MATHEMATICS  PROBLEM  127

 

Evaluate by hand (without calculator or computer) the following iterated integral, showing every step in detail to obtain an exact value, not a numerical approximation!

 

 

 

Each midshipman submitting a correct solution with a correct explanation to Problem 127 by 1700 Thursday 12 December 2002 will win a cookie.  Submit solutions to Prof. Wardlaw at mathprob@usna.edu (please no attachments!) or via his mailbox in Chauvenet 301.

         

Correct solutions to Mathematics Problem 126 Part One were submitted by Midn Justin Carlson, Nathan Fleischaker, Stephen McMath, Ralph Rogers, Andrew Watts, and Matthew Welch and by Professor Richard Davis.  Professor Davis also submitted a solution to Part Two.  A number of submissions were nearly correct, but had small errors or missing parts.  My solution to Mathematics Problem 126 is posted on the board and is on the reverse side of this sheet.

 

MATHEMATICS  PROBLEM  126

 

            Find the value of the following infinite continued fraction, assuming that it converges:

                                  1                                    

  1 +   1                            

          1   +    1                 

                      1  +   1        

                               1  + 

                                          

(For two more cookies:  Interpret the above as an infinite sequence and show that it converges.)

 

            Solution.  Assume that the continued fraction converges to the number  x.  Then  x = 1/(1+x)  follows by substitution, so multiplication by  1+ x  gives  x + x²  =  1  and hence  x² + x – 1 = 0 .  The quadratic formula gives two solutions,  x = ( -1 ±)/2.  However, it is clear that the value must be positive, so the value of the continued fraction is   x = ( -1 +)/2.

            The continued fraction can be interpreted mathematically as the limit of the sequence  (x_n)  defined by  x₀  =  0  and  x_n  =  1/(1 + x_n-1)  for  n > 0.  The first four values are  x₀  =  0,  x₁  =  1,  x₂  =  ½,  and  x₃  =  2/3,  so we have

           0    =     x₀     <     x₂       <       x₃     <       x₁       =     1.

Now

           0    <     x_2k     <    x_2k+2   <      x_2k+3   <     x_2k+1    <     1      (1)

implies

           1    <  1 + x_2k  < 1 + x_2k+2 < 1 + x_2k+3 < 1 + x_2k+1  <     2

implies upon taking reciprocals and using   x_n  =  1/(1 + x_n-1) :

           1     >     x_2k+1  >    x_2k+3    >    x_2k+4    >    x_2k+2    >     ½,

or

           0     <     x_2k+2   <   x_2k+4    <     x_2k+5   <     x_2k+3    <    1.

         

This inductive step shows that  (1) is true for all nonnegative integers  k.  Thus the sequence  (x_2k)  of terms with even subscripts is strictly increasing and bounded above by  1, and so it must converge, by the Monotone Convergence Theorem.  Also, the sequence  (x_2k+1)  of terms with odd subscripts is strictly decreasing and bounded below by  0, so it too must converge.  Let us suppose that the sequence  (x_2k)  of even subscripted terms has limit  y,  and that the sequence  (x_2k+1)  with odd subscripts converges to  z.  Now  x_2k+2 = 1/(1 + x_2k+1) = 1/(1 + 1/(1 + x_2k))= (1 + x_2k)/(2 +  x_2k).  Letting    k ¥  gives  y = (1 + y)/(2 + y), so  2y + y² = 1 + y  implies  y² + y – 1 = 0  and  (since  y > 0)  y = (-1 +)/2.  Similarly,  x_2k+3 =  (1 + x_2k+1)/(2 +  x_2k+1)  impies that  z = (1 + z)/(2 + z)  and 

z = (-1 + )/2.   Thus it follows that the sequence  (x_n)  converges to  x = y = z = (-1 +)/2.  This completes the proof.  (There are many other interesting ways to prove this.)