Approximation of $x^2-x-1 = 0$ .

Question

An exercise in Calculus With Applications by Peter D. Lax confused me, here is the original text:

1.34. Solve $x^2 − x − 1 = 0$ as follows. Restate the equation as $x = 1 + 1/x$, which suggests the sequence of approximations $$x_0 = 1,\quad x_1 = 1+ 1/x_0,\quad x_2=1 + 1/x_1, ...$$

Explain the following items to prove that the sequence converges to a solution.

(a) $x_0 < x_2 < x_1$

(b) $x_0 < x_2 < x_4 < ··· < x_5 < x_3 < x_1$

(c) The even sequence $x_{2k}$ increases to a limit $L$, and the odd sequence $x_{2k+1}$ decreases to a limit $R ≥ L$.

(d) The distances $(x_{2k+3} −x_{2k+2})$ satisfy $(x_{2k+3} −x_{2k}+2) < 1/x_4^2(x_{2k+1} -x_{2k}).$

(e) $R=L=\lim\limits_{k\to∞}x_k$ is a solution to $x^2 −x−1 = 0.$

For (c) I have trouble explaining the increment of the even-numbered sequence, although it is obvious by listing a few figures. I can tell that the even terms are bounded, but I can't verify that the even terms are increasing. Please help me please

Answer 1

This is a proof for step (b):

First we notice that $$ x_{i+2}=1+1/x_{i+1}=1+\frac{1}{1+1/x_i}=1+\frac{x_i}{x_i+1}=\frac{2x_i+1}{x_i+1} $$ and $x_i>0$. Next, we try to prove that the two-step is strictly increasing $$ x_{i+2}>x_i\\ \Leftrightarrow \frac{2x_i+1}{x_i+1}>x_i\\ \Leftrightarrow x_i+1>x_i^2 $$ So in order to show that the two step is strictly increasing we need to prove that the last inequality holds true for all $i$. We do this by induction

• Initial step: $x_0=1\Rightarrow1+1>1$ (Here we use that we are actually talking about the even indices)
• Induction step: assume that $x_i+1>x_i^2$, now show that $x_{i+2}+1>x_{i+2}^2$. $$ x_{i+2}+1>x_{i+2}^2\\ \Leftrightarrow \frac{2x_i+1}{x_i+1}+1>\frac{(2x_i+1)^2}{(x_i+1)^2}\\ \Leftrightarrow \frac{3x_i+2}{x_i+1}>\frac{(2x_i+1)^2}{(x_i+1)^2}\\ \Leftrightarrow (3x_i+2)(x_i+1)>(2x_i+1)^2\\ \Leftrightarrow x_i+1>x_i^2\\ $$

The proof for the decreasing odd indices works the same way.

To show that the even indices are always strictly below the odd indices we observe that $$ x_{2i}<x_{2i+1}\\ \Leftrightarrow x_{2i}<1+1/x_{2i} \Leftrightarrow x_{2i}^2<x_{2i}+1 $$ which we have already proven.