Starting with x1 = 2, find the third approximation x3 to the root of t | Mathematics I 2075

Mathematics I20755 marksNumerical: Newton-Raphson approximationNewton's method

Starting with $x_1 = 2$ , find the third approximation $x_3$ to the root of the equation $x^3 - 2x - 5 = 0$ . [5]

Starting with $x_1 = 2$, find the third approximation $x_3$ to the root of the equation $x^3 - 2x - 5 = 0$. [5]

Equation: $f(x) = x^3 - 2x - 5$ - Initial approximation: $x1 = 2$ - Required: third approximation $x3$ Iteration formula: $x{n+1} = xn - \frac{f(xn)}{f'(xn)}, \qquad f'(x) = 3x^2 - 2$ $f(2) = 8 - 4 - 5 = -1$ $f'(2) = 3(4) - 2 = 10$ $x2 = 2 - \frac{-1}{10} = 2 + 0.1 = 2.1$ $$f(2.1) = (2....

Equation: $f(x) = x^3 - 2x - 5$ - Initial approximation: $x1 = 2$ - Required: third approximation $x3$ Iteration formula: $$x{n+1} = xn - \frac{f(xn)}{f'(xn)}, \qquad f'(x) = 3x^2 - 2$$ $$f(2) = 8 - 4 - 5 = -1$$ $$f'(2) = 3(4) - 2 = 10$$ $$x2 = 2 - \frac{-1}{10} = 2 + 0.1 = 2.1$$ $$f(2.1) = (2....

Starting with x1 = 2, find the third approximation x3 to the root of the equation x3 - 2x - 5 = 0.