Chapter 9.3, Problem 34E

a.

To determine

To write: The first four nonzero terms of the Taylor series for f about $x=0$ .

Answer to Problem 34E

The first four non-zero terms of the Taylor series for f about $x=0$ : $1+\frac{1}{2}{x}^2+\frac{1}{24}{x}^4-\frac{121}{720}{x}^6$

Explanation of Solution

Given information:

  $f(x)={\sin }^{2}x+\cos x$

Concept used:

Taylor series for f centered at $x=0$

(Maclaurin series):

  ${\displaystyle \sum _{k=0}^{\infty }\frac{f^{\left(k\right)}\left(0\right)}{k!}}{x}^k=f\left(0\right)+f^{\prime }\left(0\right)+\frac{f\text{'}\text{'}\left(0\right)}{2!}+\mathrm{...}+\frac{f^{\left(n\right)}\left(0\right)}{n!}+\mathrm{...}$

Compute the value of $f\left(0\right)$ .

  $\begin{array}{c}f\left(0\right)={\sin }^2\left(0\right)+\cos 0\\ =1\end{array}$

Find the first derivative.

  $\begin{array}{c}f\text{'}\left(x\right)=\left({\sin }^{2}x+\cos x\right)\text{'}\\ =2x\cos x^2-\sin x\end{array}$

Compute the value of $f\text{'}\left(0\right)$ .

  $\begin{array}{c}f\text{'}\left(0\right)=2\text{ sin 0}\cos 0-\sin \text{ 0}\\ \text{=0}\end{array}$

Find the second derivative.

  $\begin{array}{c}f\text{'}\text{'}\left(x\right)=\left(2x\cos x^2-\sin x\right)\text{'}\\ =-4x^2\sin x^2-\cos x+2\cos x^2\end{array}$

Compute the value of $f\text{'}\text{'}\left(0\right)$ .

  $\begin{array}{c}f\text{'}\text{'}\left(0\right)=-4{\left(0\right)}^2\sin 0-\cos 0+2\cos 0^2\\ \text{=}-1+2\\ =1\end{array}$

Find the third derivative.

  $\begin{array}{c}{f}^{\left(3\right)}\left(x\right)=\left(-4x^2\sin x^2-\cos x+2\cos x^2\right)\text{'}\\ =-8x^3\cos x^2-12x\sin x^2+\sin x\end{array}$

Compute the value of $f^{\left(3\right)}\left(0\right)$ .

  $f^{\left(3\right)}\left(0\right)=0$

Find the fourth derivative.

  $\begin{array}{c}{f}^{\left(4\right)}\left(x\right)=\left(-8x^3\cos x^2-12x\sin x^2+\sin x\right)\text{'}\\ =16x^4\sin x^2-48x^2\cos x^2-12\sin x^2+\cos x\end{array}$

Compute the value of $f^{\left(4\right)}\left(0\right)$ .

  $f^{\left(4\right)}\left(0\right)=1$

Find the fifth derivative.

  $\begin{array}{c}{f}^{\left(5\right)}\left(x\right)=\left(16x^4\sin x^2-48x^2\cos x^2-12\sin x^2+\cos x\right)\text{'}\\ =32x^5\cos x^2+160x^3\sin x^2-120x\cos x^2-\sin x\end{array}$

Compute the value of $f^{\left(5\right)}\left(0\right)$ .

  $f^{\left(5\right)}\left(0\right)=0$

Find the sixth derivative.

  $\begin{array}{c}{f}^{\left(6\right)}\left(x\right)=\left(32x^5\cos x^2+160x^3\sin x^2-120x\cos x^2-\sin x\right)\text{'}\\ =-64x^6\sin x^2+480x^4\cos x^2+720x^2\sin x^2-\cos x-120\cos x^2\end{array}$

Compute the value of $f^{\left(6\right)}\left(0\right)$ .

  $f^{\left(6\right)}\left(0\right)=-121$

Use the calculated values to get a polynomial.

  $f(x)\approx \frac{1}{0!}{x}^0+\frac{0}{1!}{x}^1+\frac{1}{2!}{x}^2+\frac{0}{3!}{x}^3+\frac{1}{4!}{x}^4+\frac{0}{5!}{x}^5+\frac{-121}{6!}{x}^6+\frac{0}{7!}{x}^7$

Simplify.

  $f(x)\approx P(x)=1+\frac{1}{2}{x}^2+\frac{1}{24}{x}^4-\frac{121}{720}{x}^6$

The first four non-zero terms of the Taylor series for f about $x=0$.: $1+\frac{1}{2}{x}^2+\frac{1}{24}{x}^4-\frac{121}{720}{x}^6$

b.

To determine

To find: The value of $f^{(6)}\left(0\right)$ .

Answer to Problem 34E

The value of $f^{(6)}\left(0\right)$ is -121.

Explanation of Solution

Given information:

  $f(x)={\sin }^{2}x+\cos x$

Since $f^{\left(6\right)}\left(x\right)=-64x^6\sin x^2+480x^4\cos x^2+720x^2\sin x^2-\cos x-120\cos x^2$

Substitute $x=0$ and simplify.

  $\begin{array}{c}{f}^{\left(6\right)}\left(0\right)=-64{\left(0\right)}^6\sin {\left(0\right)}^2+480{\left(0\right)}^4\cos {\left(0\right)}^2+720{\left(0\right)}^2\sin {\left(0\right)}^2-\cos \left(0\right)-120\cos {\left(0\right)}^2\\ =0+0+0-1-120\left(1\right)\\ =-121\end{array}$

Hence, the value of $f^{(6)}\left(0\right)=-121$

c.

To determine

To show: $\left|P_4\left(\frac{1}{4}\right)-f\left(\frac{1}{4}\right)\right|<\frac{1}{3000}$ by using the information from the graph of $y=\left|f^{\left(5\right)}\left(x\right)\right|$ .

Answer to Problem 34E

It is shown that $\left|P_4\left(\frac{1}{4}\right)-f\left(\frac{1}{4}\right)\right|<\frac{1}{3000}$ .

Explanation of Solution

Given information:

  $f(x)={\sin }^{2}x+\cos x$

Let $P_4\left(x\right)$ be the fourth-order Taylor polynomial for f about x=0.

The graph of $y=\left|f^{\left(5\right)}\left(x\right)\right|$ is shown.

  

Concept used:

The partial sum $P_n\left(x\right)={\displaystyle \sum _{k=0}^{\infty }\frac{f^{\left(k\right)}\left(0\right)}{k!}}{x}^k$ is the Taylor polynomial of order n for f at $x=0$ .

Lagrange error: $\left|P_n\left(x\right)-f\left(x\right)\right|=\left|R_n\left(x\right)\right|$

According to the definition of Lagrange error, $\left|P_4\left(\frac{1}{4}\right)-f\left(\frac{1}{4}\right)\right|=\left|R_4\left(\frac{1}{4}\right)\right|$

From the graph of $y=\left|f^{\left(5\right)}\left(x\right)\right|$ the value of $\left|f^{(5)}\left(\frac{1}{4}\right)\right|<40$ .

Thus,

  $\begin{array}{c}\left|R_4\left(\frac{1}{4}\right)\right|\le \left|\frac{f^{(5)}\left(\frac{1}{4}\right)}{5!}{\left(\frac{1}{4}\right)}^5\right|\\ <\frac{40}{120}\cdot \frac{1}{4^5}\\ =\frac{1}{3072}<\frac{1}{3000}\end{array}$

Therefore,

  $\left|P_4\left(\frac{1}{4}\right)-f\left(\frac{1}{4}\right)\right|<\frac{1}{3000}$