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ConsciousnessLearn · Explore · Joseph Anady

Binet’s formula for Fibonacci integers

The closed form combines the two roots of r²=r+1: φ and −1/φ. Subtracting their nth powers and dividing by √5 produces the standard Fibonacci integer Fₙ. It explains the ratio limit, although direct floating-point evaluation can lose exact integer accuracy for large n.

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eq-339a4f258fc80fmathematical expression
\fib{n} = \frac{\golden^n - (-\golden)^{-n}}{\sqrt{5}}

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The closed form combines the two roots of r²=r+1: φ and −1/φ. Subtracting their nth powers and dividing by √5 produces the standard Fibonacci integer Fₙ. It explains the ratio limit, although direct floating-point evaluation can lose exact integer accuracy for large n.

Terms & definitions

TermMeaning in this source
F_n

nth Fibonacci integer

golden

Golden ratio φ

n

Nonnegative integer index

How it works

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Original context

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We provide additional detail for Theorem~\ref{thm:convergence}.

\textbf{Fibonacci Ratio Convergence.} The ratio $\fib{n}/\fib{n-1}$ converges to $\golden$ exponentially fast. Using Binet's formula:
\begin{equation}
\fib{n} = \frac{\golden^n - (-\golden)^{-n}}{\sqrt{5}}
\end{equation}

For large $n$, $(-\golden)^{-n} \to 0$, so:
\begin{equation}
\frac{\fib{n}}{\fib{n-1}} \approx \frac{\golden^n}{\golden^{n-1}} = \golden
\end{equation}
Exact original notation
\fib{n} = \frac{\golden^n - (-\golden)^{-n}}{\sqrt{5}}

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Sources & provenance

01
manuscript.tex
line 752 · Sep 22, 2026
\fib{n} = \frac{\golden^n - (-\golden)^{-n}}{\sqrt{5}}
02
manuscript.tex
lines 749–751
\fib{n} = \frac{\golden^n - (-\golden)^{-n}}{\sqrt{5}}
03
manuscript.tex
lines 751–753 · Sep 22, 2026
\fib{n} = \frac{\golden^n - (-\golden)^{-n}}{\sqrt{5}}
04
manuscript.tex
lines 749–751
\fib{n} = \frac{\golden^n - (-\golden)^{-n}}{\sqrt{5}}

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