📂Number Theory

Inverse of Dirichlet Products

Definition ¹

An arithmetic function $f^{-1}$ is said to be the (Dirichlet) inverse of another arithmetic function $f$ if there exists a unique arithmetic function $f^{-1}$ satisfying the following equation $f$. $$ f \ast\ f^{-1} = f^{-1} \ast\ f = I $$

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• Here, $I$ is the identity function with respect to convolution.

Theorem

• [1]: If an arithmetic function $f$ is $f(1) \ne 0$, then its inverse $f^{-1}$ uniquely exists and is represented by the following recursive function. $$ f^{-1}(n) = \begin{cases} \displaystyle {{1} \over {f(1)}} &,n=1 \\ \displaystyle {{-1} \over {f(1)}} \sum_{d \mid n , d < n } f \left( {{ n } \over { d }} \right) f^{-1}(d) &, n > 1\end{cases} $$
• [2]: If for two arithmetic functions $f$ and $g$, $f(1) \ne 0$ and $g(1) \ne 0$ are satisfied, then $$ (f \ast\ g)^{-1} = g^{-1} \ast\ f^{-1} = f^{-1} \ast\ g^{-1} $$

Explanation

Unlike the inverse functions commonly dealt with in most mathematics, the Dirichlet inverse is not an inverse in the sense of mapping but in an algebraic sense. This naturally leads one to think of algebraic structures, and the existence and uniqueness of such an inverse become questions of interest. Fortunately, the existence of an inverse is guaranteed by satisfying a very simple condition.

Especially for multiplicative arithmetic functions, the existence of an inverse is assured for the following reasons: $$ f(1) = f(1 \cdot 1 ) = f(1) f(1) = 1 \ne 0 $$ From these facts, the condition for the set of arithmetic functions to form an Abelian group becomes $f(1) \ne 0$.

Proof

[1]

Since $\displaystyle I(n) = \left[ {{ 1 } \over { n }} \right]$, when $n=1$, $\left( f \ast\ f^{-1} \right)(1) = I(1) = 1$, and therefore, $\displaystyle f^{-1}(1) = {{ 1 } \over { f(1) }}$. Because $f(1) \ne 0$, $f^{-1}(1)$ is also unique. If $n >1$, then $\left( f \ast\ f^{-1} \right)(n) = I(n) = 0$, thus $$ \sum_{d \mid n} f \left( {{ n } \over { d }} \right) f^{-1} (d) = 0 $$ Removing terms that are $n=d$, $$ f(1) f^{-1}(n) + \sum_{d \mid n \\ d < n} f \left( {{ n } \over { d }} \right) f^{-1} (d) = 0 $$ Summarizing, $$ f^{-1}(n) = {{-1} \over {f(1)}} \sum_{d \mid n \\ d < n } f \left( {{ n } \over { d }} \right) f^{-1}(d) $$ Since it has been shown that $f^{-1}(1) \ne 0$ uniquely exists for the case of $n=1$, by mathematical induction, $f^{-1}(n)$ also uniquely exists.

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[2]

Strategy: Once existence is proved, direct calculation is possible due to the properties of convolution.

Properties of convolution:

• (1) Associative law: $$ \left( f \ast g \right) \ast k = f \ast (g \ast k) $$
• (2) Commutative law: $$ f \ast\ g = g \ast\ f $$

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Based on the above [1], since $f(1) \ne 0$ and $g(1) \ne 0$, $f^{-1}$ and $g^{-1}$ uniquely exist. Likewise, $$ (f \ast\ g) (1) = \sum_{ d \mid 1} f(d)g(1/d) = f(1) g(1) \ne 0 $$ Therefore, $\left( f \ast g \right)^{-1}$ also uniquely exists. Then by the associative law of convolution, $$ \begin{align*} (f \ast\ g) \ast ( g^{-1} \ast\ f^{-1} ) =& f \ast (g * g^{-1} ) \ast\ f^{-1} \\ =& f \ast\ I \ast\ f^{-1} \\ =& f * f^{-1} \\ =& I \end{align*} $$ Since $\left( f \ast g \right)^{-1}$ is unique, $\left( f \ast g \right)^{-1} = g^{-1} \ast\ f^{-1}$ must be true. Additionally, by the commutative law of convolution, $$ (f \ast\ g)^{-1} = g^{-1} \ast\ f^{-1} = f^{-1} \ast\ g^{-1} $$

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