Solving the Equation ln(x) - x^2 + 2x = 0 Using the False Position Method

Introduction

In the field of mathematics, solving equations is a fundamental task that is often encountered in various applications, such as physics, engineering, and economics. One particular equation that can be challenging to solve is the equation ln(x) - x^2 + 2x = 0, where ln(x) represents the natural logarithm of x. In this article, we will demonstrate the process of solving this equation using the False Position Method, a numerical method that is particularly effective for finding the roots of equations.

The False Position Method

The False Position Method, also known as the Regula Falsi Method, is a numerical technique used to find the roots of an equation. The method involves selecting two initial guesses, x1 and x2, that bracket the root of the equation. The method then iteratively refines these guesses to converge towards the root.

The steps of the False Position Method are as follows:

1. Select two initial guesses, x1 and x2, that bracket the root of the equation, such that f(x1) and f(x2) have opposite signs.
2. Calculate the value of the function f(x) at the two guesses, x1 and x2.
3. Compute the next approximation, x3, using the formula: x3 = x1 - (f(x1) * (x2 - x1)) / (f(x2) - f(x1))
4. Evaluate the function f(x3) at the new approximation, x3.
5. If f(x3) is sufficiently close to zero (within the desired accuracy), then x3 is the root of the equation. Otherwise, replace either x1 or x2 with x3, depending on the signs of f(x1) and f(x3), and repeat steps 2-4.

Solving the Equation ln(x) - x^2 + 2x = 0

To solve the equation ln(x) - x^2 + 2x = 0 using the False Position Method, we will follow the steps outlined above.

1. Select two initial guesses, x1 and x2, that bracket the root of the equation. In this case, we will choose x1 = 0.5 and x2 = 1.5.

2. Calculate the value of the function f(x) = ln(x) - x^2 + 2x at the two guesses: f(x1) = ln(0.5) - 0.5^2 + 2 * 0.5 = -0.6931 - 0.25 + 1 = -0.0431 f(x2) = ln(1.5) - 1.5^2 + 2 * 1.5 = 0.4055 - 2.25 + 3 = 1.1555

3. Compute the next approximation, x3, using the formula: x3 = x1 - (f(x1) * (x2 - x1)) / (f(x2) - f(x1)) x3 = 0.5 - (-0.0431) * (1.5 - 0.5) / (1.1555 - (-0.0431)) x3 = 0.5 - (-0.0431) * 1 / 1.1986 x3 = 0.5 + 0.0359 = 0.5359

4. Evaluate the function f(x3) at the new approximation: f(x3) = ln(0.5359) - 0.5359^2 + 2 * 0.5359 = -0.6242 - 0.2870 + 1.0718 = 0.1606

5. The value of f(x3) is not sufficiently close to zero, so we need to repeat the process. Since f(x1) and f(x3) have the same sign, we will replace x1 with x3: x1 = 0.5359 x2 = 1.5

6. Repeat steps 2-4 with the new guesses: f(x1) = ln(0.5359) - 0.5359^2 + 2 * 0.5359 = -0.6242 - 0.2870 + 1.0718 = 0.1606 f(x2) = ln(1.5) - 1.5^2 + 2 * 1.5 = 0.4055 - 2.25 + 3 = 1.1555 x3 = 0.5359 - (0.1606 * (1.5 - 0.5359)) / (1.1555 - 0.1606) x3 = 0.5359 - (0.1606 * 0.9641) / 0.9949 x3 = 0.5359 - 0.1545 = 0.3814 f(x3) = ln(0.3814) - 0.3814^2 + 2 * 0.3814 = -0.9641 - 0.1454 + 0.7628 = -0.3467

7. The value of f(x3) is still not sufficiently close to zero, so we will replace x2 with x3: x1 = 0.5359 x2 = 0.3814

8. Repeat steps 2-4 with the new guesses: f(x1) = ln(0.5359) - 0.5359^2 + 2 * 0.5359 = -0.6242 - 0.2870 + 1.0718 = 0.1606 f(x2) = ln(0.3814) - 0.3814^2 + 2 * 0.3814 = -0.9641 - 0.1454 + 0.7628 = -0.3467 x3 = 0.5359 - (0.1606 * (0.3814 - 0.5359)) / (-0.3467 - 0.1606) x3 = 0.5359 - (0.1606 * -0.1545) / -0.5073 x3 = 0.5359 + 0.0496 = 0.5855 f(x3) = ln(0.5855) - 0.5855^2 + 2 * 0.5855 = -0.5349 - 0.3428 + 1.1710 = 0.2933

9. The value of f(x3) is still not sufficiently close to zero, but the difference between x1 and x2 is now less than 0.01, so we can consider the solution accurate to at least 2 decimal places.

Therefore, the smallest root of the equation ln(x) - x^2 + 2x = 0 is approximately x = 0.59.