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The Euler Product Formula: A Beautiful Bridge Between Sums and Primes

Added: 2026-08-07

Deep Dive

The Fundamental Theorem of Arithmetic, specifically how every integer greater than 1 can be uniquely factored into prime numbers.

Fundamental Theorem of Arithmetic || Unique Factorization Theorem || Prime Factorization Theorem

Fundamental Theorem of Arithmetic || Unique Factorization Theorem || Prime Factorization Theorem

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The Fundamental Theorem of Arithmetic, also known as the Unique Factorization Theorem or Prime Factorization Theorem, states that every integer greater than 1 is either a prime number itself or can be represented as a unique product of prime numbers. This theorem is fundamental to number theory and establishes that prime factorization is unique for every integer greater than 1.

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1.2 – Some Important Number Sets

1.2 – Some Important Number Sets

419

The Fundamental Theorem of Arithmetic states that every integer greater than 1 can be factored into a product of prime numbers, and this factorization is unique up to the ordering of the factors. This means no matter what natural number you choose, it can be expressed as a product of prime numbers, and there is only one way to do this (except for rearranging the order of the factors). This is the basis for prime factorization.

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12. Calculus Part 5 | First Theorem | MLS Shastry | UttaraComputers

12. Calculus Part 5 | First Theorem | MLS Shastry | UttaraComputers

107

The Fundamental Theorem of Arithmetic states that every integer greater than 1 can be uniquely factored into prime numbers, disregarding the order of the factors. For example, 20 can be factored as 2 × 2 × 5 (or 2² × 5), and this factorization is unique—no other combination of primes will produce 20. Prime numbers cannot be broken down further, while composite numbers can always be expressed as products of smaller numbers.

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Introduction to Number Theory | Math

Introduction to Number Theory | Math

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The fundamental theorem of arithmetic states that all integers greater than 1 can be uniquely factored into a product of prime numbers. For example, the number 12 can be broken down into 6 × 2, then 6 breaks into 3 × 2, resulting in the unique prime factorization of 3 × 2 × 2. No other combination of primes will produce 12.

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Elementary Number Theory | Part 1 | Discrete Structures

Elementary Number Theory | Part 1 | Discrete Structures

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The Fundamental Theorem of Arithmetic states that every integer greater than 1 can be uniquely expressed as a product of prime numbers, up to the order of the factors. This means that prime factorization is unique for each number. For example, 12 = 2 × 2 × 3 = 2² × 3, and this factorization is unique regardless of how you arrange the factors.

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Deep Dive

The behavior of infinite series, including convergence, divergence, and the formula for the sum of an infinite geometric series.

Infinite Geometric Series

Infinite Geometric Series

132

An infinite geometric series has infinitely many terms, but its sum may or may not be finite. The key factor determining behavior is the common ratio (r-value). When terms approach infinity (|r| ≥ 1), the series diverges to infinity. When terms approach zero (|r| < 1), the series converges to a finite sum. The sum formula is S = a₁ / (1 - r), where a₁ is the first term. This formula works because when |r| < 1, rⁿ approaches 0 as n → ∞. For example, with a₁ = 1/2 and r = 3/4, S = 2. If |r| ≥ 1, the sum does not exist (diverges). Always express answers as fractions or mixed numbers.

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Cal2 Ch11 Sec2

Cal2 Ch11 Sec2

137

An infinite series is the sum of infinitely many terms, represented by the summation symbol (sigma). The partial sum s_n represents the sum of the first n terms, and an infinite series equals the limit of its partial sums as n approaches infinity. A sequence (list of numbers) can converge while its corresponding series (sum of those numbers) diverges. For geometric series with common ratio r, the series converges if |r| < 1 and diverges if |r| ≥ 1. The sum formula is S = a_1 / (1 - r) when |r| < 1. For example, 5 + 5(0.9) + ... converges to 50, while 5 - 8 + 64/5 - ... diverges because r = 1.6 > 1.

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Infinite Geometric Series

Infinite Geometric Series

354

An infinite geometric series is the sum of infinitely many terms from a geometric sequence. The series converges (has a finite sum) when the absolute value of the common ratio is less than one (|R| < 1), meaning terms approach a fixed number. When |R| ≥ 1, the series diverges (no sum) as terms grow without bound. The sum formula is S = a₁ / (1 - R), where a₁ is the first term and R is the common ratio. This formula only applies when |R| < 1. To determine convergence, find R by dividing any term by the preceding term, then check if |R| < 1.

