Answer :
Answer:
-0.201640
Step-by-step explanation:
Iteration is a numerical method used to approximate solutions to equations where exact solutions are difficult or impossible to obtain analytically.
The given iteration formula and the starting value (x₁) is:
[tex]x_{n+1}=\dfrac{\left(x_n\right)^3-1}{5}\quad \textsf{and}\quad x_1=-2[/tex]
To find x₂, substitute x₁ = -2 into the iteration formula:
[tex]x_{2}=\dfrac{\left(x_1\right)^3-1}{5}\\\\\\x_{2}=\dfrac{\left(-2\right)^3-1}{5}\\\\\\x_{2}=\dfrac{-8-1}{5}\\\\\\x_{2}=-\dfrac{9}{5}\\\\\\\boxed{x_2=-1.8}[/tex]
To find x₃, substitute x₂ = -1.8 into the iteration formula:
[tex]x_{3}=\dfrac{\left(x_2\right)^3-1}{5}\\\\\\x_{3}=\dfrac{\left(-1.8\right)^3-1}{5}\\\\\\x_{3}=\dfrac{-5.832-1}{5}\\\\\\x_{3}=\dfrac{-6.832}{5}\\\\\\\boxed{x_{3}=-1.3664}[/tex]
Continue this way until the solutions are the same when rounded to 6 decimal places:
[tex]x_4 = -0.710227\\\\x_5=-0.271651\\\\x_6=-0.204009\\\\x_7=-0.201698\\\\x_8=-0.201641\\\\x_9=-0.201640\\\\x_{10}=-0.201640\\\\x_{11}=-0.201640[/tex]
Therefore, the approximate solution is -0.201640, rounded to 6 decimal places.
[tex]\dotfill[/tex]
Additional information
The quickest way to compute each value in the iteration sequence is to enter the value of x₁ into the calculator and press "=" so that the value of x₁ becomes the "answer". Then type the iteration formula into the calculator, replacing xₙ with "answer":
[tex]\dfrac{\left(\text{Ans}\right)^3-1}{5}[/tex]
Each time you press "=", the next value of xₙ in the sequence is computed.
To find an approximate solution to the equation \(x^3 - \frac{1}{5}\) using the iterative process, we can start with the initial value \(x_1 = -2\) and apply the iteration formula:
\[x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}\]
where \(f(x) = x^3 - \frac{1}{5}\).
First, we need to find the derivative of \(f(x)\), which is \(f'(x) = 3x^2\).
Now, we can apply the iteration formula:
\[x_{n+1} = x_n - \frac{x_n^3 - \frac{1}{5}}{3x_n^2}\]
Let's plug in the initial value \(x_1 = -2\) to find \(x_2\):
\[x_2 = -2 - \frac{(-2)^3 - \frac{1}{5}}{3(-2)^2}\]
\[x_2 = -2 - \frac{-8 - \frac{1}{5}}{12}\]
\[x_2 = -2 - \frac{-\frac{41}{5}}{12}\]
\[x_2 = -2 + \frac{41}{60}\]
\[x_2 = -2 + 0.683333...\]
\[x_2 \approx -1.316666...\]
We can continue this process iteratively to find \(x_3\), \(x_4\), and so on until we reach the desired accuracy of 6 decimal places. However, the iterative process can be quite lengthy to perform manually, so it's often done using computational methods or software.
