Problem 43 - Calculus Early Transcendentals: Pearson New International Edition

Question

Use a CAS to find the first four nonzero terms in the Maclaurin series for each of the following. Check Problems 43-48 to see that you get the same answers using the methods of Section 9.7. $$ 3 \sin x-2 \exp x $$

Step-by-Step Solution

Answer

The first four nonzero terms are \( 2 + 5x + x^2 + \frac{x^3}{6} \).

Step 1: Recall Maclaurin Series Expansions

A Maclaurin series is the Taylor series expansion of a function about 0. The general form for a Maclaurin series of a function \( f(x) \) is \( f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \dots \). We will use these formulas to expand both \( \sin x \) and \( \exp x \).

Step 2: Find the Maclaurin series for \( \sin x \)

The Maclaurin series for \( \sin x \) is \( x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots \). For our purposes of finding the first four nonzero terms, we get \( \sin x \approx x - \frac{x^3}{6} \). Multiply this by 3 to match the expression in the problem, giving \( 3\sin x \approx 3x - \frac{x^3}{2} \).

Step 3: Find the Maclaurin series for \( \exp x \)

The Maclaurin series for \( \exp x \) is \( 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots \). From this, the first four nonzero terms are \( 1 + x + \frac{x^2}{2} + \frac{x^3}{6} \). Multiply this series by -2 to align with the expression in the problem, yielding \( -2\exp x \approx -2 - 2x - x^2 - \frac{x^3}{3} \).

Step 4: Combine the series for \( 3 \sin x - 2 \exp x \)

Subtract the series found in Step 3 from that found in Step 2: \( (3x - \frac{x^3}{2}) - (-2 - 2x - x^2 - \frac{x^3}{3}) \). Simplify this expression: \( 3x - \frac{x^3}{2} + 2 + 2x + x^2 + \frac{x^3}{3} \). When simplifying, combine like terms to yield: \( 2 + 5x + x^2 + \frac{x^3}{6} \).

Step 5: Final Answer

The first four nonzero terms of the Maclaurin series for \( 3 \sin x - 2 \exp x \) are \( 2 + 5x + x^2 + \frac{x^3}{6} \). These terms accurately represent the series expansion needed for the problem.

Key Concepts

Taylor Series Expansion

The Taylor series is a powerful tool in calculus, used for approximating functions by polynomials. When we expand a function around zero, it is specifically called a Maclaurin series. This is essentially a special case of the Taylor series. The series expansion for a function \( f(x) \) around zero is expressed as:

• \( f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \dots \)

This formula allows us to break down complex functions into simpler polynomial components. For practical purposes, such as the given exercise, we often focus on only the first few nonzero terms of such series. This is because higher order terms contribute less and less to the value of the function, especially as \( x \) becomes smaller. As seen in our exercise, we applied this by finding the expansions of \( \sin x \) and \( \exp x \). This approximation technique is a cornerstone in mathematical analysis, greatly aiding in computation and understanding of functions.

Sine Function

The sine function, \( \sin x \), is one of the fundamental trigonometric functions. Its behavior can be smoothly approximated using a Maclaurin series, which allows us to represent it as an infinite sum of terms. The series is given by:

• \( \sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots \)

For most applications, especially when \( x \) is relatively small, only the first few terms are needed to achieve significant accuracy. In our exercise, the primary terms considered are \( x \) and \( -\frac{x^3}{6} \). When multiplied by 3, as needed for the exercise, this becomes \( 3x - \frac{x^3}{2} \). This polynomial provides a close approximation to \( 3\sin x \), helping in further calculations within the problem.

Exponentials in Series

The exponential function \( \exp x \) can vividly be expressed as a Maclaurin series. This series expansion helps us approximate the exponential function using simpler terms. The series is expressed as:

• \( \exp x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots \)

For tasks like the one in our problem, only the first few terms are leveraged. In this case, \( 1 + x + \frac{x^2}{2} + \frac{x^3}{6} \) are the leading terms. When multiplied by \(-2\), fitting the problem statement, it looks like this: \(-2 - 2x - x^2 - \frac{x^3}{3}\). This step is crucial as it breaks down \( \exp x \) into a format compatible with polynomial arithmetic, enabling us to better manage and simplify the expression in combination with other parts.

Series Simplification

In mathematical problem-solving, especially when dealing with series, simplification is key. After finding the series representations for \( 3 \sin x \) and \( -2 \exp x \), combining them effectively is crucial. Beginning with:

• \( 3x - \frac{x^3}{2} + 2 + 2x + x^2 + \frac{x^3}{3} \)

We then simplify by consolidating like terms:

• Collect \( x \) terms: \( 3x + 2x = 5x \)
• Constant terms: \( 2 \)
• \( x^2 \) term: \( x^2 \)
• Combine \( x^3 \) fractions: \(-\frac{x^3}{2} + \frac{x^3}{3} = \frac{x^3}{6} \)

This yields the clean simplified series: \( 2 + 5x + x^2 + \frac{x^3}{6} \). This simplification step saves time and reduces potential errors in calculations, enabling a direct and accurate representation of complex expressions.