# quick sequence question

• May 21st 2010, 01:28 AM
slevvio
quick sequence question
I was wondering if i could get some help with this question:

Define a sequence $\{x_n\}_{n=1}^{\infty}$ by $x_1 = 1, x_2 = \frac{1}{2}$ and $x_n = \frac{2x_{n-1} + x_{n-2}}{4}$for $n \ge 3.$ Use the Monotone Convergence Theorem to show that $\{x_n\}_{n=1}^{\infty}$ converges and find its limit.

Ok it is not too hard to prove by induction that $0 \le x_n \le 1$. But I am having trouble proving the sequence is decreasing. I have

$x_{n+1} - x_{n} = \frac{2x_n + x_{n-1}}{4} - x_n = \frac {x_{n-1} - 2x_n}{4}$, but I have no guarantee that this is less than or equal to zero!

Any help would be appreciated!
• May 21st 2010, 02:13 AM
kompik
Quote:

Originally Posted by slevvio
I was wondering if i could get some help with this question:

Define a sequence $\{x_n\}_{n=1}^{\infty}$ by $x_1 = 1, x_2 = \frac{1}{2}$ and $x_n = \frac{2x_{n-1} + x_{n-2}}{4}$for $n \ge 3.$ Use the Monotone Convergence Theorem to show that $\{x_n\}_{n=1}^{\infty}$ converges and find its limit.

Ok it is not too hard to prove by induction that $0 \le x_n \le 1$. But I am having trouble proving the sequence is decreasing. I have

$x_{n+1} - x_{n} = \frac{2x_n + x_{n-1}}{4} - x_n = \frac {x_{n-1} - 2x_n}{4}$, but I have no guarantee that this is less than or equal to zero!

Any help would be appreciated!

$\frac{x_n}2 \le x_{n+1} \le x_n$
by induction?
• May 21st 2010, 02:14 AM
slevvio
thank you, i will try this!:)
• May 21st 2010, 02:33 AM
chisigma
Let's write the difference equation as...

$x_{k} - \frac{x_{k-1}}{2} - \frac{x_{k-2}}{2} =0$ (1)

... so that its a constant coefficients homogeneous linear difference equation, and its solution is on the form...

$x_{k} = c_{1}\cdot u_{k} + c_{2}\cdot v_{k}$ (2)

The correponding characteristic equation is...

$\xi^{2} - \frac{\xi}{2} - \frac{1}{2} =0$ (3)

... the solutions of which are $\xi=1$ and $\xi= -\frac{1}{2}$, so that is...

$u_{k} = 1^{k} =1$

$v_{k} = (-\frac{1}{2})^{k}$ (4)

The 'initial condition' $x_{0}=1$ and $x_{1} = \frac{1}{2}$ give us $c_{1}=\frac{2}{3}$ and $c_{2}=\frac{1}{3}$ so that the solution of (1) is...

$x_{k} = \frac{2}{3} + \frac{1}{3}\cdot (-\frac{1}{2})^{k}$ (5)

... and is also $\lim_{k \rightarrow \infty} x_{k} = \frac{2}{3}$ ...

Kind regards

$\chi$ $\sigma$
• May 21st 2010, 02:37 AM
kompik
Quote:

Originally Posted by chisigma
Let's write the difference equation as...

$x_{k} - \frac{x_{k-1}}{2} - \frac{x_{k-2}}{2} =0$ (1)

I think that (1) should be:
$x_{k} - \frac{x_{k-1}}{2} - \frac{x_{k-2}}{4} =0$,
i.e. 4 instead of 2 in the denominator.
• May 21st 2010, 02:38 AM
slevvio
shouldnt the third term in your equation be over 4? cos the limit is definitely zero
• May 21st 2010, 02:44 AM
slevvio
discovered this is a simple way to show this. Just subsitute the values for n and n+1 into one recurrence relation and subtract
• May 21st 2010, 03:45 AM
chisigma
Quote:

Originally Posted by kompik
I think that (1) should be:
$x_{k} - \frac{x_{k-1}}{2} - \frac{x_{k-2}}{4} =0$,
i.e. 4 instead of 2 in the denominator.

The difference equation is then...

$x_{k} - \frac{x_{k-1}}{2} - \frac{x_{k-2}}{4} =0$ (1)

... and the characteristic equation...

$\xi^{2} - \frac{\xi}{2} - \frac{1}{4}$ (2)

... the solution of which are $\xi= \frac{1 \pm \sqrt{5}}{4}$, so that the solution of (1) is...

$x_{k} = c_{1}\cdot (\frac{1+\sqrt{5}}{4})^{k} + c_{2}\cdot (\frac{1-\sqrt{5}}{4})^{k}$ (3)

At this point it doesn't matter which are the 'initial conditions' because in any case is...

$\lim_{k \rightarrow \infty} x_{k} = 0$ (4)

Kind regards

$\chi$ $\sigma$