347 lines
9.9 KiB
Markdown
347 lines
9.9 KiB
Markdown
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title: "Use differential equation method and matrix method to find Fibonacci sequence general formula"
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time: "2023-11-20"
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tags: ["mathematics"]
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pin: true
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summary: "This article gives two methods to derive Fibonacci sequence: matrix method and difference equation method"
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---
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Here is the translation of the provided article into English:
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In Fibonacci's work _The Book of Calculation_ the Fibonacci sequence is defined as follows:
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$$
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F_n = \begin{cases}
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0 & \text{if } n = 0 \\
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1 & \text{if } n = 1 \\
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F_{n-1} + F_{n-2} & \text{if } n \geq 2
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\end{cases}
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$$
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It can be proven that its closed-form formula is:
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$$
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F_n = \frac{1}{\sqrt{5}}\left(\left(\frac{1+\sqrt{5}}{2}\right)^n - \left(\frac{1-\sqrt{5}}{2}\right)^n\right)
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$$
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With my current knowledge, I can only comprehend two proof methods as follows:
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## Matrix Method
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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.
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First, here is what we know:
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$$
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F_{n-1} + F_{n-2} = F_{n}
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$$
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We can add the equation $F_{n-1} + 0 \cdot F_{n-2} = F_{n-1}$ to form a system of linear equations:
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$$
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\begin{cases}
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F_{n-1} + F_{n-2} = F_{n} \\
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F_{n-1} + 0 \cdot F_{n-2} = F_{n-1}
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\end{cases}
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$$
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This can be transformed into matrix form as follows:
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$$
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\begin{bmatrix}
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1 & 1 \\
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1 & 0
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\end{bmatrix}
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\begin{bmatrix}
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F_{n-1} \\
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F_{n-2}
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\end{bmatrix}
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=
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\begin{bmatrix}
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F_{n} \\
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F_{n-1}
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\end{bmatrix}
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$$
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Now, we can iterate this process:
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$$
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\begin{align}
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\begin{bmatrix}
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F_{n} \\
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F_{n-1}
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\end{bmatrix}
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&=
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\begin{bmatrix}
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1 & 1 \\
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1 & 0
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\end{bmatrix}
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\begin{bmatrix}
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F_{n-1} \\
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F_{n-2}
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\end{bmatrix} \\
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&=
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\begin{bmatrix}
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1 & 1 \\
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1 & 0
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\end{bmatrix}^2
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\begin{bmatrix}
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F_{n-2} \\
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F_{n-3}
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\end{bmatrix} \\
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&=\cdots \\
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&=
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\begin{bmatrix}
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1 & 1 \\
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1 & 0
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\end{bmatrix}^{n-1}
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\begin{bmatrix}
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F_{1} \\
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F_{0}
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\end{bmatrix}
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\end{align}
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$$
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We denote the matrix $\boldsymbol{A} = \begin{bmatrix}
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1 & 1 \\
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1 & 0
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\end{bmatrix}$. So, the problem becomes finding $\boldsymbol{A}^{n-1}$, and then we can calculate $\boldsymbol{A}^{n}$ and replace all instances of $n$ with $n-1$.
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Notice that matrix $\boldsymbol{A}$ is a square matrix, and we can utilize the eigenvalues and eigenvectors of matrices.
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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:
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$$
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\boldsymbol{A}\boldsymbol{x} = \lambda\boldsymbol{x}\tag{1}
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$$
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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 $\boldsymbol{A}$.
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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:
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$$
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\boldsymbol{P}^{-1}\boldsymbol{A}\boldsymbol{P} = \rm{diag}\{\lambda_1, \lambda_2\}
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$$
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Using matrix multiplication properties:
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$$
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(\boldsymbol{P}^{-1}\boldsymbol{A}\boldsymbol{P})^n = \boldsymbol{P}^{-1}\boldsymbol{A}(\boldsymbol{P}\boldsymbol{P}^{-1})\boldsymbol{A}(\boldsymbol{P}\cdots\boldsymbol{P}^{-1})\boldsymbol{A}\boldsymbol{P} = \boldsymbol{P}^{-1}\boldsymbol{A}^n\boldsymbol{P}\tag{2}
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$$
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This allows us to calculate $\boldsymbol{A}^{n}$.
