Unit 6: Möbius Inversion, Average Orders & Ramanujan Sums
Advanced arithmetical function transforms and asymptotic distribution theory: the Möbius function $\mu(n)$ as the Dirichlet inverse of the constant function, the Möbius Inversion Formula and its multiplicative analogue, Ramanujan's trigonometric sums $c_q(n)$ and their structural identities, the von Mangoldt function $\Lambda(n)$ and Chebyshev's prime counting functions, and Dirichlet's hyperbola method for the average order of the divisor function.
§6.1 The Möbius Function mu(n) & The Möbius Inversion Formula
1. The Möbius Function
August Ferdinand Möbius introduced this fundamental function in 1832.
Definition 6.1 (The Möbius Function): The Möbius function $\mu: \mathbb{N} \to \{-1, 0, 1\}$ is defined by:
Theorem 6.1 (Fundamental Divisor Sum Identity of $\mu$): For every positive integer $n \ge 1$:
In the language of Dirichlet convolution, $\mu$ is the Dirichlet inverse of the constant function $u(n) = 1$:
Proof: For $n = 1$: $\sum_{d \mid 1} \mu(d) = \mu(1) = 1$. For $n > 1$: let $n = p_1^{a_1} p_2^{a_2} \cdots p_k^{a_k}$ with $k \ge 1$. Any divisor $d \mid n$ with a square factor contributes $\mu(d) = 0$. Thus, only square-free divisors $d = p_{i_1} p_{i_2} \cdots p_{i_j}$ contribute non-zero values $\mu(d) = (-1)^j$. The number of such divisors having exactly $j$ prime factors is $\binom{k}{j}$. Therefore, summing over all possible counts of prime factors $j \in \{0, 1, \dots, k\}$:
By the Binomial Theorem:
This establishes the identity completely. $\blacksquare$
2. The Möbius Inversion Formula
Theorem 6.2 (The Möbius Inversion Formula): Let $f$ and $F$ be arithmetical functions. Then:
Proof (via Dirichlet Convolution): The relation $F(n) = \sum_{d \mid n} f(d)$ is expressed concisely as:
Convolving both sides with the Möbius function $\mu$:
By associativity and commutativity of Dirichlet convolution (Theorem 5.8):
Thus:
The reverse implication follows symmetrically by convolving with $u$. $\blacksquare$
§6.2 Applications to Euler's Function & Multiplicative Inversion
1. Inversion of Euler's Totient Identity
Theorem 6.3 (Gauss's Totient Divisor Sum): For every positive integer $n \ge 1$:
Proof: Partition the set of integers $\{1, 2, \dots, n\}$ into classes based on their greatest common divisor with $n$:
Notice that $\gcd(k, n) = d \iff \gcd(k/d, n/d) = 1$. Letting $k' = k/d$, the number of such integers is the number of $1 \le k' \le n/d$ with $\gcd(k', n/d) = 1$, which is precisely $\phi(n/d)$. Summing the sizes of all disjoint subsets $S_d$:
Applying the Möbius Inversion Formula to $F(n) = n$ and $f(n) = \phi(n)$:
Theorem 6.4 (Möbius Inversion Formula for $\phi(n)$):
2. Multiplicative Form of Möbius Inversion
For multiplicative relationships involving products rather than sums:
Theorem 6.5 (Product Form of Möbius Inversion): Let $g$ and $G$ be functions from $\mathbb{N}$ to $\mathbb{C}^\times$. Then:
Proof: Take the complex logarithm on both sides: $\ln G(n) = \sum_{d \mid n} \ln g(d)$. Applying standard additive Möbius Inversion to $\ln G$ and $\ln g$:
Exponentiating both sides establishes the result. $\blacksquare$
§6.3 Ramanujan's Trigonometric Sums cq(n) & Explicit Closed Forms
1. Definition of Ramanujan's Sum
Srinivasa Ramanujan introduced these sums in his 1918 paper On Certain Trigonometrical Sums and their Applications in the Theory of Numbers.
Definition 6.2 (Ramanujan's Trigonometric Sum): For positive integers $q, n \in \mathbb{N}$, Ramanujan's sum $c_q(n)$ is defined as the sum of the $n$-th powers of the primitive $q$-th roots of unity:
(the imaginary sine components cancel out by symmetry).
2. The Fundamental Identity of Ramanujan's Sum
Theorem 6.6 (Ramanujan's Divisor Sum Identity): For all positive integers $q$ and $n$:
Proof: Recall the fundamental character orthogonality sum for roots of unity:
Every residue $a \in \{1, 2, \dots, q\}$ satisfies $\gcd(a, q) = d$ for some unique divisor $d \mid q$. Writing $a = d a'$ and $q = d q'$ with $\gcd(a', q') = 1$:
Applying Möbius Inversion (Theorem 6.2) with respect to the variable $q$:
Since $\eta_d(n) = d$ when $d \mid n$ and $0$ otherwise, the sum runs only over divisors $d$ that divide both $q$ and $n$, that is, $d \mid \gcd(q, n)$:
Corollary 6.1 (Special Values):
- When $n = 1$: $c_q(1) = \mu(q)$.
