Use a calculator or program to compute the first 10 iterations of Newton's method when it is applied to the following function given with the initial approximation. Make a table containing: i, xᵢ, f(xᵢ), and f'(xᵢ).
f(x) = 3sinx-2; x₀=1
To compute the first 10 iterations of Newton's method for the given function f(x) = 3sin(x) - 2 with the initial approximation x₀ = 1, we need to follow these steps:
1. Evaluate f(x₀): Plug x₀ = 1 into the function f(x) = 3sin(x) - 2 to find f(x₀).
f(x₀) = 3sin(1) - 2 ≈ -1.08
2. Evaluate f'(x₀): Compute the derivative of f(x), which is f'(x), and then plug x₀ = 1 into f'(x) to find f'(x₀).
f'(x) = 3cos(x)
f'(x₀) = 3cos(1) ≈ 2.53
3. Calculate the next approximation x₁ using the formula:
x₁ = x₀ - f(x₀)/f'(x₀)
4. Repeat steps 1-3 for the next iterations, updating the value of x using the previous approximation.
i.e., for each iteration i, calculate xᵢ₊₁ = xᵢ - f(xᵢ)/f'(xᵢ).
Using these steps, we can compute the table for the first 10 iterations:
| i | xᵢ | f(xᵢ) | f'(xᵢ) |
|:---:|:----:|:-----:|:------:|
| 0 | 1 | -1.08 | 2.53 |
| 1 | 1.43 | -0.47 | 1.86 |
| 2 | 1.44 | 0.02 | 1.86 |
| 3 | 1.44 | 0.00 | 1.86 |
| 4 | 1.44 | 0.00 | 1.86 |
| 5 | 1.44 | 0.00 | 1.86 |
| 6 | 1.44 | 0.00 | 1.86 |
| 7 | 1.44 | 0.00 | 1.86 |
| 8 | 1.44 | 0.00 | 1.86 |
| 9 | 1.44 | 0.00 | 1.86 |
| 10 | 1.44 | 0.00 | 1.86 |
To calculate the values beyond the tolerance of the iteration, we notice from the table that the value of xᵢ remains constant at approximately 1.44, and f(xᵢ) and f'(xᵢ) become very close to zero. This suggests that we have reached the approximate root of f(x) = 3sin(x) - 2, which is located near x ≈ 1.44.
there are several handy Newton's Method calculators online.
google 'em up.