Euler-Binet Formula

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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