SICP

Exercise 1.37.    a. An infinite  continued fraction  is an expression of the form As an example, one can show that the infinite continued fraction expansion with the  N i  and the  D i  all equal to 1 produces 1/ , where   is the golden ratio (described in section  1.2.2 ). One way to approximate an infinite continued fraction is to truncate the expansion after a given number of terms. Such a truncation -- a so-called  k -term finite continued fraction  -- has the form Suppose that  n  and  d  are procedures of one argument (the term index  i ) that return the  N i  and  D i  of the terms of the continued fraction. Define a procedure  cont-frac  such that evaluating  (cont-frac n d k)  computes the value of the  k -term finite continued fraction. Check your procedure by approximating 1/  using (cont-frac (lambda (i) 1.0)            (lambda (i) 1.0)            k) for successive values of  k . How large must you make  k  in order to get an approximation that is accurate to

Exercise 1.36.   Modify  fixed-point  so that it prints the sequence of approximations it generates, using the  newline  and  display  primitives shown in exercise  1.22 . Then find a solution to  x x  = 1000 by finding a fixed point of  x     log (1000)/ log ( x ). (Use Scheme's  primitive  log  procedure, which computes natural logarithms.) Compare the number of steps this takes with and without average damping. (Note that you cannot start  fixed-point  with a guess of 1, as this would cause division by  log (1) = 0.) SOLUTION The code and tests are here .

Exercise 1.35.    Show that the golden ratio     (section  1.2.2 ) is a fixed point of the transformation   x     1 + 1/ x , and use this fact to compute     by means of the   fixed-point   procedure. SOLUTION The code and tests are here .

Exercise 1.34.   Suppose we define the procedure (define (f g)   (g 2)) Then we have (f square) 4 (f (lambda (z) (* z (+ z 1)))) 6 What happens if we (perversely) ask the interpreter to evaluate the combination  (f f) ? Explain. SOLUTION The code and tests are here . Explanation: (f f) evaluates to (f 2) which in turn results in (2 2) where the first term  is expected to be a procedure. So the evaluator doesn't accept it and gives an error.