$$ F_n = \frac{\varphi^n}{\sqrt{5}} - \frac{(1 - \varphi)^n}{\sqrt{5}} \qquad \bigl(\text{Binet's formula}\bigr) $$
$$ F_n = \frac{\varphi^n}{\sqrt{5}} - \frac{(1 - \varphi)^n}{\sqrt{5}} \qquad \bigl(\text{Binet's formula}\bigr) $$
We determine that the golden ratio \((\varphi)\) is the the limit of two consecutive terms of the Fibonacci sequence is worth:
Demonstrations
The Fibonacci sequence is expressed in a recurrent way as:
$$ \forall n \in \mathbb{N}, \hspace{2em} F_{n + 2} = F_{n + 1} + F _n $$
$$ \text{with } : \Bigl \{ F_0 = 0, \ F_1 = 1 \Bigr \} $$
We know that the general term of a sequence of order 2, linear combination of its preceding terms can be determined by doing this:
$$ u_{n + 2} = p.u_{n + 1} + q.u_n \Longrightarrow u_n = A. \alpha^n + B.\beta^n
\hspace{2em} \text{with } \left \{ \begin{gather*}
(p, q) \in \bigl[\mathbb{R}^*\bigr]^2 \\ \\
A \text{ and } B \text{ to determine according to } u_0 \text{ and } u_1 \\ \\
\alpha \text{ and } \beta \text{ the two roots of } \Bigl[ r^2 - pr - q = 0 \Bigr]
\end{gather*} \right \}
$$
In our case: \((p = q = 1)\).
So,
$$ F_{n + 2} = F_{n + 1} + F _n \Longrightarrow F_n = A. \alpha^n + B.\beta^n $$
Where the coefficients \((\alpha, \ \beta)\) are solutions for \((E_c)\):
$$ r^2 - r - 1 = 0 \qquad (E_c) $$
-
Détermination of coefficients \((\alpha, \ \beta)\)
So, we have to solve this:
$$ r^2 - r - 1 = 0 \qquad (E_c) $$
Since we are dealing with a quadratic equation, we first calculate the discriminant \((\Delta)\):
$$ \Delta = (-1)^2 - 4 \times 1 \times (-1) $$
$$ \Delta = 1 + 4 $$
$$ \Delta = 5 $$
As \((\Delta > 0)\), we do have two distinct roots:
So, the general term of the sequence \((F_n)\) can be written as:
$$ F_n = A.(1 - \varphi)^n + B.\varphi^n $$
-
Détermination of coefficients \((A, \ B)\)
Let's now determine the coefficients \((A, \ B)\) thanks to the first two terms of the sequence. We know that:
$$ \Bigl \{ F_0 = 0, \ F_1 = 1 \Bigr \} $$
So,
Which leads us to solve the following system \((\mathcal{S})\):
By subtracting the values of the two lines, we get:
$$ A \times (2 \varphi - 1) = -1 $$
$$ A = -\frac{1}{\sqrt{5}} $$
Now, taking the first part of the original system, we inject the value of \(B\):
$$ -\frac{1}{\sqrt{5}} + B = 0 $$
$$ B = \frac{1}{\sqrt{5}} $$
-
Modeling the general term \(F_n\) \(F_n\)
Hence, the general term of the Fbonacci sequence \((F_n)\) is worth:
$$ F_n = -\frac{(1 - \varphi)^n}{\sqrt{5}} + \frac{\varphi^n}{\sqrt{5}} $$
$$ F_n = \frac{\varphi^n}{\sqrt{5}} - \frac{(1 - \varphi)^n}{\sqrt{5}} \qquad \bigl(\text{Binet's formula}\bigr) $$
$$ \text{with } \left( \varphi = \frac{1 + \sqrt{5}}{2} \right) $$
Retrieving the previous Binet's formula:
$$ F_n = \frac{\varphi^n}{\sqrt{5}} - \frac{(1 - \varphi)^n}{\sqrt{5}} \qquad \bigl(\text{Binet's formula}\bigr) $$
If we now compute the quotient of two consecutive terms of this sequence, we do have:
$$ \frac{F_{n + 1}}{F_n} = \frac{\frac{\varphi^{n + 1} - (1 - \varphi)^{n + 1}}{\sqrt{5}}}{\frac{\varphi^n - (1 - \varphi)^n}{\sqrt{5}}} $$
$$ \frac{F_{n + 1}}{F_n} = \frac{\varphi^{n + 1} - (1 - \varphi)^{n + 1}}{\varphi^n - (1 - \varphi)^n} $$
But, as \(\varphi\) is solution for \(E_c\):
$$ \varphi^2 = \varphi + 1 \qquad (\varphi^2) $$
Dividing by \(\varphi\), we do have now:
$$ \frac{\varphi^2}{\textcolor{#6F79AB}{\varphi}} = \frac{\varphi}{\textcolor{#6F79AB}{\varphi}} + \frac{1}{\textcolor{#6F79AB}{\varphi}} $$
Therefore:
$$ \varphi = 1 + \frac{1}{\varphi} $$
$$ 1 - \varphi = -\frac{1}{\varphi} $$
So, replacing now \((1 - \varphi)\) by its value:
