Problem 7 - Calculus for the Life Sciences: A Modeling Approach Volume II

Question

Compute \(Q_{2}, Q_{3},\) and \(Q_{4}\) for (a) \(\quad Q_{0}=1 \quad Q_{1}=1 \quad Q_{t+2}-2 Q_{t-1}+Q_{t}=0\) (b) \(\quad Q_{0}=10 \quad Q_{1}=5 \quad Q_{t+2}-1.2 Q_{t+1}+0.32 Q_{t}=0\)

Step-by-Step Solution

Answer

(a) Q2=1, Q3=1, Q4=1; (b) Q2=2.8, Q3=1.76, Q4=1.216

Step 1: Understand the Recurrence Relation

The exercise provides us with a second-order linear homogeneous recurrence relation for two distinct parts. The general form of this recurrence relation is \(Q_{t+2} = aQ_{t+1} + bQ_{t}\) where the constants \(a\) and \(b\) differ in each part.

Step 2: Solve Part (a) for Q2, Q3, Q4

Given the relation \(Q_{t+2} - Q_{t} + 2Q_{t-1} = 0\), which simplifies to \(Q_{t+2} = 2Q_{t-1} - Q_{t}\). Start by computing from the known values \(Q_{0} = 1\) and \(Q_{1} = 1\).

Step 3: Calculate Q2 for Part (a)

Use the relation \(Q_{2} = 2Q_{0} - Q_{1}\). Substitute the known values: \[ Q_{2} = 2(1) - 1 = 2 - 1 = 1 \]

Step 4: Calculate Q3 for Part (a)

Use the relation \(Q_{3} = 2Q_{1} - Q_{2}\). Substitute the known values:\[ Q_{3} = 2(1) - 1 = 2 - 1 = 1 \]

Step 5: Calculate Q4 for Part (a)

Use the relation \(Q_{4} = 2Q_{2} - Q_{3}\). Substitute the known values:\[ Q_{4} = 2(1) - 1 = 2 - 1 = 1 \]

Step 6: Conclusion for Part (a)

For part (a), the values are \(Q_{2} = 1\), \(Q_{3} = 1\), \(Q_{4} = 1\).

Step 7: Solve Part (b) for Q2, Q3, Q4

Given the relation \(Q_{t+2} - 1.2Q_{t+1} + 0.32Q_{t} = 0\), simplify and evaluate using the known values \(Q_{0} = 10\) and \(Q_{1} = 5\).

Step 8: Calculate Q2 for Part (b)

Use the relation solved from \(Q_{t+2} = 1.2Q_{t+1} - 0.32Q_{t}\):\[ Q_{2} = 1.2(5) - 0.32(10) = 6 - 3.2 = 2.8 \]

Step 9: Calculate Q3 for Part (b)

Use the relation to find:\[ Q_{3} = 1.2(Q_{2}) - 0.32(Q_{1}) = 1.2(2.8) - 0.32(5) = 3.36 - 1.6 = 1.76 \]

Step 10: Calculate Q4 for Part (b)

Use the relation to find:\[ Q_{4} = 1.2(Q_{3}) - 0.32(Q_{2}) = 1.2(1.76) - 0.32(2.8) = 2.112 - 0.896 = 1.216 \]

Step 11: Conclusion for Part (b)

For part (b), the values are \(Q_{2} = 2.8\), \(Q_{3} = 1.76\), \(Q_{4} = 1.216\).

Key Concepts

second-order linear homogeneous recurrence

A second-order linear homogeneous recurrence relation is a type of equation used to generate sequences. It predicts the next term in a sequence based on the previous two terms. This kind of recurrence is termed "homogeneous" because it equals zero when expressed in its standard form. For example, the general form might look like this:

• \[ Q_{t+2} = aQ_{t+1} + bQ_{t} \]

Here, \(a\) and \(b\) are constants. They are crucial because they determine how each new term is related to those before it. Understanding this form is the key to solving the sequences in recurrence relations, as it provides a systematic way to move from one term to another utilizing simple arithmetic.
When dealing with such relations, it's essential to first identify the order (which in this case is two) and establish that the equation truly is linear and homogeneous. You can check linearity by looking at the coefficients and verifying that there's no constant term added to the equation.

sequence computation

Sequence computation revolves around calculating terms of sequences, often using a given recurrence relation or initial values. For students, the task is to begin with initial values or known terms and apply the recurrence relation to compute subsequent terms in the sequence.
For example, in Part (a), we start with \( Q_{0} = 1 \) and \( Q_{1} = 1 \). Each subsequent term like \( Q_{2}, Q_{3}, \) and \( Q_{4} \) is calculated using the rule provided:

• \( Q_{t+2} = 2Q_{t-1} - Q_{t} \)

By substituting the known terms into this rule, we find:

• \( Q_{2} = 1 \)
• \( Q_{3} = 1 \)
• \( Q_{4} = 1 \)

This step-by-step computation is logical and straightforward, emphasizing the importance of correctly substituting values and applying the mathematical operations precisely.

mathematical modeling

Mathematical modeling with recurrence relations helps to simplify and solve real-world problems by finding patterns that predict future behavior. A recurrence relation can model diverse processes, such as population growth or financial forecasting.
In exercises like the one presented, understanding the recurrence relation's behavior aids in forming a mathematical model of sequential change. For instance, in the problem's part (b), the relation

• \( Q_{t+2} = 1.2Q_{t+1} - 0.32Q_{t} \)

reveals a modeling scenario where each subsequent term (future prediction) relies on a scaled sum of the past two terms, adjusting for decreases or growth captured by constants \(1.2\) and \(0.32\).
Such relations allow for computational predictions in a manner that, once established, requires minimal information beyond the initial conditions to forecast future states precisely and effectively.

recursive sequences

Recursive sequences are sequences of numbers where each term is defined as a function of one or more preceding terms. They are a central concept in recursion, establishing a base (or initial) conditions and a recursive rule providing a multiplier effect on generating the sequence terms. Think of it like setting a domino chain- each domino (term) depends on the fall of those before it.
In part (a) of the problem, the sequence can be defined recursively by:

• Starting terms: \( Q_{0} = 1 \) and \( Q_{1} = 1 \)
• Recurrence relation: \( Q_{t+2} = 2Q_{t-1} - Q_{t} \)

Recursive sequences allow us to approach sequence prediction systematically. We follow a pattern based on our recursive formula; however, they also demand attention to initial conditions as this sets the entire progression in motion.
In learning and applying recursive sequences, it is useful to ensure understanding of both the computational process and the broader implications, such as how small changes in initial values can cascade through the entire sequence.