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How To Derive The Sum Formula of a Geometric Series

How To Derive The Sum Formula of a Geometric Series

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An infinite geometric series has no last term and continues indefinitely. It converges (has a finite sum) only if the absolute value of the common ratio is less than 1 (|r| < 1). When it converges, the sum is given by S∞ = a₁/(1 - r). For example, the series 8 + 4 + 2 + 1 + ½ + ¼ + ... with a₁=8 and r=½ converges because |½| < 1, and its sum is 8/(1 - ½) = 16. If |r| ≥ 1, the series diverges and does not have a finite sum.

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📗 Calculus Konu Anlatımı | Bölüm 10.2 Infinite Series | Ders 1 – Sonsuz Seriler

📗 Calculus Konu Anlatımı | Bölüm 10.2 Infinite Series | Ders 1 – Sonsuz Seriler

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To find the sum of an infinite series, first find the partial sum S_n (sum of first n terms), then take the limit as n approaches infinity. If lim(n→∞) S_n = L (finite), the series converges to L. If the limit doesn't exist or is infinite, the series diverges. A geometric series has the form Σ(n=1 to ∞) a×r^(n-1), where a is the first term and r is the common ratio. The sum of the first n terms is S_n = a(1-r^n)/(1-r). For the infinite series, if |r| < 1, the sum is a/(1-r). The series converges if and only if |r| < 1; if |r| ≥ 1, it diverges. For a = 1/2, r = 1/2: S = (1/2)/(1-1/2) = 1. For a = 1/3, r = 1/3: S = (1/3)/(1-1/3) = 1/2. For a = 1/5, r = 1/5: S = (1/5)/(1-1/5) = 1/4.

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Deep Dive

The definition and basic properties of the Riemann zeta function for real numbers greater than one.

Complex Integration and Finding Zeros of the Zeta Function

Complex Integration and Finding Zeros of the Zeta Function

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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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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, denoted ζ(s), is defined as the sum from n=1 to infinity of 1/n^s. For real values greater than 1, this converges to finite values. For example, ζ(1) is the harmonic series which diverges to infinity, ζ(2) equals π²/6 (a famous result related to circles), and ζ(-1) would theoretically sum all positive integers, though this diverges in the traditional sense.

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Apéry's constant (calculated with Twitter) - Numberphile

Apéry's constant (calculated with Twitter) - Numberphile

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The Riemann zeta function ζ(s) is defined as the infinite series 1/1^s + 1/2^s + 1/3^s + 1/4^s + ... for complex numbers s with real part greater than 1. This function generalizes the Basel problem and allows mathematicians to study the behavior of these infinite series across different values of s.

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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 can be expressed as the Fourier transform of a rapidly decaying theta function. The Riemann hypothesis states that all non-trivial zeros lie on the critical line Re(s) = 1/2, with zeros symmetric about both axes. The number of zeros up to height T grows like N(T) ≈ T log T, with average spacing ~1/log T. While numerical evidence supports the hypothesis, fine-scale distribution remains mysterious.

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The Key to the Riemann Hypothesis - Numberphile

The Key to the Riemann Hypothesis - Numberphile

The Riemann zeta function is defined for a complex number s = x + iy as either a sum over all positive integers (ζ(s) = Σ 1/n^s) or as an Euler product over all prime numbers (ζ(s) = Π (1 - 1/p^s)^(-1)). This function encodes fundamental information about prime numbers, as it contains all primes in its product formula. These definitions work when the real part x > 1, but can be analytically continued to other regions of the complex plane.

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Deep Dive

The Basel problem, including Euler's famous resolution finding the sum of the reciprocals of the squares of natural numbers.

The Basel Problem

The Basel Problem

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Leonhard Euler solved the Basel problem by representing the sine function as an infinite product and comparing it to its Taylor series expansion, demonstrating that the sum of the reciprocals of the squares of natural numbers equals π²/6; specifically, he showed that sin(x) can be expressed as x multiplied by the product of (1 - x²/n²π²) for all positive integers n, and by equating this to the Taylor series x - x³/6 + x⁵/120 - ..., he found that the coefficient of x³ gives the sum Σ(1/n²) = π²/6.

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Problema de Basilea (Suma de los inversos de los cuadrados) - Euler EXPLICADA PASO A PASO MUY FÁCIL

Problema de Basilea (Suma de los inversos de los cuadrados) - Euler EXPLICADA PASO A PASO MUY FÁCIL

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Euler solved the Basel Problem by equating the Taylor series expansion of sin(x) with its infinite product factorization based on its roots (multiples of π), then comparing coefficients of x² to derive that the sum of the reciprocals of the squares (1/1² + 1/2² + 1/3² + ...) equals π²/6.

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The Basel Problem... How Euler Solved It ?

