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exercises_1_3_3.scm
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exercises_1_3_3.scm
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;;; Exercise 1.35
;;;
;;; Show that the golden ratio is a fixed point of x -> 1 + 1/x
(define tolerance 0.00001)
(define (fixed-point f first-guess)
(define (close-enough? v1 v2)
(< (abs (- v1 v2)) tolerance))
(define (try guess)
(let ((next (f guess)))
(if (close-enough? guess next)
next
(try next))))
(try first-guess))
(fixed-point (lambda (x) (+ 1 (/ 1 x))) 1.0)
;;; Exercise 1.36
;;;
;;; Change fixed-point to display iterations. Then find a solution to
;;; x^x == 1000 via fixed point of x <- log(1000)/log(x)
;;; With average damping
(define (print-iter-fixed-point f first-guess)
(define (close-enough? v1 v2)
(< (abs (- v1 v2)) tolerance))
(define (try guess)
(newline)
(display guess)
(let ((next (f guess)))
(if (close-enough? guess next)
next
(try (/ (+ next guess) 2))))) ;;; dampen here
(try first-guess))
(print-iter-fixed-point (lambda (x) (/ (log 1000) (log x))) 100.0)
;;; Without average damping.
(define (print-iter-fixed-point f first-guess)
(define (close-enough? v1 v2)
(< (abs (- v1 v2)) tolerance))
(define (try guess)
(newline)
(display guess)
(let ((next (f guess)))
(if (close-enough? guess next)
next
(try next)))) ;;; just get the next one
(try first-guess))
(print-iter-fixed-point (lambda (x) (/ (log 1000) (log x))) 100.0)
;;; Exercise 1.37
;;;
;;; Write an iterative and recursive procedure for computing a k-term finite
;;; continued fraction (continued fraction truncated at some point). Check by
;;; correctly evaluting (cont-frac (lambda (i) 1.0) (lambda (i) 1.0) k) which
;;; should converge to 1/φ = 0.61803398875
;;; recursive
(define (cont-frac n d max_iterations)
(define (cont-frac-recur iterations)
(if (< iterations max_iterations)
(/ (n iterations) (+ (d iterations) (cont-frac-recur (+ iterations 1))))
(/ (n iterations) (d iterations))
)
)
(cont-frac-recur 1))
(cont-frac (lambda (i) 1.0) (lambda (i) 1.0) 50)
;;; iterative
(define (cont-frac n d max_iterations)
(define (cont-frac-iter iterations accumulator)
(if (= iterations 0)
accumulator
(cont-frac-iter (- iterations 1) (/ (n iterations) (+ (d iterations) accumulator)))
))
(cont-frac-iter max_iterations (/ (n max_iterations) (d max_iterations))))
(cont-frac (lambda (i) 1.0) (lambda (i) 1.0) 50)
;;; Exercise 1.38
;;;
;;; Use cont-frac procedure to approximate e using Euler's expansion. Euler's
;;; expansion is a k-term continued fraction where D_i == 1 and N_i is the
;;; sequence 1, 2, 1, 1, 4, 1, 1, 6...
;;; e = 2.71828182845904
;;; e - 2 = 0.71828182845904
(define (seq i)
(define (seq-accum step accumulated)
(if (= i step)
(if (= (modulo step 3) 2) (+ accumulated 2) 1)
(if (= (modulo step 3) 2)
(seq-accum (+ step 1) (+ accumulated 2))
(seq-accum (+ step 1) accumulated))
))
(seq-accum 1 0))
(cont-frac (lambda (i) 1.0) seq 100)
;;; Exercise 1.39
;;;
;;; Implement cont-frac implementation of Lambert's tangent function
(define (seed-n x)
(lambda (i)
(if (= i 1)
x
(- (* x x)))))
(define (d i)
(+ 1 (* (- i 1) 2)))
(tan 2.0)
(cont-frac (seed-n 2.0) d 10)