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

Question

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.

Answer 1

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

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

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

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

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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 $0 \ge x_{n} \ge -2013/2$:

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

Answer 2

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 $J = [-3, \infty)$.