1
$\begingroup$

The problem seems slightly wrong in it assumptions. Please, check this proof.

For $a>0$ and $x_0\ge a$ prove that the sequence defined as $x_{n+1}=(x_n+a/x_n)/2$ decreases and converges to $\sqrt a$.

Then we have that $a/x_0\le1$ and then

$$x_1=\frac{x_0}2+\frac{a}{x_0}\frac12\le\frac{x_0}2+\frac12$$

but then

$$x_1\le\frac{x_0}2+\frac12<x_0\iff x_0>1$$

The exercise seems to be incorrect here, to continue I assume that $x_0>1$. Now I assume the induction hypothesis, i.e. $x_{n+1}<x_n$, then I will check if it is true that $x_{n+2}<x_{n+1}$. We have

$$x_{n+2}=(x_{n+1}+a/x_{n+1})/2=\frac{(x_{n+1})^2+a}{2x_{n+1}}<x_{n+1}$$

then $$(x_{n+1})^2+a<2(x_{n+1})^2\iff\sqrt a<x_{n+1}$$

but the induction hypothesis $x_{n+1}<x_n$ implies too that $\sqrt a<x_n$, then

$$\sqrt a<x_{n+1}\implies 2\sqrt a<\frac{(x_n)^2+a}{x_n}\iff 0<x_n^2-2\sqrt ax_n+a$$

To solve the inequality of second order first we solve the equation, where we get that the unique solution is $x=\sqrt a$. Then I substitute $x_n=\epsilon\sqrt a$ with $\epsilon>0$ to study the behavior of the inequality

$$0<x_n^2-2\sqrt ax_n+a\iff 0<\epsilon^2-2\epsilon+\epsilon\iff 0<(\epsilon-1)^2$$

Then for $x_0>1$ the sequence decreases.

Because the above we can see that $\sqrt a$ is a lower bound of the sequence, then (and because is decreasing) is a convergent sequence. Because the sequence converges then

$$\lim x_{n+1}=\lim \frac{(x_n)^2+a}{2x_n}\implies x=\frac{x^2+a}{2x}\iff x=\sqrt a$$


Trying to see what happen if $0<x_0\le 1$ I observe that is not monotone. What I know is that if $x_n<\sqrt a$ then $x_{n+1}\ge\sqrt a$.

The sequence seems oscillate around $\sqrt a$. I know that converges to $\sqrt a$ because the above sequence is a way to compute square roots, but I dont know how show it with these tools.

$\endgroup$
1
$\begingroup$

The sequence being decreasing means $x_{n+1}<x_n$, for all $n$. This becomes $$ x_n+\frac{a}{x_n}<2x_n $$ that is, $x_n^2>a$, because it is obvious that $x_n>0$ for all $n$. I believe that the statement should have $x_0^2>a$. For instance, if $a=1/4$ and $x_0=1/3$, we have $x_0^2<a$, which implies $x_1>x_0$.

So, I'll go with $x_0^2>a$ as the hypothesis, which is the base step in the induction proof.

Suppose $x_n^2>a$. Then $x_{n+1}^2-a$ becomes $$ x_{n+1}^2-a=\frac{1}{4}\left(x_n+\frac{a}{x_n}\right)^{\!2}-a= \frac{1}{4}\left(x_n^2+2a+\frac{a^2}{x_n^2}-4a\right)= \frac{1}{4}\left(x_n-\frac{a}{x_n}\right)^{\!2}>0 $$


If the hypothesis is $x_0>a$, then the sequence need not be decreasing, as shown before. Of course, if we assume $a>1$, then $x_0>a$ implies $x_0^2>a$, because from $x_0>a>1$ we get $$ x_0^2>ax_0>x_0>a $$

So the statement should have either $x_0^2>a$ or $a>1$. In the case $x_0>a$ and $0<a\le1$ we may indeed obtain an oscillating sequence.

$\endgroup$
1
$\begingroup$

Let $f(x)=\frac{1}{2}(x+\frac{a}{x})$

$f$ has two fixed points

$x=\sqrt{a}$ and $x=-\sqrt{a}$

those are possible limits of the sequence $(x_n)$ such that

$x_0\geq a>0$ and

$x_{n+1}=f(x_n)$.

by induction we can easily prove that

$\forall n\in \mathbb N \; x_n>0$

thus if $(x_n)$ converges, the limit will be $\sqrt{a}$ as $f$ is continuous at $(0,+\infty)$.

on the other hand,

$f'(x)=\frac{1}{2}\frac{x^2-a}{x^2}$.

$f(x)-x=\frac{1}{2}\frac{a-x^2}{x}$.

now let us discuss three cases.

FIRST

$x_0=\sqrt{a}$. in this case, the sequence is constant and convergent.

SECOND

$x_0>\sqrt{a}$ by induction we prove that in this case $\forall n \; x_n \in (\sqrt{a},+\infty)$ as $f$ is increasing.

thus

$f(x_0)-x_0<0 \implies x_1<x_0$

and also by induction we get $x_{n+1}<x_n$

finally $(x_n)$ decreases and $\sqrt{a}<x_n \implies $(x_n)$ converges to $\sqrt{a}$.

THIRD CASE

$x_0<\sqrt{a}$ this case is little more difficult since $f$ is decreasing and i think that your hypothesis should be $x_0\geq\sqrt{a}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.