I keep getting the same answer, and I know it's wrong, but I'm not seeing where I've gone wrong.
I'm supposed to find the first 4 terms of $e^{-x} \cos(x)$ using Taylor series expansion and multiplying the terms.
The answer should be: $$1-x+\frac{1}{3}x^3-\frac{1}{6}x^4$$
But I keep getting: $$1-x+\frac{1}{3}x^3-\frac{5}{24}x^4$$
The expansion for $e^{-x}$ and $\cos(x)$ are: $$e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+...\rightarrow e^{-x}=1-x+\frac{x^2}{2!}-\frac{x^3}{3!}+...$$ $$\cos(x)=1-\frac{x^2}{2!}+\frac{x^4}{4!}-...$$
When I multiply and combine terms I get: $$1\left ( 1-\frac{x^2}{2!}+\frac{x^4}{4!} \right )-x\left ( 1-\frac{x^2}{2!} \right )+\frac{x^2}{2!}\left ( 1-\frac{x^2}{2!} \right )-\frac{x^3}{3!}\left ( 1 \right )$$ $$\rightarrow \left ( 1-\frac{x^2}{2} +\frac{x^4}{24}\right )+\left ( -x+\frac{x^3}{2} \right )+\left ( \frac{x^2}{2}-\frac{x^4}{4} \right )+\left ( -\frac{x^3}{6} \right )$$ $$\rightarrow 1-x+\left ( -\frac{1}{2}+\frac{1}{2} \right )x^2+\left ( \frac{1}{2}-\frac{1}{6} \right )x^3+\left ( \frac{1}{24}-\frac{1}{4} \right )x^4$$ $$\rightarrow 1-x+\left ( 0 \right )x^2+\left ( \frac{2}{6} \right )x^3+\left ( -\frac{5}{24} \right )x^4$$ $$\rightarrow 1-x+\frac{1}{3}x^3-\frac{5}{24}x^4$$
You are missing the term $\frac{x^4}{4!} \cdot 1$ from $(\cos x)(e^{-x})$. Everything else is correct.