Problem 17 - Essential Calculus Early Transcendentals

Question

Find a power series representation for the function and determine the radius of convergence. $$ f(x)=\frac{x}{(1+4 x)^{2}} $$

Step-by-Step Solution

Answer

The power series is \( \sum_{n=1}^{\infty} n \cdot 4^n x^n \) with radius of convergence \( R = \frac{1}{4} \).

Step 1: Series Representation of Base Function

Start by considering the geometric series \( \frac{1}{1-u} = \sum_{n=0}^{\infty} u^n \) for \( |u| < 1 \). In our case, let \( u = 4x \) so the series becomes \( \sum_{n=0}^{\infty} (4x)^n = \sum_{n=0}^{\infty} 4^n x^n \).

Step 2: Differentiate the Series

Differentiate the series from Step 1 to account for the squared denominator: \( \frac{d}{dx} \left[ \frac{1}{1+u} \right] = \frac{1}{(1+u)^2} = \sum_{n=1}^{\infty} n(4x)^{n-1} \times 4 = \sum_{n=1}^{\infty} n \cdot 4^n x^{n-1} \).

Step 3: Multiply by x

Multiply the differentiated series by \( x \):\( f(x) = x \cdot \sum_{n=1}^{\infty} n \cdot 4^n x^{n-1} = \sum_{n=1}^{\infty} n \cdot 4^n x^n \).

Step 4: Determine Radius of Convergence

The series \( \sum_{n=0}^{\infty} 4^n x^n \), or its derivative, converges when \( |4x| < 1 \). Therefore, we have \( |x| < \frac{1}{4} \), making the radius of convergence \( R = \frac{1}{4} \).

Key Concepts

Geometric Series

A geometric series is a series with a constant ratio between successive terms. The classic form of a geometric series is \( \frac{1}{1-u} = \sum_{n=0}^{\infty} u^n \) where \( |u| < 1 \). This means that each term in the series is multiplied by the same factor, \( u \), to get the next term.
In the given exercise, the function \( \frac{x}{(1+4x)^2} \) is restructured by first understanding \( 1 + 4x \) as part of a geometric series and letting \( u = 4x \).
When you substitute \( u = 4x \) into the geometric series formula, the function \( \frac{1}{1-4x} \) becomes \( \sum_{n=0}^{\infty} (4x)^n \), or \( \sum_{n=0}^{\infty} 4^n x^n \). This step sets the stage for transforming the given problem into a series representation, simplifying the complexity of the original function using the properties of geometric series.

Differentiation of Series

To differentiate a power series, you simply differentiate each term of the series, treating the series like a polynomial. Given a series representation \( \sum_{n=0}^{\infty} a_n x^n \), its derivative is \( \sum_{n=1}^{\infty} n \cdot a_n x^{n-1} \).
In the original solution, the series \( \sum_{n=0}^{\infty} 4^n x^n \) was differentiated to mirror a change in the initial function's denominator from \( (1+4x) \) to \( (1+4x)^2 \). This entails deriving a series such that \( \frac{1}{(1+4x)^2} \) corresponds to the derivative \( \sum_{n=1}^{\infty} n \cdot 4^n x^{n-1} \), which manages the squared terms.
After obtaining this series for the denominator, it is then multiplied by \( x \) to form the series representation for \( f(x) \). Following differentiation, what's achieved is a new power series that accurately reflects the original function's conditions and alterations.

Radius of Convergence

The radius of convergence is a measure of the interval within which a power series converges to a limit. Specifically, it tells you how far from the center point \( x=0 \), the series will converge. This can be found using the formula \( R = \frac{1}{|L|} \), where \( L \) is the coefficient of \( x \) in the term raised to the nth power.
For the series \( \sum_{n=0}^{\infty} 4^n x^n \), the condition for convergence is \( |4x| < 1 \). Solving this inequality gives us \( |x| < \frac{1}{4} \), and thus, the radius of convergence is \( R = \frac{1}{4} \). This means the function behaves nicely and converges to a limit only when the values of \( x \) lie within this range, specifically between \( -\frac{1}{4} \) and \( \frac{1}{4} \). Understanding this concept is crucial because it defines the boundaries within which the power series representation is valid and reliable.