(define (product term a next b) (if (> a b) 1 (* (term a) (product term (next a) next b))))
factorial function, in terms of product function above, can be written as below.
(define (identity x) x) (define (next x) (+ x 1)) (define (factorial n) (product identity 1 next n))
approximations to pi using john wallis' formula, can be found by defining a new term to be used by product function as below
(define (pi-term n) (if (even? n) (/ (+ n 2) (+ n 1)) (/ (+ n 1) (+ n 2))))
And it can be used as follows.
(* (product pi-term 1 next 6) 4) ;;= 3.3436734693877552 (* (product pi-term 1 next 100) 4) ;;= 3.1570301764551676
(define (product term a next b) (define (iter a res) (if (> a b) res (iter (next a) (* (term a) res)))) (iter a 1))
Another approach would be to recognize the series as 4*4*6*6*.../3*3*5*5*... The leading factor of 2 in the numerator must be dealt with.
(define (pi n) (define (term x) (* x x)) (define (next x) (+ x 2)) ; since we are increasing the numbers by two on every iteration (define limit (* n 2)) ; upper term: - 2 always goes first, start building product from 4 ; - as 2 numbers are skipped, the limit must respect that, too ; - since we are squaring one time too often at the end, ; we have to divide that back out of the result ; lower term: start with 3, which is 1 more than the upper term ; -> so increase limit by 1 ; (* 4 (/ (/ (* 2 (product term 4 next (+ limit 2))) (+ limit 2)) (product term 3 next (+ limit 1)))))
(pi 6) ;;= 3.255721745 (pi 100) ;;= 3.149378473
n defines steps in terms of pairs of factors. So where the above solution does 6 steps when n=6, this solution does 12 steps when n=6:
(pi 3) ;;= 3,343673469 (pi 50) ;;= 3,157030176
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