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Math Help - quick sequence question

  1. #1
    Senior Member slevvio's Avatar
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    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!
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  2. #2
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    Quote Originally Posted by slevvio View Post
    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!
    What about trying to show
    \frac{x_n}2 \le x_{n+1} \le x_n
    by induction?
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  3. #3
    Senior Member slevvio's Avatar
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    thank you, i will try this!
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  4. #4
    MHF Contributor chisigma's Avatar
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    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
    Last edited by chisigma; May 21st 2010 at 02:47 AM.
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  5. #5
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    Quote Originally Posted by chisigma View Post
    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.
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  6. #6
    Senior Member slevvio's Avatar
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    shouldnt the third term in your equation be over 4? cos the limit is definitely zero
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  7. #7
    Senior Member slevvio's Avatar
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    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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  8. #8
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by kompik View Post
    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.
    ops! ... very sorry! ...

    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
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