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Math Help - Recurrence relation

  1. #1
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    Recurrence relation

    Can you please help me to obtain the generating function of the recurrence relation

    ar =ar-l + ar-z with ao = 0, al = 2, az = 3.
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  2. #2
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    Quote Originally Posted by bond View Post
    Can you please help me to obtain the generating function of the recurrence relation

    ar =ar-l + ar-z with ao = 0, al = 2, az = 3.
    Is ar-1 meant to be a_{r-1} ?

    What is ar-z meant to be? Do you mean a_{r-2} ?

    Is az = 3 meant to be a_2 = 3 ?

    Note: z is not the same as 2.
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  3. #3
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    I think it looks like this

    ar =ar-l + ar-z

    with ao = 0, al = 2, az = 3
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  4. #4
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    Quote Originally Posted by bond View Post
    I think it looks like this

    ar =ar-l + ar-z

    with ao = 0, al = 2, az = 3
    Writing the same thing bigger doesn't clarify anything.
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  5. #5
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    Please see the attachment

    Please see the attachment, I have typed in the question in the MS Word file.

    Thanks.
    Attached Files Attached Files
    Last edited by bond; April 19th 2009 at 12:07 AM. Reason: subscripts now showing properly
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  6. #6
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    Hello, bond!

    Does it really help to write things in a different color and in
    a smaller font?

    And if you have no idea what's going on, why are you working on this problem?


    Can you please help me to obtain the generating function of the recurrence relation

    ar =ar-l + ar-z with ao = 0, al = 2, az = 3.
    . . . this is wrong!
    I assume this is a Fibonacci-type sequence . . .

    a_n \:=\:a_{n-1} + a_{n-2}\quad\text{ with }a_0 = 0,\;a_1= 2 \quad \hdots\text{ and }a_2\text{ is }not\text{ equal to 3 !}
    . . Each term is the sum of the preceding two terms.

    The sequence is: . 0,2,2,4,6,10,16, 26, \hdots


    The terms are exactly twice that of the original Fibonacci sequence

    . . whose formula is: . F_n \:=\:\frac{(1+\sqrt{5})^n - (1-\sqrt{5})^n}{2^n\sqrt{5}}

    Therefore: . a_n \;=\;2\cdot\frac{(1+\sqrt{5})^n - (1-\sqrt{5})^n}{2^n\sqrt{5}}


    Does that help?

    I didn't think so . . .

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