my attempt to do the exercises in sicp.

Tuesday, January 4, 2011

sicp exercise 3.77



;; Exercise 3.77.  The integral procedure used above was analogous to the ``implicit'' definition of the infinite stream of integers in section 3.5.2. Alternatively, we can give a definition of integral that is more like integers-starting-from (also in section 3.5.2):

;(define (integral integrand initial-value dt)
;  (cons-stream initial-value
;               (if (stream-null? integrand)
;                   the-empty-stream
;                   (integral (stream-cdr integrand)
;                             (+ (* dt (stream-car integrand))
;                                initial-value)
;                             dt))))

;When used in systems with loops, this procedure has the same problem as does our original version of integral. Modify the procedure so that it expects the integrand as a delayed argument and hence can be used in the solve procedure shown above.


(define (add-streams s1 s2)
  (stream-map + s1 s2))

(define (scale-stream stream factor)
  (stream-map (lambda (x) (* x factor)) stream))

(define (integral delayed-integrand initial-value dt)
  (cons-stream initial-value
               (let ((integrand (force delayed-integrand)))
                 (if (stream-null? integrand)
                     the-empty-stream
                     (integral (delay (stream-cdr integrand))
                               (+ (* dt (stream-car integrand))
                                  initial-value)
                               dt)))))


(define (solve f y0 dt)
  (define y (integral (delay dy) y0 dt))
  (define dy (stream-map f y))
  y)

(newline)
(display (stream-ref (solve (lambda (y) y) 1 0.001) 1000))


No comments: