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 are \( x + x^2 + \frac{1}{12}x^4 \).

Step 1: Recall the Maclaurin Series Formula

The Maclaurin series of a function \( f(x) \) is the Taylor series expansion of \( f(x) \) about 0. It is given by \[ f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \cdots \] To find the first four nonzero terms, we need to calculate the derivatives of the function \( f(x) = (\sin x)(\exp x) \).

Step 2: Determine the Derivatives

For \( f(x) = (\sin x)(\exp x) \), we will use the product rule: \( (uv)' = u'v + uv' \). The derivatives are: 1. \( f'(x) = (\cos x)(\exp x) + (\sin x)(\exp x) \) 2. \( f''(x) = (-\sin x)(\exp x) + 2(\cos x)(\exp x) \) 3. \( f'''(x) = (-\cos x)(\exp x) - 2(\sin x)(\exp x) + (\cos x)(\exp x) \) (Continue this process if more derivatives are needed).

Step 3: Evaluate the Derivatives at x=0

Calculate each derivative at \( x = 0 \):1. \( f(0) = (\sin 0)(\exp 0) = 0 \)2. \( f'(0) = (\cos 0)(\exp 0) + (\sin 0)(\exp 0) = 1 \)3. \( f''(0) = (-\sin 0)(\exp 0) + 2(\cos 0)(\exp 0) = 2 \)4. \( f'''(0) = (-\cos 0)(\exp 0) - 2(\sin 0)(\exp 0) + (\cos 0)(\exp 0) = 0 \)5. \( f^{(4)}(0) = 2 \). (Additional derivatives calculated as needed)

Step 4: Substitute into Maclaurin Series

Using the evaluated derivatives, substitute into the Maclaurin series formula:\[( ext{Terms}) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \frac{f^{(4)}(0)}{4!}x^4 + \cdots\]Substitute the values and calculate:\( 0 + 1x + \frac{2}{2}x^2 + 0x^3 + \frac{2}{24}x^4 + \cdots \)This simplifies to \( x + x^2 + \frac{1}{12}x^4 + \cdots \).

Step 5: Provide the First Four Nonzero Terms

The first four nonzero terms of the Maclaurin series for \( (\sin x)(\exp x) \) are:\[ x + x^2 + \frac{1}{12}x^4 \]

Key Concepts

Taylor series expansion

The Taylor series expansion is a crucial mathematical concept used to approximate functions as a sum of simpler power terms. When we apply the Taylor series expansion at zero, it becomes the Maclaurin series. This series expands a function around the value of 0.
The general formula for the Maclaurin series is:

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

Here, \(f(0), f'(0), f''(0),\) and so on are the derivatives of the function evaluated at \(x=0\). The series allows us to express functions like \(\sin x\) or \(\exp x\) as an infinite sum, which can be particularly useful for computations and approximations.
In the Maclaurin series, each subsequent term gives a better approximation of the function, especially near \(x=0\). For example, a physical scenario or engineering problem demanding quick computation might use only the first few terms of the series.
To apply the Maclaurin series to \((\sin x)(\exp x)\), we derive the series from the product of these functions, evaluating derivatives at \(x=0\) up to the necessary order. This involves calculating several derivatives and then substituting their values into the Maclaurin formula.

derivatives calculation

Derivatives are the backbone of finding the Maclaurin series by providing the coefficients for each term in the series. When dealing with functions like \((\sin x)(\exp x)\), calculating derivatives correctly is essential.
Each derivative represents the slope of the function at a point, more specifically at \(x=0\) in the case of the Maclaurin series.
For the function \((\sin x)(\exp x)\), we need to compute as many derivatives as required to achieve our desired number of terms. First, calculate:

• First derivative: \(f'(x) = (\cos x)(\exp x) + (\sin x)(\exp x)\)
• Second derivative: \(f''(x) = (-\sin x)(\exp x) + 2(\cos x)(\exp x)\)
• Third derivative: \(f'''(x) = (-\cos x)(\exp x) - 2(\sin x)(\exp x) + (\cos x)(\exp x)\)
• And continue as necessary.

Evaluating these derivatives at \(x=0\) gives us their values which we then use to form the series.
This process can seem repetitive but it's a systematic method ensuring that every component of the function is accommodated in the final series.

product rule

The product rule is a fundamental rule in calculus used when differentiating the product of two functions. Given two functions \(u(x)\) and \(v(x)\), the product rule states:
\[(uv)' = u'v + uv'\]In practical terms, if you want to differentiate the product of two functions, derivatives are taken for each function individually while keeping the other function constant, then the results are summed.
For the function \((\sin x)(\exp x)\), we apply the product rule to find its derivative:

• First, identify \(u = \sin x\) and \(v = \exp x\)
• Calculate \(u' = \cos x\) and \(v' = \exp x\)
• Substitute into the product rule: \(f'(x) = (\cos x)(\exp x) + (\sin x)(\exp x)\)

This step is crucial for determining the first derivative, which then feeds into more complex derivatives needed for the series expansion. It may seem simple, but missing or misapplying the product rule fundamentally affects the analysis and conclusions derived from calculus.