Chapter 10.2, Problem 2E

To determine

To find:The fourth-order Taylor polynomial at $x=0$ for the function

  $f(x)=e^{2x}$

Answer to Problem 2E

  $P_4(x)=1+2x+2x^2+\frac{4}{3}{x}^3+\frac{2}{3}{x}^4$

Explanation of Solution

Given information: $f(x)=e^{2x}$

Calculation:The Taylor series generated by $f$ at $x=a$ is,

  $f(a)+f\text{'}(a)(x-a)+\frac{f\text{'}\text{'}(a)}{2!}{(x-a)}^2+\mathrm{...}+\frac{f^{(n)}(a)}{n!}{(x-a)}^n+\mathrm{...}$

Here, $f(x)=e^{2x}$

Then,

  $\begin{array}{l}f(0)=e^{2x}=1\\ f\text{'}(0)={\left[2e^{2x}\right]}_{x=0}=2\\ f\text{'}\text{'}(0)={\left[4e^{2x}\right]}_{x=0}=4\\ f\text{'}\text{'}\text{'}(0)={\left[8e^{2x}\right]}_{x=0}=8\\ f\text{'}\text{'}\text{'}\text{'}(0)={\left[16e^{2x}\right]}_{x=0}=16\end{array}$

Thus, the fourth-order Taylor polynomial at $x=0$ for the function $f(x)=e^{2x}$ is,

  $\begin{array}{l}{P}_4(x)=1+2(x-0)+\frac{4}{2!}{(}^x+\frac{8}{3!}{(}^x+\frac{16}{4!}{(}^x\\ \Rightarrow P_4(x)=1+2x+\frac{4}{2!}{x}^2+\frac{8}{3!}{x}^3+\frac{16}{4!}{x}^4\\ \Rightarrow P_4(x)=1+2x+2x^2+\frac{4}{3}{x}^3+\frac{2}{3}{x}^4\end{array}$