20 hours ago
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))
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