Let $a_0,a_1,a_2 \dots a_n, \dots$ be an infinite sequence of real numbers, defined by $$a_0 = c$$$$a_{n+1} = {a_n}^2+\frac{a_n}{2}+c$$for some real $c > 0$. Find all values of $c$ for which the sequence converges and the limit for those values.
Problem
Source: Bulgarian Autumn Math Tournament 12.1
Tags: algebra, Sequence, limit
pco
30.11.2024 13:27
topologicalsort wrote: Let $a_0,a_1,a_2 \dots a_n, \dots$ be an infinite sequence of real numbers, defined by $$a_0 = c$$$$a_{n+1} = {a_n}^2+\frac{a_n}{2}+c$$for some real $c > 0$. Find all values of $c$ for which the sequence converges and the limit for those values. Let $f(x)=x^2+\frac x2+c$ so that $a_{n+1}=f(a_n)$ 1) Necessary condition : If $a_n$ converges, we have $c\le\frac 1{16}$
If $a_n$ has a finite limit $u$, we must have $u=f(u)$ and since discriminant of tesulting quadratic is $\frac 14-4c$, we need indeed $c\le\frac 1{16}$
Q.E.D.
2) Sufficient condition : If $c\le\frac 1{16}$, then $a_n$ converges and its limit is $\frac 14-\sqrt{\frac 1{16}-c}$
Let $u_1=\frac 14-\sqrt{\frac 1{16}-c}$ and $u_2=\frac 14+\sqrt{\frac 1{16}-c}$
We easily have $0<c<u_1<u_2$
We also have $\forall x\in[c,u_1]$ $u_1\ge f(x)\ge x$
So $_n$ is a nondecreasing sequence upperbounded and so has a limit $l\le u_1$
And since the only possible limits are $u_1$ and $u_2$, the limit is $u_1$
Q.E.D
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