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Analytic Number Theory Lecture 3: Dirichlet Series Convergence

Added: 2026-08-07

Deep Dive

Fundamentals of complex analysis, specifically uniform convergence of complex functions and properties of holomorphic functions.

Complex analysis: Locally uniform convergence

Complex analysis: Locally uniform convergence

13.8K

This section establishes three convergence notions for holomorphic functions: pointwise convergence (each z has gₙ(z)→g(z)), uniform convergence (sup_z |gₙ-g|→0), and uniform convergence on compact sets. Pointwise convergence is too weak—it fails to preserve continuity and integrability, as shown by spike examples where integrals don't commute with limits. Uniform convergence is stronger but rarer. The key insight is that uniform convergence preserves continuity and allows integral interchange on finite intervals. This foundation enables proving that uniform limits of holomorphic functions are holomorphic, solving the central problem of showing series converge to holomorphic functions.

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Sequences of Complex Functions -- Complex Analysis 16

Sequences of Complex Functions -- Complex Analysis 16

8.1K

In complex analysis, a sequence of complex-valued functions converges uniformly to a limit function if, for every ε > 0, there exists an N such that for all n ≥ N and for all z in the domain, |f(z) - fₙ(z)| < ε; this stronger form of convergence preserves important properties like continuity and allows interchange of limits with integrals, unlike pointwise convergence, and if the functions are analytic and converge uniformly on a domain, the limit function is also analytic, with derivatives converging uniformly as well.

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Convergence of Infinite Products of Holomorphic Functions

Convergence of Infinite Products of Holomorphic Functions

299

If a sequence of holomorphic functions $ f_n(z) $ on a region $ \Omega $ satisfies that the series $ \sum_{n=1}^{\infty} (f_n(z) - 1) $ converges absolutely and uniformly on compact subsets of $ \Omega $, then the infinite product $ \prod_{n=1}^{\infty} f_n(z) $ also converges absolutely and uniformly to a holomorphic function $ f(z) $ on $ \Omega $; moreover, $ f(z) $ vanishes at a point $ \zeta $ if and only if only finitely many functions $ f_n $ vanish at $ \zeta $.

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HSST MATHS WORKSHOP QUESTION SERIES COMPLEX ANALYSIS

HSST MATHS WORKSHOP QUESTION SERIES COMPLEX ANALYSIS

304

A function f(z) = u + iv is holomorphic (analytic) if and only if u and v are harmonic conjugates, meaning they satisfy the Cauchy-Riemann equations. To check if f is holomorphic: (1) Verify Cauchy-Riemann equations are satisfied. (2) Verify partial derivatives are continuous. Key properties of analytic functions: (1) f'(z) = ∂u/∂x + i∂v/∂x. (2) If u is constant, f is constant. (3) If v is constant, f is constant. (4) If |f| is constant, f is constant. These properties are essential for solving competitive exam problems.

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Complex analysis: Holomorphic functions

Complex analysis: Holomorphic functions

41.4K

A complex function f(z) is holomorphic at a point if it is complex differentiable there, meaning it can be approximated by a linear function f(z) ≈ f(z₀) + a(z - z₀) + error, where the error divided by |z - z₀| tends to zero. For a function w = u + iv of z = x + iy to be holomorphic, its real and imaginary parts must satisfy the Cauchy-Riemann equations: ∂u/∂x = ∂v/∂y and ∂u/∂y = -∂v/∂x. These equations ensure that the Jacobian matrix of partial derivatives corresponds to multiplication by a complex number a = ∂u/∂x + i∂v/∂x. Holomorphic functions are also called analytic functions in complex analysis, and they possess remarkable properties such as being infinitely differentiable and having convergent power series expansions.

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Deep Dive

Basic arithmetic functions and the concept of multiplicativity (e.g., multiplicative and completely multiplicative functions).

Multiplicative and completely multiplicative arithmetic functions,. Number Theory, Edler. Lec-38

Multiplicative and completely multiplicative arithmetic functions,. Number Theory, Edler. Lec-38

8.4K

An arithmetic function f: N → C is multiplicative if f(mn) = f(m)f(n) whenever gcd(m,n) = 1, and completely multiplicative if this property holds for all natural numbers m and n regardless of their gcd. A key property is that for any multiplicative function, f(1) = 1 or f(n) = 0 for all n ∈ N. This distinction between multiplicative and completely multiplicative functions is fundamental in number theory, as completely multiplicative functions satisfy the multiplicative property universally, making them particularly useful for studying prime factorizations and constructing important number-theoretic functions.

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Teoria dei Numeri, Lezione 2, prima parte, 30.9.2020

Teoria dei Numeri, Lezione 2, prima parte, 30.9.2020

676

This section defines multiplicative functions (f(1) = 1 and f(mn) = f(m)f(n) for coprime m,n) and completely multiplicative functions (f(1) = 1 and f(mn) = f(m)f(n) for all m,n). The instructor explains that completely multiplicative functions are a subset of multiplicative functions. Examples include φ(n) and d(n) as multiplicative but not completely multiplicative, while id(n) and 1(n) are completely multiplicative. The condition f(1) = 1 is included to exclude the identically zero function. The set of completely multiplicative functions is closed under Dirichlet convolution, while the set of multiplicative functions is not.

