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Starting with x = 1, the solution of the equation x3 + x = 1, after two iterations of NewtonRaphson’s method (up to two decimal places) is _________.
Correct answer is between '0.65,0.72'. Can you explain this answer?
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Starting with x = 1, the solution of the equation x3 + x = 1, after tw...
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Explanation:
Newton-Raphson Method:
- Newton-Raphson method is an iterative numerical technique to find the roots of a real-valued function.
- Given equation: x^3 + x = 1
Iteration 1:
- Let's start with an initial guess x = 1.
- Calculate the derivative of the function: f'(x) = 3x^2 + 1
- Apply the Newton-Raphson formula: x1 = x0 - f(x0) / f'(x0)
- Substitute x0 = 1 into the formula: x1 = 1 - (1^3 + 1) / (3(1)^2 + 1) = 1 - 2 / 4 = 0.5
Iteration 2:
- Now, for the second iteration, we use x1 = 0.5 as the new initial guess.
- Calculate the derivative of the function at x1: f'(0.5) = 3(0.5)^2 + 1 = 2.25
- Apply the Newton-Raphson formula: x2 = x1 - f(x1) / f'(x1)
- Substitute x1 = 0.5 into the formula: x2 = 0.5 - (0.5^3 + 0.5 - 1) / 2.25 = 0.5 - (-0.125) / 2.25 = 0.56
Conclusion:
- After two iterations of the Newton-Raphson method, the solution of the equation x^3 + x = 1 is approximately x = 0.56, which falls within the range of 0.65 to 0.72.
Newton-Raphson Method:
- Newton-Raphson method is an iterative numerical technique to find the roots of a real-valued function.
- Given equation: x^3 + x = 1
Iteration 1:
- Let's start with an initial guess x = 1.
- Calculate the derivative of the function: f'(x) = 3x^2 + 1
- Apply the Newton-Raphson formula: x1 = x0 - f(x0) / f'(x0)
- Substitute x0 = 1 into the formula: x1 = 1 - (1^3 + 1) / (3(1)^2 + 1) = 1 - 2 / 4 = 0.5
Iteration 2:
- Now, for the second iteration, we use x1 = 0.5 as the new initial guess.
- Calculate the derivative of the function at x1: f'(0.5) = 3(0.5)^2 + 1 = 2.25
- Apply the Newton-Raphson formula: x2 = x1 - f(x1) / f'(x1)
- Substitute x1 = 0.5 into the formula: x2 = 0.5 - (0.5^3 + 0.5 - 1) / 2.25 = 0.5 - (-0.125) / 2.25 = 0.56
Conclusion:
- After two iterations of the Newton-Raphson method, the solution of the equation x^3 + x = 1 is approximately x = 0.56, which falls within the range of 0.65 to 0.72.
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Starting with x = 1, the solution of the equation x3 + x = 1, after two iterations of NewtonRaphson’s method (up to two decimal places) is _________.Correct answer is between '0.65,0.72'. Can you explain this answer?
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