Finding the Taylor Expansion for \(f(x) =\cos x\) about \(\frac{\pi }{2}\)

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 \((-\infty ,\infty).\)