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Случайные мультипликативные функции. Лекция 1, А.Б.Калмынин, П.Кучерявый

Случайные мультипликативные функции. Лекция 1, А.Б.Калмынин, П.Кучерявый

187

A function f of natural numbers is multiplicative if f(ab) = f(a)f(b) for coprime a,b, and completely multiplicative if this holds for all a,b. Key examples include Euler's totient function φ(n), the Möbius function μ(n), and Dirichlet characters. The Legendre symbol (n/p) is a multiplicative periodic function taking values 1, -1, or 0. Values of multiplicative functions are completely determined by their values on prime powers, allowing such functions to be defined by specifying f(p^k) for all primes p and k ≥ 1.

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03-01: Arithmetic and Multiplicative Functions

03-01: Arithmetic and Multiplicative Functions

657

An arithmetic function is a function with domain as positive integers and codomain as a ring (typically real or integers). A function is completely multiplicative if f(nm) = f(n)f(m) for all positive integers n and m, while a function is multiplicative if this property holds only when n and m are relatively prime. Key properties include: if a multiplicative function is not constantly zero, then f(1) must equal 1; products of multiplicative functions are multiplicative; compositions of completely multiplicative functions with multiplicative functions yield multiplicative functions; however, sums of multiplicative functions and scalar multiples (except by 1) are generally not multiplicative.

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Terence Tao: The Erdős Discrepancy Problem

Terence Tao: The Erdős Discrepancy Problem

447.6K

A completely multiplicative function f: ℕ → S satisfies f(nm) = f(n)f(m) for all n, m. Examples include the constant function f(n)=1, Dirichlet characters (periodic completely multiplicative functions), and the Liouville function λ(n) which equals (-1)^k where k is the number of prime factors of n (counting multiplicity). These functions form a special class that is easier to analyze than arbitrary sequences, yet they represent some of the hardest cases for discrepancy problems.

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Deep Dive

The theory of infinite series and infinite products, including absolute and conditional convergence criteria.

Theorems/// Fourth Semester MSc Mathematics /// Calicut university

Theorems/// Fourth Semester MSc Mathematics /// Calicut university

622

This segment covers fundamental convergence criteria for infinite series and products. The first result establishes that Σ log(1 + a_n) converges absolutely if and only if Σ a_n converges absolutely, provided a_n > -1. The second result addresses uniform convergence of exponential functions: if f_n(x) converges uniformly to f(x) on X and Re(f(x)) ≤ a for all x, then e^(f_n(x)) converges uniformly to e^(f(x)). The third result concerns infinite products: if g_n(x) converges absolutely and uniformly on X, then ∏(1 + g_n(x)) converges absolutely and uniformly, with the product equaling zero if and only if g_n(x) = -1 for some n.

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Домрина А.В. | Лекция 6 по математическому анализу II | ВМК МГУ 2021-2022

Домрина А.В. | Лекция 6 по математическому анализу II | ВМК МГУ 2021-2022

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An infinite product ∏pₙ converges to non-zero limit iff ∑log(pₙ) converges. If product converges to π, series converges to log(π). For absolute convergence, ∏(1+aₙ) converges absolutely iff ∑|aₙ| converges. If aₙ eventually maintains constant sign, product converges iff ∑aₙ converges. For alternating aₙ, use Taylor expansion: log(1+aₙ) = aₙ - aₙ²/2 + O(aₙ³). The product converges iff both ∑aₙ and ∑aₙ² converge. If ∑aₙ converges but ∑aₙ² diverges, the product diverges. Example: ∏(1 + (-1)ⁿ⁻¹/nᵖ) converges absolutely for p > 1 and conditionally for 1/2 < p ≤ 1.

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Series. Convergencia ABSOLUTA y CONDICIONAL.

Series. Convergencia ABSOLUTA y CONDICIONAL.

792

To determine whether an infinite series converges absolutely or conditionally, first apply the series of absolute values (modules) to the original series. If the series of absolute values converges, the original series is absolutely convergent. If the series of absolute values diverges, apply the Leibniz rule to the original series: if it converges, the series is conditionally convergent; if it diverges, the original series is divergent. This method allows the use of standard convergence criteria (such as the ratio test) on the transformed series of positive terms.

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24. Absolute and Conditional Convergence of a Series | Complete Concept & Problem#1| Infinite Series

24. Absolute and Conditional Convergence of a Series | Complete Concept & Problem#1| Infinite Series

104.8K

A series is absolutely convergent if the series of absolute values of its terms converges, which implies the original series also converges; a series is conditionally convergent if it converges but the series of absolute values diverges. To determine absolute convergence, apply the Comparison Test to the series of absolute values; if this converges, the original series is absolutely convergent. If the original series converges but the absolute value series diverges, the series is conditionally convergent.

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Косухин О.Н.-Математический анализ.Часть 1.Семинары - 26.Произведение рядов.Бесконечные произведения

Косухин О.Н.-Математический анализ.Часть 1.Семинары - 26.Произведение рядов.Бесконечные произведения

1.3K

For ∏(1+aₙ) with aₙ > 0, convergence is equivalent to ∑aₙ convergence. For aₙ < 0, convergence requires ∑|aₙ| convergence. Absolute convergence of the product means ∑|ln(1+aₙ)| converges. These criteria allow systematic study of infinite products by analyzing corresponding series, providing clear conditions for convergence.

