Ramanujan's Phi function

Phi function with ${\displaystyle n\to \infty }$

The Ramanujan Phifunktion after Srinivasa Ramanujan by ${\displaystyle f\colon \mathbb {R} \times \mathbb {N} \to \mathbb {R} }$

${\displaystyle \varphi (a,n)=1+2\sum _{k=1}^{n}{1 \over (ak)^{3}-ak}}$

with , , and defined. ${\displaystyle a\in \mathbb {R} }$ ${\displaystyle a,k\neq 0}$ ${\displaystyle ak\neq \pm 1}$ ${\displaystyle n\geq 1}$

For the series it results explicitly:

${\displaystyle \varphi (a,n)=1+2{1 \over a^{3}-a}+2{1 \over 8a^{3}-2a}+2{1 \over 27a^{3}-3a}+\ldots +2{1 \over (an)^{3}-a\cdot n}}$

Representation through the harmonic function

Let the harmonic function be defined using the function . As a result, the Ramanujan phi function can be represented by: ${\displaystyle H_{n}=\sum _{k=1}^{n}{1 \over k}=\gamma +\psi _{0}(n+1),n\geq 1}$

${\displaystyle \varphi (a,n)=1-{1 \over a}(H_{-1/a}+H_{1/a}+2H_{n}-H_{n-1/a}-H_{n+1/a})}$

limit

Let be the limit of the Ramanujan phi function for . In simple terms: ${\displaystyle \varphi (a)}$ ${\displaystyle n\to \infty }$

${\displaystyle \varphi (a)=\lim _{n\to \infty }\varphi (a,n)=1+2\sum _{k=1}^{\infty }{1 \over (ak)^{3}-ak}=-{1 \over a}(\psi _{0}({1 \over a})+\psi _{0}(1-{1 \over a})+2\gamma )=1-{1 \over a}(H_{1/a}+H_{1/a})}$.

Here is the digamma function and the Euler-Mascheroni constant . ${\displaystyle \psi _{0}}$ ${\displaystyle \gamma }$

Values ​​for the Ramanujan phi function

Function values ​​of the Ramanujan phi function for : ${\displaystyle 1<A\leq 6}$

a	${\displaystyle \varphi (a)}$
2	${\displaystyle 2\ln(2)}$
3	${\displaystyle \ln(3)}$
4th	${\displaystyle {3 \over 2}\ln(2)}$
5	${\displaystyle {1 \over 5}{\sqrt {5}}\ln(\phi )+{1 \over 2}\ln(5)}$
6th	${\displaystyle {1 \over 2}\ln(3)+{2 \over 3}\ln(2)}$

Here is the golden ratio . ${\displaystyle \phi ={1+{\sqrt {5}} \over 2}}$

Individual evidence

1. ↑ Eric W. Weisstein: Harmonic Number. Retrieved May 30, 2019 . 
2. ↑ ^{a b} Eric W. Weisstein: Ramanujan phi-Function. Retrieved May 30, 2019 .