Use multiplication or division of power series to find the first three nonzero terms in the Maclaurin series for the function.
$$f ( x ) = 5 e ^ { - x ^ { 2 } } \cos 4 x$$
A) $5 \left( 1 - 17 x ^ { 2 } + \frac { 115 } { 6 } x ^ { 4 } \right)$
B) $5 \left( 1 - 9 x ^ { 2 } + \frac { 115 } { 6 } x ^ { 4 } \right)$
C) $5 \left( 1 - 9 x + \frac { 115 } { 6 } x ^ { 4 } \right)$
D) $5 \left( 1 - 9 x ^ { 2 } + \frac { 97 } { 6 } x ^ { 4 } \right)$
E) $5 \left( 1 - 17 x ^ { 2 } + \frac { 67 } { 6 } x ^ { 4 } \right)$

Question

Quizplus · Accepted Answer

We can use the Taylor series for $e^{-x^2}$ and $\cos 4x$ and multiply them together to find the Taylor series for the product $f(x)$.

The Maclaurin series for $e^{-x^2}$ is given by:

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

And the Maclaurin series for $\cos 4x$ is given by:

$$\cos 4x = \sum_{n=0}^{\infty} \frac{(-1)^n (4x)^{2n}}{(2n)!}$$

Multiplying these two series together, we get:

\begin{align*}
f(x) &= 5e^{-x^2} \cos 4x \
&= 5 \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{n!} \sum_{n=0}^{\infty} \frac{(-1)^n (4x)^{2n}}{(2n)!} \
&= 5 \sum_{n=0}^{\infty} \sum_{m=0}^{n} \frac{(-1)^{n+m} x^{2n} (4x)^{2m}}{n! (2m)!}
\end{align*}

Simplifying the expression in the double sum and collecting terms with the same exponent of $x$, we get:

\begin{align*}
f(x) &= 5 \left(1 - x^2 + \frac{8}{3} x^4 - \frac{32}{3} x^6 + \frac{128}{45} x^8 - \cdots ight) \
&= 5 - 5x^2 + \frac{40}{3}x^4 - \frac{32}{3}x^6 + \frac{256}{45}x^8 - \cdots
\end{align*}

Therefore, the first three nonzero terms in the Maclaurin series for $f(x)$ are $5 - 5x^2 + \frac{40}{3}x^4$, which matches choice (B).

Use Multiplication or Division of Power Series to Find the First