Problem 21 - Thomas Calculus

Question

Use power series operations to find the Taylor series at \(x=0\) for the functions in Exercises \(13-30\) . $$\frac{x^{2}}{1-2 x}$$

Step-by-Step Solution

Answer

The Taylor series for \( \frac{x^2}{1-2x} \) is \( \sum_{m=0}^{\infty} 2^{m-2} x^m \) starting effectively from \( x^2 \).

Step 1: Identify the Basic Series

Recognize that we need to find a Taylor series representation of the given function \( \frac{x^2}{1-2x} \). The function \( \frac{1}{1-2x} \) resembles the geometric series \( \frac{1}{1-r} = \sum_{n=0}^{\infty} r^n \) with \( r = 2x \).

Step 2: Write Down the Geometric Series

The geometric series for \( \frac{1}{1-2x} \) is: \[ \frac{1}{1-2x} = \sum_{n=0}^{\infty} (2x)^n = \sum_{n=0}^{\infty} 2^n x^n \] This series is valid for \( |2x| < 1 \) or \( |x| < \frac{1}{2} \).

Step 3: Multiply by \(x^2\)

To find the series for \( \frac{x^2}{1-2x} \), multiply the entire series \( \sum_{n=0}^{\infty} 2^n x^n \) by \( x^2 \):\[ \frac{x^2}{1-2x} = x^2 \sum_{n=0}^{\infty} 2^n x^n = \sum_{n=0}^{\infty} 2^n x^{n+2} \]

Step 4: Rewrite with Matching Indices

Re-index the series from Step 3 so that the power of \( x \) starts from zero. Let \( m = n + 2 \), then \( n = m - 2 \):\[ \sum_{n=0}^{\infty} 2^n x^{n+2} = \sum_{m=2}^{\infty} 2^{m-2} x^m \]Change the index to start from \( m = 0 \):\[ = \sum_{m=0}^{\infty} 2^{m-2} x^m \] only for \( m \geq 2 \).

Step 5: Separate Initial Terms

Account for the initial terms separately since the series only starts effectively contributing at \( x^2 \):For \( m = 0 \) and \( m = 1 \), the coefficients of \( x^0 \) and \( x^1 \) are zero because the series starts at \( m = 2 \).Add these "zero" terms back:\[ = 0 + 0x + \sum_{m=2}^{\infty} 2^{m-2} x^m = \sum_{m=0}^{\infty} 2^{m-2} x^m \]

Step 6: Final Taylor Series Representation

The Taylor series at \( x = 0 \) for \( \frac{x^2}{1-2x} \) in summation notation is:\[ \boxed{x^2 + 2x^3 + 4x^4 + 8x^5 + \cdots} \]This matches:\[ \sum_{m=2}^{\infty} 2^{m-2} x^m \] or equivalent simply starting from the power 2 terms in expanded form.

Key Concepts

Power Series

Power series are fundamental in understanding complex functions since they offer a way to express these functions as an infinite sum of powers of a variable. The general form of a power series is:

• \( \sum_{n=0}^{\infty} a_n (x - c)^n \)

Here, \( a_n \) represents the coefficients, and \( c \) is the center of the series. In the context of Taylor series, this is especially useful because it provides an approximation of a function around a specific point, which in many exercises is \( x=0 \). This is known as a Maclaurin series.
In solving the given problem, recognizing the function \( \frac{x^2}{1-2x} \) as a form of a power series allows us to apply known manipulations and simplifies the process of finding its Taylor series. Understanding power series forms helps us leverage established series, like the geometric series, to construct solutions efficiently.

Geometric Series

A geometric series is a type of power series where each term is a constant multiple, called the common ratio, of the previous term. Its standard form is:

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

where \( r \) is the common ratio, and the series is valid when \( |r| < 1 \). This representation is incredibly useful because it allows us to express fractions in a simpler form, as seen in the original problem.
For the function \( \frac{1}{1-2x} \), the common ratio is \( 2x \), and the geometric series becomes:

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

By substituting \( x^2 \) into the geometric series, we adapt it to find the desired Taylor series, maintaining its convergence properties for \( |x| < \frac{1}{2} \).

Re-indexing Series

Re-indexing a series is an essential technique when aligning terms of a series to fit a specific functional form. It's often used to shift the starting index or match the powers of the variable involved in the series. In our exercise, we move from

• \( \sum_{n=0}^{\infty} 2^n x^{n+2} \)

to an index that begins with zero powers, fitting into a Maclaurin series format.
By letting \( m = n + 2 \), we find that:

• \( n = m - 2 \)

and adjust the series to:

• \( \sum_{m=2}^{\infty} 2^{m-2} x^m \)

The re-indexing aligns the series to start from zero, helping us represent the Taylor series correctly from \( m = 2 \) onward.

Multiplication of Series

The multiplication of series is an approach used to combine two series into a new one, usually resulting in a more complex series representation. This involves distributing terms and ensuring all powers are properly accounted for, which can sometimes mean adjusting indices as seen with re-indexing.
For the given function \( \frac{x^2}{1-2x} \), this step was crucial. We multiplied each term of the geometric series \( \sum_{n=0}^{\infty} 2^n x^n \) by \( x^2 \), transforming the series to:

• \( \sum_{n=0}^{\infty} 2^n x^{n+2} \)

This operation shifts each term by two powers of \( x \), effectively aligning our series expressions for further manipulation like re-indexing. Understanding this concept ensures we can handle similar operations on other series effectively, providing solutions to a wider variety of problems.