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Deep Dive

An introduction to the Riemann Zeta function defined as a Dirichlet series for Real(s) > 1.

One of my favorite identities.

One of my favorite identities.

38.6K

The Riemann zeta function ζ(s) = ∑(m=1 to ∞) 1/m^s converges for Re(s) > 1. A remarkable property is that at even natural numbers, ζ(2n) equals a rational number multiplied by π^(2n). For example, ζ(2) = π²/6 (the Basel problem), ζ(4) = π⁴/90, and ζ(6) = π⁶/945. This pattern breaks down at higher even integers like ζ(10). These values suggest interesting sums when we isolate the rational coefficients by dividing out powers of π.

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🔴 HIPÓTESE DE RIEMANN | EU RESOLVI?!

🔴 HIPÓTESE DE RIEMANN | EU RESOLVI?!

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The Riemann zeta function ζ(s) is defined by the infinite series Σ(1/n^s) for n from 1 to infinity, which converges when the real part of s exceeds 1. This function is central to number theory and is one of the seven Millennium Prize Problems, carrying a one million dollar prize for its solution. A common misconception is that ζ(-1) equals 1 + 2 + 3 + ... = -1/12, but this is incorrect—the series diverges. The value -1/12 comes from analytic continuation, a technique that extends the function beyond its original domain. The function has 'trivial zeros' at negative even integers (-2, -4, -6, ...) and 'non-trivial zeros' that are the subject of the Riemann Hypothesis.

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Дзета-функция: все значения при чётных положительных аргументах!

Дзета-функция: все значения при чётных положительных аргументах!

26.8K

The Riemann zeta function ζ(s) = ∑(n=1 to ∞) 1/n^s converges for Re(s) > 1. Riemann extended this function to the entire complex plane. The video introduces the cotangent partial fraction decomposition: cot(x) = 1/x + ∑(n=1 to ∞)[1/(x-nπ)+1/(x+nπ)], which leads to πx·cot(πx) = 1 - 2x²∑(n=1 to ∞)1/(n²-x²). This establishes a connection between trigonometric functions and number theory through infinite series manipulation.

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Terence Tao: Vaporizing and freezing the Riemann zeta function

Terence Tao: Vaporizing and freezing the Riemann zeta function

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The Riemann zeta function ζ(s) = ∑(n=1 to ∞) 1/n^s connects number theory to complex analysis. For Re(s) > 1, it converges and encodes prime distribution through the Euler product formula: ζ(s) = ∏(p prime) 1/(1 - 1/p^s). This product shows ζ(s) has no zeros in this region. To extend ζ(s) to the entire complex plane, mathematicians multiply by gamma factors to create the completed zeta function ξ(s), which has superior analytic properties and can be expressed as the Fourier transform of a rapidly decaying theta function.

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The Return of -1/12 - Numberphile

The Return of -1/12 - Numberphile

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The Riemann zeta function ζ(s) = 1/1^s + 1/2^s + 1/3^s + ... converges only when Re(s) > 1. Through analytic continuation, this extends to all complex numbers except s = 1. Zeta regularization assigns values to divergent series by evaluating ζ at negative integers: ζ(-1) = -1/12 corresponds to 1 + 2 + 3 + ... This powerful technique connects number theory, analysis, and physics, providing finite values to otherwise meaningless infinite sums.

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Dirichlet Series Intro

Zeta Function Abscissa

Alternating Series Case

General Theory Return

L-Functions Defined

Multiplicativity Concept

Euler Product & Tau

Dirichlet Series Intro

0:00

Playing Section

1
Introduces Dirichlet series and the core problem of finding convergence zones.
2
Uses the Riemann zeta function as the primary example for explanation.
3
Identifies the abscissa of convergence and absolute convergence as key concepts.

Deep Dive

Analytic continuation and the functional equation of the Riemann Zeta function and other Dirichlet series.

The Riemann Hypothesis

The Riemann Hypothesis

597.4K

Riemann extended Euler's zeta function to all complex numbers using analytic continuation. He defined an alternating series converging for Re(s) > 0, then used the functional equation ζ(s) = 2^s π^(s-1) sin(πs/2) Γ(1-s) ζ(1-s) to reflect values across Re(s)=1/2. This allows computation of ζ(-1) = -1/12 by reflecting ζ(2) = π²/6. The trivial zeros at negative even integers arise from the sine function in this formula.

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Devenir RICHE grâce aux maths (La fonction Zeta de Riemann)

Devenir RICHE grâce aux maths (La fonction Zeta de Riemann)

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The Riemann Zeta function is initially defined only for complex numbers with real part greater than 1. Through analytic continuation using the functional equation ζ(s) = 2^s π^(s-1) sin(πs/2) Γ(1-s) ζ(1-s), the function is extended to the entire complex plane except for a simple pole at s=1 with residue 1. This powerful technique allows mathematicians to assign meaningful values to divergent series: ζ(0) = -1/2 and ζ(-1) = -1/12. While these values don't make sense physically (summing 1 infinitely gives infinity), they represent the most coherent values mathematically.

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Ramanujan: Making sense of 1+2+3+... = -1/12 and Co.

Ramanujan: Making sense of 1+2+3+... = -1/12 and Co.

