Answer :
Final Answer
(a) The updating function for this dynamical system is x_(t+1) = 0.5x_t.
(b) If the initial number of mites is 10, the number of mites at times t=1 and t=2 are 5 and 2.5, respectively.
(c) The equilibrium (fixed) point of this dynamical system is x = 0.
(d) If the initial number of mites is 100, the number of mites at time t=10 is 0, and at time t=250, it remains at 0.
Explanation
(a) The updating function represents how the number of mites changes from one time step to the next. In this case, it's a simple exponential decay function where x_(t+1) is half of x_t. This means that each time step, half of the mites present in the previous step will remain.
(b) Starting with an initial population of 10 mites, at t=1, you have 0.5 * 10 = 5 mites. At t=2, you have 0.5 * 5 = 2.5 mites. This reflects the exponential decrease in mite population over time.
(c) The equilibrium point is where the population of mites no longer changes; it remains constant. In this case, when x_t is 0, x_(t+1) is also 0. Therefore, the equilibrium point is at x = 0.
(d) If you start with 100 mites, the population will continue to decrease exponentially. At t=10, it reaches 0 mites, and it remains at 0 for subsequent time steps, including t=250.
In summary, this dynamical system models the decline of mite populations over time due to the updating function x_(t+1) = 0.5x_t. Starting with 10 or 100 mites, the population ultimately reaches an equilibrium of 0.
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Final answer:
(a) The updating function for this system is xt+1 = 0.5xt + 1. (b) If the initial number of mites is 10, the number of mites at times t=1 and t=2 are 6 and 4, respectively. (c) The equilibrium (fixed) point of the system is x = 2. (d) If the initial number of mites is 100, the number of mites at time t=10 is approximately 2.002 and at time t=250 is also approximately 2.002.
Explanation:
(a) The updating function for this dynamical system is given by:
xt+1 = 0.5xt + 1
(b) If the initial number of mites is 10, we can find the number of mites at times t=1 and t=2 by plugging in the values:
For t = 1:
x1 = 0.5(10) + 1 = 6
For t = 2:
x2 = 0.5(6) + 1 = 4
(c) To find the equilibrium (fixed) point of the dynamical system, we set xt+1 = xt:
x = 0.5x + 1
0.5x - x = -1
-0.5x = -1
x = 2
Therefore, the equilibrium (fixed) point is x = 2.
(d) If the initial number of mites is 100, we can find the number of mites at time t=10 and t=250 by repeatedly applying the updating function:
For t = 10:
x10 = 0.5(0.5(0.5(0.5(0.5(0.5(0.5(0.5(0.5(0.5(100+1)+1)+1)+1)+1)+1)+1)+1)+1)+1)+1) ≈ 2.002
For t = 250:
x250 ≈ 2.002
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