$$ \frac{F_{n + 1}}{F_n} = \frac{\varphi^{n + 1} - \left(-\frac{1}{\varphi}\right)^{n + 1}}{\varphi^n - \left(-\frac{1}{\varphi}\right)^n} $$
$$ \frac{F_{n + 1}}{F_n} = \frac{\varphi^{n + 1} - \left(-\frac{1}{\varphi^{n + 1}}\right)}{\varphi^n - \left(-\frac{1}{\varphi^n}\right)} $$
Finally, taking the limit when \(n\) goes to infinity, we obtain that:
$$ \lim_{n \to \infty} \frac{F_{n + 1}}{F_n} = \lim_{n \to \infty} \frac{\varphi^{n + 1} - \left(-\frac{1}{\varphi^{n + 1}}\right)}{\varphi^n - \left(-\frac{1}{\varphi^n}\right)} $$
Both quotient obviously tend towards zero and:
$$ \lim_{n \to \infty} \frac{F_{n + 1}}{F_n} = \lim_{n \to \infty} \frac{\varphi^{n + 1}}{\varphi^n} $$
$$ \lim_{n \to \infty} \frac{F_{n + 1}}{F_n} = \varphi $$
$$ \text{with } \left( \varphi = \frac{1 + \sqrt{5}}{2} \right) $$
-
With \(\varphi\) only
Starting the characteristic equation \((E_c)\) seen above:
$$ r^2 - r - 1 = 0 \qquad (E_c) $$
As the golden ratio \((\varphi)\) is solution, we can write the following relation for \(\varphi^2\) :
$$ \varphi^2 = \varphi + 1 \qquad (\varphi^2) $$
Furthermore, it is true that \((1 - \varphi)\) is also a solution.
$$ (1 - \varphi)^2 = (1 - \varphi) + 1 $$
But upon further development, we realize that it amounts to the same thing:
$$ 1 -2 \varphi + \varphi^2 = 2 - \varphi $$
$$ \varphi^2 = \varphi + 1 \qquad (\varphi^2) $$
Multiplying both members by \(\varphi\), we do obtain that:
$$ \varphi^3 = \varphi^2 + \varphi \qquad (\varphi^3) $$
We can see that the recurrence relation is quite obvious.
Let us then attempt to demonstrate, using a double induction, that:
$$ \forall n \in \mathbb{N}, $$
$$ \varphi^{n + 2} = \varphi^{n + 1} + \varphi^{n} \qquad (P_n) $$
-
Computation of the two first terms
For \((k = 0) \), it is the previous formula, which has already been proven \((\varphi^2)\):
$$ \varphi^2 = \varphi + 1 \qquad (P_0) $$
For \((k = 1) \), we do want to prove that \((P_1)\) is true:
$$ \varphi^3 = \varphi^2 + \varphi \qquad (P_1) $$
So, both propositions \((P_0)\) and \((P_1)\) are true.
-
Heredity
From the proposition \((P_k)\):
$$ \varphi^{k + 2} = \varphi^{k + 1} + \varphi^{k} \qquad (P_k) $$
The recurrence relation is direct when multiplying by \(\varphi\):
So, both propositions \((P_{k + 1})\) et \((P_{k + 2})\) are true.
-
Conclusion
The proposition \((P_n)\) is true for its two first terms, \(n_0 = 0\) and \(n_1 = 1\), and it is hereditary from terms to terms for all \(k \in \mathbb{N}\).
By the double recurrence principle, that statement is true for all \(n \in \mathbb{N}\).
Hence,
$$ \forall n \in \mathbb{N}, $$
$$ \varphi^{n + 2} = \varphi^{n + 1} + \varphi^{n} $$
-
With \(\varphi\) and \((F_n)\)
Retrieving the expression \((\varphi^3)\), which has been checked above:
$$ \varphi^3 = \varphi^2 + \varphi \qquad (\varphi^3) $$
If we do replace \(\varphi^2\) by its other value:
$$ \varphi^3 = (\varphi + 1) + \varphi $$
$$ \varphi^3 = 2\varphi + 1 \qquad (\varphi^3)^* $$
Similarly, if we continue in this way with \(\varphi^4\):
$$ \varphi^4 = 2\varphi^2 + \varphi $$
$$ \varphi^4 = 2(\varphi + 1) + \varphi $$
$$ \varphi^4 = 2\varphi + 2 + \varphi $$
$$ \varphi^4 = 3\varphi + 2 \qquad (\varphi^4)^* $$
The terms of the Fibonacci sequence then appears:
$$ \Bigl \{ 0, \ 1, \ 1 , \ 2, \ 3, \ 5, \ 8 ...etc. \Bigr \} $$
Specifically, in \((\varphi^3)^*\) and \((\varphi^4)^*\) where:
A second time, one is tempted to demonstrate that recurrence is such that:
$$ \forall n \in \mathbb{N}, $$
$$ \varphi^{n + 1} = \varphi F_{n + 1} + F_n \qquad (Q_n) $$
-
Computation of the two first terms
For \((k = 0) \):
$$ \varphi^{1} = \varphi F_{1} + F_0 \qquad (Q_0) $$
So, the proposition \((Q_0)\) is true.
