Problem 62 - Thomas Calculus

Question

Use the Taylor series for 1\(/\left(1-x^{2}\right)\) to obtain a series for 2\(x /\left(1-x^{2}\right)^{2}.\)

Step-by-Step Solution

Answer

The series is \( \sum_{n=1}^{\infty} 4n x^{2n} \).

Step 1: Identify the Taylor series for \( \frac{1}{1-x^2} \)

We start by finding the Taylor series for the function \( \frac{1}{1-x^2} \). The geometric series formula states that for \(-1 < x < 1\), \( \frac{1}{1-x} = \sum_{n=0}^{\infty} x^n \). Substitute \(x^2\) for \(x\) in this series:\[ \frac{1}{1-x^2} = \sum_{n=0}^{\infty} x^{2n}. \]

Step 2: Derive the series for \( \frac{1}{(1-x^2)^2} \)

We take the derivative of the series \( \frac{1}{1-x^2} \) with respect to \(x\) to find the series for \( \frac{1}{(1-x^2)^2} \). The derivative of \( \frac{1}{1-x^2} \) is \( \frac{2x}{(1-x^2)^2} \).Differentiating the series term by term:\[ \frac{d}{dx} \left( \sum_{n=0}^{\infty} x^{2n} \right) = \sum_{n=1}^{\infty} 2n x^{2n-1}. \]Thus, \( \frac{1}{(1-x^2)^2} = \sum_{n=1}^{\infty} 2n x^{2n-1}. \)

Step 3: Multiply the series by \( 2x \)

To find the series for \( \frac{2x}{(1-x^2)^2} \), we multiply the series obtained from Step 2 by \(2x\):\[ 2x \cdot \sum_{n=1}^{\infty} 2n x^{2n-1} = \sum_{n=1}^{\infty} 4n x^{2n}. \]Now we have the series representation for \( \frac{2x}{(1-x^2)^2} \), which is \( \sum_{n=1}^{\infty} 4n x^{2n}. \)

Key Concepts

Geometric Series

A geometric series is one of the fundamental concepts in mathematics.Understanding this helps unravel the Taylor series, like in the exercise given.A geometric series is a sum of terms where each term is a constant multiple (called the common ratio) of the previous.For instance, the series: \( a, ar, ar^2, ar^3, \dots \) forms a geometric series.This type of series can be summed using the formula:

• \( S = \frac{a}{1 - r} \)

where \(a\) is the first term and \(r\) is the common ratio.
In the context of the Taylor series, we use the geometric series formula as follows:

• For \(-1 < x < 1\), the geometric series for \( \frac{1}{1-x} \) is \( \sum_{n=0}^{\infty} x^n \)

By substituting \(x^2\) for \(x\), we achieve the series for \( \frac{1}{1-x^2} \):

• \( \frac{1}{1-x^2} = \sum_{n=0}^{\infty} x^{2n} \)

This manipulation is crucial because it lays the foundation for more complex series manipulations.

Derivative of a Series

Once we have a series, finding its derivative can offer deeper insights and new formulas.In this exercise, the series \( \frac{1}{1-x^2} = \sum_{n=0}^{\infty} x^{2n} \) is differentiated with respect to \(x\).By doing so, we aren't just developing a new series, but also applying calculus to see how the function behaves.
The differentiation is straightforward:

• Differentiate term by term to get \( \sum_{n=1}^{\infty} 2n x^{2n-1} \)

This gives the series for the derivative \( \frac{d}{dx}( \frac{1}{1-x^2}) = \frac{2x}{(1-x^2)^2} \).
In essence, taking the derivative transforms our original problem into a new form, helping us create a series for a more complex function like \( \frac{1}{(1-x^2)^2} \).This is key in developing the new series representation through calculus.

Power Series Representation

Power series representation is a way of expressing a function as an infinite sum of terms with powers of a variable.This concept is critical in the exercise as it breaks down functions into manageable series.Having derived \( \frac{1}{(1-x^2)^2} \) as \( \sum_{n=1}^{\infty} 2n x^{2n-1} \), the next step is crucial.
We find the power series representation for \( \frac{2x}{(1-x^2)^2} \) by multiplying our series by \(2x\).This process involves only simple multiplication:

• Multiply each term by \(2x\) to get \( \sum_{n=1}^{\infty} 4n x^{2n} \)

Power series representation allows functions to be expressed in an infinite sum, facilitating easy computation and estimation.
Functions distilled into a power series are more straightforward to explore, giving insights into their behavior at different points. This makes power series representation a powerful tool for mathematicians and students alike.