perplexus.info :: Just Math : Sine recurrence

Sine recurrence (Posted on 2019-01-08)

If a₀ = sin² (π/45) and a_n+1 = 4a_n(1 - a_n) for n >= 0, find the smallest positive integer n such that a_n = a₀

No Solution Yet Submitted by Danish Ahmed Khan

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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 2019-01-08 08:08:51

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