This is the place to offer *easier and harder variations for the puzzle below:

What is the value of $\frac{1}{1+\frac{1}{1 + \frac{1}{1 + \dotsb}}}$ ?

A month ago Dave Marain ran a discussion of $\sqrt{2 + \sqrt{2 + \sqrt{2 + \dotsb}}}$ over here.

I know this one has been done many many times before. But he generated a nice discussion.

There are also places for:

What makes these two problems so similar?

*For my money, writing easier variations is the greatest challenge that I am offering. It’s easy to make something harder….

April 4, 2007 am30 12:19 am 12:19 am

A supplement could be:
Show that
$\frac{1}{1+\frac{1}{1 + \frac{1}{1 + \dotsb}}}$
is the reciprocal of
$1+\frac{1}{1+\frac{1}{1 + \frac{1}{1 + \dotsb}}}$

2. April 4, 2007 am30 12:48 am 12:48 am

How would you do that, without finding the values of each?

April 4, 2007 am30 4:06 am 4:06 am

I don’t know, lol.
I just thought of that supplement as a variation of the original problem.
What is the value of
$\frac{1}{1+\frac{1}{1 + \frac{1}{1 + \dotsb}}}$

Imagine giving this problem to a student.
That new problem is harder and could be confusing, at least it would have been for me, because

$\frac{1}{1+\frac{1}{1 + \frac{1}{1 + \dotsb}}}$
and
$\frac{1}{1+\frac{1}{1 + \frac{1}{1 + \dotsb}}}$

resemble each other and differ only by +1. I would have thought:
“Heh, how does X become the reciprocal of 1+X ?”

The student first has to understand the problem and recognize
that he must first evaluate the continued fraction.

But it could also be that the student’s first idea is to calculate:

$\left( \frac{1}{1+\frac{1}{1 + \frac{1}{1 + \dotsb}}} \right)^{-1}= 1+\frac{1}{1+\frac{1}{1 + \frac{1}{1 + \dotsb}}}$

He might think that he has to manipulate the left hand side in order
to get to the right hand side, without evaluating the continued fraction.

————————

Another variation:
If one is given the information “X is the reciprocal of 1+X” (where X is the continued fraction), one can solve the equation
$X = \frac{1}{1+X}$
and the solution is
$X = \frac{1}{2}(\sqrt{5}-1)$
(X>0)

April 4, 2007 pm30 6:13 pm 6:13 pm

Another nice variant I have seen:

a=sqrt(2)

Find a^a^a^a^a…….