Explaining the Continued Fraction Identity for arctan(x)

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SUMMARY

The discussion focuses on the equivalence of the continued fraction identity for arctan(x) and its multiplicative form. The identity is expressed as a_{0} + a_{1} + a_{1}a_{2} + a_{1}a_{2}a_{3} + ... being equivalent to a_{0} + \frac{a_{1}}{1 - \frac{a_{2}}{1 + a_{2} - \frac{a_{3}}{1 + a_{3} - ...}}}. The participants emphasize that the Taylor series for arctan(x) converges slowly, while the continued fraction converges significantly faster. The discussion also suggests defining the nth convergent by setting a_{n+1}, a_{n+2}, etc., to zero for a rigorous understanding.

PREREQUISITES
  • Understanding of continued fractions
  • Familiarity with Taylor series, specifically for arctan(x)
  • Knowledge of recursive definitions in mathematical analysis
  • Basic algebraic manipulation skills
NEXT STEPS
  • Study the properties of continued fractions in mathematical analysis
  • Explore the convergence rates of Taylor series versus continued fractions
  • Learn about Machin's formula for calculating π using arctan
  • Investigate recursive sequences and their applications in convergents
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Mathematicians, students studying advanced calculus, and anyone interested in the convergence properties of series and continued fractions.

rbzima
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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!
 
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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}-...}}}} =<br /> 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.
 
D H said:
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}-...}}}} =<br /> 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.

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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