Euler-Binet Formula

Theorem

The Fibonacci numbers have a closed-form solution:

$F_n = \dfrac {\phi^n - \paren {1 - \phi}^n} {\sqrt 5} = \dfrac {\phi^n - \paren {-1 / \phi}^n} {\sqrt 5} = \dfrac {\phi^n - \paren {-1}^n \phi^{-n} } {\sqrt 5} = \dfrac {\phi^n - \paren {1 - \phi}^n} {\phi - \paren {1 - \phi}}$

where $\phi$ is the golden mean.

Putting $\hat \phi = 1 - \phi = -\dfrac 1 \phi$ this can be written:

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

From Definition 2 of Golden Mean: $\phi = \dfrac {1 + \sqrt 5} 2$

Therefore, substituting $\sqrt 5 = 2\phi - 1 = \phi - \paren {1 - \phi} = \phi - \hat \phi$, the above can be written as:

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

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:

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

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:

$(1): \quad \forall n \in \Z_{>1}: A^n = \begin{bmatrix} F_{n + 1} & F_n \\ F_n & F_{n - 1} \end{bmatrix}$

Recall from Eigenvalues of Real Square Matrix are Roots of Characteristic Equation, we can find the eigenvalues of $\mathbf A$ by solving the equation $\map \det {\mathbf A - \lambda \mathbf I} = 0$

Therefore, taking the determinant of the square matrix below:

$\mathbf A - \lambda \mathbf I = \begin{bmatrix} 1 - \lambda & 1 \\ 1 & -\lambda \end{bmatrix}$

We obtain the equation:

 $\ds 0$ $=$ $\ds -\lambda \paren {1 - \lambda} - 1$ Determinant of Matrix Product $\ds$ $=$ $\ds \lambda^2 - \lambda - 1$ $\ds$ $=$ $\ds \lambda^2 - \lambda + \paren {\lambda \phi - \lambda \phi} + \paren {\phi - \phi} - 1$ adding 0 $\ds$ $=$ $\ds \lambda^2 - \lambda + \paren {\lambda \phi - \lambda \phi} + \phi - \paren {\phi + 1}$ regrouping $\ds$ $=$ $\ds \lambda^2 - \lambda + \paren {\lambda \phi - \lambda \phi} + \phi - \phi^2$ Square of Golden Mean equals One plus Golden Mean $\ds$ $=$ $\ds \lambda^2 - \lambda \paren {1 - \phi} - \lambda \phi + \phi \paren {1 - \phi}$ $\ds$ $=$ $\ds \paren {\lambda - \phi} \paren {\lambda - \paren {1 - \phi} }$

Therefore, $A$ has the eigenvalues $\phi$ and $\paren {1 - \phi}$.

We will now determine the eigenvectors of $A$

From the Definition of Eigenvector of Real Square Matrix, we have:

A non-zero vector $\mathbf v \in \R^n$ is an eigenvector corresponding to $\lambda$ if and only if:

$\mathbf A \mathbf v = \lambda \mathbf v$

For $\lambda = \phi$, we have:

 $\ds \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix} \times \begin{pmatrix} v_1 \\ v_2 \end{pmatrix}$ $=$ $\ds \begin{pmatrix} \phi v_1 \\ \phi v_2 \end{pmatrix}$ $\ds \leadsto \ \$ $\ds \begin{pmatrix} v_1 + v_2 \\ v_1 \end{pmatrix}$ $=$ $\ds \begin{pmatrix} \phi v_1 \\ \phi v_2 \end{pmatrix}$ Definition of Matrix Product (Conventional)

We see from the above, when $v_2 = 1$ then $v_1 = \phi$ and we have:

 $\ds \begin{pmatrix} \phi + 1 \\ \phi \end{pmatrix}$ $=$ $\ds \begin{pmatrix} \phi^2 \\ \phi \end{pmatrix}$ $\ds$ $=$ $\ds \phi \begin{pmatrix} \phi \\ 1 \end{pmatrix}$

For $\lambda = \paren {1 - \phi}$, we have:

