Certainly! Let's apply Newton's Method step-by-step to the function [tex]\( f(x) = 1 - 2x\sin(x) \)[/tex] with the initial guess [tex]\( x_0 = 7 \)[/tex].
Firstly, we need to find the derivative of the function [tex]\( f(x) \)[/tex]. The derivative, [tex]\( f'(x) \)[/tex], is calculated as:
[tex]\[ f'(x) = \frac{d}{dx} (1 - 2x\sin(x)) = -2(\sin(x) + x\cos(x)). \][/tex]
Now, using Newton's Method formula:
[tex]\[ x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}, \][/tex]
we proceed iteratively.
### Step 1: Calculate [tex]\( x_1 \)[/tex]
Given [tex]\( x_0 = 7 \)[/tex]:
[tex]\[ f(x_0) = 1 - 2 \cdot 7 \cdot \sin(7), \][/tex]
[tex]\[ f'(x_0) = -2 (\sin(7) + 7 \cdot \cos(7)). \][/tex]
Using Newton's Method:
[tex]\[ x_1 = 7 - \frac{1 - 2 \cdot 7 \cdot \sin(7)}{-2 (\sin(7) + 7 \cdot \cos(7))} \approx 6.309286. \][/tex]
### Step 2: Calculate [tex]\( x_2 \)[/tex]
Given [tex]\( x_1 \approx 6.309286 \)[/tex]:
[tex]\[ f(x_1) = 1 - 2 \cdot 6.309286 \cdot \sin(6.309286), \][/tex]
[tex]\[ f'(x_1) = -2 (\sin(6.309286) + 6.309286 \cdot \cos(6.309286)). \][/tex]
Using Newton's Method:
[tex]\[ x_2 = 6.309286 - \frac{1 - 2 \cdot 6.309286 \cdot \sin(6.309286)}{-2 (\sin(6.309286) + 6.309286 \cdot \cos(6.309286))} \approx 6.362236. \][/tex]
### Step 3: Calculate [tex]\( x_3 \)[/tex]
Given [tex]\( x_2 \approx 6.362236 \)[/tex]:
[tex]\[ f(x_2) = 1 - 2 \cdot 6.362236 \cdot \sin(6.362236), \][/tex]
[tex]\[ f'(x_2) = -2 (\sin(6.362236) + 6.362236 \cdot \cos(6.362236)). \][/tex]
Using Newton's Method:
[tex]\[ x_3 = 6.362236 - \frac{1 - 2 \cdot 6.362236 \cdot \sin(6.362236)}{-2 (\sin(6.362236) + 6.362236 \cdot \cos(6.362236))} \approx 6.361860. \][/tex]
In summary, the approximations are:
[tex]\[
\begin{array}{l}
x_1 \approx 6.309286 \\
x_2 \approx 6.362236 \\
x_3 \approx 6.361860 \\
\end{array}
\][/tex]
