Chapter 9.3, Problem 33E

(a)

To determine

To find: The first four non zero terms and general terms.

Answer to Problem 33E

The first four non-zero terms are $2x-2x^3+2x^5-2x^7$ and general term is ${\left(-1\right)}^{n}2x^{2n+1}$

Explanation of Solution

Given:

The function is $f(x)=\frac{2x}{1+x^2}$

Calculation:

Taylor series is a n^thpolynomial series in which a function is expanded in a power series with an infinite number of terms.

The Taylor series can be written as:

  $f(x)=f(a)+f\text{'}(a)\frac{{(x-a)}^2}{2!}+f"(a)\frac{{(x-a)}^3}{3!}+\mathrm{...}$

The function given is: $f(x)=\frac{2x}{1+x^2}$

This function can be written as: $f(x)=2x.\frac{1}{1+x^2}$

Now, the Taylor series for this function is:

  $\begin{array}{l}\frac{1}{1+x^2}=1-x^2+x^4-x^6+\mathrm{....}{\left(-x^2\right)}^n+\mathrm{....}\\ \frac{2x}{1+x^2}=2x-2x^3+2x^5-2x^7+\mathrm{...}+{\left(-1\right)}^{n}2x^{2n+1}+\mathrm{....}\end{array}$

The first four non zero terms and general terms are calculated.

Conclusion:

The first four non-zero terms are $2x-2x^3+2x^5-2x^7$ and general term is ${\left(-1\right)}^{n}2x^{2n+1}$

(b)

To determine

To find: The series converges at f (1) or not.

Answer to Problem 33E

The series does not converge at f (1).

Explanation of Solution

Given:

The function is $f(x)=\frac{2x}{1+x^2}$

Calculation:

The Taylor series of the function can be written as:

  $f(x)={\displaystyle \sum _{n=0}^{\infty }{(-1)}^{n}2x^{2n+1}}$

Now, the ratio test can be done to find whether the series converge to zero or not when $x=1$ .

  $\rho ={\lim }_{n\to \infty }\left|\frac{2x^{2n+3}}{2x^{2n+1}}\right|={\lim }_{n\to \infty }\left|x^2\right|<1$

So, for the series to converge $-1<x<1$ .

Now, when $x=1$ then,

  $\begin{array}{l}{\displaystyle \sum _{n=0}^{\infty }{(-1)}^{n}2}\\ {\lim }_{n\to \infty }\left|{(-1)}^{n}2\right|={\lim }_{n\to \infty }2\ne 0\end{array}$

Hence, when $x=1$ the series does not converge to zero because the terms does not become zero when $x=1$ .

Conclusion:

The series does not converge at f (1).

(c)

To determine

To find: The first four non-zero terms for the function $y=\ln (1+x^2)$ .

Answer to Problem 33E

The first four non-zero terms are ${\text{x}}^2-\frac{1}{2}{x}^4+\frac{1}{3}{x}^6-\frac{1}{4}{x}^8$

Explanation of Solution

Given:

The function is $y=\ln (1+x^2)$

Calculation:

  $\begin{array}{l}\ln (1+x^2)={\displaystyle \underset{0}{\overset{x}{\int }}\frac{2t}{1+t^2}dt}\\ \text{ = }{\displaystyle \underset{0}{\overset{x}{\int }}(2t-2t^3+2t^5-2t^7+\mathrm{....})dt}\\ {\text{ = x}}^2-\frac{1}{2}{x}^4+\frac{1}{3}{x}^6-\frac{1}{4}{x}^8+\mathrm{....}\end{array}$

Conclusion:

The first four non-zero terms are ${\text{x}}^2-\frac{1}{2}{x}^4+\frac{1}{3}{x}^6-\frac{1}{4}{x}^8$

(d)

To determine

To find: The rational number.

Answer to Problem 33E

The rational number satisfy the given equation.

Explanation of Solution

Given:

The function is $y=\ln (1+x^2)$

Calculation:

The Taylor series can be written as:

  $\begin{array}{l}\ln \left(\frac{5}{4}\right)=\ln \left(1+x^2\right)\\ \ln \left(\frac{5}{4}\right)=\ln \left(1+\frac{1}{4}\right)\end{array}$

Now, this can be expanded in Taylor series as:

  $\text{ln}\left(\frac{5}{4}\right)\text{ =}{\left(\frac{1}{2}\right)}^2-\frac{1}{2}{\left(\frac{1}{2}\right)}^4+\frac{1}{3}{\left(\frac{1}{2}\right)}^6-\frac{1}{4}{\left(\frac{1}{2}\right)}^8+\mathrm{.....}$

Now, let the value of A be:

  $\begin{array}{l}\text{ A =}{\left(\frac{1}{2}\right)}^2-\frac{1}{2}{\left(\frac{1}{2}\right)}^4\\ \text{A}=\frac{7}{32}\end{array}$

As, the series is converging as well as alternating therefore, absolute values of the terms in series decreases to zero.

  $\left|A-\ln \left(\frac{5}{4}\right)\right|<\frac{1}{3}.\frac{1}{64}<\frac{1}{100}$

Conclusion:

The rational number satisfy the given equation.