10
$\begingroup$

I've been given the following sequence: \begin{align*} &a_0 = 0; \\ &a_{n+1} = (a_n)^2+\frac{1}{4}. \end{align*}

I also have to prove that whatever I come up with is correct, but that will likely be the easy part.

Here are the first few values:

\begin{align} &a_0 = 0 \\ &a_1 = \frac{1}{4}\\ &a_2 = \frac{5}{16}\\ &a_3 = \frac{89}{256} \\ &a_4 = \frac{24305}{65536} \end{align}

I've managed to to determine that the denominators are of the form $2^{2^n}$. I've tested up to one million terms of this sequence and it appears that $\lim_{n\rightarrow\infty}a_n = \frac{1}{2}$. I spent a while trying to find something of the form $a_n = \frac{P(n)}{Q(n)}$. I haven't had any luck with this, so I started looking into some sums. I've found that \begin{align*} a_2 = \frac{1}{4} + \frac{1}{16} = \frac{5}{16} \end{align*} and, \begin{align*} a_3 = \frac{1}{4} + \frac{1}{16} + \frac{1}{32} + \frac{1}{256} = \frac{89}{256} \end{align*}

But now, \begin{align*} a_4 = \frac{1}{4} + \frac{1}{16} + \frac{1}{32} + \frac{1}{64} + \frac{1}{128} + \frac{1}{512} + \frac{1}{1024} + \frac{1}{2048} + \frac{1}{4096} + \frac{1}{65536}= \frac{24305}{65536}. \end{align*}

So it seems that there is some type of sum involving negative powers of 2 going on, but it isn't clear to me that there is even a pattern here. Any hints/help would be appreciated!

$\endgroup$
  • $\begingroup$ The sums over powers of 2 amount to expansions in base 2 of the fractions. They would all terminate if you showed the denominators are $2^{2^n}$ as you say, but the "pattern" of the expressions in terms of negative powers of 2 is just whatever happens to be the digits of the $a_n$ in base 2, and at least for the first four I don't see an obvious pattern in those digits. $\endgroup$ – coffeemath Jul 8 '13 at 10:40
  • $\begingroup$ The numerators seem to be those of the sequence at OEIS number A167424, which is a slightly different sequence, but only by a factor of 1/2 as I recall (just looked briefly at the sequence obtained...). $\endgroup$ – coffeemath Jul 8 '13 at 10:47
  • $\begingroup$ hmm, that page on OEIS is making me suspicious that there isn't a closed form for this recurrence relation... $\endgroup$ – stochasm Jul 8 '13 at 11:11
  • 1
    $\begingroup$ At least, the closed form of the limit is 1/2... $\endgroup$ – Julien Jul 8 '13 at 12:04
  • 2
    $\begingroup$ See this answer ... math.stackexchange.com/a/114949/442 $\endgroup$ – GEdgar Jul 8 '13 at 12:58
7
$\begingroup$

I just checked OEIS sequence A167424 for which $$f(1)=1/2 \\ f(n+1)=[f(n)^2+1]/2.$$ If this $f$ is divided by 2 you get your sequence $a_1,a_2,\cdots,$ since then it gives correctly $a_1=1/4$ and $a_{n+1}=a_n^2+1/4.$ The recursion on the $a_n=f(n)/2$ follows on dividing through the recursion $f(n+1)=[f(n)^2+1]/2$ by 2 to obtain $$\frac{f(n+1)}{2}=\left( \frac{f(n)}{2}\right) ^2 +\frac{1}{4}.$$

So anything like closed forms etc. should be extractable from the OEIS page, if it's there.

Side note: The Mandlebrot set intersects the real axis in the interval $[-2,1/4]$. Since it consists of those points $c$ such that the orbit of $0$ under the iteration of $f(x)=x^2+c$ is bounded, we see that your sequence is precisely the (bounded) orbit of the rightmost real point of the Mandlebrot set. (In this sense it seems not surprising there isn't a closed form, as typically Mandlebrot iterations bounce around the set unpredictably. There are periodic points, but it seems $1/4$ is not one of them. (Just thought this connection might be of interest.)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.