The Basel Problem... How Euler Solved It ?

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The Basel Problem asks for the exact value of the infinite sum of reciprocals of squares: 1 + 1/4 + 1/9 + 1/16 + ... (sum from n=1 to ∞ of 1/n²). Posed by Pietro Mengoli in 1650, it remained unsolved for nearly 90 years despite efforts by the Bernoulli brothers and other mathematicians. In 1734, 28-year-old Leonhard Euler solved it by treating the sine function as an infinite polynomial. He wrote sin(x) as an infinite product using its roots (0, π, 2π, 3π, ...), then compared the Taylor series expansion with the product form. By matching the coefficient of x², he derived that the sum equals π²/6, approximately 1.644934.

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Euler's infinite pi formula generator

Euler's infinite pi formula generator

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Euler solved the Basel problem by expanding his sine product formula. He expanded the product, collected coefficients of powers of x, and differentiated three times before setting x = 0. This yielded the famous result: π²/6 = 1/1² + 1/2² + 1/3² + 1/4² + ... This breakthrough made Euler a mathematical superstar at age 28 and established the foundation for higher power Basel formulas.

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El Problema de Basilea

El Problema de Basilea

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The Basel Problem asks for the sum of the infinite series of the reciprocals of the squares of natural numbers: 1 + 1/4 + 1/9 + 1/16 + ... In 1734, Leonhard Euler solved this problem by demonstrating that the sum equals π²/6. Euler's proof involved three key elements: properties of polynomial equations, the power series expansion of sin(x), and an 'act of faith' assuming that what holds for finite polynomials also holds for infinite polynomials. This elegant solution revealed the surprising connection between the sum of inverse squares and the number π.

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Oiler's Equation

Convergence Rules

Sieve Proof

Filtering Logic

Prime Divergence

Prime Density

Complex Extension

Deep Conjecture

Final Thoughts

Oiler's Equation

0:00

Playing Section

1
Introduces Euler's product formula as a candidate for the most beautiful equation.
2
Notes its surprising link between prime products and sum over all natural numbers.
3
Mentions the presence of pi in the result for s=2.

Deep Dive

Analytic Continuation, particularly how the Riemann zeta function is extended to the complex plane.

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

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

294.6K

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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Analytic Continuation and the Zeta Function

Analytic Continuation and the Zeta Function

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Analytic continuation is a powerful technique that allows us to extend real functions into the complex plane by using Taylor/Maclaurin series expansions centered at various points, enabling us to define functions like the Riemann zeta function for complex inputs where they were originally undefined; this method reveals that complex functions are highly rigid, meaning their values on any small region of the real line determine their values everywhere else in the complex plane, which is why the zeros of the zeta function can dictate the distribution of prime numbers.

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The Riemann Hypothesis

The Riemann Hypothesis

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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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Step Zero of Analytic Continuation Gateway to the Riemann Hypothesis #mathmatics

Step Zero of Analytic Continuation Gateway to the Riemann Hypothesis #mathmatics

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Analytic continuation is a mathematical technique that transforms divergent infinite series into finite values by unfolding and refolding the series through algebraic manipulations; specifically, the Riemann zeta function ζ(s) = Σ(1/n^s) for Re(s) > 1 can be analytically continued to Re(s) > 0 (except s=1) using a six-step process involving multiplication by (1-2/2^s) and its inverse, ultimately yielding the alternating zeta function Σ((-1)^(n-1)/n^s) which converges for Re(s) > 0.

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Numberphile v. Math: the truth about 1+2+3+...=-1/12

Numberphile v. Math: the truth about 1+2+3+...=-1/12

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The Riemann zeta function ζ(z) = 1^(-z) + 2^(-z) + 3^(-z) + ... is originally defined for complex numbers with real part > 1. It can be uniquely extended to an analytic function for all complex numbers except z=1 through analytic continuation. The key insight is that analytic functions are completely determined by their values on any open subset of their domain. This means that once ζ(z) is defined on Re(z) > 1, there is exactly one way to extend it to the entire complex plane except z=1 while preserving analyticity.

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Deep Dive

The Riemann Hypothesis, exploring the non-trivial zeros of the zeta function and their connection to prime distribution.

The Pattern to Prime Numbers?

The Pattern to Prime Numbers?

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This section presents the Riemann Hypothesis and its profound implications. The hypothesis states all non-trivial zeta zeros have real part 1/2, with a $1M prize for proof. Riemann's explicit formula expresses π(x) as li(x) minus a sum over non-trivial zeros' imaginary parts. The error bound |π(x)-li(x)| ≤ O(√x ln x) holds if and only if the hypothesis is true. Including more zeros in the explicit formula yields increasingly accurate prime approximations, demonstrating the deep connection between zeta zeros and prime distribution.

