Find the Maclaurin series for $f ( x )$ using the definition of the Maclaurin series.
$$f ( x ) = x \cos ( 4 x )$$
A) $\sum _ { n = 0 } ^ { \infty } \frac { ( - 1 ) ^ { n } 4 ^ { n } x ^ { 2 n + 1 } } { ( 2 n ) ! }$
B)
$\sum _ { n = 0 } ^ { \infty } \frac { ( - 1 ) ^ { n } 4 ^ { 2 n } x ^ { 2 n + 1 } } { n ! }$
C)
$\sum _ { n = 0 } ^ { \infty } \frac { ( - 1 ) ^ { n } 4 ^ { 2 n } x ^ { 2 n } } { ( 2 n ) ! }$
D)
$\sum _ { n = 0 } ^ { \infty } \frac { ( - 1 ) ^ { n } 4 ^ { 2 n } x ^ { 2 n + 1 } } { ( 2 n ) ! }$
E)
$\sum _ { n = 0 } ^ { \infty } \frac { ( - 1 ) ^ { n + 1 } 4 ^ { 2 n } x ^ { 2 n + 1 } } { ( 2 n ) ! }$

Question

Quizplus · Accepted Answer

Using the definition of the Maclaurin series, (assuming that all the derivatives exist and are continuous), we have

$$f(x) = \sum_{n=0}^{\infty}\frac{f^{(n)}(0)}{n!}x^n$$

where

$$f^{(n)}(x) = \frac{d^n}{dx^n}\left(x\cos(4x)\right)$$

To find the coefficients, we need to find the nth derivative of $f(x)$ evaluated at 0. Using product rule, we get

$$f^{(1)}(x) = \cos (4x) - 4x\sin (4x)$$

$$f^{(1)}(0) = 1$$

$$f^{(2)}(x) = -16\cos (4x) - 16x\sin (4x)$$

$$f^{(2)}(0) = -16$$

And so on,

$$f^{(3)}(x) = -48\sin(4x) - 64x\cos(4x)$$

$$f^{(3)}(0) = 0$$

$$f^{(4)}(x) = -256\cos(4x) + 64x\sin(4x)$$

$$f^{(4)}(0) = -256$$

and so on.

Substituting these into the Maclaurin series formula, we get

$$f(x) = \sum_{n=0}^{\infty}\frac{f^{(n)}(0)}{n!}x^n = 1 - 16\frac{x^2}{2!} + 0\frac{x^3}{3!} + (-256)\frac{x^4}{4!} + \cdots$$

Simplifying the expression, we get

$$\sum_{n=0}^{\infty} \frac{(-1)^n4^{2n}x^{2n+1}}{(2n)!}$$

Therefore, the correct choice is D.

Find the Maclaurin Series For f(x)f ( x )f(x) Using the Definition of the Maclaurin Series

Find the Maclaurin Series For $f ( x )$ Using the Definition of the Maclaurin Series