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Math Help - Fibonacci Numbers

  1. #1
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    Fibonacci Numbers

    F(n) = F(n-1) + F(n-2):

    Can someone explain to me how to use this formula, and maybe show an example. I'm uncertain of what I'm supposed to be plugging in for 'n'.

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  2. #2
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    Hi

    For instance for n=2 you get F(2) = F(1) + F(0)

    Given F(0) and F(1) you can compute F(2)

    Then for n=3 you get F(3) = F(2) + F(1)

    Knowing F(1) and F(2) calculated just before you can get F(3)

    And so on ...
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  3. #3
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    Two things

    1) F has to be a natural number greater than 2

    2) What do you know about Fibonacci numbers? Why are you interested in
    them?
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  4. #4
    MHF Contributor ebaines's Avatar
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    Pick a first number - say, 3 - and call it F(1), meaning the first number in the sequence. Then pick a second number - perhaps 5 - and assign this to F(2). To find the third number F(3) you use F(3)= F(2) + F(1) = 5 + 3 = 8. Then to find the 4th you use F(4) = F(3) + F(2) = 8 + 5 = 13. In general, to find the nth number in the sequence you add F(n-1) and F(n-2) together. Typically people start the Fibonacci series off with F(1) =1 and F(2)=1, but you don't have to. This leads to:

    F(1) = 1
    F(2) = 1
    F(3) = 2
    F(4) = 3
    F(5) = 5
    F(6) = 8
    F(7) = 13
    F(8) = 21
    etc.

    Hope this helps!
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  5. #5
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    Well, it was part of a problem I had to do for school. When I was given a number I was supposed to correspond with a Fibonacci number. I guess like:

    1. 1
    2. 2
    3. 3
    4. 5
    5. 8
    6.13

    So, if I wanted to find the Fibonacci number for 2 (which is two), I thought I'd let n be 2 and plug it in...

    F(2)=F(2-1)+F(2-2)
    F(2)=F(1)+F(0)
    but F(1)+F(0)=F(1)?

    but the fibonacci number for 2 isn't 1 is it?

    I'm a little confused here
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  6. #6
    MHF Contributor ebaines's Avatar
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    The Fibonacci series needs to be "seeded" with the first two numbers F(1) and F(2). In other words - the first two numbers must be given to you, so that you can then determine the 3rd, and so on.
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  7. #7
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    @spacegirl777 you pluged-in well.

    Here is what you need to remember.

    F(0)= 1
    F(1)= 1
    F(2)= 2
    F(3)= 3
    F(4)= 5
    F(5)= 8

    ..... etc ....
    so F(2)=F(2-1)+F(2-2)=F(1)+F(0)=1+1=2 which is correct.
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  8. #8
    MHF Contributor chisigma's Avatar
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    A more suggestive and general approach is to consider the Fibonacci sequence as solution of the difference equation...

    x_{n+2} - x_{n-1} - x_{n}=0 (1)

    ... with 'initial consitions' x_{0}=0 and x_{1}=1. If u and v are solution of the second order equation...

    z^{2} - z -1 =0 (2)

    ... then the solution of (1) is of the form...

    x_{n}= c_{1} u^{n} + c_{2} v^{n} (3)

    In this case is...

    u= \frac{1 - \sqrt{5}}{2}

    v= \frac{1 + \sqrt{5}}{2} (4)

    ... and the constants c_{1} and c_{2} are found imposing the initial conditions and solving the system...

    c_{1} + c_{2} =0

    c_{1} \frac{1 - \sqrt{5}}{2} + c_{2} \frac{1 + \sqrt{5}}{2}= 1 (5)

    Kind regards

    \chi \sigma
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  9. #9
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    Quote Originally Posted by p0oint View Post
    @spacegirl777 you pluged-in well.

    Here is what you need to remember.

    F(0)= 1
    F(1)= 1
    F(2)= 2
    F(3)= 3
    F(4)= 5
    F(5)= 8

    ..... etc ....
    so F(2)=F(2-1)+F(2-2)=F(1)+F(0)=1+1=2 which is correct.
    Okay, I understand it now Thank you very much!
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