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I want to study the sequence defined by $u_0=1/2$ and the recurrence relation $$u_{n+1}=1-u_n^2\qquad n\ge0.$$ I calculated sufficient terms to understand that this sequence does not converge because its odd and even subsequences converge to different limits.

In particular we have $$\lim_{n\to\infty}a_{2n+1}=0\text{ and}\lim_{n\to\infty}a_{2n+2}=1.$$

Now I can also prove that $u_n\leq1$ because the subsequent terms are related in this way $$f\colon x\mapsto1-x^2.$$

In order to prove that the subsequences converge I need to prove either that they are bounded (in order to use Bolzano-Weierstrass) or that they are bounded and monotonic for $n>N$. Proved that the subsequences converge to two different limits, I will be able to prove that our $u_n$ diverges.

Can you help me? Have you any other idea to study this sequence?

Thank you.

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The function $f$ decreases from $f(0)=1$ to $f(1)=0$ and meets the first diagonal at $a=\frac12(\sqrt5-1)$ hence $f$ exchanges the intervals $[0,a]$ and $[a,1]$ and $f\circ f$ is increasing and leaves each of them invariant. Furthermore, $f\circ f(x)\lt x$ for every $x$ in $(0,a)$ and $f\circ f(x)\gt x$ for every $x$ in $(a,1)$.

Thus, every sequence defined by $x_0$ in $[0,1]$ and $x_{n+1}=f\circ f(x_n)$ for every $n\geqslant0$, is either constant if $x_0=0$ or $a$ or $1$, or decreasing to $0$ if $x_0$ is in $(0,a)$, or increasing to $1$ if $x_0$ is in $(a,1)$.

In your case, $u_0=\frac12$ is in $(0,a)$ hence $u_1$ is in $(a,1)$. Thus, applying the above to $x_0=u_0$ shows that $(u_{2n})$ decreases to $0$, and, applying the above to $x_0=u_1$ shows that $(u_{2n+1})$ increases to $1$.

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To look at the odd and even subsequences, it suffices to "unfold" the recurrence one step: $$u_{n+2} = 1-u_{n+1}^2 = 1 - (1 - u_n^2)^2 = 2u_n^2 - u_n^4.$$ Then consider the fixed points of this recurrence for $u \in [0,1]$ by solving $u = 2u^2 - u^4$. Finally, look at the inequalities for those intervals bounded by those fixed points. This gives you an idea of where the subsequence will decrease or increase, and from there you can try to formally show the limiting behavior of the subsequences.

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The function $f:[0,1]\longrightarrow[0,1]$ has a unique fixed point but is repellent. And we can't reach it in a finite number of steps because is irrational.

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