Sure, let's use Newton's Method step by step to approximate a zero of the function [tex]\( f(x) = x^2 - 2 \)[/tex] with the given initial guess [tex]\( x_1 = 1.4 \)[/tex].
Newton’s Method is defined by the formula:
[tex]\[
x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}
\][/tex]
### Step-by-Step Calculation:
#### Iteration 1
Let's start with the initial guess [tex]\( x_1 = 1.4 \)[/tex].
1. Calculate [tex]\( f(x_1) \)[/tex]:
[tex]\[
f(x_1) = (1.4)^2 - 2 = 1.96 - 2 = -0.04
\][/tex]
2. Calculate [tex]\( f'(x_1) \)[/tex] (the derivative of [tex]\( f(x) \)[/tex] at [tex]\( x_1 \)[/tex]):
[tex]\[
f'(x_1) = 2 \times 1.4 = 2.8
\][/tex]
3. Update [tex]\( x_2 \)[/tex] using Newton's Method:
[tex]\[
x_2 = x_1 - \frac{f(x_1)}{f'(x_1)} = 1.4 - \frac{-0.04}{2.8} = 1.4 + 0.0142857143 \approx 1.4143
\][/tex]
So, after the first iteration, [tex]\( x_2 \approx 1.4143 \)[/tex].
#### Iteration 2
Now, using [tex]\( x_2 = 1.4143 \)[/tex]:
1. Calculate [tex]\( f(x_2) \)[/tex]:
[tex]\[
f(x_2) = (1.4143)^2 - 2 \approx 2.00004049 - 2 \approx 0.00004049
\][/tex]
2. Calculate [tex]\( f'(x_2) \)[/tex]:
[tex]\[
f'(x_2) = 2 \times 1.4143 = 2.8286
\][/tex]
3. Update [tex]\( x_3 \)[/tex] using Newton's Method:
[tex]\[
x_3 = x_2 - \frac{f(x_2)}{f'(x_2)} = 1.4143 - \frac{0.00004049}{2.8286} \approx 1.4143 - 0.00001434 \approx 1.4142
\][/tex]
So, after the second iteration, [tex]\( x_3 \approx 1.4142 \)[/tex].
### Summary of Results
[tex]\[
\begin{array}{|c|c|c|c|c|c|c|}
\hline n & x & & f(x_n) & f^{\prime}(x_n) & \frac{f(x_n)}{f^{\prime}(x_n)} & x_n - \frac{f(x_n)}{f^{\prime}(x_n)} \\
\hline 1 & 1.4 & & -0.04 & 2.8 & -0.0143 & 1.4143 \\
\hline 2 & 1.4143 & & 0.00004049 & 2.8286 & 0.00001434 & 1.4142 \\
\hline
\end{array}
\][/tex]
Thus, after two iterations of Newton's Method, the approximations are [tex]\( x_2 \approx 1.4143 \)[/tex] and [tex]\( x_3 \approx 1.4142 \)[/tex].
