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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?

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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.

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  • $\begingroup$ Is $B_{-1}$ always equal to $0$? I mean, is it the same in parts b) and c)? $\endgroup$
    – cbc bc
    Commented Sep 20, 2019 at 20:34
  • $\begingroup$ Yes, just like $B_0$ is always equal to $1$. $\endgroup$ Commented Sep 20, 2019 at 20:35

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