Even And Odd Partition?

Calculus Level 4

x = π 2 1 2 3 4 3 4 5 6 5 6 7 8 7 8 9 x = \left \lfloor \pi \cdot{} \frac{2}{1} \cdot{} \frac{2}{3} \cdot{} \frac{4}{3} \cdot{} \frac{4}{5} \cdot{} \frac{6}{5} \cdot{} \frac{6}{7} \cdot{} \frac{8}{7} \cdot{} \frac{8}{9} \cdots \right \rfloor

Find x x .

Notation : \lfloor \cdot \rfloor denotes the floor function .


The answer is 4.

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1 solution

Chew-Seong Cheong
Apr 17, 2016

Let x = π P x = \lfloor \pi P \rfloor , then we have:

P = 2 1 2 3 4 3 4 5 6 5 6 7 8 7 8 9 . . . = lim m k = 1 m 2 k 2 k 1 2 k 2 k + 1 = lim m 2 m m ! 2 m m ! ( 2 m ) ! 2 m m ! 2 m m ! ( 2 m + 1 ) ! = lim m 2 4 m ( m ! ) 4 ( 2 m ) ! ( 2 m + 1 ) ! Using Stirling’s approximation n ! 2 π e n n n + 1 2 = lim m 2 4 m ( 2 π e m m m + 1 2 ) 4 2 π e 2 m ( 2 m ) 2 m + 1 2 2 π e 2 m 1 ( 2 m + 1 ) 2 m + 3 2 = lim m 2 2 m + 1 2 m 2 m + 3 2 π e ( 2 m + 1 ) 2 m + 3 2 = lim m ( 2 m ) 2 m + 3 2 π e 2 ( 2 m + 1 ) 2 m + 3 2 = lim m π e 2 ( 1 + 1 2 m ) 2 m + 3 2 = π e 2 e = π 2 \begin{aligned} P & = \frac{2}{1} \cdot{} \frac{2}{3} \cdot{} \frac{4}{3} \cdot{} \frac{4}{5} \cdot{} \frac{6}{5} \cdot{} \frac{6}{7} \cdot{} \frac{8}{7} \cdot{} \frac{8}{9} ... \cdot{} \infty \\ & = \lim_{m \to \infty} \prod_{k=1}^m \frac{2k}{2k-1} \cdot{} \frac{2k}{2k+1} \\ & = \lim_{m \to \infty} \frac{2^m m! \cdot{} 2^m m!}{(2m)!} \cdot{} \frac{2^m m! \cdot{} 2^m m!}{(2m+1)!} \\ & = \lim_{m \to \infty} \frac{2^{4m} (m!)^4}{(2m)!(2m+1)!} \quad \quad \small \color{#3D99F6}{\text{Using Stirling's approximation } n! \approx \sqrt{2\pi} e^{-n} n^{n+\frac{1}{2}}} \\ & = \lim_{m \to \infty} \frac{2^{4m} \left(\sqrt{2\pi} e^{-m} m^{m+\frac{1}{2}}\right)^4}{\sqrt{2\pi} e^{-2m} (2m)^{2m+\frac{1}{2}} \sqrt{2\pi} e^{-2m-1} (2m+1)^{2m+\frac{3}{2}}} \\ & = \lim_{m \to \infty} \frac{2^{2m+\frac{1}{2}} m^{2m+\frac{3}{2}} \cdot{} \pi e}{(2m+1)^{2m+\frac{3}{2}}} \\ & = \lim_{m \to \infty} \frac{(2m)^{2m+\frac{3}{2}} \cdot{} \pi e}{2(2m+1)^{2m+\frac{3}{2}}} \\ & = \lim_{m \to \infty} \frac{\pi e}{2\left(1+\frac{1}{2m} \right)^{2m+\frac{3}{2}}} \\ & = \frac{\pi e}{2 e} = \frac{\pi}{2} \end{aligned}

x = π P = π π 2 = 4.9348 = 4 \Rightarrow x = \lfloor \pi P \rfloor = \left \lfloor \pi \cdot{} \dfrac{\pi}{2} \right \rfloor = \lfloor 4.9348 \rfloor = \boxed{4}

Great demonstration of the Wallis Product!

Mateo Matijasevick - 5 years, 1 month ago

Great solution

Aakash Khandelwal - 5 years, 1 month ago

Wow, what an amazing proof of wallis product.

Syed Shahabudeen - 5 years ago

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