Euler-Binet Formula
Theorem
The Fibonacci numbers have a closed-form solution:
- $F_n = \dfrac {\phi^n - \paren {1 - \phi}^n} {\sqrt 5} = \dfrac {\phi^n - \paren {-1 / \phi}^n} {\sqrt 5} = \dfrac {\phi^n - \paren {-1}^n \phi^{-n} } {\sqrt 5}$
where $\phi$ is the golden mean.
Putting $\hat \phi = 1 - \phi = -\dfrac 1 \phi$ this can be written:
- $F_n = \dfrac {\phi^n - \hat \phi^n} {\sqrt 5}$
Corollary 1
- $F_n = \dfrac {\phi^n} {\sqrt 5}$ rounded to the nearest integer
Corollary 2
For even $n$:
- $F_n < \dfrac {\phi^n} {\sqrt 5}$
For odd $n$:
- $F_n > \dfrac {\phi^n} {\sqrt 5}$
Negative Index
Let $n \in \Z_{< 0}$ be a negative integer.
Let $F_n$ be the $n$th Fibonacci number (as extended to negative integers).
Then the Euler-Binet Formula:
- $F_n = \dfrac {\phi^n - \hat \phi^n} {\sqrt 5}$
continues to hold.
Proof 1
Proof by induction:
For all $n \in \N$, let $\map P n$ be the proposition:
- $F_n = \dfrac {\phi^n - \hat \phi^n} {\sqrt 5}$
Basis for the Induction
$\map P 0$ is true, as this just says:
- $\dfrac {\phi^0 - \hat \phi^0} {\sqrt 5} = \dfrac {1 - 1} {\sqrt 5} = 0 = F_0$
$\map P 1$ is the case:
\(\ds \frac {\phi - \hat \phi} {\sqrt 5}\) | \(=\) | \(\ds \frac {\paren {\frac {1 + \sqrt 5} 2} - \paren {\frac {1 - \sqrt 5} 2} } {\sqrt 5}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\paren {1 - 1} + \paren {\sqrt 5 + \sqrt 5} } {2 \sqrt 5}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 1\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds F_1\) |
This is our basis for the induction.
Induction Hypothesis
Now we need to show that, if $\map P j$ is true for all $0 \le j \le k + 1$, then it logically follows that $\map P {k + 2}$ is true.
So this is our induction hypothesis:
- $\forall 0 \le j \le k + 1: F_j = \dfrac {\phi^j - \hat \phi^j} {\sqrt 5}$
Then we need to show:
- $F_{k + 2} = \dfrac {\phi^{k + 2} - \hat \phi^{k + 2} } {\sqrt 5}$
Induction Step
This is our induction step:
We observe that we have the following two identities:
- $\phi^2 = \paren {\dfrac {1 + \sqrt 5} 2}^2 = \dfrac 1 4 \paren {6 + 2 \sqrt 5} = \dfrac {3 + \sqrt 5} 2 = 1 + \phi$
- $\hat \phi^2 = \paren {\dfrac {1 - \sqrt 5} 2}^2 = \dfrac 1 4 \paren {6 - 2 \sqrt 5} = \dfrac {3 - \sqrt 5} 2 = 1 + \hat \phi$
This can also be deduced from the definition of the golden mean: the fact that $\phi$ and $\hat \phi$ are the solutions to the quadratic equation $x^2 = x + 1$.
Thus:
\(\ds \phi^{k + 2} - \hat \phi^{k + 2}\) | \(=\) | \(\ds \paren {1 + \phi} \phi^k - \paren {1 + \hat \phi} \hat \phi^k\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren {\phi^k - \hat \phi^k} + \paren {\phi^{k + 1} - \hat \phi^{k + 1} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \sqrt 5 \paren {F_k + F_{k + 1} }\) | Induction Hypothesis | |||||||||||
\(\ds \) | \(=\) | \(\ds \sqrt 5 F_{k + 2}\) | Definition of Fibonacci Numbers |
The result follows by the Second Principle of Mathematical Induction.
