Introduction Notes Theory HOWTO Examples
Engineering Error Questions Matlab Maple
Example 1
As an example of the secant method, suppose we wish to find a root of the function f(x) = cos(x) + 2 sin(x) + x2. A closed form solution for x does not exist so we must use a numerical technique. We will use x0 = 0 and x1 = -0.1 as our initial approximations. We will let the two values εstep = 0.001 and εabs = 0.001 and we will halt after a maximum of N = 100 iterations.
We will use four decimal digit arithmetic to find a solution and the resulting iteration is shown in Table 1.
Table 1. The secant method applied to f(x) = cos(x) + 2 sin(x) + x2.
| n | xn − 1 | xn | xn + 1 | |f(xn + 1)| | |xn + 1 - xn| |
|---|---|---|---|---|---|
| 1 | 0.0 | -0.1 | -0.5136 | 0.1522 | 0.4136 |
| 2 | -0.1 | -0.5136 | -0.6100 | 0.0457 | 0.0964 |
| 3 | -0.5136 | -0.6100 | -0.6514 | 0.0065 | 0.0414 |
| 4 | -0.6100 | -0.6514 | -0.6582 | 0.0013 | 0.0068 |
| 5 | -0.6514 | -0.6582 | -0.6598 | 0.0006 | 0.0016 |
| 6 | -0.6582 | -0.6598 | -0.6595 | 0.0002 | 0.0003 |
Thus, with the last step, both halting conditions are met, and therefore, after six iterations, our approximation to the root is -0.6595 .
Copyright ©2005 by Douglas Wilhelm Harder. All rights reserved.
