"By multiplying the appropriate Taylor series about $c=0$, compute the first four terms of the Taylor series about $c=0$ for $f(x)=e^{-x}\cos x$."
Seems straightforward enough but when I break up $e^{-x}$ and $\cos(x)$ then multiply their terms together I am getting something different than an online Taylor Series calculator is coming up with. Here's what I'm getting for the terms of the two functions and then their product:
$$e^{-x} \longrightarrow 1 - x + \frac{x^2}{2} - \frac{x^3}{6}\\ \cos(x) \longrightarrow 1 - \frac{x^2}{2} + \frac{x^4}{24}$$
Now, I only used the first two terms of $\cos(x)$ when I multiplied and I came up with:
$$1 - x +\frac{x^3}{2} \underbrace{- \frac{x^4}{4}}$$ The underbrace denotes difference between my answer and the online calculator.
The online calculator is getting the following:
$$1 - x +\frac{x^3}{3} - \frac{x^4}{6}$$
If I'm making a basic multiplication mistake I have missed it several times. Can anyone help point to my mistake? Many thanks..