Question

Suppose a population of bacterial doubles every hour, but that 1×10^6

individuals are removed before reproduction to be converted into valuable biological byproducts. Suppose the population begins with b=3×10^6 bacteria.
(a) Find the population after 1, 2, and 3 hours.
(b) Write the updating function which describes the discrete-time dynamical
system

Answer 1

To find the population after 1, 2, and 3 hours, we need to consider that the population doubles every hour and that 1x10^6 individuals are removed before reproduction. The updating function for this system is N(t) = 2[N(t-1) - 1x10^6].

Step-by-step explanation:

To find the population after 1, 2, and 3 hours, we need to consider that the population doubles every hour and that 1x10^6 individuals are removed before reproduction. Starting with an initial population b=3x10^6 bacteria, after 1 hour, the population will be 2b - 1x10^6, after 2 hours it will be 2(2b - 1x10^6) - 1x10^6, and after 3 hours it will be 2(2(2b - 1x10^6) - 1x10^6) - 1x10^6. Simplifying these expressions will give you the population at each time point.

The updating function for this discrete-time dynamical system can be written as follows:

N(t) = 2[N(t-1) - 1x10^6]

where N(t) represents the population at time t and N(t-1) represents the population at the previous time point.