Chapter 9.3, Problem 45E

a.

To determine

To state: The approximate value of $P+\sin P$ using a calculator that is correct to at least $3n$ decimal places, assume that $P$ an approximation of $\pi$ accurate to n decimal places.

Answer to Problem 45E

The resultant answer is $P+\sin P\approx 3.41592653$ .

Explanation of Solution

Given information:

It is given that $P$ an approximation of $\pi$ accurate to n decimal places.

Consider the given information:

P is an approximation of $\pi$ accurate to n decimal places.

Let’s take approximate value of $\pi$ accurate to two decimal places.

That is $P=3.14$ .

Now use a calculator to find the value of $P+\sin P$ .

Therefore, the value is $P+\sin P\approx 3.41592653$ which is accurate to more than 6 decimals places.

b.

To determine

To state: What happened in part (a) using the alternating series error bound and the Maclaurin series for $\sin x$ .

Answer to Problem 45E

The resultant answer is $\left(P+\sin P\right)-\pi <\frac{{\left|x\right|}^3}{6}$ .

Explanation of Solution

Given information:

From part (a) $P+\sin P\approx 3.41592653$ .

Let $P=\pi +x$ where x the error in the original estimate.

  $\begin{array}{c}P+\sin P=\left(\pi +x\right)+\sin \left(\pi +x\right)\\ P+\sin P=\pi +x-\sin x\\ \left(P+\sin P\right)-\pi =x-\sin x\cdots \left(1\right)\end{array}$

The given function is $f\left(x\right)=\sin x$ . Then $P_1\left(x\right)=x$ using Remainder Estimation Theorem for $n=2$ . Since $P_2\left(x\right)=x$ .

Now $f^{\left(3\right)}\left(x\right)=-\cos x$ .

  $\left|f^{\left(3\right)}\left(x\right)\right|=\left|-\cos x\right|\le 1$ for all real x .

It can be used as $M=r=1$ which gives $\left|R_2\left(x\right)\right|<\frac{{\left|x\right|}^3}{3!}$ where $\left|R_2\left(x\right)\right|=\left|x-\sin x\right|$ .

Then, $\left|x-\sin x\right|<\frac{{\left|x\right|}^3}{6}\cdots \left(2\right)$ .

Therefore, from equations (1) and (2) $\left(P+\sin P\right)-\pi <\frac{{\left|x\right|}^3}{6}$ .