1. ## monotone convergence theorem

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.

2. Firstly lets find the limit by solving

$\displaystyle x = x/2 + 2$

gives x = 4, so limit is 4

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

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

hope this helps and makes sense!

cheers Nobby

3. Originally Posted by Nobby
Firstly lets find the limit by solving

$\displaystyle x = x/2 + 2$

gives x = 4, so limit is 4

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

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.

4. 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?

5. I guess the formal argument is by induction.

$\displaystyle x_2< x_1$ you can calculate these first.

then

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

which we showed was the same as

$\displaystyle x_{n+1}< x_n$

Half of something positive is less then the original

clear?

6. Originally Posted by Jhevon
by induction.

can you continue?
i think soo...

7. 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 $\displaystyle x_{n + 1} < x_n$ for all $\displaystyle n$), and hence is bounded above by the first term, namely 8

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

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

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

adding 2 to both sides we get

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

but that means

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