I'm having difficulty understanding how to interpret the order of more complicated Taylor series.
Such as
$$\frac{1}{\cos(x)}=1+\frac{x^2}{2!}+\frac{5}{24}x^4 +\cdot \cdot \cdot$$
This can be formed up to the term $\frac{x^2}{2!}$ by combining the 2nd order (so up to 3rd term) expansions of simpler functions $\cos(x)$ and $\frac{1}{1-x}$. This is done like:
$$\frac{1}{cos(x)} = \frac{1}{1-(1-\cos(x))}= 1 + (1-1+\frac{x^2}{2!}+ o(x^2))+(1-1+\frac{x^2}{2!} + o(x^2))^2$$
and since the very last term is certainly $o(x^2)$ $$= 1 + \frac{x^2}{2!} + o(x^2)$$
But is this result the 2nd or 1st order expansion of $\frac{1}{\cos(x)}$? Should I look for the number of terms (so in the above there's the 0th term and the 1st term) when interpreting order or the highest power $k$ of $x^k$?
My definition for Taylor series uses the convention that in $f(a)+f'(a)(x-a)+\frac{f''(a)}{2}(x-a)^2 + \cdot \cdot \cdot$ the first term (no derivative) is 0th order and the second is 1st order and so on. The order also corresponds to the $k$ of the following $x^k$ of each term.
I'm unfamiliar with the exact definition of the "order" of an expansion, but it's pretty obvious here that you won't have any odd degrees in the polynomial, so the expansion will look something like
$$\sum_{n=0}^{\infty} f(n)x^{2n} = 1$$ So I'm guessing that what you found is the 2nd order expansion, because you found the first two terms (although you could also call it a first order expansion if you start counting from zero).