Advanced Algebra

Date: 09/23/97 at 19:42:53
From: Courtney Bailey
Subject: Advanced Algebra

Dear Dr. Math:

My advanced algebra teacher has given us this problem for extra 
credit. So far no one in the class has gotten it! Can you help?  
The problem looks like this:

1 + 1
     ______
      1 + 1
           ____
            1 + 1
                 _____
                  1 + 1    
                       ____
                       1 + 1
                           .
                             .
                               .

My teacher said it had something to do with the golden ratio. I turned 
in an answer of 1 because I thought the bigger the denominator got, 
the smaller the fraction and the fraction would eventually be = 0.  
He said that was wrong. I hope you can help me, not just with the 
answer but help me to understand the solution!

Thanks!
Courtney Bailey

Date: 09/23/97 at 20:16:04
From: Doctor Pete
Subject: Re: Advanced Algebra

Hi,

To see what this value is, call it x.  So

     x = 1+1/(1+1/(1+1/(1+1/1+...))).

Subtract 1 from both sides:

     x-1 = 1/(1+1/(1+1/(1+1/1+...))).

"Flip" the numerator and denominator of both sides (invert):

     1/(x-1) = 1+1/(1+1/(1+1/1+...))).

But notice that the righthand side is still x, because the fraction 
continues infinitely (compare with what we had written in the first 
line).  Therefore,

     1/(x-1) = x,

or x(x-1) = 1. From here, it is easy to solve for x, and you will see 
why the hint was "the golden ratio."

For another related problem, what is the value of

     1+Sqrt[1+Sqrt[1+Sqrt[1+Sqrt[1+...]]]]?

(Sqrt[x] means the square root of x.)  Hint: Use the same method as in 
the previous problem.  If you find the correct answer, prepare to be 
surprised!

Now, why is it that your reasoning was flawed? Well, because you are 
assuming that the value underneath the fraction gets larger and 
larger, when in fact it does not.  From the solution I have provided, 
you may realize that what happens is that the stuff under the first 
fraction is the very same value you wish to seek, and therefore is not 
infinite! Similarly, in the second problem I posed to you, the stuff 
under the first square root is the very same value you wish to find.

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