# Example 1

Consider finding the root of f(*x*) = *x*^{2} - 3. Let
ε_{step} = 0.01, ε_{abs} = 0.01 and start with the interval [1, 2].

Table 1. False-position method applied to f(*x*) = *x*^{2} - 3.

a |
b |
f(a) |
f(b) |
c |
f(c) |
Update | Step Size |
---|---|---|---|---|---|---|---|

1.0 | 2.0 | -2.00 | 1.00 | 1.6667 | -0.2221 | a = c | 0.6667 |

1.6667 | 2.0 | -0.2221 | 1.0 | 1.7273 | -0.0164 | a = c | 0.0606 |

1.7273 | 2.0 | -0.0164 | 1.0 | 1.7317 | 0.0012 | a = c | 0.0044 |

Thus, with the third iteration, we note that the last step 1.7273 → 1.7317 is less than 0.01 and |f(1.7317)| < 0.01, and therefore we chose b = 1.7317 to be our approximation of the root.

Note that after three iterations of the false-position method, we have an acceptable answer (1.7317 where f(1.7317) = -0.0044) whereas with the bisection method, it took seven iterations to find a (notable less accurate) acceptable answer (1.71344 where f(1.73144) = 0.0082)

# Example 2

Consider finding the root of f(*x*) = e^{-x}(3.2 sin(*x*) - 0.5 cos(*x*)) on the interval [3, 4],
this time with ε_{step} = 0.001, ε_{abs} = 0.001.

Table 2. False-position method applied to f(*x*) = e^{-x}(3.2 sin(*x*) - 0.5 cos(*x*)).

a |
b |
f(a) |
f(b) |
c |
f(c) |
Update | Step Size |
---|---|---|---|---|---|---|---|

3.0 | 4.0 | 0.047127 | -0.038372 | 3.5513 | -0.023411 | b = c | 0.4487 |

3.0 | 3.5513 | 0.047127 | -0.023411 | 3.3683 | -0.0079940 | b = c | 0.1830 |

3.0 | 3.3683 | 0.047127 | -0.0079940 | 3.3149 | -0.0021548 | b = c | 0.0534 |

3.0 | 3.3149 | 0.047127 | -0.0021548 | 3.3010 | -0.00052616 | b = c | 0.0139 |

3.0 | 3.3010 | 0.047127 | -0.00052616 | 3.2978 | -0.00014453 | b = c | 0.0032 |

3.0 | 3.2978 | 0.047127 | -0.00014453 | 3.2969 | -0.000036998 | b = c | 0.0009 |

Thus, after the sixth iteration, we note that the final step, 3.2978 → 3.2969 has a size less than 0.001 and |f(3.2969)| < 0.001 and therefore we chose b = 3.2969 to be our approximation of the root.

In this case, the solution we found was not as good as the solution we found using the bisection method (f(3.2963) = 0.000034799) however, we only used six instead of eleven iterations.

Copyright ©2005 by Douglas Wilhelm Harder. All rights reserved.