Chapter 8.8, Pg 496 Question 7 Find a formula for the \(n^{th}\) term of the Taylor series for
\(f(x)\).
\(f(x) = cosx,\ c = \pi/2\)
Taylor Series is defined as \[ \begin{split} f(x) &= \sum\limits_{n=0}^{\infty}\frac{f^{(n)}(a)}{n!}(x-a)^n \\ &= f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \frac{f'^v(a)}{4!}(x-a)^4 + \ ... \end{split} \]
Find the first 5 derivatives of f(x):
\[
\begin{split}
f(x) &= cosx; \ => f^0(\pi/2) = cos(\pi/2) = 0, \\
f'(x) &= -sinx;\ => f'(\pi/2) = -sin(\pi/2) = -1, \\
f''(x) &= -cosx;\ => f''(\pi/2) = -cos(\pi/2) =
0, \\
f'''(x) &= sinx;\ => f'''(\pi/2) =
sin(\pi/2) = 1, \\
f'^v(x) &= cosx;\ => f'^v(\pi/2) = cos(\pi/2) = 0, \\
f^v(x) &= -sinx;\ => f^v(\pi/2) = -sin(\pi/2) = -1
\end{split}
\]
Following the definition of the Taylor’s series given above,
\[
\begin{split}
f(x) &= \frac{f^0(c)}{0!}(x-c)^0 + \frac{f'(c)}{1!}(x-c)^1 +
\frac{f''(c)}{2!}(x-c)^2 + \frac{f'''(c)}{3!}(x-c)^3
+ \frac{f'^v(c)}{4!}(x-c)^4 +
\frac{f^v(c)}{5!}(x-c)^5 + \ ...\\
&= \frac{0}{0!}(x-\pi/2)^0 + \frac{-1}{1!}(x-\pi/2)^1 +
\frac{0}{2!}(x-\pi/2)^2 + \frac{1}{3!}(x-\pi/2)^3
+ \frac{0}{4!}(x-\pi/2)^4 +
\frac{-1}{5!}(x-\pi/2)^5 + \ ...\\
&= - (x-\pi/2) + \frac{(x-\pi/2)^3}{3!} - \frac{(x-\pi/2)^5}{5!} +
\ ... \\
&= \sum\limits_{n=0}^{\infty} (-1)^{n+1}
\frac{(x-\pi/2)^{2n+1}}{(2n+1)!}
\end{split}
\]
Therefore, the formula for the \(n^{th}\) term of the Taylor series for
\(cosx\) where \(c = \pi/2\) is given by:
\(f(n) = \sum\limits_{n=0}^{\infty} (-1)^{n+1}
\frac{(x-\pi/2)^{2n+1}}{(2n+1)!}\)