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 :

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

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:

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:

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:

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

We will now determine the eigenvectors of $A$

Recall the definition of Eigenvector of Real Square Matrix:

Let $\mathbf A$ be a square matrix of order $n$ over $\R$.

Let $\lambda \in \R$ be an eigenvalue of $\mathbf A$.

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:

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

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

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

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:

Then we notice that:

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

Putting all of the pieces together, we obtain:

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:

has the closed-form solution:

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

where:

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:

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 is also known as Binet's formula.

However, there is more than one theorem named such, so, in order to reduce ambiguity, the longer form is preferred on $\mathsf{Pr} \infty \mathsf{fWiki}$.

Source of Name

This entry was named for Leonhard Paul Euler and Jacques Philippe Marie Binet.

Historical Note

The , 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): $5$
• 2021: Richard Earl and James Nicholson: The Concise Oxford Dictionary of Mathematics (6th ed.) ... (previous) ... (next): closed-form expression