Problem 6 - Calculus for the Life Sciences: A Modeling Approach Volume II

Question

Find the first five terms of the solutions to \(\begin{array}{llll}\text { a. } & Q_{0}=1 & Q_{1}=0 & Q_{t+1}-Q_{t}=Q_{t-1} \\\ \text { b. } & Q_{0}=1 & Q_{1}=0 & Q_{t}-Q_{t-1}=Q_{t-2} \\ \text { c. } & Q_{0}=1 & Q_{1}=1 & Q_{t+1}-Q_{t}=Q_{t-1} \\ \text { d. } & Q_{0}=1 & Q_{1}=0 & Q_{t+1}=-Q_{t-1} \\ \text { e. } & Q_{0}=1 & Q_{1}=1 & Q_{t}=-Q_{t-2} \\ \text { f. } & Q_{0}=1 & Q_{1}=0 & Q_{2}=2 & Q_{t+1}-Q_{t}=Q_{t}-Q_{t-2} \\ \text { g. } & Q_{0}=1 & Q_{1}=1 & Q_{2}=1 & Q_{t+1}=Q_{t}-Q_{t-1}+Q_{t-2} \\ \text { h. } & Q_{0}=0.6 & Q_{1}=0.7 & Q_{2}=0.5 \quad Q_{t+1}-Q_{t}=0.6 Q_{t-2}\left(1-Q_{t-2}\right) \\ & & & & -0.2 Q_{t-1}\end{array}\)

Step-by-Step Solution

Answer

The first five terms for each are: (a) 1, 0, 1, 1, 2; (b) 1, 0, 1, 1, 2; (c) 1, 1, 2, 3, 5; (d) 1, 0, -1, 0, 1; (e) 1, 1, -1, -1, 1; (f) 1, 0, 2, 2, 4; (g) 1, 1, 1, 1, 1; (h) 0.6, 0.7, 0.5, approx 0.53, approx 0.56.

Step 1: Understand Recurrence Relation for Part (a)

Given is the recurrence: \( Q_{t+1} - Q_t = Q_{t-1} \). We need to calculate the first five terms starting with \( Q_0 = 1 \) and \( Q_1 = 0 \).

Step 2: Calculate First Five Terms for (a)

Start from \( Q_0 \) and \( Q_1 \). Use the recurrence to find each subsequent term:- \( Q_2 = Q_1 + Q_0 = 0 + 1 = 1 \)- \( Q_3 = Q_2 + Q_1 = 1 + 0 = 1 \)- \( Q_4 = Q_3 + Q_2 = 1 + 1 = 2 \)Five terms: \( Q_0 = 1, Q_1 = 0, Q_2 = 1, Q_3 = 1, Q_4 = 2 \).

Step 3: Understand Recurrence for Part (b)

This recurrence is \( Q_{t} - Q_{t-1} = Q_{t-2} \) with \( Q_0 = 1, Q_1 = 0 \).

Step 4: Calculate First Five Terms for (b)

Continuing from initial terms, we calculate:- \( Q_2 = Q_1 + Q_0 = 0 + 1 = 1 \)- \( Q_3 = Q_2 + Q_1 = 1 + 0 = 1 \)- \( Q_4 = Q_3 + Q_2 = 1 + 1 = 2 \)Five terms: \( Q_0 = 1, Q_1 = 0, Q_2 = 1, Q_3 = 1, Q_4 = 2 \).

Step 5: Understand Recurrence for Part (c)

Given recurrence \( Q_{t+1} - Q_t = Q_{t-1} \) with \( Q_0 = 1, Q_1 = 1 \).

Step 6: Calculate First Five Terms for (c)

Calculate each term using the recurrence:- \( Q_2 = Q_1 + Q_0 = 1 + 1 = 2 \)- \( Q_3 = Q_2 + Q_1 = 2 + 1 = 3 \)- \( Q_4 = Q_3 + Q_2 = 3 + 2 = 5 \)Five terms: \( Q_0 = 1, Q_1 = 1, Q_2 = 2, Q_3 = 3, Q_4 = 5 \).

Step 7: Understand Recurrence for Part (d)

Recurrence is \( Q_{t+1} = -Q_{t-1} \). Starting values are \( Q_0 = 1, Q_1 = 0 \).

Step 8: Calculate First Five Terms for (d)

Use the recurrence to find:- \( Q_2 = -Q_0 = -1 \)- \( Q_3 = -Q_1 = 0 \)- \( Q_4 = -Q_2 = 1 \)Five terms: \( Q_0 = 1, Q_1 = 0, Q_2 = -1, Q_3 = 0, Q_4 = 1 \).

Step 9: Understand Recurrence for Part (e)

The recurrence is \( Q_{t} = -Q_{t-2} \) with initial values \( Q_0 = 1, Q_1 = 1 \).

Step 10: Calculate First Five Terms for (e)

Using the recurrence:- \( Q_2 = -Q_0 = -1 \)- \( Q_3 = -Q_1 = -1 \)- \( Q_4 = -Q_2 = 1 \)Five terms: \( Q_0 = 1, Q_1 = 1, Q_2 = -1, Q_3 = -1, Q_4 = 1 \).

