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

(x₂)³-1
and x₁ = -1
4
Xn+1 =
4/34
a) (i) Work out the value of x₂
(ii) Work out the value of x3
b) Work out the solution to 6 decimal places.

User Sagibb
by
3.2k points

1 Answer

22 votes
22 votes

Answer:

a) i) -0.5

ii) -0.28125

b) -0.254102 (6 d.p.)

Explanation:

Given iteration formula:


x_(n+1)=(\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 & =(\left(x_1\right)^3-1)/(4)\\\\& =(\left(-1\right)^3-1)/(4)\\\\ & = (-1-1)/(4)\\\\ & = (-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 & =(\left(x_2\right)^3-1)/(4)\\\\& =(\left(-0.5\right)^3-1)/(4)\\\\ & = (-0.125-1)/(4)\\\\ & = (-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.).

User Vivek Patel
by
2.8k points