Euler-Binet Formula/Proof 3

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:

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

Proof

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 $m = 1$.

$\blacksquare$

Source of Name

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

Also known as

The Euler-Binet Formula is also known as Binet's formula.