sicp每日一题[1.39]
Exercise 1.39
A continued fraction representation of the tangent function was published in 1770 by the German mathematician J.H. Lambert:
tan x = x 1 − x 2 3 − x 2 5 − . . . \tan x = \frac{x}{1-\frac{x^2}{3-\frac{x^2}{5-...}}} tanx=1−3−5−...x2x2x
where x x x is in radians. Define a procedure ( t a n − c f x k ) (tan-cf\ x\ k) (tan−cf x k) that computes an approximation to the tangent function based on Lambert’s formula. k k k specifies the number of terms to compute, as in Exercise 1.37.
1.38难度不大，再加上今天周末，时间充足，我决定再做一道😀
1.39乍一看，并不很难，但是我想了半天也没做出来。直到我忽然发现
tan x = x 1 − x 2 3 − x 2 5 − . . . \tan x = \frac{x}{1-\frac{x^2}{3-\frac{x^2}{5-...}}} tanx=1−3−5−...x2x2x 其实等价于 x 2 1 − x 2 3 − x 2 5 − . . . ÷ x \frac{x^2}{1-\frac{x^2}{3-\frac{x^2}{5-...}}} \div x 1−3−5−...x2x2x2÷x
然后题目就变得简单了，只要把最后的结果除以 x x x 即可
; 注意 n 和 d 都是 procedure 而不是数字,k 表示要计算的项数
(define (cont-frac n d radians k); iterative implementation(define (frac-iter i pre)(if (= i 1)pre; 这里要注意计算的是 (d (- i 1))，而不是 (d i)，否则会漏掉 1-... 这一项(frac-iter (- i 1) (/ (n i) (- (d (- i 1)) pre))))) ; recurative implementation(define (frac-recur i)(if (= i k)(/ (n i) (d i))(/ (n i) (- (d i) (frac-recur (+ i 1))))))(frac-iter k (square radians))) ; 注意初始值是 x^2;(frac-recur 1)); 注意到原式最上面的分子其实可以写成 x^2 / x，这样它的结构就可以保持一致，只要在最后把结果除以 x 即可
(define (tan-cf x k)(let ((radians (/ (* pi x) 180)))(/ (cont-frac (lambda (i) (square radians))(lambda (i) (- (* 2 i) 1))radiansk)radians))); tan 30° ≈ 0.577，tan 45° = 1，tan 60° ≈ 1.732
(tan-cf 30 10)
(tan-cf 45 10)
(tan-cf 60 10); result
0.5773502691896257
1.0
1.7320508075688845