It is given that
f(x)=x3−6x−2f(x) = x^3 - 6x - 2f(x)=x3−6x−2
Show that f(x)=0f(x) = 0f(x)=0 has a root α \alpha\,α in the interval [2.6,2.7][2.6, 2.7][2.6,2.7].
Show that the equation f(x)=0f(x) = 0f(x)=0 can be rearranged into the form
x=6x+23x = \sqrt[3]{6x + 2}x=36x+2
Use the iteration formula
xn+1=6xn+23x_{n+1} = \sqrt[3]{6x_n + 2}xn+1=36xn+2
with x0=2.6x_0 = 2.6x0=2.6 to find, to four decimal places, the values of x1x_1x1, x2 x_2\,x2 and x3x_3x3.
A student instead uses the iteration formula
xn+1=xn3−26\displaystyle x_{n+1} = \frac{x_n^3 - 2}{6}xn+1=6xn3−2
with x0=2.6x_0 = 2.6x0=2.6, and finds that the values do not approach α\alphaα.
Explain why this iteration fails to converge to α\alphaα.
125 exam-style questions on AQA A Level Maths 1.12 I: Numerical methods (A-level only), covering 1.12.1 Locating roots by change of sign (A-level only), 1.12.2 Iterative methods and Newton-Raphson (A-level only), 1.12.3 Numerical integration (A-level only), and 1.12.4 Numerical methods in context (A-level only). Each one has a worked solution and a mark scheme showing where the marks go.