# Thread: Multiplying/Dividing Power Series

1. ## Multiplying/Dividing Power Series

Use multiplication or division of power series to find the first 3 non-zero terms in the Maclaurin series for:

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

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

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

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

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

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

I multiplied and got:

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

which is not right in webassign. Should I have divided? Or what am I doing wrong?

2. looks like a mistake in your expansion for $\displaystyle \cos(4x)$ ... forget the factorial?

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

$\displaystyle \cos(4x) \approx 1 - 8x^2 + \frac{32x^4}{3}$

3. in your expansion for cos(4x) the coefficient for the x^4 term should be 32/3

your method is correct

,

### division of two power series

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