Quick Method to Calculate the Maclaurin Series of $\frac{1}{\sqrt{\cos{x}}} $

Question

I am supposed to calculate the maclaurin series for $\frac{1}{\sqrt{\cos{x}}} $ but I can't seem to figure out an efficient way to go about doing this.

Answer 1

Hint. You may use, as $x \to 0$, $$ \cos x =1-\frac{x^2}{2!}+\frac{x^4}{4!}+O(x^6) \tag1 $$ and, as $u \to 0$, $$ \frac 1{\sqrt{1-u}} =1+\frac{u}{2}+\frac{3}{8}u^2+O(u^3) \tag2 $$ giving $$ \begin{align} \frac 1{\sqrt{\cos x}}&=\frac 1{\sqrt{1-\frac{x^2}{2!}+\frac{x^4}{4!}+O(x^6)}}\\\\ &=1+\frac12\left(\frac{x^2}{2!}-\frac{x^4}{4!}+O(x^6) \right)+\frac{3}{8}\left(\frac{x^2}{2!}-\frac{x^4}{4!}\right)^2+O(x^6)\\\\ &=1+\frac{x^2}{4}+\frac{7 x^4}{96}+O(x^6) \end{align} $$ $(1)$ and $(2)$ have been obtained by Taylor expansions near $0$.

Answer 2

Just to give another way which will be very close to Olivier Oloa's answer, I would have done $$\cos(x)=1-\frac{x^2}{2}+\frac{x^4}{24}+O\left(x^6\right)$$ Then, long division to get $$\frac{1}{\cos(x)}=1+\frac{x^2}{2}+\frac{5 x^4}{24}+O\left(x^6\right)$$ and then apply $$\sqrt{1+u}=1+\frac{u}{2}-\frac{u^2}{8}+O\left(u^3\right)$$ so $$\frac{1}{\sqrt{\cos{x}}}=1+\frac 12\Big(\frac{x^2}{2}+\frac{5 x^4}{24}+\cdots\Big)-\frac 18\Big(\frac{x^2}{2}+\frac{5 x^4}{24}+\cdots\Big)^2=1+\frac{x^2}{4}+\frac{7 x^4}{96}+\cdots$$