Iterative Solution: X^3 - 19x = 33 (Approx. To 2 D.p.)

In the realm of mathematics, finding solutions to equations is a fundamental task. While some equations can be solved directly using algebraic methods, others require numerical techniques to approximate the solutions. One such technique is the iterative method, which involves repeatedly applying a formula to generate a sequence of values that converge towards the solution. In this article, we'll explore how to find an approximate solution to the equation $x^3 - 19x = 33$ using an iterative formula. This method is particularly useful when dealing with equations that are difficult or impossible to solve analytically. So, let’s dive in and see how this works!

Understanding Iterative Formulas

Before we tackle the specific equation, let's briefly discuss what iterative formulas are and how they work. An iterative formula is a mathematical expression that allows us to generate a sequence of approximations to a solution. It typically takes the form:

$x_{n+1} = f(x_n)$

where:

• $x_{n+1}$ is the next approximation in the sequence.
• $x_n$ is the current approximation.
• $f$ is a function that defines the iterative process.

The process starts with an initial guess, $x_1$, and then the formula is applied repeatedly. Each application of the formula produces a new approximation, and hopefully, these approximations get closer and closer to the actual solution. The key is to choose an appropriate function $f$ that will lead to convergence. Now, let's focus on our given equation and the iterative formula provided.

The Equation and the Iterative Formula

We are given the equation:

$x^3 - 19x = 33$

And the iterative formula:

$x_{n+1} = \sqrt[3]{19x_n + 33}$

This formula is derived from the original equation by rearranging it to isolate x on one side. Specifically, we can rewrite the equation as:

$x^3 = 19x + 33$

Then, taking the cube root of both sides gives us the iterative formula. This rearrangement is crucial because it sets up the iterative process where each new value $x_{n+1}$ is calculated based on the previous value $x_n$. To start the process, we are given an initial value:

$x_1 = 5$

Our goal is to use this iterative formula to find an approximate solution to the equation, accurate to 2 decimal places. This means we will keep iterating until the value of x converges to a stable value when rounded to 2 decimal places. Let's proceed with the iterations and see how the solution unfolds.

Applying the Iterative Formula

Now, let's apply the iterative formula step-by-step, starting with $x_1 = 5$:

Iteration 1:

$x_2 = \sqrt[3]{19x_1 + 33} = \sqrt[3]{19(5) + 33} = \sqrt[3]{95 + 33} = \sqrt[3]{128} \approx 5.03968$

So, after the first iteration, we get $x_2 \approx 5.03968$. We'll keep this value for the next iteration.

Iteration 2:

$x_3 = \sqrt[3]{19x_2 + 33} = \sqrt[3]{19(5.03968) + 33} = \sqrt[3]{95.75392 + 33} = \sqrt[3]{128.75392} \approx 5.04926$

After the second iteration, we have $x_3 \approx 5.04926$. Notice that the value has changed slightly, but we are getting closer to a potential solution.

Iteration 3:

$x_4 = \sqrt[3]{19x_3 + 33} = \sqrt[3]{19(5.04926) + 33} = \sqrt[3]{95.93594 + 33} = \sqrt[3]{128.93594} \approx 5.05174$

Now, $x_4 \approx 5.05174$. The changes in the values are becoming smaller, indicating we are converging towards a solution.

Iteration 4:

$x_5 = \sqrt[3]{19x_4 + 33} = \sqrt[3]{19(5.05174) + 33} = \sqrt[3]{95.98306 + 33} = \sqrt[3]{128.98306} \approx 5.05236$

We find that $x_5 \approx 5.05236$. The difference between $x_4$ and $x_5$ is quite small now.

Iteration 5:

$x_6 = \sqrt[3]{19x_5 + 33} = \sqrt[3]{19(5.05236) + 33} = \sqrt[3]{95.99484 + 33} = \sqrt[3]{128.99484} \approx 5.05251$

We get $x_6 \approx 5.05251$. The values are converging even more closely.

Iteration 6:

$x_7 = \sqrt[3]{19x_6 + 33} = \sqrt[3]{19(5.05251) + 33} = \sqrt[3]{95.99769 + 33} = \sqrt[3]{128.99769} \approx 5.05255$

Here, $x_7 \approx 5.05255$. The values are changing very little now.

Determining the Approximate Solution to 2 Decimal Places

We need to find the solution accurate to 2 decimal places. Let's look at our last few iterations:

• $x_5 \approx 5.05236$
• $x_6 \approx 5.05251$
• $x_7 \approx 5.05255$

When we round these values to 2 decimal places:

• $x_5 \approx 5.05$
• $x_6 \approx 5.05$
• $x_7 \approx 5.05$

Since the values have stabilized to 5.05 when rounded to 2 decimal places, we can conclude that the approximate solution to the equation $x^3 - 19x = 33$ is 5.05.

Conclusion

In this article, we successfully found an approximate solution to the equation $x^3 - 19x = 33$ using an iterative formula. We started with an initial guess of $x_1 = 5$ and applied the formula $x_{n+1} = \sqrt[3]{19x_n + 33}$ repeatedly. Through several iterations, the values converged to a stable solution. By rounding the values to 2 decimal places, we determined that the approximate solution is 5.05.

Iterative methods are powerful tools for solving equations that cannot be easily solved algebraically. They provide a way to approximate solutions to a desired level of accuracy. Understanding and applying these methods can be incredibly useful in various fields of mathematics, science, and engineering.

For further reading on iterative methods and numerical solutions, you can explore resources like Numerical Methods - Iterative Methods.