Chapter 8.6, Problem 12E

a.

To determine

To find: A power series representation for $f(x)=\ln (1-x)$ and then to find the radius of convergence.

Answer to Problem 12E

  $\ln (1-x)=-{\displaystyle \sum _{n=1}^{\infty }\frac{x^n}{n}},\left|x\right|<1$ and radius of convergence of this series is $R=1$ .

Explanation of Solution

Given Information: Given power series is $f(x)=\ln (1-x)$ and $\frac{1}{1-x}=1+x+x^2+x^3+\mathrm{...}={\displaystyle \sum _{n=0}^{\infty }{x}^n},\left|x\right|<1$ .

Calculation:

  $\frac{1}{1-x}=1+x+x^2+x^3+\mathrm{...}(\left|x\right|<1)$

Integrating both sides with respect to $x$ ,

  ${\displaystyle \int \frac{1}{1-x}dx={\displaystyle \int (1+x+x^2+x^3+\mathrm{...})dx}}$

  $\Rightarrow \frac{\ln (1-x)}{-1}=\left(x+\frac{x^2}{2}+\frac{x^3}{3}+\mathrm{...}\right)+c$ , where ‘ $c$ ’ is a constant.

  $\begin{array}{l}\Rightarrow \ln (1-x)=-\left(x+\frac{x^2}{2}+\frac{x^3}{3}+\mathrm{...}\right)-c\\ \Rightarrow \ln (1-x)=-{\displaystyle \sum _{n=1}^{\infty }\frac{x^n}{n}}-c,\left|x\right|<1\end{array}$

Putting $x=0$ ,

  $\begin{array}{l}\ln 1=0-c\\ \Rightarrow c=0\end{array}$

So, $\ln (1-x)=-{\displaystyle \sum _{n=1}^{\infty }\frac{x^n}{n}},\left|x\right|<1$

Thus, $\ln (1-x)=-{\displaystyle \sum _{n=1}^{\infty }\frac{x^n}{n}},\left|x\right|<1$ and radius of convergence of this series is $R=1$ .

b.

To determine

To find: A power series representation for $f(x)=x\ln (1-x)$ .

Answer to Problem 12E

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

Explanation of Solution

Given Information: Power series from part (a).

Calculation:

From part (a),

  $\ln (1-x)=-{\displaystyle \sum _{n=1}^{\infty }\frac{x^n}{n}},\left|x\right|<1$

So,

  $\begin{array}{l}f(x)=x\ln (1-x)\\ =x(-{\displaystyle \sum _{n=1}^{\infty }\frac{x^n}{n}})\\ =-{\displaystyle \sum _{n=1}^{\infty }\frac{x^{n+1}}{n}}\end{array}$

Thus, $f(x)=x\ln (1-x)=-{\displaystyle \sum _{n=1}^{\infty }\frac{x^{n+1}}{n}}$ .

c.

To determine

To express: $\ln 2$ as the sum of an infinite series.

Answer to Problem 12E

  $\ln \left(\frac{1}{2}\right)=-{\displaystyle \sum _{n=1}^{\infty }\frac{1}{2^{n}n}}$

Explanation of Solution

Given Information: Result from part (a).

Calculation:

From part (a),

  $\ln (1-x)=-{\displaystyle \sum _{n=1}^{\infty }\frac{x^n}{n}},\left|x\right|<1$

Putting $x=\frac{1}{2}$ ,

  $\begin{array}{l}\ln \left(\frac{1}{2}\right)=-{\displaystyle \sum _{n=1}^{\infty }\frac{{\left(\frac{1}{2}\right)}^n}{n}}\\ =-{\displaystyle \sum _{n=1}^{\infty }\frac{1}{2^{n}n}}\end{array}$

Thus, $\ln \left(\frac{1}{2}\right)=-{\displaystyle \sum _{n=1}^{\infty }\frac{1}{2^{n}n}}$ .