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Value that a sequence tends to- please explain!

  1. Nov 3, 2009 #1
    1. The problem statement, all variables and given/known data

    Two sequences are defined by

    Xn+1 = 3Xn + 4Yn ; Yn+1 = 2Xn + 3Yn

    What is the approximate value of Xn/Yn as n increases?

    2. Relevant equations

    none

    3. The attempt at a solution

    My solution was to say that as n tends to infinity, both Xn, Yn tend to infinity therefore Xn/Yn tends to 7/5... on checking by calculating values this is completely wrong, infact Xn/Yn tends very quickly to root 2. Why??

    EDIT: ok, not completely wrong, 1.4 isn't that far off root 2, but root 2 is such a neat number it feels like it should be possible to reach it using algebra.
     
    Last edited: Nov 3, 2009
  2. jcsd
  3. Nov 3, 2009 #2

    Mark44

    Staff: Mentor

    I might be way off here, but this looks like a matrix diagonalization problem to me.
    You have:
    x_n+1 = [3 4][x_n]
    y_n+1 = [2 3][y_n]

    One way to untangle this is to find a diagonal matrix D = P-1AP, where P is the matrix whose columns are the eigenvectors of the matrix above, and P-1 is its inverse. The diagonal matrix will have the eigenvalues on its main diagonal. These are 3 +/- 2sqrt(2).

    Does any of this sound familiar to you? I can't think of any other way to approach this problem.
     
  4. Nov 3, 2009 #3
    Hmmmn. Yeah I have done diagonalization of matrices, (though not in relation to sequences), but I know its definetely not on the syllabus of this particular exam. So I guess they just want the answer 1.4!

    I can see how that matrix eq relates in the first place, though not exactly how I'd get from the eigenvalues to the value the sequence tends to.. however the exam is tomorrow and as I say this method isn't expected so I won't panic :P
    Thanks anyway
     
  5. Nov 3, 2009 #4

    Mark44

    Staff: Mentor

    Once you get to a diagonal matrix, you'll have x_(n+1) in terms of x_n only, and y_(n+1) in terms of y_n only, so you could calculate the ratio of x_(n+1) to y_(n+1).

    The fact that the eigenvalues I found are 3 +/- 2sqrt(2) might have something to do with what you found as the limit...
     
  6. Nov 3, 2009 #5
    Since you're asked only for an approximate value for [itex]\frac{X_n}{Y_n}[/itex], perhaps the following reasoning will suffice:

    Write

    [tex]\frac{X_{n+1}}{Y_{n+1}} = \frac{3X_n + 4Y_n}{2X_n + 3Y_n}[/tex]

    Now divide both the numerator and the denominator of the right side of the above equation by [itex]Y_n[/itex] and simplify. Then, assuming that the limit as n goes to infinity of [itex]\frac{X_n}{Y_n}[/itex] exists, what can you say?

    Please let me know if this isn't clear, or if you'd like another hint.

    Petek
     
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