2

Question.

Let $(z_n)$ be a complex sequence such that $$ z_0 = 0,\quad z_1=1, \quad z_{n}=\frac13(z_{n-1}+z_{n-2}+z_{n-1}z_{n-2})\quad (n>1)\tag{1} $$ Prove that the limit of the sequence exists.

Observations.

  1. This is a complex sequence; one cannot expect to apply the monotone convergence theorem directly.

    (Added notes: Thanks to paw88789's comment below, this is actually a real sequence. The $z_n$'s are originally coefficients of a complex power series (in the linked question mentioned later) which could be complex. But I overlooked the assumption that it is immediate from the definition that they are actually all real. One could generalize this problem by replacing $z_1$ with some non-real complex number though. But that's not the current focus. Since this sequence is actually real and bounded from below, one can directly apply the monotone convergence theorem if one can show that the sequence is "eventually" decreasing.)

  2. One straightforward idea is to show that the sequence is Cauchy. However, it seems difficult to directly estimate $|z_m-z_n|$ using (1).

  3. The recurrence in (1) is very close to linear, where one can use the characteristic function techniques. But it is not very clear how to pass from "linear" to "nonlinear".

Remarks.

This is a follow-up question to another one of mine. The question above is an essential step in one approach to establish the calculation in the linked question. I find this isolated step itself is interesting. Since I figured out a detailed proof enlighted by a sketchy comment under the linked post, I will post one proof as an answer below.

I would also like to see other approaches to the question.

  • 1
    Calling this a complex sequence seems a bit of a stretch, as all the terms will be pure real (imaginary part $0$); or am I missing something? – paw88789 Dec 05 '21 at 15:09
  • @paw88789: Aha, indeed! The $z_n$'s are original coefficients of a complex power series (in the linked question) which could be complex. But I overlooked the assumption that it is immediate from the recursive definition that they are actually all real. // Now one can generalize this problem by replacing $z_1$ with some non-real complex number though. –  Dec 05 '21 at 15:40

2 Answers2

3

Updated. Thanks to Philipp's comment below, I realize that the "base step" is missing in the original argument; I find a way to fix it as below. The parameter $k$ is the original argument is also not needed.

Proof. It suffices to show that

for sufficiently large $n$, $0\le a_n\le q^n$ for some $0<q<1$ where $a_n:=|z_n|$.

By considering the function $f(x)=3x^2-x-1$ one can find that for all $n\ge 7$, $$ q^n\le 3q^2-q-1,\quad q:=0.9\tag{3} $$ which is equivalent to $$ \frac13(q^n+q^{n-1}+q^{2n-1})\le q^{n+1}\tag{4} $$ Let $P(m)$ be the statement that $a_m\le q^m$. We now show by induction that $P(m)$ is true for all $m\ge 7$.

Base step. $P(7)$ is true by direct calculation by hand. (One can also write a program with a short loop to do that.)

Induction step. Suppose $P(m)$ is true for all $m$ with $7\le m\le n$ ($n>7$), we show that $P(n+1)$ is true. This follows from the induction hypothesis and (4): $$ a_{n+1}\le \frac13(|z_n|+|z_{n-1}|+|z_n|\cdot |z_{n-1}|)\le \frac{1}{3}(q^n+q^{n-1}+q^{2n-1})\le q^{n+1}~. $$

Q.E.D.

[Original proof where the base step is missing.]

It suffices to show that

for sufficiently large $n$, $0\le a_n\le kq^n$ for some $k>0$ and $0<q<1$ where $a_n:=|z_n|$.

A straightforward induction shows that $0\le a_n< 1$ for $n\ge 2$. Indeed, in the induction step, by the triangle inequality: $$ 0\le a_{n+1}=|z_{n+1}|< \frac13(|z_n|+|z_{n-1}|+|z_nz_{n-1}|)<\frac13(1+1+1)=1\;. $$

Now let us find a possible pair of $k$ and $q$ that works. Assume that $a_m\le kq^m$ for all $m\le n$. One must have $$ a_{n+1}\le \frac13(|z_n|+|z_{n-1}|+|z_nz_{n-1}|)\le \frac{k}{3}(q^n+q^{n-1}+kq^{2n-1})~. $$ So if $$ \frac{k}{3}(q^n+q^{n-1}+kq^{2n-1})\le kq^{n+1}\tag{1} $$ then by induction $a_n\le kq^n$ for all $n$. Note that (1) is equivalent to

$$ kq^n\le 3q^2-q-1\tag{2} $$ Consider the function $f(x)=3x^2-x-1$. Note that $f(0)=-1<0$ and $f(1)=1>0$. Hence, by the intermidiate value theorem, there exists $r\in(0,1)$ such that $f(r)=\frac12$. Let $q=r, k=1$ in (2). Then (2) is true for all sufficiently large $n$.

