0
$\begingroup$

Problem:

Let $x$ and $y$ be positive real numbers. Let $a_0 =y$. Show that $$a_n = \frac12\left(\frac{x}{a_{n-1}} + a_{n-1}\right)$$ is a decreasing sequence.

Solution Attempt:

I have tried showing that the difference is positive $$a_{n-1} - a_{n} > 0,$$ that the ratio is greater than 1 $$\frac{a_n}{a_{n-1}} < 1,$$ and I have tried to use induction by considering $a_0$ and $a_1$. Somewhere, I must be doing something wrong. I keep getting the result $a_n > \frac{1}{2}a_{n-1},$ which doesn't seem very useful.

$\endgroup$
3
  • $\begingroup$ i think induction will help you $\endgroup$ Feb 20, 2017 at 16:28
  • $\begingroup$ When $a_0 = y = 0.1, x = 2$ we get $a_1 = 10.05$, which is contradicting to the claim right? $\endgroup$
    – rookie
    Feb 20, 2017 at 16:32
  • $\begingroup$ @stud_iisc You're right. We need $y >\sqrt{x}$. See my answer. $\endgroup$
    – S.C.B.
    Feb 20, 2017 at 16:36

2 Answers 2

3
$\begingroup$

Your claim is not true. We need a condition that $a_{0}=y \ge \sqrt{x}$.

We can use that $$a_{n}=\frac{1}{2} \left(a_{n-1}+\frac{x}{a_{n-1}} \right) \ge \sqrt{x}$$ By $\text{AM-GM}$. So $a_{n} \ge \frac{x}{a_{n}} $ for all $n$. Now note that $$a_{n+1}= \frac{1}{2} \left(a_{n}+\frac{x}{a_{n}} \right) \le \frac{1}{2} (a_{n}+a_{n})=a_{n} $$

From $(1)$. So $a_{n+1} \le a_{n}$. The sequence is monotonically decreasing.

$\endgroup$
8
  • $\begingroup$ What does AM-GM stand for? $\endgroup$ Feb 20, 2017 at 16:47
  • $\begingroup$ @EternusVia I added a link. $\endgroup$
    – S.C.B.
    Feb 20, 2017 at 16:51
  • $\begingroup$ It seems like your proof doesn't need the condition $y \geq \sqrt{x}$. Where does this assumption come into play? Is it a sort of implicit induction? $\endgroup$ Feb 20, 2017 at 17:00
  • $\begingroup$ @EternusVia We need to prove that that $a_{0}>a_{1}$. $\endgroup$
    – S.C.B.
    Feb 20, 2017 at 17:00
  • $\begingroup$ I get it now. Thanks! $\endgroup$ Feb 20, 2017 at 17:01
1
$\begingroup$

Your claim is NOT true in general. It is only true if $y\ge\sqrt{x}$.

However, if $y<\sqrt{x}$, then one can show the following: $$ a_0<\sqrt{x}<a_n<\cdots<a_2<a_1 $$ and that $a_n\to\sqrt{x}$.

$\endgroup$
3
  • $\begingroup$ How were you able to determine these conditions on $y$? $\endgroup$ Feb 20, 2017 at 16:48
  • $\begingroup$ This seems untrue. For example, if $a_{0}=1$, $x=2$, then we have that $a_{1}=\frac{3}{2}$, $a_{2}=\frac{17}{12}>\sqrt{2}$. So you're incorrect. $\endgroup$
    – S.C.B.
    Feb 20, 2017 at 16:49
  • $\begingroup$ I have corrected my answer accordingly. Thanx. $\endgroup$ Feb 20, 2017 at 16:56

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .