Solving Continued Fractions

Date: 08/08/98 at 18:43:53
From: kristy
Subject: Solving continued fractions

I've read the archives and determined that the problem in my review 
packet is in fact a continued fraction. I also know, from reading the 
archives, that it is equal to sqrt(2).  My question is how in the 
world do you get sqrt(2) from:

        1
   ----------------
   2 +  1
       ------------
       2 +  1
           --------
           2 + ....

Any help on your part would be greatly appreciated.

Date: 08/09/98 at 20:15:05
From: Doctor Pete
Subject: Re: Solving continued fractions

Hi,

The key in solving continued fractions is to use a bit of algebra.  
Say the above continued fraction is x. Then notice that:

   x = 1/(2 + x)

This is because the continued fraction contains a copy of itself. So 
solving this equation for x, we find:

   x(2+x) = 1

or x^2 + 2x - 1 = 0. The quadratic formula gives:

   x = (-2 + Sqrt[8])/2 or (-2 - Sqrt[8])/2

Thus:
 
   x = Sqrt[2] - 1 or -Sqrt[2] - 1

Since the second solution is negative, and it is obvious that the 
continued fraction is positive, it follows that x = Sqrt[2] - 1.

Another example:

              1
   1 + ---------------
                1
       2 + -----------
                  1
           1 + -------
                    1
               2 + ---
                    1 ...

Here, the 1's and 2's alternate. Then we see that:

              1
   x = 1 + -------
               1
          2 + --- 
               x

Again solving for x, we find:

   x = (3x+1)/(2x+1)
   2x^2 - 2x - 1 = 0
  
Thus:

   x = (1 + Sqrt[3])/2 or (1 - Sqrt[3])/2

and discarding the negative root gives x = (1 + Sqrt[3])/2.

- Doctor Pete, The Math Forum
Check out our web site! http://mathforum.org/dr.math/