In Exercises 41–48 find the fourth Taylor polynomial $P_4\left(x\right)$for the specified function and the given value of $x_0$.

$\text{47.}\cos 2x,\frac{\pi }{4}$

We have the given function $f\left(x\right)=\cos \left(2x\right)$with a derivative of order$4$ at $x=\frac{\mathrm{\pi }}{4}$.

The fourth taylor polynomial for $x=\frac{\mathrm{\pi }}{4}$is given by,

$\begin{array}{rl}{P}_4\left(x\right)=& f\left(\frac{\pi }{4}\right)+f^{\mathrm{\prime }}\left(\frac{\pi }{4}\right)(x-\frac{\pi }{4})+\frac{f^{\prime \prime }\left(\frac{\pi }{4}\right)}{2!}{(x-\frac{\pi }{4})}^2+\frac{f^{\prime \prime \mathrm{\prime }}\left(\frac{\pi }{4}\right)}{3!}{(x-\frac{\pi }{4})}^3+\frac{f^{\prime \prime \prime \prime }\left(\frac{\pi }{4}\right)}{4!}{(x-\frac{\pi }{4})}^4\end{array}$

Therefore, we have to find the value of the function along with $f^{\text{'}}\left(x\right),f^{\text{'}\text{'}}\left(x\right),f^{\text{'}\text{'}\text{'}}\left(x\right)$and$f\text{'}\text{'}\text{'}\text{'}\left(x\right)$at $x=\frac{\mathrm{\pi }}{4}$.

The value of the function at $x=\frac{\mathrm{\pi }}{4}$is,

$\begin{array}{rl}f\left(\frac{\pi }{4}\right)& =\cos (2\cdot \frac{\pi }{4})\\ & =\cos \left(\frac{\pi }{2}\right)\\ & =0\end{array}$

The derivatives of the function, $f\left(x\right)=\cos \left(2x\right)$

$\begin{array}{rl}{f}^{\mathrm{\prime }}\left(x\right)& =\frac{d}{dx}\left[\cos 2x\right]\\ & =-2\sin 2x\end{array}$$\begin{array}{rl}{f}^{\mathrm{\prime }}\left(x\right)& =\frac{d}{dx}\left[\cos 2x\right]\\ & =-2\sin 2x\end{array}$

So, at $x=\frac{\pi }{4}$

$\begin{array}{rl}{f}^{\mathrm{\prime }}\left(0\right)& =-2\sin (2\cdot \frac{\pi }{4})\\ & =-2\sin \left(\frac{\pi }{2}\right)\\ & =-2\cdot 1\\ & =-2\end{array}$

$\begin{array}{rl}{f}^{\prime \prime }\left(x\right)& =\frac{d}{dx}[-2\sin 2x]\\ & =-2\frac{d}{dx}\left[\sin 2x\right]\\ & =-4\cos 2x\end{array}$

So, at $x=\frac{\pi }{4}$

$\begin{array}{rl}{f}^{\prime \prime }\left(0\right)& =-4\cos (2\cdot \frac{\pi }{4})\\ & =-4\cos \left(\frac{\pi }{2}\right)\\ & =-4\cdot 0\\ & =0\end{array}$

$\begin{array}{rl}{f}^{\prime \prime \mathrm{\prime }}\left(x\right)& =-4\frac{d}{dx}\left[\cos 2x\right]\\ & =-4(-2\sin 2x)\\ & =8\sin 2x\end{array}$

So, at $x=\frac{\pi }{4}$

$\begin{array}{rl}{f}^{\prime \prime \mathrm{\prime }}\left(0\right)& =8\sin (2\cdot \frac{\pi }{4})\\ & =8\sin \left(\frac{\pi }{2}\right)\\ & =8\cdot 1\\ & =8\end{array}$

$\begin{array}{rl}{f}^{\prime \prime \prime \prime }\left(x\right)& =\frac{d}{dx}\left[8\sin 2x\right]\\ & =8\frac{d}{dx}\left[\sin 2x\right]\\ & =16\cos 2x\end{array}$

So, at $x=\frac{\pi }{4}$

$\begin{array}{rl}{f}^{\prime \prime \prime \prime }\left(0\right)& =16\cos (2\cdot \frac{\pi }{4})\\ & =16\cos \left(\frac{\pi }{2}\right)\\ & =16\cdot 0\\ & =0\end{array}$

Hence, the fourth Taylor polynomial of the function $f\left(x\right)=\cos 2x$at$x=\frac{\pi }{4}$

$\begin{array}{rl}{P}_4\left(x\right)& =0+-2\cdot (x-\frac{\pi }{4})+\frac{0}{2!}{(x-\frac{\pi }{4})}^2+\frac{8}{3!}{(x-\frac{\pi }{4})}^3+\frac{0}{4!}{(x-\frac{\pi }{4})}^4\\ & =-2(x-\frac{\pi }{4})+\frac{8}{3!}{(x-\frac{\pi }{4})}^3\\ & =-2(x-\frac{\pi }{4})+\frac{4}{3}{(x-\frac{\pi }{4})}^3\end{array}$