mini challenge: find the smallest cosine of an integer - Printable Version +- HP Forums (https://www.hpmuseum.org/forum) +-- Forum: HP Calculators (and very old HP Computers) (/forum-3.html) +--- Forum: General Forum (/forum-4.html) +--- Thread: mini challenge: find the smallest cosine of an integer (/thread-17595.html) Pages: 1 2 |
RE: mini challenge: find the smallest cosine of an integer - Albert Chan - 10-24-2021 11:42 PM (10-24-2021 06:17 PM)Albert Chan Wrote: 10 P0=0 @ P1=1 @ P=P0 @ S=2 It is not obvious (at least for me), why smaller |sin(P)| signaled a convergent. Say, 2 neighboring convergents, p1/q1 > π > p2/q2 We write this to show p2/q2 is the better estimate for π: p1/q1 > (p1+p2)/(q1+q2) > π > p2/q2 (p1+p2) > (q1+q2)*π p1 - q1*π > q2*π - p2 > 0 |p1 - q1*π| > |p2 - q2*π| Both sides are tiny. Within this range, sin is increasing function, so take sin of both side. sin(|p1 - q1*π|) > sin(|p2 - q2*π|) |sin(p1)| > |sin(p2)| Thus, when |sin(P)| get smaller, we found a convergent. |