Newton's method is an iterative technique to approximate the roots of a real-valued function. Given a function [tex]\( f(x) \)[/tex] and its derivative [tex]\( f'(x) \)[/tex], the method generates a sequence of approximations using the formula:
[tex]\[
x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}
\][/tex]
Let's start with the given equation [tex]\( e^{0.3 x^2} = 4 - x \)[/tex].
First, we rewrite the equation in the form [tex]\( f(x) = 0 \)[/tex]:
[tex]\[
f(x) = e^{0.3 x^2} - 4 + x
\][/tex]
We also need the derivative [tex]\( f'(x) \)[/tex]:
[tex]\[
f'(x) = \frac{d}{dx} \left( e^{0.3 x^2} - 4 + x \right) = 0.6 x e^{0.3 x^2} + 1
\][/tex]
We are given the initial approximation [tex]\( x_1 = 1 \)[/tex]. Now we will calculate the second and third approximations.
Step 1: Calculate [tex]\( x_2 \)[/tex]
[tex]\[
x_2 = x_1 - \frac{f(x_1)}{f'(x_1)}
\][/tex]
First, evaluate [tex]\( f(x_1) \)[/tex] at [tex]\( x_1 = 1 \)[/tex]:
[tex]\[
f(1) = e^{0.3 \cdot 1^2} - 4 + 1 = e^{0.3} - 3
\][/tex]
Then, evaluate [tex]\( f'(x_1) \)[/tex] at [tex]\( x_1 = 1 \)[/tex]:
[tex]\[
f'(1) = 0.6 \cdot 1 \cdot e^{0.3 \cdot 1^2} + 1 = 0.6 e^{0.3} + 1
\][/tex]
Now, calculate [tex]\( x_2 \)[/tex]:
[tex]\[
x_2 = 1 - \frac{e^{0.3} - 3}{0.6 e^{0.3} + 1}
\][/tex]
After performing the calculation, we find:
[tex]\[
x_2 \approx 1.9117228891949392
\][/tex]
Step 2: Calculate [tex]\( x_3 \)[/tex]
[tex]\[
x_3 = x_2 - \frac{f(x_2)}{f'(x_2)}
\][/tex]
First, evaluate [tex]\( f(x_2) \)[/tex] at [tex]\( x_2 = 1.9117228891949392 \)[/tex]:
[tex]\[
f(1.9117228891949392) = e^{0.3 \cdot (1.9117228891949392)^2} - 4 + 1.9117228891949392
\][/tex]
Then, evaluate [tex]\( f'(x_2) \)[/tex] at [tex]\( x_2 = 1.9117228891949392 \)[/tex]:
[tex]\[
f'(1.9117228891949392) = 0.6 \cdot 1.9117228891949392 \cdot e^{0.3 \cdot (1.9117228891949392)^2} + 1
\][/tex]
Now, calculate [tex]\( x_3 \)[/tex]:
[tex]\[
x_3 = 1.9117228891949392 - \frac{f(1.9117228891949392)}{f'(1.9117228891949392)}
\][/tex]
After performing the calculation, we find:
[tex]\[
x_3 \approx 1.7075712731011243
\][/tex]
Therefore, the second approximation [tex]\( x_2 \)[/tex] is approximately [tex]\( 1.9117228891949392 \)[/tex], and the third approximation [tex]\( x_3 \)[/tex] is approximately [tex]\( 1.7075712731011243 \)[/tex].