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Let's first find the eigenvalues. We can rewrite equation $(1)$ as:
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$$
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(\boldsymbol{A}-\lambda\boldsymbol{E})\boldsymbol{x} = \boldsymbol{0}\tag{3}
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$$
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Here, $\boldsymbol{E}$ is the identity matrix, and we can compute that $\boldsymbol{A}-\lambda\boldsymbol{E} = \begin{bmatrix}
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1-\lambda & 1 \\
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1 & -\lambda
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\end{bmatrix}$. To ensure that there is a non-zero solution, we solve for:
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$$
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\left| \boldsymbol{A}-\lambda\boldsymbol{E} \right| = \begin{vmatrix}
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1-\lambda & 1 \\
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1 & -\lambda
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\end{vmatrix} = \lambda^2 - \lambda - 1 = 0
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$$
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Solving this equation, we obtain $\lambda_1 = \frac{1+\sqrt{5}}{2}$ and $\lambda_2 = \frac{1-\sqrt{5}}{2}$.
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Therefore, the diagonal matrix is:
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$$
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\rm{diag}\{\lambda_1, \lambda_2\} = \begin{bmatrix}
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\frac{1+\sqrt{5}}{2} & 0 \\
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0 & \frac{1-\sqrt{5}}{2}
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\end{bmatrix}
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$$
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Assuming the eigenvector is $\boldsymbol{x} = \begin{bmatrix}
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x & y
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\end{bmatrix}^T$, we can substitute $\lambda_1$ and $\lambda_2$ into equation $(2)$ to obtain two systems of equations:
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$$
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\begin{cases}
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(1-\lambda_1)x + y = 0 \\
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x - \lambda_1y = 0
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\end{cases},
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\begin{cases}
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(1-\lambda_2)x + y = 0 \\
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x - \lambda_2y = 0
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\end{cases}
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$$
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Setting $y = 1$ in both systems of equations gives us the two eigenvectors of the matrix:
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$$
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\boldsymbol{x}_1 = \begin{bmatrix}
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\frac{1+\sqrt{5}}{2} & 1
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\end{bmatrix}^T,
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\boldsymbol{x}_2 = \begin{bmatrix}
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\frac{1-\sqrt{5}}{2} & 1
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\end{bmatrix}^T
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$$
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So, the invertible matrix $\boldsymbol{P}$ is formed by these two eigenvectors:
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$$
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\boldsymbol{P} = \begin{bmatrix}
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\frac{1+\sqrt{5}}{2} & \frac{1-\sqrt{5}}{2} \\
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1 & 1
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\end{bmatrix}
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$$
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Why is it like this? Let $\boldsymbol{P} = \begin{bmatrix}
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x_1 & x_2 \\
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y_1 & y_2
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\end{bmatrix}$ (where $x_1, y_1, x_2, y_2$ are the components of eigenvectors $\boldsymbol{x}_1, \boldsymbol{x}_2$ respectively).
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Now, if we calculate $\boldsymbol{P}\rm{diag}\{\lambda_1, \lambda_2\}$, it exactly equals $\begin{bmatrix}
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\lambda_1x_1 & \lambda_2x_2 \\
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\lambda_1y_1 & \lambda_2y_2
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\end{bmatrix}$. 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, $\boldsymbol{P} = \begin{bmatrix}
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x_1 & x_2 \\
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y_1 & y_2
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\end{bmatrix}$ is reasonable.
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Its inverse matrix is easy to calculate:
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$$
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\boldsymbol{P}^{-1} = \frac{\boldsymbol{P}^*}{\left|\boldsymbol{P}\right|} = \frac{1}{\sqrt{5}}\begin{bmatrix}
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1 & \frac{1-\sqrt{5}}{2} \\
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1 & \frac{1+\sqrt{5}}{2}
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\end{bmatrix}
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$$
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Substituting into equation $(2)$:
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$$
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\boldsymbol{A}^n = \boldsymbol{P}(\boldsymbol{P}^{-1}\boldsymbol{A}\boldsymbol{P})^n\boldsymbol{P}^{-1} = \boldsymbol{P}\rm{diag}^n\{\lambda_1, \lambda_2\}\boldsymbol{P}^{-1}
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$$
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Where the diagonal matrix is:
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$$
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\rm{diag}^n\{\lambda_1, \lambda_2\} = \begin{bmatrix}
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\left(\frac{1+\sqrt{5}}{2}\right)^n & 0 \\
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0 & \left(\frac{1-\sqrt{5}}{2}\right)^n
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\end{bmatrix}
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$$
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Therefore,
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$$
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\boldsymbol{A}^n = \frac{1}{\sqrt{5}}\begin{bmatrix}
