Find series expansion of 1/cosx Find the series expansion of 1/cosx from basic series expansions.
I tried to find 1/cosx from the expansion of cosx but was unsure how to continue. When I found 1/cosx from the basic formula for finding series expansions I didn't get the same answer.
 A: $\sec\left(x\right)=\dfrac{1}{\cos \left(x\right)}$ is well-behaved about $x=0$, so assume 
$$
\frac{1}{\cos \left(x\right)} = a_0 + a_1 x + a_2 x^2 + a_3 x^3 + \ldots 
$$ 
Also, $\cos \left(x\right)$ is 
$$
\cos \left(x\right) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \ldots  
$$
So, on performing multiplication of the two series,
$$
1 \equiv a_0 + a_1 x +  \left(a_2 - \frac{1}{2!}\right)x^2 + \left(a_3 - \frac{a_1}{2}\right)x^3 + \left(a_4 - \frac{a_2}{2} + \frac{1}{24}\right)x^4 \ldots
$$
which, \left(on equating the coefficients on powers of x on the left and right hand sides\right) gives the first few terms as 
$a_0=1, a_1=0, a_2=\dfrac{1}{2}, a_3=0, a_4=\dfrac{1}{4}-\dfrac{1}{24} = \dfrac{5}{24}$. You can proceed to get more terms or a general formula for the nth term.
A: (going to the fifth term for an example purpose)
Using the basic expansions of cos(x) gives us
$$
\frac{1}{\cos(x)} = \frac{1}{1-\frac{x^2}{2}+\frac{x^4}{24} + \cdots}
$$
of the form  $ \frac{1}{1-X} $ which has a known and easy expansion :
$$
1+X+X^2+X^3+X^4+X^5+\cdots
$$
where $ X = \frac{x^2}{2} + \frac{x^4}{24} $ (approching $0$ when $x$ approaches $0$). The smaller $x$ term is $x^2$, so we don't need to take more terms than $X^2$ in the above expansion (otherwise terms would exceed $x^5$ and be negligeable).
Hence,
$$
\frac{1}{\cos(x)} = 1+(\frac{x^2}{2}−\frac{x^4}{24})+(\frac{x^2}{2})^2+o_{x\to0}(x^5)
$$
$$
= 1+\frac{x^2}{2}+\frac{5x^4}{24} 
$$
A: $\displaystyle\dfrac1{\cos x}=\sum_{n=0}^\infty a_{n}\dfrac{x^{2n}}{(2n)!},~$ where $a_n$ form this sequence. See Euler number or zigzag number for 
more details.
