Problem Analysis:

We analyze each part of the problem systematically to determine the answers.

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(a) Evaluating the Statements:

The logistic discrete-time dynamical system is: $x_{t+1} = \frac{9}{10} x_t (1 - x_t),$ where the updating function is $f(x) = \frac{9}{10} x (1 - x)$.

1. Per capita production: Per capita production is defined as $\frac{f(x)}{x} = \frac{9}{10}(1 - x)$ (for $x \neq 0$). This is a linear function of $x$, not a constant function. Thus, statement (i) is False.

2. Derivative of the updating function: Compute $f'(x)$: $f'(x) = \frac{d}{dx} \left( \frac{9}{10} x (1 - x) \right) = \frac{9}{10} (1 - 2x).$ The derivative is a decreasing linear function because $1 - 2x$ decreases as $x$ increases. Thus, statement (ii) is False.

Answer for Part (a):

2,2

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(b) Non-Zero Equilibrium:

To find equilibria, solve $x_{t+1} = x_t$, i.e., $f(x) = x$: $\frac{9}{10} x (1 - x) = x.$ Factorizing: $\frac{9}{10} x - \frac{9}{10} x^2 = x \implies \frac{9}{10} x - x - \frac{9}{10} x^2 = 0 \implies x \left( -\frac{1}{10} - \frac{9}{10} x \right) = 0.$ The solutions are: $x = 0 \quad \text{and} \quad x = -\frac{1}{9/10} = \frac{10}{9}.$ Since $x$ represents a fraction of a population (non-negative), only $x = \frac{10}{9}$ is relevant.

Answer for Part (b):

Non-zero equilibrium: $x = \frac{10}{9}$.

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(c) Stability of Equilibria:

To determine the stability, compute $|f'(x)|$ at each equilibrium point.

1. At $x = 0$: $f'(x) = \frac{9}{10} (1 - 2x) \implies f'(0) = \frac{9}{10}.$ Since $|f'(0)| < 1$, $x = 0$ is a stable equilibrium.

2. At $x = \frac{10}{9}$: $f'(x) = \frac{9}{10} (1 - 2x) \implies f'\left(\frac{10}{9}\right) = \frac{9}{10} \left(1 - 2 \cdot \frac{10}{9}\right) = \frac{9}{10} \left(1 - \frac{20}{9}\right) = \frac{9}{10} \cdot \left(-\frac{11}{9}\right) = -\frac{99}{90}.$ Since $|f'(\frac{10}{9})| = \frac{99}{90} > 1$, $x = \frac{10}{9}$ is an unstable equilibrium.

Answer for Part (c):

(H) The non-zero equilibrium is unstable; 0 is a stable equilibrium.

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Final Answers:

1. (a): $2,2$
2. (b): $x = \frac{10}{9}$
3. (c): $\text{H: The non-zero equilibrium is unstable; 0 is a stable equilibrium.}$

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Would you like further explanation on any part of the analysis or derivation?

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Related Questions:

1. How does the derivative of a logistic function affect equilibrium stability?
2. What happens to the population $x_t$ if $x_0 > \frac{10}{9}$?
3. What is the biological interpretation of stability in this model?
4. Can a logistic map with a different growth factor ($r$) have multiple stable equilibria?
5. How is the concept of per capita growth applied in real-world models?

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Tip:

When analyzing stability, always compute $|f'(x)|$ at the equilibrium points to determine whether they are stable ($|f'(x)| < 1$) or unstable ($|f'(x)| > 1$).