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\left(\frac{1+\sqrt{5}}{2}\right)^{n+1} + \left(\frac{1-\sqrt{5}}{2}\right)^{n+1} & \left(\frac{1+\sqrt{5}}{2}\right)^{n+1}\left(\frac{1-\sqrt{5}}{2}\right) + \left(\frac{1-\sqrt{5}}{2}\right)^{n+1}\left(\frac{1+\sqrt{5}}{2}\right) \\
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\left(\frac{1+\sqrt{5}}{2}\right)^n + \left(\frac{1-\sqrt{5}}{2}\right)^n & \left(\frac{1+\sqrt{5}}{2}\right)^{n}\left(\frac{1-\sqrt{5}}{2}\right) + \left(\frac{1-\sqrt{5}}{2}\right)^{n}\left(\frac{1+\sqrt{5}}{2}\right)
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\end{bmatrix}
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$$
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Then,
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$$
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\begin{bmatrix}
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F_{n} \\
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F_{n-1}
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\end{bmatrix}
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= \boldsymbol{A}^{n-1}
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\begin{bmatrix}
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F_1 \\
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F_0
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\end{bmatrix}
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= \frac{1}{\sqrt{5}}\begin{bmatrix}
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\left(\frac{1+\sqrt{5}}{2}\right)^{n} + \left(\frac{1-\sqrt{5}}{2}\right)^{n} \\
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\left(\frac{1+\sqrt{5}}{2}\right)^{n-1} + \left(\frac{1-\sqrt{5}}{2}\right)^{n-1}
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\end{bmatrix}
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$$
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Considering only the first row of the matrices on both sides of the equation, we obtain the closed-form formula for the Fibonacci sequence:
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$$
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F_n = \frac{1}{\sqrt{5}}\left(\left(\frac{1+\sqrt{5}}{2}\right)^n - \left(\frac{1-\sqrt{5}}{2}\right)^n\right)
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$$
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Substituting $n=0$ and $n=1$ to verify, we find that they both satisfy the equation.
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---
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## Difference Equation Method
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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:
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$$
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\Delta^2 a_n = \Delta a_{n+1} - \Delta a_n = a_{n+2} - 2a_{n+1} + a_n
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$$
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Further, we can define the $m$-th order difference:
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$$
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\Delta^m a_n = \Delta^{m-1} a_{n+1} - \Delta^{m-1} a_n = \sum_{i=0}^{m} (-1)^i C_m^i a_{n+m-i}
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$$
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And
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$$
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F(a_n, \Delta a_n,
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\Delta a_n, \Delta^2 a_n, \Delta^3 a_n, \ldots)
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$$
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The Fibonacci sequence, defined as $F_n = F_{n-1} + F_{n-2}$, has a second-order difference that is constant:
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$$
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\Delta^2 F_n = F_{n+2} - 2F_{n+1} + F_n = F_{n+1} + F_{n+1} - 2F_{n+1} + F_n = F_{n+1} - F_n = F_n
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$$
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Now, we can write the second-order linear homogeneous difference equation for $F_n$:
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$$
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\Delta^2 F_n - F_n = 0
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$$
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This is a characteristic equation, and we can solve it by assuming $F_n = r^n$:
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$$
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r^2 - 1 = 0
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$$
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Solving this equation gives us two solutions: $r = 1$ and $r = -1$. Therefore, the general solution for the homogeneous difference equation is:
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$$
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F_n = c_1 \cdot 1^n + c_2 \cdot (-1)^n
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$$
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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:
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$$
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\begin{align*}
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F_0 &= c_1 \cdot 1^0 + c_2 \cdot (-1)^0 = c_1 + c_2 = 0 \\
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F_1 &= c_1 \cdot 1^1 + c_2 \cdot (-1)^1 = c_1 - c_2 = 1
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\end{align*}
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$$
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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:
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$$
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F_n = \frac{1}{2} \cdot 1^n - \frac{1}{2} \cdot (-1)^n
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$$
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This can be simplified further by noting that $(-1)^n$ is equal to $(-1)^{n-1} \cdot (-1) = -(-1)^{n-1}$:
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$$
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F_n = \frac{1}{2} - \frac{1}{2} \cdot (-1)^{n-1}
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$$
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This is indeed the closed-form formula for the Fibonacci sequence:
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$$
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F_n = \frac{1}{\sqrt{5}}\left(\left(\frac{1+\sqrt{5}}{2}\right)^n - \left(\frac{1-\sqrt{5}}{2}\right)^n\right)
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$$
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So, we have successfully derived the same result using the difference equation method.
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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.
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