quick sequence question

Oct 2007
347
17
I was wondering if i could get some help with this question:

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

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

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

Any help would be appreciated!
 
Mar 2010
116
41
Bratislava
I was wondering if i could get some help with this question:

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

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

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

Any help would be appreciated!
What about trying to show
\(\displaystyle \frac{x_n}2 \le x_{n+1} \le x_n\)
by induction?
 
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chisigma

MHF Hall of Honor
Mar 2009
2,162
994
near Piacenza (Italy)
Let's write the difference equation as...

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

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

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

The correponding characteristic equation is...

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

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

\(\displaystyle u_{k} = 1^{k} =1\)

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

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

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

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

Kind regards

\(\displaystyle \chi\) \(\displaystyle \sigma\)
 
Last edited:
Mar 2010
116
41
Bratislava
Let's write the difference equation as...

\(\displaystyle x_{k} - \frac{x_{k-1}}{2} - \frac{x_{k-2}}{2} =0\) (1)
I think that (1) should be:
\(\displaystyle x_{k} - \frac{x_{k-1}}{2} - \frac{x_{k-2}}{4} =0\),
i.e. 4 instead of 2 in the denominator.
 
Oct 2007
347
17
shouldnt the third term in your equation be over 4? cos the limit is definitely zero
 
Oct 2007
347
17
discovered this is a simple way to show this. Just subsitute the values for n and n+1 into one recurrence relation and subtract
 
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chisigma

MHF Hall of Honor
Mar 2009
2,162
994
near Piacenza (Italy)
I think that (1) should be:
\(\displaystyle x_{k} - \frac{x_{k-1}}{2} - \frac{x_{k-2}}{4} =0\),
i.e. 4 instead of 2 in the denominator.
ops! (Doh)... very sorry!(Headbang) ...

The difference equation is then...

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

... and the characteristic equation...

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

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

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

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

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

Kind regards

\(\displaystyle \chi\) \(\displaystyle \sigma\)