Wikipedia:Reference desk/Archives/Mathematics/2019 March 18

= March 18 =

Intuition behind &zeta;(3)3 and &zeta;(9) being algebraically independent
For the zeta function of even arguments, we have $$\dfrac{\zeta(2an)^b}{\zeta(2bn)^a}\in\mathbb Q.$$ Now, I possess some basic intuition for why the following statements (proven or conjectural), are most likely true: So, in light of the above, I am hardly shocked by the fact that &zeta;(3)2 and &zeta;(6), for instance, are algebraically independent of one another, but, at the same time, I am rather startled as to why &zeta;(3)3 and &zeta;(9), for example, are algebraically independent as well; or, to put it in more general terms, I am at a loss for grasping, intuitively, why zeta functions whose arguments are powers of the same prime, appear to be algebraically independent. — 86.123.9.238 (talk) 06:46, 18 March 2019 (UTC)
 * $$\dfrac{\zeta(2n)}{\pi^{2n}}\in\mathbb Q.$$
 * $$\dfrac{\zeta(2n+1)}{\pi^{2n+1}}\not\in\mathbb Q.$$
 * $$\dfrac{\zeta(p)^q}{\zeta(q)^p}\not\in\mathbb Q,~$$ for distinct prime values of p and q.
 * Not really sure what kind of answer you're looking for, but I think the short one is that there's no (apparent) relationship between ζ(3) and ζ(9) is there's no reason for there to be one. The reductions that allow you prove ζ(2n) is a rational multiple of a power of π don't work for ζ(2n+1). Similarly, there are reductions that allow you to prove that
 * $$ \int_0^\infty x^z e^{-x}\,dx$$
 * is an integer when z is an integer (namely z!), but they don't apply when z is not an integer. To me the more surprising fact is that ζ(2) does have a simple expression. If you look at the history (see Basel problem) it took nearly a century for Euler to find this, so it certainly wasn't 'intuitive' for mathematicians at the time. --RDBury (talk) 17:53, 20 March 2019 (UTC)
 * &zeta;(2n) = a2n &pi;2n were initially defined as infinite sums of squares. Sums of squares also appear in, say, $$\int_0^\infty\frac{dx}{1+x^2}=\int_{-1}^1\sqrt{1-x^2}~dx,~$$ the latter being intrinsically linked to the well-known equation x2 + y2 = r2, whose constant is &Gamma;2(1/2) = &pi;: This is the intuition that I had in mind. Personally, I expected all &zeta;(3n) to be expressed in terms of &Gamma;(1/3). — 84.232.135.229 (talk) 02:22, 22 March 2019 (UTC)