Chapter 9.3, Problem 32E

(a)

To determine

To find: P(x) by computing the coefficients.

Answer to Problem 32E

P(x) is $P(x)=\frac{\sqrt{2}}{2}+\frac{5\sqrt{2}}{2}x-\frac{25\sqrt{2}}{2\left(2!\right)}{x}^2-\frac{125\sqrt{2}}{2\left(3!\right)}{x}^3$

Explanation of Solution

Given:

The function is $f(x)=\sin \left(5x+\frac{\pi }{4}\right)$

Calculation:

  $\begin{array}{l}f(0)=\sin \left(5(0)+\frac{\pi }{4}\right)\\ f(0)=\sin \left(\frac{\pi }{4}\right)=\frac{\sqrt{2}}{2}\end{array}$

  $\begin{array}{l}f\text{'}(0)=5\cos \left(5(0)+\frac{\pi }{4}\right)\\ f\text{'}(0)=5\cos \left(\frac{\pi }{4}\right)=\frac{5\sqrt{2}}{2}\end{array}$

  $\begin{array}{l}f\text{'}\text{'}(0)=-25\sin \left(5(0)+\frac{\pi }{4}\right)\\ f\text{'}\text{'}(0)=-25\sin \left(\frac{\pi }{4}\right)=\frac{-25\sqrt{2}}{2}\end{array}$

  $\begin{array}{l}f\text{'}\text{'}\text{'}(0)=-125\cos \left(5(0)+\frac{\pi }{4}\right)\\ f\text{'}\text{'}\text{'}(0)=-125\cos \left(\frac{\pi }{4}\right)=\frac{-125\sqrt{2}}{2}\end{array}$

Hence, P(x) can be written as:

  $P(x)=\frac{\sqrt{2}}{2}+\frac{5\sqrt{2}}{2}x-\frac{25\sqrt{2}}{2\left(2!\right)}{x}^2-\frac{125\sqrt{2}}{2\left(3!\right)}{x}^3$

Conclusion:

The given function increases in the interval $(0,2]$ .

(b)

To determine

To find: The coefficient for $x^{22}$

Answer to Problem 32E

The coefficient for $x^{22}$ is $\frac{-5^{22}\sqrt{2}}{2\left(22!\right)}$ .

Explanation of Solution

Given:

The function is $f(x)=\sin \left(5x+\frac{\pi }{4}\right)$ .

Calculation:

From the Taylor series, the value of $x^{22}$ is:

  $\frac{-5^{22}\sqrt{2}}{2\left(22!\right)}$

Conclusion:

The coefficient for $x^{22}$ is $\frac{-5^{22}\sqrt{2}}{2\left(22!\right)}$ .

(c)

To determine

To show: $\left|f\left(\frac{1}{10}\right)-P_3\left(\frac{1}{10}\right)\right|<\frac{1}{100}$ .

Answer to Problem 32E

The given quantity $\left|f\left(\frac{1}{10}\right)-P_3\left(\frac{1}{10}\right)\right|<\frac{1}{100}$ is showed.

Explanation of Solution

Given:

The function is $f(x)=\sin \left(5x+\frac{\pi }{4}\right)$ .

Calculation:

  $\left|f\left(\frac{1}{10}\right)-P_3\left(\frac{1}{10}\right)\right|=\left|\frac{f^{(4)}(c){(0.1)}^4}{4!}\right|$

The value of varies from $0\le c\le 0.1$ , then

  $f^{(4)}(x)=5^4\sin \left(5x+\frac{\pi }{4}\right)$

This shows; $f^{(4)}(c)\le 5^4$

Hence, it can be said that:

  $\left|f\left(\frac{1}{10}\right)-P_3\left(\frac{1}{10}\right)\right|=\left|\frac{f^{(4)}(c){(0.1)}^4}{4!}\right|<\frac{5^4{(0.1)}^4}{4!}<\frac{1}{100}$

Conclusion:

The given quantity $\left|f\left(\frac{1}{10}\right)-P_3\left(\frac{1}{10}\right)\right|<\frac{1}{100}$ is showed.

(d)

To determine

The third order Taylor polynomial for G.

Answer to Problem 32E

The third order Taylor polynomial for G is $\frac{\sqrt{2}}{2}x+\frac{5\sqrt{2}}{4}{x}^2-\frac{25\sqrt{2}}{12}{x}^3$ .

Explanation of Solution

Given:

  $G(x)={\displaystyle \underset{0}{\overset{x}{\int }}f(t)dt}$ and centered at x=0.

Calculation:

  $\begin{array}{l}Atx=0={\displaystyle \underset{0}{\overset{x}{\int }}\left(\frac{\sqrt{2}}{2}+\frac{5\sqrt{2}}{2}t-\frac{25\sqrt{2}}{4}{t}^2\right)dt}\\ =\frac{\sqrt{2}}{2}x+\frac{5\sqrt{2}}{4}{x}^2-\frac{25\sqrt{2}}{12}{x}^3\end{array}$

Conclusion:

The third order Taylor polynomial for G is $\frac{\sqrt{2}}{2}x+\frac{5\sqrt{2}}{4}{x}^2-\frac{25\sqrt{2}}{12}{x}^3$ .