Problem 2
Question
Equation 1.30 $$ P_{t+1}-P_{t}=b $$ represents a large number of equations $$ \begin{array}{r} P_{1}-P_{0}=b \\ P_{2}-P_{1}=b \\ P_{3}-P_{2}=b \\ \vdots \quad \vdots \quad \vdots \quad \vdots \\ P_{n-1}-P_{n-2}=b \\ P_{n}-P_{n-1}=b \end{array} $$ Add these equations to obtain $$ P_{n}=P_{0}+n b $$ Substitute \(t\) for \(n\) to obtain $$ P_{t}=P_{0}+t b $$
Step-by-Step Solution
Verified Answer
The equation sums up to \( P_{t} = P_{0} + tb \).
1Step 1: Understanding the Sequence of Equations
The given set of equations \( P_{t+1} - P_{t} = b \) represents a change by a constant \( b \) from one time period to the next. It describes how each subsequent element \( P_{t} \) is generated by adding \( b \) to the previous element.
2Step 2: Setting Up the Summation
To express \( P_{n} \) in terms of \( P_{0} \), we add all the equations from \( P_{1}-P_{0} = b \) to \( P_{n} - P_{n-1} = b \). This action accounts for every change over \( n \) time steps from the start.
3Step 3: Adding the Equations Together
If we sum up the series of differences:\[ \begin{align*} (P_{1} - P_{0}) + (P_{2} - P_{1}) + (P_{3} - P_{2}) + \, \cdots \, + (P_{n} - P_{n-1}) &= b + b + b + \, \cdots \, + b \end{align*} \]The left side of the equation telescopes to \( P_{n} - P_{0} \), and the right adds up to \( nb \), since \( b \) is added \( n \) times.
4Step 4: Simplifying the Equation
After adding the equations, we get:\[ P_{n} - P_{0} = nb \]Solving for \( P_{n} \) gives us \( P_{n} = P_{0} + nb \).
5Step 5: Generalizing with Variable Substitution
To make this formula applicable for any time period \( t \), simply replace \( n \) with \( t \). This yields:\[ P_{t} = P_{0} + t b \]
Key Concepts
Difference EquationLinear Growth ModelSummation of Series
Difference Equation
When you see the term "difference equation," think of how quantities change over time. In this context, it refers to the difference between elements in a sequence, highlighted in the equation \( P_{t+1} - P_{t} = b \). This tells us that each new term in the sequence is derived by adding a fixed number \( b \) to the previous term. This is why we call it a difference equation - it shows a pattern of change.
Difference equations are crucial because they model how a situation progresses. They help predict future values by providing a mathematical way to calculate subsequent terms based on a consistent rule. In our example, that rule is simply adding \( b \) each time. Notice how the constant difference signified by \( b \) predicts a uniform evolution in terms of change over each period.
Difference equations are crucial because they model how a situation progresses. They help predict future values by providing a mathematical way to calculate subsequent terms based on a consistent rule. In our example, that rule is simply adding \( b \) each time. Notice how the constant difference signified by \( b \) predicts a uniform evolution in terms of change over each period.
Linear Growth Model
A linear growth model is a straight line if you were to graph it. It means that growth happens at a constant rate. Here, the linear growth is depicted by the equation \( P_{t} = P_{0} + t b \). This formula implies that \( P_{0} \), the initial term, is being increased by the repeated addition of \( b \) times \( t \), which is the time period count.
The elegance of a linear growth model lies in its simplicity:
Knowing that each period \( t \) adds the same \( b \), we see a clear and predictable pattern of growth that stays linear over time. This characteristic makes linear models useful and dependable for forecasting under conditions of stable growth.
The elegance of a linear growth model lies in its simplicity:
- Constant Rate: The same additive value \( b \) for each time interval.
- Predictability: Easily project future values based on initial conditions.
- Simplicity: Only involves straightforward multiplication and addition.
Knowing that each period \( t \) adds the same \( b \), we see a clear and predictable pattern of growth that stays linear over time. This characteristic makes linear models useful and dependable for forecasting under conditions of stable growth.
Summation of Series
Summation essentially means adding up a group of numbers to get a total. In the context of arithmetic sequences, like the one we're examining, summation is about gathering all the incremental steps to see the overall impact. When we summed the initial series of difference equations, the rule was simple: keep adding \( b \).
Here's what's important about this operation:
Summation in this setup reveals the simple structure of growth or change across multiple periods by effectively gathering incremental changes into a singular expression, in our case resulting in \( P_{n} = P_{0} + nb \), showcasing how an arithmetic sequence can be managed as a summation of series.
Here's what's important about this operation:
- Telescoping Effect: The left side of the summation, \((P_{1} - P_{0}) + (P_{2} - P_{1}) + \ldots + (P_{n} - P_{n-1})\), beautifully collapses into \( P_{n} - P_{0} \).
- Clarity of Pattern: Since \( b \) is added consistently, the total can be easily calculated as \( n \times b \), simplifying our understanding of change over time.
Summation in this setup reveals the simple structure of growth or change across multiple periods by effectively gathering incremental changes into a singular expression, in our case resulting in \( P_{n} = P_{0} + nb \), showcasing how an arithmetic sequence can be managed as a summation of series.
Other exercises in this chapter
Problem 1
Determine the doubling times of the following exponential equations. (a) \(y=2^{t}\) (b) \(y=2^{3 t}\) (c) \(y=2^{0.1 t}\) (d) \(y=10^{t}\) (e) \(y=10^{3 t}\) (
View solution Problem 2
Two kilos of a fish poison that does not decompose are mixed into a lake that has a volume of \(100 \times 20 \times 2=4000\) cubic meters. A stream of clean wa
View solution Problem 2
Show that \(A_{t}=0\) for all \(t\) is a solution to Equation 1.18 $$ A_{t+1}-A_{t}=2 \pi k \sqrt{A_{t}} $$ This means that for every \(t,\) if \(A_{t}=0\) and
View solution Problem 2
Show that the doubling time of \(y=A B^{t}\) is \(1 /\left(\log _{2} B\right)\)
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