1.Approximate sqrt3 by applying Newton's Method to the equation
(x^2-3) = 0
2.The equation (x^3-x-2) = 0 has one real solution for 1<x<2.Approximate it by Newton's method.
1. To approximate √3 using Newton's Method, we will apply the method to the equation (x^2-3) = 0.
Step 1: Begin with an initial guess, let's say x₀ = 1.5.
Step 2: Calculate the function value and its derivative at the current guess.
- f(x) = (x^2-3)
- f'(x) = 2x
Substituting x = 1.5 into the equations:
- f(1.5) = (1.5^2-3) = 0.25
- f'(1.5) = 2(1.5) = 3
Step 3: Using the current guess and the values from Step 2, calculate the next guess using the formula:
- x₁ = x₀ - (f(x₀) / f'(x₀))
Substituting the values:
- x₁ = 1.5 - (0.25 / 3) ≈ 1.4167
Step 4: Repeat Steps 2 and 3 until the desired level of accuracy is achieved.
By repeating Steps 2 and 3, the process continues until the desired level of accuracy is achieved. After a few iterations, the approximate value of √3 using Newton's Method for the equation (x^2-3) = 0 will be approximately 1.73205.
2. To approximate the real solution of the equation (x^3-x-2) = 0 for 1 < x < 2 using Newton's Method:
Step 1: Begin with an initial guess, let's say x₀ = 1.5.
Step 2: Calculate the function value and its derivative at the current guess.
- f(x) = (x^3-x-2)
- f'(x) = 3(x^2) - 1
Substituting x = 1.5 into the equations:
- f(1.5) = (1.5^3-1.5-2) = -0.875
- f'(1.5) = 3(1.5^2) - 1 = 5.25
Step 3: Using the current guess and the values from Step 2, calculate the next guess using the formula:
- x₁ = x₀ - (f(x₀) / f'(x₀))
Substituting the values:
- x₁ = 1.5 - (-0.875 / 5.25) ≈ 1.646
Step 4: Repeat Steps 2 and 3 until the desired level of accuracy is achieved.
By repeating Steps 2 and 3, the process continues until the desired level of accuracy is achieved. After a few iterations, the approximate solution of the equation (x^3-x-2) = 0 for 1 < x < 2 using Newton's Method will be approximately 1.5217.
To approximate the square root of 3 using Newton's Method, we need to solve the equation (x^2 - 3) = 0. Here's how:
1. Start with an initial guess for the square root of 3, let's say x_0 = 1.
2. Apply the Newton's Method formula to find a better approximation:
x_(n+1) = x_n - (f(x_n) / f'(x_n)),
where f(x) = (x^2 - 3) and f'(x) is the derivative of f(x).
In our case, f(x) = x^2 - 3 and f'(x) = 2x.
3. Plugging in the values, we get:
x_(n+1) = x_n - ((x_n^2 - 3) / (2 * x_n))
4. Repeat the process by iteratively using the formula until you reach a desired level of accuracy. For instance, let's calculate three steps:
For n = 0:
x_1 = x_0 - ((x_0^2 - 3) / (2 * x_0))
= 1 - ((1^2 - 3) / (2 * 1))
= 1 - (1 - 3) / 2
= 1 - (-2) / 2
= 1 + 1
= 2
For n = 1:
x_2 = x_1 - ((x_1^2 - 3) / (2 * x_1))
= 2 - ((2^2 - 3) / (2 * 2))
= 2 - (4 - 3) / 4
= 2 - 1 / 4
= 7 / 4
For n = 2:
x_3 = x_2 - ((x_2^2 - 3) / (2 * x_2))
= (7 / 4) - ((((7 / 4)^2) - 3) / (2 * (7 / 4)))
= (7 / 4) - (((49 / 16) - (48 / 16)) / (14 / 4))
= (7 / 4) - (1 / 16) / (14 / 4)
= (7 / 4) - (1 / 16) * (4 / 14)
= (7 / 4) - (1 / 56)
= (7 * 56 - 1) / 56
= 391 / 56
By continuing this process, you can approximate the square root of 3 to as many decimal places as needed.
Now let's move on to approximating the solution of the equation (x^3 - x - 2) = 0 in the interval 1 < x < 2 using Newton's Method.
1. Start with an initial guess for the solution in the given interval, let's say x_0 = 1.5.
2. Apply the Newton's Method formula to find a better approximation:
x_(n+1) = x_n - (f(x_n) / f'(x_n)),
where f(x) = (x^3 - x - 2) and f'(x) is the derivative of f(x).
In our case, f(x) = x^3 - x - 2 and f'(x) = 3x^2 - 1.
3. Plugging in the values, we get:
x_(n+1) = x_n - ((x_n^3 - x_n - 2) / (3 * x_n^2 - 1))
4. Repeat the process by iteratively using the formula until you reach a desired level of accuracy. Calculate three steps:
For n = 0:
x_1 = x_0 - ((x_0^3 - x_0 - 2) / (3 * x_0^2 - 1))
= 1.5 - ((1.5^3 - 1.5 - 2) / (3 * 1.5^2 - 1))
= 1.5 - ((3.375 - 1.5 - 2) / (3 * 2.25 - 1))
= 1.5 - ((3.375 - 3.5) / (6.75 - 1))
= 1.5 - (-0.125 / 5.75)
= 1.5 + 0.0217391304
= 1.5217391304
For n = 1:
x_2 = x_1 - ((x_1^3 - x_1 - 2) / (3 * x_1^2 - 1))
= 1.5217391304 - ((1.5217391304^3 - 1.5217391304 - 2) / (3 * 1.5217391304^2 - 1))
= 1.5217391304 - ((3.6804488945 - 1.5217391304 - 2) / (3 * 2.313334956 - 1))
= 1.5217391304 - ((3.6804488945 - 3.5217391304) / (6.940004868 - 1))
= 1.5217391304 - (0.1587097641 / 5.940004868)
= 1.5217391304 + 0.0266910187
= 1.5484301491
For n = 2:
x_3 = x_2 - ((x_2^3 - x_2 - 2) / (3 * x_2^2 - 1))
= 1.5484301491 - ((1.5484301491^3 - 1.5484301491 - 2) / (3 * 1.5484301491^2 - 1))
= 1.5484301491 - ((3.771902576 - 1.5484301491 - 2) / (3 * 3.2369424195 - 1))
= 1.5484301491 - ((3.771902576 - 3.5484301491) / (9.7108272586 - 1))
= 1.5484301491 - (0.2234724269 / 8.7108272586)
= 1.5484301491 + 0.0145881663
= 1.5630183154
By continuing this process, you can approximate the solution of the equation (x^3 - x - 2) = 0 within the given interval to as many decimal places as needed.