A math course explaining the Riemann zeta function
Curated by: Zeta Explained (121 videos)
This is the 78th video in a series explaining the Riemann zeta function. The idea of the series is to start with basics and eventually work our way to advanced topics. Today we follow "The Riemann Zeta-Function" by Aleksandar Ivić. We also reference the paper Balasubramanian, R., Ramachandra, K., & Subbarao, M. V. (1988). On the error function in the asymptotic formula for the counting function of k-full numbers. Acta Arithmetica, 50(2), 107-118. https://eudml.org/doc/206125 This particular video discusses powerful, or squareful, numbers, which are numbers where the exponent of every prime factor is at least 2. The number 1 is considered powerful, despite not matching this definition. We mainly cover the result that if f_2(n) is the indicator function that is equal to 1 if n is powerful and 0 otherwise, then ∑_(n=1 to ∞) f_2(n)/n^s = ζ(2s)ζ(3s)/ζ(6s). We also show the result for cubeful numbers, which are numbers where the exponent of every prime factor is at least 3. We get an infinite product of zeta functions. Finally, we show very informally that the number of powerful numbers up to x is approximately [ζ(3/2)/ζ(3)]x^(1/2) + [ζ(2/3)/ζ(2)]x^(1/3). We also mention that the Riemann Hypothesis is related to the bound of the error term of this approximation. 00:00 - Intro 04:00 - Theorem 1: ∑_(n=1 to ∞) f_2(n)/n^s = ζ(2s)ζ(3s)/ζ(6s) 08:58 - Corollary 1: Cubeful Numbers: ∑_(n=1 to ∞) f_3(n)/n^s = ζ(3s)ζ(4s)ζ(5s).../[ζ(8s)ζ(9s)ζ(10s)...] 18:11 - Theorem 2: The number of powerful numbers up to x is approximately [ζ(3/2)/ζ(3)] x^(1/2) + [ζ(2/3)/ζ(2)] x^(1/3)