Exercise 1.10

(define (A x y) 
  (cond ((= y 0) 0) 
        ((= x 0) (* 2 y)) 
        ((= y 1) 2) 
        (else (A (- x 1) 
                 (A x (- y 1)))))) 
(A 1 10) => 1024 
(A 2 4) => 65536 
(A 3 3) => 65536 


(define (f n) (A 0 n)) 
(define (g n) (A 1 n)) 
(define (h n) (A 2 n)) 
(define (k n) (* 5 n n))

(f n) computes \(2n\), since (A 0 n) => (* 2 n).

(g n) computes \(2^n\) since (A 1 n) => (A 0 (A 1 (- n 1))) => (f (g (- n 1))).

(h n) computes \(^n{2}\): (A 2 n) => (A 1 (A 2 (- n 1))) => (g (h (- n 1))).

(k n) computes \(5n^2\), as stated in the exercise.