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Continued fractions

  1. Jun 29, 2006 #1
    I am stuck how to determine a general formula for tn+1 in terms of tn given that the 'infinite fraction' as a sequence of terms tn is :

    t1 = 1+1

    t2 = 1 + __1__
    1+1

    t3 = 1 + ___1___
    1 + __1__
    1+1

    It involves the fibonacci sequence.
     
  2. jcsd
  3. Jun 29, 2006 #2

    Hurkyl

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    If you can't figure it out from the continued fraction, then write it in a different form!
     
  4. Jun 29, 2006 #3
    the generalized fomula for tn+1 will have to be :

    1 + _1_
    tn

    One of the fibonacci formulas that I think should be used is

    F_n=((1+sqrt(5))^n-(1-sqrt(5))^n)/(2^nsqrt(5))

    but I am not sure how to combine the two to get a generalized formula for tn+1 in terms of tn?
     
  5. Jun 29, 2006 #4

    Hurkyl

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    Don't worry about trying to apply a "Fibonacci formula" -- if you can do such a thing, it will be clear what to do when you can do it. Until then, just work the problem.


    I'm sort of confused by the rest of your statement. You said you know that the general formula is:

    [tex]
    t_{n+1} = 1 + \frac{1}{t_n}
    [/tex]

    So I'm confused when you say you don't know how to get a formula for [itex]t_{n+1}[/itex] in terms of [itex]t_n[/itex].


    P.S. if you don't want to use latex, you can use [ code ] tags to format your expressions. e.g.
    Code (Text):

    1 + _1_
        tn
     
     
  6. Jun 29, 2006 #5
    Im not sure if that is the right formula or not :

    [tex]t_{n+1} = 1 + \frac{1}{t_n}[/tex]
     
  7. Jun 29, 2006 #6

    Hurkyl

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    Well, I think we both agree that it looks like sequence you've been given. Unless you were given that sequence in a rigorous way, we'll have to settle for "looks like".

    Wait a sec -- this is a thread on continued fractions. I suppose you mean to look at the continued fraction [1;1, 1, 1, ...], and the [itex]t_n[/itex] are the partial convergents? If so, then take the definition of partial convergences, and see if you can prove they satisfy the recursion you gave!
     
  8. Jun 29, 2006 #7

    StatusX

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    You can solve for the limit of this series L by setting noting that L=1+1/L. (do you see why?). This should give a familiar result related to the Fibonacci sequence which should give you a clue as to the relation between the nth terms of both series.
     
  9. Jun 29, 2006 #8
    Im not too sure i follow your L=1+1/L. could you explain it a bit more please
     
  10. Jun 29, 2006 #9

    StatusX

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    Well, if the series converges to a limit, tnand tn+1 must get very close to each other (and the limit) as n gets very large, and so you can constrain what this limit may be by letting tn=tn+1 and solving the resulting equation. Of course, this doesn't always determine the limit uniquely, nor does it guarantee one exists. But if there is a limit, it must be one of the solutions to this equation. Assuming you know this series does converge, you can get an idea which solution is the correct limit by seeing what the first few terms of the series look like.
     
    Last edited: Jun 29, 2006
  11. Jun 29, 2006 #10
    But the series is an 'infinite fraction' what exactly does that mean?
     
  12. Jun 29, 2006 #11

    StatusX

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    It doesn't really mean anything. If you just think of each tn as a number, and that you get the next term in the series by tn+1 =1+1/tn, then you don't have to worry about gigagntic mosnter nested fractions. For example, t1 =1+1=2, t2 =1+1/2=3/2, t3 =1+1/(3/2)=1+2/3=5/3, etc. These are the convergents that Hurkyl was talking about before.
     
  13. Jun 29, 2006 #12
    Im sure I heard my teacher mention using a fibonacci formula and phi etc, any ideas what to come up with?
     
  14. Jun 29, 2006 #13

    StatusX

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    I already mentioned one way to find the connection. Another would be to write out the first few terms of the series, as I started in the last post, and see if that gives you any ideas.
     
  15. Jun 29, 2006 #14

    Hurkyl

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    He's jumping ahead! :smile:

    Suppose you are right -- that [itex]t_{n+1} = 1 + 1/t_n[/itex] is a correct recursion. If you assume [itex]L = \lim_{n \rightarrow+\infty} t_n[/itex], then what happens if you take the limit of both sides of your recursion?
     
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