Power series expansion

Lets suppose that the McLaurin expansion of $\frac{1}{\cos x}$ exists and write...

$\frac{1}{\cos x} = \sum_{n=0}^{\infty} a_{n}\cdot x^{n}$ (1)

In this case we can derive the $a_{n}$ from the McLaurin expansion of $\cos x$ ...

$\cos x = \sum_{k=0}^{\infty} b_{k}\cdot x^{k}$ (2)

... where $b_{k} = \frac{(-1)^{\frac{k}{2}}}{k!}$ for $k$ even and $b_{k}=0$ for $k$ odd. If we write the product of the series (1) and (2) we obtain...

$\sum_{n=0}^{\infty} a_{n}\cdot x^{n} \cdot \sum_{k=0}^{\infty} b_{k}\cdot x^{k} = \sum_{n=0}^{\infty} x^{n}\cdot \sum_{k=0}^{n} a_{k}\cdot b_{n-k} = 1$ (3)

... and from (3) we obtain...

$a_{0}\cdot b_{0}= 1 \rightarrow a_{0} = 1$

$a_{0}\cdot b_{1} + a_{1}\cdot b_{0} = 0 \rightarrow a_{1} = 0$

$a_{0}\cdot b_{2} + a_{1}\cdot b_{1} + a_{2}\cdot b_{0} = 0 \rightarrow a_{2}= \frac{1}{2}$

$a_{0}\cdot b_{3} + a_{1}\cdot b_{2} + a_{2}\cdot b_{1} + a_{3}\cdot b_{0} = 0 \rightarrow a_{3} = 0$

$a_{0}\cdot b_{4} + a_{1}\cdot b_{3} + a_{2}\cdot b_{2} + a_{3}\cdot b_{1} + a_{4}\cdot b_{0} = 0 \rightarrow a_{4} = - \frac{5}{24}$

The McLaurin expansion of $\frac{1}{\cos x}$ with maximum exponent 4 is then...

$\frac{1}{\cos x} = 1 + \frac{1}{2}\cdot x^{2} - \frac{5}{24}\cdot x^{4} + \dots$ (4)

Kind regards

$\chi$ $\sigma$