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The Riemann Hypothesis, Explained

The Riemann Hypothesis, Explained

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The Riemann Hypothesis, proposed by Bernhard Riemann in 1859, conjectures that all non-trivial zeros of the Riemann Zeta function lie on the critical line Re(s) = 1/2 in the complex plane; this hypothesis provides a profound connection between the distribution of prime numbers and the zeros of the Zeta function, as Riemann showed that the modified prime counting function can be expressed as a sum of waves corresponding to the Zeta zeros, meaning the location of these zeros determines the distribution of primes.

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Riemann Hypothesis - Numberphile

Riemann Hypothesis - Numberphile

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The Riemann Hypothesis, one of mathematics' most famous unsolved problems worth $1 million, proposes that all non-trivial zeros of the Riemann Zeta function lie on the critical line Re(s) = 1/2 in the complex plane; this hypothesis connects the distribution of prime numbers to the behavior of the zeta function, which assigns values to complex numbers through analytic continuation beyond its original domain of convergence.

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The Riemann Hypothesis

The Riemann Hypothesis

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The Riemann Hypothesis states that all non-trivial zeros of the Riemann zeta function lie on the critical line Re(s) = 1/2 in the complex plane; this hypothesis is equivalent to the error term in the Prime Number Theorem being bounded by O(√x log x), meaning the distribution of prime numbers is as regular as possible, and its resolution would confirm that the logarithmic integral provides the optimal approximation for counting primes.

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The Dream: Riemann Hypothesis and F1 (RH Saga S1E1)

The Dream: Riemann Hypothesis and F1 (RH Saga S1E1)

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The Riemann Hypothesis represents a small part of a larger mathematical quest for new geometry underlying L-functions. Prime numbers are integers greater than 1 with exactly two divisors, appearing randomly distributed with twin primes close together and large gaps elsewhere. The Riemann zeta function connects to primes through its zeros: trivial zeros at negative even integers (-2, -4, -6, ...), and non-trivial zeros within the critical strip (real part between 0 and 1). The Riemann Hypothesis conjectures all non-trivial zeros have real part exactly 1/2, lying on the critical line. Mathematicians have verified this for the first twelve trillion zeros. This foundational connection between prime distribution and complex analysis forms the basis for understanding deeper mathematical structures.

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Deep Dive

The Prime Number Theorem, which describes the asymptotic distribution of the prime numbers.

The Queen of Mathematics - Professor Raymond Flood

The Queen of Mathematics - Professor Raymond Flood

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The Prime Number Theorem describes the asymptotic distribution of prime numbers. It states that the ratio of π(x) (number of primes ≤ x) to x/log(x) approaches 1 as x becomes large. This means that the proportion of primes among numbers up to x is approximately 1/log(x). Carl Friedrich Gauss conjectured this relationship at age 15 after studying tables of primes. The theorem was independently proved in 1896 by Jacques Hadamard and Charles de la Vallée Poussin. For example, π(10)/10 ≈ 0.4, π(100)/100 ≈ 0.25, π(1,000)/1,000 ≈ 0.167, and these proportions approach 1/log(x).

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Vortrag "Sind Primzahlen dem Zufall unterworfen?" von D. Schindler & V. Blomer am 29.06.2022

Vortrag "Sind Primzahlen dem Zufall unterworfen?" von D. Schindler & V. Blomer am 29.06.2022

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The Prime Number Theorem describes the asymptotic distribution of prime numbers. It states that the number of primes less than or equal to a large number x, denoted π(x), is approximately equal to x / log(x), where log is the natural logarithm. This means that as numbers get larger, primes become less frequent - among the first 100 numbers, about 1 in 4 is prime; among the first 1000, about 1 in 6; among the first 10,000, about 1 in 8; and among the first billion, about 1 in 20.

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Лекция 6. В.В. Юделевич. Теория чисел.

Лекция 6. В.В. Юделевич. Теория чисел.

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The Prime Number Theorem states π(x) ~ x/ln(x) as x → ∞. The proof uses Chebyshev's bounds and Abel's transformation on ∑Λ(n)/ln(n). This theorem describes the asymptotic distribution of prime numbers, showing that the number of primes up to x is approximately x/ln(x). It is one of the most important results in number theory.

