Calculating Value of Continued Fraction

[Hypercase]

hi u guys .
I was hoping u could show me how to calculate the value of a continued fraction.
1/(6+(1/(6+1/...))

 

[outandbeyond2004]

Hmmm ...
$$\frac{1}{(6 + \frac{1}{(6 + \frac{1}{(...)})})}$$

Hmm...

 

[matt grime]

let x be the fraction. what's 1/x?

 

[Bob3141592]

Originally posted by Hypercase
hi u guys .
I was hoping u could show me how to calculate the value of a continued fraction.
1/(6+(1/(6+1/...))

I love continued fraction, they're just so cool.

This one is easy to solve, since it's a simple continued fraction (all numerators are 1) and the denominator doesn't vary. That gives us a similarity in the continued fraction that we can use to our advantage.

To write the equation out graphically:

$$X = \frac{1}{6 + \frac{1}{6 + \frac{1}{6 + \frac{1}{6 + \frac{1}{6 + ...}}}}}$$

Let's use the variable X to represent the continued fraction in the denominator, that is

$$X = \frac{1}{(6 + X)}$$

See, those are equivalent statements, and this second form is easy to solve with basic algebra, yielding $X^2 + 6X - 1 = 0$ which gives X a numeric value of $\sqrt{10} - 3$, or in decimal form .162277660168... which is just a little bit less than $\frac{1}{6}$ as we'd expect.

Note that we throw away the negative root because the result has to be positive.

See? That's all it takes.

 

[Bob3141592]

This topic of continued fractions got me thinking, and I have a two part question about them. The first is “simple” and the second might be “hard” At least, hard for me to get a handle on.

First, some preliminaries.

The approach of self-similarity can be used to solve for any simple continued fractions with repeating patterns. For simplicity of notation, instead of the graphical form

$$X = n + \frac{1}{a + \frac{1}{b + \frac{1}{c + \frac{1}{d + \frac{1}{e + ...}}}}}$$

we’ll use the standard text notation for simple continued fractions of (n;a,b,c,d,e,...)

where the first number before the semicolon can be any value and the numbers after it are non-negative integers).

From the original post we considered the simplest case of X = (0;6,6,6,6,6,…), and we got $\sqrt{10}$ – 3 as a result. In general, if we use (n;a,…) (how can I use Tex to put a bar above the repeating values?) we get a general solution of

$$(Eq 1) X = n + \frac{-a}{2} + \frac{\sqrt{a^2 + 4}}{2}$$

by a fairly simple application of algebra.

The same approach with a little more algebraic manipulation applied to (n;a,b,…) yields

$$(Eq 2) X = n + \frac{(-b)}{2} + \frac{\sqrt{(a^2b^2 + 4ab)}}{2a}$$

A quick check shows that Eq 2 does degenerate into Eq 1 when b = a.

Going to (n;a,b,c…) yields the rather ugly looking

$$(Eq 3) X = n + \frac{b-a-c-abc}{2(ab + 1)} +$$

$$\frac{\sqrt{a^2b^2c^2 + 2a^2bc + 2ab^2c + 2abc^2 + a^2 + b^2 + c^2 + 2ab + 2ac + 2bc + 4}}{2(ab + 1)}$$

(My browser truncates that last expression. If yours does too, the terms under the root go "2ac + 2bc + 4")

It's ugly, but there is a certain rhythm to it. Again, checking, Eq 3 degenerates into Eq 1 when c = b = a. Unfortunately, setting c = b or b = a doesn't help us any, since the repeat still takes three steps, so Eq 3 doesn't degenerate into Eq 2.

I leave it to others with more patience to come up with the general solutions for (n;a,b,c,d…) and (n;a,b,c,d,e…) and so on. To simplify the notation, we can rewrite these continued fractions using subscripted variables instead of sequential letters, such as

$$(n;a_0,a_1,a_2,a_3,a_4, a_5...)$$

I don’t know the normal terminology for this, but we’ll say Eq 1 shows a first order repetition, Eq 2 a second order repetition, and so on. Then a repetition is of order k when $a_{i+k} = a_i$. I don’t see a pattern to it as we go from one order of repetition to the next, but I suspect there must be one. I do note, however, that for any repeating continued fraction, no matter what the order is, the solution involves at most a square root and a constant. Higher orders of repetitions do not make for higher order roots. So, the first part of my question is this: Is there a general form by which we can generate the solution for a continued fraction of an “n-th order repetition”? I’m betting the answer must be yes, and probably involves some variant of binomial expansion, but it’s messy enough to elude me. If anyone can shed some light on this, I’d really appreciate it.

Now, here’s my hard question. I understand there is no analytic method for solving continued fractions in general. And yet, any non-repeating continued fraction can be approximated by a repeating continued fraction, and the approximation should get better for increasing orders. Does this imply there might be some method, maybe related to or just analagous to Fourier transforms, where the shape of a continued fraction could be expressed in a different form that is amenable to analytic solution? Would this give a handle on an otherwise intractable problem? Wouldn't that be neat?