The largest root of a recursively defined polynomial Suppose that for all $x \in \mathbb{R}$, $f_1(x)=x^2$ and for all $k \in \mathbb{N}$,
$$
  f_{k+1}(x) = f_k(x) - f_k'(x) x (1-x).
$$
Let $\underline{x}_k$ denote the largest root of $f_k(x)=0$.
I want to prove the following conjectures:


*

*$0=\underline{x}_1 < \underline{x}_2 < \underline{x}_3 < \dots < 1$.

*$\lim_{n \to \infty} \underline{x}_n = 1$.

*$f_k'(\underline{x}_k) \geq 0$ for all $k$.


These conjectures are the missing part of a larger proof and it seems to hold for any $f_k$ that I can compute by hand or numerically. The first few polynomials:


*

*$f_1(x)=x^2$, so that $\underline{x}_1=0$ and $f_1'(\underline{x}_1) = 0$

*$f_2(x)=x^2 (2x-1)$, so that $\underline{x}_2=\frac{1}{2} \in (\underline{x}_1,1)$ and $f_2'(1/2)=1/2>0$ 

*$f_3(x)=x^2 (6x^2-6x+1)$, so that $\underline{x}_3 = \frac{1}{2} + \frac{1}{2 \sqrt{3}} \approx 0.7887 \in (\underline{x}_2,1)$ and $f_3'(\underline{x}_3) \approx 2.1547>0$

*$f_4(x)=x^2 (2x-1) (12 x^2-12 x+1)$, so that $\underline{x}_4=\frac{1}{2}+\frac{1}{\sqrt{6}} \approx 0.9082 \in (\underline{x}_3,1)$ and $f_4'(\underline{x}_4) \approx 6.5993 > 0$


The polynomials $f_k$ have many properties that should help. For example it is easy to show that $f_k(1)=1$ for all $k$ and $f_k(x) > 1$ for all $x>1$ for all $k$. Therefore all real roots must be strictly below $1$.
 A: We can start by proving that for all $k\ge1$
$$
f_k(x)=k!\cdot(x-x_0)\cdots(x-x_k)
$$
where
$$
0=x_0=x_1<x_2<\cdots<x_k<1.
$$
Let's assume this holds for $f_k$ and prove it for $f_{k+1}$.
First, we may note that the derivative
$$
f_k'(x)=(k+1)!\cdot(x-y_1)\cdots(x-y_k)
$$
where
$$
0=x_0=y_1=x_1<y_2<\cdots<y_k<x_k<1.
$$
The $y_1=0$ follows from the double root at zero; the $x_{j-1}<y_j<x_j$ for $j>1$ follows from $f_k(x_{j-1})=f_k(x_j)=0$, alternatively from observing that $f_k'(x_{j-1})$ and $f_k'(x_j)$ have different signs.
The rest simply boils down to observing that the sign of $x(1-x)f_k'(x)$ at the points $x_j$ alternates between positive and negative. Since $f_k(x)$ is zero at all $x_j$, $f_{k+1}(x_j)=x_j(x_j-1)f_k'(x_j)$ also has alternating sign along the points $x_2,\ldots,x_k,1$: ie, $f_{k+1}(1)>0$, $f_{k+1}(x_k)<0$, etc. Thus, $f_{k+1}(x)$ must have zeroes between the points $0=x_1,x_2,\ldots,x_k,1$: say at $z_j$ where
$$
0=z_0=x_0=z_1=x_1<z_2<x_2<\cdots<x_k<z_{k-1}<1
$$
where I have included the double root at zero which is easy to check. There can be no more roots than this due to the degree of $f_{k+1}$.
There's a little extra checking for $y_2$ and $z_2$ because of the double root at zero which makes the sign argument fail there. Alternatively, you could redo the entire argument for $0=x_0<x_1<\cdots$ and just take the limit as $x_1\rightarrow0$.
This proves points (1) and (3). What remains is point (2): that the top root converges to 1.
You have already observed that $f_k(1)=1$ for all $k$. This makes
$$
(1-x_2)\cdots(1-x_k)=\frac{f_k(1)}{k!}=\frac{1}{k!}
$$
where I ignore the two zero-roots. Since $0<x_j<x_k$ for $j=2$ to $k-1$, this makes $(1-x_k)^{k-1}<1/k!$, which makes $1-x_k<(k!)^{-1/(k-1)}$ which drops towards zero as $k$ increases. 

Previous proof of (2) in case of interest:
Recall that, in addition to the two roots at zero, $f_{k+1}$ had roots $z_2,\ldots,z_k,z_{k+1}$ where $z_j<x_j$ for $j=2$ to $k$, so this makes
$$
\frac{1}{(k+1)!}=(1-z_2)\cdots(1-z_{k+1})
>(1-x_2)\cdots(1-x_k)(1-z_k)=\frac{1-z_{k+1}}{k!}
$$
which makes $1-z_{k+1}<1/(k+1)$. So in general, we get that the top root satisfies $\underline{x}_n>1-1/n$.

Origin of proof
In case someone wonder which hat I pulled this rabbit out of. I was already familiar with this method for proving that polynomials defined recursively like this have distinct roots (in a case where we didn't have the double root at zero). I may have made a slip-up from thinking, throughout writing the main part of the proof, that was what point (1) was.
Thinking about it, I suppose when the only interest is the top root, the argument could be simplified to address only this, not all the lower roots as well.
A: From
$f_{k+1}(x) = f_k(x) - f_k'(x) x (1-x)
$,
let
$f_k(x)
=x^2 g_k(x)
$,
so
$f_k'(x)
=x^2 g_k'(x)+2xg_k(x)
=x(x g_k'(x)+2g_k(x))
$.
Then
$\begin{array}\\
x^2g_{k+1}(x) 
&= x^2g_k(x) -  x (1-x)x(x g_k'(x)+2g_k(x))\\
&= x^2(g_k(x) -   (1-x)(x g_k'(x)+2g_k(x)))\\
\text{so}\\
g_{k+1}(x) 
&= g_k(x) -   (1-x)(x g_k'(x)+2g_k(x))\\
&= g_k(x) -   (1-x)x g_k'(x)-2(1-x)g_k(x)\\
&= (1-2(1-x))g_k(x) -   (1-x)x g_k'(x)\\
&= (2x-1)g_k(x) -   (1-x)x g_k'(x)\\
&= (2x-1)g_k(x) + x(x-1) g_k'(x)\\
\end{array}
$
As a check,
since
$g_1(x) = 1$,
$g_2(x)
=2x-1
$
and
$\begin{array}\\
g_3(x)
&=(2x-1)(2x-1)+x(x-1)(2)\\
&=4x^2-4x+1+2x(x-1)\\
&=6x^2-6x+1\\
\end{array}
$
Note that
$g_{k+1}(x)
= (2x-1)g_k(x) + x(x-1) g_k'(x)
=((x^2-x)g_k(x))'
$
so that
$\begin{array}\\
g_{k+2}(x)
&=((x^2-x)g_{k+1}(x))'\\
&=((x^2-x)((2x-1)g_k(x) + x(x-1) g_k'(x))'\\
&=((x^2-x)(2x-1)g_k(x) + x^2(x-1)^2 g_k'(x))'\\
&=((2x^3-3x^2-x)g_k(x) + (x^4-2x^3+x^2) g_k'(x))'\\
&= (6 x^2-6 x-1) g_k(x)+x (x (x-1)^2 g_k''(x)+(6 x^2-9 x+1) g_k'(x))\\
\end{array}
$
Let
$G(x, y)
=\sum_{k=1}^{\infty} y^kg_k(x)
$.
Then
$G_x(x, y)
=\sum_{k=1}^{\infty} y^kg_k'(x)
$
and
$\begin{array}\\
\dfrac1{y}G(x, y)
&=\sum_{k=1}^{\infty} y^{k-1}g_k(x)\\
&=\sum_{k=0}^{\infty} y^{k}g_{k+1}(x)\\
&=g_1(x)+\sum_{k=1}^{\infty} y^{k}g_{k+1}(x)\\
&=1+\sum_{k=1}^{\infty} y^{k}((2x-1)g_k(x) + x(x-1) g_k'(x))\\
&=1+\sum_{k=1}^{\infty} y^{k}(2x-1)g_k(x) + \sum_{k=1}^{\infty}x(x-1)y^k g_k'(x)\\
&=1+(2x-1)\sum_{k=1}^{\infty} y^{k}g_k(x) + x(x-1)\sum_{k=1}^{\infty}y^k g_k'(x)\\
&=1+(2x-1)\sum_{k=1}^{\infty} y^{k}g_k(x) + x(x-1)\sum_{k=1}^{\infty}y^k g_k'(x)\\
&=1+(2x-1)G(x, y) + x(x-1)G_x(x, y)\\
\text{or}\\
x(x-1)G_x(x, y)
&=G(x, y)(\dfrac1{y}-(2x-1))-1
\end{array}
$
Not sure where to go from here,
so I'll leave it at this.
