monotone

**bigb** · September 25th 2008, 02:12 PM

a) X1=8 and Xn+1=.5Xn +2 for n within naturals. Show that Xn is bounded and monotone and find the limit?

b) X1> and Xn+1=2-1/Xn for n within naturals. Show that Xn is bounded and monotone and find the limit

for part a i can say x1>x2, thus the sequence is decreasing.

**Nobby** · September 25th 2008, 05:13 PM

Firstly lets find the limit by solving

$ x = x/2 + 2 $

gives x = 4, so limit is 4

lets rewrite
$ x_n = 4 + \delta_n $
$ x_n+1 = (4 + \delta_n) /2 + 2 = 4 + \delta_n / 2 $
we can see that if $x_n > 4$ then $x_n+1 > 4$ and also that $x_n+1 < x_n$

so we can conclude that the sequence decreases and has a lower bound of 4.

hope this helps and makes sense!

cheers Nobby

**bigb** · September 25th 2008, 05:20 PM

Originally Posted by Nobby

Firstly lets find the limit by solving

$ x = x/2 + 2 $

gives x = 4, so limit is 4

lets rewrite
$ x_n = 4 + \delta_n $
$ x_n+1 = (4 + \delta_n) /2 + 2 = 4 + \delta_n / 2 $
we can see that if $x_n > 4$ then $x_n+1 > 4$ and also that $x_n+1 < x_n$

so we can conclude that the sequence decreases and has a lower bound of 4.

hope this helps and makes sense!

cheers Nobby

can you explain to me how to find out the sequence is decreasing. I know how to solve for the limit.

**Jhevon** · September 25th 2008, 05:28 PM

Originally Posted by bigb

can you explain to me how to find out the sequence is decreasing. I know how to solve for the limit.

by induction.

can you continue?

**Nobby** · September 25th 2008, 05:29 PM

I guess the formal argument is by induction.

$x_2< x_1$ you can calculate these first.

then

$4 + \delta/2 < 4 + \delta$

which we showed was the same as

$x_{n+1}< x_n$

Half of something positive is less then the original

clear?

**bigb** · September 25th 2008, 05:37 PM

Originally Posted by Jhevon

by induction.

can you continue?

i think soo...

**Jhevon** · September 25th 2008, 10:17 PM

Originally Posted by bigb

i think soo...

here's the outline to a more straightforward induction proof. i leave it to you to formalize it. we use induction to show that the sequence is monotonically decreasing (that is $x_{n + 1} < x_n$ for all $n$ ), and hence is bounded above by the first term, namely 8

Note that $x_2 = \frac 12 \cdot 8 + 2 = 6$ so that $x_2 < x_1$

Now assume $x_{n + 1} < x_n$

then $\frac 12 x_{n + 1} < \frac 12 x_n$

adding 2 to both sides we get

$\underbrace{\frac 12 x_{n + 1} + 2}_{\text{note this is }x_{n + 2}} < \underbrace{\frac 12 x_n + 2}_{\text{note this is }x_{n + 1}}$

but that means

$x_{n + 2} < x_{n + 1}$

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