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Start with any polynomial of degree $n$ with complex coefficients, e.g., $$z^3+z^2+2 z+3 \;.$$ Find its $n$ roots, and list them in order of their modulus: $$-1.28, (0.14\pm 1.53 i)$$ Now form a new polynomial with leading coefficient $1$ and the other coefficients the $n$ roots: $$ z^3 -1.28\, z^2 +(0.14 -1.53 i)\, z +(0.14 +1.53 i) \;. $$ Here I used the sorting convention that $(a-b i) < (a + b i)$. A bit more formally, let $P$ be $$ P \;:\; z^n + a_{n-1}z^{n-1} + \cdots + a_0 \;, $$ with roots $$r_{n-1}, \ldots ,r_0 \;,$$ sorted so that $|r_i| \le |r_{i-1}|$. Define $P_1$ as $$ P_1 \;:\; z^n + r_{n-1}z^{n-1} + \cdots + r_0 \;. $$ My question is:

Q1. Which are the polynomials whose roots are its coefficients, in the sense that $r_i=a_i$, $i=0,\ldots,n-1$, i.e., $P_1 = P$?

Here is one fixed point: $$ z^3 -(0.782599 +0.521714 i)\, z^2 +(0.884646 -0.589743 i)\, z +(0.680552 +1.63317 i) $$

When this process is iterated, it appears to fall into cycles, not always of length $1$.

Q2. Does iteration of this process always fall into a cycle?

(Added. 3Sep15.) The question suggested by Igor Rivin is also interesting, especially in light of Richard Stanley's remark concerning real polynomials:

Q3. Which are the polynomials whose roots are its coefficients, in any order? (This is an unordered version of Q1.)

(Added. 4Sep15.) Concerning Q2, which I realize is less interesting than the other questions, here is an example of a cycle of length $11$. I plot the magnitude of the vector of $3$ roots versus iteration #, starting from a random cubic polynomial. The last vector of roots is $$ (-1.52+0.88 i,-0.48-0.90 i,1.00 +0.02 i) $$ with magnitude $2.27$:


         


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    $\begingroup$ The notation $P'$ is somewhat unfortunate. $\endgroup$ – Benoît Kloeckner Sep 3 '15 at 15:07
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    $\begingroup$ Do you really want the coefficients in some goofy sorting order. A permutation of the roots seems much more natural. $\endgroup$ – Igor Rivin Sep 3 '15 at 15:54
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    $\begingroup$ As a complete outsider, I wonder whether it would make better sense to go in the opposite direction, from roots to coefficients, giving you the map defined by taking $n$ indeterminates and evaluating them at the $n$ symmetric functions. $\endgroup$ – Lubin Sep 3 '15 at 16:46
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    $\begingroup$ Lubin's suggestion gives a continuous map, which is nice. The process outlined here can't be made continuous (e.g. to define it for $z^2 + e^{i\theta}$ you essentially have to choose branches of the square root function for the coefficients). On its own, that seems unnatural, but it makes sense as picking a branch of the inverse of Lubin's map. Igor's idea of looking at all permutations then corresponds to taking the full inverse. $\endgroup$ – Martin M. W. Sep 3 '15 at 18:26
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    $\begingroup$ Again as an outsider, I wonder whether it’s of any use or significance that, going in the direction I suggested, the square of the Jacobian determinant is the discriminant of the polynomial (maybe up to sign). $\endgroup$ – Lubin Sep 4 '15 at 13:01
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It is known (though I don't have a reference except for the statement at http://ocw.mit.edu/courses/mathematics/18-s34-problem-solving-seminar-fall-2007/assignments/roots.pdf, #2) that the only real polynomials with nonzero constant term whose roots equal their coefficients are $x^2+x-2$, $x^3+x^2-x-1$, and (approximately) $$ x^3 + .56519772x^2 - 1.76929234x + .63889690 $$ $$ x^4 + x^3 - 1.7548782x^2 - .5698401x + .3247183. $$ Update. I found a reference: P. Stein, On polynomial equations with coefficients equal to their roots, Amer. Math. Monthly 73 (1966), 272-274.

