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ex1.45
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ex1.45
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(define tolerance 0.00001)
(define (average a b) (/ (+ a b) 2.0))
(define (close-enough? a b)
(< (abs (- a b)) tolerance))
(define (fixed-point f first-guess)
(define (try guess)
(let ((next (f guess)))
(if (close-enough? guess next)
next
(try next))))
(try first-guess))
(define (average-damping f)
(lambda (x) (average x (f x))))
(define (exp b n)
(if (= n 0)
1
(* b (exp b (- n 1)))))
(define (compose f g)
(lambda (x) (f (g x))))
(define (repeated f n)
(if (= n 1)
f
(compose f (repeated f (- n 1)))))
;; (define (repeated2 f n)
;; (define (iter counter result)
;; (if (= counter 1)
;; result
;; (iter (- counter 1) (compose f result))))
;; (iter n f))
;; By default the base of log is e
;; 1 ]=> (log 2)
;;
;; ;Value: .6931471805599453
;;
;; 1 ]=> (log 3)
;Value: 1.0986122886681098
;; Above result is of base e
;; if we have to find log having base b
;; devide result by (log b)
;; 1 ]=> (/ (log 2) (log 2)) ;; value of (log 2) having base 2
;; 1 ]=> (/ (log 3) (log 2)) ;; value of (log 3) having base 2
(define (logB n b)
(/ (log n) (log b)))
;; (nth-root x n)
;; if n = 1 ==> average-damping require 0 times
;; if n = 2 ==> average-damping require 1 times
;; if n = 3 ==> average-damping require 1 times
;; if n = 4 ==> average-damping require 2 times
;; if n = 5 ==> average-damping require 2 times
;; if n = 6 ==> average-damping require 2 times
;; if n = 7 ==> average-damping require 2 times
;; if n = 8 ==> average-damping require 3 times
;; if n = 2^m ==> average-damping require m times
;; How to find m?
;; n = 2^m
;; log(n) = log(2^m)
;; log(n) = m.log(2)
;; m = log(n) / log(2)
;; if n = 8 ==> n = 2^3 ==> m = 3
;; if base = 2 ==> m = log(8)
;; ottherwise ==> m = log(8) / log(2)
(define (nth-root x n)
(let ((counter (floor (/ (log n) (log 2)))))
(if (= counter 0)
x
(fixed-point ((repeated average-damping counter) (lambda (y) (/ x (exp y (- n 1)))))
1.0))))