Therefore:
- $\forall n \in \N: F_n = \dfrac {\phi^n - \hat \phi^n} {\sqrt 5}$
$\blacksquare$
Proof 2
Let $A = \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}$.
First by the lemma to Cassini's Identity:
- $(1): \quad \forall n \in \Z_{>1}: A^n = \begin{bmatrix} F_{n + 1} & F_n \\ F_n & F_{n - 1} \end{bmatrix}$
Next is is demonstrated that $A$ has the eigenvalues $\phi$ and $\hat \phi$, where $\hat \phi = 1 - \phi$.
Now we have that:
\(\text {(2)}: \quad\) | \(\ds A \begin{pmatrix} \phi \\ 1 \end{pmatrix}\) | \(=\) | \(\ds \begin{pmatrix} \phi + 1 \\ \phi \end{pmatrix}\) | |||||||||||
\(\ds \) | \(=\) | \(\ds \phi \begin{pmatrix} \phi \\ 1 \end{pmatrix}\) | since $\phi$ is a solution of $x^2 - x - 1 = 0$ | |||||||||||
\(\ds A \begin{pmatrix} \hat \phi \\ 1 \end{pmatrix}\) | \(=\) | \(\ds \begin{pmatrix} \hat \phi + 1 \\ \hat \phi \end{pmatrix}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \hat \phi \begin{pmatrix} \hat \phi \\ 1 \end{pmatrix}\) | since $\hat \phi$ is a solution of $x^2 - x - 1 = 0$ |
This shows that:
- $\begin{pmatrix} \phi \\ 1 \end{pmatrix}$ is an eigenvector of $A$ with eigenvalue $\phi$
and:
- $\begin{pmatrix} \hat \phi \\ 1 \end{pmatrix}$ is an eigenvector of $A$ with eigenvalue $\hat \phi$.
Thus:
- $\begin{pmatrix} \frac \phi {\sqrt 5} \\ \frac 1 {\sqrt 5} \end{pmatrix}$ is an eigenvector of $A$ with eigenvalue $\phi$
and:
- $\begin{pmatrix} \frac {\hat \phi} {\sqrt 5} \\ \frac 1 {\sqrt 5} \end{pmatrix}$ is an eigenvector of $A$ with eigenvalue $\hat \phi$.
By Eigenvalue of Matrix Powers we get for a positive integer $n$:
\(\ds \phi^n \begin{pmatrix} \frac \phi {\sqrt 5} \\ \frac 1 {\sqrt 5} \end{pmatrix}\) | \(=\) | \(\ds A^n \begin{pmatrix} \frac \phi {\sqrt 5} \\ \frac 1 {\sqrt 5} \end{pmatrix}\) | ||||||||||||
\(\ds {\hat \phi}^n \begin{pmatrix} \frac {\hat \phi} {\sqrt 5} \\ \frac 1 {\sqrt 5} \end{pmatrix}\) | \(=\) | \(\ds A^n \begin{pmatrix} \frac {\hat \phi} {\sqrt 5} \\ \frac 1 {\sqrt 5} \end{pmatrix}\) |
From $(1)$ we get:
- $A^n \begin{pmatrix} 1 \\ 0 \end{pmatrix} = \begin{pmatrix} F_{n + 1} \\ F_n \end{pmatrix}$
Substituting:
- $\begin{pmatrix} 1 \\ 0 \end{pmatrix} = \begin{pmatrix} \frac \phi {\sqrt 5} \\ \frac 1 {\sqrt 5} \end{pmatrix} - \begin{pmatrix} \frac {\hat \phi} {\sqrt 5} \\ \frac 1 {\sqrt 5} \end{pmatrix}$
we get:
\(\ds \begin{pmatrix} F_{n + 1} \\ F_n \end{pmatrix}\) | \(=\) | \(\ds \frac 1 {\sqrt 5} A^n \left({\begin{pmatrix} \phi \\ 1 \end{pmatrix} - \begin{pmatrix} \hat \phi \\ 1 \end{pmatrix} }\right)\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds A^n \begin{pmatrix} \frac \phi {\sqrt 5} \\ \frac 1 {\sqrt 5} \end{pmatrix} - A^n \begin{pmatrix} \frac {\hat \phi} {\sqrt 5} \\ \frac 1 {\sqrt 5} \end{pmatrix}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \phi^n \begin{pmatrix} \frac \phi {\sqrt 5} \\ \frac 1 {\sqrt 5} \end{pmatrix} - \hat \phi^n \begin{pmatrix} \frac {\hat \phi} {\sqrt 5} \\ \frac 1 {\sqrt 5} \end{pmatrix}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 {\sqrt 5} \begin{pmatrix} \phi^n \cdot \phi - \hat \phi^n \cdot \hat \phi \\ \phi^n - \hat \phi^n \end{pmatrix}\) |
Hence the result.
