Chapter 9, Problem 61RE

$\text{(a)}$

To determine

To find: The first four non zero terms and the general term of the Taylor series.

Answer to Problem 61RE

The answer:

Explanation of Solution

Given information:

The equation is $f(x)=e^{-2x^2}$

The Taylor series generated by $f(x)$ at $x=0$ .

Calculation:

The Maclaurin series, i.e. the Taylor series at $x=0$:

  

Based on this, the Taylor series generated by $f$ at $x=0$ can be obtained by replacing $x$ to $-2x^2$ :

  $\begin{array}{c}{e}^{-2x^2}=1-2x+\frac{{\left(-2x^2\right)}^2}{2!}-\frac{{\left(-2x^2\right)}^3}{3!}+\dots +\frac{{\left(-2x^2\right)}^n}{n!}+\dots \\ 1-2x+2x^4-\frac{4}{3}{x}^6+\dots +{(}^{-}\frac{2^n{x}^{2n}}{n!}+\dots \end{array}$

Therefore, the required first four non zero terms and the general term of the Taylor series is:

  $f(x)=1-2x+2x^4-\frac{4}{3}{x}^6+\dots +{(-1)}^n\frac{2^n{x}^{2n}}{n!}+\dots$

$\text{(b)}$

To determine

To find: The interval of convergence for the series found in part $\text{(a)}$ .

Answer to Problem 61RE

The answer: The interval of convergence is $ℝ$

Explanation of Solution

Given information:

The equation is $f(x)=e^{-2x^2}$

The Taylor series generated by $f(x)$ at $x=0$ .

Calculation:

Use the Ratio Test solely for series with non-negative terms; as a result, consider the series' absolute value of terms,

Take the absolute value of the terms in the series since you can only use the Ratio Test for series with non-negative terms, and then look at the following limit:

  

Based on the previous step, the series converges absolutely, if the limit less than but for every , hence the series converges (absolutely) for all real i.e. the interval of convergence is

To determine

To prove: The statement $|f(x)-g(x)|<0.02$ for $-0.6\le x\le 0.6$ .

Explanation of Solution

Given information:

The equation is $f(x)=e^{-2x^2}$

The Taylor series generated by $f(x)$ at $x=0$ .

Calculation:

  $g(x)=1-2x+2x^4-\frac{4}{3}{x}^6$

The difference $|f(x)-g(x)|$ means the truncation error for the approximation of $f(x)$ by $P_6(x)$ estimate this error by applying the Alternating Series Bound Theorem, because the series alternates.

  $\begin{array}{c}\text{error }<\frac{2^4|x{|}^8}{4!}\le \frac{2^4\cdot {0.6}^8}{4!}\\ =0.0112<0.02\end{array}$