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Fibonacci Sequence

  1. Sep 4, 2011 #1
    Let f[itex]_{n}[/itex] be the Fibonacci sequence and let [itex]x_{n}[/itex] = [itex]f_{n+1}[/itex]/[itex]f_{n}[/itex]. Given that lim[itex](x_{n}[/itex])=L exist determine L.

    Ok so I know that the limit is [itex]\frac{1+\sqrt{5}}{2}[/itex] from previous experience with the sequence, but I am not sure how do you show that with out writing out a lot of terms and then noticing what I all ready know it is. How do you find the limit of a sequence to a number if your not given any numbers to work with?
     
  2. jcsd
  3. Sep 4, 2011 #2

    tiny-tim

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    Hi Punkyc7 ! :smile:

    If there is a limit, then you can assume that xn = xn-1 :wink:
     
  4. Sep 4, 2011 #3
    true, but how do you ever get a number when you are dealing with f[itex]_{n}[/itex] and f[itex]_{n+1}[/itex]. How can you just make a jump and say there is a [itex]\sqrt{5}[/itex] in there
     
  5. Sep 4, 2011 #4

    tiny-tim

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    quadratic equation? :wink:
     
  6. Sep 4, 2011 #5
    Ok im not sure how you got there but this is what I have so far


    let [itex]x_{n}[/itex] = [itex]f_{n+1}[/itex]/[itex]f_{n}[/itex] and let lim[itex](x_{n}[/itex])=L. From here is where I get stuck. I know that every sub sequence of a convergent sequence converges to the same number by some theorem, but I am not sure how that is at all helpful.

    Would you do something like
    [itex]x_{n}[/itex] = [itex]f_{n+1}[/itex]/[itex]f_{n}[/itex]
    [itex]x_{n-1}[/itex] = [itex]f_{n}[/itex]/[itex]f_{n-1}[/itex]
     
  7. Sep 4, 2011 #6

    tiny-tim

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    put xn = xn-1
     
  8. Sep 4, 2011 #7
    Ok so you get

    = [itex]f_{n}[/itex]/[itex]f_{n-1}[/itex] = [itex]f_{n+1}[/itex]/[itex]f_{n}[/itex]

    I think I see where you got the quadratic equation now

    [itex]f_{n}[/itex] ^2= [itex]f_{n-1}[/itex] [itex]f_{n+1}[/itex]=[itex]f_{n}[/itex] ^2 - [itex]f_{n-1}([/itex] [itex]f_{n+1}[/itex])


    to use the quadratic equation is this [itex]f_{n-1}([/itex] [itex]f_{n+1}[/itex]) considered b or c?

    and that doesnt look very pretty to solve...
     
    Last edited: Sep 4, 2011
  9. Sep 4, 2011 #8

    tiny-tim

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    erm :redface:

    you'll also need fn+1 = … ? :wink:
     
  10. Sep 4, 2011 #9
    Ok

    [itex]f_{n+1}[/itex]=[itex]\frac{f_{n}}{f_{n-1}}[/itex]


    [itex]f_{n}[/itex] ^2 - f[itex]_{n-1}[/itex]* [itex]\frac{f_{n}}{f_{n-1}}[/itex]

    [itex]f_{n}[/itex] ^2 -[itex]f_{n}[/itex]=0

    is that right?
     
  11. Sep 4, 2011 #10

    tiny-tim

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    This is the Fibonacci sequence!

    So fn+1 = ? :smile:
     
  12. Sep 4, 2011 #11
    oh so
    f[itex]_{n+1}[/itex]= f[itex]_{n}[/itex] +f[itex]_{n-1}[/itex]


    do I use that for f[itex]_{n+1}[/itex]
    Also how do you know when to use what?
     
  13. Sep 4, 2011 #12

    tiny-tim

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    yes … that should give you your quadratic equation :wink:
    You're told it's a Fibonacci sequence, so you obviously have to use that information somewhere! :smile:

    and now i'm off to bed :zzz:
     
  14. Sep 4, 2011 #13
    let [itex]x_{n}[/itex] = [itex]f_{n+1}[/itex]/[itex]f_{n}[/itex] and let lim[itex](x_{n}[/itex])=L. Since we know the sequence converges we can say

    [itex]x_{n}[/itex] =[itex]x_{n-1}[/itex] Which Implies

    = [itex]f_{n}[/itex]/[itex]f_{n-1}[/itex] = [itex]f_{n+1}[/itex]/[itex]f_{n}[/itex]

    [itex]f_{n}[/itex] ^2= [itex]f_{n-1}[/itex] [itex]f_{n+1}[/itex]=[itex]f_{n}[/itex] ^2 - [itex]f_{n-1}([/itex] [itex]f_{n+1}[/itex])=0

    [itex]f_{n}[/itex] ^2 - f[itex]_{n}[/itex]f[itex]_{n-1}[/itex]-f[itex]_{n-1}[/itex]^2=0


    How do you hammer this into the quadratic equation I am thinking the a=1 b=not sure c=not sure ? Also how do you get numbers from this when we don't have a single number?
     
    Last edited: Sep 4, 2011
  15. Sep 4, 2011 #14

    eumyang

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    ...which also equals L:

    [itex]\frac{f_n}{f_{n-1}} = \frac{f_{n+1}}{f_n} = L[/itex]

    Now take this portion:
    [itex]\frac{f_{n+1}}{f_n} = L[/itex]

    Replace the numerator with its equivalent, and then rewrite as a sum of two fractions. A substitution can be made, and you will end up with an expression on the left side with NO f's. Soon you will see a quadratic equation in terms of L. Solve for L.
     
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