Problem 8 - Advanced Engineering Mathematics

Question

In Problems 7 and 8 , the given function is analytic at \(x=0\). Find the first four terms of a power series in \(x\). Perform the long division by hand or use a CAS, as instructed. Give the open interval of convergence. $$ \frac{1-x}{2+x} $$

Step-by-Step Solution

Answer

The first four terms of the power series are \( \frac{1}{2} - \frac{3}{4}x + \frac{7}{8}x^2 - x^3 + \cdots \). Interval of convergence: \(|x| < 1\).

Step 1: Express the function in a more manageable form

Given the function \( \frac{1-x}{2+x} \), notice that it doesn't readily lend itself to a power series expression about \( x = 0 \). We begin by rewriting the denominator in a form that facilitates expansion: rewrite it as \( 2(1 + \frac{x}{2}) \). The function then becomes \( \frac{1-x}{2\left(1+\frac{x}{2}\right)} \).

Step 2: Split the fraction using partial fraction method

Rewriting \( \frac{1-x}{2+x} \) in the form \( \frac{1-x}{2(1+\frac{x}{2})} = \frac{1}{2}(1-x)(1-\frac{x}{2})^{-1} \) gives us a form suitable for a geometric series expansion.

Step 3: Expand the expression as a geometric series

Using the geometric series formula \( (1-u)^{-1} = 1 + u + u^2 + u^3 + \ldots \) for \( |u| < 1 \), let \( u = -\frac{x}{2} \). Therefore, the expanded series becomes:\[ 1 - \frac{x}{2} + (\frac{x}{2})^2 - (\frac{x}{2})^3 + \cdots \]This gives us the series: \[ 1 - \frac{x}{2} + \frac{x^2}{4} - \frac{x^3}{8} + \cdots \]

Step 4: Multiply each term by \( \frac{1-x}{2} \) to find the power series of the original function

Now, multiply each term by \( \frac{1-x}{2} \). First expand \( 1-x \) as a series:\[ 1 - x = 1 - x \]Multiply:\[ \frac{1}{2} \cdot (1 - x)(1 - \frac{x}{2} + \frac{x^2}{4} - \frac{x^3}{8}) \]Expanding and collecting terms gives:\[ \frac{1}{2} \cdot \left[(1-x) - \frac{x}{2}(1-x) + \frac{x^2}{4}(1-x) - \frac{x^3}{8}(1-x)\right] \]\[ = \frac{1}{2} \cdot (1 - x + x^2 - x^3) - \frac{x}{4} + \frac{x^2}{8} - \frac{x^3}{16} \]\[ = \frac{1}{2} - \frac{3}{4}x + \frac{7}{8}x^2 - x^3 + \cdots \]

Step 5: Identify interval of convergence

For the geometric series \((1-x)^{-1}\), the series converges for \(|x| < 1\). Since the transformations applied (scalars and additional terms) do not affect the point of convergence at \(x=0\), this remains the interval of convergence of the original function.

Key Concepts

Power Series

A power series is essentially an infinite sum of terms, each consisting of a coefficient multiplied by a power of a variable, typically written as:

• \( a_0 + a_1x + a_2x^2 + a_3x^3 + \cdots \)

For analytic functions, a power series can be used to represent these functions around a certain point, often called the center of the series, which in many cases is \( x = 0 \). This makes them incredibly useful for approximating complicated functions with polynomials.

The power series representation of a function can provide insight into the function's behavior. It allows us to approximate functions and perform operations like differentiation and integration term by term, which can greatly simplify calculations. In essence, power series transform challenging calculus problems into more manageable algebra problems.

Interval of Convergence

The interval of convergence is the range of values for which a power series converges to a finite value. In simpler terms, it's where the series makes sense and gives reliable approximations.

When we talk about power series, each series converges at different intervals based on its characteristics. To find this interval, we often use methods such as the ratio test or the root test. For the given function, after expanding it into a series:

• For the series derived from the geometric series expansion, it will converge for \(|x| < 1\).

This means within this range, substituting values of \( x \) into the series will yield finite and meaningful results. Understanding the interval of convergence is crucial as it defines where the power series is valid.

Geometric Series

A geometric series is a series where each term is a product of the previous one by a constant factor. The classic form of a geometric series looks like this:

• \( a + ar + ar^2 + ar^3 + \cdots \)

Where \( a \) is the first term, and \( r \) is the common ratio.

Such series converges if and only if the absolute value of the ratio \( |r| \) is less than one, that is, \(|r| < 1\). This classic property forms the basis for representing many functions as power series by manipulating them into a suitable form.

In our specific example, turning the denominator into a geometric series form allows us to easily express the given function as a power series through expansion, ultimately making analysis simpler.

Partial Fraction Decomposition

Partial fraction decomposition is a method used to express a rational function—the division of two polynomials—as the sum of simpler fractions. This technique simplifies integration and other operations by breaking down complex fractions.

By using partial fractions, you can convert a polynomial into a sum of fractions with lower degree denominators. It’s particularly handy when trying to integrate a seemingly complex rational function or when you want to work with simpler expressions.

• Consider the function \( \frac{1-x}{2+x} \). By expressing it in terms of simpler fractions, we set the stage for further expansion.
• This breaks the expression down and matches it to forms that are straightforward to handle in power series expansions, like the geometric series.

Partial fraction decomposition is a powerful tool in calculus for dealing with rational expressions, letting us transform them into a series-friendly format.