text: In this post we solve the Fibonacci sequence using linear algebra. The Fibonacci equation is a second-order difference equation which is a particular type of sequence.
  
  ## Sequences
  
  A sequence is a (possibly infinite) set of numbers with a defined order.
  
  $a_n = \frac{1}{n}, \textit{ for } n = 0, 1, 2, ...,$
  
  ### Difference Equations
  
  A difference equation is a type of sequence. A difference equation expresses the value of the sequence relative to the value of another point in the sequence.
  
  $x_{n+1} = 1.5x_n, \textit{ for } n = 0, 1, 2, ...,$
  
  To evaluate this equation at $n=3$ we need to know the initial value $x_0$, and evaluate $n=1, 2, 3$ in succession. For example, if $x_0=100$:
  
  \begin{align} x_0 & && = 100 \\ x_1 & = 1.5 \times 100 && = 150\\ x_2 & = 1.5 \times 150 && = 225\\ x_3 & = 1.5 \times 225 && = 337.5 \end{align}
  
  In the example above, we evaluate the equation recursively. It would be better to be able to evaluate the equation at $n = 100$ without performing $100$ calculations. Note that we can also write the equation above in this form:
  
  $x_3 = 100 \times 1.5 \times 1.5 \times 1.5 = 337.5$
  
  Solving a difference equation means writing the same equation such that it can be evaluated at a given $n$ without first evaluating $n=1,2,etc$. The solution to the equation above is:
  
  $x_{n} = 1.5^{n} \times x_0, \textit{ for } n = 0, 1, 2, ...,$
  
  ### Second Order Difference Equations
  
  The previous example is a first order difference equation because it expresses the value of $x_n$ as a function of one other sample, $x_{n-1}$. A second order difference equation is a function of two other samples, for example
  
  $x_{n+2} = 1.5x_n + 1.4x_{n+1}, \textit{ for } n = 0, 1, 2, ...,$
  
  ## Fibonacci Numbers
  
  The easiest way to describe the Fibonacci sequence is with a second order difference equation. Assume $F_0 = 0$ and $F_1 = 1$
  
  $F_{k+2} = F_{k+1} + F_{k}$
  
  The first eight Fibonacci numbers are:
  
  $F_k=0,1,1,2,3,5,8,13,...$
  
  We can solve the sequence with linear algebra. Let us start our sequence with a vector, $u_0=\begin{bmatrix}1\\0\end{bmatrix}$. Find the next vector in our sequence, $u_1$ like this:
  
  $u_{1} = \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} 1 \\ 0 \end{bmatrix} = \begin{bmatrix} 1 \\ 1 \end{bmatrix}$
  
  More generally:
  
  $u_{k+1} = \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix} u_k$
  
  Note that $u_0$ contains $\begin{bmatrix}F_1\\F_0\end{bmatrix}$.
  
  The key to solving the sequence is in the eigenvalues and eigenvectors of $\begin{bmatrix}1&1\\1&0\end{bmatrix}$. We can write $u_3$ as
  
  $u_3 = \begin{bmatrix}1 & 1 \\ 1 & 0\end{bmatrix} \begin{bmatrix}1 & 1 \\ 1 & 0\end{bmatrix} \begin{bmatrix}1 & 1 \\ 1 & 0\end{bmatrix} \begin{bmatrix}1 \\ 0\end{bmatrix} = \begin{bmatrix}3 \\ 2\end{bmatrix} = \begin{bmatrix}F_4 \\ F_3\end{bmatrix}$
  
  import numpy as np
  fib = np.matrix([[1,1],[1,0]])
  u_0 = np.matrix([[1],[0]])
  
  fib * fib * fib * u_0 # The * operator is dot product for np.matrix (but not np.array)
  matrix([[3],
  [2]])
  
  ### Eigenvector Solution
  
  The general case of the previous example is:
  
  \begin{align} u_k &= \begin{bmatrix}1 & 1 \\ 1 & 0\end{bmatrix}^k\begin{bmatrix}1 \\ 0\end{bmatrix}\\ u_k &= A^ku_0 \end{align}
  