- When $q \mid n$: $c_q(n) = \phi(q)$.
- For any prime $p$:
§6.4 The von Mangoldt Function & Chebyshev's Counting Functions
1. The von Mangoldt Function $\Lambda(n)$
Definition 6.3 (The von Mangoldt Function): The von Mangoldt function $\Lambda: \mathbb{N} \to \mathbb{R}$ is defined by:
Theorem 6.7 (Logarithmic Divisor Sum Identity): For every positive integer $n \ge 1$:
Proof: Let $n = p_1^{a_1} p_2^{a_2} \cdots p_k^{a_k}$ be the prime factorization. The only divisors $d \mid n$ for which $\Lambda(d) \ne 0$ are the prime powers $p_i^j$ ($1 \le j \le a_i$). Therefore:
By Möbius Inversion:
2. Chebyshev's Functions $\psi(x)$ and $\theta(x)$
Pafnuty Chebyshev introduced these functions to study the distribution of primes.
Definition 6.4 (Chebyshev Functions): For $x \ge 1$:
- Chebyshev's theta function: $\theta(x) = \sum_{p \le x} \ln p$
- Chebyshev's psi function: $\psi(x) = \sum_{n \le x} \Lambda(n) = \sum_{p^k \le x} \ln p$
Theorem 6.8 (Equivalence with Prime Number Theorem): The Prime Number Theorem $\pi(x) \sim \frac{x}{\ln x}$ is logically equivalent to:
§6.5 Average Orders of Arithmetical Functions & Dirichlet's Hyperbola Method
1. The Concept of Average Order
Many arithmetical functions (like $\tau(n)$ or $\mu(n)$) fluctuate wildly from point to point. To understand their macroscopic behavior, number theorists study their average order:
We say $f(n)$ has average order $g(n)$ if $\sum_{n \le x} f(n) \sim \sum_{n \le x} g(n)$.
2. Dirichlet's Divisor Problem & The Hyperbola Method
Consider the summatory function of the divisor function:
This counts the total number of integer lattice points $(u, v) \in \mathbb{N}^2$ lying under the hyperbola:
Theorem 6.9 (Dirichlet's Divisor Asymptotic, 1849): As $x \to \infty$:
where $\gamma \approx 0.57721566...$ is the Euler-Mascheroni constant.
Proof (Dirichlet's Hyperbola Method): Consider the hyperbolic region $\mathcal{R} = \{(u, v) \in \mathbb{R}^2 : u \ge 1, v \ge 1, uv \le x\}$. By symmetry across the line $u = v$, split the region into three pieces:
- Region I: $u \le \sqrt{x}$ and $v \le x/u$.
- Region II: $v \le \sqrt{x}$ and $u \le x/v$.
- Overlap (square): $u \le \sqrt{x}$ and $v \le \sqrt{x}$.
Counting lattice points by inclusion-exclusion:
Replace $\lfloor x/u \rfloor = x/u - \{x/u\}$:
Recall the asymptotic for harmonic numbers $H_N = \sum_{k=1}^N \frac{1}{k} = \ln N + \gamma + O(1/N)$. Setting $N = \lfloor \sqrt{x} \rfloor$:
Multiplying by $2x$:
Now subtract the overlap square $\lfloor \sqrt{x} \rfloor^2 = (\sqrt{x} + O(1))^2 = x + O(\sqrt{x})$:
Step-by-step rigorous derivations with unskipped proofs, categorized into Foundational Concepts, Advanced Structural Analysis, and Honors / Proof Challenge tiers.
Consider the integer $n = 30$:
- List all positive divisors $d$ of $30$.
- Compute the value of the Möbius function $\mu(d)$ for each divisor.
- Apply the Möbius inversion identity $\phi(n) = \sum_{d \mid n} d \, \mu(n/d)$ to explicitly compute $\phi(30)$ and verify that it matches the product formula.
