Problem 10 - Calculus for the Life Sciences: A Modeling Approach Volume II

Question

Find solutions for both \(A_{t}\) and \(B_{t}\) satisfying $$ \begin{array}{lll} \text { a. } & A_{0}=10 & A_{t+1}=0.50 A_{t}+0.2 B_{t} \\ & B_{0}=0 & B_{t+1}=0.15 B_{t}+0.7 B_{t} \\ \text { b. } & A_{0}=5 & A_{t+1}=0.6 A_{t}+0.1 B_{t} \\ & B_{0}=10 & B_{t+1}=0.2 A_{t}+0.7 B_{t} \end{array} $$

Step-by-Step Solution

Answer

For part (a), \(A_t\) halves every time and \(B_t=0\). For part (b), \(A_t\) and \(B_t\) decrease gradually and converge.

Step 1: Understanding the Recurrences

We have two separate problems (a) and (b). Both involve recursive equations that define sequences \(A_t\) and \(B_t\). Our task is to compute \(A_t\) and \(B_t\) over time based on the given initial values and update rules.

Step 2: Solving Part (a)

In part (a), we have the initial values \(A_0 = 10\) and \(B_0 = 0\). The update rules are:\(A_{t+1} = 0.50 A_t + 0.2 B_t\) and \(B_{t+1} = 0.15 B_t + 0.7 B_t = 0.85 B_t\). First, calculate \(A_1\) and \(B_1\): - \(A_1 = 0.50 \times 10 + 0.2 \times 0 = 5\)- \(B_1 = 0.85 \times 0 = 0\).

Step 3: Solving Part (b)

In part (b), we start with \(A_0 = 5\) and \(B_0 = 10\). The equations are \(A_{t+1} = 0.6 A_t + 0.1 B_t\) and \(B_{t+1} = 0.2 A_t + 0.7 B_t\). For the first iteration (\(t=0\) to \(t=1\)), calculate:- \(A_1 = 0.6 \times 5 + 0.1 \times 10 = 3 + 1 = 4\)- \(B_1 = 0.2 \times 5 + 0.7 \times 10 = 1 + 7 = 8\).

Step 4: Iterate Further for Part (a)

Continue solving iteratively for part (a). For \(t=1\) to \(t=2\), calculate:- \(A_2 = 0.50 \times 5 + 0.2 \times 0 = 2.5\)- \(B_2 = 0.85 \times 0 = 0\).Notice that \(B_t = 0\) for all iterations since \(B_0 = 0\), leading to no change in \(B_t\).Therefore, \(A_t\) halves with each iteration because \(B_t\) remains zero.

Step 5: Iterate Further for Part (b)

Continue for part (b). For \(t=1\) to \(t=2\), compute new values:- \(A_2 = 0.6 \times 4 + 0.1 \times 8 = 2.4 + 0.8 = 3.2\)- \(B_2 = 0.2 \times 4 + 0.7 \times 8 = 0.8 + 5.6 = 6.4\).Continue iterating in similar fashion to observe any patterns.

Step 6: Observing Patterns and Convergence

By observing the iterations computed, note that for part (a) \(A_t\) reduces by half each time while \(B_t\) remains zero. In part (b), both \(A_t\) and \(B_t\) decrease less sharply but approach steady values, suggesting a convergence pattern. Repeating these updates reveals the long-term behavior of the sequences.

Key Concepts

Recurrence Relations

Recurrence relations are equations that define sequences using a formula to express each term based on its preceding terms. In the exercise, the sequences \( A_t \) and \( B_t \) are defined by such relations. The recurrence relations describe how to obtain each next term from the ones before it:

• For part (a): \( A_{t+1} = 0.50 A_t + 0.2 B_t \) and \( B_{t+1} = 0.85 B_t \)
• For part (b): \( A_{t+1} = 0.6 A_t + 0.1 B_t \) and \( B_{t+1} = 0.2 A_t + 0.7 B_t \)

These equations demonstrate how each term in the sequence depends on both current terms \( A_t \) and \( B_t \).
Unlike simple geometric or arithmetic sequences that have straightforward pattern rules, recursive sequences require solving these relations step by step, revealing interesting behaviors like convergence or periodic patterns as time progresses.

Initial Values

Initial values are the foundation of any recursive sequence. They provide the starting terms from which all subsequent terms are derived. In our problem, the recurrence relations use the initial terms specified as:

• Part (a): \( A_0 = 10 \) and \( B_0 = 0 \)
• Part (b): \( A_0 = 5 \) and \( B_0 = 10 \)

These values are crucial because even a small change can lead to entirely different sequences.
Understanding initial values is like knowing the launch pad for a rocket; it determines the trajectory the sequence takes. Therefore, identifying and respecting these values is essential when solving problems involving recurrence relations.

Iterative Process

The iterative process in solving recursive sequences involves repeatedly applying the recurrence relations using initial values, slowly unraveling the behavior of the sequence over time. Each step requires calculating the next term using the previous ones by substituting them back into the recurrence formula.
For instance:

Example of Iteration for Part (a):

• Start with \( A_0 = 10 \) and \( B_0 = 0 \)
• Calculate \( A_{1} = 0.5 \times 10 + 0.2 \times 0 = 5\)
• Then calculate \( B_{1} = 0.85 \times 0 = 0\)

The iterations continue to derive \( A_2 \), \( B_2 \), and so forth.
This meticulous step-by-step approach, akin to unwrapping the layers of an onion, reveals the evolution and dynamic nature of the sequence. Over time, this process allows us to detect patterns, convergence, or eventual steady states in the sequence, crafting a complete narrative of its behavior.