x = ⌊ π ⋅ 1 2 ⋅ 3 2 ⋅ 3 4 ⋅ 5 4 ⋅ 5 6 ⋅ 7 6 ⋅ 7 8 ⋅ 9 8 ⋯ ⌋
Find x .
Notation : ⌊ ⋅ ⌋ denotes the floor function .
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Great demonstration of the Wallis Product!
Great solution
Wow, what an amazing proof of wallis product.
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Let x = ⌊ π P ⌋ , then we have:
P = 1 2 ⋅ 3 2 ⋅ 3 4 ⋅ 5 4 ⋅ 5 6 ⋅ 7 6 ⋅ 7 8 ⋅ 9 8 . . . ⋅ ∞ = m → ∞ lim k = 1 ∏ m 2 k − 1 2 k ⋅ 2 k + 1 2 k = m → ∞ lim ( 2 m ) ! 2 m m ! ⋅ 2 m m ! ⋅ ( 2 m + 1 ) ! 2 m m ! ⋅ 2 m m ! = m → ∞ lim ( 2 m ) ! ( 2 m + 1 ) ! 2 4 m ( m ! ) 4 Using Stirling’s approximation n ! ≈ 2 π e − n n n + 2 1 = m → ∞ lim 2 π e − 2 m ( 2 m ) 2 m + 2 1 2 π e − 2 m − 1 ( 2 m + 1 ) 2 m + 2 3 2 4 m ( 2 π e − m m m + 2 1 ) 4 = m → ∞ lim ( 2 m + 1 ) 2 m + 2 3 2 2 m + 2 1 m 2 m + 2 3 ⋅ π e = m → ∞ lim 2 ( 2 m + 1 ) 2 m + 2 3 ( 2 m ) 2 m + 2 3 ⋅ π e = m → ∞ lim 2 ( 1 + 2 m 1 ) 2 m + 2 3 π e = 2 e π e = 2 π
⇒ x = ⌊ π P ⌋ = ⌊ π ⋅ 2 π ⌋ = ⌊ 4 . 9 3 4 8 ⌋ = 4