Problem 9 - 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. $$f(x)=e^{-x}$$

Step-by-Step Solution

Answer

Answer: The Maclaurin series for the function \(f(x) = e^{-x}\) can be written as \(f(x) = \sum_{n=0}^{\infty}\frac{(-1)^n}{n!}x^n\), and its interval of convergence is \((-\infty, \infty)\), or the entire real number line.

Step 1: Find the first four non-zero terms of the Maclaurin series

Find the first few derivatives of \(f(x) = e^{-x}\) and evaluate them at \(x=0\): 1. \(f'(x) = -e^{-x}\), \(f'(0) = -1\) 2. \(f''(x) = e^{-x}\), \(f''(0) = 1\) 3. \(f'''(x) = -e^{-x}\), \(f'''(0) = -1\) 4. \(f^{(4)}(x) = e^{-x}\), \(f^{(4)}(0) = 1\) Now, plug these derivatives into the Maclaurin series formula: $$f(x) ≈ f(0) + \frac{f'(0)}{1!}x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \dots$$ $$f(x) ≈ e^0 - x + \frac{1}{2}x^2 - \frac{1}{6}x^3$$

Step 2: Write the power series using summation notation

Using the derivatives found in step 1, we can express the Maclaurin series in summation notation as: $$f(x) = \sum_{n=0}^{\infty}\frac{(-1)^n}{n!}x^n$$

Step 3: Determine the interval of convergence of the series

To find the interval of convergence, we will use the Ratio Test: $$\lim_{n \to \infty} \frac{\big|\frac{(-1)^{n+1}}{(n+1)!}x^{n+1}\big|}{\big|\frac{(-1)^n}{n!}x^n\big|} = \lim_{n \to \infty} \frac{(n+1)!}{n!} \frac{n!}{(n+1)!} |x| = \lim_{n \to \infty} \frac{1}{n+1} |x|$$ Since the limit approaches 0 as \(n \to \infty\), the power series converges for all values of x. Thus, the interval of convergence is \((-\infty, \infty)\), or the entire real number line. So, the final answers are: a. \(f(x) ≈ e^0 - x + \frac{1}{2}x^2 - \frac{1}{6}x^3\) b. \(f(x) = \sum_{n=0}^{\infty}\frac{(-1)^n}{n!}x^n\) c. The interval of convergence is \((-\infty, \infty)\).