Answer:
a) i) -0.5
ii) -0.28125
b) -0.254102 (6 d.p.)
Step-by-step explanation:
Given iteration formula:
[tex]x_{n+1}=\dfrac{\left(x_n\right)^3-1}{4} \quad \textsf{and} \quad x_1=-1[/tex]
Part (a)(i)
Substitute the value of x₁ into the formula and solve for x₂ :
[tex]\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}[/tex]
Part (a)(ii)
Substitute the value of x₂ into the formula and solve for x₃ :
[tex]\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}[/tex]
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.
[tex]\implies x_4=-0.2555618286...[/tex]
[tex]\implies x_5=-0.2541728038...[/tex]
[tex]\implies x_6=-0.2541051331...[/tex]
[tex]\implies x_7=-0.2541018552...=-0.254102\:\: \sf (6 \:d.p.)[/tex]
[tex]\implies x_8=-0.2541016964...=-0.254102\:\: \sf (6 \:d.p.)[/tex]
[tex]\implies x_9=-0.2541016888...=-0.254102\:\: \sf (6 \:d.p.)[/tex]
Therefore, the solution is -0.254102 (6 d.p.).
