Euler the magician

Calculus Level 5

p P ( 1 + sin ( π p 2 ) p ) \large \prod_{p \in \mathfrak{P}} \left ( 1 + \frac{\sin (\frac{\pi p}{2})}{p} \right)

Let P \mathfrak{P} be the set of prime numbers. Then, what is the value of the above expression?


The answer is 0.63661977.

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1 solution

Mark Hennings
Feb 28, 2018

If χ \chi is the nontrivial Dirichlet character modulo 4 4 , so that χ ( n ) = sin n π 2 = { 1 n 1 ( m o d 4 ) 1 n 3 ( m o d 4 ) 0 o . w . \chi(n) \; = \; \sin \tfrac{n\pi}{2} \; = \; \left\{ \begin{array}{lll} 1 & \hspace{1cm} & n \equiv 1 \pmod{4}\\ -1 & & n \equiv 3 \pmod{4} \\ 0 & & \mathrm{o.w.} \end{array} \right. then we are interested in the product P + = p > 2 ( 1 + χ ( p ) p ) P_+ \; = \; \prod_{p > 2} \left(1 + \tfrac{\chi(p)}{p}\right) It is a well-known result that the product P = p > 2 ( 1 χ ( p ) p ) P_- \; = \; \prod_{p > 2}\left(1 - \tfrac{\chi(p)}{p}\right) is such that P 1 = 1 4 π P_-^{-1} \; = \; \tfrac14\pi On the other hand P + P = p > 2 ( 1 1 p 2 ) = 4 3 p ( 1 1 p 2 ) = 4 3 ζ ( 2 ) 1 = 8 π 2 P_+P_- \; =\; \prod_{p > 2} \left(1 - \tfrac{1}{p^2}\right) \; = \; \tfrac43\prod_p \left(1 - \tfrac{1}{p^2}\right) \; = \; \tfrac43 \zeta(2)^{-1} \; = \; \tfrac{8}{\pi^2} and hence P + = 8 π 2 × 1 4 π = 2 π = 0.63661977 P_+ \; = \; \tfrac{8}{\pi^2} \times \tfrac14\pi \; = \; \tfrac{2}{\pi} \; = \; \boxed{0.63661977}

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