Question

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

$\cos x,\frac{\mathrm{\pi }}{2}$

Short answer

Ans: The fourth Taylor polynomial for the specified function is$=-\left(x-\frac{\pi }{2}\right)+\frac{1}{6}{\left(x-\frac{\pi }{2}\right)}^3$.

Step-by-step solution

Step 1. Given information:

$f\left(x\right)=\cos x$

Step 2. The fourth Taylor polynomial:

Since for any function $f$with a derivative of order 4 at $x=\frac{\pi }{2}$, the fourth Taylor polynomial for $x=\frac{\pi }{2}$is given by

$$\begin{gathered} P_4\left(x\right)=f\left(\frac{\pi }{2}\right)+f^{\text{'}}\left(\frac{\pi }{2}\right)\left(x-\frac{\pi }{2}\right)+\frac{f^{\text{'}\text{'}}\left(\frac{\pi }{2}\right)}{2!}{\left(x-\frac{\pi }{2}\right)}^2 \\ +\frac{f^{\text{'}\text{'}}\left(\frac{\pi }{2}\right)}{3!}{\left(x-\frac{\pi }{2}\right)}^3+\frac{f^{\text{'}\text{'}\text{'}}\left(\frac{\pi }{2}\right)}{4!}{\left(x-\frac{\pi }{2}\right)}^4 \end{gathered}$$

Therefore, first, 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 $x=\frac{\pi }{2}$

Therefore, first, 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 $x=\frac{\pi }{2}$

Step 3. Finding the fourth Taylor polynomial through derivative of order 4:

$$\begin{gathered} Thus,thevalueofthethefunctionstx=\frac{\mathrm{\pi }}{2}is \\ f\left(\frac{\mathrm{\pi }}{2}\right)=\cos \left(\frac{\mathrm{\pi }}{2}\right) \\ =0 \\ Thederivativesofthefunctionf\left(x\right)=\cos \left(x\right)are \\ f\text{'}\left(x\right)=\frac{d}{dx}\left[\cos x\right] \\ =-\sin x \\ So,atx=\frac{\mathrm{\pi }}{2} \\ f\text{'}\left(\frac{\mathrm{\pi }}{2}\right)=-\sin \left(\frac{\mathrm{\pi }}{2}\right) \\ =-1 \\ Also, \\ f\text{'}\text{'}\left(x\right)=\frac{d}{dx}[-\sin x] \\ =-\cos x \\ So,atx=\frac{\mathrm{\pi }}{2} \\ f\text{'}\left(\frac{\mathrm{\pi }}{2}\right)=-\cos \left(\frac{\mathrm{\pi }}{2}\right) \\ =0 \\ Again, \\ f\text{'}\text{'}\text{'}\text{'}\left(x\right)=\frac{d}{dx}\left[\sin x\right] \\ =\cos x \\ So,atx=\frac{\mathrm{\pi }}{2} \\ f\text{'}\left(\frac{\mathrm{\pi }}{2}\right)=\cos \left(\frac{\mathrm{\pi }}{2}\right) \\ =0 \end{gathered}$$

Step 4. Substituting the derivative of order 4 in the fourth Taylor polynomial :

Therefore, the fourth Taylor polynomial for the function$f\left(x\right)=\cos x$ is

$$\begin{gathered} P_4\left(x\right)=0+-1·\left(x-\frac{\pi }{2}\right)+\frac{0}{2!}{\left(x-\frac{\pi }{2}\right)}^2+\frac{1}{3!}{\left(x-\frac{\pi }{2}\right)}^3+\frac{0}{4!}{\left(x-\frac{\pi }{2}\right)}^4 \\ =-\left(x-\frac{\pi }{2}\right)+\frac{1}{3!}{\left(x-\frac{\pi }{2}\right)}^3 \\ =-\left(x-\frac{\pi }{2}\right)+\frac{1}{6}{\left(x-\frac{\pi }{2}\right)}^3 \end{gathered}$$