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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

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    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

    [tex]a_1(1+a_2(1+a_3(1+a_4(\cdots[/tex]

    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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