Euler-Binet Formula/Proof 1

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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:

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


Proof

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:

$\forall n \in \N: F_n = \dfrac {\phi^n - \hat \phi^n} {\sqrt 5}$

$\blacksquare$


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.