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Math Help - differencial equation system

  1. #1
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    differencial equation system

    Now ... I ve made it. Files are smaller.
    I need help with question 2 part (b)
    many thanks
    Attached Thumbnails Attached Thumbnails differencial equation system-assign2p1.jpg   differencial equation system-assign2p2.jpg  
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  2. #2
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    Quote Originally Posted by simfonija View Post
    Now ... I ve made it. Files are smaller.
    I need help with question 2 part (b)
    many thanks
    I'm curious. If all you need is 2 (b) why did you post the first page?

    Anyway, we have a matrix equation:
    x^{(k)} = Ax^{(k-1)}
    and an initial population distribution: x^{(0)} = a_1 x_1 + ... + a_n x_n
    where \{ x_n \} is the set of eigenvectors of A.

    Let \{ \lambda _n \} be the set of eigenvalues corresponding to the (respective) eigenvector.

    Then
    x^{(1)} = Ax^{(0)} = A(a_1x_1 + ... + a_n x_n)

    x^{(1)} = a_1 \lambda _1 x_1 + ... + a_n \lambda_n x_n

    Iterating this process gives:
    x^{(k)} = a_1 ( \lambda _1 )^k x_1 + ... + a_n ( \lambda_n  )^k x_n

    Now, if \lambda _1 is the dominant eigenvalue (I think this said the first eigenvalue. It was hard to read.) then we may assume
    \lambda _1 >> \lambda _k for all k \neq 1. (Obviously only a positive eigenvalue can dominate the population.) This means that for large k there is effectively only one term in the x^{(k)} that is significant: the first term. So
    x^{(k)} \approx a_1 ( \lambda_1 )^k x_1 for large k.

    This means the population is stable in the state represented by eigenvector x_1 and has a growth rate represented by \lambda _1.

    If there is no dominant eigenvalue then the population will fluctuate in a series of eigenstates with no specific growth rate dominating. However, for large enough k the population should stabilize around a vector that is the sum of eigenvectors with the largest eigenvalues. For example, say that we know that \lambda _1 = \lambda _2 > \lambda _k for k \neq 1, 2. Then, given a large enough k we would have:
    x^{(k)} \approx a_1 ( \lambda _1)^k x_1 + a_2 ( \lambda _2)^k x_2
    So the population would settle into a growth state vector a_1 x_1 + a_2 x_2 with (in this case) a growth of \lambda _1 = \lambda _2.

    -Dan
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  3. #3
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    Thank you so much for the detail answer. The first page was mistake, previously I saw there is a part of question 2 (b) on first page. Once again thank you, was great help!
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