Euler-Binet Formula

Theorem

The Fibonacci numbers have a closed-form solution:

$F_n = \dfrac {\phi^n - \left({1 - \phi}\right)^n} {\sqrt 5} = \dfrac {\phi^n - \left({-1 / \phi}\right)^n} {\sqrt 5} = \dfrac {\phi^n - \left({-1}\right)^n\phi^{-n} } {\sqrt 5}$

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 $P(n)$ be the proposition:

$F_n = \dfrac {\phi^n - \hat \phi^n} {\sqrt 5}$

Basis for the Induction

$P(0)$ is true, as this just says:

$\dfrac {\phi^0 - \hat \phi^0} {\sqrt 5} = \dfrac {1 - 1} {\sqrt 5} = 0 = F_0$

$P(1)$ is the case:

 $\displaystyle \frac {\phi - \hat \phi} {\sqrt 5}$ $=$ $\displaystyle \frac {\left({\frac {1 + \sqrt 5} 2}\right) - \left({\frac {1 - \sqrt 5} 2}\right)} {\sqrt 5}$ $\displaystyle$ $=$ $\displaystyle \frac {\left({1 - 1}\right) + \left({\sqrt 5 + \sqrt 5 }\right)} {2 \sqrt 5}$ $\displaystyle$ $=$ $\displaystyle 1$ $\displaystyle$ $=$ $\displaystyle F_1$

This is our basis for the induction.

Induction Hypothesis

Now we need to show that, if $P(j)$ is true for all $0 \le j \le k + 1$, then it logically follows that $P(k + 2)$ 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:

$\phi^2 = \left({\dfrac {1 + \sqrt 5} 2}\right)^2 = \dfrac 1 4 \left({6 + 2 \sqrt 5}\right) = \dfrac {3 + \sqrt 5} 2 = 1 + \phi$
$\hat \phi^2 = \left({\dfrac {1 - \sqrt 5} 2}\right)^2 = \dfrac 1 4 \left({6 - 2 \sqrt 5}\right) = \dfrac {3 - \sqrt 5} 2 = 1 + \hat \phi$

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:

 $\displaystyle \phi^{k + 2} - \hat \phi^{k + 2}$ $=$ $\displaystyle \left({1 + \phi}\right) \phi^k - \left({1 + \hat \phi}\right) \hat \phi^k$ $\displaystyle$ $=$ $\displaystyle \left({\phi^k - \hat \phi^k}\right) + \left({\phi^{k + 1} - \hat \phi^{k + 1} }\right)$ $\displaystyle$ $=$ $\displaystyle \sqrt 5 \left({F_k + F_{k+1} }\right)$ Induction Hypothesis $\displaystyle$ $=$ $\displaystyle \sqrt 5 F \left({k+2}\right)$ 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:

 $(2):\quad$ $\displaystyle A \begin{pmatrix} \phi \\ 1 \end{pmatrix}$ $=$ $\displaystyle \begin{pmatrix} \phi + 1 \\ \phi \end{pmatrix}$ $\displaystyle$ $=$ $\displaystyle \phi \begin{pmatrix} \phi \\ 1 \end{pmatrix}$ since $\phi$ is a solution of $x^2 - x - 1 = 0$ $\displaystyle A \begin{pmatrix} \hat \phi \\ 1 \end{pmatrix}$ $=$ $\displaystyle \begin{pmatrix} \hat \phi + 1 \\ \hat \phi \end{pmatrix}$ $\displaystyle$ $=$ $\displaystyle \hat \phi \begin{pmatrix} \hat \phi \\ 1 \end{pmatrix}$ since $\hat \phi$ is a solution of $x^2 - x - 1 = 0$

This shows that:

$\displaystyle \begin{pmatrix} \phi \\ 1 \end{pmatrix}$ is an eigenvector of $A$ with eigenvalue $\phi$

and:

$\displaystyle \begin{pmatrix} \hat \phi \\ 1 \end{pmatrix}$ is an eigenvector of $A$ with eigenvalue $\hat \phi$.

