Answer :
To approximate the value of [tex]\(\sqrt[8]{550}\)[/tex] using Newton's method and achieve an accuracy correct to eight decimal places, let's proceed with the following steps:
1. Define the function and its derivative:
We need to define a function whose root we are trying to find. In this case, we are solving for [tex]\(x\)[/tex] in the equation:
[tex]\[ f(x) = x^8 - 550 \][/tex]
The derivative of the function [tex]\(f(x)\)[/tex] is:
[tex]\[ f'(x) = 8x^7 \][/tex]
2. Initial guess:
We need to make an initial guess for [tex]\(x\)[/tex]. Let's choose [tex]\(x_0 = 2.0\)[/tex].
3. Newton's iteration formula:
Newton's method uses the following iterative process to approximate the root:
[tex]\[ x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} \][/tex]
where [tex]\(x_n\)[/tex] is the current approximation, and [tex]\(x_{n+1}\)[/tex] is the next approximation.
4. Iteration process:
Repeat the iterative process until the difference between consecutive approximations is less than the desired tolerance. For eight decimal places, a tolerance of [tex]\(1 \times 10^{-9}\)[/tex] is appropriate.
5. Convergence:
Continue the iterations until [tex]\(|x_{n+1} - x_n| < 1 \times 10^{-9}\)[/tex].
Let's go through the Newton's iteration steps in detail:
1. Define the functions:
[tex]\[ f(x) = x^8 - 550, \quad f'(x) = 8x^7 \][/tex]
2. Initial guess:
[tex]\[ x_0 = 2.0 \][/tex]
3. Iterative calculations (we will describe a few iterations):
First iteration:
[tex]\[ x_1 = x_0 - \frac{f(x_0)}{f'(x_0)} = 2.0 - \frac{(2.0)^8 - 550}{8 \times (2.0)^7} = 2.0 - \frac{256 - 550}{2048} = 2.0 + 0.1435546875 \approx 2.1435546875 \][/tex]
Second iteration:
[tex]\[ x_2 = x_1 - \frac{f(x_1)}{f'(x_1)} = 2.1435546875 - \frac{(2.1435546875)^8 - 550}{8 \times (2.1435546875)^7} \][/tex]
Calculating the values, we get:
[tex]\[ f(2.1435546875) \approx (2.1435546875)^8 - 550 \approx 775.219 - 550 = 225.219 \][/tex]
[tex]\[ f'(2.1435546875) \approx 8 \times (2.1435546875)^7 \approx 8 \times 163.83 = 1310.64 \][/tex]
[tex]\[ x_2 \approx 2.1435546875 - \frac{225.219}{1310.64} \approx 2.1435546875 - 0.171769 \approx 2.168233 \][/tex]
Continue iterating in a similar fashion until the approximation meets the desired tolerance level.
For the sake of brevity, after several iterations following the same pattern as outlined, the root converges to approximately:
[tex]\[ \sqrt[8]{550} \approx 2.2006214214590667 \][/tex]
Therefore, using Newton's method, the value of [tex]\(\sqrt[8]{550}\)[/tex] approximated to eight decimal places is:
[tex]\[ \boxed{2.20062142} \][/tex]
1. Define the function and its derivative:
We need to define a function whose root we are trying to find. In this case, we are solving for [tex]\(x\)[/tex] in the equation:
[tex]\[ f(x) = x^8 - 550 \][/tex]
The derivative of the function [tex]\(f(x)\)[/tex] is:
[tex]\[ f'(x) = 8x^7 \][/tex]
2. Initial guess:
We need to make an initial guess for [tex]\(x\)[/tex]. Let's choose [tex]\(x_0 = 2.0\)[/tex].
3. Newton's iteration formula:
Newton's method uses the following iterative process to approximate the root:
[tex]\[ x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} \][/tex]
where [tex]\(x_n\)[/tex] is the current approximation, and [tex]\(x_{n+1}\)[/tex] is the next approximation.
4. Iteration process:
Repeat the iterative process until the difference between consecutive approximations is less than the desired tolerance. For eight decimal places, a tolerance of [tex]\(1 \times 10^{-9}\)[/tex] is appropriate.
5. Convergence:
Continue the iterations until [tex]\(|x_{n+1} - x_n| < 1 \times 10^{-9}\)[/tex].
Let's go through the Newton's iteration steps in detail:
1. Define the functions:
[tex]\[ f(x) = x^8 - 550, \quad f'(x) = 8x^7 \][/tex]
2. Initial guess:
[tex]\[ x_0 = 2.0 \][/tex]
3. Iterative calculations (we will describe a few iterations):
First iteration:
[tex]\[ x_1 = x_0 - \frac{f(x_0)}{f'(x_0)} = 2.0 - \frac{(2.0)^8 - 550}{8 \times (2.0)^7} = 2.0 - \frac{256 - 550}{2048} = 2.0 + 0.1435546875 \approx 2.1435546875 \][/tex]
Second iteration:
[tex]\[ x_2 = x_1 - \frac{f(x_1)}{f'(x_1)} = 2.1435546875 - \frac{(2.1435546875)^8 - 550}{8 \times (2.1435546875)^7} \][/tex]
Calculating the values, we get:
[tex]\[ f(2.1435546875) \approx (2.1435546875)^8 - 550 \approx 775.219 - 550 = 225.219 \][/tex]
[tex]\[ f'(2.1435546875) \approx 8 \times (2.1435546875)^7 \approx 8 \times 163.83 = 1310.64 \][/tex]
[tex]\[ x_2 \approx 2.1435546875 - \frac{225.219}{1310.64} \approx 2.1435546875 - 0.171769 \approx 2.168233 \][/tex]
Continue iterating in a similar fashion until the approximation meets the desired tolerance level.
For the sake of brevity, after several iterations following the same pattern as outlined, the root converges to approximately:
[tex]\[ \sqrt[8]{550} \approx 2.2006214214590667 \][/tex]
Therefore, using Newton's method, the value of [tex]\(\sqrt[8]{550}\)[/tex] approximated to eight decimal places is:
[tex]\[ \boxed{2.20062142} \][/tex]
