Find the Taylor expansion for $f(x) =\cos x$ about $\frac{\pi}{2}.$

Solution To express $f(x) =\cos x$ as a Taylor expansion about $\frac{\pi}{2}$, we evaluate $f$ and its derivatives at $\frac{\pi}{2}.$ $$\begin{array}{rl@{\qquad}rll} f(x) &=\cos x & f\!\!\left(\frac{\pi }{2}\right)& =0 \\ f^\prime (x) &=-\sin x & f^\prime \!\!\left( \frac{\pi }{2}\right) &=-1 \\ f^{\prime \prime} (x) &=-\cos x & f^{\prime \prime} \!\!\left( \frac{\pi }{2}\right) &=0 \\ f^{\prime \prime \prime} (x) &=\sin x & f^{\prime \prime \prime} \!\!\left( \frac{\pi }{2}\right) &=1 \end{array}$$

For derivatives of odd order, $f^{(2n+1) }\left( \frac{\pi }{2}\right) =(-1) ^{n+1}.$ For derivatives of even order, $f^{(2n) }\left( \frac{\pi }{2}\right) =0.$ The Taylor expansion for $f(x) =\cos x$ about $\frac{\pi }{2}$ is

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$$\begin{eqnarray*} f(x) &=&\cos x=f\!\!\left( \frac{\pi }{2}\right) +f^\prime \!\!\left( \frac{\pi }{2}\right) \left( x-\frac{\pi }{2}\right) +\frac{f^{\prime \prime} \!\!\left( \frac{\pi }{2}\right) }{2!}\left( x-\frac{\pi }{2}\right) ^{2}\\[6pt] &&+\, \frac{f^{\prime \prime \prime} \!\!\left( \frac{\pi }{2}\right) }{3!}\left( x-\frac{\pi }{2}\right) ^{3}+\cdots \\[6pt] &=&-\left( x-\frac{\pi }{2}\right) +\frac{1}{3!}\left( x-\frac{\pi }{2} \right) ^{3}-\frac{1}{5!}\left( x-\frac{\pi }{2}\right) ^{5}+\text{ }\cdots \\[6pt] &=& \sum\limits_{k\,=\,0}^{\infty }\frac{(-1) ^{k+1}}{(2k+1) !}\left( x-\frac{\pi }{2}\right) ^{2k+1} \end{eqnarray*}$$

The radius of convergence is $\infty ;$ the interval of convergence is $(-\infty ,\infty).$