Problem 17 - 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. $$f(x)=3^{x}$$

Step-by-Step Solution

Answer

a) The first four nonzero terms of the Maclaurin series for the given function are: $$f(x) = 1 + (\ln(3))x + \frac{(\ln^2(3))x^2}{2!} + \frac{(\ln^3(3))x^3}{3!}$$ b) The power series can be written using summation notation as: $$f(x) = \sum_{n=0}^{\infty} \frac{(\ln^n(3))x^n}{n!}$$ c) The interval of convergence for this power series is: $$(-\infty, \infty)$$

Step 1: Find the first four nonzero terms of the Maclaurin series

First, obtain the functional derivatives. \(f'(x)=(3^x)'=\ln(3)*3^x\) \(f''(x)=(\ln(3)*3^x)'=\ln^2(3)*3^x\) \(f'''(x)=(\ln^2(3)*3^x)'=\ln^3(3)*3^x\) \(f''''(x)=(\ln^3(3)*3^x)'=\ln^4(3)*3^x\) Now, evaluate each derivative at x = 0 to get their respective coefficients: \(f(0)=3^0=1\) \(f'(0)=\ln(3)*3^0=\ln(3)\) \(f''(0)=\ln^2(3)*3^0=\ln^2(3)\) \(f'''(0)=\ln^3(3)*3^0=\ln^3(3)\) The first four nonzero terms of the Maclaurin series are: $$f(x) = 1 + (\ln(3))x + \frac{(\ln^2(3))x^2}{2!} + \frac{(\ln^3(3))x^3}{3!}$$

Step 2: Write the power series using summation notation

We can write the power series as a summation: $$f(x) = \sum_{n=0}^{\infty} \frac{(\ln^n(3))x^n}{n!}$$

Step 3: Determine the interval of convergence

To determine the interval of convergence, we will use the ratio test: $$\lim_{n\to\infty} \frac{a_{n+1}}{a_n} = \lim_{n\to\infty} \left(\frac{x\ln(3)}{n+1}\right)$$ The ratio test states that when this limit is less than 1, the series converges. Since this function is an exponential function, it will converge for all x: $$\lim_{n\to\infty} \left|\frac{x\ln(3)}{n+1}\right| < 1$$ So, the interval of convergence is: $$(-\infty, \infty)$$