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68

answers:

1

procedure accumulate is defined like this:

(define (accumulate combiner null-value term a next b)
  (if (> a b) null-value
      (combiner (term a)
                (accumulate combiner null-value term (next a) next b))))

problem 1: x^n ;Solution: recursive without accumulate

(define (expon x n)
  (if (> n 0) (* x
                 (expon x (- n 1))
              )
              1))

problem 2: x + x^2 + x^4 + x^6 + ...+ ,calculate for given n the first n elements of the sequence.

problem 3: 1 + x/1! + x^2/2! + ... + x^n/n!; calculate the sum for given x,n possibly incorrect solution:

(define (exp1 x n)
 (define (term i)
   (define (term1 k) (/ x k))
   (accumulate * 1 term1 1 1+ i))
  (accumulate + 0 term 1 1+ n))

why the previous code is incorrect:

(exp1 0 3) -> 0 ; It should be 1 (exp1 1 1) -> 1 ; It should be 2

+1  A: 

First off, I would say that your EXP1 procedure is operating at too low a level in being defined in terms of ACCUMULATE, and for the sake of perspicacity rewrite it instead in terms of sums and factorials:

(define (sum term a b)
  (accumulate + 0 term a 1+ b))

(define (product term a b)
  (accumulate * 1 term a 1+ b))

(define (identity x) x)

(define (fact n)
  (if (= n 0)
      1
      (product identity 1 n)))

(define (exp1 x n)
  (define (term i)
    (/ (expon x i) (fact i)))
  (sum term 1 n))

Now to your question: the reason you are getting (EXP1 0 3) → 0 is no more than that you forgot to add the 1 at the start of the series, and are just computing x/1! + x^2/2! + ... + x^n/n!

Changing EXP1 to include the missing term works as expected:

(define (exp1 x n)
    (define (term i)
          (/ (expon x i) (fact i)))
    (+ 1 (sum term 1 n)))

=> (exp1 0 3)
1
=> (exp1 1 1)
2
Cirno de Bergerac