Consequently, we have $a_n\le r^n$ for all sufficiently large $n$.
  • The order of your arguments is a bit confusing and makes it hard to follow your idea. Let me add a few remarks. First, your induction hypothesis: "$x_m\leq kq^m$ for all $m\leq n$" is slightly flawed. This becomes clear if you write down the proof a bit more thoroughly. Second, we don't need $k$, instead we look for a $n_0$ and a $q$ with $0<q<1$ such that $$ \frac{q^n+q^{n-1}+q^{2n-1}}{3}\leq q^{n+1}\iff q^n\le 3q^2-q-1~~~~~(1) $$ holds for all $n>n_0$. The existence of such $n_0$ and $q$ is clear. – Philipp Dec 18 '21 at 18:14
  • Then, we try to find the base case where we would like to start the induction, i.e. some index $i$, usually $i=0$ or $i=1$. As we have already assumed $n>n_0$ our base case must have an index $i$ such that $i>n_0$. To be rigorous, one must show that $x_i\leq q^i$ for that particular $i$. Then, we assume the induction hypothesis "$x_n\leq q^n$" for a specific $n$ with $n>n_0+1$, combine it with $(1)$ and conclude $$ x_{n+1}=\frac{x_n+x_{n-1}+x_nx_{n-1}}{3}<\frac{q^n+q^{n-1}+q^{2n-1}}{3}\leq q^{n+1}. $$ Hence, $x_n\leq q^n$ for all $n\geq i$. – Philipp Dec 18 '21 at 18:15
  • @Philipp: Oops, the base step is indeed missing in the original argument. Thank you for the comments. I edited the proof accordingly. –  Dec 27 '21 at 16:59
  • There is still a little mistake. The induction hypothesis must be $a_m\le q^m$ and $a_{m-1}\le q^{m-1}$ because you use the indices $n$ and $n-1$ in your induction step. So you have to add the case $n+2$ to your induction step. Then, the induction hypothesis is shown for $n+1$ and $n+2$ which shows monotonicity for all $n>7$. – Philipp Dec 28 '21 at 14:57
  • @Philipp: I am using the strong version of induction. The induction hypothesis is (for any fixed $n>7$) "Suppose () is true for all $m$ with $7\le m\le n$". In particularly, $P(n)$ and $P(n-1)$ are both true. –  Dec 28 '21 at 17:18
  • 1
    Yes you are right, please ignore my last comment, my bad. – Philipp Dec 29 '21 at 13:59
0

Although ripples answer is pretty clever, I would like add another one where we use the monotone convergence theorem of real sequences.

The boundedness is clear. Then we do a short proof by induction to show monotonicity.

The base case/the induction starts at the index $6$ as $z_6-z_5<0$ and $z_6-z_4<0$.

Then, we assume the induction hypothesis "$z_{n}<z_{n-1}$ and $z_n<z_{n-2}$ with $n>5$".

The induction step $n\to n+1$ can be shown as follows:

$$ z_{n+1}-z_n=\frac{z_{n}+z_{n-1}+z_{n}z_{n-1}}{3}-\frac{z_{n-1}+z_{n-2}+z_{n-1}z_{n-2}}{3}\\=\frac{z_n-z_{n-2}+z_{n-1}(z_n-z_{n-2})}{3}=\frac{(1+z_{n-1})(z_n-z_{n-2})}{3}\underset{\text{hypothesis}}{\underset{\text{induction}}{<}}0 $$ and $$ z_{n+1}-z_{n-1}<z_{n+1}-z_n\underset{\text{above}}{\underset{\text{see}}{<}}0. $$ Hence, the real valued sequence $(z_n)$ is monotone from the $5$-th member onward. Due to monotone convergence theorem the sequence is convergent.

Philipp
  • 4,483
  • 2
  • 10
  • 22