  We can use eigenvectors and eigenvalues to solve the matrix power (see Eigenvalues and Eigenvectors). First, find the eigenvalues of $A$:
  
  \begin{align} det\begin{vmatrix} 1 - \lambda & 1\\ 1 & - \lambda \end{vmatrix} & = 0\\ -\lambda(1-\lambda) - 1 & = 0\\ \lambda{}^2 - \lambda - 1 & = 0 \end{align}
  
  Solve using the quadratic formula $\frac{-b \pm \sqrt{b^2-4ac}}{2a} \implies \frac{1 \pm \sqrt{5}}{2} \approx \frac{1 \pm 2.236}{2}\\$, which gives the eigenvalues for our matrix: \begin{align} \lambda_1 & \approx &1.618 \\ \lambda_2 & \approx &-0.618 \end{align}
  
  Now we plug our constant eigenvalues ($\lambda_1$ and $\lambda_2$) into $(A-\lambda I)\vec{x}=0$ to find the eigen vectors.
  
  $\begin{bmatrix}1-\lambda_1 & 1 \\ 1 & -\lambda_1 \end{bmatrix}\vec{x}_1 = 0 \implies \vec{x}_1 = \begin{bmatrix}\lambda_1 \\ 1\end{bmatrix}$
  
  Note that this solution is not necessarily intuitive. This is not a valid solution for any arbitrary value of $\lambda$: our values, $\lambda_1$ and $\lambda_2$ are valid solutions. We can use the same process to solve for $\lambda_2$, revealing eigenvectors $\begin{bmatrix}\lambda_1 \\1\end{bmatrix}$ and $\begin{bmatrix}\lambda_2 \\1\end{bmatrix}$. Now create the eigenvector matrix, $S$ and eigenvalue matrix, $\Lambda$.
  
  \begin{align} S &= \begin{bmatrix}\lambda_1 & \lambda_2 \\ 1 & 1\end{bmatrix}\\ \Lambda &= \begin{bmatrix}\lambda_1 & 0 \\ 0 & \lambda_2\end{bmatrix} \end{align}
  
  Remeber that $A^k = S^{-1}\Lambda^kS$. To proceed, invert $S$. Remember this shortcut for inverting a $2 \times 2$ matrix.
  
  \begin{align} \begin{bmatrix} a & b \\ c & d \end{bmatrix}^{-1} =&& \frac{1}{ad-cb} &\begin{bmatrix} d & -b \\ -c & a \end{bmatrix} \\ S^{-1} =&& &\begin{bmatrix}\lambda_1 & \lambda_2 \\ 1 & 1\end{bmatrix}^{-1} \\ S^{-1} =&& \frac{1}{\lambda_1-\lambda_2} &\begin{bmatrix}1 & -\lambda_2\\ -1 & \lambda_1 \end{bmatrix}\\ S^{-1} =&& \tfrac{1}{\sqrt{5}} &\begin{bmatrix}1 & -\lambda_2\\ -1 & \lambda_1 \end{bmatrix} \end{align}
  
  Using the components $(A, \Lambda, S, S^{-1},u_0)$:
  
  \begin{align} u_k &= A^ku_0 \\ u_k &= S\Lambda^kS^{-1}u_o \\ u_k &= S\Lambda^k\frac{1}{\sqrt{5}}\begin{bmatrix}1 & -\lambda_2\\ -1 & \lambda_1 \end{bmatrix}u_0 \\ u_k &= \frac{1}{\sqrt{5}}S\Lambda^k\begin{bmatrix}1 & -\lambda_2\\ -1 & \lambda_1 \end{bmatrix}u_0 \\ u_k &= \tfrac{1}{\sqrt{5}} \begin{bmatrix}\lambda_1 & \lambda_2 \\ 1 & 1\end{bmatrix} \begin{bmatrix}\lambda_1 & 0 \\ 0 & \lambda_2\end{bmatrix}^k \begin{bmatrix}1 & -\lambda_2\\ -1 & \lambda_1 \end{bmatrix} u_0 \\ u_k &= \tfrac{1}{\sqrt{5}} \begin{bmatrix}\lambda_1 & \lambda_2 \\ 1 & 1\end{bmatrix} \begin{bmatrix}\lambda_1^k & 0 \\ 0 & \lambda_2^k\end{bmatrix} \begin{bmatrix}1 & -\lambda_2\\ -1 & \lambda_1 \end{bmatrix} \begin{bmatrix}1\\0\end{bmatrix} \\ u_k &= \tfrac{1}{\sqrt{5}} \begin{bmatrix}\lambda_1 & \lambda_2 \\ 1 & 1\end{bmatrix} \begin{bmatrix}\lambda_1^k & 0 \\ 0 & \lambda_2^k\end{bmatrix} \begin{bmatrix}1 \\ -1 \end{bmatrix}\\ u_k &= \tfrac{1}{\sqrt{5}} \begin{bmatrix}\lambda_1 & \lambda_2 \\ 1 & 1\end{bmatrix} \begin{bmatrix}\lambda_1^k \\ -\lambda_2^k\end{bmatrix} \end{align}
  
