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Math Help - Recurrence Relations: Constraint Conditions

  1. #1
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    Recurrence Relations: Constraint Conditions

    Consider the recurrence relation b_n = 3b_{n-1} + 10b_{n-2}, b_0 = x. b_1 = y
    Under what conditions on x and y are the constants in the general solution to the recurrence integers?

    Assume this is correct:

    b_n = C_1 * (5)^n + C_2 * (-2)^n
    then

    b_0 = x = C_1 + C_2
    b_1 = y = 5C_1 - 2C_2

    I have figured out by trial and error that
    x and y can be any integer that when modded with 7 is congruent to 0.
    or
    If x and y are divided by 7 and they leave no remainder,
    then that will produce the constants C_1 and C_2

    Can someone please explain to me why this happens?
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Celcius View Post
    Consider the recurrence relation b_n = 3b_{n-1} + 10b_{n-2}, b_0 = x. b_1 = y
    Under what conditions on x and y are the constants in the general solution to the recurrence integers?

    Assume this is correct:

    b_n = C_1 * (5)^n + C_2 * (-2)^n
    then

    b_0 = x = C_1 + C_2
    b_1 = y = 5C_1 - 2C_2

    I have figured out by trial and error that
    x and y can be any integer that when modded with 7 is congruent to 0.
    or
    If x and y are divided by 7 and they leave no remainder,
    then that will produce the constants C_1 and C_2

    Can someone please explain to me why this happens?
    Have you tried solving the simulate nous equations??
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  3. #3
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    Quote Originally Posted by Drexel28 View Post
    Have you tried solving the simulate nous equations??
    Sure, sort of.

    C_2 = (-y + 5x)/7
    and
    C_1 = (y + 2x)/7

    I kind of see something there, but still not 100% sure.
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