sicp 1.31 另一个关于 pi 的计算公式 -- john wallis pi 的另一种形式理解

john_wallis_pi.scm

(define (product term a next b)
    (if (> a b)
        1
        (* (term a)
           (product term (next a) next b)
        )
    )
)

(define (product term a next b)
    (define (iter a result)
        (if (> a b)
            result
            (iter (next a) (* (term a) result))
        )
    )

    (iter a 0)
)

; 阶乘函数:  n! = n * (n - 1) * (n - 2) * ... 3 * 2 * 1

(define (factorial n)
    (define (term x) x)

    (define (next x) (+ 1 x))

    (product term 1 next n)
)

; 规律: (1 - 1/3) * (1 + 1/3) * (1 - 1/5) * (1 + 1/5)
; item: 2N+1

(define (wallis-pi n)
    (define (wallis)
        (define (term x)
            (define i (/ 1.0 (+ (* 2 x) 1)))
            (* (- 1 i) (+ 1 i))
        )
        (define (next x)
            (+ x 1)
        )

        (product term 1 next n)
    )

    (* 4 (wallis))
)

#|
> (wallis-pi 100)
3.1493784731685985
> (wallis-pi 1000)
3.1423773650938758
> (wallis-pi 10000)
3.1416711865344302
> (wallis-pi 100000)
3.141600507501392
> (wallis-pi 1000000)
3.141593438980562
> (wallis-pi 10000000)
3.1415927321150257
|#

ref: http://community.schemewiki.org/?sicp-ex-1.31
ref: https://en.wikipedia.org/wiki/Wallis_product

和社区答案, 对于公式的形式化理解有差异.

我的第一反应是拆成: (1 - 1/3) * (1 + 1/3) * (1 - 1/5) * (1 + 1/5) * ...