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If $P(x)=x^3-2x+1$, define $z_n$ as the number of real roots of the polynomial $P^{\circ n}(x)$, where the superscript denotes $n$-fold composition. Can we find a general formula for $z_n$, or perhaps a recurrence of some sort? The sequence begins $3,7,15,27,47,...$ and is not in the OEIS.

I have solved this problem for the polynomial $Q(x)=x^3-3x+1$, and determined that the number of real zeroes of $Q^{\circ n}$ is equal to $2^{n+1}-1$. However, this problem was much easier because the maximum values of $Q$ occur at integer values of $x$ and the zeroes of $Q$ are irrational, allowing one to break the real line into intervals of the form $[k,k+1]$ and kept track of which intervals $Q$ maps onto one another.

Can anyone figure out how to do this with $P(x)$? This problem has puzzled me for a while, so I am willing to offer a $+50$ bounty for a satisfactory answer or analysis of the problem (as soon as the rules of MSE will allow me to offer it).

It would also be helpful if anyone could provide a large list of values of $z_n$, since all of the values I have were obtained by counting by hand.

Cheers!

EDIT: Should a closed-form formula or recurrence elude any potential answerers, it would also be nice to obtain a (proven) asymptotic formula for $z_n$ instead of a closed-form formula.

Franklin Pezzuti Dyer
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  • For $1\le n \le 20$, the counts are $$3, 7, 15, 27, 47, 75, 117, 173, 253, 357, \503, 691, 951, 1283,1739, 2319, 3113, 4121, 5497, 7241$$ – quasi Oct 19 '18 at 23:48
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    If $n$ is even, with $n\ge 4$, the data appears to support the relation $$z_n = 3z_{n-1}-3z_{n-2}+z_{n-3}$$ – quasi Oct 20 '18 at 00:39
  • I have $z_2=5$ did I misunderstand something. I calculated that two of the roots are complex and 5 are real. That's 7 total complex roots.... but only 5 real ones. – user351797 Oct 20 '18 at 01:41
  • @Clclstdnt: Note that $\deg(f)=3$, hence $\deg(f{\circ}f)=9$, so $f{\circ}f$ has $9$ complex roots. Of those $9$ complex roots, $7$ of them are real. Thus, $z_2=7$. – quasi Oct 20 '18 at 04:57
  • There's no need to count by hand: there are free symbolic algebra packages which can count for you. E.g. with Sage this program will produce counts for 1 to 20, given enough time and memory. With a bit more effort a substantial speed-up can be produced, because you don't actually ask for the locations of the roots so separation into square-free parts and then application of Sturm sequences would do the trick. – Peter Taylor Oct 20 '18 at 10:47
  • For $21\le n\le 30$, the counts are $$9619, 12631, 16735, 21931, 29005\ 37957, 50141, 65557, 86531, 113063$$ – quasi Oct 20 '18 at 17:12
  • @quasi Perhaps these are the convergents of the continued fraction of some familiar irrational number... that might explain the recurrence you found. – Franklin Pezzuti Dyer Oct 20 '18 at 17:23
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    The obvious observations are:1) Whenever $x$ is a real zero of the $n$th iterate it will produce 3 zeros for the next iterate whenever $x$ is between $P(\pm\sqrt{2/3})$, and a single real zero otherwise. 2) The zeros of the $n$th iterate are also zeros of the $(n+2)$nd iterate. It may be tough to analyze it further. – Jyrki Lahtonen Oct 22 '18 at 21:37
  • I'm trying to analyze the problem in the following way. For the simpler polynomial $Q(x)=x^3-3x+1$ we have the formula $$Q(2\cos\alpha)=2\cos 3\alpha+1.$$ This might (should?) become useful in getting from the real zeros of the $n$the iterate to those of iterate number $n+1$. Possibly using a heuristic that the $\alpha$'s might (?) be uniformly distributed asymptotically. We do need to factor in the effect that not all zeros will be in the interval $[-2,2]$, and I can't wrap my head around it at this time. – Jyrki Lahtonen Oct 23 '18 at 07:59
  • With $P(x)$ the equivalent formula in terms of cosine triplication formula reads $$P(\sqrt{\frac83}\cos\alpha)=\frac23\sqrt{\frac83}\cos3\alpha+1.$$ Looks messy. I'm not sure this would lead to anything useful. At first I could explain the asymptotic ratio $z_{n+1}/z_n\to 2$ for $Q(x)$ using this, but the zeros bleeding out of the range of $2\cos \alpha$ ruined it. – Jyrki Lahtonen Oct 23 '18 at 08:04
  • ^ ... I thought I could explain... I didn't claim it was a good idea :-/ – Jyrki Lahtonen Oct 23 '18 at 12:05

1 Answers1

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A sketch immediately reveals what is going on.

Polynomial_Iterated_1

The polynomial $$ y = x^{\,3} - 2x + 1 $$ is a depressed cubic, with three real zeros at $x=\{-\phi, 1/\phi, 1\}$.

It has a local maximum $(x_{max},y_{max})=( - \sqrt{2/3}, \; 1+4*\sqrt{6}/9)$ and a local minimum $(x_{min},y_{min})=( \sqrt{2/3}, \; 1-4*\sqrt{6}/9)$.

The range $[y_{min},\,y_{max}]$ includes the two zeros $1/\phi$ and $1$, but not the lower at $-\phi$.

Then the sketch shows that at the first iteration $$ y_{\,2} (y_{\,1} (x)) $$ we will have that:
- the lower zero will remain, while the upper two will be replicated $3$ times;
- the maximum will remain, while the minimum will be replicated $3$ times;
- in between each triple $zero,min,zero$ a new maximum will appear;

However it is not easy to predict what the value of the two new maxima will be, and thus how many of the zeros $1/\phi$ and $1$ they will encompass in the following iteration, specially in the long run.
And that confirms @Jirky comment.

G Cab
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    I have downvoted this answer because it does not answer my question. The analysis in this answer goes no deeper than the discussion in the comments. – Franklin Pezzuti Dyer Oct 29 '18 at 14:10
  • @Frpzzd: sorry that my post did not meet your expectations. That's true, it is not an answer actually: it is just a (hopefully) clear explanation of where the difficulty to provide an answer lies. – G Cab Oct 29 '18 at 14:29