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Continued fraction problem

  1. Nov 19, 2006 #1
    Hello, I was not sure if this belonged in the precalculus or calculus section so i hope no one minds that i posted it here.

    This is a problem that has completely lost me.


    T1 = 1 + 1

    T1 = 1+ (1/1+1)

    T3 = 1+ (1/(1+1/1+1)

    and so on

    I have been able to draw a generalized formula for Tn+1 in terms of Tn and that is:


    i realized that as n increases to infinity, (Tn+1) - Tn converges towards 0

    There i wrote Tn = Tn+1

    and inserting this into


    you get



    An exact value for the continued fraction will be considering Tn = x :

    x^2 - x - 1 = 0

    so the exact value is:





    For any value of k, I am supposed to determine a generalized statement for the exact value of any such continued fraction. For which values of k does this hold true and how do i know?

    Can anyone please help me with this, i got so far and do not want to give up. Any tips? anything.
  2. jcsd
  3. Nov 19, 2006 #2


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    You haven't finished the first problem!

    You've shown that if the limit exists, then it is either equal to [itex](1 + \sqrt{5}) / 2[/itex] or [itex](1 - \sqrt{5})/2[/itex]. You haven't shown that the limit does, in fact, exist... nor have you figured out which of those values it equals.

    As for the second problem... have you tried doing exactly the same thing that you did for the first problem?
  4. Nov 19, 2006 #3
    NO! dont tell me i did not finish the first problem... haha. So how can i show that the limit exists or which of the values it equals?
  5. Nov 19, 2006 #4


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    Have you actually looked (numerically) at the terms in your sequence? How would you describe them?
  6. Nov 19, 2006 #5
    well the sequence for the first 10 terms goes like this

    1/2, 3/2, 5/3, 8/5, 13/8, 21/13, 34/21, 55/35, 89/55, 144/89

    from what i can tell these number are converging towards a number and since the numerator is always greater than the denominator, the number they are converging towards must be > 1 .

    So that eliminates the [itex](1 - \sqrt{5}) / 2[/itex] solution to the problem and leaves [itex](1 + \sqrt{5}) / 2[/itex].

  7. Nov 19, 2006 #6


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    Importantly, it's a increasing sequence, right? And increasing, bounded sequences converge!

    You have a guess as to what the upper bound is -- so all you have to do is prove it starts below that upper bound, and then each time you iterate, the value increases, but stays below that upper bound!
  8. Nov 21, 2006 #7
    it looks to me as if the sequence is decreasing and moving towards the asymptote which i guess is [itex](1 + \sqrt{5}) / 2[/itex]
  9. Nov 21, 2006 #8


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    Er, wait. Silly mistake. It's an alternating sequence! I knew that increasing sounded wrong.

    So you just need to show the magnitude of the differences is a decreasing sequence.

    In fact, that's the essential property of a continued fraction: the terms in any such sequence (called convergents) form an alternating sequence whose differences are decreasing. Wikipedia says a lot about them.
  10. Nov 22, 2006 #9
    I did some calculations and found that the only numbers for k that make the fraction invalid are 0 and -1.

    What do you say?
  11. Nov 22, 2006 #10
    k=-1 has a well defined limit, while k=0 does not (it oscillates), although both will probably be considered invalid.
  12. Nov 22, 2006 #11


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    Okay, so you got that it works for positive integers k, which agrees with the usual theory, so that's good!

    The sequences is, indeed, not well defined for k=0. The first few convergents would be:

    so every other term is undefined.

    For k = -1, they are:
    -1 / 1
    0 / -1
    -1 / 2
    1 / -3
    -2 / 5
    3 / -8
    which appear to be converging to [itex]-((-1 + \sqrt{5}) / 2)^2[/itex].

    I'm pretty sure the theory in the Wikipedia link I gave will actually work for any domain (not just the integers). Well, I broke down and grinded it out! Let r and s be the two roots to:

    x^2 - kx - 1

    then you can rigorously prove that the n-th convergent is

    (r^(n+1) - s^(n+1)) / (r^n - s^n)

    If |r| = |s| (i.e. when k = 0), the denominator is zero, which gives you problems. Otherwise, assume |r| > |s|, (and so |r| > 1) so the n-th convergent is

    (r^(2n+1) - (-1)^n s) / (r^(2n) - (-1)^n)

    which converges to r.
  13. Nov 23, 2006 #12
    I dont quite think i understand this.

    As i see it,

    for k = -1

    1st term = -1 + 1 = 0
    2nd term = -1+1/-1+1 = undefined
    3rd term = undefined
    4th term undefined

    This happens since the bottom part is always -1 + 1 = 0 and you can't divide by 0.
    What you guys are saying is considering what I posted right? not any other continued fraction?

    What is this convergents and things you guys are talking about? Sorry, but I dont understand the theorems. This is the first time i see a continued fraction hehe. Maybe you could explain it a bit more? I need to understand this stuff for Monday...

  14. Nov 23, 2006 #13


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    Ah, that's the problem. I usually see the the continued fraction defined by this sequence:

    k + 1/k
    k + 1/(k + 1/k)
    k + 1/(k + 1/(k + 1/k)
  15. Nov 23, 2006 #14
    You could reach a compromise of

    In this way, the values for k=-1 become:
    [tex]-1,0,-2,[\tex]und[tex],-1.5,\to-1,-1.\overbar{6},\to-2,-1.6,\to-1.5,-1.625,\to-1.\overbar{6},...[/tex], tending towards [tex]-\frac{1+\sqrt5}{2}[/tex]
    And for k=0,
    Last edited: Nov 23, 2006
  16. Dec 9, 2006 #15
    so what is the general formula?? i need help quick!!

    and i dont understand how you guys went from x^2 - x - 1=0 to (root5 +1)/2

    help please
  17. Dec 17, 2006 #16
    The general formula is (K+sqrt((K^2)+4))/2 for k>0 and (k-sqrt((K^2)+4))/2 for k<0. Let x be the continued fraction.
    1=x^2 - xk
    0=x^2 - xk - 1
    Using the quadratic formula, x= (k+sqrt((k^2)+4))/2
    If k<0, x<0 so you subtract the sqrt instead of adding.
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