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Sine recurrence (
Posted on 2019-01-08
)
If a
0
= sin
2
(π/45) and a
n+1
= 4a
n
(1 - a
n
) for n >= 0, find the smallest positive integer n such that a
n
= a
0
No Solution Yet
Submitted by
Danish Ahmed Khan
No Rating
Comments: (
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re: Full solution...............kudos
Comment 6 of 6 |
(In reply to
Full solution
by Jer)
Brilliant logic!
I've got the same result - using EXCELL spread sheet.
Saw no point of posting it after Charlie's solution.
Posted by
Ady TZIDON
on 2019-01-09 03:23:54
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