I have a sequence defined as $u_{n+1}=f(u_n)=(u_n^2+3)/4$, and with $u_0\geq0$. I am asked to determine what are the possible limits of $u_n$ by studying $f(x)$ and the sign of $f(x)-x$. $$ \begin{array}{c|ccccccccc} x & 0 & & 1 & & 2& & 3 & & \infty\\\hline f'(x) & & & & & +& & & &\\\hline f(x) &3/4 & & & &\nearrow& & & & \infty\\\hline s\left(f(x)-x\right)&& + & | & & - & & | & + & \end{array} $$ Suppose $0\leq u_0\leq1$, then $\frac{3}{4}\leq \frac{u_0^2+3}{4}=u_1\leq1$, which means that $u_n$ will always stay in $[0, 1]$.
The only thing I am sure of is that $\lim_{n\to\infty}u_n\leq1$, but I don't see how I can determine a possible limit other than $1$ here.

Suppose $1\leq u_0\leq 3$, then $1\leq u_1\leq3$, which means that $u_n$ will always stay in $[1, 3]$.
Since $f(x)-x\leq0$, and because of the sentence above, I am tempted to say that $u_n$ will go backwards until it reaches $1$, but I also don't see how I can put this mathematically, like the first case.

Suppose $3\leq u_0\leq k$, then $3\leq u_1\leq\frac{k^2+3}{4}$. Using the sign of $f(x)-x$, I know that $k\leq\frac{k^2+3}{4}$, and $u_0\leq u_1$. The upper bound keeps getting bigger, and the difference in two consecutive terms of the sequence also does, so there is no limit.

I would gladly accept any help for the first two parts, and a verification for the last one. Thank you very much!


1 Answer 1


Consider first $0 \le u_0 \le 1$. If $u_0 = 1$, you'll have $u_i = 1$ for all $i \ge 0$. Otherwise, as you showed, the upper bound for the values is $1$. Also, note that you have, with $x = u_n \lt 1$, that

$$\begin{equation}\begin{aligned} f(x) - x & = \frac{u_n^2 + 3}{4} - u_n \\ & = \frac{u_n^2 + 3 - 4u_n}{4} \\ & = \frac{(u_n - 1)(u_n - 3)}{4} \\ & \gt 0 \end{aligned}\end{equation}\tag{1}\label{eq1A}$$

Thus, you have a strictly increasing sequence that has an upper bound. To determine it's limit point $L$, note that the differences between $u_{n}$ and $u_{n+1}$, as well as with $L$, become arbitrarily small. As such, you can determine what $L$ is by replacing $u_{n}$ and $u_{n+1}$ with $L$ and then solving, i.e.,

$$L = \frac{L^2 + 3}{4} \implies L^2 - 4L + 3 = (L - 3)(L - 1) = 0 \tag{2}\label{eq2A}$$

Thus, the limit would be $L = 1$.

You can do likewise for the case of $1 \le u_0 \le 3$, where if $k = 3$, then $u_i = 3$ for all $i \ge 0$, so the limit point would be $3$. Otherwise, for $u_0 \lt 3$, then \eqref{eq1A} shows in this case that $f(x) - x \lt 0$, so you have a strictly decreasing sequence which, once again, has a lower bound of $1$. Using \eqref{eq2A}, you can determine the limit is once again $L = 1$.

Finally, starting with $u_0 \gt 3$, as you've noted, it becomes an increasing sequence. Also, as you state, and \eqref{eq1A} confirms, as $u_n$ increases, the value of $f(x) - x$ also increases, so the difference keeps increasing faster, and thus the values, with no upper bound and, thus, no limit.

  • $\begingroup$ Thank you very much, especially for $(2)$. I've first wrote that the difference becomes arbitrarily small, and wrote it, but then dismissed it as obvious and haven't though of considering the actual limit as a value. Thank you again! $\endgroup$
    Mar 21, 2020 at 7:29

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