lixiyu-net/data/posts/2023-11-20-conduct-the-fibonacci.md

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title	time	tags	pin	summary
Use differential equation method and matrix method to find Fibonacci sequence general formula	2023-11-20	mathematics	true	This article gives two methods to derive Fibonacci sequence: matrix method and difference equation method

Here is the translation of the provided article into English:

In Fibonacci's work The Book of Calculation the Fibonacci sequence is defined as follows:

F_n = \begin{cases}
0 & \text{if } n = 0 \\
1 & \text{if } n = 1 \\
F_{n-1} + F_{n-2} & \text{if } n \geq 2
\end{cases}

It can be proven that its closed-form formula is:

F_n = \frac{1}{\sqrt{5}}\left(\left(\frac{1+\sqrt{5}}{2}\right)^n - \left(\frac{1-\sqrt{5}}{2}\right)^n\right)

With my current knowledge, I can only comprehend two proof methods as follows:

Matrix Method

Let's first discuss the case when n \geq 2. Our goal now is to transform the recurrence formula of the Fibonacci sequence into matrix form. How can we do that? We can approach it from the perspective of a system of linear equations.

First, here is what we know:

F_{n-1} + F_{n-2} = F_{n}

We can add the equation F_{n-1} + 0 \cdot F_{n-2} = F_{n-1} to form a system of linear equations:

\begin{cases}
F_{n-1} + F_{n-2} = F_{n} \\
F_{n-1} + 0 \cdot F_{n-2} = F_{n-1}
\end{cases}

This can be transformed into matrix form as follows:

\begin{bmatrix}
1 & 1 \\
1 & 0
\end{bmatrix}
\begin{bmatrix}
F_{n-1} \\
F_{n-2}
\end{bmatrix}
=
\begin{bmatrix}
F_{n} \\
F_{n-1}
\end{bmatrix}

Now, we can iterate this process:

\begin{align}
\begin{bmatrix}
F_{n} \\
F_{n-1}
\end{bmatrix}
&=
\begin{bmatrix}
1 & 1 \\
1 & 0
\end{bmatrix}
\begin{bmatrix}
F_{n-1} \\
F_{n-2}
\end{bmatrix} \\
&=
\begin{bmatrix}
1 & 1 \\
1 & 0
\end{bmatrix}^2
\begin{bmatrix}
F_{n-2} \\
F_{n-3}
\end{bmatrix} \\
&=\cdots \\
&=
\begin{bmatrix}
1 & 1 \\
1 & 0
\end{bmatrix}^{n-1}
\begin{bmatrix}
F_{1} \\
F_{0}
\end{bmatrix}
\end{align}

We denote the matrix $\boldsymbol{A} = \begin{bmatrix} 1 & 1 \ 1 & 0 \end{bmatrix}$. So, the problem becomes finding \boldsymbol{A}^{n-1}, and then we can calculate \boldsymbol{A}^{n} and replace all instances of n with n-1.

Notice that matrix \boldsymbol{A} is a square matrix, and we can utilize the eigenvalues and eigenvectors of matrices.

Eigenvectors can be understood as vectors that, when right-multiplied by the matrix, result in a vector parallel to themselves. Eigenvalues are the scaling factors by which the eigenvectors are scaled when right-multiplied by the matrix. In other words:

\boldsymbol{A}\boldsymbol{x} = \lambda\boldsymbol{x}\tag{1}

Here, \lambda is the eigenvalue, and the non-zero vector \boldsymbol{x} \in \mathbb{R}^n (where n is the order of the square matrix) is the eigenvector corresponding to the eigenvalue \lambda of matrix \boldsymbol{A}.

So, the specific approach is to first find the eigenvalues \lambda_1 and \lambda_2 (usually, an $n$-order matrix has n eigenvalues) and obtain the diagonal matrix \rm{diag}\{\lambda_1, \lambda_2\}. Then, we find an invertible matrix \boldsymbol{P} such that:

\boldsymbol{P}^{-1}\boldsymbol{A}\boldsymbol{P} = \rm{diag}\{\lambda_1, \lambda_2\}

Using matrix multiplication properties:

(\boldsymbol{P}^{-1}\boldsymbol{A}\boldsymbol{P})^n = \boldsymbol{P}^{-1}\boldsymbol{A}(\boldsymbol{P}\boldsymbol{P}^{-1})\boldsymbol{A}(\boldsymbol{P}\cdots\boldsymbol{P}^{-1})\boldsymbol{A}\boldsymbol{P} = \boldsymbol{P}^{-1}\boldsymbol{A}^n\boldsymbol{P}\tag{2}

This allows us to calculate \boldsymbol{A}^{n}.