Analytic continuation extends functions defined on restricted domains to larger domains while preserving analytic properties. The Dirichlet eta function η(z) = 1 - 1/2^z + 1/3^z - ... converges for Re(z) > 0 and extends to all complex numbers except z=1. At z=0, Cesàro summation gives η(0) = 1/2, and second-order Cesàro gives η(0) = 1/4. The Riemann zeta function ζ(s) = 1 + 1/2^s + 1/3^s + ... converges for Re(s) > 1 and extends to all complex numbers except s=1, with ζ(-1) = -1/12. Ramanujan's sophisticated summation method matches this analytic continuation value.

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The Generalized Riemann Hypothesis (RH Saga S1E3)

The Generalized Riemann Hypothesis (RH Saga S1E3)

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L-functions are defined by coefficient sequences a₁,a₂,a₃,... forming Dirichlet series L(s) = Σaₙ/n^s. These converge absolutely only in a right half-plane (Re(s) > 1 for zeta). Outside this region, series may diverge or become conditionally convergent. Analytic continuation extends L-functions to the entire complex plane, except isolated poles. Many L-functions have known continuations, especially those from automorphic objects—a key motivation for Langlands reciprocity. The Riemann zeta function demonstrates this with an integral formula valid for all s ≠ 1, enabling computation of values like ζ(2) or ζ(-1) through numerical integration.

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Complex Integration and Finding Zeros of the Zeta Function

Complex Integration and Finding Zeros of the Zeta Function

234.7K

The Riemann zeta function ζ(s) is initially defined by the series ζ(s) = Σ 1/n^s for real numbers s > 1. Through analytic continuation using Taylor series, this definition extends uniquely to all complex numbers except s = 1. The function has trivial zeros at negative even integers and non-trivial zeros confined to the critical strip 0 < Re(s) < 1, symmetric about Re(s) = 1/2. The Riemann Hypothesis conjectures that all non-trivial zeros lie exactly on this central line.

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Deep Dive

Dirichlet L-functions, Dirichlet characters, and their application in proving Dirichlet's Theorem on Arithmetic Progressions.

Dirichlet Biography: Genius of Analytic Number Theory (1805–1859)

Dirichlet Biography: Genius of Analytic Number Theory (1805–1859)

3K

Dirichlet L-functions are infinite sums encoding arithmetic properties of number sequences. For studying primes in arithmetic progressions, he used special coefficients called Dirichlet characters—periodic, multiplicative sequences that activate for numbers in target progressions and deactivate otherwise. By weaving characters into infinite series, he created functions uniquely sensitive to primes in specific progressions. The entire proof of his theorem boiled down to determining whether the L-function equals zero at the value one. If nonzero, the progression contains infinitely many primes. This translation of discrete arithmetic questions into continuous analytic properties was revolutionary. The Riemann zeta function (simplest L-function) and the entire Langlands program are built upon generalized versions of L-functions.

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L-functions in the theory of numbers by Ritabrata Munshi

L-functions in the theory of numbers by Ritabrata Munshi

26.8K

Dirichlet revolutionized number theory by introducing characters χ: Z/bZ → C*, which are completely multiplicative homomorphisms from the multiplicative group modulo b to complex numbers. For integers m not coprime to b, χ(m) = 0; for m coprime, χ(m) uses the representative modulo b. These characters enable defining Dirichlet L-functions L(s, χ) = ∑(χ(n)/n^s) = ∏(1 - χ(p)/p^s)^(-1). The key insight is that summing over all characters and dividing by φ(b) yields a characteristic function for arithmetic progressions. This framework allowed Dirichlet to prove there are infinitely many primes in any arithmetic progression a + nb with gcd(a,b)=1, extending Euclid's method through sophisticated analytic machinery.

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А.Б. Калмынин. Распределение степенных вычетов. Семинар 2

А.Б. Калмынин. Распределение степенных вычетов. Семинар 2

111

The orthogonality relation states that ∑χ(n) = φ(Q) if n≡1(mod Q), else 0. The Dirichlet L-function L(s, χ) = ∑χ(n)/n^s has Euler product ∏(1-χ(p)/p^s)^(-1). Taking logarithm gives log L(s, χ) = ∑∑χ(p^k)/(k p^(ks)). This series converges absolutely for Re(s) > 1. Using orthogonality, ∑χ̄(a)L(s, χ) equals ∑_{p≡a(mod Q)} φ(Q)/(p^s) + O(1). This shows that the character sum method can detect primes in arithmetic progressions. The error term is bounded because higher powers of p contribute less. This is the key step in proving Dirichlet's theorem on primes in arithmetic progressions.

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Introduction to number theory lecture 52. Nonvanishing of L series at s=1.

Introduction to number theory lecture 52. Nonvanishing of L series at s=1.

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To complete the proof of Dirichlet's theorem on primes in arithmetic progressions, we must show that Dirichlet L-functions do not vanish at s=1. The key technique involves taking the product of all L-functions over characters χ, which has a logarithm with non-negative coefficients. By Landau's lemma, this product has a singularity at s=1, so it cannot be zero there. If any individual L(χ,1) were zero, it would cancel the pole from the principal character, contradicting the non-vanishing of the product. This proves L(χ,1) ≠ 0 for all non-principal characters χ, completing Dirichlet's theorem.