  ## Verify Solution in Python
  
  import numpy as np
  
  lmbd_1 = (1+np.sqrt(5))/2
  lmbd_2 = (1-np.sqrt(5))/2
  
  a       = np.matrix([[1,      1],      [1,  0]])
  s       = np.matrix([[lmbd_1, lmbd_2], [1,  1]])
  s_inv   = np.matrix([[1,     -lmbd_2], [-1, lmbd_1]]) * (1/np.sqrt(5))
  eig_val = np.matrix([[lmbd_1, 0],      [0,  lmbd_2]])
  u_0     = np.matrix([[1],              [0]])
  
  def eig_k(k):
  return eig_val ** k
  
  def u_k(k):
  return s * eig_k(k) * s_inv * u_0
  
  def fib(n):
  f_n = u_k(n)[1,0]
  return int(round(f_n))
  
  [fib(n) for n in range(15)]
  [0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377]
explanation: The document provides a detailed explanation of solving the Fibonacci sequence using linear algebra. It covers the basics of sequences, difference equations, and the specific case of the Fibonacci sequence. The text includes the derivation of the eigenvalues and eigenvectors of the matrix used to solve the Fibonacci sequence, and it also includes a Python code snippet to verify the solution.
  
  The document does not contain a specific question or a question and answer. It is primarily an explanatory text.
distilabel_metadata.content: Evaluate the user given text extract and determine if it contains a math question and optionally and answer to this question.
  
  Output rules:
  
  1. Briefly describe the document and mention if it contains a question or a question and answer.
  2. Finalize the output with: Label: no-question,question,question-and-answer,multiple-question,multiple-question-and-answer
distilabel_metadata.role: system
distilabel_metadata.content: In this post we solve the Fibonacci sequence using linear algebra. The Fibonacci equation is a second-order difference equation which is a particular type of sequence.
  
  ## Sequences
  
  A sequence is a (possibly infinite) set of numbers with a defined order.
  
  $a_n = \frac{1}{n}, \textit{ for } n = 0, 1, 2, ...,$
  
  ### Difference Equations
  
  A difference equation is a type of sequence. A difference equation expresses the value of the sequence relative to the value of another point in the sequence.
  
  $x_{n+1} = 1.5x_n, \textit{ for } n = 0, 1, 2, ...,$
  
  To evaluate this equation at $n=3$ we need to know the initial value $x_0$, and evaluate $n=1, 2, 3$ in succession. For example, if $x_0=100$:
  
  \begin{align} x_0 & && = 100 \\ x_1 & = 1.5 \times 100 && = 150\\ x_2 & = 1.5 \times 150 && = 225\\ x_3 & = 1.5 \times 225 && = 337.5 \end{align}
  
  In the example above, we evaluate the equation recursively. It would be better to be able to evaluate the equation at $n = 100$ without performing $100$ calculations. Note that we can also write the equation above in this form:
  
  $x_3 = 100 \times 1.5 \times 1.5 \times 1.5 = 337.5$
  
  Solving a difference equation means writing the same equation such that it can be evaluated at a given $n$ without first evaluating $n=1,2,etc$. The solution to the equation above is:
  
  $x_{n} = 1.5^{n} \times x_0, \textit{ for } n = 0, 1, 2, ...,$
  
  ### Second Order Difference Equations
  
  The previous example is a first order difference equation because it expresses the value of $x_n$ as a function of one other sample, $x_{n-1}$. A second order difference equation is a function of two other samples, for example
  
  $x_{n+2} = 1.5x_n + 1.4x_{n+1}, \textit{ for } n = 0, 1, 2, ...,$
  
  ## Fibonacci Numbers
  
  The easiest way to describe the Fibonacci sequence is with a second order difference equation. Assume $F_0 = 0$ and $F_1 = 1$
  
