Question P.174: Let x0 be a rational number, and let (xn)n≥0 be the sequence......
Let x_0 be a rational number, and let (x_n)_{n≥0} be the sequence defined recursively by
x_{n+1} =|\frac{2x_n^3}{3x_n^2 − 4}|.
Prove that this sequence converges, and find its limit as a function of x_0.
Note that all x_n are actually rational numbers, since when x_n is rational, 3x^2_n − 4 will not be zero and so x_{n+1} will again be rational. Also that after the initial term, the sequence starting with x_0 is the same as the sequence starting with −x_0, so we can assume x_0 ≥ 0.
Let f(x) = 2x^3/(3x^2 − 4),~so~that~x_{n+1} = |f(x_n)|. We can use standard techniques to sketch the graph of f (shown in the figure for x ≥ 0 only). From this graph it seems reasonable to suspect that 2 < f(x)< x for x > 2, and it is straightforward to show this. Thus, if x_n > 2 for some n, the sequence will thereafter decrease, but stay above 2. Since any bounded decreasing sequence has a limit, L = \lim_{n→\infty} x_n will then exist. Taking limits on both sides of x_{n+1} = f(x_n) and using the continuity of f for x ≥ 2, we then have L = f(L), so L(3L² −4) = 2L³ and L(L² −4) = 0. Of the three roots, L = 0, L = ±2, only L = 2 is possible in this case, since x_n > 2 for large enough n. In particular, we see that L = 2 if x_0 > 2. We can also see that L = 2 if x_0 = 2 (in this case, x_n = 2 for all n) and even that L = 2 if 2\sqrt{3}< x_0 < 2, for in that case x_1 = f(x_0) > 2.
Now consider what happens if 0 ≤ x_0 < 2/\sqrt{3}.~If~x_0 = 0,~then~x_n = 0 for all n, so L = 0. From the graph, we expect that for x close to 0,
will be even closer to 0, while for x close enough to 2/\sqrt{3}, |f(x)| becomes large and thus |f(x)| > x.
A simple calculation shows that 2x³/(4 − 3x²) = x for x = 0 and for x = ±2/\sqrt{5}. If 0 < x_n < 2/\sqrt{5} at any point in the sequence, then it is easily shown that 0 < x_{n+1} = |f(x_n)| < x_n, and so forth, so the sequence will decrease (and be bounded) from that point on. Thus, L will exist; as before, we have L = |f(L)| = 2L³/(4 − 3L²), which implies L = 0 or L = ±2/\sqrt{5} . However, once 0 < x_n < 2/ \sqrt{5} and (x_n) is decreasing, we must have L = 0.
Since x_0 is rational, x_0 = 2/\sqrt{5} is impossible, so we are left with the case 2/\sqrt{5} < x_0 < 2/ \sqrt{3}. In this case, x_1 = |f(x_0)| > x_0, so the sequence increases at first. Suppose that x_n < 2/\sqrt{3} for all n. In this case, the sequence is bounded and increasing, so L exists. But from L = |f(L)| and 0 < L ≤ 2/\sqrt{3} we again get L = 0 or L = ±2/\sqrt{5}, a contradiction since x_0 > 2/\sqrt{5} and (x_n) is increasing. It follows that there is an n with x_n > 2/ \sqrt{3}. But then x_{n+1} ≥ 2, so L = 2, and we are done.
