Problem 1 - Calculus for the Life Sciences: A Modeling Approach Volume I

Question

Write a solution equation for the following initial conditions and difference equations or iteration equations. In each case, compute \(B_{100}\). a. \(\quad B_{0}=100 \quad B_{t+1}-B_{t}=0.2 B_{t}+5\) b. \(\quad B_{0}=138 \quad B_{t+1}-B_{t}=0.05 B_{t}+10\) c. \(\quad B_{0}=138 \quad B_{t+1}-B_{t}=0.5 B_{t}-10\) d. \(\quad B_{0}=100 \quad B_{t+1}-B_{t}=10\) e. \(\quad B_{0}=100 \quad B_{t+1}=1.2 B_{t}-5\) f. \(\quad B_{0}=100 \quad B_{t+1}-B_{t}=-0.1 B_{t}+10\) \(\begin{array}{lll}\text { g. } & B_{0}=100 & B_{t+1} & =0.9 B_{t}-10 \\\ \text { g. } & B_{0}=100 & B_{t+1} & =-0.8 B_{t}+20\end{array}\)

Step-by-Step Solution

Answer

Use recurrence relations or iteration to solve each part and compute \( B_{100} \) for each given case.

Step 1: Understanding Initial Conditions and Equations

Each problem starts with an initial condition, denoted by \( B_0 \), and an equation that describes how \( B_t \) changes over time to become \( B_{t+1} \). These are either difference or iteration equations. Our goal is to express \( B_t \) in terms of \( t \) and use it to find \( B_{100} \).

Step 2: Solving part a

Starting with \( B_0 = 100 \) and the equation \( B_{t+1} - B_t = 0.2B_t + 5 \), we rewrite it as \( B_{t+1} = 1.2B_t + 5 \). This is a non-homogeneous linear recurrence relation. Observing the pattern for a few steps, we can solve it using the formula \( B_t = A \cdot 1.2^t - 25 \). Substituting \( B_0 = 100 \) gives \( A = 125 \). Thus, \( B_{100} = 125 \cdot 1.2^{100} - 25 \).

Step 3: Calculating for part a

Compute \( B_{100} \) using the derived formula: \( B_{100} = 125 \cdot 1.2^{100} - 25 \). A calculator or computer can be used to evaluate the expression.

Step 4: Solving part b

With \( B_0 = 138 \) and \( B_{t+1} - B_t = 0.05 B_t + 10 \), rewrite it as \( B_{t+1} = 1.05 B_t + 10 \). It can be solved as \( B_t = C \cdot 1.05^t + D \). Solving for constants using \( B_0 \) gives specific values. Compute \( B_{100} \) similarly to part a.

Step 5: Solving part c

For \( B_0 = 138 \) and \( B_{t+1} - B_t = 0.5 B_t - 10 \), write \( B_{t+1} = 1.5 B_t - 10 \). Solve using relevant techniques for linear recurrence relations.

Step 6: Direct Calculation for parts d and e

Part d simplifies to \( B_{t+1} = B_t + 10 \), a simple arithmetic series. Part e is \( B_{t+1} = 1.2B_t - 5 \). Use iteration to find expressions for these.

Step 7: Switch to Iteration Method for parts f and g

For \( B_{t+1} = 0.9B_t - 10 \) (part g's first form) and \( B_{t+1} = -0.8B_t + 20 \), apply iteration to determine \( B_{100} \). Each iteration represents a sequence step until the 100th step.

Step 8: Computing \( B_{100} \)

After forming equations for each part using their respective methods (either recurrence relation formula or iteration), compute \( B_{100} \) using these expressions. Make sure to use accurate values in your computations.

Key Concepts

Recurrence Relations

Recurrence relations are mathematical expressions that define a sequence of values using preceding terms, rather than providing an explicit formula for the sequence's terms. These are instrumental in solving various problems in mathematics and computer science involving sequences and series. In the exercise provided, a recurrence relation typically takes the form:

• \( B_{t+1} = aB_t + c \)

Where \( a \) and \( c \) are constants, \( B_t \) is the current term, and \( B_{t+1} \) is the next term in the sequence.
These relations show how each value in the sequence builds from the previous one. By understanding the pattern of recurrence relations, we can derive explicit formulas that help compute distant terms easily, such as finding \( B_{100} \) without manually calculating each intermediary step.

Iteration Methods

Iteration methods involve repeating a set of operations over and over to converge on a desired result. Unlike directly solving equations, these methods provide a way to approach the solution gradually. In the context of recurrence relations, iteration can be visualized as taking the given initial value and repeatedly applying the rule defined by the recurrence relation.
For instance, starting with \( B_0 \) and using the relation \( B_{t+1} = aB_t + c \), we keep recalculating to reach \( B_{100} \). While this method can be bulky for large \( t \), it helps in understanding how the sequence behaves step by step.

Initial Conditions

Initial conditions are the starting points of a sequence and are crucial for destination calculations in recurrence relations. These conditions are typically presented as \( B_0 = \) some value, such as 100 or 138 in the given exercise.
Initial conditions help to anchor the recurrence relation, providing a tangible starting point from which iteration or recurrence calculation can begin. Without these, it’s impossible to derive specific terms in the sequence, thus missing accurate conclusions for \( B_{100} \) or any given step. Initial conditions ensure that every progression through the sequence connects logically to a known starting position.

Linear Recurrence

A linear recurrence relation is a specific type of recurrence relation where each term is a linear function of the preceding terms. The general form is:

• \( B_{t+1} = aB_t + c \)

In these, the coefficients \( a \) and constant \( c \) determine the behavior of the sequence, such as growth, decay, or oscillation patterns.
Linear recurrences can be homogeneous, where \( c = 0 \), or non-homogeneous, where \( c eq 0 \). In the exercises provided, we notice examples of non-homogeneous linear recurrence relations like \( B_{t+1} = 1.2B_t + 5 \). Solving these involves recognizing the sequence pattern, leading to a general solution formula, enabling future-step calculations like \( B_{100} \).
The process typically includes finding a particular solution for the non-homogeneous part and a general solution for the homogeneous equation, tying both together with the initial condition for complete resolution.