Question

In Exercises 41–48 in Section 8.2, you were asked to find the fourth Taylor polynomial $P_4\left(x\right)$for the specified function and the given value of x 0. Here give Lagrange’s form for the remainder role="math" localid="1650650220589" $R_4\left(x\right)$.

$\sin x,\mathrm{\pi }$

Short answer

Ans:$R_4\left(x\right)=\frac{\cos c}{120}(x-\pi {)}^5$

Step-by-step solution

Step 1. Given information.

given,

$\sin x,\mathrm{\pi }$

Step 2. Consider the given function,

The derivatives of the function $f\left(x\right)=\sin x$are

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

Also,

$\begin{array}{r}{f}^{\text{'}\text{'}}\left(x\right)=\frac{d}{dx}[\cos x]\\ =-\sin x\end{array}$

Again,

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

Also

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

Implies that

$f^{\left(4\right)}\left(x\right)=\sin x$

Finally,

$\begin{array}{r}{f}^{\left(5\right)}\left(x\right)=\frac{d}{dx}[\sin x]\\ =\cos x\end{array}$

Step 3. Now,

by the Lagrange's form for the remainder, if f is a function that can be differentiated n+1 times in some open interval / containing the point $x_0$and $R_n\left(x\right)$ be the nth remainder for f at $x=x_0$. Then there exists at least one c between $x_0$and x such that

$R_n\left(x\right)=\frac{f^{(n+1)}\left(c\right)}{(n+1)!}{\left(x-x_0\right)}^{n+1}$

Since role="math" localid="1650651004890" $f^{\left(5\right)}\left(x\right)=\cos x\text{and}{x}_0=\mathrm{\pi }\text{then}$

$R_4\left(x\right)=\frac{f^5\left(c\right)}{5!}(x-\pi {)}^5$

That is,

$R_4\left(x\right)=\frac{\cos c}{120}(x-\pi {)}^5$