Find the Maclaurin expansion for $f(x) =\cos x.$

Solution We apply the differentiation property of power series to the Maclaurin expansion for $\sin x.$ $$\begin{eqnarray*} \frac{d}{dx}\sin x &=& \frac{d}{dx}\!\left( x-\frac{x^{3}}{3!}+\frac{x^{5}}{5!} -\cdots +(-1)^{n}\frac{x^{2n+1}}{(2n+1)!}+\cdots \right) \\[5pt] &=&\frac{d}{dx}\sum\limits_{k\,=\,0}^{\infty }\,(-1) ^{k}\frac{x^{2k+1}}{ (2k+1) !} \end{eqnarray*}$$

Then $$\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt]{ \cos x=1-\frac{x^{2}}{2!}+\frac{x^{4}}{4!}-\cdots +(-1)^{n}\frac{x^{2n}}{(2n)!}+\cdots =\sum\limits_{k\,=\,0}^{\infty }\,(-1) ^{k}\frac{x^{2k}}{(2k) !}}}$$

for all numbers $x$.