Wikipedia:Reference desk/Archives/Mathematics/2013 May 30

= May 30 =

Generalized totient function
Is there such a thing as a generalized totient function? Specifically, a function &phi;i(n) that is the count of positive integers less than n with exactly i prime divisors. This means that Euler's totient function &phi;(n) is &phi;1(n) in the more general form (i.e., it is the number of positive integers less than n with only one prime divisor). Likewise, &phi;2(n) is the number of positive composite integers less than n with only two prime divisors, e.g., numbers from the set {4,6,9,10,14,15,21,22,25,...,n}, which includes all the squares of primes. &phi;3(n) includes {8,12,18,27,28,30,...,n}, and so forth. I dimly recall seeing something about Ramanujan studying something similar to this(?). Perhaps such a thing might also be related to the Riemann hypothesis? — Loadmaster (talk) 17:24, 30 May 2013 (UTC)


 * Do you mean prime counting function rather than totient function? Otherwise I'm very confused.  Sławomir Biały  (talk) 17:29, 30 May 2013 (UTC)


 * Yes, that's what I meant. So (replacing &phi; above with &pi;): Is there such a thing as a generalized prime counting function? Specifically, a function &pi;i(n) that is the count of positive integers less than n with exactly i prime divisors. — Loadmaster (talk) 18:48, 30 May 2013 (UTC)
 * Yes, but these can be expressed in terms of the functions pi[n^(1/i)] using Mobius inversion. Count Iblis (talk) 19:54, 30 May 2013 (UTC)

Elliptic integrals and Gaussian integrals
With your help, I would like to find out a mathematical relationship between complete elliptic integrals of the first kind
 * $$K(n)\ =\ \int_0^{\pi\over2} \frac{d\theta}{\sqrt{1-n^2 \sin^2\theta}}\ =\ \int_0^1 \frac{dt}{\sqrt{(1-t^2)(1-n^2 t^2)}}$$

and gaussian integrals
 * $$G(n)\ =\ \int_0^\infty{e^{-x^n}}dx$$

all of which are known to possess the following property
 * $$\frac{G(m) \cdot G(n)}{G({m\ \cdot\ n \over m\ +\ n})}\ =\ \int_0^1{\sqrt[m]{1-x^n}}dx\ =\ \int_0^1{\sqrt[n]{1-x^m}}dx$$

where
 * $$\tfrac{m\ n}{m+n}$$ is half of the harmonic mean between m and n, and the entire above expression is equal to the product between 1 + $$\tfrac1m$$ + $$\tfrac1n$$ and the beta function of arguments 1 + $$\tfrac1m$$ and 1 + $$\tfrac1n$$.

It also goes on without saying that the factorial of every positive number is the gaussian integral of its reciprocal or multiplicative inverse
 * $$n!\ =\ G(\tfrac1n)\ =\ \int_0^\infty{e^{-\sqrt[n]x}}dx$$

— 79.118.171.165 (talk) 18:33, 30 May 2013 (UTC)