Solution 86: Double Factorial and Roots of Unity Filter (Proof)

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We prove the roots of unity filter and apply it to evaluate a fascinating summation.

Congratulations to Gabriel N., Essentials of Math, Peter, fmakofmako, Hiren Bavaskar, Paco Libre, aby p, santosh tripathy, Proof by Meme, and Mushishi2872 for successfully solving this math challenge question! Gabriel N. was the first person to solve the question.

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I love it so much when I learn new techniques. And it comes with a well explained proof <3

Risuchan

This inspired me to try to find the general solution for any S(a) = Sum[1/(2an)!!, {n, 0, Infinity}] where a is a positive integer.

Turns out with some cleaver manipulation of the sum after applying the filter it generalizes to:

S(a) = (1/a) Sum[e^((1/2) cos (k theta)) e^((i/2) sin (k theta)), {k, 0, a-1}] = (1/2) Sum[f(k), {k, 0, a-1}] (where theta is 2 pi/a)

which simplifies by noticing that f(k) + f(a-k) = 2 e^((1/2) cos(k theta)) * cos[(1/2) sin (k theta)]

thus (after manipulating and shifting the index of the sum)

S(a) = (1/a)(e^1/2) + (2/a) Sum[e^((1/2) cos(m theta)) * cos((1/2) sin(m theta)), {m, 1, k}]

where if a is even, k = (a/2), and if a is odd k = (1/2)(a-1)
and theta = 2 pi/a

Plugging in a = 3 (as in the video) we get

S(3) = (1/3)(e^(1/2)) + (2/3) e^(-1/4) cos(sqrt(3)/4)

jeffreyhersh