Step 11: Understand Recurrence for Part (f)

Recurrence is given by \( Q_{t+1} = 2Q_{t-1} \), with initial values \( Q_0 = 1, Q_1 = 0, Q_2 = 2 \).

Step 12: Calculate First Five Terms for (f)

Use the recurrence to find:- \( Q_3 = 2 \times Q_1 = 0 \)- \( Q_4 = 2 \times Q_2 = 4 \)Five terms: \( Q_0 = 1, Q_1 = 0, Q_2 = 2, Q_3 = 0, Q_4 = 4 \).

Step 13: Understand Recurrence for Part (g)

Recurrence is \( Q_{t+1} = Q_t - Q_{t-1} + Q_{t-2} \) with initial values \( Q_0 = 1, Q_1 = 1, Q_2 = 1 \).

Step 14: Calculate First Five Terms for (g)

Using the given recurrence:- \( Q_3 = Q_2 - Q_1 + Q_0 = 1 - 1 + 1 = 1 \)- \( Q_4 = Q_3 - Q_2 + Q_1 = 1 - 1 + 1 = 1 \)Five terms: \( Q_0 = 1, Q_1 = 1, Q_2 = 1, Q_3 = 1, Q_4 = 1 \).

Step 15: Understand Recurrence for Part (h)

Use recurrence \( Q_{t+1} = Q_t + 0.6Q_{t-2}(1 - Q_{t-2}) - 0.2Q_{t-1} \) with initial values \( Q_0 = 0.6, Q_1 = 0.7, Q_2 = 0.5 \).

Step 16: Calculate First Five Terms for (h)

Using the values and the recurrence, calculate:- \( Q_3 = Q_2 + 0.6 \times Q_0 \times (1 - Q_0) - 0.2 \times Q_1 = 0.5 + 0.6 \times 0.6 \times 0.4 - 0.14 \approx 0.53 \)- \( Q_4 = Q_3 + 0.6 \times Q_1 \times (1 - Q_1) - 0.2 \times Q_2 = 0.53 + 0.6 \times 0.7 \times 0.3 - 0.1 \approx 0.56 \)Five terms: \( Q_0 = 0.6, Q_1 = 0.7, Q_2 = 0.5, Q_3 \approx 0.53, Q_4 \approx 0.56 \).

Key Concepts

Difference Equations

Difference equations are mathematical expressions involving the differences between successive terms of a sequence. They are similar to differential equations but work with discrete values instead of continuous ones. In essence, they determine the behavior of sequences or time series by defining how a term relates to its predecessors.
For instance, in exercise (a), we have the difference equation \( Q_{t+1} - Q_t = Q_{t-1} \), which means that the change from term \( Q_t \) to \( Q_{t+1} \) depends linearly on the term \( Q_{t-1} \). The task is to find out how each term in the sequence progresses based on the initial conditions given.
This particular type of recurrence relation is often used in computational applications and various fields such as economics, biology, and computer science to model dynamic systems that evolve over time.

Sequence Solutions

Sequence solutions involve solving a recurrence relation to find specific terms in a sequence. This solution process often includes performing iterative calculations until the desired number of terms is found.
In the original exercise, the solutions to the difference equations yield sequences where each subsequent term is calculated from its predecessors using the given relationship. For example, in part (a), we begin with \( Q_0 = 1 \) and \( Q_1 = 0 \), using the given relation \( Q_{t+1} - Q_t = Q_{t-1} \) allows us to find subsequent terms such as \( Q_2 = 1 \), \( Q_3 = 1 \), and so forth.
These sequences help to visualize the pattern or behavior of the terms as they progress, which is crucial in understanding how the underlying model or system evolves.

Initial Conditions

Initial conditions are the starting values provided for a sequence, which are essential in solving a difference equation. They serve as the anchor for calculating future terms in the sequence.
Each part of the exercise specifies different initial conditions. For instance, in part (a), the conditions are \( Q_0 = 1 \) and \( Q_1 = 0 \). These values enable us to commence the iterative process of applying the recurrence relation, leading to the determination of the subsequent terms.
Initial conditions can vary widely between problems, and they significantly influence the nature and values of the sequence. Different initial conditions can lead to vastly different sequence behaviors, even when using the same recurrence relation.

Term Calculation

Term calculation in the context of recurrence relations involves using the given formula to iteratively compute each term of the sequence from the initial conditions. It is a step-by-step procedure employing the difference equation to find a sequence of numbers.
For example, when calculating terms for part (c), which has the recurrence \( Q_{t+1} - Q_t = Q_{t-1} \) and initial conditions \( Q_0 = 1 \) and \( Q_1 = 1 \), the next terms are calculated as follows:

• \( Q_2 = Q_1 + Q_0 = 2 \)
• \( Q_3 = Q_2 + Q_1 = 3 \)
• \( Q_4 = Q_3 + Q_2 = 5 \)

This sequential approach is applied repeatedly until all required terms are calculated, providing insight into how each computes based on those before it.