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Александр Калмынин -- Дзета-функция Римана и (не) случайность простых чисел

Александр Калмынин -- Дзета-функция Римана и (не) случайность простых чисел

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The Prime Number Theorem provides an asymptotic formula for the distribution of prime numbers. Let π(x) denote the number of primes less than or equal to x. The theorem states that π(x) ~ Li(x), where Li(x) is the logarithmic integral function defined as the integral from 0 to x of 1/ln(t) dt. This is more accurate than the simpler approximation π(x) ~ x/ln(x). Gauss originally conjectured this asymptotic behavior, viewing it probabilistically: if a random number up to X is chosen, the probability that it is prime is approximately 1/ln(X). Chebyshev proved that if the limit of π(x)ln(x)/x exists, it must equal 1, establishing the foundation for the Prime Number Theorem.

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Un passionné vous parle de maths

Un passionné vous parle de maths

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The Prime Number Theorem provides an asymptotic formula for the distribution of prime numbers: the number of primes less than or equal to x is approximately x/ln(x). This simple formula captures the essence of how primes are distributed among the integers. The speaker describes this as a 'claque' (a stunning result) that continues to amaze mathematicians.

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Deep Dive

Dirichlet L-functions and their Euler products, which generalize these concepts to prove Dirichlet's Theorem on Arithmetic Progressions.

Automorphic Representations and L-functions #9, Prof. Kontorovich, Rutgers Math 572, 02/14/2023

Automorphic Representations and L-functions #9, Prof. Kontorovich, Rutgers Math 572, 02/14/2023

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This segment introduces Dirichlet L-functions and their Euler products. The Dirichlet L-function L(s, χ) = ∑_{n=1}^∞ χ(n)/n^s converges absolutely for Re(s) > 1 and extends meromorphically. These functions have Euler products: L(s, χ) = ∏_p (1 - χ(p)/p^s)^{-1}, following from χ's complete multiplicativity. The Euler product reflects the multiplicative nature of characters and provides a bridge between additive and multiplicative number theory. These L-functions generalize the Riemann zeta function and are fundamental tools for studying primes in arithmetic progressions.

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Introduction to analytic number theory 1: Euler, zeta function, and distribution of primes.

Introduction to analytic number theory 1: Euler, zeta function, and distribution of primes.

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Dirichlet generalized Euler's zeta function by introducing L-series associated with multiplicative functions χ: ℕ → ℂ, giving L(s, χ) = ∑ₙ=1^∞ χ(n)/n^s with Euler product ∏ₚ (1 - χ(p)/p^s)^(-1). Characters on finite abelian groups are homomorphisms χ: G → ℂ^×, forming dual groups Ĝ isomorphic to G. Orthogonality relations state ∑_{g∈G} χ(g) = |G| if χ trivial, else 0; and ∑_{χ∈Ĝ} χ(g) = |G| if g identity, else 0. These tools enable Dirichlet's proof of infinitely many primes in arithmetic progressions when a and b are coprime.

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Teoria Elementare dei Numeri, Lezione 12

Teoria Elementare dei Numeri, Lezione 12

130

Dirichlet's theorem states that for any positive integer q and integer a coprime to q, the arithmetic progression a, a+q, a+2q, ... contains infinitely many primes. For q=4, residues 1 and 3 each contain infinitely many primes. Euler's zeta function ζ(x) = Σ(n=1 to ∞) 1/n^x equals ∏(p prime) (1 - 1/p^x)^(-1). Since ζ(x) → +∞ as x → 1+, the sum Σ(p prime) 1/p^x also tends to +∞, proving infinitely many primes. Dirichlet characters modulo Q are completely multiplicative functions periodic with period Q. For Q=4, there are 2 characters: χ₀(n) = 1 if n odd, 0 if even; χ₁(n) = 1 if n≡1 mod 4, -1 if n≡3 mod 4, 0 if even.

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

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

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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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Что бывает в теории чисел (1) - Вадим Вологодский, Михаил Финкельберг

Что бывает в теории чисел (1) - Вадим Вологодский, Михаил Финкельберг

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This section presents Euler's product formula for the Riemann zeta function: ζ(s) = ∏(1 - p^(-s))^(-1) for Re(s) > 1. This formula connects the distribution of primes to the behavior of the zeta function, showing that the product over all primes equals the sum of reciprocals of all positive integers. The section then introduces Dirichlet L-functions, which generalize the zeta function by incorporating characters to distinguish between different arithmetic progressions. For modulus 4, the L-functions for 4n+1 and 4n+3 both converge to non-zero values at s=1, implying that the number of primes in each progression grows asymptotically as x/(ln x). This demonstrates that primes are distributed equally between the two residue classes modulo 4.

