Problem 45 - Calculus for Biology and Medicine

Question

In Problems , find the population sizes for \(t=0,1,2, \ldots, 5\) for each recursion. $$ N_{t+1}=\frac{1}{5} N_{t} \text { with } N_{0}=31250 $$

Step-by-Step Solution

Answer

The population sizes are 31250, 6250, 1250, 250, 50, 10 for \( t=0 \) to \( t=5 \).

Step 1: Understand the recursion formula

The given recursion formula is \( N_{t+1} = \frac{1}{5} N_{t} \). This indicates that each population size at a time \( t+1 \) is one-fifth of the population size at time \( t \).

Step 2: Apply the initial condition

We start with \( N_0 = 31250 \). This is the initial population size at time \( t = 0 \). We'll use this to find subsequent population sizes.

Step 3: Calculate \( N_1 \)

Using the recursion formula, \( N_1 = \frac{1}{5} N_0 = \frac{1}{5} \times 31250 = 6250 \).

Step 4: Calculate \( N_2 \)

Continuing with the recursion, \( N_2 = \frac{1}{5} N_1 = \frac{1}{5} \times 6250 = 1250 \).

Step 5: Calculate \( N_3 \)

Using the recursion rule, \( N_3 = \frac{1}{5} N_2 = \frac{1}{5} \times 1250 = 250 \).

Step 6: Calculate \( N_4 \)

Following the same formula, \( N_4 = \frac{1}{5} N_3 = \frac{1}{5} \times 250 = 50 \).

Step 7: Calculate \( N_5 \)

Finally, \( N_5 = \frac{1}{5} N_4 = \frac{1}{5} \times 50 = 10 \).

Key Concepts

Understanding Population Size

Population size refers to the number of individuals within a specified population at a given time. In our exercise, population size is determined recursively for each time step. By tracking the population size, we can observe how the population changes over time. This can be affected by various factors, such as birth rates, death rates, and immigration/emigration rates. In our specific problem, the size of the population is calculated based on a given recursion formula. This allows us to see how the population diminishes step-by-step as we move forward in time.

The Initial Condition Explained

The initial condition is the starting point for calculating future population sizes in a recursive sequence. It is a crucial aspect of the problem, as it defines the value of the population at the beginning of the observation period, denoted here as when time \( t = 0 \). In our situation, the initial condition is given as \( N_0 = 31250 \), meaning the population size at time \( t = 0 \) is 31,250.

• This value serves as the base from which we calculate subsequent population sizes using the recursion formula.
• Any change in this initial condition will affect all future calculations and consequently the resulting sequence of population sizes.

Exploring the Recursive Sequence

A recursive sequence is a sequence in which each term is derived from one or more preceding terms by some rule or formula. In this exercise, the recursive formula provided is \( N_{t+1} = \frac{1}{5} N_t \). This means that each new population size (\( N_{t+1} \)) is calculated as one-fifth of the previous size (\( N_t \)).

• This specific recursive formula shows a decay sequence, where the population decreases over time.
• Recursive sequences are often used to model situations where changes are proportional to current values, like population decay, interest in financial accounts, or radioactive decay.

The recursive nature of the formula allows us to determine future population sizes efficiently by using the result from the prior step.

A Step-by-Step Guide to the Solution

A step-by-step solution helps break down the process of solving the recursion problem into manageable and understandable parts.

• Step 1: Identify the recursion formula. Here, it's \( N_{t+1} = \frac{1}{5} N_t \), which informs us about the population's rate of change.
• Step 2: Apply the initial condition, where \( N_0 = 31250 \). This is our starting point for the calculation.
• Step 3 to 7: Calculate subsequent terms iteratively, relying on previous results:
• \( N_1 = \frac{1}{5} \times 31250 = 6250 \)
• \( N_2 = \frac{1}{5} \times 6250 = 1250 \)
• \( N_3 = \frac{1}{5} \times 1250 = 250 \)
• \( N_4 = \frac{1}{5} \times 250 = 50 \)
• \( N_5 = \frac{1}{5} \times 50 = 10 \)

This step-by-step process ensures that each future term is correctly derived from the one before it, providing clarity and consistency in the solution approach.