Problem 16 - Calculus

Question

expand the quantity about 0 in terms of the variable given. Give four nonzero terms. $$\frac{1}{a-r} \text { in terms of } \frac{r}{a}$$

Step-by-Step Solution

Answer

The first four non-zero terms of the expansion are \( \frac{1}{a}, \frac{r}{a^2}, \frac{r^2}{a^3}, \frac{r^3}{a^4} \).

Step 1: Identify the Function and Variable

We are given the function \( \frac{1}{a-r} \) and need to expand it in terms of \( \frac{r}{a} \). This means we need to express our function in a form that makes it possible to expand around 0 for the variable \( \frac{r}{a} \).

Step 2: Rewrite the Function

Express \( \frac{1}{a-r} \) in terms of \( \frac{r}{a} \). Use the substitution \( r = a \cdot x \), where \( x = \frac{r}{a} \). This gives us \( \frac{1}{a(1-x)} \), which is \( \frac{1}{a} \cdot \frac{1}{1-x} \).

Step 3: Use the Geometric Series Expansion

Recall that the geometric series \( \frac{1}{1-x} \) can be expanded as \( 1 + x + x^2 + x^3 + \ldots \) for \( |x| < 1 \). Here, \( x = \frac{r}{a} \), so our series becomes \( 1 + \frac{r}{a} + \left(\frac{r}{a}\right)^2 + \left(\frac{r}{a}\right)^3 + \ldots \).

Step 4: Multiply Through by \( \frac{1}{a} \)

Factor out \( \frac{1}{a} \) from each term of the series expansion: \( \frac{1}{a}(1) + \frac{1}{a}\left(\frac{r}{a}\right) + \frac{1}{a}\left(\frac{r}{a}\right)^2 + \frac{1}{a}\left(\frac{r}{a}\right)^3 + \ldots \).

Step 5: Write the First Four Non-Zero Terms

Simplify to get the series: \( \frac{1}{a} + \frac{r}{a^2} + \frac{r^2}{a^3} + \frac{r^3}{a^4} + \ldots \). The first four non-zero terms are \( \frac{1}{a} + \frac{r}{a^2} + \frac{r^2}{a^3} + \frac{r^3}{a^4} \).

Key Concepts

Geometric Series

A geometric series is a series of numbers where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio. In the context of a Taylor series, understanding the geometric series is crucial.
The geometric series formula, for a simplified case with a common ratio of less than 1, is given by:

• First term: 1
• Second term: x
• Third term: x²
• Fourth term: x³, and so on

These terms follow the pattern:\[ 1 + x + x^2 + x^3 + \ldots \]This series converges when the absolute value of the common ratio is less than 1, \( |x| < 1 \). This condition ensures that the sum of an infinite number of terms is finite.
This property makes geometric series particularly useful for expanding functions into infinite series around a point, like in our exercise with the Taylor series expansion.

Series Expansion

Series expansion is a technique used in mathematics to write a complex function as a sum of simpler terms. Often, these simpler terms are polynomials, which makes them easier to work with in analysis and computation. The Taylor series is a common method of series expansion.
Expanding functions into a series allows mathematicians and scientists to approximate values and analyze behaviors near specific points.
In our exercise, the function \( \frac{1}{a-r} \) is expanded using a substitution to express it in terms of a simpler geometric series format. Here's the transformation:

• Start with \( \frac{1}{a-r} \)
• Use substitution: \( r = a \cdot x \) and \( x = \frac{r}{a} \)
• Now, we have \( \frac{1}{a(1-x)} \)

The next step is to expand this using the geometric series formula:\[ \frac{1}{1-x} = 1 + x + x^2 + x^3 + \ldots \]This approach allows complex functions to be rewritten as polynomials, which can significantly simplify the analysis and calculations.

Algebraic Manipulation

Algebraic manipulation involves rearranging and simplifying expressions to solve problems or get them into a desired form. In the context of series expansion, it is often necessary to perform algebraic manipulations to set up expressions for effective series application.
For the given exercise, manipulation was essential to convert \( \frac{1}{a-r} \) into a suitable form for expansion:

• Substitute terms to express the function in terms of \( \frac{r}{a} \) as \( \frac{1}{a(1-x)} \)
• Factor out constants, in this case, \( \frac{1}{a} \)
• The process simplifies each term: \( 1, \frac{r}{a}, \left(\frac{r}{a}\right)^2, \left(\frac{r}{a}\right)^3, \ldots \)
• Multiply each term by \( \frac{1}{a} \) to reflect the original function's form

Algebraic manipulation is a crucial step that supports the transition from a complex function to a simple, manageable series. It bridges the gap between abstract mathematical concepts and practical implementation.