Question

In the following exercises, find the power series for the given function using term-by-term differentiation or integration. $$ f(x)=\frac{x}{\left(2+x^{2}\right)^{2}} $$

Short answer

The power series of \( f(x) \) is \( \frac{x}{4} - \frac{x^3}{8} + \frac{3x^5}{64} - \ldots \).

Step-by-step solution

Understand the Function

The given function is \( f(x) = \frac{x}{(2 + x^2)^2} \). Our goal is to represent this function as a power series, possibly using a known series as a starting point.

Express Denominator in Standard Form

The denominator \( (2 + x^2)^2 \) suggests that using a geometric or binomial series expansion might be useful. Recognize that \( \frac{1}{(2 + x^2)^2} \) can be addressed by first simplifying \( \frac{1}{1 - (-x^2/2)} = 1 + \frac{x^2}{2} + \frac{(x^2)^2}{4} + \ldots \).

Identify Related Series

The strategy is to first find the series for \( \frac{1}{2 + x^2} \):Use the geometric series \[ \frac{1}{1-u} = \sum_{n=0}^{\infty} u^n \] where \( u = -\frac{x^2}{2} \) and expand to get:\[ \frac{1}{2 + x^2} = \frac{1}{2} (1 - \frac{x^2}{2} + (\frac{x^2}{2})^2 - \ldots ) \].

Derive the Series for the Square

To find the series of \( \frac{1}{(2 + x^2)^2} \), differentiate the power series:\[ \frac{1}{(2 + x^2)^2} = \frac{d}{dx} \left( \frac{-1}{x} \frac{1}{2 + x^2} \right) \].Apply term-by-term differentiation to each term of the series obtained from the previous step.

Multiply by x

As \( f(x) = \frac{x}{(2 + x^2)^2} \), multiply the resulting series from Step 4 by \( x \) to obtain the complete series for \( f(x) \):\( f(x) = x \cdot \sum_{n=0}^{\infty} c_n x^n = \sum_{n=0}^{\infty} c_n x^{n+1} \). Where each \( c_n \) comes from the differentiated series.

Write the Power Series

Substitute the coefficients \( c_n \) into the expanded series and simplify if possible, representing \( f(x) \) as a power series centered at zero: \[ f(x) = x \left( \frac{1}{4} - \frac{x^2}{8} + \frac{3x^2}{64} - \ldots \right) \].Thus, the power series is:\[ f(x) = \frac{x}{4} - \frac{x^3}{8} + \frac{3x^5}{64} - \ldots \]

Key concepts

Term-by-Term Differentiation

Term-by-term differentiation allows us to differentiate a function's power series by working with each term separately. This technique is especially useful when dealing with power series expansions of functions that cannot be easily differentiated as a whole.
When applying term-by-term differentiation, we look at the series representation of a function and take the derivative of each term individually.

• Start with the power series representation of a function.
• Differentiate each term in the series separately.
• Ensure the new series still converges at the original points.

This technique aligns well with our function f(x) , as we initially found the corresponding power series and then targeted each component as a distinct term. This method allows us to manage complex functions, like those with rational expressions, smoothly.

Geometric Series

The concept of a geometric series is central to simplifying and expressing functions as power series. A geometric series takes a form like: \[\frac{1}{1-u} = \sum_{n=0}^{\infty} u^n\]
where u is a constant ratio between successive terms. When translating a function into a geometric series, we need to strategically choose and manipulate u so that it helps us approximate the function adequately.

• Identify a suitable u value, often replacing parts of the function to fit the geometric series structure.
• Expand the expression using the geometric series formula.
• Conveniently tailor the series to match the specific needs of the problem.

Here, u was chosen as \(-x^2/2\) to fit the denominator into a geometric series. This helped us derive the initial power series expansion necessary for further operations.

Differentiation Techniques

Differentiation techniques in power series expansions help tackle complex derivatives step by step. Understanding multiple ways to take derivatives is beneficial, especially for series that may not look straightforward.
Before proceeding, it's important to remember essential differentiation rules:

• Product Rule: For functions multiplied together, \((uv)' = u'v + uv'\).
• Chain Rule: For nested functions, \((f(g(x)))' = f'(g(x))g'(x)\).
• Term-by-term: Simply deriving each term independently as needed.

In the given solution, differentiating \(\frac{1}{(2 + x^2)^2}\) was essential. By carefully applying the differentiated rules, the specific term-by-term derivation allowed the series to be explored more deeply and accurately, moving us closer to the expanded power series for f(x) .

Binomial Series Expansion

The binomial series expansion is an invaluable tool for simplifying expressions that involve powers. Based on the formula: \[ (1 + x)^n = \sum_{k=0}^{\infty} \binom{n}{k} x^k \] this approach expands a power of a binomial into a series.
This formula can be customized to accommodate fractional or negative powers, allowing a vast range of functions to be broken down into manageable parts.

• Identify the binomial pattern: Find components that can be rewritten in the form of \( (1 + x)^n \).
• Apply the appropriate formula: Use the expansion formula and plug in the necessary values.
• Tailor the expansion: Adjust the series for negative or fractional exponents as needed.

The original function had a denominator suggestive of a binomial form (\((2 + x^2)^2)\), guiding us to derive an adapted binomial series. This adjustment made it feasible to decompose seemingly complex denominators, leading to a straightforward series outcome.