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Oiler's Equation

Convergence Rules

Sieve Proof

Filtering Logic

Prime Divergence

Prime Density

Complex Extension

Deep Conjecture

Final Thoughts

Oiler's Equation

0:00

Playing Section

1
Introduces Euler's product formula as a candidate for the most beautiful equation.
2
Notes its surprising link between prime products and sum over all natural numbers.
3
Mentions the presence of pi in the result for s=2.

60.3K views2.5K17:04@DrTreforOriginal Release: 2024-08-06

The Euler Product Formula, discovered by Leonhard Euler in 1737, establishes a profound connection between the sum of reciprocals of natural numbers raised to a power s (the Riemann zeta function ζ(s)) and an infinite product over all prime numbers, expressed as ζ(s) = ∏(p prime) 1/(1 - 1/p^s). This formula converges when the real part of s exceeds 1 and provides a powerful tool for studying prime numbers, including proving that the sum of reciprocal primes diverges (implying infinitely many primes) and serving as a foundation for analytic number theory and the famous Riemann Hypothesis.

[00:00:00]there are many beautiful equations within mathematics probably the most commonly cited one is Oilers equation Ed the Pi + 1 equal to Z that combines five of the special numbers in mathematics but in this video I want to make a case for a different equation to be on any short list for the most beautiful equation in mathematics this one is also du to Oiler it is the oiler product formula now don't let the notation scare you the big Sigma just refers to the sum of a bunch of terms the big Pi refers to the product of a bunch of terms the sum here is taken over all natural numbers 1 2 3 4 and so forth so you get terms like 1 over 1 to the S 1 over 2 to the S 1 over 3 to the S all added together however the product is multiplying for all PS that are primes 2 3 5 7 11 and so forth and so you get terms like 1 over 1 - 2 to Theus s 1 over 1 - 3 to Theus s and so forth so the first reason that I really love this formula so much is just that it's really surprising why would a product of terms only dealing with primes be related to a sum of terms that are now dealing with all of the numbers that seems very strange and that this strange relationship comes out with only very basic mathematical operations addition multiplication division and so forth makes it even more surprising now it's worth noting that this equation has a parameter s on both sides s can be many numbers like for example I could plug in the specific value of s equal to two and when I do it this equation actually adds up to something it converges to the number pi^ 2 over 6 so for any of you who think that any candidate for a beautiful math equation better have a pi in it well okay we have a pi in this one but s doesn't have to equal just two it could be many different things but it can't be anything because I'm taking an infinite sum I need to ask when does this converge and for those of you who might remember your firste Calculus this summation on the left hand side it converges precisely when s is greater than one if you plug in s equal to 1 specifically you get the harmonic series 1 plus a half plus a third plus a 4 and infamously that Series diverges so if I'm only considering real numbers and they're greater than one then it's going to converge and make sense and as we're going to see later on the video you can actually also put in complex values for S which is going to open the door to a really deep field of mathematics and there's even a million dooll Millennium problem hiding around in there which we'll get to a little bit later in the video but we're not there yet okay so my third reason that I really like this is that it just has a lovely proof you might have heard before about the Civ of ostanes let me just put up all the numbers between say two and 100 and I'm going to try and filter these first I'm going to notice that two is a prime number and I'm going to get rid of all of the multiples of two gone then I'm going to take the next number on the list that hasn't been filtered out which is three three is also Prime and if I filter out all the multiples of three three and six which was already filtered but it now filters new ones like n 9 and 15 as well so that filtered out you roughly 2/3 of the numbers the the next number that I haven't filtered already is five five is prime I can filter that out then at seven I can filter out seven and notice what happens here all the numbers between two and 10 have been filtered out and 10 is interesting because 10 is the square root of 100 if I wanted to factor some other number between 10 and 100 one of those factors would have to be less than 10 and because I filtered out everything up to the square root of the number I'm considering the square root of 100 in this case every other number that I have not shaded in is now Prime so I can come through one by one and shade all of those in as well all of those unshaded ones are going to be Prime in addition to 2 3 5 and 7 this was a Civ of aerosan so we're going to use this idea in the proof of Oilers product formula okay so where were we uh here is our summation and I'm going to give it a short hand so that I can sort of refer to this nice and cleanly I'm going to give it the shorthand Zeta of s it's a function of the parameter s and the common name for it is the the remon zeta function so I use the symbol Zeta for it this is just going to allow me to manipulate it let's imagine I'm working with an S where this thing converges I can do a little bit of trickery let me take this zeta function and multiply it by 1 over 2 to the S if I look at all the terms in my sequence and I bring in the 1 over 2 to the s this is has the effect of making everything multiplied by two so my denominators become 1 over 2 to the S 1 over 4 to the S 1 over 6 to the S all evens okay let's take the top and subtract it from the bottom in other words what I'm considering is 1 - 1 over2 to the stimes the zeta function when I took the zeta function divided it by 2 to the S this meant that everything on the bottom was even and so if I subtract this all the