a. Use the definition of a Taylor series to find the first four nonzero terms of the Taylor series for the given function centered at a.

The value of \(f(x)=\cos x\) and its derivatives at \(x=\pi\) are computed in the below table,

\(\begin{array}{|c|c|}\hline k&f^k(x)&f^k(\pi)\\\hline0&f(x)=\cos x&f(\pi)=-1\\\hline1&f'(x)=-\sin x&f'(\pi)=0\\\hline2&f''(x)=-\cos x&f''(\pi)=1\\\hline3&f'''(x)=\sin x&f'''(\pi)=0\\\hline4&f^4(x)=\cos x&f^4(\pi)=-1\\\hline5&f^5(x)=-\sin x&f^5(\pi)=0\\\hline6&f^6(x)=-\cos x&f^6(\pi)=1\\\hline\end{array}\)

The Taylor series centered at \(a=\pi\) is computed as follows,

\(f(x)=f(\pi)+f'(\pi)(x)+\frac{f''(\pi)(x^2)}{2!}+\frac{f'''(\pi)(x^3}{3!}+...\)

Plug the required values from the table in the above,

\(f(x)=-1+0(x-\pi)+\frac{(1)}{2!}(x-\pi)^2+\frac{0}{3!}(x-\pi)^3+\frac{(-1)}{4!}(x-\pi)^4+\frac{0}{5!}(x-\pi)^5+\frac{(1)}{6!}(x-\pi)^6+...\)

\(=-1+0+\frac{(1)}{2!}(x-\pi)^2+0+\frac{(-1)}{4!}(x-\pi)^4+0+\frac{(1)}{6!}(x-\pi)^6+...\)

\(=-1+\frac12(x-\pi)^2-\frac{1}{24}(x-\pi)^4+\frac{1}{720}(x-\pi)^6+...\)

Therefore, the first four nonzero terms of the Taylor series centered at \(a=\pi\) is

\(-1+\frac12(x-\pi)^2-\frac{1}{24}(x-\pi^4)+\frac{1}{720}(x-\pi)^6+...\)

b). From part (a), the first four nonzero terms of the Taylor series centered at \(a=\pi\) is

\(-1+\frac12(x-\pi)^2-\frac{1}{24}(x-\pi^4)+\frac{1}{720}(x-\pi)^6+...\)

Here, the series is alternative and the general kth term can be written as \(\frac{x^{2k}}{(2k)!}(x-\pi)^{2k}\)

\(-1+\frac12(x-\pi)^2-\frac{1}{24}(x-\pi^4)+\frac{1}{720}(x-\pi)^6+...=\sum_{k=0}^\infty\frac{x^{k+1}}{(2k)!}(x-\pi)^{2k}\)

Therefore, the summation notation of the series is \(\sum_{k=0}^\infty\frac{x^{k+1}}{(2k)!}(x-\pi)^{2k}\)

The value of \(f(x)=\cos x\) and its derivatives at \(x=\pi\) are computed in the below table,

\(\begin{array}{|c|c|}\hline k&f^k(x)&f^k(\pi)\\\hline0&f(x)=\cos x&f(\pi)=-1\\\hline1&f'(x)=-\sin x&f'(\pi)=0\\\hline2&f''(x)=-\cos x&f''(\pi)=1\\\hline3&f'''(x)=\sin x&f'''(\pi)=0\\\hline4&f^4(x)=\cos x&f^4(\pi)=-1\\\hline5&f^5(x)=-\sin x&f^5(\pi)=0\\\hline6&f^6(x)=-\cos x&f^6(\pi)=1\\\hline\end{array}\)

The Taylor series centered at \(a=\pi\) is computed as follows,

\(f(x)=f(\pi)+f'(\pi)(x)+\frac{f''(\pi)(x^2)}{2!}+\frac{f'''(\pi)(x^3}{3!}+...\)

Plug the required values from the table in the above,

\(f(x)=-1+0(x-\pi)+\frac{(1)}{2!}(x-\pi)^2+\frac{0}{3!}(x-\pi)^3+\frac{(-1)}{4!}(x-\pi)^4+\frac{0}{5!}(x-\pi)^5+\frac{(1)}{6!}(x-\pi)^6+...\)

\(=-1+0+\frac{(1)}{2!}(x-\pi)^2+0+\frac{(-1)}{4!}(x-\pi)^4+0+\frac{(1)}{6!}(x-\pi)^6+...\)

\(=-1+\frac12(x-\pi)^2-\frac{1}{24}(x-\pi)^4+\frac{1}{720}(x-\pi)^6+...\)

Therefore, the first four nonzero terms of the Taylor series centered at \(a=\pi\) is

\(-1+\frac12(x-\pi)^2-\frac{1}{24}(x-\pi^4)+\frac{1}{720}(x-\pi)^6+...\)

b). From part (a), the first four nonzero terms of the Taylor series centered at \(a=\pi\) is

\(-1+\frac12(x-\pi)^2-\frac{1}{24}(x-\pi^4)+\frac{1}{720}(x-\pi)^6+...\)

Here, the series is alternative and the general kth term can be written as \(\frac{x^{2k}}{(2k)!}(x-\pi)^{2k}\)

\(-1+\frac12(x-\pi)^2-\frac{1}{24}(x-\pi^4)+\frac{1}{720}(x-\pi)^6+...=\sum_{k=0}^\infty\frac{x^{k+1}}{(2k)!}(x-\pi)^{2k}\)

Therefore, the summation notation of the series is \(\sum_{k=0}^\infty\frac{x^{k+1}}{(2k)!}(x-\pi)^{2k}\)