The Fibonacci sequence \((F_n)\) and the golden ratio \((\varphi)\)

Binet's formula

So,

Where the coefficients \((\alpha, \ \beta)\) are solutions for \((E_c)\):

1. 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:

   $$ \alpha = \frac{1 - \sqrt{5}}{2} $$

   $$ \beta = \frac{1 + \sqrt{5}}{2} $$

   For the sake of simplicity, let us set down the golden ratio :

   $$ \varphi = \frac{1 + \sqrt{5}}{2} \qquad (\varphi) $$

   $$ 1 - \varphi = \frac{1 - \sqrt{5}}{2} \qquad (1 - \varphi) $$

   So that:

   $$ \varphi = \beta $$

   $$ 1 - \varphi = \alpha $$

   So, the general term of the sequence \((F_n)\) can be written as:

   $$ F_n = A.(1 - \varphi)^n + B.\varphi^n $$

2. 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,

   $$ F_0 = A.(1 - \varphi)^0 + B.\varphi^0 = 0 $$

   $$ F_1 = A.(1 - \varphi)^1 + B.\varphi^1 = 1 $$

   Which leads us to solve the following system \((\mathcal{S})\):

   $$ (\mathcal{S}) \ \left \{ \begin{gather*} A + B = 0 \\ \\ A.(1 - \varphi) + B.\varphi = 1 \end{gather*} \right \} $$

   $$ \Longleftrightarrow $$

   $$ \left \{ \begin{gather*} A.\textcolor{rgb(118 139 240)}{\varphi} + B.\textcolor{rgb(118 139 240)}{\varphi} = 0 \\ \\ A.(1 - \varphi) + B.\varphi = 1 \end{gather*} \right \} $$

   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 \(B\):

   $$ -\frac{1}{\sqrt{5}} + B = 0 $$
   $$ B = \frac{1}{\sqrt{5}} $$

3. Modeling the general term \(F_n\) \(F_n\)

   Hence, the general term of the Fbonacci sequence \((F_n)\) is worth:

   $$ F_n = -\frac{(1 - \varphi)^n}{\sqrt{5}} + \frac{\varphi^n}{\sqrt{5}} $$
   $$ F_n = \frac{\varphi^n}{\sqrt{5}} - \frac{(1 - \varphi)^n}{\sqrt{5}} \qquad \bigl(\text{Binet's formula}\bigr) $$
   $$ \text{with } \left( \varphi = \frac{1 + \sqrt{5}}{2} \right) $$

Golden ratio \((\varphi)\): quotient of two consecutive term of the Fibonacci sequence

Retrieving the previous Binet's formula :

If we now compute the quotient of two consecutive terms of this sequence, we do have:

So, replacing now \((1 - \varphi)\) by its value:

Finally, taking the limit when \(n\) goes to infinity, we obtain that:

Both quotient obviously tend towards zero and:

Expressions of powers of the golden ratio under a recurrent form

1. 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:

   $$ \forall n \in \mathbb{N}, $$
   $$ \varphi^{n + 2} = \varphi^{n + 1} + \varphi^{n} \qquad (P_n) $$

   1. 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:

      $$ \varphi^3 = \varphi^2 + \varphi \qquad (P_1) $$
      $$ \varphi^3 = \varphi \cdot \varphi^2 $$

      But, \(\varphi^2 = \varphi + 1\), donc :

      $$ \varphi^3 = \varphi \cdot (\varphi + 1) $$
      $$ \varphi^3 = \varphi^2 + \varphi \qquad (P_1) $$

      We definitely proved that \((P_1)\) is true:

      $$ \varphi^3 = \varphi^2 + \varphi \qquad (P_1) $$

      So, both propositions \((P_0)\) and \((P_1)\) are true.

   2. Inductive Step

      Assume that statements \((P_k)\) and \((P_{k+1})\) are true (Double Induction Hypothesis). We therefore have:

      $$ \varphi^{k + 2} = \varphi^{k + 1} + \varphi^{k} \qquad (P_k) $$
      $$ \varphi^{k + 3} = \varphi^{k + 2} + \varphi^{k + 1} \qquad (P_{k + 1}) $$

      Let us show that statement \((P_{k+2})\) is also true at the next rank, meaning that:

      $$ \varphi^{k + 4} = \varphi^{k + 3} + \varphi^{k + 2} \qquad (P_{k + 2}) $$

      Starting from \((P_{k+1})\), multiplying the entire equation by \(\varphi\):

      $$\varphi \cdot \varphi^{k + 3} = \varphi \cdot (\varphi^{k + 2} + \varphi^{k + 1})$$

      We easily arrive at our target statement \((P_{k+2})\):

      $$ \varphi^{k + 4} = \varphi^{k + 3} + \varphi^{k + 2} \qquad (P_{k + 2}) $$

      Thus, the inductive step is proven at rank \((k + 2)\).

