問題解答

問2.56

(define (make-exponentiation base exp)
  (cond ((= exp 0) 1)
        ((= exp 1) base)
        (else (list '** base exp))))

(define (exponentiation? x)
  (if (eq? (car x) '**) #t #f))

(define (base x)
  (cadr x))

(define (exponent x)
  (caddr x))

(define (deriv exp var)
  (cond ((number? exp) 0)
        ((variable? exp)
         (if (same-variable? exp var) 1 0))
        ((sum? exp)
         (make-sum (deriv (addend exp) var)
                   (deriv (augend exp) var)))
        ((product? exp)
         (make-sum
          (make-product (multiplier exp)
                        (deriv  (multiplicand exp) var))
          (make-product (deriv (multiplier exp) var)
                        (multiplicand exp))))
        ((exponentiation? exp)
         (make-product
          (make-product (exponent exp) (make-exponentiation (base exp) (make-sum (exponent exp) -1)))
          (deriv (base exp) var)))
        (else
         (error "unknown expression type -- DERIV" exp))))


(deriv '(** x 2) 'x)
(deriv '(** x 3) 'x)

問題だけ見るとなんじゃこりゃこんなの解けないよ?って思うんですが、ヒントを元にやってみると簡単で驚きます。こんな簡単な方法で記号微分って表現できるんですね。すごい。