Q. 39 - Calculus

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 $R_4\left(x\right)$.  

       $\sin x, \mathrm{\pi }$

Step-by-Step Solution

Answer

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

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  $f^{\left(5\right)}\left(x\right)=\cos x\text{ and }{x}_0=\mathrm{\pi }\text{ then }$

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

 That is,

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