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Introduction to number theory lecture 51. Proof of Dirichlet's theorem

Introduction to number theory lecture 51. Proof of Dirichlet's theorem

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Dirichlet's theorem states that there are infinitely many primes in any arithmetic progression a, a+d, a+2d, ... where a and d are coprime. The proof uses Dirichlet characters and their associated L-functions: by constructing a linear combination of characters that isolates a specific residue class b modulo n, and showing that the logarithm of the corresponding L-series is finite (which requires that L(χ,1) ≠ 0 for all non-principal characters χ), one can demonstrate that the sum of reciprocals of primes congruent to b modulo n diverges, thereby proving there are infinitely many such primes.

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Deep Dive

Perron's Formula and Tauberian theorems for extracting asymptotic estimates of arithmetic sums from Dirichlet series.

A simple proof of the Wiener-Ikehara Tauberian Theorem

A simple proof of the Wiener-Ikehara Tauberian Theorem

322

This section presents the complete proof strategy for extracting asymptotic formulas from Dirichlet series. The proof analyzes two cases based on parameter U relative to log X. For small U (U ≤ (log X)^ε), the Laurent series G(s) = C_k/(s-1)^k + ... applies directly since s approaches 1. For large U, non-negativity bounds the series by G(1 + 1/log X), yielding ∑a_n/n ≪ (log X)^k × negligible integral. Rapid decay of the Fourier transform ensures large-U contributions vanish faster than any negative power of log X. To extend partial results from restricted regions to full integrals, the proof adds and subtracts tails, showing they contribute negligibly (smaller than any negative power of log X). Applying the iterated integral identity converts Fourier transform integrals to function integrals, yielding ∫₀^∞ f(t) t^{J-1} dt = 1/J when f approximates [0,1]. Combined with factorial denominators, this produces the main asymptotic term. Final partial summation combines both main terms, yielding ∑_{n≤X} a_n ~ C_k/(k-1)! X(log X)^{k-1} with error O((log X)^{k-2}), achieving the classical Wiener-Ikehara asymptotic with logarithmic savings.

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Teoremas Tauberianos para Series de Dirichlet y aplicaciones

Teoremas Tauberianos para Series de Dirichlet y aplicaciones

185

Perron's formula is the most basic Tauberian theorem, relating a Dirichlet series to an integral. For a Dirichlet series with coefficients a_n and abscissa of convergence σ₀, the sum of coefficients up to x equals (1/2πi) times the integral of the series times x^s/s along a vertical line. This formula serves as the foundation for deriving other Tauberian theorems, including Landau's and Selberg's, by analyzing the integral and applying appropriate conditions.

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1. Elia Gorokhovsky (Harvard): Tauberian theorems

1. Elia Gorokhovsky (Harvard): Tauberian theorems

108

Tauberian theorems provide a powerful framework for deriving asymptotic formulas for partial sums of arithmetic sequences by analyzing the analytic properties of their associated Dirichlet series; specifically, if a Dirichlet series A(s) has a meromorphic continuation to a half-plane with a single pole at s=α of order m, then the partial sums of the sequence exhibit an asymptotic expansion with a leading term determined by the residue of the pole, and additional analytic continuation to the left of this pole yields improved error bounds with power-saving estimates.

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Ila Varma, Counting quartic number fields and predicting asymptotics

Ila Varma, Counting quartic number fields and predicting asymptotics

578

For abelian Galois groups, Wright and Maki derived general formulas: a_d,G = 1/d × ∏(1 - 1/p), where p ranges over primes dividing d; b_d,G = (n_p - 1)/(p - 1), where n_p is the number of elements of order p in G. These formulas come from analyzing Dirichlet series with simple poles using Tauberian theorems and Perron's formula.

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Programa de Doutorado - Teoria Analítica dos Números - Aula 15

Programa de Doutorado - Teoria Analítica dos Números - Aula 15

968

Perron's formula expresses the Chebyshev function ψ(x) as an integral: ψ(x) = (1/2πi) ∫_{c-i∞}^{c+i∞} -ζ'(s)/ζ(s) x^s/s ds, where c > 1. This integral captures the contribution of the zeros of ζ(s) to the distribution of primes. Perron's approximate inversion formula states that for c > 1 and T > 0, the integral equals Δ(x) + O(x^c/T log x) for x ≠ 1, where Δ(x) is the indicator function: Δ(x) = 0 for x < 1, Δ(x) = 1/2 for x = 1, and Δ(x) = 1 for x > 1. As T → ∞, the error term tends to zero for fixed x ≠ 1.

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Deep Dive

The analytic proof of the Prime Number Theorem using the non-vanishing of the Riemann Zeta function on the line Real(s) = 1.

What is the Riemann Hypothesis REALLY about?

What is the Riemann Hypothesis REALLY about?

718.6K

The Prime Number Theorem was proven using the fact that there are no zeros of the zeta function on the line Re(s) = 1 or to its right. This zero-free region result guarantees that the pole at s=1 is the dominant factor in approximating π(x), leading to the theorem that π(x) ~ Li(x). The critical strip zero-free region was essential for this proof.

