Exercise 4.21: Amazingly, Louis’s intuition in
Exercise 4.20 is correct.  It is indeed possible to specify recursive
procedures without using letrec (or even define), although the
method for accomplishing this is much more subtle than Louis imagined.  The
following expression computes 10 factorial by applying a recursive factorial
procedure:231


((lambda (n)
   ((lambda (fact) (fact fact n))
    (lambda (ft k)
      (if (= k 1)
          1
          (* k (ft ft (- k 1)))))))
 10)


 Check (by evaluating the expression) that this really does compute factorials.
Devise an analogous expression for computing Fibonacci numbers.

 Consider the following procedure, which includes mutually recursive internal
definitions:


(define (f x)
  (define (even? n)
    (if (= n 0)
        true
        (odd? (- n 1))))
  (define (odd? n)
    (if (= n 0)
        false
        (even? (- n 1))))
  (even? x))

Fill in the missing expressions to complete an alternative definition of
f, which uses neither internal definitions nor letrec:


(define (f x)
  ((lambda (even? odd?)
     (even? even? odd? x))
   (lambda (ev? od? n)
     (if (= n 0) 
         true 
         (od? ⟨??⟩ ⟨??⟩ ⟨??⟩)))
   (lambda (ev? od? n)
     (if (= n 0) 
         false 
         (ev? ⟨??⟩ ⟨??⟩ ⟨??⟩)))))