Problem 18
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. $$f(x)=\log _{3}(x+1)$$
Step-by-Step Solution
Verified Answer
Answer: The interval of convergence for the Maclaurin series of the function \(f(x)=\log_3(x+1)\) is \(-1\leq x\leq1\).
1Step 1: 1. Find the first four derivatives of the function
First, we must rewrite the function f(x) with base e in order to differentiate it: \(f(x)=\frac{\ln(x+1)}{\ln(3)}\). Then, find the first four derivatives at x=0:
\(f(x) = \frac{\ln(x+1)}{\ln(3)}\)
\(f'(x) = \frac{1}{(x+1)\ln(3)}\)
\(f''(x) = -\frac{1}{(x+1)^2\ln(3)}\)
\(f'''(x) = \frac{2}{(x+1)^3\ln(3)}\)
\(f^{(4)}(x) = -\frac{6}{(x+1)^4\ln(3)}\)
2Step 2: 2. Evaluate the derivatives at x=0
Now we'll evaluate the first four derivatives at x=0:
\(f(0) = \frac{\ln(1)}{\ln(3)} = 0\)
\(f'(0) = \frac{1}{(0+1)\ln(3)} = \frac{1}{\ln(3)}\)
\(f''(0) = -\frac{1}{(0+1)^2\ln(3)} = -\frac{1}{\ln(3)}\)
\(f'''(0) = \frac{2}{(0+1)^3\ln(3)} = \frac{2}{\ln(3)}\)
\(f^{(4)}(0) = -\frac{6}{(0+1)^4\ln(3)} = -\frac{6}{\ln(3)}\)
3Step 3: 3. Write the first four nonzero terms of the Maclaurin series
Use the derivatives evaluated at x=0 to write the first four nonzero terms of the Maclaurin series:
\(f(x) \approx \frac{1}{\ln(3)}x -\frac{1}{2\ln(3)}x^2 +\frac{2}{3! \ln(3)}x^3 -\frac{6}{4!\ln(3)}x^4\)
4Step 4: 4. Write the power series using summation notation
Express the function in terms of summation notation starting from n=1 since the first term is 0:
\(f(x) = \sum_{n=1}^\infty \frac{(-1)^{n-1}}{n\ln(3)} x^n\)
5Step 5: 5. Determine the interval of convergence of the series
To determine the interval of convergence, we'll use the Ratio Test:
\(\lim_{n\to\infty} \frac{\frac{1}{(n+1)\ln(3)}x^{n+1}}{\frac{1}{n\ln(3)}x^n} = \lim_{n\to\infty} \frac{n}{n+1} |x|\)
As \(n \to \infty\), the expression becomes \(|x|\). Since the Ratio Test requires the limit to be less than 1, we have:
\(|x| < 1 \Rightarrow -1
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