2
$\begingroup$

Let $\{X_n\}$ be the sequence defined recursively by $x_1=2$ and $x_{(n+1)}=(x_n/2)+(5/x_n)$. Prove that $\{x_n\}$ converges and find the limit of the sequence.

I understand the definition of a Cauchy Sequence, but I am having trouble getting my foot in the door with this one. I have been working on this problem for several hours without any luck, so I was just wondering if someone could help me determine a way to start this problem?

$\endgroup$
  • $\begingroup$ Find the possible limit(s) first: If $x = x/2 +5/x$ then $x=?$ $\endgroup$ – Winther Nov 4 '14 at 0:09
6
$\begingroup$

We start with $$ x_{n+1}=\frac{10+x_n^2}{2x_n}\tag{1} $$ and subtracting $\sqrt{10}$, we get $$ \begin{align} x_{n+1}-\sqrt{10} &=\frac{10-2x_n\sqrt{10}+x_n^2}{2x_n}\\ &=\frac{(x_n-\sqrt{10})^2}{2x_n}\\[6pt] &\ge0\tag{2} \end{align} $$ After that, we have $x_{n+1}\ge\sqrt{10}$ and therefore, $$ \begin{align} x_{n+2}-x_{n+1} &=\frac{10-x_{n+1}^2}{2x_{n+1}}\\[6pt] &\le0\tag{3} \end{align} $$ Thus, for $n\gt1$, $x_n$ is decreasing $(3)$ and bounded below $(2)$. Therefore, $\lim\limits_{n\to\infty}x_n=\inf\limits_{n\gt1}x_n$.

Furthermore, taking the limit of $(1)$ yields $$ \lim_{n\to\infty}x_n=\frac{\lim\limits_{n\to\infty}x_n}2+\frac5{\lim\limits_{n\to\infty}x_n}\tag{4} $$ Thus, $(2)$ and $(4)$ imply $$ \lim_{n\to\infty}x_n=\sqrt{10}\tag{5} $$

$\endgroup$
2
$\begingroup$

Hints:

Always start by calculating a few terms:

$$x_2 = 3.50000$$ $$x_3 = 3.17857$$ $$x_4 = 3.16231$$ $$x_5 = 3.16227$$

We see that $x_2>x_3>x_4>x_5$ so it is probably true that $x_{n} > x_{n+1}$. Rewrite the recurence relation to see if we can prove this:

$$x_{n} - x_{n+1} = \frac{x^2_n-10}{2x_n} $$

So we see that if we can prove $x_n^2 > 10$ then we automatically get $x_{n} > x_{n+1}$. To do this rewrite the recurence relation as

$$x_{n+1} - \sqrt{10} = \frac{(x_n - \sqrt{10})^2}{2x_n}$$

$\endgroup$
1
$\begingroup$

First, recall that a sequence of real numbers is Cauchy if and only if it converges.

Next, recall that any eventually monotonically decreasing sequence that is bounded below converges. Therefore, we'll first set out to prove that $\{x_n\}$ is monotonically decreasing for all sufficiently large $n$. In other words, we show that $x_{n+1} - x_n < 0$:

$$x_{n+1} - x_n = \frac{x_n}{2} + \frac{5}{x_n} - x_n$$

$$= \frac{-x_n^2 + 10}{2x_n}$$

Note that this is less than $0$ $\iff$ $x_n > \sqrt{10}$. Therefore, to show it is monotonically decreasing beyond a certain point, we simply need to show that $x_n$ is bounded below by $\sqrt{10}$ for all sufficiently large $n$ (killing two birds with one stone). This can be done by induction, and I'll leave that to you. Hint: It appears that $x_n > \sqrt{10}$ for all $n \geq 2$.

Once we have proven that the limit exists, then notice that $\displaystyle \lim_{n \rightarrow \infty} \left( x_{n+1} \right) = \lim_{n \rightarrow \infty} \left( x_n \right)$ since the former is a subsequence of the latter. Therefore, if we let $L = \displaystyle \lim_{n \rightarrow \infty} \left( x_{n} \right)$, we get the following from our recursive formula by taking a limit of both sides:

$$\lim \left( x_{n+1} \right) = \lim \left( \frac{x_n}{2} + \frac{5}{x_n} \right)$$

Which reduces using limit laws:

$$L = \frac{L}{2} + \frac{5}{L}$$

And now one simply needs to find $L$.

$\endgroup$
0
$\begingroup$

First, I suggest finding the limit via this method:

Take $\lim\limits_{n\rightarrow \infty} x_{n+1} =\lim\limits_{n\rightarrow \infty}\frac{x_n}{2}+\lim\limits_{n\rightarrow \infty}\frac{5}{x_n}$. Let $\lim\limits_{n\rightarrow \infty} x_n = x$. Also, observe that $\lim\limits_{n\rightarrow \infty} x_{n+1} = \lim\limits_{n\rightarrow \infty} x_n = x$. Then we have

\begin{gather*} x=x/2 + 5/x\\ x^2 = \frac{x^2}{2} +5\\ 2x^2=x^2+10\\ x^2=10\\ x=\pm\sqrt{10} \end{gather*}

But clearly the sequence is always positive, so we may discard the negative value.

