Chapter 8.7, Problem 69E

(a)

To determine

To Prove: The given expression for the given range.

Explanation of Solution

Given:

The given equation expression is,

  $g^{\prime }\left(x\right)=\frac{kg\left(x\right)}{1+x}$

The given range is,

  $-1<x<1$

Calculation:

Consider the series is,

  $\begin{array}{l}g\left(x\right)={\displaystyle \sum _{n=0}^{\infty }\left(\begin{array}{c}k\\ n\end{array}\right)x^n}\\ g^{\prime }\left(x\right)={\displaystyle \sum _{n=0}^{\infty }\left(\begin{array}{c}k\\ n\end{array}\right)n{x}^{n-1}}\\ \left(1+x\right)g^{\prime }\left(x\right)=\left(1+x\right){\displaystyle \sum _{n=0}^{\infty }\left(\begin{array}{c}k\\ n\end{array}\right)n{x}^{n-1}}\\ \left(1+x\right)g^{\prime }\left(x\right)={\displaystyle \sum _{n=0}^{\infty }\left(\begin{array}{c}k\\ n\end{array}\right)n{x}^{n-1}}+{\displaystyle \sum _{n=0}^{\infty }\left(\begin{array}{c}k\\ n\end{array}\right)n{x}^n}\end{array}$

Then, solve as,

  $\begin{array}{c}\left(1+x\right)g^{\prime }\left(x\right)={\displaystyle \sum _{n=0}^{\infty }\left(n+1\right)\frac{k\left(k-1\right)\left(k-2\right)\mathrm{....}\left(k-n+1\right)}{\left(n+1\right)!}}{x}^n+{\displaystyle \sum _{n=0}^{\infty }\left(n\right)\frac{k\left(k-1\right)\left(k-2\right)\mathrm{....}\left(k-n+1\right)}{\left(n\right)!}}{x}^n\\ ={\displaystyle \sum _{n=0}^{\infty }\frac{\left(n+1\right)k\left(k-1\right)\left(k-2\right)\mathrm{....}\left(k-n+1\right)}{\left(n\right)!}}{x}^n\\ =k{\displaystyle \sum _{n=0}^{\infty }\left(\begin{array}{c}k\\ n\end{array}\right)x^n}\\ =kg\left(x\right)\end{array}$

Since,

  $\begin{array}{l}\left(1+x\right)g^{\prime }\left(x\right)=kg\left(x\right)\\ g^{\prime }\left(x\right)=\frac{kgx}{\left(1+x\right)}\end{array}$

Hence, the formula is proved.

(b)

To determine

To Prove: The expression $h^{\prime }\left(x\right)=0$ .

Explanation of Solution

Given:

The given expression is $h\left(x\right)={\left(1+x\right)}^{-k}g\left(x\right)$ .

Calculation:

Consider the given expression is,

  $h\left(x\right)={\left(1+x\right)}^{-k}g\left(x\right)$

Then, differentiate the above equation as,

  $h^{\prime }\left(x\right)=-k{\left(1+x\right)}^{-k-1}g\left(x\right)+{\left(1+x\right)}^{-k}{g}^{\prime }\left(x\right)$

Since, $g^{\prime }\left(x\right)=\frac{kgx}{\left(1+x\right)}$ then,

  $\begin{array}{c}{h}^{\prime }\left(x\right)=-k{\left(1+x\right)}^{-k-1}g\left(x\right)+{\left(1+x\right)}^{-k}\frac{kg\left(x\right)}{\left(1+x\right)}\\ =-k{\left(1+x\right)}^{-k-1}g\left(x\right)+k{\left(1+x\right)}^{-k-1}g\left(x\right)\\ =0\end{array}$

Hence, proved.

(c)

To determine

To Prove: The expression $g\left(x\right)={\left(1+x\right)}^k$ .

Explanation of Solution

Calculation:

Consider $h^{\prime }\left(x\right)=0$ , it can be deduced that $h\left(x\right)$ is constant for all $x$ in $\left(-1,1\right)$ ,

Let us take $h\left(x\right)=c$ , since $h\left(0\right)=0$ . Then $c=1$ and from part (b) $h\left(x\right)={\left(1+x\right)}^{-k}g\left(x\right)$ .

Thus, the value of $h\left(x\right)=1$ for $x\in \left(-1,1\right)$ .

Then,

  $\begin{array}{l}1={\left(1+x\right)}^{-k}g\left(x\right)\\ g\left(x\right)={\left(1+x\right)}^k\end{array}$

That is $g\left(x\right)={\left(1+x\right)}^k$ for $x\in \left(-1,1\right)$ .

Hence, Proved.