Fibonacci, Lucas, and the square root of five

Discussion in 'Recreational Math' started by walt, Apr 3, 2005.

  1. walt

    walt Guest

    I have little formal training in mathematics, but I would guess that
    the answer to this question is already well known. Kindly clue me
    in if you understand the following:

    The square-root of five can be calculated to any arbitrary precision
    on a common spreadsheet by doing the following:

    1) In column one of a spreadsheet create the Fibonacci numbers
    by adding the previous two numbers to make the next in the series.
    Start with 1 + 1 = 2, and continue with 1 + 2 = 3, etc.

    2) In column two, create the so-called Lucas numbers, which are
    closely related to the Fibonacci numbers except that the two
    starting numbers are 1 and 3 instead of 1 and 1.

    You should now have the following in your spreadsheet:

    1 1
    1 3
    2 4
    3 7
    5 11
    8 18
    13 29
    21 47
    34 76
    55 123

    and so on...

    Now, the part that really puzzles me is this: in the third column,
    divide column 2 by column 1 and you will see that the result converges
    on the square root of five (2.23606........)

    My question, of course, is WTF is special about the square root of five?

    The Fibonacci ratio is given by phi = (Sqrt(5) + 1)/2 so I'm not too
    surprised to see it appear in my puzzle, but I just can't figure out
    why the Lucas numbers should be related to the Fibonacci numbers by
    exactly that ratio.
     
    walt, Apr 3, 2005
    #1
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  2. walt

    Stan Liou Guest

    The reccurence relation is F(n+2) = F(n+1) + F(n+1) in both
    cases, only subject to different initial values of F(0) and
    F(1), so the characteristic polynomial is the same in both
    cases: x^2 - x - 1. The roots of this determine the closed-
    form solution.

    One can also think of it in terms of limits of n->infinity:
    L = lim[F(n+1)/F(n) ]
    = lim[F(n+2)/F(n+1) ]
    = lim[F(n+1)/F(n+1) + F(n)/F(n+1)]
    = 1 + L
    thus L^2 - L - 1 = 0. Notice that the initial values of
    F(0), F(1) are completely irrelevant here. Any sequence
    defined by this recurrence relation, be it Fibonacci or
    Lucas or whatever, has the same ratio limit.
     
    Stan Liou, Apr 3, 2005
    #2
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  3. walt

    Stan Liou Guest

    Pardon, L = 1 + 1/L.
     
    Stan Liou, Apr 3, 2005
    #3
  4. walt

    Proginoskes Guest

    That should be: F(n+2) = F(n+1) + F(n).
    Any sequence which satisfies F(n+2) = F(n+1) + F(n) is of the
    form F(n) = A ((1 + sqrt(5))/2)^n + B ((1 - sqrt(5))/2)^n,
    for certain constants A and B.
    That last line should be 1 + 1/L.
    --- Christopher Heckman
     
    Proginoskes, Apr 3, 2005
    #4
  5. walt

    night Guest

    I don't know, but here's something interesting.

    If you start the first sequence with 1, 1, ... and the second sequence
    with 1, N, ..., then the limit of the ratio is a root of the quadratic:

    x^2 + (N-3) x + (1+N-N^2) = 0

    The positive root of which is:

    ((sqrt(5) - 1)/2) N + (3 - sqrt(5))/2

    Or, where phi = (sqrt(5) - 1)/2:

    (N-1)phi + 1

    When the second sequence is the Lucas numbers, N=3, so the limit of the
    ratio is 2phi + 1 = sqrt(5).
     
    night, Apr 3, 2005
    #5
  6. -
    Expanding some:

    The Fibonacci and Lucas sequences are given by the rule

    f_n = f_{n-1}+f_{n-2}

    This formula has the characteristic equation

    r^2-r-1 = 0

    Solving gives (1/2)*(+/-1+Sqrt[5]). Setting

    A = (1/2)*(1+Sqrt[5]) (the Golden Ratio)
    B = (1/2)*(1-Sqrt[5]) (the negative of the reciprocal of the Golden Ratio),

    any sequences defined by the above rule is given by

    c_1*A^n+c_2*B^n for some choice of c_1, c_2.

    For the Fibonacci sequence, we set f_1 = f_2 = 1 and find have c_1 =
    1/Sqrt[5], c_2 = -1/Sqrt[5]. For the Lucas sequence, we have c_1 = c_2 = 1.

    Let's consider the ratio of a given Lucas number L_n and a given
    Fibonacci number F_n. Explicitly,

    L_n/F_n = (A^n+B^n)/(1/Sqrt[5]*A^n-1/Sqrt[5]*B^n)

    |B| < 1, so as n gets very large, B^n approaches zero. So for large n,
    we have

    L_n/F_n ~ A^n/(1/Sqrt[5]*A^n) = Sqrt[5]

    Cheers,
    Travis
     
    Travis Willse, Apr 3, 2005
    #6
  7. walt

    walt Guest

    Please don't think I'm ignoring your answer -- I'm still trying to make
    an intelligent response. According to my wife you can expect to wait
    another thirteen years...no, wait, I think it's fourteen...
     
    walt, Apr 5, 2005
    #7
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