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A proof of the Prime Number Theorem

A proof of the Prime Number Theorem

21.7K

This major segment develops the Riemann zeta function ζ(s) as the central tool connecting complex analysis to number theory. It begins with the definition ζ(s) = ∑n=1^∞ 1/n^s for Re(s) > 1 and derives the Euler product formula ζ(s) = ∏p (1 - 1/p^s)^(-1), revealing the fundamental connection between additive and multiplicative structures in integers. The segment then constructs the analytic continuation of ζ(s) to Re(s) > 0 using the integral representation ζ(s) = 1/(s-1) + ∑n=1^∞ [n^(1-s) - (n+1)^(1-s)]/s, which is analytic except at s = 1. The critical step proves that g(s) = -s/ζ'(s) - 1/(s-1) is analytic on Re(s) ≥ 1 by showing singularities cancel at s = 1. The segment culminates in proving ζ(s) has no zeros on Re(s) = 1 through contradiction: assuming a root of order a leads to impossible negative limits when examining εκ(1+ε±iα) as ε → 0+, thereby eliminating all potential zeros and completing the proof.

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Introduction to number theory lecture 48 Proof of the prime number theorem

Introduction to number theory lecture 48 Proof of the prime number theorem

The Prime Number Theorem states that the number of primes less than x is approximately x/log(x); its proof involves showing the Riemann zeta function has no zeros on Re(s) = 1 using a clever inequality, applying Newman's Tauberian theorem to show the integral of ψ(x) - x converges, and deducing ψ(x) ~ x, which then implies π(x) ~ x/log(x) because ψ(x) ≈ log(x)·π(x) and log(x) is approximately constant for large x.

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Week6Lecture5: The Prime Number Theorem

Week6Lecture5: The Prime Number Theorem

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The Prime Number Theorem states that the prime counting function π(x), which counts the number of primes less than or equal to x, is asymptotically equivalent to x divided by the natural logarithm of x (π(x) ~ x/ln(x)), meaning their ratio approaches 1 as x approaches infinity; this deep result in number theory was first proved using complex analysis by Hadamard and de la Vallée Poussin in 1896, who showed that the Riemann zeta function has no zeros on the critical line Re(s) = 1, building upon Euler's discovery that the zeta function can be expressed as an infinite product over all prime numbers (ζ(s) = ∏_{p prime} 1/(1 - p^{-s})), and the Riemann Hypothesis remains equivalent to the best possible bound on the error term in the Prime Number Theorem.

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Zeta Explained #28: The Prime Number Theorem (2/2)

Zeta Explained #28: The Prime Number Theorem (2/2)

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The Riemann zeta function has a pole at s=1, trivial zeros at negative even integers, and non-trivial zeros in the critical strip 0 < Re(s) < 1. The argument principle states that ∫ f'(z)/f(z) dz equals +1 for zeros and -1 for poles inside the contour. These properties enable residue calculations for the Prime Number Theorem proof.

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Dirichlet Series Intro

Zeta Function Abscissa

Alternating Series Case

General Theory Return

L-Functions Defined

Multiplicativity Concept

Euler Product & Tau

Dirichlet Series Intro

0:00

Playing Section

1
Introduces Dirichlet series and the core problem of finding convergence zones.
2
Uses the Riemann zeta function as the primary example for explanation.
3
Identifies the abscissa of convergence and absolute convergence as key concepts.

744 views1332:24@masoudkhalkhali3940Original Release: 2023-05-18

The convergence domain of a Dirichlet series ∑a_n/n^s is determined by its abscissa of convergence ρ and absolute convergence abscissa ρ'; for example, the Riemann zeta function ζ(s) = ∑1/n^s has ρ = ρ' = 1, while the alternating series ∑(-1)^n/n^s has ρ = 0 and ρ' = 1, demonstrating conditional convergence. A key proposition states that if coefficients a_n are bounded, the series converges absolutely for Re(s) > 1, ensuring holomorphicity in that region. Multiplicative functions, such as Dirichlet characters, yield L-functions that share similar convergence properties, while fully multiplicative functions satisfy f(mn) = f(m)f(n) for all m, n, unlike regular multiplicative functions which require m and n to be coprime.

