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Puzzle: one over one plus … harder and easier variations

April 2, 2007 pm30 6:59 pm

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:

  1. different methods of solution
  2. Questions about the problem.

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….

6 Comments leave one →
  1. Edgardo permalink
    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?

  3. Edgardo permalink
    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.
    Instead of just asking:
    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)

  4. Clueless permalink
    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…….

  5. Clueless permalink
    April 4, 2007 pm30 6:16 pm 6:16 pm

    A somwhat more interesting problem:

    What is the maximum value of a for which a^a^a^a…. is finite?

  6. April 4, 2007 pm30 9:09 pm 9:09 pm

    In this case, the easier problems are easy to find: just terminate the fraction, putting some n under the last bar. The more levels, the more challenge for the students—and you can replace some of the 1’s with other numbers, too. We did several of these with my Math Counts (6-8th grade) class this spring. It’s good practice for kids who are just getting comfortable with fractions and multi-step calculations.

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