$\blacksquare$
Proof 3
This follows as a direct application of the first Binet form:
- $U_n = m U_{n - 1} + U_{n - 2}$
where:
\(\ds U_0\) | \(=\) | \(\ds 0\) | ||||||||||||
\(\ds U_1\) | \(=\) | \(\ds 1\) |
has the closed-form solution:
- $U_n = \dfrac {\alpha^n - \beta^n} {\Delta}$
where:
\(\ds \Delta\) | \(=\) | \(\ds \sqrt {m^2 + 4}\) | ||||||||||||
\(\ds \alpha\) | \(=\) | \(\ds \frac {m + \Delta} 2\) | ||||||||||||
\(\ds \beta\) | \(=\) | \(\ds \frac {m - \Delta} 2\) |
where $m = 1$.
$\blacksquare$
Proof 4
From Generating Function for Fibonacci Numbers, a generating function for the Fibonacci numbers is:
- $\map G z = \dfrac z {1 - z - z^2}$
Hence:
\(\ds \map G z\) | \(=\) | \(\ds \dfrac z {1 - z - z^2}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac 1 {\sqrt 5} \paren {\dfrac 1 {1 - \phi z} - \dfrac 1 {1 - \hat \phi z} }\) | Partial Fraction Expansion |
where:
- $\phi = \dfrac {1 + \sqrt 5} 2$
- $\hat \phi = \dfrac {1 - \sqrt 5} 2$
By Sum of Infinite Geometric Sequence:
- $\dfrac 1 {1 - \phi z} = 1 + \phi z + \phi^2 z^2 + \cdots$
and so:
- $\map G z = \dfrac 1 {\sqrt 5} \paren {1 + \phi z + \phi^2 z^2 + \cdots - 1 - \hat \phi z - \hat \phi^2 z^2 - \cdots}$
By definition, the coefficient of $z^n$ in $\map G z$ is exactly the $n$th Fibonacci number.
That is:
- $F_n = \dfrac {\phi^n - \hat \phi^n} {\sqrt 5}$
$\blacksquare$
Also known as
The Euler-Binet Formula is also known as Binet's formula.
Source of Name
This entry was named for Jacques Philippe Marie Binet and Leonhard Paul Euler.
Historical Note
The Euler-Binet Formula, derived by Binet in $1843$, was already known to Euler, de Moivre and Daniel Bernoulli over a century earlier.
However, it was Binet who derived the more general Binet Form of which this is an elementary application.
Sources
- 1722: Abraham de Moivre: De Fractionibus Algebraicis Radicalitate Immunibus ad Fractiones Simpliciores Reducendis, Deque Summandis Terminis Quarumdam Serierum Aequali Intervallo a Se Distantibus (Phil. Trans. Vol. 32: pp. 162 – 178) www.jstor.org/stable/103594
- 1728: Daniel Bernoulli: Obseruationes de seriebus recurrentibus (Commentarii Acad. Sci. Imp. Pet. Vol. 3: pp. 85 – 100)
- 1986: David Wells: Curious and Interesting Numbers ... (previous) ... (next): $5$
- 1992: David Wells: Curious and Interesting Puzzles ... (previous) ... (next): Liber Abaci: $88$
- 1997: David Wells: Curious and Interesting Numbers (2nd ed.) ... (previous) ... (next): $5$