To find [tex]\( x_3 \)[/tex], the third approximation to the solution of the equation [tex]\(\frac{2}{x} - x^2 + 1 = 0\)[/tex] using Newton's method, we follow these steps:
1. Define the function and its derivative:
[tex]\[
f(x) = \frac{2}{x} - x^2 + 1
\][/tex]
The derivative of [tex]\( f(x) \)[/tex] is:
[tex]\[
f'(x) = -\frac{2}{x^2} - 2x
\][/tex]
2. Initial approximation:
[tex]\[
x_1 = 2
\][/tex]
3. Iteration 1 to find [tex]\( x_2 \)[/tex]:
Newton's method formula is given by:
[tex]\[
x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}
\][/tex]
For [tex]\( x = x_1 \)[/tex]:
[tex]\[
f(x_1) = \frac{2}{2} - 2^2 + 1 = 1 - 4 + 1 = -2
\][/tex]
[tex]\[
f'(x_1) = -\frac{2}{2^2} - 2 \cdot 2 = -\frac{2}{4} - 4 = -0.5 - 4 = -4.5
\][/tex]
Now, calculate [tex]\( x_2 \)[/tex]:
[tex]\[
x_2 = x_1 - \frac{f(x_1)}{f'(x_1)} = 2 - \frac{-2}{-4.5} = 2 - \frac{2}{4.5} = 2 - 0.4444 \approx 1.5556
\][/tex]
4. Iteration 2 to find [tex]\( x_3 \)[/tex]:
For [tex]\( x = x_2 \)[/tex]:
[tex]\[
f(x_2) = \frac{2}{1.5556} - (1.5556)^2 + 1 \approx 1.2862 - 2.4203 + 1 \approx -0.1341
\][/tex]
[tex]\[
f'(x_2) = -\frac{2}{(1.5556)^2} - 2 \cdot 1.5556 \approx -\frac{2}{2.4203} - 3.1112 \approx -0.8263 - 3.1112 = -3.9375
\][/tex]
Now, calculate [tex]\( x_3 \)[/tex]:
[tex]\[
x_3 = x_2 - \frac{f(x_2)}{f'(x_2)} = 1.5556 - \frac{-0.1341}{-3.9375} \approx 1.5556 - 0.0341 \approx 1.5215
\][/tex]
So, the third approximation [tex]\( x_3 \)[/tex] to the solution of the equation [tex]\(\frac{2}{x} - x^2 + 1 = 0\)[/tex] is:
[tex]\[
x_3 \approx 1.5215
\][/tex]
This is the result rounded to four decimal places.
