1. Consider the following discrete-time dynamical system: xt+1 = x2t + 7xt + 8 (

Question

1. Consider the following discrete-time dynamical system:

xt+1 = x2t + 7xt + 8 (a) Find the equilibria algebraically.

(b) Apply the Stability Test/Slope Criterion to each of the equilibria you found in (a). What can you conclude?

(c) The updating function is graphed below, along with the diagonal. Cobweb for at least 3 steps starting from the initial condition x0 = 3.5. Describe in detail the long-term behavior of this solution.

1. Consider the following discrete-time dynamical system: at+1 (a) Find the equilibria algebraically. (b) Apply the Stability Test/Slope Criterion to each of the equilibria you found in (a). What can you conclude? (c) The updating function is graphed below, along with the diagonal. Cobweb for at least 3 steps starting from the initial condition ro 3.5. Describe in detail the long-term behavior of this solution. f(z)

Explanation / Answer

The given discrete-time dynamical system is:

x(t+1) = x^2(t) + 7x(t) + 8

a) Finding equilibria :

We must know that an equilibrium is the simplest possible solution to a dynamical system where the state variable is a constant value (means, the variable doesn't change with time)

Now here we need to find a solution x(t) that doesn't depend on time t.

Let x(t) = E is an equilibrium solution, then x(t) = E for all values of t, means x(1) = E, x(2) = E and x(10) = E and so on.

It implies further that x(t+1) = E whenever x(t) = E.

So now plugging these values in our discrete-time dynamical system, we get

E = E^2 + 7E + 8   =>   E^2 + 6E + 8 = 0

Upon solving this quadratic equation, we get => E^2 + 4E + 2E + 8 = 0

(E + 4)(E + 2) = 0

So,the solutions are E = -4 & -2 implies that the equilibria are E = -4 and E = -2 and the equilibria are the solutions x(t) = -4 and x(t) = -2 to the dynamical system.

b)

We will apply for this, the derivative criterion for the test of stability of equilibria we have found to the dynamical system.

For an updating function under discrete-time dynamical system, x(t+1) = f [x(t)], which has an equilibrium at x = a,

The equilibrium is said to be Stable if and only if, 1 < f(a) < 1
The equilibrium is said to be Unstable if, f(a) =< 1 or f(a) >= 1

In our question, we have

x(t+1) = x^2(t) + 7x(t) + 8

or f(x) = x^2 + 7x + 8   and     f’(x) = 2x + 7

We have found two equilibria x = -4 and -2 to this dynamical system, so we have :

f’(-4) = 2*(-4) + 7 = -1 which is = -1, so x(t) = -4 is unstable

f’(-2) = 2*(-2) + 7 = +3 which > 1, so x(t) = -2 is unstable

c)

We have x(t+1) = x^2(t) + 7x(t) + 8 with the initial condition x(0) = 3.5

Now,

x(1) = x^2(0) + 7x(0) + 8 = (-3.5)^2 + 7(-3.5) + 8 = 12.25 – 24.5 + 8 = -4.25

x(2) = x^2(1) + 7x(1) + 8 = (-4.25)^2 + 7(-4.25) + 8 = 18.0625 – 29.75 + 8 = -3.69

x(3) = x^2(2) + 7x(2) + 8 = (-3.69)^2 + 7(-3.69) + 8 = 13.6161 – 25.83 + 8 = -4.21

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