1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Continued fractions

  1. Jul 21, 2007 #1
    Hi, my question is given the recurrence relation for the convergents, could we construct a continued fraction so..

    [tex] \alpha = a_{0}+ \frac{b_{0}}{a_{1}+\frac{b_{1}}{a_{2}}+.... [/tex]

    all the coefficients a's and b's are equal to a certain integer ?

    for example if all the coefficients (numerators and denomiators)

    * are one we have just the Fibonacci (Golden ratio) constant [tex] \frac{2}{\sqrt 5 -1} [/tex]

    * are two we have exactly [tex] \sqrt 2 +1 [/tex]

    i the sense that expanding the 2 numbers above their continued fraction is made only by 1 or 2, but can we construct a general continued fraction with all the numbers equal to 3,4,5,6,......
     
  2. jcsd
  3. Jul 26, 2007 #2
    Something doesn't seem right. If making all the a's and b's 1 gives the Fibonacci (golden ratio) constant, wouldn't making all the a's and b's 2 simply be 2 times the Golden ration rather than [tex] \sqrt 2 + 1 [/tex].

    Postscript. On the other hand: It seemed to me that any similar manner of construction a continued fraction will give an irrational number of some fixed value which would be a multiple of the continued fraction comprising only ones, but then n/n = 1 so I guess I made a mistake. Anyway at least the numbers are completely predefined.
     
    Last edited: Jul 26, 2007
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Continued fractions
  1. Continued fractions (Replies: 4)

  2. Continued fractions (Replies: 3)

Loading...