Problem 54 - Calculus for Biology and Medicine

Question

$$ \text { In Problems } \text { , write } N_{t} \text { as a function of } t \text { for each recursion } $$ $$ N_{t+1}=\frac{1}{2} N_{t} \text { with } N_{0}=2300 $$

Step-by-Step Solution

Answer

The function is \(N_t = 2300 \left(\frac{1}{2}\right)^t\).

Step 1: Understanding the Recursion Formula

The recursion formula provided is \(N_{t+1} = \frac{1}{2}N_t\). This means that each term after \(N_0\) is half the previous term. We also know the initial value is \(N_0 = 2300\).

Step 2: Writing the First Few Terms

To find \(N_1\), plug \(N_0\) into the formula: \(N_1 = \frac{1}{2} \times 2300 = 1150\). \(N_2 = \frac{1}{2} \times 1150 = 575\). Notice the pattern of multiplying by \(\frac{1}{2}\).

Step 3: Deriving the General Formula

From the pattern observed, each \(N_t\) can be written as \(N_t = 2300 \times \left(\frac{1}{2}\right)^t\). This formula comes from multiplying the initial term by \(\frac{1}{2}\) raised to the power of the number of steps forward in time \(t\).

Step 4: Writing the Function

Express \(N_t\) as a function of \(t\) using the pattern found: \(N_t = 2300 \left(\frac{1}{2}\right)^t\). This function describes \(N_t\) completely for any non-negative integer \(t\).

Key Concepts

Recursion formula

Recursion formulas are a wonderful way to express sequences where each term is defined based on the previous one. They are succinct and often reveal intriguing patterns. In the given problem, the recursion formula is \( N_{t+1} = \frac{1}{2} N_t \). This formula tells us that every subsequent term is half of the prior term.
For recursion, you start with an initial term; here we know that \( N_0 = 2300 \). From this starting point, you use the recursion formula to generate each term.
To put it simply:

• Plug the previous term into the formula.
• Perform the calculation to find the next term.
• Repeat until you reach the desired term.

This method efficiently calculates sequences, especially those following consistent patterns.

Geometric sequence

Geometric sequences are a fascinating type of progression where each term is a constant multiple of the previous one. In our exercise, the sequence forms as \( N_t = 2300 \left(\frac{1}{2}\right)^t \).
The number \( \frac{1}{2} \) is known as the common ratio. The sequence here sharply reduces as each term halves successively.

• Start with the initial value \( N_0 = 2300 \).
• Continue multiplying by \( \frac{1}{2} \) for each subsequent term.
• Notice how the sequence decreases: \( 2300, 1150, 575, \ldots \)

The general formula for a geometric sequence like this is \( N_t = a \times r^t \), where \( a \) is the first term, \( r \) is the common ratio, and \( t \) is the term number. It brilliantly captures both growth and decay scenarios.

Population modeling

In mathematics, population modeling with recursion and geometric sequences is often applied to understand population dynamics.
This exercise is a great example of modeling a declining population. Starting with a population of 2300, every year, the population decreases to half of its previous size. Such a model can simulate scenarios like radioactive decay, extinction processes, or resource depletion where entities halve over regular intervals.

• \( N_0 = 2300 \) represents your starting population.
• Every subsequent population count thrives on the relationship \( N_{t+1} = \frac{1}{2} N_t \).
• The population diminishes exponentially over time.

The function formulated as \( N_t = 2300 \left(\frac{1}{2}\right)^t \) delineates this population's steep descent through time. It is a simple yet powerful way to predict future population sizes based on recursive relationships.