x^3-x-2=0 correct to six decimal places by newton raphson method
there are some good online calculators for this
To solve the equation x^3 - x - 2 = 0 correct to six decimal places using the Newton-Raphson method, you need to follow these steps:
1. First, rewrite the equation in the form f(x) = 0, which in this case is f(x) = x^3 - x - 2.
2. Next, take the derivative of f(x) to find f'(x), which gives us f'(x) = 3x^2 - 1.
3. Choose an initial guess for the root of the equation. Let's say x0 = 1.
4. Use the formula x1 = x0 - f(x0)/f'(x0) to find the next approximation x1.
Substitute x0 into f(x) and f'(x), we get:
f(x0) = 1^3 - 1 - 2 = -2
f'(x0) = 3(1)^2 - 1 = 2
x1 = 1 - (-2)/(2) = 2
5. Repeat step 4 using x1 as the new approximation, to find x2.
Substitute x1 into f(x) and f'(x), we get:
f(x1) = 2^3 - 2 - 2 = 6
f'(x1) = 3(2)^2 - 1 = 11
x2 = 2 - (6)/(11) ≈ 1.454545
6. Continue this iterative process until you reach the desired level of accuracy or precision (in this case, six decimal places).
Substituting x2 into f(x) and f'(x), we iterate again:
f(x2) = (1.454545)^3 - 1.454545 - 2 ≈ -0.191301
f'(x2) = 3(1.454545)^2 - 1 ≈ 6.636363
x3 = 1.454545 - (-0.191301)/(6.636363) ≈ 1.401987
7. Repeat step 6 until the desired level of accuracy is achieved. Continue iterating using the latest approximation until the difference between two consecutive iterations is within the desired precision.
Substituting x3 into f(x) and f'(x), we iterate again:
f(x3) ≈ -0.009476
f'(x3) ≈ 5.246150
x4 ≈ 1.401987 - (-0.009476)/(5.246150) ≈ 1.401953
8. Keep repeating step 7 until the desired level of accuracy is reached. Iterate using the latest approximation until the difference between two consecutive iterations is within the desired precision of six decimal places.
Continuing with the iterations, we find:
x5 ≈ 1.401953
x6 ≈ 1.401953
Therefore, the root of the equation x^3 - x - 2 = 0, correct to six decimal places, is approximately 1.401953.