Chapter 11.8, Problem 29WE

(a)

To determine

To Calculate: The first four terms in the expansion ${\left(1+x\right)}^{\frac{1}{2}}$

Answer to Problem 29WE

The first four term of the expansion series is $1+\frac{x}{2}-\frac{x^2}{8}+\frac{x^3}{16}$ .

Explanation of Solution

Given Information:

Considering the binomial theorem continues to hold when $n$ is a positive rational number that is not integer. The expansion coefficients will not be integers and the expansion will be an infinite series,

  ${(a+b)}^n=a^n+\frac{n}{1}{a}^{n-1}b+\frac{n(n-1)}{1\cdot 2}{a}^{n-2}{b}^2+\mathrm{.....}$

Calculation:

The 4^th term of the expansion will be $\frac{n(n-1)(n-2)}{3!}{a}^{n-3}{b}^3$ So now plugging in the stuff ${(1+x)}^{\frac{1}{2}}$ where we see that ,

  $\begin{array}{l}a=1\\ b=x\\ n=\frac{1}{2}\end{array}$

Therefore, $\begin{array}{c}$ {(1+x)}^{\frac{1}{2}}=1^{1/2}+\frac{1/2}{1}{(1)}^{-1/2}x+\frac{1/2(1/2-1)}{2}{(1)}^{3/2}{x}^2+\frac{1/2(1/2-1)(1/2-2)}{6}{(1)}^{-5/2}{x}^3$=1+\frac{x}{2}-\frac{x^2}{8}+\frac{x^3}{16}\end{array}$

Hence, the first four term of the series is $1+\frac{x}{2}-\frac{x^2}{8}+\frac{x^3}{16}$ .

(b)

To determine

To Find: The numerical estimate of $\sqrt{2}$ by using the prove(a) with consideration of $x=1$ .

Answer to Problem 29WE

The required numerical estimate of $\sqrt{2}$ is 1.4375

Explanation of Solution

Given Information:

Considering the binomial theorem continues to hold when $n$ is a positive rational number that is not integer. The expansion coefficients will not be integers and the expansion will be an infinite series,

  ${(a+b)}^n=a^n+\frac{n}{1}{a}^{n-1}b+\frac{n(n-1)}{1\cdot 2}{a}^{n-2}{b}^2+\mathrm{.....}$

According to the given statement considering $x=1$ , and we know that $\sqrt{2}$ is approximately equal to 1.4142 so we probably need more terms to get into a better estimate.

From the given pattern the next term will be possibly $\frac{-1}{32},\frac{1}{64},\frac{-1}{128}\mathrm{....}$

Now,

  $\begin{array}{c}$ {(1+x)}^{\frac{1}{2}}=1+\frac{x}{2}-\frac{-x^2}{8}+\frac{x^3}{16}$If,\text{}\text{}x=1\\ $ {(1+x)}^{\frac{1}{2}}=1+\frac{1}{2}-\frac{1}{8}+\frac{1}{16}$=1.4375\end{array}$

Hence, numerical estimate of $\sqrt{2}$ is 1.4375.