Question

Represent the function \(f(x)=\frac{x^{3}}{2-x}\) using a power series and find the interval of convergence.

Short answer

The power series is \(\sum_{n=0}^{\infty} \frac{x^{n+3}}{2^{n+1}}\) with interval of convergence \((-2, 2)\).

Step-by-step solution

Rewrite the Function

First, express the function in a form suitable for geometric series expansion. Note that power series expansions work well with functions of the form \( \frac{1}{1 - u} \). We rewrite \( f(x) = \frac{x^3}{2-x} \) as \( x^3 \cdot \frac{1}{2-x} \).

Geometric Series Formula

Recall the geometric series formula, \( \frac{1}{1-u} = \sum_{n=0}^{\infty} u^n \), which is valid for \( |u| < 1 \). We want to express \( \frac{1}{2-x} \) in this form.

Manipulate the Denominator

Rewrite the denominator \( \frac{1}{2-x} = \frac{1}{2(1 - \frac{x}{2})} \). Here, \( u = \frac{x}{2} \).

Expand Using the Series Formula

Apply the geometric series expansion formula: \[ \frac{1}{2-x} = \frac{1}{2} \sum_{n=0}^{\infty} \left(\frac{x}{2}\right)^n. \] This simplifies to \[ \sum_{n=0}^{\infty} \frac{x^n}{2^{n+1}}. \]

Final Expansion

Multiply each term by \( x^3 \) to find the power series for \( f(x) \): \[ f(x) = x^3 \cdot \sum_{n=0}^{\infty} \frac{x^n}{2^{n+1}} = \sum_{n=0}^{\infty} \frac{x^{n+3}}{2^{n+1}}. \]

Find the Interval of Convergence

Since the geometric series \( \sum_{n=0}^{\infty} \left( \frac{x}{2} \right)^n \) converges for \( \left| \frac{x}{2} \right| < 1 \), the series for \( f(x) \) converges when \( |x| < 2 \). Thus, the interval of convergence is \((-2, 2)\).

Key concepts

Interval of Convergence

The interval of convergence is a crucial concept when dealing with power series. It tells us the set of values for which the series converges to a function. In other words, within this interval, the series provides meaningful and predictable output.

To find the interval of convergence for a power series, examine the absolute value condition derived from the geometric series formula. Typically, the series \( \sum_{n=0}^{\infty} a_n (x-c)^n \) converges when \(|x-c| < R\), where \(R\) is the radius of convergence and \(c\) is the center of the series. However, in exercises like the one given, you often start from a specific function.

In this case, the transformation used the geometric series concept to modify the denominator as \( \frac{1}{2-x} = \frac{1}{2(1 - \frac{x}{2})} \), identifying \( u = \frac{x}{2} \). The geometric series converges when \(|u| < 1\). That translates to \( |\frac{x}{2}| < 1 \), which simplifies to \(|x| < 2\).

So, the interval of convergence for the power series representation of \(f(x)\) is \((-2, 2)\). This range means that any \(x\) value within -2 and 2 (not including -2 and 2 themselves) will make the series converge neatly without diverging.

Geometric Series

The geometric series is one of the simplest yet most powerful types of series. It provides a way to represent functions in terms of sums. A geometric series is typically in the form of \( \frac{1}{1-u} = \sum_{n=0}^{\infty} u^n \), converging for \(|u| < 1\).

The power of a geometric series lies in its ability to represent certain functions that otherwise seem difficult to break down. Notably, it's instrumental in expressing repeating decimal expansions and rational functions as infinite series.

In our original exercise, we expressed the function \( \frac{x^3}{2-x} \) using the geometric series concept. By rewriting the denominator as \( \frac{1}{2(1-\frac{x}{2})} \), we utilized the geometric series \( \frac{1}{1-u} = \sum_{n=0}^{\infty} u^n \) to break it down. This strategic manipulation made it simple to express the function as a sum of terms, which is much easier to integrate or differentiate if needed.

Understanding the geometric series formula and its applications can considerably simplify many calculus problems, especially when working with convergence and function representation.

Function Representation in Series

Representing functions using power series is a powerful calculus technique. It allows for the approximation of complex functions through more manageable series forms. Such representations enable mathematicians and scientists to perform easier calculations and derive approximate solutions for real-world problems.

To represent a function by a series, we often start by expressing it in terms that match known series formulas. The geometric series formula \( \frac{1}{1-u} = \sum_{n=0}^{\infty} u^n \) is frequently helpful, as seen in our exercise where \( f(x) = \frac{x^3}{2-x} \) was expanded through series manipulation.

The approach often involves identifying a suitable substitution or transformation to match the function with a series form. Once in a suitable series format, we perform algebraic manipulations such as factoring or distribution to express the entire function as a power series.

This expansion results in a series that can be easily examined for convergence and used in calculations. By multiplying the series by appropriate terms, like the \(x^3\) in the example, we obtain a complete, expanded form. This makes analysis and computation significantly simpler and opens up a host of tools in calculus to further explore and exploit these functions. Understanding these techniques is invaluable for anyone looking to master advanced calculus concepts.