# Show the sequence is decreasing

• Oct 2nd 2008, 12:20 PM
universalsandbox
Show the sequence is decreasing
I have attached a picture of the question. How would you go about showing this? Thanks.

http://lighthousegraphicdesign.com/images/real2.png
• Oct 2nd 2008, 02:14 PM
CaptainBlack
Quote:

Originally Posted by universalsandbox
I have attached a picture of the question. How would you go about showing this? Thanks.

http://lighthousegraphicdesign.com/images/real2.png

Is/are there an extra condition/s that is/are missing here?

RonL
• Oct 2nd 2008, 02:22 PM
universalsandbox
Quote:

Originally Posted by CaptainBlack
Is/are there an extra condition/s that is/are missing here?

RonL

Ex:

let a=2, xknot = 1

then

x(sub1) = (1/2)(xknot + 2/xknot) = 3/2

x(sub2) = (1/2)( x(sub1) + 2/(xsub1) ) = 17/2 etc.
• Oct 2nd 2008, 02:38 PM
icemanfan
I work out that if $\displaystyle x_n > \frac{a}{x_n}$,

$\displaystyle x_n + \frac{a}{x_n} > \frac{2a}{x_n}$

$\displaystyle 0.5\left(x_n + \frac{a}{x_n}\right) > \frac{a}{x_n}$

$\displaystyle x_{n+1} > \frac{a}{x_n}$.

As long as $\displaystyle x_n > \frac{a}{x_n}$, then the function will be decreasing, but I can't find a way to show that statement inductively.
• Oct 2nd 2008, 09:24 PM
CaptainBlack
Quote:

Originally Posted by universalsandbox
Ex:

let a=2, xknot = 1

then

x(sub1) = (1/2)(xknot + 2/xknot) = 3/2

x(sub2) = (1/2)( x(sub1) + 2/(xsub1) ) = 17/2 etc.

Compare what happens when $\displaystyle x_0>\sqrt{a}$ with what happens when $\displaystyle x_0<\sqrt{a}$ with what happens when $\displaystyle x_0=\sqrt{a}.$

RonL