# Increasing derivatives of recursively defined polynomials

Consider recursively defined polynomials $f_0(x)=x$ and $f_{n+1}(x)=f_n(x)−f_n'(x) x (1−x)$.

These polynomials have some special properties, for example $f_n(0)=0$, $f_n(1)=1$, and all $n+1$ roots of $f_n$ are in $[0,1)$. Let $x_n$ denote the largest root of $f_n$. Then $f_n(x_n)=0$, and $f_n'(x_n)>0$. Moreover, $0=x_1 < x_2 < \dots < x_n < 1$.

I want to prove the following conjecture: $f_n'(x)>f_{n-1}'(x)$ for all $x \in [x_{n+1},1)$.

Some of the first polynomials are:

1. $f_1(x)=x^2$, $f_1'(x)=2x$, $x_1=0$
2. $f_2(x)=2 x^3−x^2$, $f_2'(x)=6 x^2−2$, $x_2=\frac{1}{2}$
3. $f_3(x)=6x^4−6x^3+x^2$, $f_3'(x)=24 x^3−18x^2+2x$, $x_3 \approx 0.7887$
4. $f_4(x)=24x^5−36x^4+14x^3−x^2$, $f_4'(x)=120x^4−144x^3+42x^2−2x$, $x_4 \approx 0.9082$
5. $f_5(x)=120 x^6-240 x^5+ 150 x^4-30 x^3 +x^2$, $f_5'(x) =720x^5-1200 x^4+600 x^3-90 x^2+ 2 x$, $x_5 = 0.9587$

Therefore

• $f_2'(x)−f_1'(x)=6x^2−4x \geq 0$ for all $x \geq \frac{2}{3}$. Note that, $x_2<\frac{2}{3} < x3$.
• $f_3'(x)−f_2'(x)=24 x^3−24x^2+4x \geq 0$ for all $x \geq 0.7887 = x_3 < x_4$.
• $f_4'(x)-f_3'(x)=120 x^4-168x^3+60 x^2-4x \geq 0$ for all $x \geq 0.8685 < x_5$
• Note oeis.org/A019538. – Tom Copeland Aug 11 '16 at 21:06
• Just some potential leads. See also oeis.org/A133314 which is a refined signed array. You should be able to at least find an e.g.f. for your sequence, which might possibly be of help to you or others in finding an answer. – Tom Copeland Aug 11 '16 at 21:22
• A potential e.g.f. is $\frac{x}{1-x (1-e^{-t})}$. – Tom Copeland Aug 11 '16 at 21:37
• So the e.g.f. for the derivatives would be $\frac{1}{[1-x (1-e^{-t})]^2}$. – Tom Copeland Aug 11 '16 at 21:51
• Wilf's generatingfunctionology – J.J. Green Aug 11 '16 at 23:44