Find the first three nonzero terms of the Maclaurin series for the given function and the values of x for which the series
converges absolutely.
-f(x) = (cos x) ln(1 + x)
A) $x - \frac { x ^ { 2 } } { 2 } - \frac { x ^ { 3 } } { 6 } + \ldots , - 1

Question

Find the first three nonzero terms of the Maclaurin series for the given function and the values of x for which the series
converges absolutely.
-f(x) = (cos x) ln(1 + x) 
A) $x - \frac { x ^ { 2 } } { 2 } - \frac { x ^ { 3 } } { 6 } + \ldots , - 1 < x < 1$
B) $x - \frac { x ^ { 2 } } { 3 } - \frac { x ^ { 3 } } { 5 } + \ldots , - 1 < x < 1$
C) $x - \frac { x ^ { 2 } } { 2 } + \frac { x ^ { 3 } } { 6 } + \ldots , - 1 < x < 1$
D) $x - \frac { x ^ { 2 } } { 3 } - \frac { x ^ { 3 } } { 5 } + \ldots , - 1 < x < 1$.

Quizplus · Accepted Answer

The Maclaurin series for cos x is $$cos x = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!} = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots$$The Maclaurin series for ln(1 + x) is $$ln(1 + x) = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} x^n = x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots$$So the Maclaurin series for f(x) = (cos x) ln(1 + x) is $$f(x) = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!} \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} x^n$$$$= \sum_{n=0}^{\infty} \sum_{k=1}^{\infty} \frac{(-1)^{n+k} x^{2n+k}}{(2n)!n}$$$$= x - \frac{x^2}{2} - \frac{x^3}{6} + \cdots$$The series converges absolutely for -1 < x < 1.

Find the First Three Nonzero Terms of the Maclaurin Series x−x22−x36+…,−1<x<1x - \frac { x ^ { 2 } } { 2 } - \frac { x ^ { 3 } } { 6 } + \ldots , - 1 < x < 1x−2x2​−6x3​+…,−1<x<1

Find the First Three Nonzero Terms of the Maclaurin Series $x - \frac { x ^ { 2 } } { 2 } - \frac { x ^ { 3 } } { 6 } + \ldots , - 1 < x < 1$