The simplest way to solve any equation - except linear or quadratic equations - is by trial and error.
A. Interpolation
To solve
Let:
| x | LHS | RHS | DIFF | |
|---|---|---|---|---|
| 0 | 1 | 0 | 1 | } 2.5 |
| 1 | .5 | 2 | -1.5 | |
| .4 | .862 | .8 | .062 | |
| .5 | …. | …. | …. |
Explanation
This document explains how to solve the equation
B. Iteration
Solve the equation for
| trial | result |
|---|---|
| 0 | |
| .5 | |
| .4 | |
| … | … |
Explanation
This example demonstrates another powerful numerical technique, known as Fixed-Point Iteration, to solve an equation. Notice that, this is the same equation as in the previous example (1 / (1 + x²) = 2x), just rearranged to isolate x on one side:
Here is a step-by-step breakdown of the process shown:
- Initial Guess (Trial 1): We start with an initial guess of
trial x = 0. We plug this into the right side of the equation:- Result =
1/2 * (1 / (1 + 0²)) = 1/2 * (1) = 0.5This result,0.5, becomes our next trial value.
- Result =
- First Iteration (Trial 2): Now, we use this result as our new
trial x = 0.5:- Result =
1/2 * (1 / (1 + 0.5²)) = 1/2 * (1 / 1.25) = 0.5 * 0.8 = 0.4
- Result =
- Second Iteration (Trial 3): We repeat the process with
trial x = 0.4:- Result =
1/2 * (1 / (1 + 0.4²)) = 1/2 * (1 / 1.16) ≈ 0.5 * 0.862 = 0.431
- Result =
The process continues this way. We would next use 0.431 as our trial x and calculate a new result. With each step, the “trial x” and “result x” get closer to each other. The solution is found when the value we put in is the same as the value that comes out (i.e., when the sequence converges).
If the equation is solved in the form
C. Newton’s Method
Solve the equation in the form

[Let’s go back to the previous example]
| x | f(x) | f’(x) |
|---|---|---|
| 0 | 1 | -2 |
| .5 | .2 | …. |
| … | …. | …. |
Explanation
The table shows the step-by-step process of applying the method.
- First Guess: You start with a simple initial guess: x₁ = 0.
- Plug in the Guess: At x=0, you calculate the values of the function and its derivative:
- f(0) = 1/(1+0²) - 2(0) = 1
- f’(0) = -2(0)/(1+0²)² - 2 = -2
- Calculate the Next Guess (x₂): Now, you use the Newton’s Method formula:
- The Next Step (Row 2): Your new, improved guess is x₂ = 0.5. The second row of the table would use this value to calculate f(0.5) and f’(0.5) in order to find an even better guess, x₃. This process continues until f(x) is extremely close to zero.
Problem: Find first positive root of