[00:00:00]okay it's coming good yeah so now okay so this is the picture that we should have in mind about uh judicial Series this is the general picture but it's good to have some examples also Yeah question yeah okay yeah that's an excellent question I don't know the answer that's good why don't you explore the question and let me know next lecture I I don't have any answer to that that's an excellent question yeah it sounds like a question that has been answered to me but okay but that's that's that's a very excellent question yeah yeah now we will see actually an example of that series those Series right away that's our example one so eg1 so what's the most uh what's the simplest type of non-trivial or whatever uh British series if the simplest example of judicial series is a zeta function it's just that doesn't get as simple as that so Zeta s equal to sum one and to the S and from one please I mean I I you we wrote that but I I hate to write z for this analytic number two and stuff I really like to write S as Riemann wrote S I believe set of s equal to this one so this is obviously the original series uh now so what is rho and what is row Prime for this one what do you think is row and row prime or discount so the abscissa of convergence uh the vertical line ah yeah the other one is two yeah I know ah they're good after two yeah huh row is zero Prime is one okay so let me write these lists rho equal to two row Prime was not committed rho equal to one oh wait a minute uh yeah yeah rho equal to one or what was what was maybe you meant Row from equal to one rho is equal to zero is that what because row param is bigger than equal to zero Prime equal to one rho equal to zero is that what you said and any other vehicles yeah that's that's relevant yeah for sure ah and it gets post one yeah indeed both are equal to one let's yeah I mean this is as I said this is an example where actually the answer to your question is positive rho is equal to row Prime equal to one let's see why you see first of all this guy is convergent here right this guy's convergent here because of t-test in fact absolutely conversion because of it because now you have to deal with as complex also no but anyhow if you prove that it is conversion on real numbers you don't you don't have to worry about absolute if it is converging on real numbers here by the term be proof you know that it is conversion all the way up and down to the right so you don't have to worry about complex issues at all okay so here but now I claim that it cannot be conversion any point here not conversion if at any point here or less than I'm saying it is Divergent at all points where real part is less than one why is that so well the proof is very easy if it is conversion that sums at zero we're a real part of Z zero less than one okay like that then we know that there's going to be a real number that this guy is going to be convergent here or s real s bigger than a real part of Z zero it should be conversion but you see that this series cannot be converged for any real less than one P test that's it so rho is equal to rho Prime is equal to one that is simple observation proves it that's the power of the theorem you see it's really it frees you from this complex plane complications it really brings you down to checking on the real line so this is very useful for example two let's look at the alternating uh version of this maybe eight of s go to 1 over 1 to the S minus 1 over 2 to the S Plus 1 over 3 to the S minus 1 over 4 to the s so on so what is rho what is rho Prime in this case in this case so row Prime is about absolute values right let's take absolute value of these things I mean these negatives becomes positive and uh it's one so row Prime is definitely one and you take absolute value on the real s is real less than one does not converge because of P test again so rho is definitely one we know that now what about row Prime okay so sorry row Prime is one I meant what is rho now rho is zero yeah rho is equal to zero in fact and the reason for that is this famous alternating test the alternating test was that if the general term goes to zero is what was it what was the thing was it decreasing is decreasing and is alternating in so on is going to zero then it is conversion so I guess by alternating test Sigma minus 1 to the n to the s and from one to Infinity converges for at least s belonging to R and as bigger than zero that's what the alternating test tells us and it's certainly not convergent for S equal to zero and for S less than zero because for S less than zero the general term isn't even doesn't go to zero or S equal to zero you have one minus one plus one minus one plus one which is direct okay so you know that then as a result row is equal to zero so that has a very interesting thing so this is like zero this is one here is the thing and here is what we call conditional conversions I believe yeah in analysis conditional it's conditionally conversion only it means this conversion but it's not absolute so next question where this function is holomorphic where this function is holomorphic after rule Prime yeah it doesn't have to be after row oh you say I'm confusing row and row Prime I meant [Music] um yeah yeah the absolute convergence was not relevant for holomorphicity everywhere yeah yeah this function is holomorphic for real part of s bigger than zero okay but the zeta function originally is holomorphic only in real part of s bigger than one of course it's a question if the zeta function is conversion at these points or not because this is kind of critical line right we know that it is diversion that all other points below the line that's for sure that we know but on the line if it is conversion or diversion is a delicate matter we cannot answer it right now maybe in 10 minutes it needs some elaborate uh rather elaborate consideration it's not too bad but you just just pay attention same now if you look at here though at real part of s equal to zero this function is totally diversion everywhere here because you get what you get 1 minus one plus one minus one is so you see different types of course for Zeta you know that that s equal to one itself is not convergent because that's one that's harmonic Series right so that's but we don't know if it is conversion at all these funny points and what is your guess just some of them diverges yeah any other yes it's very hard to guess here really it's not a fair question I I agree I mean it's kind of how can we so it's like I have no clue just by looking at the question uh you know because these complex powers can be very very complicated arranging themselves like that and to the power one plus I so one plus t i for different values of T right that that and yeah that series over here I mean this series over here is uh it's like this series yeah I mean this is um here but what did this we know for sure that's the whole of the zeta function yeah s equal to one but what what happens yeah we'll see that actually it converges at all these points that's quite interesting oh my God yeah I think I'm what's that this point no no no it's not removable at all that's that's that that's a real Singularity we'll see we'll see okay I'm just teasing your appetite for things to come um now okay so now let's go back to generalities again you find some ordinary I think sir calls it ordinary the original series so this is what most people really consider when they think about the original series they think about such things and longing to see these are the type of delicious series are be um will be most be working with so so now the proposition is that if a n or bounded so here's a useful proposition so let's call it f of s if a n is less than equal to n for all n then F of s is convergent for real parts of sphere than one and of course holomorphic