Q. 81 - Calculus

Question

In Exercises 77–82,

(a) Use appropriate Maclaurin series to express the quantities
in series form.

(b) Use Lagrange’s form for the remainder to bound the error
in using the 5th degree Maclaurin polynomial to approximate the given quantity.

(c) Find the smallest value of  so that the nth degree Maclaurin polynomial approximation to the given quantity is guaranteed to be accurate to within 

${10}^{-6}$

$\sin 1$

Step-by-Step Solution

Answer

(a) The function in series form is $\sin 1=\sum _{k=1}^{\infty }\frac{(-1{)}^2}{(2k+1)!}$

(b) $\left|R_3\left(x\right)\right|\le \frac{1}{720}$

(c) the smallest value of  so that the the degree Maclaurin polynomial approximation to the given quantity is guaranteed to be accurate to within  is .

Step 1 Given Information

Consider the function $f\left(x\right)=\sin x$

Part (a) Step 1

The objective is to use the appropriate Maclaurin series to express the term in series form.
The Maclaurin series for the function$f\left(x\right)=\sin x$ is $\sin x=\sum _{i=1}^{\infty }\frac{(-1{)}^2}{(2k+1)!}{x}^{2k+1}$.
Therefore, the function $\sin x$ for $x=1$ in series form is $\sin 1=\sum _{1=1}^{\infty }\frac{(-1{)}^2}{(2k+1)!}(1{)}^{2k+1}$.
Thus, the given function in series form is $\sin 1=\sum _{k=1}^{\infty }\frac{(-1{)}^2}{(2k+1)!}$.

Part (b) Step 1

Consider the function $f\left(x\right)=\sin x$.
The objective is to use the Lagrange's form for the remainder to bound the error in using the$5^{\text{th }}$ degree Maclaurin polynomial to approximate the above function.
The Lagrange's form for the remainder is $R_n\left(x\right)=\frac{f^{(n+1)}\left(c\right)}{(n+1)!}{x}^{n+1}$.
Therefore, the Lagrange's form for the remainder to bound the error in using the $5^{\text{th }}$ degree Maclaurin polynomial to approximate the given quantity is to find $R_5\left(x\right)$, that is
$\left|R_5\left(x\right)\right|\le \left|\frac{f^{\left(6\right)}\left(c\right)}{6!}{x}^6\right|$

Part (b) Step 2

The derivatives of the function $f\left(x\right)=\sin x$ is calculated as follows:
$\begin{array}{rcl}{f}^{\text{'}}\left(x\right)& =& \frac{d}{dx}\left[\sin x\right]\\ & =& \cos x\end{array}$

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

$\begin{array}{rcl}{f}^3\left(x\right)& =& \frac{d}{dx}[-\sin x]\\ & =& -\cos x\end{array}$

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

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

$\begin{array}{rcl}{f}^{\left(6\right)}\left(x\right)& =& \frac{d}{dx}\left[\cos x\right]\\ & =& -\sin x\end{array}$

Part (b) Step 3

Substitute$x=1$ into the sixth derivative of the function$f\left(x\right)=\sin x$ to calculate the bound for $\left|R_5\left(x\right)\right|$ as follows:
$f^{\left(\theta \right)}\left(1\right)=-\sin 1$
Substitute the value of $f^{\left(\theta \right)}\left(1\right)=-\sin 1$ into $\left|R_4\left(x\right)\right|\le \left|\frac{f^{\left(\theta \right)}\left(c\right)}{6!}{x}^6\right|$.
$\left|R_5\left(x\right)\right|\le \left|\frac{-\sin 1}{720}(1{)}^6\right|$
As $\sin 1\approx 1$, therefore, $\left|R_3\left(x\right)\right|\le \frac{1}{720}$.

Part (c) Step 1

Consider the function $f\left(x\right)=\sin x$.
The objective is to find the smallest value $n$ so that the approximation by Maclaurin series is accurate within ${10}^{-6}$.
As $\sin x=x-\frac{x^3}{3!}+\frac{x^3}{5!}-\frac{x^7}{7!}+\cdots$, therefore, $\sin 1=1-\frac{1^3}{3!}-\frac{1^3}{5!}-\cdots$
Hence, by trial and error, the smallest value of $n$ so that the $n$the degree Maclaurin polynomial approximation to the given quantity is guaranteed to be accurate to within ${10}^{-6}$ is $11$ .