Bell series

In mathematics, the Bell series is a formal power series used to study properties of arithmetical functions. Bell series were introduced and developed by Eric Temple Bell.

Given an arithmetic function f and a prime p, define the formal power series f_p(x), called the Bell series of f modulo p as:

:f_p(x)=\sum_{n=0}^\infty f(p^n)x^n.

Two multiplicative functions can be shown to be identical if all of their Bell series are equal; this is sometimes called the uniqueness theorem: given multiplicative functions f and g, one has f=g if and only if:

:f_p(x)=g_p(x) for all primes p.

Two series may be multiplied (sometimes called the multiplication theorem): For any two arithmetic functions f and g, let h=f*g be their Dirichlet convolution. Then for every prime p, one has:

:h_p(x)=f_p(x) g_p(x).\,

In particular, this makes it trivial to find the Bell series of a Dirichlet inverse.

If f is completely multiplicative, then formally:

:f_p(x)=\frac{1}{1-f(p)x}.

Examples

The following is a table of the Bell series of well-known arithmetic functions.

  • The Möbius function \mu has \mu_p(x)=1-x.
  • The Mobius function squared has \mu_p^2(x) = 1+x.
  • Euler's totient \varphi has \varphi_p(x)=\frac{1-x}{1-px}.
  • The multiplicative identity of the Dirichlet convolution \delta has \delta_p(x)=1.
  • The Liouville function \lambda has \lambda_p(x)=\frac{1}{1+x}.
  • The power function Idk has (\textrm{Id}_k)_p(x)=\frac{1}{1-p^kx}. Here, Idk is the completely multiplicative function \operatorname{Id}_k(n)=n^k.
  • The divisor function \sigma_k has (\sigma_k)_p(x)=\frac{1}{(1-p^kx)(1-x)}.
  • The constant function, with value 1, satisfies 1_p(x) = (1-x)^{-1}, i.e., is the geometric series.
  • If f(n) = 2^{\omega(n)} = \sum_{d|n} \mu^2(d) is the power of the prime omega function, then f_p(x) = \frac{1+x}{1-x}.
  • Suppose that f is multiplicative and g is any arithmetic function satisfying f(p^{n+1}) = f(p) f(p^n) - g(p) f(p^{n-1}) for all primes p and n \geq 1. Then f_p(x) = \left(1-f(p)x + g(p)x^2\right)^{-1}.
  • If \mu_k(n) = \sum_{d^k|n} \mu_{k-1}\left(\frac{n}{d^k}\right) \mu_{k-1}\left(\frac{n}{d}\right) denotes the Möbius function of order k, then (\mu_k)_p(x) = \frac{1-2x^k+x^{k+1}}{1-x}.

See also

References