2.57

练习 2.57 请扩充求导程序，使之能处理任意项（两项或者更多项）的和与乘积。这样，上面最后一个例子就可以表示为：

(deriv '(* x y (+ x 3)) 'x)

设法通过只修改和与乘积的表示，而完全不修改过程 deriv 的方式完成这一扩充。例如，让一个和式的addend是它的第一项，而其augend是和式中的其余项。

先复用一下前面的代码：

 
(define (=number? v n)
  (if (number? v)
      (= v n)
      #f
      )
  )
​
(define (error msg v)
  msg
  )
​
(define variable? symbol?)
(define (same-variable? v1 v2)
  (and (variable? v1) (variable? v2) (eq? v1 v2))
  )
(define (make-sum a1 a2) 
  (cond 
   ((=number? a1 0) a2)
   ((=number? a2 0) a1)
   ((and (number? a1) (number? a2)) (+ a1 a2))
   (else (list '+ a1 a2))
   )
  )
(define (make-product m1 m2) 
  (cond
   ((=number? m1 0) 0)
   ((=number? m2 0) 0)
   ((=number? m1 1) m2)
   ((=number? m2 1) m1)
   (else (list '* m1 m2))
   )
  )
(define (sum? x) (and (pair? x) (eq? (car x) '+)))
(define addend cadr)
(define augend caddr)
(define (product? x) (and (pair? x) (eq? (car x) '*)))
(define multiplier cadr)
(define multiplicand caddr)
​
(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-product 
     (multiplier exp)
     (deriv (multiplicand exp) var)
     )
    (make-product
     (deriv (multiplier exp) var)
     (multiplicand exp)
     )
    )
   ((and (pair? exp) (= 1 (len exp)))
    (deriv (car exp) var)
    )
   (else
    (error exp)
    )
   )
  )
​
; 期待返回 1
(deriv '(+ x 3) 'x)
​x
 
1

 
; 期待得到 3x^2 的结果
(deriv '(** x 3) 'x)
 
Error: execute: unbound symbol: "len" [deriv]true

定义 exponentiation?：

 
(define (exponentiation? exp)
  (and (pair? exp) (eq? (car exp) '**))
  )
​
; 期待返回 true
(exponentiation? '(** x 3))
 
true

定义 base：

 
(define base cadr)
​
; 期待返回 x
(base '(** x 3))
 
'x

定义 exponent：

 
(define exponent caddr)
​
; 期待返回 3
(exponent '(** x 3))
 
3

定义 make-exponentiation：

 
(define (make-exponentiation m1 m2) (list '** m1 m2))
​
; 期待返回 (** x 3)
(make-exponentiation 'x 3)
 
('** 'x 3)

增加一个子句后的 deriv：

 
(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)
    (cond
     ((= (exponent exp) 0) 0)
     ((= (exponent exp) 1) 1)
     (else
      (make-product
       (exponent exp)
       (make-exponentiation 
        (base exp) 
        (- (exponent exp) 1)
        )
       )
      )
     )
    )
   ((and (pair? exp) (= 1 (length exp)))
    (deriv (car exp) var)
    )
   (else
    (error "unknown expression type -- DERIV" exp)
    )
   )
  )
​
(deriv '(** x 0) 'x)
 
0

 
(define addend cadr)
(define (augend exp)
  (if (= 1 (length (cddr exp)))
      (caddr exp)
      (cons '+ (cddr exp))
      )
  )
(define multiplier cadr)
(define (multiplicand exp)
  (if (= 1 (length (cddr exp)))
      (caddr exp)
      (cons '* (cddr exp))
      )
  )
​
; 期待得到 2xy + 3y 的结果
(deriv '(* x y (+ x 3)) 'x)
 
('+ ('* 'x 'y) ('* 'y ('+ 'x 3)))