I don't even have to say that now oof okay well I mean if to see that you just use the absolute values right it's just uh what is it I mean Sigma absolute value of a n Over N to the s I mean this is less than equal to sum I mean this guy's m hmm and we know that this is of course convergent for a real part of s bigger than one why this is like the estimate the use for Zeta functions and all that so this is convergent so first series that has this condition that uh a ends are bounded you're always at least holomorphic in real part of a speed number one thank you okay so now we can introduce Zeta functions and L functions we have introduced already zeta function so this is deliciously L function or L series so let's Okay so again let's fix n bigger than equal to two let I be a character a modulo m so we are going back to the beginning of today's lecture so there's a character on GN right Etc so I mean this is a this is a judicially character modulo n then L of s and Chi let's define it to be sum Pi n divided by m to the S right and from one Infinity so this is the dirichlet L function or L Series attached to this character modulo n for any n you can do such things and of course by this thing that we proved this guy is holomorphic in a real part of s bigger than one at least right right okay so one thing that we have to recall is that this guy is fully multiplicative I.E Chi of NM is equal to y n for all and and in Z right this is what we I mentioned at least that they enjoy and the multiplicative condition was just one condition it was that Chi of an m is equal to cry and because I am only if n and M are relatively prime only for those pairs if and on um okay so so we have two concepts we have to be very careful to distinguish them fully multiplicative arithmetic functions and multiplicative audit functions multiplicative is a weaker condition fully multiplicative is the natural condition I would call it multiplicative the other one weekly marketed but some people so here are some examples uh let's look at the Euler's file function is this um multiplicative or fully multiplicative or none what can you say about this forever huh not fully multiplicative you're saying but multiplicative okay that's that's an answer any other answers no answers yeah indeed this is correct answer it's uh multiplicative but not fully multiplicative is more publicative but not fully so proves so what we want to show we want to show that Phi of MN is equal to Phi of M y of n if M and N or relatively prime how do you prove that okay oh yeah yeah well I mean so driving the general formula really depends on this also to show that it is multiplicative any so this is a basic algebra I thought algebra last semester so yeah last semester I think so I'm still fresh on that huh before last semester okay yeah so I already forgot what it is yes that's right yeah so how do you prove this though that there is an elementary proof for sure but a good proof is to use Chinese remainder theorem so what does Chinese Remain the term say you see so let's let's see we know that fireban okay is the order of the multiplicative group of Z mod and Z right is the order of multiplicative group of Discord but if n and M are relatively prime you can decompose uh Z over m z cross Z over NZ only if M and N or relatively prime not otherwise so what do we have then we have this isomorphism of rings this is emotional rings so what does it mean it means that if I want to take invertible elements there some element is invertible if and only if the first component is invertible and second component is immersible because they are just multiplying pairwise right so this is going to be this across that and then of course follows right and you can convince yourself that this is not fully multiplicative by just giving example like fire 4 is not equal to Phi of 2 squared you know that y of 2 is 1 but Phi of four is two it's not the square of five and so on um by the way you know that Phi of p is equal to a minus 1 and for this point of p n what you mean p n minus p n minus one yeah is that what that's yeah that's what I tried to say I don't know if that's what I did say oh yeah so I I almost heard you yeah the N minus p n minus 1. maybe it is correct but I don't know I have to check I can I can't is this correct okay so you are responsible for this you know if some bridges break down tomorrow you know based on this I don't know I won't be responsible so maybe I just put with some caution but that's I believe you this is correct so using that then we know everything right because so you can just write Phi of any number as P one and one p two and two so on you can write it as then E1 and 1 minus P1 and 1 minus 1 times using uh multiplicativity each multiplicity so that's basically now there is a result which is a Euler product formula for such things that I want to mention and then we are finishing in five minutes that's that's it yeah um the last thing we want to do today now that we are added oh there is another one which is uh maybe I mentioned so that's kind of interesting so this is another example this is ramanu Jones uh Tau function so it is it is defined from sum Tau and Q to the n from 1 to Infinity is equal to Q times product of 1 minus q and n squared and from 1 to Infinity so you might say okay what I mean okay so um first of all this right hand side is convergent or absolute value of Q less than one you know why the right hand side is conversion for absolute value of Q less than one it is an is an infinite product right so you have to use the theory of infinite products it's an infinite product so you better know some Criterion for convergence of infinite products that's important and uh what is that okay so I don't expect you to remember but from you can take yeah yeah indeed the proof of stars by taking log of these terms and writing estimates based on that but it's good to know the final result somehow right I mean yeah so the final result is that indeed this product of 1 minus alpha n it's convergent so imagine these guys are less than equal to one in absolute value so they're never never equal to one if and only if the sigma is conversion right oh no no that's not Sigma yeah yeah that's not what uh product yeah uh n equal to 1 to Infinity conversion if and only if sum alpha n this is uh something maybe you have seen in complex analysis by distress infinite products expansions this sort of thing so this is gonna so this is I believe this is correct and uh no it is correct and it's extremely useful because working with infinite products is kind of unintuitive it's uh yeah but the series you have humans have better understanding of series than products somehow okay so this is obviously then you can apply this criteria and the test this infinite product Sigma qn is conversion for absolute value of Q less than one so this is conversion so this conversion is a holomorphic function in Q so you can expand it perhaps the value of Q lesson in Taylor series and this is what so that's a roundabout way of defining these numbers Tau ends so ramonesian numbers there are other definitions for them but now the thing is this is a multiplicative function not fully multiplicative for this multiple I don't know I didn't check sir what he says maybe he just gives a proof I'm not sure uh maybe he just mentions it or maybe this is just I got it from in my notes from some other sources I'm not sure does sir say anything about it yeah I don't know because if he mentions it at one stage certainly I'm almost sure that he will he would he would give a proof otherwise he wouldn't talk about it just mentioned it okay so I think we can prove it at one stage yeah all right so I think it's a good time to stop now and I just stop the video let me get uh oh yeah

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