Taylor expansion of $\dfrac{e^x}{\cos x}$ Let $g\left(x\right) = \dfrac{e^x}{\cos x}$ and find the first four terms of its taylor polynomial. 
The first four terms are: $$1 + x + x^2 + \frac{2x^3}{3} + \frac{x^4}{2}$$ 
Why do you have to solve this problem term by term? Why does just dividing the terms of the series work out? 
Why doesn't the division of the series of $e^x$ and $\cos (x)$ which gives $\sum \limits_{n=0}^{\infty} \dfrac{x^n (2n)!}{n! (-1)^{n}(x^{2n})} $ give the correct series. 

I hate denominators. 
First write ${e^x\over\cos(x)}=a_0+a_1x+a_2x^2+a_3x^3+\cdots$ 
Then, since I hate denominators, I consider $e^x=\cos(x)(a_0+a_1x+a_2x^2+a_3x^3+\cdots)$
Let's work from the right hand side, and take an intuitive guess that we only need expand $\cos(x)$ to a third degree polynomial: 
$$\begin{align*}
\cos(x)(a_0+a_1x+a_2x^2+a_3x^3+\cdots) & = (1-x^2/2+\cdots)(a_0+a_1x+a_2x^2+a_3x^3+\cdots) \\
%& = (a_0+a_1x+a_2x^2+a_3x^3+\cdots)-(x^2/2)(a_0+a_1x+a_2x^2+a_3x^3+\cdots)+\cdots\\
& = a_0+a_1x+(a_2-a_0/2)x^2+(a_3-a_2/2)x^3+\cdots\\
& = e^x\\
& = 1+x+x^2/2+x^3/6
\end{align*}$$
Thus, $a_0=1$, $a_1=1$, $a_2=1$, $a_3=2/3$
$${e^x\over\cos(x)}=1+x+x^2+{2\over3}x^3+\cdots$$
So remember, when dividing power series, "I hate denominators". 
