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Continued Fractions

  1. Apr 28, 2012 #1
    Suppose we can write a real number x as a continued fraction like this

    x=a0+1/(a1+1/(a2+1/(a2+...=[a0; a1, a2, a3, a4, ... an...].

    Is there a binary operation f(i,x) so that f(i,x)=ai? I was wondering if there was a formula which gives the ith item in the sequence of integers which is connected to every x in the context of this expansion.

    Every rational number has a unique continued fraction expansion so I think this is a valid question. Moreover, every irrational number has a unique, infinite continued fraction expansion.

    My first guess was to combine the inputs a, b in the Euclidean algorithm from which the continued fraction expansion arises but I don't know how to extract the ith item in the sequence of quotients. Any thoughts?
     
  2. jcsd
  3. Apr 29, 2012 #2
  4. May 1, 2012 #3
    Thanks a lot for answering my question. It's a big step in the right direction.

    My question was how to compute the continued fraction expansion of a fraction. For example, using the Euclidean algorithm, we have
    [tex]
    \frac{7}{10}=0+\cfrac{1}{1+\cfrac{1}{2+\cfrac{1}{3}}}
    [/tex]
    Therefore, $$a_0=0, a_1=1, a_2=3, a_3=3, a_4=0, a_5=0, a_6=0,\dots$$

    Is there a way to compute $$a_n$$ without using the euclidean algorithm?
     
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