Q. 85 - Calculus

Question

Newton's approach will also not work if $\left|x_{k+1}-x_k\right|,$ the difference between subsequent approximations, does not diminish as $k$ rises.

(a) Demonstrate that when you select $x_0=0$, this occurs for the function  $f\left(x\right)=\sqrt[3]{x-2}$

(b) What does $f\left(x\right)=\sqrt[3]{x-2}$ have as its root?

Step-by-Step Solution

Answer

1. The Newton's method fail because the successive terms are not approaching.
2. The root of the function is $2.$ 

Part(a) Step 1: Given information

The function  $f\left(x\right)=\sqrt[3]{x-2}$ and $x_0=0.$

Part(a) Step 2: Calculation

Explaining why the Newton's technique fails when the difference between two subsequent approximations exists is the goal $\left|x_{k+1}-x_k\right|$ does not get smaller as k gets bigger.

Given the function $f\left(x\right)=\sqrt[3]{x-2}.$

Rewrite the function $f\left(x\right)=(x-2{)}^{\frac{1}{3}}$

Given  the equation $x_{k+1}=x_k-\frac{f\left(x_k\right)}{f^{\text{'}}\left(x_k\right)}\dots \dots \left(1\right)$

Now, $f\left(x_k\right)={\left(x_k-2\right)}^{\frac{1}{3}}$

And $f^{\text{'}}\left(x_k\right)=\frac{1}{3}{\left(x_k-2\right)}^{\frac{-2}{3}}$

Substituting the values in first equation

$$\begin{gathered} x_{k+1}=x_k-\frac{{\left(x_k-2\right)}^{\frac{1}{3}}}{\frac{1}{3}{\left(x_k-2\right)}^{\frac{-2}{3}}} \\ =x_k-3{\left(x_k-2\right)}^{\frac{1}{3}+\frac{2}{3}} \\ =x_k-3\left(x_k-2\right) \\ =x_k-3x_k+6 \\ =-2x_k+6 \\ {x}_{k+1}=-2x_k+6 \end{gathered}$$

Part(a) Step 3: Further Calculation

Now, to find the value of $x_1,$ substitute $k=0$ in second equation

$$\begin{gathered} x_{k+1}=-2x_k+6 \\ {x}_1=-2x_0+6 \end{gathered}$$

Substitute $x_0=0$

$x_1=-2\left(0\right)+6$

      $$\begin{gathered} =-2+6 \\ =4 \end{gathered}$$

Hence, $x_1=4$

Now, to find the value of $x_2$, substitute $k=1$ in second equation

$x_2=-2x_1+6$

Substitute $x_1=4$

$$\begin{gathered} x_2=-2\left(4\right)+6 \\ =-8+6 \\ =-2 \end{gathered}$$

Hence, $x_2=-2$

Now, to find the value of $x_3,$ substitute $k=2$ in second equation

$x_3=-2x_2+6$

Substitute $x_2=-2$

$$\begin{gathered} x_3=-2(-2)+6 \\ =4+6 \\ =10 \end{gathered}$$

Hence, $x_3=10$

Now, to find the value of $x_4$, substitute $k=3$ in second equation

$x_4=-2x_3+6$

Substitute $x_3=10$

$$\begin{gathered} x_4=-2\left(10\right)+6 \\ =-20+6 \\ =-14 \end{gathered}$$

Hence, $x_4=-14$

Every phrase in $x_k$  has an alternating sign.

Therefore, the value of  $\left|x_{k+1}-x_k\right|$ will never decrease.

The Newton's method therefore fails since the succeeding words are not getting closer.

Part(b) Step 1: Given information

The function $f\left(x\right)=\sqrt[3]{x-2}.$

Part(b) Step 2: Calculation

The goal is to find the root of $f\left(x\right)=\sqrt[3]{x-2}.$

To find the root of $f\left(x\right)=\sqrt[3]{x-2},$equate $f\left(x\right)$ with $0$ .

Hence,

$$\begin{gathered} \sqrt[3]{x-2}=0(x-2{)}^{\frac{1}{3}}=0 \\ x-2=0 \\ x=2 \end{gathered}$$

Hence, the root of the function is $2$ .