Chapter 4.7, Problem 31E

To determine

To calculate: The roots of the curve,

  $y={(x-1)}^2$ .

Answer to Problem 31E

The roots are

  $x_1\approx 0.4$

  $x_2\approx 0.410126$

  $x_3\approx 0.410245$

Explanation of Solution

Given information:

The parabola $y={(x-1)}^2$ that is closest to the origin .

Formula used:

Newton’s Method:

We seek a solution of $f(x)=0$ , starting from an initial estimate $x=x_1$ .

For $x=x_n$ , compute the next approximation $x_{n+1}$ by

  $x_{n+1}=x_n-\frac{f(x_n)}{f\text{'}(x_n)}$ and so on.

Distance Formula :- $D=\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}$

Calculation:

Consider the parabola ,

  $y={(x-1)}^2$

Let the point on the parabola is $(x,{(x-1)}^2)$ .

The distance formula from point $(x,{(x-1)}^2)$ to origin $(0,0)$ is

  $\begin{array}{l}D=\sqrt{{(x-0)}^2+{({(x-1)}^2-0)}^2}\\ D=\sqrt{{(x)}^2+{({(x-1)}^2)}^2}\\ D=\sqrt{x^2+{(x-1)}^4}\end{array}$

The minimum distance without the square root will be same as the minimum distance with square root.

  $\begin{array}{l}=D=x^2+{(}^x\\ =D\text{'}=2x+4{(}^x\end{array}$

Therefore, $g(x)=2x+4{(x-1)}^4$ used to find out roots using Newton’s Method.

  $\begin{array}{l}g(x)=2x+4{(}^x\\ g\text{'}(x)=2+12{(}^x\end{array}$

Now, let initial approximation be $x_1=0.4$

For $n=1$

  $x_2=x_1-\frac{g(x_1)}{g\text{'}(x_1)}$

  $x_2=x_1-\frac{2x_1+4{(x_1-1)}^3}{2+12{(x_1-1)}^2}$

  $\begin{array}{l}{x}_2=(0.4)-\frac{2(0.4)+4{(0.4-1)}^3}{2+12{(0.4-1)}^2}\\ x_2=(0.4)-\frac{0.8-0.864}{2+4.32}\end{array}$

  $\begin{array}{l}{x}_2=0.4-\frac{(-0.064)}{6.32}\\ x_2=0.4+0.010126582\\ x_2=0.410126582\end{array}$

The second approximation is $x_2=0.410126$ .

For $n=2$

  $x_3=x_2-\frac{g(x_2)}{g\text{'}(x_2)}$

  $x_3=x_2-\frac{2x_2+4{(x_2-1)}^3}{2+12{(x_2-1)}^2}$

  $\begin{array}{l}{x}_3=(0.410126)-\frac{2(0.410126)+4{(0.410126-1)}^3}{2+12{(0.410126-1)}^2}\\ x_3=(0.410126)-\frac{0.820252-0.820989785}{2+4.175416031}\end{array}$

  $\begin{array}{l}{x}_3=0.410126-\frac{(-0.000737785)}{6.175416031}\\ x_3=0.410126+0.000119471303\\ x_3=0.410245471\end{array}$

The third approximation is $x_3=0.410245$ .

Hence , the roots of the curve are :-

  $x_1\approx 0.4$

  $x_2\approx 0.410126$

  $x_3\approx 0.410245$