Now that we have a suggested limit, let's attempt to show that the sequence is Cauchy.


Proof. We wish to show $x_n<x_{n-1}$ for $n>3$. \begin{gather*} x_n<x_{n-1}\\ x_n-x_{n-1}<0\\ \frac{x_{n-1}^2+10}{2x_{n-1}}-\frac{2x_{n-1}^2}{2x_{n-1}}<0\\ \frac{-x_{n-1}^2+10}{2x_{n-1}}<0\\ -x_{n-1}^2+10<0\\ x_{n-1}^2>10\\ |x_{n-1}|>\sqrt{10} \end{gather*} So, $x_n<x_{n-1}$ if and only if $|x_{n-1}|>\sqrt{10}$.

It suffices to show $\sqrt{10}$ is a lower bound for the sequence for $n\geq 2$. We do so through manipulation of the definition of this sequence. Observe \begin{gather*} x_{n}-\sqrt{10}=\frac{10-2x_n\sqrt{10}+x^2_n}{2x_n}\\ x_n-\sqrt{10}= \frac{(x_n-\sqrt{10})^2}{2x_n}\\ \end{gather*} Thus, $x_n-\sqrt{10}>0$ if and only if $x_n$ is positive.

We prove $x_n$ is positive by induction on $n$ (note: this is a little overkill but, you get the point). Observe $x_1=2>0$. Now assume this is true for $n=k$. We will show $x_{k+1}>0$ follows from this assumption. By the definition of the sequence we have $x_{k+1}=\frac{x_k}{2}+\frac{5}{x_k}$. Since $x_k,\frac{1}{2}\in \mathbb{R}^+$ and multiplication in $\mathbb{R}^+$ is closed, $\frac{x_k}{2}>0$. Furthermore, $\frac{1}{x_k}\in \mathbb{R}^+$ since it is the inverse of $x_k$, $5\in \mathbb{R}^+$ and by the same argument $\frac{5}{x_k}>0$. Thus, $x_{k+1}=\frac{x_k}{2}+\frac{5}{x_k}>0$. Therefore, by the first principle of mathematical induction,$\forall n\in \mathbb{N}$, $x_n>0$.

This not only shows, $\forall n\geq 3$, $x_n<x_{n-1}$ but it also shows $\sqrt{10}$ is a lower bound for the sequence $\forall n\geq 3$. So, by the monotone convergence theorem $(x_n)_{n=3}^\infty$ is a convergent sequence and therefore $(x_n)^\infty_{n=1}$ is a convergent sequence. Then the work in the beginning has value, and $\lim\limits_{n\rightarrow\infty}x_n=\sqrt{10}$.

$\endgroup$
  • $\begingroup$ I will fill out a more thorough proof in a few hours if you wish to wait for that. $\endgroup$ – Eoin Nov 4 '14 at 0:13
  • 2
    $\begingroup$ Typo: $x=\pm\sqrt{10}$ $\endgroup$ – Winther Nov 4 '14 at 0:13
  • $\begingroup$ @Winther Thank you haha $\endgroup$ – Eoin Nov 4 '14 at 0:14
  • $\begingroup$ Thank you so much for the quick response! How are we to be sure that the sequence is monotonically increasing? By writing out a few of the terms of the sequence, I obtain 2, 3.5, 3.18, 3.16232, 3.16288. That is why I jumped to trying to prove that this sequence is a Cauchy sequence $\endgroup$ – Jimmy Nov 4 '14 at 0:23
  • $\begingroup$ It isn't! Apparently it is decreasing for $n> 1$. That's okay, it still works. We just change $\sqrt{10}$ to a lower bound and use the same theorem. $\endgroup$ – Eoin Nov 4 '14 at 2:42
0
$\begingroup$

First note that if the sequence has a limit $x$, it should satisfy: $$x = \frac{x}{2}+\frac{5}{x}.$$ Solving for $x$ gives: $$x=\pm\sqrt{10}.$$ If we look at the difference between subsequent terms we find: $$x_{n+1}-x_n = \frac{x_n}{2}+\frac{5}{x_n}-x_n = \frac{5}{x_n}-\frac{x_n}{2}.$$ We could have skipped the first part and put $x_{n+1}-x_n=0$, this should give the same limit value(s). However, this equation also tells us that $x_{n+1}-x_n>0$ when $0<x_n<\sqrt{10}$ and $x_{n+1}-x_n<0$ when $x_n>\sqrt{10}$. Since the starting value is $x_1=2$ this means the sequence will tend to $\sqrt{10}$.

$\endgroup$

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.