-
Heredity
$$ \forall n \in \mathbb{N}, $$
$$ \varphi^{k + 1} = \varphi F_{k + 1} + F_k \qquad (Q_k) $$
We wish to establish the following relationship:
$$ \varphi^{k + 2} = \varphi F_{k + 2} + F_{k + 1} \qquad (Q_{k + 1}) $$
Let's start from \((Q_k)\) and let us multiply by \(\varphi\):
$$ \textcolor{#6F79AB}{\varphi}\varphi^{k + 1} = \textcolor{#6F79AB}{\varphi}\varphi F_{k + 1} + \textcolor{#6F79AB}{\varphi}F_k \qquad (\textcolor{#6F79AB}{\varphi} Q_k) $$
$$ \varphi^{k + 2} = \varphi^2 F_{k + 1} + \varphi F_k $$
$$ \varphi^{k + 2} = (\varphi + 1) F_{k + 1} + \varphi F_k $$
$$ \varphi^{k + 2} = \varphi F_{k + 1} + F_{k + 1} + \varphi F_k $$
$$ \varphi^{k + 2} = \varphi (F_{k + 1} + F_k) + F_{k + 1} $$
$$ \varphi^{k + 2} = \varphi (F_{k + 2}) + F_{k + 1} \qquad (Q_{k + 2}) $$
We have clearly shown that \((Q_{k})\) is true, then \((Q_{k + 1})\) was too.
-
Conclusion
The proposition \((P_n)\) is true for its first term \(n_0 = 0\), and it is hereditary from terms to terms for all \(k \in \mathbb{N}\).
By the recurrence principle, that statement is true for all \(n \in \mathbb{N}\).
Hence,
$$ \forall n \in \mathbb{N}, $$
$$ \varphi^{n + 1} = \varphi F_{n + 1} + F_n $$
-
With square roots
By retrieving the previous expression of \(\varphi^2\), which has already been proved:
$$ \varphi^2 = \varphi + 1 \qquad (\varphi^2) $$
Taking the square root of it:
$$ \varphi = \sqrt{\varphi + 1} \qquad (1) $$
-
Under the form of infinite square roots
As \(\varphi\) is expressed according to itself in \((1)\), it can be self-injected:
$$ \varphi = \sqrt{\sqrt{\varphi + 1} + 1} $$
$$ \varphi = \sqrt{\sqrt{\sqrt{\varphi + 1} + 1} + 1} $$
And so on...
$$ \varphi = \sqrt{\sqrt{\sqrt{\sqrt{1 + \dots} + 1} + 1} + 1} $$
-
Under the form of a recurrent sequence
We can also express \((1)\) as a recurrent sequence:
$$ a_{n + 1} = \sqrt{1 + a_{n}} $$
Hence,
$$ \varphi = \lim_{n \to \infty} a_{n} $$
$$
\text{where } \Biggl \{ \begin{gather*}
\text{the sequence } a_n \text{ is expressed under a recurrent form } : a_{n + 1} = \sqrt{1 + a_{n}} \\
\text{is first term } a_0 = 0
\end{gather*}
$$
-
With fractions
As well as above, we start again from \(\varphi^2\):
$$ \varphi^2 = \varphi + 1 \qquad (\varphi^2) $$
Dividing by \(\varphi\), we do have now:
$$ \frac{\varphi^2}{\textcolor{#6F79AB}{\varphi}} = \frac{\varphi}{\textcolor{#6F79AB}{\varphi}} + \frac{1}{\textcolor{#6F79AB}{\varphi}} $$
Hence:
$$ \varphi = 1 + \frac{1}{\varphi} \qquad (2) $$
-
Under the form of infinite fractions
Again, \(\varphi\) is expressed according to itself in \((2)\), so we can auto-inject it:
$$ \varphi = 1 - \frac{1}{1 - \frac{1}{\varphi}} $$
$$ \varphi = 1 - \frac{1}{1 - \frac{1}{1 - \frac{1}{\varphi}}} $$
And so on...
$$ \varphi = 1 - \frac{1}{1 - \frac{1}{1 - \frac{1}{1 - \frac{1}{1 - \dots}}}} $$
-
Under the form of a recurrent sequence
We can also express \((2)\) as a recurrent sequence:
$$ b_{n + 1} = 1 + \frac{1}{b_n} $$
Ainsi,
$$ \varphi = \lim_{n \to \infty} b_{n} $$
$$
\text{où } \Biggl \{ \begin{gather*}
\text{la suite } b_n \text{ est exprimée sous forme récurrente } : b_{n + 1} = 1 + \frac{1}{b_n} \\
\text{son premier terme } : b_1 = 1
\end{gather*}
$$
Go to the top of the page
