# Continued Fractions

1. Feb 26, 2007

### aznkangaroo3

1. The problem statement, all variables and given/known data

Give the formula in terms of tn+1 for the continued fraction:

and so on....
2. Relevant equations

3. The attempt at a solution
I got the recursive formula:tn+1=1+(1/tn), but I need the explicit formula

2. Feb 26, 2007

### Tedjn

I am not brilliant at math, so I don't know if there is a systematic way to solve these problems. You know that $$t_1 = 1$$, so when you calculate $$t_2$$ and so on, you would plug 1 in for $$t_1$$. Don't. Treat all cases of $$t_1$$ as if it were a symbol from another planet. Write out a few terms in terms of $$t_1$$ and see if you can find a pattern.

3. Feb 26, 2007

### Dick

Ok. After much confusion, I'm finally getting this straight. If you start writing out your t_n's you'll see that you are getting ratios of consecutive fibonacci numbers. This suggests that you look for solutions of the form t_n=F_(n+1)/F_n. Put this in and find a recurrence relation for the F's. (It should look familiar). Now you want to look for solutions to this recurrence relation of the form C*a^n. Using your recurrence relation, what value(s) can 'a' have? Can you add two solutions of this form? Do you have enough constants to reproduce t1=1, t2=2?

Last edited: Feb 27, 2007
4. Mar 2, 2007

### tim_lou

ha, yeah, you got the right idea. You need to solve a recursion.

actually, let me tell you something about a more general recursion:
$$x_{n+1}=\frac{a+bx_n}{c+dx_n}$$

How would you solve it?
well our goal is add constants to both sides to make
$$x_{n+1}+\alpha=a'\frac{1+\alpha x_n}{c+dx_n}$$
and
$$x_{n+1}+\beta=b'\frac{1+\beta x_n}{c+dx_n}$$

so that when I divide the first equation by the second equation, I'll get:
$$\frac{x_{n+1}+\alpha}{x_{n+1}+\beta}=\frac{a'}{b'}\left (\frac{1+\alpha x_n}{1+\beta x_n}\right )$$

a' and b' can be expressed in a,b,c,d, and alpha and beta respectively. Basically, all you need to do is find two solutions to a quadratic equation, if there are two distinct roots, then you are in business. The original problem can be solved using the exact method.

and of course, you can just guess and check and prove that the formula works by induction, and it will turn out to be some Fibonacci numbers...

Last edited: Mar 2, 2007