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Math Help - solving recurrences - I keep gettin' alpha = some fraction???

  1. #1
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    solving recurrences - I keep gettin' alpha = some fraction???

    ok, so I have this recurrence:

    a(n) = 2(a(n-1)) + 3(a(n-2))

    so the degree is k, & k=2, & I used the formula r^2 - 2r - 3, to get the roots r=1 & r= -3.

    then I used the general solution technique

    a(n) = alpha1(r1^n) + alpha2(r2^n)

    & subbed in my "r" values to get:

    a(n) = alpha1[(1)^n] + alpha2[(-3)^n]

    then I used the initial conditions a(0) = 0 & a(1) = 2 (subbed in n=0 & n=1) to the equation:

    a(0) = alpha1[(1)^0] + alpha2[(-3)^0] = 0

    a(1) = alpha1[(1)^1] + alpha2[(-3)^1] = 2

    to try & solve for alpha1 & alpha2, but I keep gettin' fractions for both alphas...(I got alpha1 = 1/2, & alpha2 = -(1/2))

    is this right?!??! b/c most of the examples in my textbook have alpha equaling an integer....

    I'm not sure if I messed up the calculations.
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  2. #2
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    Hello amyu2005
    Quote Originally Posted by amyu2005 View Post
    ok, so I have this recurrence:

    a(n) = 2(a(n-1)) + 3(a(n-2))

    so the degree is k, & k=2, & I used the formula r^2 - 2r - 3, to get the roots r=1 & r= -3.

    then I used the general solution technique

    a(n) = alpha1(r1^n) + alpha2(r2^n)

    & subbed in my "r" values to get:

    a(n) = alpha1[(1)^n] + alpha2[(-3)^n]

    then I used the initial conditions a(0) = 0 & a(1) = 2 (subbed in n=0 & n=1) to the equation:

    a(0) = alpha1[(1)^0] + alpha2[(-3)^0] = 0

    a(1) = alpha1[(1)^1] + alpha2[(-3)^1] = 2

    to try & solve for alpha1 & alpha2, but I keep gettin' fractions for both alphas...(I got alpha1 = 1/2, & alpha2 = -(1/2))

    is this right?!??! b/c most of the examples in my textbook have alpha equaling an integer....

    I'm not sure if I messed up the calculations.
    Thanks for showing us all your working. You just made a simple slip in solving the quadratic. The roots are -1, 3. The solution is then

    a_n = (-1)^n - 3^n

    Grandad
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  3. #3
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    Quote Originally Posted by Grandad View Post
    Hello amyu2005Thanks for showing us all your working. You just made a simple slip in solving the quadratic. The roots are -1, 3. The solution is then

    a_n = (-1)^n - 3^n

    Grandad
    thanks~~~
    Last edited by amyu2005; April 6th 2009 at 07:46 AM. Reason: I found my mistake, so this reply isn't really necessary
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  4. #4
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    I used the correct roots, but I'm still gettin' alpha as 0.5 & -0.5.....

    is this possible?
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  5. #5
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    Hello amyu2005

    Sorry, I got a sign wrong in my working as well! I think it should be:

    r = -1, 3 (That bit was correct!)

    n= 0 : a_0 = 0 = \alpha_1 + \alpha_2

    n=1: a_1 = 2 = -\alpha_1 +3\alpha_2 (I originally had 2 = -\alpha_1 -3\alpha_2 here, which gave me my first answer.)

    \Rightarrow 4\alpha_2 = 2

    \Rightarrow \alpha_2 = \tfrac12

    \Rightarrow \alpha_1 = -\tfrac12

    \Rightarrow a_n = \tfrac12\Big(3^n - (-1)^n\Big)

    Check this out to see that the original equation a_n = 2a_{n-1} + 3a_{n-2} is satisfied:

    a_0 = 0,\, a_1 = 2,\, a_2 = 4 = 2a_1 + 3a_0,\, a_3 = 14 = 2a_2 +3a_1,\, a_4 = 40 = 2a_3+3a_2,\,...

    Grandad
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