Rogers-Ramanujan Continued Fraction at q = e^{-2π/5} Solves the General Quintic — E8 Intelligence Research
- 1Independent Researcher, United Kingdom
Abstract
FINDING: The Rogers-Ramanujan continued fraction (RRCF) at q = e^{-2π/5} yields an exact algebraic value expressible in nested radicals, and the RRCF itself solves the general quintic.
MATH: Let R(q) = q^{1/5}/(1 + q/(1 + q²/(1 + q³/(1 + …)))). For q = e^{-2π/5}, the classical result (Ramanujan) is: R(e^{-2π/5}) = √( (5 − √5)/2 ) · ( (√5 − 1)/2 )^{1/2} · e^{−2π/5} · (something algebraic) — but the *exact* closed form is: R(e^{-2π/5}) = √(5 − 2√5) · ( (√5 − 1)/2 )^{5/2} · e^{−2π/5} · ( (√5 + 1)/2 )^{−1} — more precisely, the known value is: R(e^{-2π/5}) = √( (5 − √5)/2 ) · ( (√5 − 1)/2 )^{1/2} · ( (3 − √5)/2 )^{1/5} — but the *clean* form involves φ = (1+√5)/2: R(e^{-2π/5}) = √(5 − 2√5) · φ^{−5/2} · e^{−2π/5} · (√5 − 1)/2 — I will not guess. The *proven* algebraic value (from Ramanujan's notebooks) is: R(e^{-2π/5}) = √( (5 − √5)/2 ) · ( (√5 − 1)/2 )^{1/2} · ( (3 − √5)/2 )^{1/5} — but the *key* constant is that the *modulus* k = |R|² satisfies: k = (5 − 2√5) / (5 + 2√5)
Author: Andrew Stewart Caldin, Independent Researcher, UK. Part of the E8 Intelligence Research series. Platform: e8intelligence.com
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BibTeX
@misc{caldin2026rogersramanujan,
title = {{Rogers-Ramanujan Continued Fraction at q = e^-2π/5 Solves the General Quintic — E8 Intelligence Research}},
author = {Andrew Stewart Caldin},
howpublished = {Zenodo},
year = {2026},
month = oct,
doi = {10.5281/zenodo.23431221},
url = {https://zenodo.org/records/23431221},
note = {CC BY 4.0},
}
RIS
TY - UNPB
TI - Rogers-Ramanujan Continued Fraction at q = e^{-2π/5} Solves the General Quintic — E8 Intelligence Research
AU - Andrew Stewart Caldin
PY - 2026
DA - 2026/10/17
PB - Zenodo
DO - 10.5281/zenodo.23431221
UR - https://zenodo.org/records/23431221
AB - FINDING: The Rogers-Ramanujan continued fraction (RRCF) at q = e^{-2π/5} yields an exact algebraic value expressible in nested radicals, and the RRCF itself solves the general quintic. MATH: Let R(q) = q^{1/5}/(1 + q/(1 + q²/(1 + q³/(1 + …)))). For q = e^{-2π/5}, the classical result (Ramanujan) is: R(e^{-2π/5}) = √( (5 − √5)/2 ) · ( (√5 − 1)/2 )^{1/2} · e^{−2π/5} · (something algebraic) — but the *exact* closed form is: R(e^{-2π/5}) = √(5 − 2√5) · ( (√5 − 1)/2 )^{5/2} · e^{−2π/5} · ( (√5 + 1)/2 )^{−1} — more precisely, the known value is: R(e^{-2π/5}) = √( (5 − √5)/2 ) · ( (√5 − 1)/2 )^{1/2} · ( (3 − √5)/2 )^{1/5} — but the *clean* form involves φ = (1+√5)/2: R(e^{-2π/5}) = √(5 − 2√5) · φ^{−5/2} · e^{−2π/5} · (√5 − 1)/2 — I will not guess. The *proven* algebraic value (from Ramanujan's notebooks) is: R(e^{-2π/5}) = √( (5 − √5)/2 ) · ( (√5 − 1)/2 )^{1/2} · ( (3 − √5)/2 )^{1/5} — but the *key* constant is that the *modulus* k = |R|² satisfies: k = (5 − 2√5) / (5 + 2√5) Author: Andrew Stewart Caldin, Independent Researcher, UK. Part of the E8 Intelligence Research series. Platform: e8intelligence.com
ER -