Problem 16 - Calculus Early Transcendentals

Question

a. Find the first four nonzero terms of the Maclaurin series for the given function. b. Write the power series using summation notation. c. Determine the interval of convergence of the series. $$f(x)=\sin 3 x$$

Step-by-Step Solution

Answer

Based on the given function \(f(x) = \sin 3x\), we found the first four nonzero terms of its Maclaurin series as: $$3x - \frac{27}{3!}x^3$$ The power series in summation notation is: $$\sin 3x = \sum_{n=0}^{\infty} (-1)^n \frac{(3x)^{2n+1}}{(2n+1)!}$$ The interval of convergence for this series is: $$-\infty < x < \infty$$

Step 1: Find the first four nonzero terms

To find the first four nonzero terms, we need to compute the derivatives of \(f(x) = \sin 3x\) until the fourth derivative. Then, we will evaluate each derivative at x = 0: 1. \(f(x) = \sin 3x\) 2. \(f'(x) = 3 \cos 3x\) 3. \(f''(x) = -9 \sin 3x\) 4. \(f'''(x) = -27 \cos 3x\) Now, evaluate these at x = 0: 1. \(f(0) = \sin(0) = 0\) 2. \(f'(0) = 3 \cos(0) = 3\) 3. \(f''(0) = -9 \sin(0) = 0\) 4. \(f'''(0) = -27 \cos(0) = -27\) The first four nonzero terms of the Maclaurin series for \(f(x) = \sin 3x\) are: $$3x - \frac{27}{3!}x^3$$

Step 2: Write the power series using summation notation

To write the power series using summation notation, recall that the sine function has a Maclaurin series that converges for all x: $$\sin x = \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+1}}{(2n+1)!}$$ We can use this general formula to express \(f(x)=\sin 3x\) in terms of summation notation: $$\sin 3x = \sum_{n=0}^{\infty} (-1)^n \frac{(3x)^{2n+1}}{(2n+1)!}$$

Step 3: Determine the interval of convergence

Since the Maclaurin series for the sine function converges for all x, the interval of convergence of the series for \(f(x) = \sin 3x\) is the same. Thus, the interval of convergence is: $$-\infty < x < \infty$$