# Real solution of $P(P(\cdots P(x))\cdots)=0$

Let $$P(x)=x^2 +2013x+1$$. Show that $$P(P(\cdots P(x))\cdots)=0$$ (i.e. $$P$$ is $$n$$-times nested) has at least one real root for any $$n$$ $$P$$.

For $$n = 1$$ this is obvious. Next, for $$n = 2$$ we get a fourth order polynomial $$x^4 + 4,026 x^3 + 4,054,184 x + 2,015$$ and by substituiting $$y = x + 2,013/2$$ can elimate the cubic term. Rearranging and standardizing yields a 4th order equation $$y^4 – 4,048,139/2 y^2 + 16,387,413,154,661/16 = 0$$ It can be solved but further substitutions quickly become intractable and don´t lead anywhere. Can anyone help? Thanks.

• If we solve by iteration it would be: $x_{n}^{2} + 2013x_{n} + 1 - x_{n - 1} = 0$ – R zu Oct 27 '18 at 20:14
• The quadratic has real solution if $2013^{2} - 4(1 - x_{n - 1}) \ge 0$ – R zu Oct 27 '18 at 20:16
• The roots of $P(x)=0$ are negative. – hamam_Abdallah Oct 27 '18 at 20:17

Let $$P_{1}(x) = P(x)$$

Let $$P_{n}(x) = P(P_{n-1}(x))$$

Let $$x_{n}$$ be one of the solution for $$P_{n}(x) = 0$$

Initiation

Both roots of $$P(x)$$ are real and negative.

We choose the less negative solution $$x_{1}$$.

$$x_{1} = \frac{-2013 + \sqrt{2013^{2} - 4(1 - 0)}}{2}$$.

Notice that $$0 \ge x_{1} \ge -2013/2$$.

Induction

From the last step of the induction, we have $$0 \ge x_{n-1} \ge -2013/2$$.

One of the solution of $$P_{n}(x) = 0$$ is the solution of:

$$P(x) = x_{n-1}$$

Again, we can choose a real and negative solution $$x_{n}$$ such that $$0 \ge x_{n} \ge -2013/2$$:

$$x_{n} = \frac{-2013 + \sqrt{2013^{2} - 4(1 - x_{n-1})}}{2}$$

• Both roots of $P(x)$ are real and negative. Are you picking a specific one of them, or doesn't it matter? – Arthur Oct 27 '18 at 20:43
• Can you explain the notation $x_n$ and $x_{n - 1}?$ – 伽罗瓦 Oct 27 '18 at 20:45
• @R zu: Thank you very much. It seems a clever approach to use iteration. But could you please explain your thoughts a bit more for a novice like me, in particular your conclusion? Much appreciated. – Parzifal Oct 27 '18 at 20:55
• should be correct and clear now. – R zu Oct 27 '18 at 21:16
• @Arthur: Choosing a more negative solution in all steps also works. I haven't thought about whether all possible choices in all step would work. – R zu Oct 27 '18 at 21:31

R zu's is a nice algebraic answer. Here is a more geometric way of seeing it. Let $$P(x) = x^2 + bx + 1$$ where $$b$$ is any real number. $$P(x)$$ tends to $$\infty$$ as $$x$$ tends to $$+\infty$$ or $$-\infty$$ and achieves its minimum where $$P'(x) = 2x + b = 0$$, i.e., at $$x = -b/2$$. The value of that minimum is $$P(-b/2) = 1 - b^2/4$$. So $$P$$ maps the interval $$I = [-b/2, \infty)$$ onto the interval $$J = [1-b^2/4, \infty)$$ (the range of the function $$P$$). Now assume $$b$$ is large enough so that $$1 - b^2/4 < -b/2 < 0$$ as is certainly the case in your example where $$b = 2013$$. Then $$I \subseteq J$$, so as $$P$$ maps $$I$$ onto $$J$$ and $$J$$ is the range of the function $$P$$, $$P$$ also maps $$J$$ onto $$J$$. Hence by induction $$P_n$$ (defined by $$P_1(x) = P(x)$$ and $$P_n(x) = P(P_{n-1}(x))$$ as in R zu's answer) maps $$J$$ onto $$J$$ and as $$0 \in J$$, this means $$P_n$$ has a root.

Just for fun, here's a plot of $$P_1$$, $$P_2$$ and $$P_3$$ for $$b = 4$$, illustrating that the the range of each of these functions is $$J = [-3, \infty)$$.

• Is this related to something about dynamical systems? What branches of mathematics study this? – R zu Oct 28 '18 at 13:58
• I don't know much about dynamical systems and ergodic theory, I'm afraid. I was just thinking about the shape of the graph and what subintervals of the domain map onto the range. Maybe someone else can comment on branches of mathematics that study generalisation of this kind of thing. – Rob Arthan Oct 28 '18 at 14:11
• This is a very interesting depiction – Parzifal Oct 28 '18 at 17:12
• @Parzifal: indeed! One interesting question it raises is "are the local maxima of $P_n(x)$ for $n > 1$ all equal to $-2$?". – Rob Arthan Oct 28 '18 at 18:58