22 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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