Starting with x1 = 1, find the third approximate x3 to the root of the | Mathematics I 2079

Mathematics I20795 marksNumerical: Newton's method rootNewton's method

Starting with $x_1 = 1$ , find the third approximate $x_3$ to the root of the equation $x^3 - x - 5 = 0$ . [5]

Starting with $x_1 = 1$, find the third approximate $x_3$ to the root of the equation $x^3 - x - 5 = 0$. [5]

Equation: $f(x) = x^3 - x - 5 = 0$ - Starting value: $x1 = 1$ - Required: $x3$ (third approximation) - Method implied: Newton-Raphson Iteration formula: $x{n+1} = xn - \frac{f(xn)}{f'(xn)}, \qquad f'(x) = 3x^2 - 1$ At $x1 = 1$ : $f(1) = 1 - 1 - 5 = -5$ $f'(1) = 3(1)^2 - 1 = 2$ $$x2 = 1 - ...

Equation: $f(x) = x^3 - x - 5 = 0$ - Starting value: $x1 = 1$ - Required: $x3$ (third approximation) - Method implied: Newton-Raphson Iteration formula: $$x{n+1} = xn - \frac{f(xn)}{f'(xn)}, \qquad f'(x) = 3x^2 - 1$$ At $x1 = 1$: $$f(1) = 1 - 1 - 5 = -5$$ $$f'(1) = 3(1)^2 - 1 = 2$$ $$x2 = 1 - ...

Starting with x1 = 1, find the third approximate x3 to the root of the equation x3 - x - 5 = 0.