Question

Use a power series to represent each of the following functions \(f\). Find the interval of convergence. a. \(f(x)=\frac{1}{1+x^{3}}\) b. \(f(x)=\frac{x^{2}}{4-x^{2}}\)

Short answer

For (a), power series is \(\sum_{n=0}^{\infty} (-1)^n x^{3n}\), with interval \((-1, 1)\). For (b), power series is \(\sum_{n=0}^{\infty} \frac{x^{2n+2}}{4^n}\), with interval \((-2, 2)\).

Step-by-step solution

Identify the Geometric Series Formula

Recognize that we can use the geometric series formula, \( \frac{1}{1-r} = \sum_{n=0}^{\infty} r^n \), where \(|r|<1\). We will transform our given functions to fit this formula.

Rewrite Function a for Power Series

For function a, \( f(x) = \frac{1}{1+x^3} \), rewrite as \( \frac{1}{1-(-x^3)} \). Here, \( r = -x^3 \) and we can apply the geometric series.

Power Series for Function a

Apply the geometric series formula: \( \frac{1}{1+x^3} = \sum_{n=0}^{\infty} (-x^3)^n = \sum_{n=0}^{\infty} (-1)^n x^{3n} \).

Interval of Convergence for Series a

The series converges when \(|-x^3| < 1\) or \(|x^3| < 1\), which simplifies to \(|x| < 1\). Thus, the interval of convergence is \(-1 < x < 1\).

Rewrite Function b for Power Series

For function b, \( f(x) = \frac{x^2}{4-x^2} \), rewrite as \( \frac{x^2}{4(1-(x^2/4))} \). Factor \(4\) in the denominator to match the geometric series form, and set \( r = \frac{x^2}{4} \).

Power Series for Function b

Using the geometric series, \( \frac{1}{1-x^2/4} = \sum_{n=0}^{\infty} \left(\frac{x^2}{4}\right)^n \). Thus, \( f(x) = x^2 \sum_{n=0}^{\infty} \left(\frac{x^2}{4}\right)^n \), and simplifying gives \( \sum_{n=0}^{\infty} \frac{x^{2n+2}}{4^n} \).

Interval of Convergence for Series b

Series converges when \(|x^2/4| < 1\), leading to \(|x^2| < 4\) or \(|x| < 2\). Therefore, the interval of convergence is \(-2 < x < 2\).

Key concepts

Geometric Series

The geometric series is a fundamental mathematical concept that forms the basis for many applications, particularly in power series expansions. It is an infinite sum, represented by the formula:

• \( \frac{1}{1-r} = \sum_{n=0}^{\infty} r^n \)

Here, \(r\) is the common ratio, and the series only converges when \(|r| < 1\). This means the series will approach a specific value as the number of terms increases without bound.
Geometric series are prevalent because they provide a simple way to represent functions as power series. By rewriting a function to fit the geometric series form of \( \frac{1}{1-r} \), a seemingly complex function can be expressed as an infinite sum.
For instance, in problem (a), \( f(x) = \frac{1}{1+x^3} \) is rewritten to \( \frac{1}{1-(-x^3)} \), thus identifying \(r = -x^3\). This allows us to express the function as \( \sum_{n=0}^{\infty} (-1)^n x^{3n} \). In problem (b), we similarly transform \(f(x) = \frac{x^2}{4-x^2}\) for series expansion.

Interval of Convergence

The interval of convergence is an essential concept regarding power series. It defines the set of values for the variable \(x\) in which the series converges to the function it represents.
For a geometric series given by \(\sum_{n=0}^{\infty} r^n\), convergence demands that \(|r| < 1\). This restriction ensures that the terms of the series get smaller and approach zero as more terms are added, allowing the series to sum up to a finite value.
Considering function (a), \( f(x) = \frac{1}{1+x^3} \), rewriting it leads us to the sequence \( \sum_{n=0}^{\infty} (-x^3)^n \), where convergence is dictated by \(|x^3| < 1\). Solving this inequality results in the interval \(-1 < x < 1\).
Similarly, function (b) \( f(x) = \frac{x^2}{4-x^2} \), is transformed, and the series converges within \(|x| < 2\), thus yielding an interval of \(-2 < x < 2\).
This interval of convergence is crucial because it establishes the exact range of \(x\) values for which the power series represents the actual function.

Function Transformation

Function transformation is a powerful technique used to rewrite functions in a form that facilitates easier computation or representation, like turning them into a geometric series.
It involves a systematic alteration of a function’s form to reveal new properties or simplify computations. The goal is often to align the function with known mathematical forms like the geometric series.
In the exercise, the functions \( f(x) = \frac{1}{1+x^3} \) and \( f(x) = \frac{x^2}{4-x^2} \) are transformed to fit the model of \( \frac{1}{1-r} \). For function (a), this transformation led to identifying \(r = -x^3\), allowing its power series expansion as \( \sum_{n=0}^{\infty} (-1)^n x^{3n} \).
Similarly, function (b) was adjusted to have \(r = \frac{x^2}{4} \), thereby enabling the representation \( x^2 \sum_{n=0}^{\infty} \left(\frac{x^2}{4}\right)^n \).
By transforming these functions, we gain insight into their behavior and make it easier to analyze them using power series.