Problem 47 - 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 \(9.7 .\) \((\sin x)(\exp x)\)

Step-by-Step Solution

Answer

The first four nonzero terms of the Maclaurin series are \( x + \frac{x^2}{2} - \frac{x^4}{24} \).

Step 1: Recall Maclaurin Series Formula

The Maclaurin series of a function \( f(x) \) is given by the formula \( f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \cdots \). We want to apply this to \( f(x) = \sin(x) \cdot e^x \).

Step 2: Find Derivatives at x = 0

Calculate the necessary derivatives of \( f(x) = (\sin x)(e^x) \) and evaluate them at \( x = 0 \). First, \( f(0) = (\sin 0)(e^0) = 0 \), second, \( f'(x) = \cos x \cdot e^x + \sin x \cdot e^x \), thus \( f'(0) = (\cos 0)(e^0) + (\sin 0)(e^0) = 1 \).

Step 3: Continue Finding Higher Order Derivatives

Calculate more derivatives: \( f''(x) = -\sin x \cdot e^x + 2 \cos x \cdot e^x + \sin x \cdot e^x \), which equals \( (\cos x)(e^x) \), and \( f''(0) = (\cos 0)(e^0) = 1 \). Next, \( f'''(x) = -\cos x \cdot e^x + (-\sin x \cdot e^x - \cos x \cdot e^x) + 2(\cos x \cdot e^x) + \sin x \cdot e^x \) which simplifies to \( -\sin x + \sin x = 0 \) at \( x = 0 \), yielding \( f'''(0) = 0 \).

Step 4: Calculate Fourth Derivative

Lastly, using the pattern, calculate \( f^{(4)}(x) = \) derivative of third derivative, and by direct computation, find \( f^{(4)}(0) = -1 \).

Step 5: Write out the Maclaurin Series

Using the derivatives, construct the series: \( \sin x \cdot e^x = 0 + 1x + \frac{1}{2!}x^2 + 0 + \frac{-1}{4!}x^4 + \cdots = x + \frac{x^2}{2} - \frac{x^4}{24} + \cdots \). The first four nonzero terms are \( x + \frac{x^2}{2} - \frac{x^4}{24} \).

Step 6: Verification Using Section 9.7

Check the result with other methods from Section 9.7 of your textbook to ensure correctness with previous computations.

Key Concepts

Understanding Derivatives

Derivatives are at the heart of calculus and are especially important when finding the Maclaurin series of a function. In simple terms, a derivative represents how a function changes as its input changes. It's the mathematical way of describing the rate of change, much like how a speedometer shows how speed changes over time.

• The first derivative of a function can be thought of as the slope of the tangent line to the function's graph at a certain point.
• The second derivative tells us how the rate of change itself is changing, and so on for higher order derivatives.

Each derivative plays a critical role in forming the terms of the Maclaurin series. For example, if you have a function like \(f(x) = \sin(x) \cdot e^x\), you calculate the derivatives to build the series. Finding each derivative requires applying rules from calculus, like the product rule, which helps differentiate functions that are multiplied together.

Exploring sin(x) in Calculus

The sine function, written as \( \sin(x) \), is a fundamental part of trigonometry and calculus.

• It represents oscillatory motion, often used in modeling waves or sound vibrations.
• The function starts at 0 when \(x = 0\), climbs to 1 at \(x = \frac{\pi}{2}\), descends back to 0 at \(x = \pi\), and forms a smooth, repeating cycle.

In calculus, \( \sin(x) \) has a cyclic pattern of derivatives, which is very useful for series expansions like the Maclaurin series. For example:- The first derivative of \( \sin(x) \) is \( \cos(x) \),- The second derivative is \(-\sin(x)\), and then it repeats the cycle. Understanding this cycle helps in calculating Maclaurin series, where evaluating these derivatives at \(x = 0\) gives the coefficients for each term.

Exponential Function and Its Significance

The exponential function, \( e^x \), is a key concept across mathematics. Known for its natural cousin, \( e \approx 2.71828\), it's unique because its rate of change is proportional to its current value. This leads to many useful properties:

• Unlike polynomials, its derivative is the same as the function itself, \( \frac{d}{dx}e^x = e^x \).
• This property makes it straightforward to develop series expansions, such as the Maclaurin series.

When \( e^x \) is part of a product with other functions like \( \sin(x) \), the differentiability remains smooth and straightforward, helping simplify complex calculations, especially when combined with other calculus rules.

The Role of Calculus in Series Expansion

Calculus provides tools to understand and quantify change, leading to the development of series expansion methods to approximate functions. The Maclaurin series, a type of Taylor series centered at \( x = 0 \), is one such method, allowing representation of complex functions as an infinite sum of terms derived from derivatives.

• Each term in the series is derived from the function's derivatives at a single point, often providing a good approximation near this point.
• However, higher-order derivatives play a significant role as you seek more accuracy.
• By taking derivatives of functions like \(f(x) = \sin(x) \cdot e^x \), you acquire the necessary terms to write a series.

This technique finds practical applications in many aspects of science and engineering, including signal processing and control systems, where approximations of periodic or exponential behaviors are often needed.