 $\ds \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix} \times \begin{pmatrix} v_1 \\ v_2 \end{pmatrix}$ $=$ $\ds \begin{pmatrix} \paren {1 - \phi} v_1 \\ \paren {1 - \phi} v_2 \end{pmatrix}$ $\ds \leadsto \ \$ $\ds \begin{pmatrix} v_1 + v_2 \\ v_1 \end{pmatrix}$ $=$ $\ds \begin{pmatrix} \paren {1 - \phi} v_1 \\ \paren {1 - \phi} v_2 \end{pmatrix}$ Definition of Matrix Product (Conventional)

We see from the above, when $v_2 = 1$ then $v_1 = \paren {1 - \phi}$ and we have:

 $\ds \begin{pmatrix} 2 - \phi \\ 1 - \phi \end{pmatrix}$ $=$ $\ds \begin{pmatrix} \paren {1 - \phi}^2 \\ \paren {1 - \phi } \end{pmatrix}$ $\ds$ $=$ $\ds \paren {1 - \phi} \begin{pmatrix} \paren {1 - \phi} \\ 1 \end{pmatrix}$

We have now demonstrated that:

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

and:

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

We now observe that:

 $\ds A^n \begin{pmatrix} 1 \\ 0 \end{pmatrix}$ $=$ $\ds \begin{bmatrix} F_{n + 1} & F_n \\ F_n & F_{n - 1} \end{bmatrix} \begin{pmatrix} 1 \\ 0 \end{pmatrix}$ from (1) above $\ds$ $=$ $\ds \begin{pmatrix} F_{n + 1} \\ F_n \end{pmatrix}$ Definition of Matrix Product (Conventional)

Then we notice that:

 $\text {(2)}: \quad$ $\ds \begin{pmatrix} 1 \\ 0 \end{pmatrix}$ $=$ $\ds \dfrac 1 {\phi - \paren {1 - \phi} } \paren {\begin{pmatrix} \phi \\ 1 \end{pmatrix} - \begin{pmatrix} \paren {1 - \phi} \\ 1 \end{pmatrix} }$

From Eigenvalue of Matrix Powers for a positive integer $n$, we have:

 $\ds A^n \begin{pmatrix} \phi \\ 1 \end{pmatrix}$ $=$ $\ds \phi^n \begin{pmatrix} \phi \\ 1 \end{pmatrix}$ $A^n \mathbf v = \lambda^n \mathbf v$ $\ds A^n \begin{pmatrix} \paren {1 - \phi} \\ 1 \end{pmatrix}$ $=$ $\ds \paren {1 - \phi}^n \begin{pmatrix} \paren {1 - \phi} \\ 1 \end{pmatrix}$ $A^n \mathbf v = \lambda^n \mathbf v$

Putting all of the pieces together, we obtain:

 $\ds \begin{pmatrix} F_{n + 1} \\ F_n \end{pmatrix}$ $=$ $\ds A^n \begin{pmatrix} 1 \\ 0 \end{pmatrix}$ $\ds$ $=$ $\ds \dfrac 1 {\phi - \paren {1 - \phi} } A^n \paren {\begin{pmatrix} \phi \\ 1 \end{pmatrix} - \begin{pmatrix} \paren {1 - \phi} \\ 1 \end{pmatrix} }$ from (2) above $\ds$ $=$ $\ds \dfrac 1 {\phi - \paren {1 - \phi} } \paren {A^n \begin{pmatrix} \phi \\ 1 \end{pmatrix} - A^n \begin{pmatrix} \paren {1 - \phi} \\ 1 \end{pmatrix} }$ $\ds$ $=$ $\ds \dfrac 1 {\phi - \paren {1 - \phi} } \paren {\phi^n \begin{pmatrix} \phi \\ 1 \end{pmatrix} - \paren {1 - \phi}^n \begin{pmatrix} \paren {1 - \phi} \\ 1 \end{pmatrix} }$ Eigenvalue of Matrix Powers $\ds$ $=$ $\ds \dfrac 1 {\phi - \paren {1 - \phi} } \paren {\begin{pmatrix} \phi^{n + 1} \\ \phi^n \end{pmatrix} - \begin{pmatrix} \paren {1 - \phi}^{n + 1} \\ \paren {1 - \phi}^n \end{pmatrix} }$ Definition of Matrix Product (Conventional)

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

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

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:

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

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.