  $F_{k+2} = F_{k+1} + F_{k}$
  
  The first eight Fibonacci numbers are:
  
  $F_k=0,1,1,2,3,5,8,13,...$
  
  We can solve the sequence with linear algebra. Let us start our sequence with a vector, $u_0=\begin{bmatrix}1\\0\end{bmatrix}$. Find the next vector in our sequence, $u_1$ like this:
  
  $u_{1} = \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} 1 \\ 0 \end{bmatrix} = \begin{bmatrix} 1 \\ 1 \end{bmatrix}$
  
  More generally:
  
  $u_{k+1} = \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix} u_k$
  
  Note that $u_0$ contains $\begin{bmatrix}F_1\\F_0\end{bmatrix}$.
  
  The key to solving the sequence is in the eigenvalues and eigenvectors of $\begin{bmatrix}1&1\\1&0\end{bmatrix}$. We can write $u_3$ as
  
  $u_3 = \begin{bmatrix}1 & 1 \\ 1 & 0\end{bmatrix} \begin{bmatrix}1 & 1 \\ 1 & 0\end{bmatrix} \begin{bmatrix}1 & 1 \\ 1 & 0\end{bmatrix} \begin{bmatrix}1 \\ 0\end{bmatrix} = \begin{bmatrix}3 \\ 2\end{bmatrix} = \begin{bmatrix}F_4 \\ F_3\end{bmatrix}$
  
  import numpy as np
  fib = np.matrix([[1,1],[1,0]])
  u_0 = np.matrix([[1],[0]])
  
  fib * fib * fib * u_0 # The * operator is dot product for np.matrix (but not np.array)
  matrix([[3],
  [2]])
  
  ### Eigenvector Solution
  
  The general case of the previous example is:
  
  \begin{align} u_k &= \begin{bmatrix}1 & 1 \\ 1 & 0\end{bmatrix}^k\begin{bmatrix}1 \\ 0\end{bmatrix}\\ u_k &= A^ku_0 \end{align}
  
  We can use eigenvectors and eigenvalues to solve the matrix power (see Eigenvalues and Eigenvectors). First, find the eigenvalues of $A$:
  
  \begin{align} det\begin{vmatrix} 1 - \lambda & 1\\ 1 & - \lambda \end{vmatrix} & = 0\\ -\lambda(1-\lambda) - 1 & = 0\\ \lambda{}^2 - \lambda - 1 & = 0 \end{align}
  
  Solve using the quadratic formula $\frac{-b \pm \sqrt{b^2-4ac}}{2a} \implies \frac{1 \pm \sqrt{5}}{2} \approx \frac{1 \pm 2.236}{2}\\$, which gives the eigenvalues for our matrix: \begin{align} \lambda_1 & \approx &1.618 \\ \lambda_2 & \approx &-0.618 \end{align}
  
  Now we plug our constant eigenvalues ($\lambda_1$ and $\lambda_2$) into $(A-\lambda I)\vec{x}=0$ to find the eigen vectors.
  
  $\begin{bmatrix}1-\lambda_1 & 1 \\ 1 & -\lambda_1 \end{bmatrix}\vec{x}_1 = 0 \implies \vec{x}_1 = \begin{bmatrix}\lambda_1 \\ 1\end{bmatrix}$
  
  Note that this solution is not necessarily intuitive. This is not a valid solution for any arbitrary value of $\lambda$: our values, $\lambda_1$ and $\lambda_2$ are valid solutions. We can use the same process to solve for $\lambda_2$, revealing eigenvectors $\begin{bmatrix}\lambda_1 \\1\end{bmatrix}$ and $\begin{bmatrix}\lambda_2 \\1\end{bmatrix}$. Now create the eigenvector matrix, $S$ and eigenvalue matrix, $\Lambda$.
  
  \begin{align} S &= \begin{bmatrix}\lambda_1 & \lambda_2 \\ 1 & 1\end{bmatrix}\\ \Lambda &= \begin{bmatrix}\lambda_1 & 0 \\ 0 & \lambda_2\end{bmatrix} \end{align}
  
  Remeber that $A^k = S^{-1}\Lambda^kS$. To proceed, invert $S$. Remember this shortcut for inverting a $2 \times 2$ matrix.
  
