Showing that $x_{n+1} = 1/2(x_n + 2/x_n)$ is a decreasing sequence.

Question

I need to show that $x_{n+1} = 1/2(x_n + 2/x_n)$ is a decreasing sequence, where $x_1 = 2$ and n = 1,2,3,....

I tried to show this with induction, where since $x_1= 2$ then $x_2 = 1.5$, hence the base case satisfies.

Assuming that $x_n \geq x_{n+1}$ , I need to prove that $x_{n+1} \geq x_{n+2}$.

I started with the statement $x_n \geq x_{n+1} \Rightarrow \frac{1}{x_n} \leq \frac{1}{x_{n+1}}$

Also, $x_n \geq x_{n+1}$ then $\frac{1}{2}(x_n + \frac{2}{x_{n+1}}) \geq \frac{1}{2}(x_{n+1} + \frac{2}{x_{n+1}})$

$= \frac{1}{2}(x_n + \frac{2}{x_{n+1}} - \frac{2}{x_n} + \frac{2}{x_n}) \geq x_{n+2} $

$= x_{n+1} - \frac{1}{x_n} + \frac{1}{x_{n+1}} \geq x_{n+2} $

since $- \frac{1}{x_n} + \frac{1}{x_{n+1}} \geq 0$

then I cannot say that $ x_{n+1} \geq x_{n+2} $

How do I proceed forward?

Edit: Since $x_1 = 2$, hence $1 \leq x_1 \leq 2$ holds for base case.

If I assume that $1 \leq x_n \leq 2 $ and show that $1 \leq x_{n+1} \leq 2 $ then by induction we can say that $1 \leq x_n \leq 2 \:\forall n$

I have shown this as:

$x_{n+1} = \frac{x_n}{2} + \frac{1}{x_n}, \: now \: since\:1 \leq x_n\leq2\:\Rightarrow x_{n+1} \leq \frac{1}{2}(2 + \frac{2}{1}) = 2$

Also by AM GM inequality, $x_n \geq \sqrt{2} \: \: \forall n$

Now all that I have to show is that $x_n \geq x_{n+1} \: \forall n?$

Yes, the @rtybase's answer does answer my question somewhat but if I want to use induction then how do I use it? — Nisarg Jain, Apr 25 '20 at 09:10

Bernard · Answer 1 · 2020-04-25T09:03:00.140

1

Hint:

Actually, you only need to prove that all $x_n$ live in an interval $I$ (to be determined) such that $$f(x)=\frac12\Bigl(x+\frac2x\Bigr)<x \quad\text{on }I.$$ So you have to determine the variations of the function $f$ and show by induction that $x_n\in I$ for all $n$.

edited Apr 25 '20 at 09:03

answered Apr 25 '20 at 08:22

Bernard

175,478

But $f(x)\leq x$ should be shown. – Jens Schwaiger Apr 25 '20 at 08:40
The sign is wrong: it should be $\frac{1}{2} (x+\frac{2}{x}) ≤ x$ since the sequence is only defined for $n ≥ 1$. – Toby Mak Apr 25 '20 at 09:00
Oh! yes, of course. Thanks for pointing the typo! One should always check before posting… – Bernard Apr 25 '20 at 09:04
If I could show that f(x) lies in interval $[\sqrt(2), 2]$ then how would that prove that it is decreasing? – Nisarg Jain Apr 25 '20 at 09:15
Because on this interval, $f(x)\le x$ (and $<x$ on the open interval). Don't forget that $f(x_n)=x_{n+1}$. – Bernard Apr 25 '20 at 09:18
My comment above is missing context how, since the answer by @Bernard has been edited later on. – Jens Schwaiger Apr 25 '20 at 10:53

score 1 · Answer 2 · answered Apr 25 '20 at 08:59

1

At first one could show that $f(x)\geq 2$ for all $x>0$, where $f(x):=\frac12\left(x+\frac2{x}\right)$. This follows immediately from $f(x)^2-4= \frac14\left(x-\frac2{x}\right)^2\geq0$. But then $f(x)\leq x$ is equivalent to $\frac2{x}\leq x$ which is always true when $x^2\geq 2$. But this holde true for $x_0=2$ by assumption, and thus by definition of $x_n$ for all $n$.

answered Apr 25 '20 at 08:59

Jens Schwaiger

4,015

But $f(x) \leq 2$ always right? – Nisarg Jain Apr 25 '20 at 09:11
Yes, since $x_0=2$ and $x_{n+1}\leq x_n$ for all $n$. But $f(x)$ may be arbitraryly large when $x$ becomes large. Note that for example $f(100)=50.01$. – Jens Schwaiger Apr 25 '20 at 10:44

Showing that $x_{n+1} = 1/2(x_n + 2/x_n)$ is a decreasing sequence.

2 Answers2