# Multiplying/Dividing Power Series

• Apr 16th 2009, 04:34 PM
mollymcf2009
Multiplying/Dividing Power Series
Use multiplication or division of power series to find the first 3 non-zero terms in the Maclaurin series for:

$e^{-9x^2} cos(4x)$

$\sum^{\infty}_{n=0} \frac{(-9x^2)^n}{n!}$

$= 1 - 9x^2 + \frac{81x^4}{2} - ...$

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

$= 1 - \frac{16x^2}{2} + \frac{256x^4}{4} - ...$

$\left(1 - 9x^2 + \frac{81x^4}{2}\right) \cdot \left(1 - \frac{16x^2}{2} + \frac{256x^4}{4}\right)$

I multiplied and got:

$1 - \frac{34x^2}{2} + \frac{706x^4}{4}$

which is not right in webassign. Should I have divided? Or what am I doing wrong?
• Apr 16th 2009, 05:13 PM
skeeter
looks like a mistake in your expansion for $\cos(4x)$ ... forget the factorial?

$\cos(4x) \approx 1 - \frac{(4x)^2}{2!} + \frac{(4x)^4}{4!}
$

$\cos(4x) \approx 1 - 8x^2 + \frac{32x^4}{3}$
• Apr 16th 2009, 05:33 PM
Calculus26
in your expansion for cos(4x) the coefficient for the x^4 term should be 32/3