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This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.
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Let a1=21 and let an+1=21+21an for integers n≥1. We want the value of n=1∏∞an.
Note that a1=cos(4π) and if an=cos(2n+1π) then an+1=21+21cos(2n+1π)=cos(2n+2π).
Therefore, by induction, we have an=cos(2n+1π) for all integers n≥1.
Let PN=n=1∏Nan=n=1∏Ncos(2n+1π). Using the identity sinθcosθ=21sin2θ repeatedly, we have:
PNsin(2N+1π)=sin(2N+1π)n=1∏Ncos(2n+1π)=2N1sin(2π)=2N1.
Therefore, PN=2N1csc(2N+1π) for all integers N≥1.
For x≈0 we have cscx=x1+O(x). Hence, PN=2N1[π2N+1+O(2N+1π)]=π2+O(2−2N).
As N→∞ we have 2−2N→0. Therefore, n=1∏∞an=N→∞limPN=π2.
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thanks a lot!. it was actually not easy to get to substitute 21 by cos(4π)
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If you don't mind, can you please share the source of this problem?
Thanks!
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Oh yeah, 2/pi, Viete's formula.