Problem 14 - Essential Calculus

Question

Find a power series for the function, centered at \(c,\) and determine the interval of convergence. $$ f(x)=\frac{4}{4+x^{2}}, \quad c=0 $$

Step-by-Step Solution

Answer

The power series representation for the given function centered at 0 is \( \sum_{n=0}^{\infty}(-1)^n \frac{x^{2n}}{4^n} \). The interval of convergence is \(-2 < x < 2\).

Step 1: Rewrite the function

Start by rewriting the given function as: \( f(x)=\frac{4}{4}\times\frac{1}{1+\frac{x^{2}}{4}} \). This can be further simplified to \( f(x)=\frac{1}{1-(\frac{-x^{2}}{4})} \). This is in the form of the geometric series \( \frac{1}{1-u} \) where \( u=-\frac{x^{2}}{4} \).

Step 2: Generate the Power Series

For a geometric series \( \frac{1}{1-u} \), the power series representation is \( \sum_{n=0}^{\infty}u^n \). We substitute \( u=-\frac{x^{2}}{4} \) to this formula and obtain the power series: \( \sum_{n=0}^{\infty} (-1)^n (\frac{x^{2n}}{4^n}) \). This can be simplified to: \( \sum_{n=0}^{\infty}(-1)^n \frac{x^{2n}}{4^n} \). This is the power series centered at 0 for the given function.

Step 3: Find the Interval of Convergence

For a geometric series, the part of the series \(u=-\frac{x^{2}}{4}\) must be less than 1 in magnitude for the series to converge. That is, \( |-x^2/4| < 1 \). Solving this inequality, we get \(-4 < x^2 < 4\), which simplifies to \(-2 < x < 2\). Therefore, the interval of convergence is \(-2 < x < 2\).

Key Concepts

Interval of Convergence

The interval of convergence is crucial when dealing with power series. It specifies the range of values for which the power series converges to the function it's meant to represent.
Determining this interval ensures that the series expression is valid and provides meaningful results.
In this problem, we encounter a necessity to find where the series converges by solving the inequality derived from the geometric representation. Specifically:

• We find that for a geometric series, the term inside \( (-\frac{x^2}{4}) \) must have a magnitude less than one.
• This leads us to solve \(|-x^2/4| < 1\).
• Ultimately, simplifying gives us \(-2 < x < 2\).

When working with power series, always begin by analyzing the term under consideration and the condition for convergence. Solve this inequality to determine the interval over which the series sum is equal to the function.Thus, in this exercise, the interval is \(-2 < x < 2\).

Geometric Series

A geometric series is a powerful tool in mathematics, often allowing complex functions to be expressed in a simpler, infinite sum form.
It's based on the series formula \( \frac{1}{1-u} \), where \( u \) is a constant ratio.
For the given function, the transformation into a geometric series involves a few key steps:

• Reformat the original function to resemble \( \frac{1}{1-u} \), where \( u = -\frac{x^2}{4} \).
• Recognize that this form allows the use of the geometric series expansion, \( \sum_{n=0}^{\infty} u^n \).
• Substitute \( u \) with \( -\frac{x^2}{4} \) to get the power series.

The beauty of geometric series lies in its ability to easily convert functions into manageable form. It's essential to note here that such conversions only hold within the interval where \( |u| < 1 \).This approach neatly expresses the given function as an infinite series of powers, which can then be manipulated and studied further.

Function Centered at a Point

When dealing with power series, the notion of a series being 'centered' at a specific point is essential. It refers to the value around which the series is expanded, often denoted by \( c \).
In some contexts, this is also called the 'point of expansion.'
In this exercise, we center the function at \( c = 0 \):

• A power series centered at a point \( c \), is expressed as \( \sum_{n=0}^{\infty} a_n(x-c)^n \).
• Our function transforms into a series where each term features powers of \( (x-0) \) or simply \( x \).
• This ensures the power series is convenient for replacements and computation about the point \( x=0 \).

The choice of center often simplifies the algebra and makes it easier to understand the behavior of the series around that specific value. In this problem, centering at zero simplifies \( x-c = x \, \) allowing a direct substitution and leading to smoother calculations.Choosing an appropriate center can significantly influence the usefulness and simplicity of the resulting power series.