Lagmental Vicfred

Möbius Inversion Recovers a Function from Its Divisor Sum by Vicfred

tags: mathematics, arithmetic functions, Dirichlet convolution, Mobius inversion

If F is the divisor sum of f, then f is the Möbius-weighted divisor sum of F. The example is deliberately concrete; it is a test of the statement, not a substitute for it.

Statement

Arithmetic functions \(f:\mathbf N\to\mathbf C\) form a commutative ring under Dirichlet convolution. Multiplicative functions are determined by their values on prime powers \(p^k\).

$$ F(n)=\sum_{d\mid n}f(d) $$

The definition determines which expressions are legal; only then does the identity become meaningful. An equality in \(\mathcal A\) may change ambient meaning, so I keep \(\mathsf D\) separate from \(\mathsf C\).

$$ f(n)=\sum_{d\mid n}\mu(d)F(n/d) $$

Worked algebra

The following line is the smallest calculation that still exercises the mechanism. It keeps nested delimiters and the order of operations explicit.

$$ \varphi(n)=\sum_{d\mid n}\mu(d)\frac nd=n\sum_{d\mid n}\frac{\mu(d)}d $$

Conceptual compression

The compact alignment is a local map of the argument: assumptions on the first row, consequence on the second. Any generalisation must preserve that dependency.

$$ \begin{aligned} \mathsf{D}\;&:\quad F(n)=\sum_{d\mid n}f(d),\\[5pt] \mathsf{C}\;&:\quad f(n)=\sum_{d\mid n}\mu(d)F(n/d). \end{aligned} $$

Caveat

Pointwise multiplication and Dirichlet convolution are different operations. Möbius inversion reverses convolution with the constant-one function, not ordinary multiplication.

$$ \boxed{\begin{gathered} \text{compact conclusion}\\[-2pt] f(n)=\sum_{d\mid n}\mu(d)F(n/d) \end{gathered}} $$

A symbolic summary is trustworthy only because the example and limitation remain visible beside it. The box compresses the conclusion without hiding its origin.

This article was posted on Mon 31 December 2012. Facts and circumstances may have changed since publication.
Please contact me before jumping to conclusions if something seems wrong or unclear.