Problem 40 - Calculus

Question

About exponential \(v^{3}\) s and \(f^{\prime}\) s. Suppose \(v_{j}=r^{j}\). Show that \(f_{j}=\left(r^{j+1}-1\right) /(r-1)\) starts from \(f_{0}=1\) and has \(f_{j}-f_{j-1}=v_{j}\), (Then this is the correct \(f_{j}=1+r+\cdots+r^{j}=\) sum of a geometric series.)

Step-by-Step Solution

Answer

\( f_j = 1 + r + \cdots + r^j \) satisfies \( f_0 = 1 \) and \( f_j - f_{j-1} = v_j = r^j \).

Step 1: Verify Initial Condition for f

First, we need to confirm that the function \( f_j \) starts with \( f_0 = 1 \). Substituting \( j=0 \) into the given formula \( f_j = \frac{r^{j+1} - 1}{r - 1} \), we get \[ f_0 = \frac{r^{0+1} - 1}{r - 1} = \frac{r^1 - 1}{r - 1} = \frac{r - 1}{r - 1} = 1. \] This verifies that the initial condition \( f_0 = 1 \) is satisfied.

Step 2: Express Difference f_j - f_{j-1}

Next, we will show that \( f_j - f_{j-1} = v_j \). Recall \( v_j = r^j \). We calculate \( f_j - f_{j-1} \) using the formula for \( f_j \). Substitute into the difference: \[ f_j = \frac{r^{j+1} - 1}{r - 1}, \quad f_{j-1} = \frac{r^j - 1}{r - 1}. \]Thus, \[ f_j - f_{j-1} = \frac{r^{j+1} - 1}{r - 1} - \frac{r^j - 1}{r - 1} = \frac{r^{j+1} - 1 - (r^j - 1)}{r - 1}. \]This simplifies to \[ \frac{r^{j+1} - r^j}{r - 1} = \frac{r^j (r - 1)}{r - 1}. \]Cancelling \( r-1 \), we have \( f_j - f_{j-1} = r^j = v_j \). Thus, it matches the given \( v_j \).

Step 3: Confirm Expression for Geometric Series

Finally, we confirm that the expression \( f_j = 1 + r + r^2 + \cdots + r^j \) is indeed equal to the sum of a geometric series.The sum of a geometric series with first term 1 and common ratio \( r \) over \( j + 1 \) terms is given by \[ S = \frac{r^{j+1} - 1}{r - 1}. \]Therefore, \[ f_j = 1 + r + r^2 + \cdots + r^j = \frac{r^{j+1} - 1}{r - 1}. \]Both expressions match, confirming \( f_j \) is the sum of the geometric series.

Key Concepts

Exponential Functions

Exponential functions are equations where a constant base is raised to a variable exponent. In the context of sequences and series, exponential functions often model growth or decay processes. For instance, a function like \( v_j = r^j \) is an exponential function with base \( r \) and exponent \( j \). This means that as \( j \) increases, the values depend exponentially on \( r \).
This type of function is pivotal because it represents how quantities can multiply over stages or time intervals. If \( r > 1 \), the sequence grows progressively larger, depicting exponential growth.
When \( 0 < r < 1 \), the sequence describes exponential decay, which tapers off as it moves further into its domain.

• The main characteristics are the rapid increase or decrease of values as the exponent \( j \) changes.
• They simplify calculations in geometric progressions, making them fundamental in various applications like finance and population modeling.

In summary, understanding exponential functions helps unlock the behavior of sequences that change multiplicatively over steps.

Geometric Series Sum

A geometric series is a series of the form \( a + ar + ar^2 + \cdots + ar^n \), where \( a \) is the first term, and \( r \) is the common ratio. When you sum these terms, it forms the geometric series sum.
The formula for the sum of a geometric series with \( (n+1) \) terms starting from \( 1 \) is:
\[ S = \frac{r^{n+1} - 1}{r - 1} \]
This equation is crucial because it allows for quick calculation of total sums without adding each term individually.

• If \( r = 1 \), this formula simplifies to simply \( a(n+1) \), since the terms do not change.
• If \( r eq 1 \), this formula beautifully encapsulates how the multiplication of the base affects the summation.

Understanding this key concept means grasping how multiplying by a common ratio affects an entire sequence and its summation.

Difference of Sequences

The difference of sequences reflects the change between successive terms in a sequence. It's a critical concept when determining patterns or rules governing the sequences.
For example, given sequences like \( f_j \) and \( f_{j-1} \), the difference \( f_j - f_{j-1} \) identifies how each term in the sequence changes as we progress.

• In the provided example, \( v_j = r^j \) represents the rate of change.
• Verifying the equality \( f_j - f_{j-1} = v_j \) helps confirm that \( f_j \) accurately captures the cumulative sum of \( v_j \).

This verification process is pivotal in many mathematical analyses, especially when validating formulas like the one for geometric series. Understanding the difference of sequences is a necessary step in perceiving how different components of mathematical sequences interplay to form coherent structures.