Chapter 8.8, Problem 15E

(a)

To determine

To Find Taylor’s polynomial with degree n at the number a

Answer to Problem 15E

Taylor polynomials $T_3(x)$

  $T_3(x)=1+{(x)}^2$

Explanation of Solution

Given information:

  $f(x)=e^{x^2},a=0,n=3,0\le x\le 0.1$

Formula Used:

Taylor Polynomials

  $T_n(x)={\displaystyle \sum _{i=0}^n\frac{f^{(i)}(a)}{i!}{(x-a)}^i}=f(a)+\frac{f^{\text{'}}(a)}{1!}(x-a)+\frac{f^{\text{'}\text{'}}(a){(x-a)}^2}{2!}\dots +\frac{f^n(a)}{n!}{(x-a)}^n$

Calculation:

  $f(x)=e^{x^2},a=0,n=3,0\le x\le 0.1$

Taylor polynomials

  $\begin{array}{l}{T}_n(x)\\ T_3(x)=f(a)+\frac{f^{\text{'}}(a)}{1!}(x-a)+\frac{f^{\text{'}\text{'}}(a){(x-a)}^2}{2!}+\frac{f^{\text{'}\text{'}\text{'}}(a){(x-a)}^3}{3!}\\ f(x)=e^{x^2},a=0\\ f(0)=1\\ f^{\text{'}}(0)=0\\ f^{\text{'}\text{'}}(0)=2\\ f^{\text{'}\text{'}\text{'}}(0)=0\\ T_3(x)=f(0)+\frac{f^{\text{'}}(0)}{1!}(x)+\frac{f^{\text{'}\text{'}}(0){(x)}^2}{2!}+\frac{f^{\text{'}\text{'}\text{'}}(0){(x)}^3}{3!}\\ T_3(x)=1+{(}^x\\ \end{array}$

Taylor polynomials $T_3(x)$

  $T_3(x)=1+{(x)}^2$

(b)

To determine

To Find Use Taylor’s Inequality to estimate the accuracy of the approximation

  $f(x)\approx T_n(x)$

Answer to Problem 15E

The approximation is accurate to within

  $5.2\times {10}^{-5}$

Explanation of Solution

Given information:

  $f(x)=e^{x^2},a=0,n=3,0\le x\le 0.1$

Formula Used:

  $T_n(x)\to f(x)$ ???? ??&#??????????;&#??????????;

And $T_n(x)$ can be used as an approximation to

  $f:f(x)=T_n(x)$

Taylor’s Inequality, which says that if

  $\left|f^{(n+1)}(x)\right|\le M$ , then

  $\left|R_n(x)\right|\le \frac{M}{(n+1)!}{\left|x-a\right|}^{n+1}$

Calculation:

Taylor’s Inequality with $a=0$ ?????? ??=3

  $\left|R_3(x)\right|\le \frac{M}{4!}{\left|x\right|}^4$

Where $\left|f^4(x)\right|\le M$

  $f^4(x)=4e^{x^2}\left(4x^4+12x^2+3\right)$

  $\begin{array}{l}M\le 12.6\\ \left|R_3(x)\right|\le \frac{12.6}{4!}{\left|0.1\right|}^4\\ \left|R_3(x)\right|\le 5.2\times {10}^{-5}\end{array}$

The approximation is accurate to within

  $5.2\times {10}^{-5}$

(c)

`

To determine

To check your result by graphing $\left|R_n(x)\right|$

Answer to Problem 15E

Error estimate from graphical method is

  $5.2\times {10}^{-5}$

Explanation of Solution

Given information:

  $f(x)=e^{x^2},a=0,n=3,0\le x\le 0.1$

  $T_3(x)=1+{(x)}^2$

Formula Used:

  $\left|R_n(x)\right|=\left|f(x)-T_n(x)\right|$

Calculation:

  $\begin{array}{l}\left|R_3(x)\right|=\left|f(x)-T_3(x)\right|\\ \left|R_3(x)\right|=\left|e^{x^2}-\left(1+x^2\right)\right|\end{array}$