# Contest problem - region in which a sequence converges

(Please do not offer a solution! I am seeking a hint; more on that below.)

Problem: For any pair $(x, y)$ of real numbers, define the sequence $a_n (x, y)$ as follows:

$$a_0 (x,y) = x$$

$$a_{n+1}(x,y) = \frac{a_n (x,y))^2 + y^2}{2}, \text{ } n \geq 0$$

Find the area of the region of coordinates $(x, y)$ in $\mathbb{R}^2$ such that $a_n$ is convergent.

I believe I have found the region, but am having trouble with the key step in the proof.

(Proposed) Solution: AM-GM shows that we must have $|y| \leq 1$ for the sequence to converge, since by induction we have $a_{n+1} \geq {a_n}|y|$ (i.e., with $|y| > 1$, the sequence telescopes into $a_0 |y|^n$, and therefore diverges). Similarly, when $\frac{x^2 + y^2}{2} \leq |x|$, the sequence is bounded below (the terms are all non-negative after $a_0$) and monotonically decreasing. Note that the box on $[-1, 1]$ (on both axes) also contains only convergent points, since the sequence is bounded above there by $1$, and is monotone. Clearly, $x \leq 2$ for convergent $a_n$....

...By the (missing!) lemma, we have that for $1 \leq |x| \leq 2,$ having $\frac{x^2 + y^2}{2} > |x|$ causes $a_n$ to diverge. (Trial computation supports this lemma.) Thus, when $1 \leq |x| \leq 2$, $\frac{x^2 + y^2}{2} \leq |x|$ is necessary and sufficient for convergence of $a_n$.

The region $\frac{x^2 + y^2}{2} \leq |x|$ is the area of twin unit circles, centered at $x = \pm 1$. Its intersection with our other area yields half unit circles (on $[-2, -1]$ and $[1, 2]$) coupled with the unit box on $[-1, 1]$. The area of the box is 4, while the combined areas of the half-unit-circles are $\pi$; therefore, the area of the region in question is $\pi + 4$.

To summarize: I want to prove that

$$\forall x \in [-2, -1] \cup [1,2], \frac{x^2 + y^2}{2} > |x| \implies a_n \text{ diverges}$$

Please offer hints (unless you discover that the lemma is false, in which case candor is appreciated).

(The source for this is 1992's Putnam Exam.)

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This answer occurred to me while working on other things.

Assuming that $a_1 = \frac{x^2 + y^2}{2} > |x|,$ then we have $a_1 = |x| + \varepsilon_0$ for $\varepsilon > 0$. Let $P(n)$ be the assertion that $a_{n+1} \geq a_{n} + \varepsilon_0$.

Suppose that $(x,y)$ is such that $P(0)$ holds. If $P(k)$ holds $\forall k \in \mathbb{N}: k \leq n$,

$$a_{n+2} = \frac{a_{n+1}^2 + y^2}{2} \geq \frac{(a_n + \varepsilon_0)^2 + y^2}{2} \geq \frac{a_n^2 + y^2}{2} + a_n\varepsilon_0 = a_{n+1} + a_n\varepsilon_0 \geq a_{n+1} + \varepsilon_0$$

i.e., $P(n+1)$ holds. Telescoping shows that, given $\frac{x^2 + y^2}{2} > |x|$, we have $a_n \geq |x| + n\varepsilon_0$, which shows that $a_n$ is divergent in this case.

From the lemma, it follows (as I had imagined) that when $1 \leq |x| \leq 2$, $a_n$ is convergent iff $\frac{x^2 + y^2}{2} \leq |x|$.

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