Question

An approximate solution to an equation is found using this iterative process.

Answer 1

Answer:

a) i) -0.5

   ii) -0.28125

b)  -0.254102 (6 d.p.)

Step-by-step explanation:

Given iteration formula:

$x_{n+1}=\dfrac{\left(x_n\right)^3-1}{4} \quad \textsf{and} \quad x_1=-1$

Part (a)(i)

Substitute the value of x₁ into the formula and solve for x₂ :

$\begin{aligned}\implies x_2 & =\dfrac{\left(x_1\right)^3-1}{4}\\\\& =\dfrac{\left(-1\right)^3-1}{4}\\\\ & = \dfrac{-1-1}{4}\\\\ & = \dfrac{-2}{4}\\\\ & = -0.5\end{aligned}$

Part (a)(ii)

Substitute the value of x₂ into the formula and solve for x₃ :

$\begin{aligned}\implies x_3 & =\dfrac{\left(x_2\right)^3-1}{4}\\\\& =\dfrac{\left(-0.5\right)^3-1}{4}\\\\ & = \dfrac{-0.125-1}{4}\\\\ & = \dfrac{-1.125}{4}\\\\ & = -0.28125\end{aligned}$

Part (b)

To find the solution to 6 decimal places, keep substituting each new value into the iteration formula until the answers are the same when rounded to the required level of accuracy.

$\implies x_4=-0.2555618286...$

$\implies x_5=-0.2541728038...$

$\implies x_6=-0.2541051331...$

$\implies x_7=-0.2541018552...=-0.254102\:\: \sf (6 \:d.p.)$

$\implies x_8=-0.2541016964...=-0.254102\:\: \sf (6 \:d.p.)$

$\implies x_9=-0.2541016888...=-0.254102\:\: \sf (6 \:d.p.)$

Therefore, the solution is -0.254102 (6 d.p.).