  \begin{align} \begin{bmatrix} a & b \\ c & d \end{bmatrix}^{-1} =&& \frac{1}{ad-cb} &\begin{bmatrix} d & -b \\ -c & a \end{bmatrix} \\ S^{-1} =&& &\begin{bmatrix}\lambda_1 & \lambda_2 \\ 1 & 1\end{bmatrix}^{-1} \\ S^{-1} =&& \frac{1}{\lambda_1-\lambda_2} &\begin{bmatrix}1 & -\lambda_2\\ -1 & \lambda_1 \end{bmatrix}\\ S^{-1} =&& \tfrac{1}{\sqrt{5}} &\begin{bmatrix}1 & -\lambda_2\\ -1 & \lambda_1 \end{bmatrix} \end{align}
  
  Using the components $(A, \Lambda, S, S^{-1},u_0)$:
  
  \begin{align} u_k &= A^ku_0 \\ u_k &= S\Lambda^kS^{-1}u_o \\ u_k &= S\Lambda^k\frac{1}{\sqrt{5}}\begin{bmatrix}1 & -\lambda_2\\ -1 & \lambda_1 \end{bmatrix}u_0 \\ u_k &= \frac{1}{\sqrt{5}}S\Lambda^k\begin{bmatrix}1 & -\lambda_2\\ -1 & \lambda_1 \end{bmatrix}u_0 \\ u_k &= \tfrac{1}{\sqrt{5}} \begin{bmatrix}\lambda_1 & \lambda_2 \\ 1 & 1\end{bmatrix} \begin{bmatrix}\lambda_1 & 0 \\ 0 & \lambda_2\end{bmatrix}^k \begin{bmatrix}1 & -\lambda_2\\ -1 & \lambda_1 \end{bmatrix} u_0 \\ u_k &= \tfrac{1}{\sqrt{5}} \begin{bmatrix}\lambda_1 & \lambda_2 \\ 1 & 1\end{bmatrix} \begin{bmatrix}\lambda_1^k & 0 \\ 0 & \lambda_2^k\end{bmatrix} \begin{bmatrix}1 & -\lambda_2\\ -1 & \lambda_1 \end{bmatrix} \begin{bmatrix}1\\0\end{bmatrix} \\ u_k &= \tfrac{1}{\sqrt{5}} \begin{bmatrix}\lambda_1 & \lambda_2 \\ 1 & 1\end{bmatrix} \begin{bmatrix}\lambda_1^k & 0 \\ 0 & \lambda_2^k\end{bmatrix} \begin{bmatrix}1 \\ -1 \end{bmatrix}\\ u_k &= \tfrac{1}{\sqrt{5}} \begin{bmatrix}\lambda_1 & \lambda_2 \\ 1 & 1\end{bmatrix} \begin{bmatrix}\lambda_1^k \\ -\lambda_2^k\end{bmatrix} \end{align}
  
  ## Verify Solution in Python
  
  import numpy as np
  
  lmbd_1 = (1+np.sqrt(5))/2
  lmbd_2 = (1-np.sqrt(5))/2
  
  a       = np.matrix([[1,      1],      [1,  0]])
  s       = np.matrix([[lmbd_1, lmbd_2], [1,  1]])
  s_inv   = np.matrix([[1,     -lmbd_2], [-1, lmbd_1]]) * (1/np.sqrt(5))
  eig_val = np.matrix([[lmbd_1, 0],      [0,  lmbd_2]])
  u_0     = np.matrix([[1],              [0]])
  
  def eig_k(k):
  return eig_val ** k
  
  def u_k(k):
  return s * eig_k(k) * s_inv * u_0
  
  def fib(n):
  f_n = u_k(n)[1,0]
  return int(round(f_n))
  
  [fib(n) for n in range(15)]
  [0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377]
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The document provides a detailed explanation of solving the Fibonacci sequence using linear algebra. It covers the basics of sequences, difference equations, and the specific case of the Fibonacci sequence. The text includes the derivation of the eigenvalues and eigenvectors of the matrix used to solve the Fibonacci sequence, and it also includes a Python code snippet to verify the solution.
  
  The document does not contain a specific question or a question and answer. It is primarily an explanatory text.
  
  Label: no-question