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 for \(\frac{2x}{(1-x^2)^2}\) is \(4x + 8x^3 + 12x^5 + \cdots\).

Step 1: Identify the Taylor Series

The function we have is \(\frac{1}{1-x^2}\). The Taylor series expansion for \(\frac{1}{1-x}\) about \(x=0\) is \(1 + x + x^2 + x^3 + \cdots\). Replacing \(x\) with \(-x^2\) gives us the series for \(\frac{1}{1-x^2}\): \[1 + x^2 + x^4 + x^6 + \cdots.\]

Step 2: Differentiate the Series

We need the series for \(\frac{2x}{(1-x^2)^2}\). Recognizing \(\frac{d}{dx}\left(\frac{1}{1-x^2}\right) = \frac{2x}{(1-x^2)^2}\), differentiate the series \(1 + x^2 + x^4 + x^6 + \cdots\) term by term: \[\frac{d}{dx}(1) = 0, \quad \frac{d}{dx}(x^2) = 2x, \quad \frac{d}{dx}(x^4) = 4x^3, \quad \cdots\] This gives us: \[0 + 2x + 4x^3 + 6x^5 + \cdots.\]

Step 3: Multiply by Constant Factor

Our goal is \(\frac{2x}{(1-x^2)^2}\), which involves multiplying the derivative series by 2: \[2\cdot(2x + 4x^3 + 6x^5 + \cdots)\] This results in: \[4x + 8x^3 + 12x^5 + \cdots.\] So, the series expansion for \(\frac{2x}{(1-x^2)^2}\) is \[\sum_{n=0}^{\infty} 4(n+1)x^{2n+1}.\]

Key Concepts

Series Differentiation

Differentiating a series means taking the derivative of each term in the series. When we want to differentiate a function expressed as a series, we simply apply the differentiation rules to each term individually. This is especially useful when working with Taylor or power series, since they break complex functions into simpler components.
When differentiating a power series like \[1 + x^2 + x^4 + x^6 + \cdots,\]we perform the differentiation term by term:

• The derivative of the constant \(1\) is \(0\).
• The derivative of \(x^2\) is \(2x\).
• The derivative of \(x^4\) is \(4x^3\).

This stepwise differentiation helps transform the series to a form that represents the derivative of the function, making complex calculus tasks more manageable.

Power Series Expansion

Power series expansion is a way of expressing a function as an infinite sum of terms, calculated from the values of its derivatives at a single point. This form is similar to a polynomial but with infinitely many terms.
For instance, the series expansion of \[\frac{1}{1-x} = 1 + x + x^2 + x^3 + \cdots\]can be modified for different functions by changing the variable or its powers. By replacing \(x\) with \(-x^2\), we transform the series to suit our new function, \(\frac{1}{1-x^2}\), yielding:\[1 + x^2 + x^4 + x^6 + \cdots.\]This process leverages calculus to reduce complex expressions into simpler, infinite series representations, useful in approximating functions and solving equations in applied mathematics.

Calculus Series

Calculus series involve using calculus techniques to manipulate and understand series. A common method is to use Taylor or Maclaurin series to approximate and analyze functions in their infinite series form.
When tackling problems like finding the series for \[\frac{2x}{(1-x^2)^2},\]we follow structured steps:

• Identify a known series, like the Taylor series for \(\frac{1}{1-x^2}\).
• Differentiate the series, handling each term individually.
• Apply necessary multipliers to match the required function form.

This approach ensures all calculations maintain consistency with calculus principles and helps convert cumbersome functions into sums of simpler terms, fostering deeper understanding and calculation efficiency.