Answer :
To approximate the solution to the equation [tex]\( x^3 + 2x - 1 = 0 \)[/tex], we will use an iterative method with the function [tex]\( x_{n+1} = \frac{1}{x_n^2 + 2} \)[/tex]. Starting with [tex]\( x_1 = 1 \)[/tex], we will apply this iterative formula until the values converge to a stable solution within a defined tolerance.
Let's break down the steps:
1. Initial value:
[tex]\[ x_1 = 1 \][/tex]
2. Define the iterative formula:
[tex]\[ x_{n+1} = \frac{1}{x_n^2 + 2} \][/tex]
3. Compute successive iterations until the change between successive values is less than a specified tolerance (let's use [tex]\( 10^{-6} \)[/tex]) or a maximum number of iterations is reached (let's use 1000 iterations as a safeguard).
Let's perform a few iterations to show how this converges:
1. Iteration 1:
[tex]\[ x_2 = \frac{1}{x_1^2 + 2} = \frac{1}{1^2 + 2} = \frac{1}{3} \approx 0.3333 \][/tex]
2. Iteration 2:
[tex]\[ x_3 = \frac{1}{x_2^2 + 2} = \frac{1}{0.3333^2 + 2} = \frac{1}{0.1111 + 2} \approx \frac{1}{2.1111} \approx 0.4736 \][/tex]
3. Iteration 3:
[tex]\[ x_4 = \frac{1}{x_3^2 + 2} = \frac{1}{0.4736^2 + 2} = \frac{1}{0.2243 + 2} \approx \frac{1}{2.2243} \approx 0.4497 \][/tex]
4. Iteration 4:
[tex]\[ x_5 = \frac{1}{x_4^2 + 2} = \frac{1}{0.4497^2 + 2} = \frac{1}{0.2022 + 2} \approx \frac{1}{2.2022} \approx 0.4541 \][/tex]
5. Iteration 5:
[tex]\[ x_6 = \frac{1}{x_5^2 + 2} = \frac{1}{0.4541^2 + 2} = \frac{1}{0.2062 + 2} \approx \frac{1}{2.2062} \approx 0.4532 \][/tex]
As we can see, the values are converging. If we continue this process until the change is less than the tolerance:
6. Iteration 6:
[tex]\[ x_7 = \frac{1}{x_6^2 + 2} \approx \frac{1}{0.4532^2 + 2} = \frac{1}{0.2054 + 2} \approx \frac{1}{2.2054} \approx 0.4534 \][/tex]
Continuing this iteration, we find that the values will stabilize around the same number.
After the iterations converge, we arrive at an approximate value for the solution. For the purposes of giving the answer to 2 decimal places, the solution is approximately:
[tex]\[ x \approx 0.45 \][/tex]
Therefore, the approximate solution to the equation [tex]\( x^3 + 2x - 1 = 0 \)[/tex] using the iteration method is:
[tex]\[ \boxed{0.45} \][/tex]
Let's break down the steps:
1. Initial value:
[tex]\[ x_1 = 1 \][/tex]
2. Define the iterative formula:
[tex]\[ x_{n+1} = \frac{1}{x_n^2 + 2} \][/tex]
3. Compute successive iterations until the change between successive values is less than a specified tolerance (let's use [tex]\( 10^{-6} \)[/tex]) or a maximum number of iterations is reached (let's use 1000 iterations as a safeguard).
Let's perform a few iterations to show how this converges:
1. Iteration 1:
[tex]\[ x_2 = \frac{1}{x_1^2 + 2} = \frac{1}{1^2 + 2} = \frac{1}{3} \approx 0.3333 \][/tex]
2. Iteration 2:
[tex]\[ x_3 = \frac{1}{x_2^2 + 2} = \frac{1}{0.3333^2 + 2} = \frac{1}{0.1111 + 2} \approx \frac{1}{2.1111} \approx 0.4736 \][/tex]
3. Iteration 3:
[tex]\[ x_4 = \frac{1}{x_3^2 + 2} = \frac{1}{0.4736^2 + 2} = \frac{1}{0.2243 + 2} \approx \frac{1}{2.2243} \approx 0.4497 \][/tex]
4. Iteration 4:
[tex]\[ x_5 = \frac{1}{x_4^2 + 2} = \frac{1}{0.4497^2 + 2} = \frac{1}{0.2022 + 2} \approx \frac{1}{2.2022} \approx 0.4541 \][/tex]
5. Iteration 5:
[tex]\[ x_6 = \frac{1}{x_5^2 + 2} = \frac{1}{0.4541^2 + 2} = \frac{1}{0.2062 + 2} \approx \frac{1}{2.2062} \approx 0.4532 \][/tex]
As we can see, the values are converging. If we continue this process until the change is less than the tolerance:
6. Iteration 6:
[tex]\[ x_7 = \frac{1}{x_6^2 + 2} \approx \frac{1}{0.4532^2 + 2} = \frac{1}{0.2054 + 2} \approx \frac{1}{2.2054} \approx 0.4534 \][/tex]
Continuing this iteration, we find that the values will stabilize around the same number.
After the iterations converge, we arrive at an approximate value for the solution. For the purposes of giving the answer to 2 decimal places, the solution is approximately:
[tex]\[ x \approx 0.45 \][/tex]
Therefore, the approximate solution to the equation [tex]\( x^3 + 2x - 1 = 0 \)[/tex] using the iteration method is:
[tex]\[ \boxed{0.45} \][/tex]
