How can we show that there is a zero of $f$ Let $$f(x)= x^2+2020x+1$$ a polynomial.
We define $F_n(x)$ to be $f(f(f...f(x)))$ applied $n$ times .
Show that for all positive integer $n\ge1$ there exists a real $x$ such that $F_n(x)=0$
Please help me, I've tried induction but it doesn't seem to work. $n=1$ is trivial but then it's hard. I don't know much about polynomials. Any ideas?
 A: First, note that for each $n$, $F_n(x)$ is a polynomial of degree $2n$ (with leading coefficient $1$). This is easy to show by induction as
$$F_1(x)=f(x)=x^2+2020x+1$$
Then if $r(x)$ is the remaining polynomial such that $F_{n-1}(x)=x^{2(n-1)}+r(x)$ (note then that the degree of $r(x)$ is less than $2(n-1)$) we have
$$F_n(x)=f(F_{n-1}(x))=(x^{2(n-1)}+r(x))^2+2020 (x^{2(n-1)}+r(x))+1$$
$$=x^{2n}+2x^{2(n-1)}r(x)+r(x)^2+2020x^{2(n-1)}+2020r(x)+1$$
Since this clearly has degree $2n$ the proposition is proved. This implies that
$$\lim_{x\to\infty}F_n(x)=\infty$$
or that there exists $X\in\mathbb{R}$ such that $x\geq X$ implies $F_n(x)\geq 1$. Of course, this implies that $F_n(X)=1$. Now, let $x_0$ be the lesser fixed point of $f(x)$. That is
$$f(x_0)=x_0$$
$$x_0^2+2020x_0+1=x_0$$
$$\Rightarrow x_0=\frac{-2019\pm\sqrt{4076357}}{2}$$
Since we want the lesser fixed point, we may conclude that
$$x_0=\frac{-2019-\sqrt{4076357}}{2}$$
But for this $x_0$ we have
$$F_1(x_0)=f(x_0)=x_0$$
$$F_2(x_0)=f(F_1(x_0))=f(x_0)=x_0$$
$$\vdots$$
$$F_n(x_0)=x_0=\frac{-2019-\sqrt{4076357}}{2}<0$$
Thus, $F_n(x_0)<0$ and $F_n(X)=1>0$. By the mean value theorem, there exists some $x\in (x_0,X)$ such that $F_n(x)=0$.
A: *

*Consider the solutions to $ f(x) = x$. These are $ x_1, x_2 = \frac{ - 2019 \pm \sqrt{ 2019^2 - 4 } } { 2 } < 0  $.

*We have $ f(x_i) = x_i$, so $ F_n(x_i) = x_i$.

*In particular, $ F_n(x_i) < 0$.

*Observe that for $x > 0$ $ f(x) > 0$ , so $ F_n(x) > 0 $ for $ x > 0 $.

*Hence by IVT, there exists a $x'$ such that $F_n(x') = 0$.

Note: $ \frac{ - 2019 + \sqrt{ 2019^2 - 4 } } { 2 } \approx -0.0005$, and so we can conclude that there is a root in $(-0.0005, 0)$.