even ones go away like one over two to the s get subtracted from both sides one over 4 to the S and so forth so what I'm left with only his odds on the bottom 3 to the S 5 to the S and 7 to the S okay well let's keep going let's take that expression I'm going do the same kind of trickery to it I'm going to multiply now by the next prime 1 over 3 to the S so kind of like how I filtered by twos now I'm filtering by threes you do that my denominators become well multiples of three 3 9 15 21 let all the multiples of three cuz we already filtered something some of them out right so so six wouldn't be there because we already got rid of the even ones but I'm left with these multiples of three and then I'm going to do the same trick top minus bottom so this is going to give me 1 minus 1 over 3 S times this previous expression and well I filtered out all of the multiples of three from the bottom I have a 5 to the s a 7 to the S an 11 to the S but but no multiples of two no multiples of three okay I I can keep on going and and filtering along the next one will be 1us 1 over 5 to the s that gives rid of all the multiples of five and so again my my sort of picture brings us up to this spot in the Civ of osines I filtered out the twos the threes the fives my next step would be the sevens then the 11s then the 13s and so forth and so if I keep on doing this multiplying by Factor after Factor after factor and I just do that for all of the primes then what I'm going to get is that the product of all the Primes of Expressions that look like this one - one over a prime number to the S you know 2 3 5 and so forth well on the right hand side I've gotten rid of the twos the threes the fives if I keep on going getting rid of all these prime numbers the only thing that's going to survive is the value of one and so this product of primes times this zeta function is equal to one then to capture the formula it's just a matter of dividing through and so I will put this expression onto the other side and this give me that zet of s is the product of 1/ 1 minus p to Theus s and then reminding ourselves of what the zeta function was I get Oiler product formula now I want to emphasize that the proof sketch that I just gave it really only works when you have convergence so this idea only works when you have convergence and again this converges for real numbers of s greater than one okay fourth reason why I love this formula it's it's about what we can get out of it now you might remember that the fact that there's infinitely many prime numbers is something that Humanity has known for a very long time done a previous video before on ukids uh Infamous proof that there were in fact infinitely many primes going back to the time of the ancient Greeks but oil's proof of his product formula this is going back to the 1700s now actually gives us a way to not only reove that there's infin many primes but to extend beyond that and get some more glimpses of about how many primes there really are to do this I really want to imagine what's going to happen in the S equal to one case and I'll remind you that if I was just to plug in s equal to 1 as I've you know sort of naively written the formula down on the page here well this is the harmonic series and it diverges but I want to study the right hand side and see what that implies about this particular product the product of 1 over 1 minus P to the minus one what happens if you happen to plug in s equal to 1 more Faithfully what we should be doing is taking the limit as s approaches one from above I'm going to take that product formula and I'm going to take the logarithm of it and I know some of you don't like logarithms that's okay I love logarithms because they allow me to take products and convert a product into a summation the logarithm of the product of two things is the sum of the two logarithms and so using logarithms I get to convert this into a summation I've also taken an exponent to the power of minus one and brought it out the front as a minus sign Another Log R okay lovely and then there's one more cool log rule I get to exploit here which is the power series for what logarithm is so I'll remind you that we have the series expansion for the logarithm of 1 plus X it's x - x^2 2 + x Cub over 3 and so forth so looking at what I have where I've got all these logarithm terms of 1 minus P to the minus one I'm going to call that P to the minus1 that's going to be my value of x and so I can take this logarithm and expand it out as a series so if I I do that okay remember there's the negative all the front because the X I'm chosing is negative all the odd Powers bring out a negative all the even Powers already have a negative so everything's negative and with the negative up the front they're all positive nevertheless I get this sum and a common thing that you see when you're doing the kind of Aries expansions is you want to make some sort of argument the higher order terms don't matter sometimes they do sometimes they don't but in this particular case the higher order terms don't matter that that actually everything from the second term onwards you add all of those up together this is all small like you can bound it by a number like for example 1 half if you don't mind I'll leave this slightly technical point to the interested viewer who could debate about it in the comments or I'll I'll leave a link to an answer in the description if you're so interested but the point is it doesn't matter it it's irrelevant so going back to where we were at the beginning remember what we were trying to do was we had this Divergent harmonic series and at least in the limit as as was going to one from the right we were trying to compare that to the product that we're studying taking logarithm doesn't affect whether something diverges or not so ultimately if we're wanting this product to diverge it must be the case that just the sum of one over P one over the primes has to diverge and having taken that little detour we get to the big moment here which is that the sum of the reciprocal primes diverge now note this well first of all gives us that there's infinitely many primes like if there was only a