Proving convergence of solution of a fixed point equation

Question

I encountered a nasty sequence $(x_n)_{n=1}^\infty $ defined as the smallest positive fixed point of the fixed point equation $ x_n = f_n(x_n) $, where $f_n$ is given by

$$ f_n(x) = \sum_{k=0}^{\lfloor \tfrac{n-1}{2}\rfloor} {n-1 \choose 2k} {2k \choose k} \alpha^k x^{2k} (1-x)^{n-1-2k}, $$

where $\alpha \in (0,1/4)$. From plotting a few values, I think this sequence converges to $0$ as $n \rightarrow \infty$ in a monotonically decreasing fashion. However, I'm unable to prove it. Does someone have an idea how it could be done? (or thinks that my hunch is incorrect?) Thanks so much!

Answer 1

Convergence to $0$ is trivial: as you noticed yourself, the coefficient is just ${n-1\choose k,k,n-1-2k}$. Splitting the rest as $(\sqrt\alpha x)^k(\sqrt\alpha x)^k(1-x)^{n-1-2k}$, we can bound $f_n(x)$ by the full trinomial sum $$ f_n(x)\le\sum_{k+l+m=n-1}{n-1\choose k,l,m}(\sqrt\alpha x)^k(\sqrt\alpha x)^l(1-x)^m=(1-x+2\sqrt\alpha x)^{n-1}\,. $$ Since your $\alpha<\frac 14$, the RHS is $(1-\beta x)^{n-1}$ with $\beta=1-2\sqrt\alpha>0$, so $f_n(x)$ tends to $0$ uniformly as $n\to\infty$ on every interval $[\delta,1]$ with $\delta>0$.

That also provides a simple but effective bound $x_n=O(\frac{\log n}{n})$ and makes the monotonicity claim quite plausible, but by no means proves it. Do you care about that part, or not really?

Answer 2

$x_n$ eventually decreases.

Denote $\beta=\sqrt\alpha$, then $0<\beta<1/2$. Then $f_n(0)=1>0$ and therefore $x_n$ is the maximal positive number for which $$f_n(x)=[t^0](\beta(t+t^{-1})x+1-x)^{n-1}\geqslant x, \, \forall x\in [0,x_n].\tag{1}$$ Here $[t^a]F(t)$ denotes a coefficient of $t^a$ in the Laurent polynomial $F$.

So, in order to prove that $x_n\leqslant x_{n-1}$, it suffices to prove that (1) holds with exponent $n-2$ instead $n-1$, that in turns follows from the inequality $$[t^0](\beta(t+t^{-1})x+1-x)^{n-1}\leqslant [t^0](\beta(t+t^{-1})x+1-x)^{n-2}\tag{2} $$ for all $x\in [0,x_n]$. (2) may be proved by applying the formula $$[t^0]A(t)=\frac1{2\pi}\int_{0}^{2\pi}A(e^{is})ds$$ for any Laurent polynomial $A(t)$ and in our situation $\beta(e^{is}+e^{-is})x+1-x=1-x(1-2\beta \cos s)\leqslant 1$ and also $1-x(1-2\beta \cos s)\geqslant 0$ provided that $x_n\leqslant 1/(1+2\beta)$, that holds for large enough $n$, as shown by fedja.