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Continued Fraction problem

  1. Oct 31, 2007 #1
    I'm having trouble understanding a simple identity and was wondering if anyone could explain it to me:

    Why is it that [tex]a_{o}+a_{1}+a_{1}a_{2}+a_{1}a_{2}a_{3}+a_{1}a_{2}a_{3}a_{4}...[/tex] is equivalent to the continued fraction in the form:[tex]a_{0}+\frac{a_{1}}{1-\frac{a_{2}}{1+a_{2}-\frac{a_{3}}{1+a_{3}-...}}}}[/tex]

    What then should I do to make arctan(x) look something like the above continued fraction. Any advice would be fantastic!
  2. jcsd
  3. Oct 31, 2007 #2

    D H

    Staff: Mentor

    The a0 term is obvious. Just focus on the continued fraction itself:

    [tex]a_{0}+\frac{a_{1}}{1-\frac{a_{2}}{1+a_{2}-\frac{a_{3}}{1+a_{3}-...}}}} =
    a_1+a_1a_2 + a_1a_2a_3 + a_1a_2a_3a_4 +\cdots[/tex]

    Rewriting the RHS as


    might help you see how this identity falls out.

    To do this rigorously, look at the identity recursively by defining the nth convergent that results by setting [itex]a_{n+1}, a_{n+2}, \cdots[/itex] to zero.

    I have to run off to work, so I can't help much more. Final note: The Taylor series for arctan(x) converges very slowly. The convergence of the continued fraction is much, much faster.
  4. Oct 31, 2007 #3
    Thanks bro! I never saw the multiplicative form before, so that helps a lot. BTW, I figured that the Taylor series for arctan(x) was a slower convergence. When you use Machin's formula for [tex]\frac{\pi}{4}[/tex], it requires 71 terms until you reach convergence with 100 decimal place accuracy.
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