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Sine recurrence (Posted on 2019-01-08) Difficulty: 3 of 5
If a0 = sin2 (π/45) and an+1 = 4an(1 - an) for n >= 0, find the smallest positive integer n such that an = a0

No Solution Yet Submitted by Danish Ahmed Khan    
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Solution computer solution | Comment 4 of 6 |
    1   kill "sineiter.txt"
    2   open "sineiter.txt" for output as #2
   10   A0=(sin(#pi/45))^2
   20   A=A0:N=0
   30   print A0
   40   for Iter=1 to 100
   50     A=4*A*(1-A)
   60     N=N+1
   70     print N;" ";A;"   ";A-A0
   71     print 2,N;" ";A;"   ";A-A0
   80   next
  200   close #2
  
The answer seems assuredly to be n=12.  
  
When the above was run with about 195-digit precision it was apparent that the difference from a0 was non-zero through a11. Truncated values are:

 n difference from a0 (truncated)
 1 0.014503186401625726583638
 2 0.071109986292572172348799
 3 0.275948460976246448815520
 4 0.802964772033614297376292
 5 0.616094982170619018822108
 6 0.936607830800248628557973
 7 0.215537582635411742461673
 8 0.682437331078741175249369
 9 0.854803934540110727219914
10 0.477684286019534671718889
11 0.993916059500697281348468

At n=12 we get

-0.000000000000000000000000000000000000000000000000000000000000000000
000000000000000000000000000000000000000000000000000000000000000000000
000000000000000000000000000000000000000000000000000000462 

as the difference from a0.

At successive multiples of 12, the rounding error accumulates:

n=24
-0.000000000000000000000000000000000000000000000000000000000000000000
000000000000000000000000000000000000000000000000000000000000000000000
000000000000000000000000000000000000000000000000001895733 

n=36
-0.000000000000000000000000000000000000000000000000000000000000000000
000000000000000000000000000000000000000000000000000000000000000000000
000000000000000000000000000000000000000000000007764925473

etc.

Edited on January 8, 2019, 11:23 am
  Posted by Charlie on 2019-01-08 11:22:15

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