Find the Maclaurin series for the function $f ( x ) = x \cos x$
A) $\sum _ { k = 0 } ^ { \infty } \frac { ( - 1 ) ^ { k } } { k ! } x ^ { 2 k + 1 }$
B) $\sum _ { k = 0 } ^ { \infty } \frac { ( - 1 ) ^ { k } } { ( 2 k ) ! } x ^ { 2 k + 1 }$
C) $\sum _ { k = 0 } ^ { \infty } \frac { ( - 1 ) ^ { k } } { ( 2 k ) ! } x ^ { 2 k }$
D) $\sum _ { k = 0 } ^ { \infty } \frac { ( - 1 ) ^ { k } } { ( 2 k ) ! } x ^ { k + 1 }$

Question

Quizplus · Accepted Answer

The Maclaurin series for $\cos x$ is $\sum_{k=0}^{\infty} \frac{(-1)^k}{(2k)!}x^{2k}$. Multiplying this by $x$ gives $\sum_{k=0}^{\infty} \frac{(-1)^k}{(2k)!}x^{2k+1}$, which matches option B.

Find the Maclaurin Series for the Function f(x)=xcos⁡xf ( x ) = x \cos xf(x)=xcosx

Find the Maclaurin Series for the Function $f ( x ) = x \cos x$