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SICP exercise 1.39
Problem
A continued fraction representation of the tangent function was published in 1770 by the German mathematician J.H. Lambert:
<math>\tan x = \frac{x}{1 - \frac{x^2}{3 - \frac{x^2}{5 - \ddots}}}</math>
where x is in radians. Define a procedure (tan-cf x k) that computes an approximation to the tangent function based on Lambert's formula. k specifies the number of terms to compute, as in exercise 1.37.
Solution
Here's one implementation of Lambert's continued fraction representation of the tangent function, using the cont-frac function from exercise 1.37:
(define (cont-frac n d k) (define (cont-frac-iter k result) (if (= k 0) result (cont-frac-iter (- k 1) (/ (n k) (+ (d k) result))))) (cont-frac-iter k 0)) (define (square x) (* x x)) (define (tan-cf x k) (define (n k) (if (= k 1) x (- (square x)))) (define (d k) (- (* 2 k) 1)) (cont-frac n d k))
We can test the procedure by comparing the results to the tan primitive provided by Chicken Scheme 3.1 on a MacBook Pro running Mac OS X 10.5:
(tan 0)
Output:
0.0
(tan-cf 0 10)
Output:
0
Here's an approximation of <math>\pi</math>:
(define pi 3.14159265) (tan (/ pi 4))
Output:
0.999999998205103
(tan-cf (/ pi 4) 10)
Output:
0.999999998205103
(tan (/ (* 3 pi) 4))
Output:
-1.00000000538469
(tan-cf (/ (* 3 pi) 4) 10)
Output:
-1.00000000539582
Looks good. Values near a multiple of <math>\pi</math> (i.e., those for which tan x is nearly or exactly zero) need more iterations:
(tan pi)
Output:
-3.58979302983757e-09
(tan-cf pi 10)
Output:
-5.48300699971479e-09
(tan-cf pi 20)
Output:
-3.58979298522367e-09