Chapter 10, Problem 61RE

(a)

To determine

To calculate: The first four non-zero terms based on f(x) at x=0.

Answer to Problem 61RE

The general term of the series is given by $T_n={\left(-1\right)}^n\frac{2^n{x}^{2n}}{n!}$

Explanation of Solution

Given information:

The given function is

Let $f\left(x\right)=e^{-2x^2}$

Formula used:

Maclaurin series is used.

Calculation:

Maclaurin series generated by function $e^x$ is given by

  $e^x=1+x+\frac{x^2}{2!}+\mathrm{....}+\frac{x^n}{n!}+\mathrm{....}$

Replacing x by -2x²$\begin{array}{l}f\left(x\right)=1+\left(-2x^2\right)+\frac{{\left(-2x^2\right)}^2}{2!}+\frac{{\left(-2x^2\right)}^3}{3!}+\mathrm{......}+\frac{{\left(-2x^2\right)}^n}{n!}+\mathrm{.....}\\ f\left(x\right)=1-2x^2+\frac{4x^4}{2!}-\frac{8x^6}{3!}+\mathrm{.....}+{\left(-1\right)}^n\frac{2^n{x}^{2n}}{n!}+\mathrm{...}\end{array}$

Therefore, the first four non-zero terms of the series are $T_n=1T_1=-2x^2{T}_2=2x^4{T}_3=-\frac{4}{3}{x}^6$

The general term of the series is given by $T_n={\left(-1\right)}^n\frac{2^n{x}^{2n}}{n!}$

Conclusion:

The general term of the series is given by $T_n={\left(-1\right)}^n\frac{2^n{x}^{2n}}{n!}$

(b)

To determine

To calculate: The interval of convergence based on f(x) at x=0.

Answer to Problem 61RE

The series is always converges for interval $\left(-\infty ,\infty \right)$

Explanation of Solution

Given information:

The given function is

Let $f\left(x\right)=e^{-2x^2}$

Formula used:

Ratio test is used.

Calculation:

For absolute convergence using ratio test.

  $\begin{array}{l}\underset{n\to \infty }{\lim }\frac{\left|a_{n+1}\right|}{\left|a_n\right|}=\underset{n\to \infty }{\lim }{\left(-1\right)}^n\frac{2^n\left|x^{2n}\right|}{n!}.\frac{\left(n-1\right)!}{{\left(-1\right)}^{n-1}{2}^{n-1}\left|x^{2n-2}\right|}\\ =\underset{n\to \infty }{\lim }\frac{1}{n}\frac{\left|-x^2\right|}{2}\\ =\frac{\left|-x^2\right|}{2}.0\\ \underset{n\to \infty }{\lim }\frac{\left|a_{n+1}\right|}{\left|a_n\right|}=0\end{array}$

Since,

  $\underset{n\to \infty }{\lim }\frac{\left|a_{n+1}\right|}{\left|a_n\right|}=0$

Therefore the series is always absolutely for all real x.

Since,

  $\underset{n\to \infty }{\lim }\frac{\left|a_{n+1}\right|}{\left|a_n\right|}<1$

Therefore, the series is always converges for interval $\left(-\infty ,\infty \right)$

Conclusion:

The series is always converges for interval $\left(-\infty ,\infty \right)$ .

(c)

To determine

To find: The value of $|f(x)-g(x)|$

Answer to Problem 61RE

The error is less than $\frac{2{\left(0.6\right)}^8}{3}$ and this error is less than 0.02.

Explanation of Solution

Given information:

The given function is

Let $f\left(x\right)=e^{-2x^2}$

Formula used:

  $g\left(x\right)=1-2x^2+\frac{4x^4}{2!}-\frac{8x^6}{3!}$

Calculation:

The function $g\left(x\right)$ is given by $g\left(x\right)=1-2x^2+\frac{4x^4}{2!}-\frac{8x^6}{3!}$ the difference between $f\left(x\right)$ and $g\left(x\right)$ is the truncation error. Since the series is an alternating series, the error is bounded by the magnitude of the fifth order.That is, $\frac{{\left(2x^2\right)}^4}{4!}=\frac{2x^8}{3}$

Since,

  $-0.6\le x\le 0.6$

The error is less than $\frac{2{\left(0.6\right)}^8}{3}$ and this error is less than 0.02.

Conclusion:

The error is less than $\frac{2{\left(0.6\right)}^8}{3}$ and this error is less than 0.02.