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XCAS wins
02-23-2019, 08:58 PM (This post was last modified: 02-24-2019 04:06 AM by Albert Chan.)
Post: #4
RE: XCAS wins
It seems this always work ... Can anyone prove it ? Huh

p(n) := {local k; product(sin(k/n * pi/2), k, 1, n-1)}

Prove p(n) = √(n) / 2n-1

Example:
p(2) = sin(pi/2) = √(2) / 2¹
p(3) = sin(1/3 pi/2) sin(2/3 pi/2) = (1/2)(√(3)/2) = √(3)/2²
p(4) = sin(1/4 pi/2) sin(2/4 pi/2) sin(3/4 pi/2) = 1/4 = √(4)/2³

And, the one from first post: p(107) = √(107)/2106

Update: got the proof online: https://math.stackexchange.com/questions...ect=1&lq=1
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Messages In This Thread
XCAS wins - yangyongkang - 02-23-2019, 01:31 PM
RE: XCAS wins - chromos - 02-23-2019, 03:05 PM
RE: XCAS wins - Eddie W. Shore - 12-16-2019, 04:01 PM
RE: XCAS wins - parisse - 02-23-2019, 06:22 PM
RE: XCAS wins - CyberAngel - 12-16-2019, 01:44 PM
RE: XCAS wins - Eddie W. Shore - 12-16-2019, 04:04 PM
RE: XCAS wins - Albert Chan - 02-23-2019 08:58 PM
RE: XCAS wins - Albert Chan - 12-16-2019, 12:50 PM
RE: XCAS wins - parisse - 12-16-2019, 07:35 PM
RE: XCAS wins - compsystems - 12-17-2019, 05:16 PM
RE: XCAS wins - Dirk.nl - 12-18-2019, 01:08 PM
RE: XCAS wins - John P - 12-18-2019, 10:50 PM
RE: XCAS wins - CyberAngel - 12-19-2019, 08:17 PM



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