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For the unordered version, the solutions for degree $3$ are $$ \eqalign{x^3&\cr x^3& {}+x^2-2\,x\cr x^3&{}+x^2-x-1\cr x^3&{}+ \left( r^2/2-1 \right) x^2+rx-r-r^2+2 \ \text{where}\ r^3 - 2 r + 2 = 0\cr} $$ The solutions for degree $4$ are $$ \eqalign{ x^4\cr x^4&{}+x^3-x^2-x\cr x^4&{}+x^3-2\; x^2 \cr x^4&{}+ \left( \frac{r^2}{2}-1 \right) x^3+rx^2+(-r^2-r+2) x \ \text{where}\ r^3-2\,r+2 = 0\cr x^4&{}-r^2 x+rx^2+x^3+r^2-2\,rx+r-x-1 \ \text{where}\ r^3+2\,r^2+r+1 = 0\cr x^4&{}+ \left( \frac{r^{12}}{4}+{\frac {53\,r^{11}}{116}}+{\frac {19\, r^{10}}{116}}-{\frac {43\,r^9}{116}}-{\frac {59\,r^8}{116}}- {\frac {9\,{r}^{7}}{29}}+{\frac {11\,{r}^{6}}{29}}+{\frac {31\,{r}^{5} }{29}}+{\frac {35\,{r}^{4}}{58}}\right.\cr &\left.{}-\frac{{r}^{3}}{29}+{\frac {2\,{r}^{2}}{29} }+{\frac {5\,r^{13}}{116}}+\frac{r}{29}-{\frac {12}{29}} \right) {x}^{3}+r{x }^{2}\cr &{}+ \left( -{\frac {49\,r^{11}}{29}}+{\frac {123\,{r}^{10}}{58}}+ {\frac {207\,{r}^{9}}{58}}+{\frac {195\,{r}^{8}}{58}}-{\frac {27\,{r}^ {7}}{58}}-{\frac {373\,{r}^{6}}{58}}-{\frac {113\,{r}^{5}}{29}}+{ \frac {77\,{r}^{4}}{29}}\right. \cr &\left.{}+{\frac {13\,{r}^{3}}{29}}-{\frac {47\,{r}^{13 }}{58}}-\frac{5\,{r}^{12}}{2}+{\frac {3\,{r}^{2}}{29}}-{\frac {100\,r}{29}}+{ \frac {98}{29}} \right) x+2\,{r}^{12}+{\frac {45\,{r}^{11}}{58}}-{ \frac {71\,{r}^{10}}{29}}\cr &{}-{\frac {82\,{r}^{9}}{29}}-{\frac {68\,{r}^{8 }}{29}}+{\frac {63\,{r}^{7}}{58}}+{\frac {329\,{r}^{6}}{58}}+{\frac { 51\,{r}^{5}}{29}}-{\frac {112\,{r}^{4}}{29}}+{\frac {21\,{r}^{13}}{29} }-{\frac {74}{29}}-{\frac {11\,{r}^{3}}{29}}\cr&{}-{\frac {7\,{r}^{2}}{29}}+ {\frac {69\,r}{29}}\ \text{where}\cr {r}^{14}&{}+2\,{r}^{13}-{r}^{12}-4\,{r}^{11}-{r}^{10}+4\,{r}^{8}+6\,{r}^{ 7}-4\,{r}^{6}-6\,{r}^{5}+4\,{r}^{4}+4\,{r}^{2}-8\,r+4 =0 }$$ Note that you can always multiply a solution by $x^k$ for positive integer $k$, obtaining $k$ more coefficients of $0$ to match $k$ more roots of $0$.

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    $\begingroup$ Can you describe (at a high level) how you reached these solutions? Thanks! $\endgroup$ – Joseph O'Rourke Sep 3 '15 at 22:00
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    $\begingroup$ Wrote out the equations, and solved in Maple. $\endgroup$ – Robert Israel Sep 3 '15 at 23:37
  • $\begingroup$ Thanks. Incidentally, the one fixed point I found is your 4th cubic equation, with $r= 0.8846 - 0.5897 i$, one of the roots of $r^3 - 2 r + 2 = 0$. $\endgroup$ – Joseph O'Rourke Sep 3 '15 at 23:56
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    $\begingroup$ @RobertIsrael: Just a little trivia. Since $1/r = -\frac{\eta(2\tau)}{\exp(\pi\,i/24)\,\eta({\tau})}$ where $\tau = \frac{1+\sqrt{-19}}{2}$, then, $$e^{\pi\sqrt{19}} \approx r^{24}-24.0003$$ $\endgroup$ – Tito Piezas III Sep 5 '15 at 2:25
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This is a partial answer to Q1. I have found the polynomials of degree 1, 2, and 3 using Vieta's formulas.

Let $|\alpha| < |\beta| < |\gamma|$ be the roots of the polynomial.

For degree 1: We get that the last coefficient is $-\alpha$ by Vieta's formula, and it is $\alpha$ by the criterium so $\alpha=0$ and the only polynomial is $P(x)=x$, which indeeds statistifies.