Thus:

$\displaystyle \begin{pmatrix} \frac \phi {\sqrt 5} \\ \frac 1 {\sqrt 5} \end{pmatrix}$ is an eigenvector of $A$ with eigenvalue $\phi$

and:

$\displaystyle \begin{pmatrix} \frac {\hat \phi} {\sqrt 5} \\ \frac 1 {\sqrt 5} \end{pmatrix}$ is an eigenvector of $A$ with eigenvalue $\hat \phi$.

By Eigenvalue of Matrix Powers we get for a positive integer $n$:

 $\displaystyle \phi^n \begin{pmatrix} \frac \phi {\sqrt 5} \\ \frac 1 {\sqrt 5} \end{pmatrix}$ $=$ $\displaystyle A^n \begin{pmatrix} \frac \phi {\sqrt 5} \\ \frac 1 {\sqrt 5} \end{pmatrix}$ $\displaystyle {\hat \phi}^n \begin{pmatrix} \frac {\hat \phi} {\sqrt 5} \\ \frac 1 {\sqrt 5} \end{pmatrix}$ $=$ $\displaystyle A^n \begin{pmatrix} \frac {\hat \phi} {\sqrt 5} \\ \frac 1 {\sqrt 5} \end{pmatrix}$

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:

 $\displaystyle \begin{pmatrix} F_{n + 1} \\ F_n \end{pmatrix}$ $=$ $\displaystyle \frac 1 {\sqrt 5} A^n \left({\begin{pmatrix} \phi \\ 1 \end{pmatrix} - \begin{pmatrix} \hat \phi \\ 1 \end{pmatrix} }\right)$ $\displaystyle$ $=$ $\displaystyle A^n \begin{pmatrix} \frac \phi {\sqrt 5} \\ \frac 1 {\sqrt 5} \end{pmatrix} - A^n \begin{pmatrix} \frac {\hat \phi} {\sqrt 5} \\ \frac 1 {\sqrt 5} \end{pmatrix}$ $\displaystyle$ $=$ $\displaystyle \phi^n \begin{pmatrix} \frac \phi {\sqrt 5} \\ \frac 1 {\sqrt 5} \end{pmatrix} - \hat \phi^n \begin{pmatrix} \frac {\hat \phi} {\sqrt 5} \\ \frac 1 {\sqrt 5} \end{pmatrix}$ $\displaystyle$ $=$ $\displaystyle \frac 1 {\sqrt 5} \begin{pmatrix} \phi^n \cdot \phi - \hat \phi^n \cdot \hat \phi \\ \phi^n - \hat \phi^n \end{pmatrix}$

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:

 $\displaystyle U_0$ $=$ $\displaystyle 0$ $\displaystyle U_1$ $=$ $\displaystyle 1$

has the closed-form solution:

$U_n = \dfrac {\alpha^n - \beta^n} {\Delta}$

where:

 $\displaystyle \Delta$ $=$ $\displaystyle \sqrt {m^2 + 4}$ $\displaystyle \alpha$ $=$ $\displaystyle \frac {m + \Delta} 2$ $\displaystyle \beta$ $=$ $\displaystyle \frac {m - \Delta} 2$

where $m = 1$.

$\blacksquare$

Proof 4

$\map G z = \dfrac z {1 - z - z^2}$

Hence:

 $\displaystyle \map G z$ $=$ $\displaystyle \dfrac z {1 - z - z^2}$ $\displaystyle$ $=$ $\displaystyle \dfrac 1 {\sqrt 5} \paren {\dfrac 1 {1 - \phi z} - \dfrac 1 {1 - \hat \phi z} }$ Partial Fraction Expansion

where:

$\phi = \dfrac {1 + \sqrt 5} 2$
$\hat \phi = \dfrac {1 - \sqrt 5} 2$
$\dfrac 1 {1 - \phi z} = 1 + \phi z + \phi^2 z^2 + \cdots$

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.