Challenge #2 Solution

And here are the solutions to Challenge #2…

Using the same technique from the solution to Challenge #1, we can reduce the first continued fraction to the equation

$x = 2 + \frac{1}{x}.$

Multiplying through by x and moving everything to one side yields the quadratic equation $x^2 - 2x - 1 = 0$ , which can be solved by the quadratic formula to yield $x = 1 \pm \sqrt{2}$ ; we want the positive solution, $x = 1 + \sqrt{2} \approx 2.41421\dots$ (Note: if I’ve skipped too many steps and you can’t see how to work this out for yourself, just ask and I would be happy to explain in more detail…)

The second problem is a little trickier, since replacing a subexpression by x yields

$x = 1 + \cfrac{1}{3 + \cfrac{1}{x}},$

which means we need to dust off our Fraction Skills. We’ll, let’s see: starting with $3 + \frac{1}{x}$ , we need to find a common denominator; x will do nicely, so we have $3 + \frac{1}{x} = \frac{3x}{x} + \frac{1}{x} = \frac{3x + 1}{x}$ . Now we recall that dividing by a fraction is the same as multiplying by its reciprocal, so we now have

$x = 1 + \frac{x}{3x + 1}$

At this point we can just multiply both sides by $3x + 1$ to get rid of the fraction. After collecting like terms and moving everything to one side we end up with the quadratic equation $3x^2 - 3x - 1 = 0$ ; solving yields $x = \frac{3 \pm \sqrt{21}}{6}$ . (Again: try working this out for yourself!)

The third problem yields the equation $x = \sqrt{1 + x}$ ; squaring both sides gives us the quadratic equation

$x^2 - x - 1 = 0$ .

And then we… wait a minute… there’s something familiar about that equation…

If you enjoyed these problems and want another (slightly more difficult) one in the same vein, try this one proposed by Matt Spencer: what is the value of

$\displaystyle \sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\dots}} }$ ?

About Brent

Associate Professor of Computer Science at Hendrix College. Functional programmer, mathematician, teacher, pianist, follower of Jesus.

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4 Responses to Challenge #2 Solution

Laura says:

April 9, 2006 at 9:22 am

For the Matt Spencer problem, I get x = 2.
Laura says:

April 27, 2006 at 9:31 pm

Hey, I was procrastinating on the Mudd Math Fun Facts page and happened across this article about the Matt Spencer problem: http://www.math.hmc.edu/funfacts/ffiles/20007.3.shtml. Looks like it’s not as straightforward as I assumed. There’s an interesting article about it in JSTOR, linked from the Mudd page.
Brent says:

April 28, 2006 at 10:32 am

Cool, thanks for the link. Another way to think about it is that 2 and 4 are both fixed points of the function $f(x) = \sqrt{2}^x$ , but 2 is an attractive fixed point and 4 is repellent. More on this in a future post, hopefully…
Brent says:

October 5, 2006 at 5:58 pm

[This comment was made in reply to another comment which has been deleted; I’ve retained it here since others may find it interesting/informative in and of itself.]

Hmm… I guess the best way to put it is this: if you keep evaluating more and more terms of the infinite continued fraction (2, then 2 + 1/2, then $2 + \cfrac{1}{2 + \cfrac{1}{2}}$ , and so on), the sequence of values you get will forever be getting closer and closer to some particular value. That value is x. I’ve written more about this in another post, too.

To calculate the 200th term would definitely be quite tedious by hand, although it would only involve simple multiplication and addition (of very large numbers!). Using a computer, however, it is easy to write a program which can calculate the 200th term very quickly. In fact, the 200th term is 30645573943232956180057972969833245887630954508753693529117371074705767728665/12693812353994620481037986488739368440399451028645237163046012909971924256728 (that’s a fraction… notice the slash buried in the middle there!), which is about equal to 2.41421356237…, already very close to $1 + \sqrt{2}$ .

Does that help?