Find p : (Report your answer to 5 places after decimal)
5 2 ⋅ 3 − ϕ 1 ⋅ s → ∞ lim 2 s 1 ⋅ k = 1 ∏ s ( Δ k ) = p 1
where, Δ m = m r o o t s 2 + 2 + 2 + ⋯ + 2 + 2 + ϕ
and, ϕ = 2 1 + 5 is the Golden Ratio.
For example: Δ 1 = 2 + ϕ , Δ 3 = 2 + 2 + 2 + ϕ , etc.
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The 2 equations you have written after "we know" is amazing. May you send proof for those Equation??
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We know,
θ sin θ = cos 2 θ ⋅ cos 4 θ ⋅ cos 8 θ ⋅ cos 1 6 θ ⋅ cos 3 2 θ ⋯
Putting θ = 5 π we get:
2 ⋅ π 5 ⋅ 3 − ϕ = cos 1 0 π ⋅ cos 2 0 π ⋅ cos 4 0 π ⋅ cos 8 0 π ⋅ cos 1 6 0 π ⋯
Also we know, cos 1 0 n π = 2 1 Δ m = 2 1 m r o o t s 2 + 2 + 2 + ⋯ + 2 + 2 + ϕ
hence,
5 2 ⋅ 3 − ϕ 1 ⋅ s → ∞ lim 2 s 1 ⋅ k = 0 ∏ s ( Δ 2 k ) = π 1 ∴ p = π ≈ 3 . 1 4 1 5 9