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Collected Papers ofSrinivasa Ramanujan

Name: Collected Papers of Srinivasa Ramanujan
Author: Srinivasa Ramanujan

Srinivasa Ramanujan · 1911 · retold from the English original

Collected Papers of Srinivasa Ramanujan

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The first paper the self-taught genius ever published, written in Madras at 23: strange patterns in the Bernoulli numbers, the numbers behind sums of powers. Most of his claims are true, some are guesses, and one is false; the lectures show which is which.

This is a sample: 1 of 37 papers.

From the lectures

Two rows of coefficient boxes for the cotangent and the hyperbolic cotangent; adding them keeps B0, B4, B8 and subtracting keeps B2, B6, B10; a final pair of bars shows half the terms against a quarter.

Some properties of Bernoulli's numbers

Every second term cancels

Section 3. Again, −½x(cot ½x + coth ½x) = −½x(cot ½x + i cot ½ix) = 2(B₀ + B₄ x⁴/4! + B₈ x⁸/8! + ⋯), by using equation one.

The divisors of 24 each get one added; the composites are struck out; the primes 2, 3, 5, 7, 13 that remain multiply to 2730.

Sections 5 to 7

Building the denominator of B24

Section 7. Next, let us consider the denominators.

Equation twenty-nine gives the tail of the reciprocal squares in powers of 1/n; then the combinations B2 − 1/6, −B4 − 1/6 + 1/5, B6 − 1/6 + 1/7, −B8 − 1/6 + 1/5, B10 − 1/6 + 1/11 all come out as 0 with the known values.

Section 12

The coefficients that must be whole

From equation twenty-seven we have d² log Γ(n + 1)/dn² = 1/(n + 1)² + 1/(n + 2)² + 1/(n + 3)² + ⋯ = 1/n − 1/(2n²) + B₂/n³ − B₄/n⁵ + B₆/n⁷ − B₈/n⁹ + ⋯ − (−1)^p B₂ₚθ/n^(2p+1), equation twenty-nine. Here B₂ₚθ/n^(2p+1) = B₂ₚ/n^(2p+1) − B₂ₚ₊₂/n^(2p+3) + ⋯.

Result thirty-five: Bₙ/Bₙ₋₂ is close to n(n − 1)/(4π²); a table of B14 to B20, each found to lie between two whole numbers and then written exactly with the prime rule.

Sections 13 to 15

Each B from the one before

Section 14. Changing n to n minus 2 in equation twenty-four and taking the ratio of the two results, we have Bₙ = (n(n − 1)/(4π²)) Bₙ₋₂ (1 − (2² − 1)/(2ⁿ − 1))(1 − (3² − 1)/(3ⁿ − 1))(1 − (5² − 1)/(5ⁿ − 1))⋯, equation thirty-four, where 2, 3, 5 and so on are prime numbers.

Equations thirty-eight to forty: each isolates one strand of Bernoulli numbers spaced ten apart — B₀, B₁₀, B₂₀, …; B₂, B₁₂, B₂₂, …; B₄, B₁₄, B₂₄, … — as a ratio of two series with coefficients 1 ± αₙ built from (1 ± √5)/2.

Sections 16 and 17

One strand per formula

Taking logarithms and differentiating both sides, we have 5(B₀ + B₁₀x¹⁰/10! + B₂₀x²⁰/20! + B₃₀x³⁰/30! + ⋯) = −x ((x⁴/4!)(1 + α₅) − (x¹⁴/14!)(1 + α₁₅) + ⋯) / ((x⁵/5!)(1 + α₅) − (x¹⁵/15!)(1 + α₁₅) + ⋯), equation thirty-eight, where αₙ = ((1 + √5)/2)ⁿ + ((1 − √5)/2)ⁿ, so that αₙαₘ = αₙ₊ₘ + (−1)ⁿαₘ₋ₙ.

Appendix notes by page: B₂ₙ for the usual Bₙ; cₖ is the binomial coefficient; equation twelve holds for odd n > 1 with three last terms; the derivations after (27) and (29) are invalid because a series diverges for x ≥ n; integral coefficients would be odd; (32) from ψ(z + 1) − ψ(½z + 1); N and D are the top and bottom of B₄₄₄.

Section 18 and the Editors' Note

The editors' notes

Editors' note, from Appendix I, the notes on paper 1. Page 1, equation one: Ramanujan writes B₂ₙ where Bₙ is usual. Page 1, equation two: cₖ = n!/(k!(n − k)!), the binomial coefficient.

How it was made

Level 1 · Wie im Original
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worked examples
Voice: Charon (Gemini)
Told in English

Examples for: a curious adult with school maths who likes number puzzles and knows Ramanujan's story; add one modern instance per idea, such as a calculator, computer algebra or cryptography

The source

The lectures retell the English text of the Cambridge 1927 edition (Internet Archive pli.kerala.rare.28155). The author died in 1920; the work is in the public domain. The retelling, the drawings and the voice are made by AI and checked against the book by a second model; nothing was edited by hand. How AI is used.

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