Question
(a) Topic-10.14-Finding Taylor or Maclaurin Series for a Function
(b) Topic-10.5-Harmonic Series and p-Series
(c) Topic-10.10-Alternating Series Error Bound
On its interval of convergence, this series converges to \( \ln(1 + x) \). Let \( f \) be the function defined by
\(
f(x) = x \ln\left(1 + \frac{x}{3}\right).
\)
(a) Write the first four nonzero terms and the general term of the Maclaurin series for \( f \).
(b) Determine the interval of convergence of the Maclaurin series for \( f \). Show the work that leads to your answer.
(c) Let \( P_4(x) \) be the fourth-degree Taylor polynomial for \( f \) about \( x = 0 \). Use the alternating series error bound to find an upper bound for \( |P_4(2) – f(2)| \).
▶️Answer/Explanation
\(\textbf{6(a)}\)
Using the Maclaurin series for \( \ln(1 + x) \) that was given:
\(
f(x) = x \ln\left(1 + \frac{x}{3}\right) = x \left[ \frac{x}{3} – \frac{\left(\frac{x}{3}\right)^2}{2} + \frac{\left(\frac{x}{3}\right)^3}{3} – \frac{\left(\frac{x}{3}\right)^4}{4} + \cdots + (-1)^{n+1} \frac{\left(\frac{x}{3}\right)^n}{n} \right].
\)
\(
= \frac{x^2}{3} – \frac{x^3}{3 \cdot 2^2} + \frac{x^4}{3 \cdot 3^3} – \frac{x^5}{4 \cdot 3^4} + \cdots + (-1)^{n+1} \frac{x^{n+1}}{n \cdot 3^n}.
\)
\(\textbf{6(b)}\)
Using the ratio test to find the interval of convergence for \( f \):
\(
\frac{(-1)^{n+2} \frac{x^{n+2}}{(n+1) \cdot 3^{n+1}}}{(-1)^{n+1} \frac{x^{n+1}}{n \cdot 3^n}} = \lim_{n \to \infty} \frac{(-1)^{n+2} \cdot x^{n+2} \cdot n \cdot 3^n}{(-1)^{n+1} \cdot x^{n+1} \cdot (n+1) \cdot 3^{n+1}}.
\)
\(
= \lim_{n \to \infty} \frac{n}{n+1} \cdot \frac{|x|}{3} = \frac{|x|}{3} < 1 \quad \text{for convergence.}
\)
So, \(-1 < \frac{x}{3} < 1 \implies -3 < x < 3\).
Now, testing the endpoints of the interval:
\(
x = 3: \quad \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n \cdot 3^n} = \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}, \quad \text{which is an alternating harmonic series, so it converges.}
\)
\(
x = -3: \quad \sum_{n=1}^\infty \frac{(-1)^{n+1}(-3)^n}{n \cdot 3^n} = \sum_{n=1}^\infty \frac{(-1)^n}{n}, \quad \text{which is a harmonic series, so it diverges.}
\)
\(
\therefore \text{The interval of convergence for } f \text{ is } -3 < x \leq 3.
\)
\(\textbf{6(c)}\)
Since \( 2 \) is within the interval of convergence for \( f \), the alternating series error bound using the 4th-degree Taylor polynomial would be the absolute value of the next term, the 5th-degree term, evaluated at \( x = 2 \).
So \(\left| P_4(2) – f(2) \right| < \frac{2^5}{4 \cdot 3^4}\).