I saw a fun math problem on reddit the other day.
find a number n, so that sin(n)=-1, where n is an integer in radians; so sin(270 degrees) doesn’t count. Obviously it will never be exactly -1 but close enough for the difference to be lost in rounding.
I need the argument of sin function to be as close to an integer as possible. Call this integer m.
Solving for leads to:
If I have a rational approximation to with an even numerator I can divide it by two get my m. I also have to make sure that the denominator is in the form of 4n+3.
It’s possible to use continued fractions to approximate real numbers. Here’s a continued fraction sequence for pi: https://oeis.org/A001203
The first rational approximation I learned in elementary school is 22/7 which is perfect.
> (sin 11)
For the others I’ll have to evaluate the continued fraction to get my approximation of a simple fraction.
> (eval-terms (list 3 7 15 1 292 1))
> (sin (/ 104348 2))
> (eval-terms (list 3 7 15 1 292 1 1 1 2 1 3 1 14 2 1))
> (sin (/ 245850922 2))
Looks like a good candidate was found.
This is the code to evaluate a continued fraction coefficients. It’s very convenient that scheme has a native rational data type.
(define (eval-terms ts)
(cond ((= (length ts) 1) (car ts))
(+ (car ts) (/ 1 (eval-terms (cdr ts)))))))
(define (eval-terms-iter ts)
(let ((rev-ts (reverse ts)))
(define (ev-terms ts frac)
(if (null? ts)
(ev-terms (cdr ts) (+ (car ts) (/ 1 frac)))))
(ev-terms rev-ts (car rev-ts))))