Problem 5 - Calculus

Question

About sums \(f_{j}\) and differences \(\boldsymbol{v}_{j}\). Show that \(f_{j}=r_{i}^{j}(r-1)\) has \(f_{j}-f_{j-1}=r^{j}\) '. Therefore the geometric series \(1+r+\cdots+r^{j-1}\) adds up to _____ (remember to subtract \(f_{0}\) ).

Step-by-Step Solution

Answer

The series sums to \( \frac{r^j - 1}{r - 1} \).

Step 1: Recap the Formula

We are given the formula for the function as \( f_j = r^j (r-1) \). We want to show that \( f_j - f_{j-1} = r^j \).

Step 2: Substitute and Expand

Substitute \( j-1 \) in the formula to get \( f_{j-1} = r^{j-1} (r-1) \). Then, calculate \( f_j - f_{j-1} \).

Step 3: Calculate the Difference

Compute \( f_j - f_{j-1} = r^j (r-1) - r^{j-1} (r-1) \). This simplifies to \((r^j - r^{j-1})(r-1) \).

Step 4: Factor Out the Common Term

Factor out \( r^{j-1} \) from the term \((r^j - r^{j-1})\) to get \( r^{j-1} (r - 1) \). Then, multiply by \( (r-1) \, \) giving \( r^j \).

Step 5: Derive the Geometric Series Sum Formula

Recognize that \( f_j = r^j (r-1) \) arose from the expression for the sum of a geometric series. Using the result, the sum of the series \( 1 + r + \cdots + r^{j-1} \) is \( \frac{r^j - 1}{r - 1} \) after subtracting \( f_0 = 0 \).

Key Concepts

Sum of Geometric Series

A geometric series is a sum of terms where each term is a constant multiple, known as the ratio, of the previous term. To find the sum of a geometric series, such as \( 1 + r + r^2 + \, \cdots + \, r^{j-1} \), you can use a handy formula. This series can be expressed in a compact, formulaic way: \( \frac{r^j - 1}{r - 1} \).
For clarity:

• \( n \) is the number of terms.
• \( r \) is the common ratio.
• \( 1 \) is the first term.

Imagine you have to sum the first \( j \) terms of this series. The formula saves you from tedious calculations, giving you a result quickly. This sum formula is crucial in various mathematical applications, such as calculating interest or predicting growth patterns.

Formula Derivation

Deriving the formula for the sum of a geometric series involves a little algebraic manipulation. Let's see how it's done. Begin with the sum:

• \( S = 1 + r + r^2 + \, \cdots + \, r^{j-1} \)

Then, multiply the entire sum \( S \) by the common ratio \( r \):

• \( rS = r + r^2 + r^3 + \, \cdots + \, r^j \)

Now, subtract the second equation from the first:

• \( S - rS = 1 - r^j \)

Factor out \( S \) to get:

• \( S (1 - r) = 1 - r^j \)

And solving for \( S \) gives:

• \( S = \frac{1 - r^j}{1 - r} \)

This is your formula for the sum of a geometric series with the first term as \( 1 \) and the ratio as \( r \). It elegantly wraps up all terms into a neat expression.

Difference Calculation

Calculating the difference between consecutive terms in a geometric series is often necessary to understand the series' progression. Let's illustrate why this is important using the given functions, \( f_j = r^j (r-1) \) and \( f_{j-1} = r^{j-1} (r-1) \).
To find the difference, subtract \( f_{j-1} \) from \( f_j \):

• \( f_j - f_{j-1} = r^j (r-1) - r^{j-1} (r-1) \)

This simplifies to:

• \( f_j - f_{j-1} = (r^j - r^{j-1}) (r-1) \)

Factoring out \( r^{j-1} \), we find:

• \( r^{j-1}(r-1)(r-1) = r^j - r^{j-1} = r^j (r-1) \)

Eventually, simplifying expression leads to \( r^j \). This process shows that the difference in terms is consistent and adheres to a pattern, underscoring the series' stability.

Sequence and Series

In mathematics, understanding the interplay between sequences and series helps grasp how numbers pattern and accumulate. A sequence is an ordered list of numbers, while a series is the sum of a sequence's terms. Geometric sequences, where each term is a constant multiple of the previous, often morph into geometric series when summed.
Key concepts to remember include:

• In a geometric sequence, each term varies by a consistent ratio \( r \).
• A geometric series is formed by summing the elements of a geometric sequence.
• The geometric series sum formula is \( \frac{r^j - 1}{r - 1} \), applicable for any \( j \) terms.

By recognizing patterns in sequences, and using series formulas, mathematicians can solve complex problems swiftly. These concepts build understanding of important mathematical behavior and relationships.