Question

Find power series representations centered at 0 for the following functions using known power series. $$f(x)=\frac{2 x}{\left(1+x^{2}\right)^{2}}$$

Short answer

Answer: The power series representation of the function is \(f(x)=\sum_{n=0}^{\infty}\binom{-2}{n}\left(-\frac{1}{x^2}\right)^n\).

Step-by-step solution

Identify the geometric series formula

The geometric series formula is given by: $$\frac{1}{1-x} = \sum_{n=0}^{\infty} x^n$$ Our goal is to manipulate the given function to resemble this formula.

Manipulate the given function to resemble the geometric series formula

We begin by dividing both the numerator and denominator of the given function by \(x^2\) to manipulate it: $$f(x)=\frac{2 x}{\left(1+x^{2}\right)^{2}} = \frac{2}{\left(1+\frac{1}{x^2}\right)^{2}}$$ Now, the function is in the form: $$f(x)=\frac{2}{(1+\frac{1}{x^2})^2} \Rightarrow f(x)=2\frac{1}{(1+\frac{1}{x^2})^2}$$

Apply the geometric series formula

We can now rewrite the function \(f(x)\) as a geometric series by letting \(y = \frac{1}{x^2}\): $$f(x)=2\frac{1}{(1+y)^2}$$ By applying the geometric series formula, we have: $$f(x)=\sum_{n=0}^{\infty}\binom{-2}{n}(-y)^n$$

Replace \(y\) with \(\frac{1}{x^2}\)​

Now that we have our function in the form of a geometric series, we must replace \(y\) with \(\frac{1}{x^2}\): $$f(x)=\sum_{n=0}^{\infty}\binom{-2}{n}\left(-\frac{1}{x^2}\right)^n$$ There you have it, the power series representation for the given function, centered at 0, is: $$f(x)=\sum_{n=0}^{\infty}\binom{-2}{n}\left(-\frac{1}{x^2}\right)^n$$

Key concepts

Geometric Series

The geometric series is a fundamental concept in mathematics where each term is a constant multiple of the previous term. The simplest form of a geometric series is represented as \( \frac{1}{1-x} = \sum_{n=0}^{\infty} x^n \). This formula describes an infinite series that converges when the absolute value of \( x \) is less than 1.

When dealing with problems like finding a power series representation, you often want to manipulate the given function to resemble this form. This allows for easier summation and understanding of the series behavior. In our problem, the objective was to transform \( f(x) = \frac{2x}{(1+x^2)^2} \) into something that looks like a geometric series. By adjusting the terms, you can use the formula to express the function as an infinite sum.

• A geometric series helps in simplifying complex functions by breaking them down into sums of simpler polynomials.
• This is because calculus operations like differentiation and integration can be easily applied term by term.
Geometric series expansions are particularly useful in solving problems involving fractions and functions that can be rewritten using simple manipulations and substitutions. In this exercise, knowing how to use a geometric series was crucial for finding the representation.

Binomial Theorem

The Binomial Theorem is a powerful tool for expanding expressions that are raised to a power. It provides the formula for expanding \((1 + x)^n\) into a series and is expressed as: \[ (1 + x)^n = \sum_{k=0}^{n} \binom{n}{k} x^k \] Here, \( \binom{n}{k} \) are the binomial coefficients, and they tell you how many ways to choose \( k \) items from \( n \) items.

In the context of our problem, we apply a form of the binomial theorem to extend the geometric series for a fraction that doesn’t have integer exponents, specifically \((1 + \frac{1}{x^2})^{-2}\). In such cases, the binomial theorem allows us to handle negative and non-integer exponents by using \( \binom{-2}{n} \).

• This type of expansion is particularly handy when dealing with rational functions.
• It can convert these functions into an easier format for integration or differentiation, as seen in power series.
The binomial theorem not only accommodates standard expansions but also those with negative powers, as shown when transforming our function into a series.

Calculus Series Expansion

Calculus series expansions, such as Taylor or Maclaurin series, are a type of power series used to represent functions. They revolve around expressing a function as an infinite sum of its derivatives evaluated at a specific point. For a function that is centered at 0, this is referred to as a Maclaurin series, expressed by: \[ f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} x^n \] This allows functions to be approximated by polynomials, providing a simple way to understand their behavior near the expansion point.

In our exercise, after transforming the function with the geometric and binomial approaches, the resulting series representation resembles the series expansion by focusing on the behavior around zero. The series expansion provides insights into both the local and global behavior of the function:

• These expansions are particularly valuable in approximating difficult-to-calculate functions within a given range.
• They are foundational for error approximation, as each additional term provides a more accurate representation of the function.
Calculus series expansions merge algebraic manipulation and calculus concepts to build effective models of functions. Understanding them enriches your calculus toolkit allowing you to deconstruct and reconstruct functions with precision.