Math Problem Statement

Consider the following logistic discrete-time dynamical system: x_(t+1) = (39/10) * x_t * (1 - x_t) (a) Analyze the following statements: (i) The per capita production of this system is a linear function. (ii) The derivative of the updating function is a decreasing function. (b) Determine the non-zero equilibrium of this system. (c) Which statement best describes the stability of the equilibria for this system?

Solution

Let's analyze and solve the given problem step-by-step:

The logistic discrete-time dynamical system is defined as:

xt+1=3910xt(1xt)x_{t+1} = \frac{39}{10} x_t (1 - x_t)

Part (a): Statements

  1. The per capita production of this system is a linear function:

    • The per capita production refers to the growth rate per individual, which is proportional to 1xt1 - x_t. This term is linear in xtx_t, so this statement is true.
  2. The derivative of the updating function is a decreasing function:

    • The updating function is f(xt)=3910xt(1xt)f(x_t) = \frac{39}{10} x_t (1 - x_t). Its derivative is: f(xt)=3910(12xt)f'(x_t) = \frac{39}{10} (1 - 2x_t)
      • f(xt)f'(x_t) is a linear function of xtx_t, not necessarily a decreasing function for all xtx_t. Therefore, this statement is false.

Answer: 1, 2


Part (b): Non-zero equilibrium

To find the equilibrium, solve: xt+1=xt=x(equilibrium point)x_{t+1} = x_t = x \quad \text{(equilibrium point)} x=3910x(1x)x = \frac{39}{10} x (1 - x) Simplify: x(13910(1x))=0x \left(1 - \frac{39}{10} (1 - x)\right) = 0 Either x=0x = 0 or: 13910+3910x=01 - \frac{39}{10} + \frac{39}{10}x = 0 3910x=2910\frac{39}{10}x = \frac{29}{10} x=2939x = \frac{29}{39}

Non-zero equilibrium: x=2939\mathbf{x = \frac{29}{39}}.


Part (c): Stability

To assess stability, calculate f(x)f'(x) at the equilibria: f(x)=3910(12x)f'(x) = \frac{39}{10} (1 - 2x) At x=0x = 0: f(0)=3910(12(0))=3910>1f'(0) = \frac{39}{10} (1 - 2(0)) = \frac{39}{10} > 1 x=0x = 0 is unstable.

At x=2939x = \frac{29}{39}: f(2939)=3910(122939)f'\left(\frac{29}{39}\right) = \frac{39}{10} \left(1 - 2 \cdot \frac{29}{39}\right) f(2939)=3910(15839)f'\left(\frac{29}{39}\right) = \frac{39}{10} \left(1 - \frac{58}{39}\right) f(2939)=3910(1939)f'\left(\frac{29}{39}\right) = \frac{39}{10} \cdot \left(-\frac{19}{39}\right) f(2939)=7413901.9f'\left(\frac{29}{39}\right) = -\frac{741}{390} \approx -1.9 1.9-1.9 lies outside the range (1,1)(-1, 1), so x=2939x = \frac{29}{39} is also unstable.


Let me know if you would like further clarification or steps! Here are some related questions to expand your understanding:

  1. How does the updating function change if the coefficient 3910\frac{39}{10} is modified?
  2. What happens to stability when the derivative f(x)f'(x) is adjusted to fall within the interval (1,1)(-1, 1)?
  3. How is the equilibrium x=0x = 0 interpreted biologically in logistic growth models?
  4. Can there be a scenario where both equilibria are stable in logistic systems?
  5. What are the implications of stability analysis in predicting long-term population behavior?

Tip: To verify stability, always check the magnitude of the derivative of the updating function at equilibrium points. Stability occurs when f(x)<1|f'(x)| < 1.

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Math Problem Analysis

Mathematical Concepts

Discrete-Time Dynamical Systems
Logistic Growth
Stability Analysis

Formulas

x_(t+1) = r * x_t * (1 - x_t)
f'(x_t) = r * (1 - 2x_t)

Theorems

Equilibrium Analysis
Stability Condition: |f'(x)| < 1

Suitable Grade Level

Undergraduate (Calculus/Dynamical Systems Course)