Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $x_0=0$ and $x_{n+1} = \sqrt{5-2{x_{n}}}$ for $n\ge0$. Show that this sequence converges and compute the limit.

Hint from prof: Show that the even terms increase and the odd term decreases


My thought:

odd terms:

$x_1=x_{0+1}=\sqrt{5-2x_0}=\sqrt5=2.23606...$

$x_3=x_{2+1}=\sqrt{5-2x_2}=1.88332...$, decreasing

even terms:

$x_2=x_{1+1}=\sqrt{5-2x_1}=0.723654...$

$x_4=x_{3+1}=\sqrt{5-2x_3}=1.11056...$, increasing

So what I need to show is when $n\to\infty$, both odd and even sequence converges to the same point, right?

Since I have no idea how to do that, I take a different approach which is to prove the difference for every two consecutive elements is getting smaller and smaller as $n\to\infty$. As even or odd element I choose doesn't matter, I will just put on a absolute value on the difference.

$x_{n} = \sqrt{5-2_{x_{n-1}}}$

$x_{n+1} = \sqrt{5-2_{x_{n}}} = \sqrt{5-2(\sqrt{5-2_{x_{n-1}}}})$

$x_{n+2} = \sqrt{5-2_{x_{n+1}}} = \sqrt{5-2(\sqrt{5-2(\sqrt{5-2_{x_{n-1}}}}))}$

Prove $|x_{n+2} - x_{n+1}|<|x_{n+1}-x_n|$

$|\sqrt{5-2(\sqrt{5-2(\sqrt{5-2_{x_{n-1}}}}))} - \sqrt{5-2(\sqrt{5-2(\sqrt{5-2_{x_{n-1}}}}))}| < |\sqrt{5-2(\sqrt{5-2(\sqrt{5-2_{x_{n-1}}}}))} - \sqrt{5-2_{x_{n-1}}}|$

Then how should I finish this...?

Thanks!!

share|improve this question
    
Hint: $|x_{n+1} - x_n| = |\sqrt{5-2x_n} - \sqrt{5-2x_{n-1}}|$ and apply mean value theorem for $n > 3$. –  achille hui Feb 5 '13 at 1:44

2 Answers 2

up vote 1 down vote accepted

Let $f(x)=\sqrt{5-2x}$, whose domain is $[5/2,+\infty)$. A quick study shows that $f$ is a decreasing bijection from $[0,5/2]$ onto $[0,\sqrt{5}]$. Since the latter is contained in the former, we see that $(x_n)$ is well-defined for all $n$, and that $0\leq x_n\leq \sqrt{5}$ for all $n$.

Now comparing $f(x)$ with $x$ on $[0,5/2]$, we see that there exists a unique fixed point for $f$ here (and on its domain too, actually): $$ l=\sqrt{6}-1. $$ Moreover, if $0\leq x<l$ then $l<f(x)\leq \sqrt{5}$. And if $l< x\leq 5/2$, then $0\leq f(x)<l$. Thus $(x_n)$ oscillates around $l$, without ever being equal to $l$.

Applying the mean value theorem on $[0,\sqrt{5}]$ we find that for all $x,y$, there exists $z$ in between such that $$ f(x)-f(y)=\frac{-1}{\sqrt{5-2z}}(x-y). $$

Applying this twice, we see that $x_{n+2}-x_n=f\circ f(x_n)-f\circ f(x_{n-2})$ and $x_{n}-x_{n-2}$ have the same sign. Looking at the intial ranks, we conclude that $(x_{2n})$ is increasing and $(x_{2n+1})$ is decreasing. So $$ x_{2n}<x_{2n+2}<l<x_{2n+3}<x_{2n+1} $$ for all $n$.

At this point, we can conclude that $(x_{2n})$ (increasing + bounded above) converges, and $(x_{2n+1})$ (decreasing + bounded below) converges as well.

It only remains to use the mean value theorem above once more to see that $$ |x_{n+1}-x_n|=|f(x_n)-f(x_{n-1})| = \frac{|x_n-x_{n-1}|}{\sqrt{5-2z}}\leq \frac{|x_n-x_{n-1}|}{\sqrt{5-x_3}} $$ for all $n\geq 4$, since both $x_n$ and $x_{n-1}$ are then smaller than $x_3$.

Since we can check that $$ \frac{1}{\sqrt{5-x_3}}<1 $$ it follows easily that $\lim_{n\rightarrow +\infty}(x_{n+1}-x_n)=0$.

Finally, we have $\lim x_{2n}=\lim x_{2n+1}$, and this has to be $l$. So $$ \lim_{n\rightarrow+\infty}x_n=\sqrt{6}-1. $$

share|improve this answer

You need to use the two results

1) Every increasing bounded above sequence is convergent,

2) Every decreasing bounded below sequence is convergent.

In your case, you have two subsequences, so you need to prove that they converge to the same limit point. To find the limit, assume $\lim_{n\to \infty} x_n=l$, then

$$\lim_{n\to \infty} x_{n+1} = \lim_{n\to \infty}\sqrt{5-2 {x_{n}}} = l=\sqrt{5-2l} = \dots. $$

share|improve this answer
    
However, how do I know $\lim_{n\to\infty}x_n$ exists? –  Paul Feb 5 '13 at 2:14
    
1) and 2) above tell you $\lim_{n\to\infty} x_{2n}$ and $\lim_{n\to\infty} x_{2n+1}$ exists. What you remain to show is they are the same limit. –  achille hui Feb 5 '13 at 2:18
    
@achillehui: Thanks for answering him. –  Mhenni Benghorbal Feb 5 '13 at 7:10

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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