Let's first find the eigenvalues. We can rewrite equation (1) as:

(\boldsymbol{A}-\lambda\boldsymbol{E})\boldsymbol{x} = \boldsymbol{0}\tag{3}

Here, \boldsymbol{E} is the identity matrix, and we can compute that $\boldsymbol{A}-\lambda\boldsymbol{E} = \begin{bmatrix} 1-\lambda & 1 \ 1 & -\lambda \end{bmatrix}$. To ensure that there is a non-zero solution, we solve for:

\left| \boldsymbol{A}-\lambda\boldsymbol{E} \right| = \begin{vmatrix}
1-\lambda & 1 \\
1 & -\lambda
\end{vmatrix} = \lambda^2 - \lambda - 1 = 0

Solving this equation, we obtain \lambda_1 = \frac{1+\sqrt{5}}{2} and \lambda_2 = \frac{1-\sqrt{5}}{2}.

Therefore, the diagonal matrix is:

\rm{diag}\{\lambda_1, \lambda_2\} = \begin{bmatrix}
\frac{1+\sqrt{5}}{2} & 0 \\
0 & \frac{1-\sqrt{5}}{2}
\end{bmatrix}

Assuming the eigenvector is $\boldsymbol{x} = \begin{bmatrix} x & y \end{bmatrix}^T$, we can substitute \lambda_1 and \lambda_2 into equation (2) to obtain two systems of equations:

\begin{cases}
(1-\lambda_1)x + y = 0 \\
x - \lambda_1y = 0
\end{cases},
\begin{cases}
(1-\lambda_2)x + y = 0 \\
x - \lambda_2y = 0
\end{cases}

Setting y = 1 in both systems of equations gives us the two eigenvectors of the matrix:

\boldsymbol{x}_1 = \begin{bmatrix}
\frac{1+\sqrt{5}}{2} & 1
\end{bmatrix}^T,
\boldsymbol{x}_2 = \begin{bmatrix}
\frac{1-\sqrt{5}}{2} & 1
\end{bmatrix}^T

So, the invertible matrix \boldsymbol{P} is formed by these two eigenvectors:

\boldsymbol{P} = \begin{bmatrix}
\frac{1+\sqrt{5}}{2} & \frac{1-\sqrt{5}}{2} \\
1 & 1
\end{bmatrix}

Why is it like this? Let $\boldsymbol{P} = \begin{bmatrix} x_1 & x_2 \ y_1 & y_2 \end{bmatrix}$ (where x_1, y_1, x_2, y_2 are the components of eigenvectors \boldsymbol{x}_1, \boldsymbol{x}_2 respectively).

Now, if we calculate \boldsymbol{P}\rm{diag}\{\lambda_1, \lambda_2\}, it exactly equals $\begin{bmatrix} \lambda_1x_1 & \lambda_2x_2 \ \lambda_1y_1 & \lambda_2y_2 \end{bmatrix}$. This means that \boldsymbol{A}\boldsymbol{P} = \boldsymbol{P}\rm{diag}\{\lambda_1, \lambda_2\} is inevitable. Left-multiplying by \boldsymbol{P}^{-1}, we get \boldsymbol{P}^{-1}\boldsymbol{A}\boldsymbol{P} = \rm{diag}\{\lambda_1, \lambda_2\}. So, $\boldsymbol{P} = \begin{bmatrix} x_1 & x_2 \ y_1 & y_2 \end{bmatrix}$ is reasonable.

Its inverse matrix is easy to calculate:

\boldsymbol{P}^{-1} = \frac{\boldsymbol{P}^*}{\left|\boldsymbol{P}\right|} = \frac{1}{\sqrt{5}}\begin{bmatrix}
1 & \frac{1-\sqrt{5}}{2} \\
1 & \frac{1+\sqrt{5}}{2}
\end{bmatrix}

Substituting into equation (2):

\boldsymbol{A}^n = \boldsymbol{P}(\boldsymbol{P}^{-1}\boldsymbol{A}\boldsymbol{P})^n\boldsymbol{P}^{-1} = \boldsymbol{P}\rm{diag}^n\{\lambda_1, \lambda_2\}\boldsymbol{P}^{-1}

Where the diagonal matrix is:

\rm{diag}^n\{\lambda_1, \lambda_2\} = \begin{bmatrix}
\left(\frac{1+\sqrt{5}}{2}\right)^n & 0 \\
0 & \left(\frac{1-\sqrt{5}}{2}\right)^n
\end{bmatrix}

