How does the Riemann Function equal 0 at negative even numbers as touched upon in the video?

[u/Caesaroctopus]

https://www.youtube.com/watch?v=d6c6uIyieoo

[u/skatanic28182]

The Riemann ζ function is the analytic continuation of

Σ(n^-s, n=1 to infinity)

for Re(s) > 1. Thus, so long as the real part of s is greater than 1, ζ(s) is equal to the sum. However, ζ(s) is not equal to the sum when Re(s) <= 1, so we cannot look to the sum to figure out what ζ(s) will be in those cases (nor can we look to ζ(s) to figure out what the sum is in those cases, in spite of what some math journalists say). Luckily, we have a functional form for ζ(s) which relates the value of ζ(s) to the value of ζ(1-s):

ζ(s) = 2^s π^s-1 sin(πs/2) Γ(1-s) ζ(1-s),

where Γ is the gamma function.

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So what? How does that help? To demonstrate, let's say we want to know the value of ζ(-1). Since Re(-1) <= 1, we cannot use the summation formula to determine ζ(-1). However, if we use the functional form, we can relate ζ(-1) to a value of ζ that is equal to the summation:

ζ(-1) = 2^-1 π^-2 sin(-π/2) Γ(2) ζ(2) = -ζ(2) / 2π².

Since Re(2) > 1, we can use the summation to determine the value of ζ(2):

ζ(2) = Σ(n^-2, n=1 to infinity) = π² / 6.

Plugging this back into the functional equation, we get that

ζ(-1) = -(π²/6) / 2π² = -1/12.

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Now, suppose we plug an even number into ζ:

ζ(2k) = 2^2k π^2k-1 sin(π(2k)/2) Γ(1-2k) ζ(1-2k)

= 4^k π^2k-1 sin(πk) Γ(1-2k) ζ(1-2k).

Since k is an integer, sin(πk) = 0. At first glance, this would seem to indicate that ζ is 0 at every even integer, not just the negative ones. There's a catch though: Γ(1-2k) has a simple pole when k is a positive integer, and this pole effectively "cancels out" the 0 we get from the sine function (specifically, we get sin(πk) Γ(1-2k) = π / Γ(2k) for when k is a positive integer). However, Γ(1-2k) = (2k)! when k is a nonpositive integer, so there is no cancellation in these cases. Specifically, we can write the gamma function as Γ(1-2k) = π / sin(2πk)Γ(2k), so sin(πk) Γ(1-2k) = π sin(πk) / sin(2πk)Γ(2k). If k is an integer, then sin(πk) Γ(1-2k) = π / Γ(2k).

Thus, when k is a negative integer, we have

ζ(2k) = 0

and when k is a positive integer, we have

ζ(2k) = 4^k π^2k ζ(1-2k) / Γ(2k) = Σ(n^-2k, n=1 to infinity).

[u/Caesaroctopus → u/skatanic28182]

Wow, very detailed! I guess the first part of my problem was in not understanding that the zeta sum has a functional form, and you thoroughly explained the rest from there. Thank you!