# Continued Fraction problem

1. Oct 31, 2007

### rbzima

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!

2. Oct 31, 2007

### Staff: Mentor

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

$$a_{0}+\frac{a_{1}}{1-\frac{a_{2}}{1+a_{2}-\frac{a_{3}}{1+a_{3}-...}}}} = a_1+a_1a_2 + a_1a_2a_3 + a_1a_2a_3a_4 +\cdots$$

Rewriting the RHS as

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

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.

3. Oct 31, 2007

### rbzima

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.