1. Divisors of $30$
The prime factorization is $30 = 2 \times 3 \times 5$. The positive divisors are:
2. Values of the Möbius Function
- $\mu(1) = 1$
- $\mu(2) = (-1)^1 = -1$
- $\mu(3) = (-1)^1 = -1$
- $\mu(5) = (-1)^1 = -1$
- $\mu(6) = \mu(2 \times 3) = (-1)^2 = 1$
- $\mu(10) = \mu(2 \times 5) = (-1)^2 = 1$
- $\mu(15) = \mu(3 \times 5) = (-1)^2 = 1$
- $\mu(30) = \mu(2 \times 3 \times 5) = (-1)^3 = -1 \quad \blacksquare
\begin{aligned} \phi(30) &= 1 \times \mu(30) + 2 \times \mu(15) + 3 \times \mu(10) + 5 \times \mu(6) \\ &\quad + 6 \times \mu(5) + 10 \times \mu(3) + 15 \times \mu(2) + 30 \times \mu(1) \\ &= 1(-1) + 2(1) + 3(1) + 5(1) + 6(-1) + 10(-1) + 15(-1) + 30(1) \\ &= -1 + 2 + 3 + 5 - 6 - 10 - 15 + 30 \\ &= 9 - 31 + 30 = 8 \end{aligned}
\phi(30) = 30 \left(1 - \frac{1}{2}\right)\left(1 - \frac{1}{3}\right)\left(1 - \frac{1}{5}\right) = 30 \times \frac{1}{2} \times \frac{2}{3} \times \frac{4}{5} = 30 \times \frac{8}{30} = 8 \quad \checkmark
Consider Ramanujan's trigonometric sum $c_6(n) = \sum_{\substack{a=1 \\ \gcd(a, 6)=1}}^6 e^{2\pi i a n / 6}$:
- Identify all reduced residues modulo $6$.
- State Ramanujan's identity $c_q(n) = \sum_{d \mid \gcd(q, n)} d \, \mu(q/d)$.
- Evaluate $c_6(n)$ explicitly for each integer $n \in \{1, 2, 3, 4, 5, 6\}$.
1. Reduced Residues Modulo $6$
The integers in $\{1, 2, 3, 4, 5, 6\}$ coprime to $6$ are:
Thus, $\phi(6) = 2$, and the sum has exactly two terms:
2. Divisor Formula
By Theorem 6.6:
3. Explicit Values for $n = 1, 2, 3, 4, 5, 6$
- For $n = 1$: $\gcd(6, 1) = 1$.
Trigonometric check: $2 \cos(\pi/3) = 2(1/2) = 1 \quad \checkmark$
- For $n = 2$: $\gcd(6, 2) = 2$. Divisors: $\{1, 2\}$.
Trigonometric check: $2 \cos(2\pi/3) = 2(-1/2) = -1 \quad \checkmark$
- For $n = 3$: $\gcd(6, 3) = 3$. Divisors: $\{1, 3\}$.
Trigonometric check: $2 \cos(\pi) = 2(-1) = -2 \quad \checkmark$
- For $n = 4$: $\gcd(6, 4) = 2$.
Trigonometric check: $2 \cos(4\pi/3) = 2(-1/2) = -1 \quad \checkmark$
- For $n = 5$: $\gcd(6, 5) = 1$.
Trigonometric check: $2 \cos(5\pi/3) = 2(1/2) = 1 \quad \checkmark$
- For $n = 6$: $\gcd(6, 6) = 6$. Divisors: $\{1, 2, 3, 6\}$.
Trigonometric check: $2 \cos(2\pi) = 2(1) = 2 \quad \checkmark$
Summary of values:
Prove Dirichlet's asymptotic formula for the sum of the divisor function:
- Express $\sum_{n \le x} \tau(n)$ as the count of integer lattice points $(u, v) \in \mathbb{N}^2$ such that $u v \le x$.
- Decompose the counting sum using Dirichlet's hyperbola symmetry across $u = v = \sqrt{x}$.
- Use the Euler-Maclaurin expansion for harmonic numbers $\sum_{u \le K} \frac{1}{u} = \ln K + \gamma + O(1/K)$ to evaluate the principal terms and rigorously establish the $O(\sqrt{x})$ error bound.
1. Lattice Point Representation
Recall that $\tau(n) = \sum_{d \mid n} 1 = \sum_{u v = n} 1$. Therefore:
This counts the total number of integer points in the first quadrant lying on or beneath the rectangular hyperbola $u v = x$. $\blacksquare$
2. Dirichlet's Hyperbola Decomposition
Let $K = \lfloor \sqrt{x} \rfloor$. The hyperbola region can be split into three parts:
- Points with $u \le K$: for each fixed $u$, $v$ can range from $1$ to $\lfloor x/u \rfloor$.
Count $= \sum_{u=1}^K \lfloor x/u \rfloor$.
- Points with $v \le K$: by symmetry between $u$ and $v$, this also equals $\sum_{v=1}^K \lfloor x/v \rfloor$.
- Points in the square $[1, K] \times [1, K]$ were counted in both (1) and (2).
Their count is $K^2 = \lfloor \sqrt{x} \rfloor^2$. By inclusion-exclusion:
3. Evaluation of the Sums and Error Bounding
Write $\lfloor x/u \rfloor = \frac{x}{u} - \left\{ \frac{x}{u} \right\}$ where $\{t\} \in [0, 1)$ denotes the fractional part:
Since $0 \le \{x/u\} < 1$:
Now use the classical asymptotic expansion of the harmonic sum:
Since $K = \sqrt{x} + O(1)$, we have:
Thus:
Multiplying by $x$:
Subtracting the fractional error:
Now multiply by $2$:
Finally, evaluate the subtracted square $K^2$:
Subtracting $K^2$:
This complete proof confirms Dirichlet's landmark result!