Explaining the Continued Fraction Identity for arctan(x)

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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 a_{o}+a_{1}+a_{1}a_{2}+a_{1}a_{2}a_{3}+a_{1}a_{2}a_{3}a_{4}... is equivalent to the continued fraction in the form:a_{0}+\frac{a_{1}}{1-\frac{a_{2}}{1+a_{2}-\frac{a_{3}}{1+a_{3}-...}}}}

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:

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

Rewriting the RHS as

a_1(1+a_2(1+a_3(1+a_4(\cdots

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 a_{n+1}, a_{n+2}, \cdots 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:

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

Rewriting the RHS as

a_1(1+a_2(1+a_3(1+a_4(\cdots

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 a_{n+1}, a_{n+2}, \cdots 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 \frac{\pi}{4}, it requires 71 terms until you reach convergence with 100 decimal place accuracy.
 
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