Problem 55 - Essential Calculus Early Transcendentals

Question

Use multiplication or division of power series to find the first three nonzero terms in the Maclaurin series for each function. $$y=e^{-x^{2}} \cos x$$

Step-by-Step Solution

Answer

The first three terms are: \(1 - \frac{3x^2}{2} + \frac{3x^4}{8}\).

Step 1: Write Down Known Series Expansions

The Maclaurin series for the exponential function \(e^{x}\) is given by \(1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots\). Therefore, \(e^{-x^2}\) will become \(1 - x^2 + \frac{(x^2)^2}{2!} - \frac{(x^2)^3}{3!} + \cdots\). Next, the Maclaurin series for \(\cos(x)\) is \(1 - \frac{x^2}{2!} + \frac{x^4}{4!} + \cdots\).

Step 2: Consider the Term-by-Term Multiplication

To obtain the Maclaurin series for \(y = e^{-x^2} \cos x\), multiply the corresponding series terms from \(e^{-x^2}\) and \(\cos(x)\). Start with the first few terms. The constant term is \(1 \cdot 1 = 1\). The \(x^2\) term is \(-x^2 \cdot 1 + 1 \cdot -\frac{x^2}{2} = -x^2 - \frac{x^2}{2}\). Simplifying gives \(-\frac{3x^2}{2}\).

Step 3: Continue the Multiplication for Higher Order Terms

To find the next term, consider \(x^4\): the terms contributing are \(\frac{x^4}{2!}\) from \(e^{-x^2}\), \(-\frac{x^2}{2} \times -\frac{x^2}{2}\), and \(1 \times \frac{x^4}{4!}\). After calculation, this is \(\frac{x^4}{2} + \frac{x^4}{4}\). Simplifying yields \(\frac{3x^4}{8}\).

Step 4: Compile the First Three Terms

The multiplication gives us the first three terms of the Maclaurin series for \(y = e^{-x^2} \cos x\): 1. Constant term: \(1\)2. Second term: \(-\frac{3x^2}{2}\)3. Third term: \(\frac{3x^4}{8}\)Therefore, the series is \(1 - \frac{3x^2}{2} + \frac{3x^4}{8} + \cdots\).

Key Concepts

Exponential Function

The exponential function, denoted as \( e^x \), is one of the most important and widely used functions in mathematics. It has key applications in calculus due to its unique properties. The Maclaurin series for \( e^x \) represents it as an infinite series:

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

This series is known for its ability to approximate the exponential function closely, even with just the first few terms.
The exponential function's inverse, \( e^{-x^2} \), is given by replacing \( x \) with \( -x^2 \) in its series form:

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

This shows how the substitution within the function affects every term in the series. It generates a new polynomial which is crucial when approximating functions that involve exponential decays.

Cosine Function

The cosine function \( \cos x \) is another fundamental function in trigonometry and calculus. Its Maclaurin series expansion is given by:

• \( \cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots \)

This periodic function can be represented as a power series to approximate its values, especially useful in mathematical analysis.
In the context of the given problem, you need the series for \( \cos x \) to be multiplied with another series, showing how series combinations can approximate more complex functions.
The alternating positive and negative signs in the series represent the cosine function's wave-like behavior, aligning coefficients with powers of \( x \) to form the polynomial.

Power Series

A power series is an infinite sum of terms in the form \( a_0 + a_1x + a_2x^2 + a_3x^3 + \cdots \), where \( a_n \) are coefficients. These series aid in approximating functions analytically within specific intervals. The general format allows them to represent many standard functions like exponentials and trigonometric functions.
This expansion helps understand function behavior without relying on graphical data.
For example, the exponential and cosine functions use power series in their Maclaurin expansion, making it easy to find their values for small \( x \).

• The coefficients account for the influence of each power of \( x \).
• By calculating enough terms, you can approximate the function to a desired accuracy.

Multiplication of Series

Multiplying two power series involves multiplying each term of one series by every term of the other, then combining like terms. This method helps derive a new series representing the product of the original functions.
In the given problem, you multiply the series for \( e^{-x^2} \) and \( \cos x \) to find the first few terms of their product.

• Start with the highest degree terms that are manageable.
• Carefully combine coefficients to construct a valid expression.

The process demonstrates how you connect separate series representations to achieve an expanded polynomial that captures the compound behavior of the original functions.
This technique is part of a broader mathematical toolkit for handling complex function forms.