If $f(x) =\frac{1}{2} (x + a/x)$ such that $r_{n+1} = f(r_n)$ show that $0 < f^{'}(r_{n+1})< \frac{1}{2}$

Question

If $f(x) =\frac{1}{2} (x + a/x)$ such that $r_{n+1} = f(r_n)$

Show that $0 < f^{'}(r_{n+1})< \frac{1}{2}$

I calculated that $f'(r_{n+1})=\frac{1}{2}(1-\frac{a}{r_{n+1}^2})$ but now I am stuck

any hint?

Answer 1

Is $r_n \in R \;\&\; r_0 \neq 0$ given? And is a given to be positive? Because I can see the inequality working only when a > 0. Let us prove this :

as you correctly calculated $f'(r_{n+1})=\frac{1}{2}(1-\frac{a}{r_{n+1}^2})$

Now set the inequality $0 < f'(r_{n+1}) < \frac{1}{2}$

Substituting $f'(r_{n+1})=\frac{1}{2}(1-\frac{a}{r_{n+1}^2})$ we get

$0 < \frac{1}{2}(1-\frac{a}{r_{n+1}^2}) < \frac{1}{2}$

Multiplying by -2 on both sides (inequality changes signs) and then adding 1 on both sides, we get :

$0 < \frac{a}{r_{n+1}^2} < 1$

Now it is easy to why a must be $> 0 \because \frac{a}{r_{n+1}^2} > 0 \;\&\; r_{n+1}^2 > 0$

So now to prove second inequality let us go back to the formula $r_{n+1} = f(r_n)$

From this we get $r_{n+1} = \frac{1}{2}(r_n + \frac{a}{r_n})$

Let us square both sides and subtract a from both sides

$\therefore r_{n+1}^2 - a = \frac{1}{4}(r_n^2 + \frac{a^2}{r_n^2} + 2 \cdot r_n \cdot \frac{a}{r_n}) - a = \frac{1}{4}(r_n^2 + \frac{a^2}{r_n^2} + 2a - 4a) = \frac{1}{4}(r_n^2 + \frac{a^2}{r_n^2} -2a) = (\frac{1}{2}(r_n -\frac{a}{r_n}))^2 > 0$

$\therefore r_{n+1}^2 - a > 0$

Here we can also see that it is strictly greater than 0 only when $r_{n+1} \neq \sqrt a \implies r_0 \neq \sqrt a$

So adding a on both sides and dividing by $r_{n+1}^2$ we get :

$\frac{a}{r_{n+1}^2} < 1$

Answer 2

As it is written, the result does not hold. The sequence $(r_n)$ corresponds to the application of Newton's method to find the positive root of $x^2-a = 0$ and will converge to $\sqrt{a}$. If you start with $r_0 = \sqrt{a}$, the sequence is constant and, in fact, $f'(r_{n+1})=0$ for all $n$.