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### A Monster of a Continued Fraction

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Date: 11/09/96 at 00:36:11
From: Patrick Beebe
Subject: How to solve an infinitely repetitious problem

I'm curious about solving a specific problem and others similar to it.

Given the forever repeating fraction shown below, what is the
solution?

1 + ...
1 + --------
3 + ...
1 + -------------
5 + ...
3 + --------
1 + ...
1 + ------------------
1 + ...
5 + --------
3 + ...
3 + -------------
5 + ...
1 + --------
1 + ...
1 + -----------------------
1 + ...
1 + --------
3 + ...
5 + -------------
5 + ...
3 + --------
1 + ...
3 + ------------------
1 + ...
5 + --------
3 + ...
1 + -------------
5 + ...
1 + --------
1 + ...

It is an absolute monster!  As I see it, there are two distinct
infinite patterns within it: the 1 + 1/3... pattern, and the
1 + 5/1... pattern.  I have tried substituting these patterns with
x and y, and then attempting to solve from there, but to no avail.
I am still stuck.  Is there another way to solve these sorts of
problems?

Thanks!
```

```
Date: 11/17/96 at 23:15:01
From: Doctor Sam
Subject: Re: How to solve an infinitely repititious problem

Patrick,

You are correct...this IS a monster. I have seen continued fractions
before, but never such a combination. I have a possible solution but
I would like someone else to verify it so that we could both be
confident that it is correct!

I liked your idea of substitution.  Here are the ones I made:

1 + ...   <<.  <<<<<<< .
1 + --------      .    .
3 + ...       .    .
1 + -------------     A    .
5 + ...      .    .
3 + --------      .    .
1 + ...  <<.    .
1 + ------------------     A
1 + ...  <<.    .
5 + --------      .    .
3 + ...      .    .
3 + -------------      B   .
5 + ...      .    .
1 + --------      .    .
1 + ...  <<.  <<<<<<<< .
1 + -----------------------
1 + ...  <<.  <<<<<<<< .
1 + --------      .    .
3 + ...      .    .
5 + -------------      X   .
5 + ...      .    .
3 + --------      .    .
1 + ...  << .   .
3 + ------------------     B
1 + ...  <<.    .
5 + --------      .    .
3 + ...      .    .
1 + -------------      Y   .
5 + ...      .    .
1 + --------      .    .
1 + ...  <<. <<<<<<<< .

So I see the numerator of this expression two ways.  It is both A and
1 + A/B.  This gives one condition on A and B.  The trick is to see
the denominator of this monster also in terms of A and B.

The entire denominator is just B.  But what are X and Y?  I can't be
sure of what all of your "..." stand for but I *think* that X is
4 + A.  (Instead of thinking of the numerator of X as 5 + ... I am
thinking of it as 4 + 1 + ...  and that makes it 4+A.)

Similarly, I think the denominator, Y, of the monster is B - 2.
(Think of the 3 + ... as -2 + 5 + ... ).

If I am correct in this interpretation, then the denomintor is both B
and 3 + (4+A)/(B-2).  This gives a second condition on A and B.  We
now have to solve the simultaneous equations:

A = 1 + A/B    and    B = 3 + (4+A)/(B-2)

I get a cubic equation in B with a single real solution of B
approximately equal to 4.84732.

Since A = 1 + A/B,  A = B/(B-1) = 1.25992

And A is also the value of the entire monster.

I think.

I hope that helps!

-Doctor Sam,  The Math Forum
Check out our web site!  http://mathforum.org/dr.math/
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Associated Topics:
High School Sequences, Series

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