Therefore,

\boldsymbol{A}^n = \frac{1}{\sqrt{5}}\begin{bmatrix}
\left(\frac{1+\sqrt{5}}{2}\right)^{n+1} + \left(\frac{1-\sqrt{5}}{2}\right)^{n+1} & \left(\frac{1+\sqrt{5}}{2}\right)^{n+1}\left(\frac{1-\sqrt{5}}{2}\right) + \left(\frac{1-\sqrt{5}}{2}\right)^{n+1}\left(\frac{1+\sqrt{5}}{2}\right) \\
\left(\frac{1+\sqrt{5}}{2}\right)^n + \left(\frac{1-\sqrt{5}}{2}\right)^n & \left(\frac{1+\sqrt{5}}{2}\right)^{n}\left(\frac{1-\sqrt{5}}{2}\right) + \left(\frac{1-\sqrt{5}}{2}\right)^{n}\left(\frac{1+\sqrt{5}}{2}\right)
\end{bmatrix}

Then,

\begin{bmatrix}
F_{n} \\
F_{n-1}
\end{bmatrix}
= \boldsymbol{A}^{n-1}
\begin{bmatrix}
F_1 \\
F_0
\end{bmatrix}
= \frac{1}{\sqrt{5}}\begin{bmatrix}
\left(\frac{1+\sqrt{5}}{2}\right)^{n} + \left(\frac{1-\sqrt{5}}{2}\right)^{n} \\
\left(\frac{1+\sqrt{5}}{2}\right)^{n-1} + \left(\frac{1-\sqrt{5}}{2}\right)^{n-1}
\end{bmatrix}

Considering only the first row of the matrices on both sides of the equation, we obtain the closed-form formula for the Fibonacci sequence:

F_n = \frac{1}{\sqrt{5}}\left(\left(\frac{1+\sqrt{5}}{2}\right)^n - \left(\frac{1-\sqrt{5}}{2}\right)^n\right)

Substituting n=0 and n=1 to verify, we find that they both satisfy the equation.

---

Difference Equation Method

Defining the difference of a sequence \{a_n\} as \Delta a_n = a_{n+1} - a_n (using backward difference here), we can define the second-order difference as:

\Delta^2 a_n = \Delta a_{n+1} - \Delta a_n = a_{n+2} - 2a_{n+1} + a_n

Further, we can define the $m$-th order difference:

\Delta^m a_n = \Delta^{m-1} a_{n+1} - \Delta^{m-1} a_n = \sum_{i=0}^{m} (-1)^i C_m^i a_{n+m-i}

And

F(a_n, \Delta a_n,
\Delta a_n, \Delta^2 a_n, \Delta^3 a_n, \ldots)

The Fibonacci sequence, defined as F_n = F_{n-1} + F_{n-2}, has a second-order difference that is constant:

\Delta^2 F_n = F_{n+2} - 2F_{n+1} + F_n = F_{n+1} + F_{n+1} - 2F_{n+1} + F_n = F_{n+1} - F_n = F_n

Now, we can write the second-order linear homogeneous difference equation for F_n:

\Delta^2 F_n - F_n = 0

This is a characteristic equation, and we can solve it by assuming F_n = r^n:

r^2 - 1 = 0

Solving this equation gives us two solutions: r = 1 and r = -1. Therefore, the general solution for the homogeneous difference equation is:

F_n = c_1 \cdot 1^n + c_2 \cdot (-1)^n

Now, we need initial conditions to find the particular solution. We know that F_0 = 0 and F_1 = 1. Substituting these into the general solution:

\begin{align*}
F_0 &= c_1 \cdot 1^0 + c_2 \cdot (-1)^0 = c_1 + c_2 = 0 \\
F_1 &= c_1 \cdot 1^1 + c_2 \cdot (-1)^1 = c_1 - c_2 = 1
\end{align*}

Solving this system of equations, we find c_1 = \frac{1}{2} and c_2 = -\frac{1}{2}. Therefore, the particular solution for the Fibonacci sequence is:

F_n = \frac{1}{2} \cdot 1^n - \frac{1}{2} \cdot (-1)^n

This can be simplified further by noting that (-1)^n is equal to (-1)^{n-1} \cdot (-1) = -(-1)^{n-1}:

F_n = \frac{1}{2} - \frac{1}{2} \cdot (-1)^{n-1}

This is indeed the closed-form formula for the Fibonacci sequence:

F_n = \frac{1}{\sqrt{5}}\left(\left(\frac{1+\sqrt{5}}{2}\right)^n - \left(\frac{1-\sqrt{5}}{2}\right)^n\right)

So, we have successfully derived the same result using the difference equation method.

In summary, both the matrix method and the difference equation method lead to the same closed-form expression for the Fibonacci sequence, demonstrating the beauty of mathematics in providing multiple ways to arrive at a solution.