sqrt(1 - 4c)               1 - sqrt(1 - 4c)
>              x1 = ----------------    and   x2 = ----------------
>                          2                              2
>
>with  x1  repelling and  x2  attracting; the basin of attraction for  x2  
>is  (-x1, x1).

For  c  outside this range (except perhaps at the endpoints),  Q  
(where  Q(x) = x^2 + c ) has no attracting fixed points.  So, for 
c = -3/2  the answer is going to be FALSE no matter what value you
write down!

A more interesting question would be:  is the orbit of  0  periodic
when  c = -3/2 ?


In article <1991Mar29.141108.8914@athena.cs.uga.edu> is@athena.cs.uga.edu (Bob Stearns) writes:

>I am neither Russian nor a powerful mathematician, but a simple test with
>my local computer indicates that the iterated function x = x^2-3/2 is
>periodic with period 254 or less.

I'm not sure what you mean by this.  Is it periodic, and if so,
what is the period?  Does anyone know?


In article <5071@network.ucsd.edu> revans@euclid.UUCP (Ron Evans) continues:
>    Perhaps the best chance for a solution lies with a powerful
>Russian mathematician, so I hope this posting reaches Harvard.

You're probably referring to Sarkovskii's theorem, but I don't
see what that has to do with the original question?  And what
does Harvard have to do with anything?

Try not to be so vague next time.

Tom Scavo
scavo@cie.uoregon.edu