   3. 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,

   $$ \forall n \in \mathbb{N}, $$
   $$ \varphi^{n + 2} = \varphi^{n + 1} + \varphi^{n} $$

2. 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:

   $$ \varphi^3 = 2\varphi + 1 \qquad (\varphi^3)^*

   \varphi^3 = \varphi F_3 + F_2 \qquad (\varphi^3)' $$

   $$ \varphi^4 = 3\varphi + 2 \qquad (\varphi^4)^*

   \varphi^4 = \varphi F_4 + F_3 \qquad (\varphi^4)' $$

   A second time, one is tempted to demonstrate that recurrence is such that:

   $$ \forall n \in \mathbb{N}, $$
   $$ \varphi^{n + 1} = \varphi F_{n + 1} + F_n \qquad (Q_n) $$

   1. Computation of the two first terms

      For \((k = 0) \):

      $$ \varphi^{1} = \varphi F_{1} + F_0 \qquad (Q_0) $$
      $$ \varphi^1 = \varphi $$

      $$ \varphi F_{1} + F_0 = \varphi \times 1 + 0

      \varphi F_{1} + F_0 = \varphi $$

      So, the proposition \((Q_0)\) is true.

   2. Inductive step

      $$ \forall n \in \mathbb{N}, $$
      $$ \varphi^{k + 1} = \varphi F_{k + 1} + F_k \qquad (Q_k) $$

      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 \(\varphi\):

      $$ \textcolor{rgb(118 139 240)}{\varphi}\varphi^{k + 1} = \textcolor{rgb(118 139 240)}{\varphi}\varphi F_{k + 1} + \textcolor{rgb(118 139 240)}{\varphi}F_k \qquad (\textcolor{rgb(118 139 240)}{\varphi} Q_k) $$
      $$ \varphi^{k + 2} = \varphi^2 F_{k + 1} + \varphi F_k $$
      $$ \varphi^{k + 2} = (\varphi + 1) F_{k + 1} + \varphi F_k $$
      $$ \varphi^{k + 2} = \varphi F_{k + 1} + F_{k + 1} + \varphi F_k $$
      $$ \varphi^{k + 2} = \varphi (F_{k + 1} + F_k) + F_{k + 1} $$
      $$ \varphi^{k + 2} = \varphi (F_{k + 2}) + F_{k + 1} \qquad (Q_{k + 2}) $$

      We have clearly shown that \((Q_{k})\) is true, then \((Q_{k + 1})\) was too.

   3. 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,

      $$ \forall n \in \mathbb{N}, $$
      $$ \varphi^{n + 1} = \varphi F_{n + 1} + F_n $$

Recursive expression of the golden ratio

1. 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 (we already set that \(\varphi = \frac{1 + \sqrt{5}}{2}\)):

   $$ \varphi = \sqrt{\varphi + 1} \qquad (1) $$

   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...

      $$ \varphi = \sqrt{\sqrt{\sqrt{\sqrt{1 + \dots} + 1} + 1} + 1} $$

   2. Under the form of a recurrent sequence

      We can also express \((1)\) as a recurrent sequence:

      $$ a_{n + 1} = \sqrt{1 + a_{n}} $$

      Hence,

      $$ \varphi = \lim_{n \to \infty} a_{n} $$

      $$ \text{where } \Biggl \{ \begin{gather*} \text{the sequence } a_n \text{ is expressed under a recurrent form } : a_{n + 1} = \sqrt{1 + a_{n}} \\ \text{is first term } a_0 = 0 \end{gather*} $$

2. 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:

   $$ \frac{\varphi^2}{\textcolor{rgb(118 139 240)}{\varphi}} = \frac{\varphi}{\textcolor{rgb(118 139 240)}{\varphi}} + \frac{1}{\textcolor{rgb(118 139 240)}{\varphi}} $$

   Hence:

   $$ \varphi = 1 + \frac{1}{\varphi} \qquad (2) $$

   1. 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...

      $$ \varphi = 1 + \frac{1}{1 + \frac{1}{1+ \frac{1}{1 + \frac{1}{1 + \dots}}}} $$

   2. 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,

      $$ \varphi = \lim_{n \to \infty} b_{n} $$

      $$ \text{où } \Biggl \{ \begin{gather*} \text{la suite } b_n \text{ est exprimée sous forme récurrente } : b_{n + 1} = 1 + \frac{1}{b_n} \\ \text{son premier terme } : b_1 = 1 \end{gather*} $$