Study Guide: Discrete Dynamical Systems, Systems of Equations, and Matrices

Discrete Dynamical Systems (DDS)

Introduction to DDS and Linear DDS

Discrete Dynamical Systems (DDS) are mathematical models that describe how a quantity evolves in discrete steps, often using recurrence relations. Linear DDS are a special class where the updating function is linear in the previous term.

• Explicit Sequence: A sequence where the nth term is given directly as a function of n, e.g., $x_n = 7(1.5)^n$.

• Recursive Sequence: A sequence where each term is defined in terms of previous terms, e.g., $x_n = 1.5x_{n-1}$.

• General Linear DDS: $x_n = f(x_{n-1}) = a x_{n-1} + b$

• Equilibrium: A value $x_e$ such that $x_e = a x_e + b$. Solving gives $x_e = \frac{b}{1-a}$ (if $a \neq 1$).

• Stability: The equilibrium is stable if $|a| < 1$, unstable if $|a| > 1$, and oscillates if $a = -1$.

Table: Equilibrium and Stability for Linear DDS

Case	Equilibrium	Stability
$a \neq 1$	$x_e = \frac{b}{1-a}$	Stable if $|a| < 1$
$a = 1, b \neq 0$	None	–
$a = 1, b = 0$	Infinitely many	–

• Explicit Solution (for $a \neq 1$): $x_n = a^n x_0 + \frac{b}{1-a}(1 - a^n)$

• Equilibrium Form: $x_n = a^n(x_0 - x_e) + x_e$

Malthusian Growth Model

• Basic Model: $P_n = (1 + r)P_{n-1}$, where $r$ is the growth rate.

• With Immigration/Emigration: $P_n = (1 + r)P_{n-1} + \mu$

• Both are linear DDS.

Non-linear DDS and the Logistic Model

Non-linear DDS have updating functions that are not linear, often leading to more complex behavior such as multiple equilibria.

• Logistic Model (Additive Form): $p_n = p_{n-1} + r p_{n-1} \left(1 - \frac{p_{n-1}}{N}\right)$

• Logistic Model (Quadratic Form): $p_n = (1 + r)p_{n-1} - \frac{r}{N}p_{n-1}^2$

• Parameters: $r$ (growth rate), $N$ (carrying capacity)

• Equilibria: $p_e = 0$ (trivial, unstable), $p_e = N$ (non-trivial, stable)

Form	Equilibrium	Stability
Additive	$p_e = 0$ (unstable), $p_e = N$ (stable)	As shown
Quadratic	$p_e = 0$ (unstable), $p_e = \frac{b-1}{a} = N$ (stable)	As shown

Systems of Equations and Gauss-Jordan Method

Solving Systems of Linear Equations

A system of equations consists of multiple equations with several variables. Solutions can be found using substitution, elimination, or matrix methods.

• Solution Form: $(x_1, x_2, ..., x_n)$

• Augmented Matrix: Represents the system in matrix form, with the last column for constants.

• Gauss-Jordan Method: Uses elementary row operations to reduce the matrix to Reduced Row Echelon Form (RREF).

• Elementary Row Operations:

  • Swap two rows

  • Multiply a row by a non-zero constant

  • Add a multiple of one row to another

• Types of Solutions:

  • Unique Solution: Identity matrix on the left; solution is the last column.

  • No Solution: Row of zeros with a non-zero constant in the last column.

  • Infinitely Many Solutions: Row of all zeros; corresponding variable is free.

Matrices

Matrix Basics and Operations

Matrices are rectangular arrays of numbers, used to represent systems of equations and perform linear transformations.

• Notation: Capital letters (e.g., A, B) denote matrices.

• Dimensions: $m \times n$ matrix has $m$ rows and $n$ columns.

• Square Matrix: $m = n$

Addition and Subtraction

• Possible only if matrices have the same dimensions.

• Add corresponding entries.

• Example:

  • $\begin{bmatrix}1 & 2 \\ 3 & 4\end{bmatrix} + \begin{bmatrix}5 & 6 \\ 7 & 8\end{bmatrix} = \begin{bmatrix}6 & 8 \\ 10 & 12\end{bmatrix}$

Multiplication

• Possible only if the number of columns in the first matrix equals the number of rows in the second.

• Resulting matrix has dimensions (rows of first) × (columns of second).

• Matrix multiplication is not commutative: $AB \neq BA$ in general.

