# Continued fractions using recursion relations

Let $$q_0, q_1, q_2$$, . . . be the terms of a continued fraction. Use the recursion relations to determine the continued fraction (as a rational number) for each of the following terms:

$$(a) 1, 2, 3, 4$$

$$(b) 3, 3, 3$$

$$(c) 3, 3, 3, 3, 3, 3$$

The recursion relations are:

$$A_n = A_{n-2}+q_n.A_{n-1}$$, $$A_0 = q_0$$

$$B_n = B_{n-2}+q_n.B_{n-1}$$, $$B_0 = 1$$

So for part a):

I got $$A_0 = 1, A_1 = 3, A_2 = 10, A_3 = 43$$ and $$B_0 = 1, B_1 = 3, B_2 = 10, B_3=43$$. So the continued fraction is $$\frac{43}{43}$$. Is this correct? And this is how I should do the other parts?

Not quite. The recursive formulas require the previous two terms, and so you need two base cases to start them off. (In other words: how did you decide that $$A_1=3$$ and $$B_1=3$$ given the data in your post?)
Here, in addition to $$A_0=q_0$$ and $$B_0=1$$, you can use $$A_{-1} = 1$$ and $$B_{-1}=0$$. This leads in part (a) to $$A_0 = 1, A_1 = 3, A_2 = 10, A_3 = 43$$ as you obtained, but instead to $$B_0 = 1, B_1 = 2, B_2 = 7, B_3=30$$, so that the fraction is $$\frac{43}{30}$$.
You can check that this is the correct answer by carrying out the Euclidean algorithm on $$43$$ and $$30$$ and recording the partial quotients; they are indeed 1, 2, 3, 4.
• Is $B_{-1}$ always equal to $0$? I mean, is it the same in parts b) and c)? – cbc bc Sep 20 at 20:34
• Yes, just like $B_0$ is always equal to $1$. – Greg Martin Sep 20 at 20:35