Use Newton's method to find sqrt[6]2, correct to five decimal places. | Mathematics I 2078

Mathematics I20785 marksNumerical: Newton's method rootNewton's method

Use Newton's method to find $\sqrt[6]{2}$ , correct to five decimal places. [5]

Use Newton's method to find $\sqrt[6]{2}$, correct to five decimal places. [5]

Target: $\sqrt[6]{2} = 2^{1/6}$ - Required accuracy: 5 decimal places Let $x = 2^{1/6}$ , so $x^6 = 2$ , giving: $f(x) = x^6 - 2 = 0, \qquad f'(x) = 6x^5$ $x{n+1} = xn - \frac{xn^6 - 2}{6xn^5} = \frac{5xn^6 + 2}{6xn^5}$ Since $1^6 = 1 < 2 < 2^6 = 64$ , and $(1.1)^6 = 1.7716$ , $(1.2)^6 = 2.986$ ...

Target: $\sqrt[6]{2} = 2^{1/6}$ - Required accuracy: 5 decimal places Let $x = 2^{1/6}$, so $x^6 = 2$, giving: $$f(x) = x^6 - 2 = 0, \qquad f'(x) = 6x^5$$ $$x{n+1} = xn - \frac{xn^6 - 2}{6xn^5} = \frac{5xn^6 + 2}{6xn^5}$$ Since $1^6 = 1 < 2 < 2^6 = 64$, and $(1.1)^6 = 1.7716$, $(1.2)^6 = 2.986$...