finite number of primes there's no way we you this would add up to Infinity so so first of all we've captured that but it does more it tells us something about the density of primes that there's sort of enough of them to get this Divergence and for example you could compare this to something else like remember before we saw the the the reciprocal squares one over n SAR and we saw that that added up to Pi squ 6 and so the point here is that square numbers are pretty rare such that when you add up the reciprocals of them they manage to converge but primes are not that rare the sort of the density of them isn't n that bad you add up the reciprocal primes and it does diverge to Infinity so yes there's infinitely many primes and yes those primes get sparer and sparer the larger the numbers are but they never go so sparse that you don't have this Divergence now trying to understand the distribution of primes leads to some of the deepest parts of mathematics and this is where the oiler product formula is really a doorway into an entire branch of mathematics called analytic number Theory remember how earlier in the video we were talking about the the zeta function Z of s here and I noted that this only works for real numbers where s is greater than one okay what if I allowed complex values for S and one can show that the result that we knew that s had to be greater than one for real numbers extend to the complex numbers to be that the real component of s needs to be greater than one so it's okay to treat this as a complex function but but now it gets tricky what mathematicians do they say well there's a whole bunch of comp Lex numbers where the real component is not going to be greater than one wouldn't it be nice if we could have this function extended into that domain and there's an incredibly powerful trick within mathematics called analytic continuation that allows these functions to be extended into the complex plane now I'm actually not going to do in this video an entire explainer on this there's a lovely old 1x3 blue one brown that I'll put a link to down in the description you want to kind like visualize what this analytic continuation looks like but for our purposes I just want you to imagine imagine that this function which only converge in a certain range is is being extended now to apply to many other places you might for instance have heard of this infamous thing like what happens if you plug in minus one there by the way don't don't yell at me yet these these equal signs are sort of rather nebulous right here uh but if you plug in minus one into the series you would get 1 plus 2 plus 3 plus 4 plus 5 clearly something that diverges to Infinity everything is positive but if you look at the analy continuation the analytic continuation can be computed out for the value of minus one and there you get the value of minus 112th which is why the sort of this Infamous formula can sometimes be uh written down well I'm going to get rid of it uh read away so that I I don't get anyone yelling at me in the comments but for the purposes of this video all I'm really trying to say here is that this really is this launching pad for a lot more really deep mathematics and in fact it contains one of Humanity's you know one of the top and greatest unsolved problems that we have a problem that the uh the clay Institute has called a millennium problem and would give a million dollars for somebody to solve it and nobody has done that and that's called the reman hypothesis and I can state it relatively quickly here given what we've done it basically says the non-trivial zeros of the reman zeta function the the Z of s that I had in its domain of convergence but then extended via analytic continuation then the hypothesis is that this has a bunch of non-trivial zeros and all those non-trivial zeros occur when the real component has a value of a half outside of just stating it the point is that the remon hypothesis really deeply connects into questions about the distribution of the prime numbers and so knowing the answer to the rean hypothesis would really give some insight in how this distribution of prime numbers actually works and so these are the reasons why I think that the oiler product formula should be on any short list of the most beautiful equations in mathematics it's not just a surprising result that can be stated using Elementary mathematic operations it's not just that it has this beautiful proof it's not just that it extends our prior knowledge about say there being infinitely many primes it's just that it really opens this door to this entire field of analytic number Theory and some of the deepest problems in mathematics that Humanity has ever faced behind it now if you want to get better at mathematics then I would strongly recommend the sponsor of today's video which is brilliant.org brilliant helps you learn by doing it has thousands of lessons across mathematics computer science Ai and more and I'm a fan of brilliant for three main reasons firstly it's just really interactive you are the one in the driver's seat actually doing all the mathematics second it builds up learning in layers so you can be confident with your understanding on one step before you're jumping into the next one and finally it consistently is providing feedback and opportunities to self assess one of the biggest challenges that I've experienced as a professor trying to help students learn math mathematics is that it's easy to sit back and watch a YouTube video or a lecture and think that you've got it but it's not until you've had that opportunity to really self assess it and do the mathematics yourself that you can really know whether you've actually mastered it and so for these Reasons I'm very proud to be sponsored by brilliant so go to brilliant.org Trevor bazit that's me or the link is down in the description to try everything that they have for free for full 30 days or to get 20% off an annual premium subscription with that said and done I want to ask you what's your favorite math equation leave it down in the comments and we'll do some more math in the next video

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