sicpsgjd/1/3/jord/exercises.org

• 1.30
• 1.31
• 1.32
• 1.33
• 1.34

1.30

  (define (sum term a next b)
    (define (iter term a next b result)
      (if (> a b)

	  (iter (
    (iter term a next b 0))

1.31

  (define (product term a next b)
    (if (> a b)
 1
 (* (term a)
    (product term (next a) next b))))

  (define (factorial x)
    (define (inc n) (+ n 1))
    (define (self x) x)
    (product self 1 inc x))
  (factorial 10)

  (define (pi-over-four a b)
    (define (term x)
      (if (= 1 (remainder x 2))
   (/ (+ x 1)
      (+ x 2))
   (/ (+ x 2)
      (+ x 1))))
    (define (next x)
      (+ x 1))
    (product term a next b))
  (* 4 (pi-over-four 1 10))

1.32

  (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))))

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

  (define (factorial x)
    (define (inc n) (+ n 1))
    (define (self x) x)
    (product self 1 inc x))

  (factorial 10)

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

  (define (integral f a b dx)
    (define (add-dx x) (+ x dx))
    (* (sum f (+ a (/ dx 2.0)) add-dx b)
       dx))

  (define (square x) (* x x))
  (integral square 0 1 0.001)

0.33433325000000047

1.33

  (define (filter-accumulate combiner null-value filter term a next b)
    (cond ((> a b) null-value)
	  ((filter a) null-value)
	  (else (combiner (term a)
			  (accumulate combiner null-value term (next a) next b)))))

  (define (sum-filter filter term a next b)
    (filter-accumulate + 0 filter term a next b))

  (define (sum-prime-squares a b)
    (define (square x) (* x x))
    (define (inc x) (+ 1 x))
    (sum-filter prime? square a inc b))

1.34

suppose we define

  (define (f g)
    (g 2))

then we can

  (define (square x) (* x x))
  (f square)

4

and

  (f (lambda (z) (* z (+ z 1))))

6

but what if we (f f)?

  (f f)

scheme says:

ERROR: Wrong type to apply: 2

          0 (_ 2)

… i dunno! applying 2 to 2?