• Example:

  • Multiply $\begin{bmatrix}0 & 1 \\ 2 & 3 \\ 4 & 5\end{bmatrix}$ (3×2) by $\begin{bmatrix}6 & 7 & 8 \\ 9 & 10 & 11\end{bmatrix}$ (2×3):

  • Result: $\begin{bmatrix}9 & 10 & 11 \\ 39 & 44 & 49 \\ 69 & 78 & 87\end{bmatrix}$ (3×3)

Applications and Example Problems

Classifying Sequences

• Explicit: $x_n = 7(1.5)^n$, $x_n = 7n + 8$

• Recursive: $x_n = 1.5x_{n-1}$, $x_n = \frac{x_{n-1}}{2} + 8$

Classifying DDS Equations

Equation	Type
$P_n = 1.3P_{n-1}$	Recursive and linear
$P_n = P_{n-1}^2 + 7$	Recursive and non-linear
$P_n = 1.3P_{n-1} + 7$	Recursive and linear
$P_n = 5n + 7$	Explicit
$P_n = 5(1.3)^n$	Explicit

Finding Equilibrium and Stability

• For $P_{n+1} = 0.8P_n + 20$:

  • Linear DDS: $a = 0.8$, $b = 20$

  • Equilibrium: $P_e = \frac{20}{1-0.8} = 100$

  • Stability: $|a| = 0.8 < 1$ (stable)

  • Explicit formula: $P_n = 0.8^n(P_0 - 100) + 100$

• For $P_{n+1} = -1.5P_n + 25$:

  • Linear DDS: $a = -1.5$, $b = 25$

  • Equilibrium: $P_e = \frac{25}{1 - (-1.5)} = 10$

  • Stability: $|a| = 1.5 > 1$ (unstable)

  • Explicit formula: $P_n = (-1.5)^n(P_0 - 10) + 10$

• For $P_{n+1} = P_n^2$:

  • Non-linear DDS (quadratic term)

  • Equilibria: $P_e = 0$ and $P_e = 1$ (since $P_e = P_e^2$)

  • Stability: $P_e = 0$ is stable for $|P_0| < 1$, $P_e = 1$ is unstable

Logistic Model Example

• Given $p_n = 2.7p_{n-1} - 0.0001p_{n-1}^2$:

• Quadratic form: $b = 2.7$, $a = 0.0001$

• Growth rate: $r = b - 1 = 1.7$

• Carrying capacity: $N = \frac{b-1}{a} = \frac{1.7}{0.0001} = 17000$

• Equilibria: $p_e = 0$ (unstable), $p_e = 17000$ (stable)

Writing Systems from Augmented Matrices

• Given $\begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix} | \begin{bmatrix}14 \\ -2 \\ 13\end{bmatrix}$, the system is:

  • $x_1 = 14$

  • $x_2 = -2$

  • $x_3 = 13$

• Given $\begin{bmatrix}3 & 0 & 2 \\ 3 & 1 & 0 \\ 5 & 0 & -3\end{bmatrix} | \begin{bmatrix}37 \\ 40 \\ 19\end{bmatrix}$, the system is:

  • $3x_1 + 0x_2 + 2x_3 = 37$

  • $3x_1 + x_2 + 0x_3 = 40$

  • $5x_1 + 0x_2 - 3x_3 = 19$

General and Specific Solutions with Free Variables

• Given $\begin{bmatrix}1 & 0 & 13 \\ 0 & 1 & 3 \\ 0 & 0 & 0\end{bmatrix} | \begin{bmatrix}39 \\ -15 \\ 0\end{bmatrix}$, the general solution is:

  • $x = 39 - 13z$

  • $y = -15 - 3z$

  • $z$ is free

  • General solution: $(x, y, z) = (39 - 13z, -15 - 3z, z)$

  • For $z = 1$: $(x, y, z) = (26, -18, 1)$

Modeling Resource Constraints with Systems

Given the following table of resources needed per acre for apples (a), broccoli (b), and carrots (c):

	Apples (a)	Broccoli (b)	Carrots (c)
Nitrogen (lb/acre)	120	150	180
Phosphate (lb/acre)	180	80	80
Labor (hr/acre)	4.97	4.45	4.65

• Constraints:

  • $120a + 150b + 180c = 21800$ (nitrogen)

  • $180a + 80b + 80c = 29600$ (phosphate)

  • $4.97a + 4.45b + 4.65c = 550$ (labor)

  • $a + b + c = 160$ (total acres)

Matrix Dimensions and Operations

• Given:

  • $A = \begin{bmatrix}1 & 2 \\ 3 & 4\end{bmatrix}$ (2×2)

  • $B = \begin{bmatrix}0 & 1 \\ 0 & 1\end{bmatrix}$ (2×2)

  • $C = \begin{bmatrix}1 & 0 & 1 \\ 0 & 1 & 0\end{bmatrix}$ (2×3)

  • $D = \begin{bmatrix}2 & 1 \\ -1 & 0 \\ 3 & 4\end{bmatrix}$ (3×2)

• Matrix operations:

  • $A + B$: Possible (both 2×2), result is 2×2

  • $B + C$: Not possible (dimensions differ)

  • $AB$: Possible (2×2 × 2×2), result is 2×2

  • $BA$: Possible (2×2 × 2×2), result is 2×2

  • $AD$: Not possible (2×2 × 3×2)

  • $DA$: Possible (3×2 × 2×2), result is 3×2

Additional info: For matrix multiplication, the inner dimensions must match; the result has the outer dimensions.