Chapter 9.2, Problem 19E

To determine

To find: The Taylor polynomials of orders $0,1,2,3$ for the given function.

Answer to Problem 19E

The Taylor’s polynomial is,

  $\begin{array}{l}{P}_0(x)=\frac{\sqrt{2}}{2}\\ P_1(x)=\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}\left(x-\frac{\pi }{4}\right)\\ P_2(x)=\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}\left(x-\frac{\pi }{4}\right)-\frac{\sqrt{2}}{4}{\left(x-\frac{\pi }{4}\right)}^2\\ P_3(x)=\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}\left(x-\frac{\pi }{4}\right)-\frac{\sqrt{2}}{4}{\left(x-\frac{\pi }{4}\right)}^2-\frac{\sqrt{2}}{12}{\left(x-\frac{\pi }{4}\right)}^3\end{array}$

Explanation of Solution

Given information:

  $f\left(x\right)=\sin x,a=\pi /4$

Calculation:

The function at $\pi /4$ corresponds to the Taylor polynomial of function $f$ of order $0$ .

  $\begin{array}{c}{P}_0(x)=\sin \frac{\pi }{4}\\ =\frac{\sqrt{2}}{2}\end{array}$

Let's differentiate the function to obtain the first-order Taylor polynomial.

  $f^{\prime }(x)=\cos x$

Now substituting,

  $f^{\prime }(\pi /4)=\sqrt{2}/2$

Thus, the first-order Taylor polynomial is

  $P_1(x)=\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}\left(x-\frac{\pi }{4}\right)$

Now differentiating $f^{\prime \prime }(x)$ to get,

  $f^{\prime \prime }(x)=-\sin x$

Now substituting,

  $f^{\prime \prime }(\pi /4)=-\sqrt{2}/2$

Thus, the second-order Taylor polynomial is

  $\begin{array}{c}{P}_2(x)=\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}\left(x-\frac{\pi }{4}\right)-\frac{\sqrt{2}}{2\cdot 2!}{\left(x-\frac{\pi }{4}\right)}^2\\ =\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}\left(x-\frac{\pi }{4}\right)-\frac{\sqrt{2}}{4}{\left(x-\frac{\pi }{4}\right)}^2\end{array}$

Now differentiating $f^{\prime \prime }{}^{\prime }(x)$ to get,

  $f^{\prime \prime }{}^{\prime }(x)=-\cos x$

Now substituting,

  $f^{\prime \prime }{}^{\prime }(\pi /4)=-\sqrt{2}/2$

Thus, the third-order Taylor polynomial is

  $\begin{array}{c}{P}_3(x)=\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}\left(x-\frac{\pi }{4}\right)-\frac{\sqrt{2}}{2\cdot 2!}{\left(x-\frac{\pi }{4}\right)}^2-\frac{\sqrt{2}}{2\cdot 3!}{\left(x-\frac{\pi }{4}\right)}^3\\ =\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}\left(x-\frac{\pi }{4}\right)-\frac{\sqrt{2}}{4}{\left(x-\frac{\pi }{4}\right)}^2-\frac{\sqrt{2}}{12}{\left(x-\frac{\pi }{4}\right)}^3\end{array}$

Therefore, the Taylor polynomial for order $0,1,2,3$ is given by,

  $\begin{array}{l}{P}_0(x)=\frac{\sqrt{2}}{2}\\ P_1(x)=\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}\left(x-\frac{\pi }{4}\right)\\ P_2(x)=\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}\left(x-\frac{\pi }{4}\right)-\frac{\sqrt{2}}{4}{\left(x-\frac{\pi }{4}\right)}^2\\ P_3(x)=\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}\left(x-\frac{\pi }{4}\right)-\frac{\sqrt{2}}{4}{\left(x-\frac{\pi }{4}\right)}^2-\frac{\sqrt{2}}{12}{\left(x-\frac{\pi }{4}\right)}^3\end{array}$