Challenge #1

Posted on March 8, 2006 by Brent

What better way to start things off than a challenge? Challenging problems are the golden road to discovering mathematical beauty: they spark interest, invite exploration, require thought and creativity, and usually beg to be tweaked, generalized, and otherwise turned into avenues for further exploration. And beauty discovered for oneself is the best sort.

So, enough rambling, on to the challenge already! It is this: can you find a value for the following expression?

\displaystyle 1 + \cfrac{1}{1 + \cfrac{1}{1 + \cfrac{1}{1 + \ddots}}}

The dots (\ddots) mean that the pattern continues forever. If you have seen this sort of problem before then it will not be too hard; but if not, the solution is not obvious and requires a certain creative leap. It does not, however, require any math more advanced than basic algebra.

(And no, the answer is not “infinity”, even though that is the value of the very similar-looking expression 1 + 1 + 1 + 1 + \dots)

Feel free to comment with guesses, questions, solutions, or whatever.

About Brent

Associate Professor of Computer Science at Hendrix College. Functional programmer, mathematician, teacher, pianist, follower of Jesus.
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8 Responses to Challenge #1

  1. jianxiu chen says:
    March 8, 2006 at 7:55 pm

    is the answer 1 + (square root of 5 – 1 )/ 2
    which is about 2.736067977 …
    is that answer right ?

  2. Brent says:
    March 8, 2006 at 8:28 pm

    close! but not quite…

  3. Anonymous says:
    March 9, 2006 at 7:42 pm

    the answer is phi, or 1.618, or (1 +/- sqrt(5))/2

  4. Ricky says:
    March 15, 2006 at 9:33 pm

    Well, I think I’m sure that it’s between 1 and 2. how’s that for a start?

  5. Brent says:
    March 16, 2006 at 7:28 am

    yup, that’s a good start!

  6. Ricky says:
    March 16, 2006 at 11:09 am

    So . . . how about a hint on this “creative leap”?

  7. Brent says:
    March 16, 2006 at 12:28 pm

    notice that the expression is self-similar — that is, it contains copies of itself as subparts.

  8. Pingback: Challenge #1 Solution « The Math Less Traveled

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