For degree 2: We get $\alpha\beta=\beta$, $-\alpha-\beta=\alpha$. From the first equation we get that $\beta=0$ or $\alpha =1$. $\beta=0$ gives $\alpha=0$. $\alpha=1$ gives $\beta=-2$.

Therefore the polynomials are $P(x)=x^2$ and $P(x)=x^2+x-2.$ They indeed statistify.


For degree 3: We get $-\alpha\beta\gamma=\gamma$, $\alpha\beta+\beta\gamma+\alpha\gamma=\beta$ and $-\alpha-\beta-\gamma=\alpha$.

From the first equation we deduce that $\gamma=0$ or $\alpha\beta=-1$. If $\gamma =0$, then $\alpha\beta=\beta$ and $-\alpha-\beta=\alpha$. Therefore we have the same equation as in the first part, thus we get the polynomials are $P(x)=x^3$ and $P(x)=x^3+x^2-2x$. Only the former statistifies $|\alpha| < |\beta| < |\gamma|$.

If $\alpha\beta=-1$, then $\alpha=-\frac{1}{\beta}$. We also have $\gamma=-2\alpha-\beta=2\frac{1}{\beta}-\beta$. Therefore $-1+2-\beta^2+1-\frac{2}{\beta^2}=\beta$. This rewrites to $\beta^4+\beta^3-2\beta^2+2=0$.

Solutions for $\beta$ are (W|A verification): $\beta=-1$, $\beta=-1.7693$, $\beta=0.88465-0.58974i$, $\beta=0.88465+0.58974i$. From this we can get the values of $\beta$, and thus the values of $\alpha$ and $\gamma$.

$$P(x)=x^3+x^2-x-1$$

$$P(x)=x^3+0.565195x^2-1.7693x+0.638909$$

$$P(x)=x^3+(-0.78260-0.52171i)x^2+(0.88465-0.58974i)x+(0.68055+1.63316i)$$

$$P(x)=x^3+(-0.78260+0.52171i)x^2+(0.88465+0.58974i)x+(0.68055-1.63316i)$$

The second violates the ordering condition, but does statistify for Q3. The other three all statistify.

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  • $\begingroup$ Nice! The one fixed point I found is your 3rd polynomial. $\endgroup$ – Joseph O'Rourke Sep 4 '15 at 10:33
  • $\begingroup$ @JosephO'Rourke I somehow screwed up the computation first time. $\endgroup$ – wythagoras Sep 4 '15 at 10:37
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This is a long comment rather than a complete answer.

This is very similar to a multi-variable Julia set question, which has the same flavour as the No wandering domain theorem.

The last sentence on wikipedia is key: "However, the result can be generalized to many situations where the functions naturally belong to a finite-dimensional parameter space, most notably to transcendental entire and meromorphic functions with a finite number of singular values."

This hints that the answer is YES for most points, in some sense.

However, I also strongly suspect that there is a "small" (measure 0) "Julia-like", repelling set of parameters, so that numerics prevent you from actually finding a chaotic sequence of steps, but this set is on the boundary of the basin of attraction of your cycles.

In conclusion, my intuition says that there is an exceptional set of starting parameters, which never converges to a cycle, but it is a small set, and you will not find it by computer experiments with your method.

However, if you manage to run your map "backwards", this exceptional set should be attracting instead. That might be a way to get an idea of what parameters to pick to find such a point.

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    $\begingroup$ Q1 has nothing to do with wandering domains. It is equivalent to asking what are the fixed points of Vieta map: (roots) to (coefficients). In general it is always better to iterate a single valued map than its inverse. $\endgroup$ – Alexandre Eremenko Sep 3 '15 at 19:22
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    $\begingroup$ @AlexandreEremenko: Yes, I was thinking of Q2. But in the case of Julia fractals, the julia set is repelling, so iterating the map forward will, together with numerical instability, push you to a limit cycle of a fixed point. So, forward iteration will make you believe every starting point is well-behaved. To find points in the julia set, I suggest to iterate backwards, even though one needs to pick a random branch in each step. $\endgroup$ – Per Alexandersson Sep 3 '15 at 23:39
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There’s certainly a whole line of periodic points of order two, namely the $n$-tuples of roots $(t,0,\cdots,0)$, giving the polynomial $X^n-tX^{n-1}$. More generally if $(\mathbf r)$ is an $n$-tuple giving a periodic point of order $m$, then it seems that $((\mathbf r),0)$ is an $(n+1)$ tuple giving a periodic point of the same order for the $(n+1)$-situation.

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