# 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*) + *x*^{2}. A
closed form solution for *x* does not exist so we must use a numerical
technique. We will use *x*_{0} = 0 and *x*_{1} = -0.1 as our initial approximations. We will let the
two values ε_{step} = 0.001 and ε_{abs} = 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*) + *x*^{2}.

n |
x_{n − 1} |
x_{n} |
x_{n + 1} |
|f(x_{n + 1})| |
|x_{n + 1} - x_{n}| |
---|---|---|---|---|---|

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.