Problem 1 - Calculus: One and Several Variables

Question

Expand \(f(x)\) in powers of \(x,\) basing your calculations on the geometric series $$\frac{1}{1-x}=1+x+x^{2}+\cdots+x^{n}+\cdots$$ $$f(x)=\frac{1}{(1-x)^{2}}$$

Step-by-Step Solution

Answer

The expanded power series representation of the function \(f(x)=\frac{1}{(1-x)^{2}}\) is given by: \[f(x) = 1 + 2x + 3x^{2}+\cdots+nx^{n-1}+\cdots\] This was derived by differentiating the geometric series formula \(\frac{1}{1-x}=1+x+x^{2}+\cdots+x^{n}+\cdots\) and equating the result with \(f(x)\).

Step 1: Differentiate the given geometric series

We are given the geometric series formula \(\frac{1}{1-x}=1+x+x^{2}+\cdots+x^{n}+\cdots\). To find the expansion of \(f(x)=\frac{1}{(1-x)^{2}}\), we'll first differentiate both sides of the geometric series equation with respect to x: \[\frac{d}{dx}(\frac{1}{1-x}) = \frac{d}{dx}(1+x+x^{2}+\cdots+x^{n}+\cdots)\]

Step 2: Evaluate the derivative on both sides

Now let's find the derivative on both sides of the equation: \[\frac{1}{(1-x)^{2}} = \frac{d}{dx}(1) + \frac{d}{dx}(x) + \frac{d}{dx}(x^{2}) + \cdots + \frac{d}{dx}(x^{n}) + \cdots\] \[\frac{1}{(1-x)^{2}} = 0 + 1 + 2x + \cdots + nx^{n-1} + \cdots\]

Step 3: Equate the given function to the resultant power series

We have found a power series representation for the derivative of the given geometric series formula, which turns out to be equal to \(f(x)\). Thus, we have: \[f(x) = \frac{1}{(1-x)^{2}} = 1 + 2x + 3x^{2}+\cdots+nx^{n-1}+\cdots\]

Step 4: Conclusion

The expanded power series representation of the function \(f(x)=\frac{1}{(1-x)^{2}}\) is: \[f(x) = 1 + 2x + 3x^{2}+\cdots+nx^{n-1}+\cdots\]

Key Concepts

Geometric Series

A geometric series is a sequence 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 formula, \( \frac{1}{1-x} = 1 + x + x^2 + x^3 + \cdots \), the series expands indefinitely as long as \( |x| < 1 \). This means it's an infinite series where each term gets progressively smaller if \( x \) is between -1 and 1.
This concept is foundational because it lets us break down complex functions into simpler parts to easily analyze or manipulate them. Instead of handling \( \frac{1}{1-x} \) as a whole, you deal with a series of terms like 1, \( x \), \( x^2 \), and so forth.
Understanding geometric series is key because many functions can be expressed as or relate to series. This method is highly effective for calculating approximations or analyzing behaviors of functions over an interval.

Differentiation

Differentiation is a core concept in calculus that involves finding the derivative of a function. The derivative represents the function's rate of change concerning one of its variables. For our exercise, differentiating the function \( \frac{1}{1-x} \) helps us find its derivative's power series expansion.
When you differentiate the whole geometric series, \( 1 + x + x^2 + \cdots \), you perform the operation term by term. For each term, it's like you perform a separate mini-calculation:

• The derivative of 1 is 0
• The derivative of \( x \) is 1
• The derivative of \( x^2 \) is \( 2x \)
• The derivative of \( x^n \) in general is \( nx^{n-1} \)

By differentiating, you transform the series into a new one. This gives rise to the series \( 1 + 2x + 3x^2 + \cdots \), which accurately represents the derivative of \( \frac{1}{(1-x)^2} \).
Understanding differentiation in this context reveals how to manipulate series to analyze more complicated functions.

Infinite Series Expansion

An infinite series expansion allows you to express a function as an infinite sum of terms. This approach is beneficial because it can simplify complex problems. In the context of power series, each term's coefficient is derived from a pattern, increasing the power of \( x \) incrementally.
For \( f(x) = \frac{1}{(1-x)^2} \), using differentiation, we find it can be written as the power series \( 1 + 2x + 3x^2 + \cdots \). Each term is derived from differentiating the geometric series. This power series is an infinite polynomial that continues indefinitely. Still, when used over the interval where \( |x| < 1 \), it offers a powerful method to understand and approximate \( f(x) \).
Infinite series expansions are critical in calculus and applied mathematics because they provide methods to approach functions that are difficult to handle directly. By understanding the power series, this exercise shows how expanded views of a function encapsulate essential behaviors and properties.