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Difference equations

  1. Mar 14, 2010 #1
    1. The problem statement, all variables and given/known data
    Suppose that {uk} and {vk} are sequences satisfying uk = Auk-1 k = 1, 2, ... and vk = Avk-1 k = 1, 2,... Show that if u0 = v0 then ui = vi for all values of i.

    2. Relevant equations

    uk --> is u subscript k
    u0 --> is u subscript 0
    uk-1 --> is u subscript k-1
    ui --> is u subscript i

    3. The attempt at a solution

    Well so far I have...
    uk = A^k(u0)
    = A^k(a1u1 + a2u2 +...+anun)
    = a1(A^k)u1 + a2(A^k)u2 +...+ an(A^k)un
    = a1(lambda1^k)u1 + a2(lambda2^k) +...+ an(lambdan^k)un

    But since u0 = v0 A^k(u0) = A^k(v0)

    But after there I get uncertain cause I think my next steps would be:
    = A^k(a1v1 + a2v2 +...+ anvn)
    = a1(A^k)v1 + a2(A^k)v2 +...+ an(A^k)vn
    = a1(lambda1^k)v1 + a2(lambda2^k)v2 +...+ an(lambdan^k)vn)

    Then conclude uk = vk?
    Is this correct?

    Thanks! :)
  2. jcsd
  3. Mar 14, 2010 #2


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    Hi simmonj7! :smile:

    (try using the X2 tag just above the Reply box :wink:)
    (what are the as and the lambdas? :confused: anyway …)

    Hint: put wk = uk - vk, for all k. :wink:
  4. Mar 14, 2010 #3
    When solving a difference equation all the way out you have to find the eigen values, then find the corresponding eigen vectors and then have to find a relationship between those eigen vectors and the vector x0. Thus x0 = a1u1 + a2u2 +...+ anun is the relationship between all the eigen vectors.

    Then you just plug in that relationship for x0 into the equation xk = (A^k)x0. From there you distribute the A^k through and then (because (A)x = (lambda)x then (A^k)x = (lambda^k)x where lambda is an eigen vector of A) you substitute that back into the equation and that is what all the a's and lambda's are.
  5. Mar 14, 2010 #4


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    But there's only one eigenvalue here (and it's A). :confused:
  6. Mar 14, 2010 #5
    Where are you getting that there is only one eigen value? There is no way to determine how many eigen values there are of A...
  7. Mar 15, 2010 #6


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    The eigenvalues of A don't matter …

    all that matters is the roots of the characteristic equation of this recurrence relation which in this case is the single root, A.

    Try using the wk I mentioned earlier. :smile:
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