Certainly! Let's work through the iteration process used to approximate the solution to the equation [tex]\(x^3 + 2x - 1 = 0\)[/tex].
1. Initial Setup:
- We start with an initial guess, [tex]\( x_1 = 1 \)[/tex].
- We use the iteration formula [tex]\( x_{n+1} = \frac{1}{x_n^2 + 2} \)[/tex].
- To determine convergence, we stop the iterations when the absolute difference between successive iterations is less than a tolerance level, which we'll set to [tex]\( 1 \times 10^{-5} \)[/tex] for accuracy to two decimal places.
2. Iteration Process:
- Compute the following sequence until the change between consecutive terms is very small:
[tex]\[
\begin{align*}
x_1 &= 1 \\
x_2 &= \frac{1}{x_1^2 + 2} = \frac{1}{1^2 + 2} = \frac{1}{3} \approx 0.3333 \\
x_3 &= \frac{1}{x_2^2 + 2} = \frac{1}{(0.3333)^2 + 2} \approx \frac{1}{0.1111 + 2} \approx \frac{1}{2.1111} \approx 0.4737 \\
x_4 &= \frac{1}{x_3^2 + 2} = \frac{1}{(0.4737)^2 + 2} \approx \frac{1}{0.2244 + 2} \approx \frac{1}{2.2244} \approx 0.4498 \\
\vdots
\end{align*}
\][/tex]
3. Checking for Convergence:
- We continue this iteration process, updating [tex]\( x_n \)[/tex] each time, until [tex]\( \left| x_{n+1} - x_n \right| \)[/tex] is less than [tex]\( 1 \times 10^{-5} \)[/tex].
4. Final Result:
- After several iterations, the sequence converges to a value that does not significantly change with further iterations. The value stabilizes and we can round to two decimal places.
- After sufficient iterations, we obtain [tex]\( x \approx 0.45 \)[/tex].
Thus, the approximate solution to the equation [tex]\( x^3 + 2x - 1 = 0 \)[/tex] is [tex]\( \boxed{0.45} \)[/tex], correct to two decimal places.
