## 2 Neural Model
### The Fibonacci's Series and Binet's Formula  
The [PERSON]'s series \(F_{n}=\{1,\ 1,\ 2,\ 3,\ 5,\ 8,\ \ldots\}\) is a numeric sequence, which owes its popularity to the fact that it reflects a ubiquitous scheme of growth in nature. The generic term of the sequence is described analytically by the following finite equation:  
\[F_{n}=F_{n-1}+F_{n-2},\ \ with\ n\geq 2 \tag{6}\]  
As can be readily demonstrated, the Fibonacci sequence can also be interpreted as the impulse response of a second-order IIR filter with unitary recursive weights, as depicted in Figure 2. The structure of this filter closely resembles the initial portion of the internal architecture of a hidden neuron, as illustrated in Figure 1b.  
As can be seen, Equation (6) makes the sequence implicit as each term is the sum of the two last terms before. Fortunately, [PERSON] (1786-1858) provided a formula that allows one to directly calculate any value of the Fibonacci series without calculating the previous ones, stating it as follows:  
_The \(n\)-th term of the Fibonacci sequence denoted as \(F_{n}\), can be expressed explicitly using [PERSON]'s formula:_  
\[F_{n}=\frac{\phi^{n}-\psi^{n}}{\sqrt{5}} \tag{7}\]  
_where \(\phi=\frac{1+\sqrt{5}}{2}\) represents the golden ratio and \(\psi=\frac{1-\sqrt{5}}{2}\) is its conjugate._  
Here, the development of that formula is briefly summarized.  
_Demonstration of [PERSON]'s formula_. As said before, the Fibonacci sequence is defined recursively as:  
\[F_{0}=0,\ F_{1}=1,\ \ldots,\ F_{n}=F_{n-1}+F_{n-2}\]  
Let us assume that an explicit function, which provides the terms of the Fibonacci series, exists, and for the \(n\)-th term, it has the general expression:  
\[F_{n}=C\cdot z^{n} \tag{8}\]  
Figure 2: IIR filter of the second order with unitary recursive weights, whose impulsive response corresponds to the Fibonacci series.  
with \(C\) and \(z\) constant values to be determined. By substituting (8) in (6) the following expression is obtained  
\[C\cdot z^{n}=C\cdot z^{n-1}+C\cdot z^{n-2}\ \ \Rightarrow\ C\cdot z^{n}-C\cdot z^{n-1}-C\cdot z^{n-2}=0 \tag{9}\]  
from which:  
\[C\cdot z^{n-2}\Big{(}z^{2}-z^{1}-1\Big{)}=0 \tag{10}\]  
Equation (10) has two trivial solutions (\(C=0\) ; \(z=0\)), which must be excluded because they could not generate the series, and two non-trivial solutions, namely the roots of the polynomial within brackets. These two solutions are:  
\[z_{1}=\tfrac{1+\sqrt{5}}{2}\quad;\quad z_{2}=\tfrac{1-\sqrt{5}}{2} \tag{11}\]  
It is worth noting that the first root \(z_{1}\) in (11) is the golden ratio value. Given that the recurrence relation (9) is linear, the sum of the two solutions is also a solution. The sought function can therefore be obtained as a linear combination of the two solutions corresponding to the two roots \(z_{1}\) and \(z_{2}\):  
\[F_{n}=C_{1}z_{1}^{n}+C_{2}z_{2}^{n} \tag{12}\]  
with \(C_{1}\) and \(C_{2}\) to be determined. To this end, we can impose the correspondence with two arbitrary values of the series, for example, the first two: \(1\), \(1\).  
\[\begin{cases}F_{1}=1\ \Leftrightarrow C_{1}\cdot z_{1}+C_{2}\cdot z_{2}=1\\ F_{2}=1\ \Leftrightarrow C_{1}\cdot z_{1}^{2}+C_{2}\cdot z_{2}^{2}=1\end{cases} \tag{13}\]  
The solution of the system (13) is \(C_{1}=\tfrac{1}{\sqrt{5}}\); \(C_{2}=-\tfrac{1}{\sqrt{5}}\) from which the following expression of [PERSON]'s formula comes:  
\[F_{n}=\frac{1}{\sqrt{5}}\Bigg{[}\Bigg{(}\frac{1+\sqrt{5}}{2}\Bigg{)}^{n}- \Bigg{(}\frac{1-\sqrt{5}}{2}\Bigg{)}^{n}\Bigg{]} \tag{14}\]