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Sine recurrence (
Posted on 20190108
)
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
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Comments: (
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)
@ DAK
 Comment 1 of 6
(SIN a(n/45) ) ^2 ??
did you mean pi/45
or
integer n , starting from n=0 ??
Edited on
January 8, 2019, 8:10 am
Posted by
Ady TZIDON
on 20190108 08:08:51
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