# Numerical Analysis/Bisection Method Worked Example

X^3=20 ,f(X), X4=4, X1=1 take four iterations .

## Example

[edit | edit source]Find an approximation of correct to within 10^{-4} by using the bisection method on . starting on [1, 2].

### Analysis of the Problem

[edit | edit source]The number of iterations we will use, *n*, must satisfy the following formula:

.

Thus, we will use 14 iterations of the bisection method.

### Iteration #1

[edit | edit source]First, we find the midpoint of the interval [1, 2]:

Then we check if is positive or negative:

The value of is negative, which means that is less than .

We therefore use as the left endpoint of our new interval and keep 2 as the right endpoint.

yes

### Iteration #2

[edit | edit source]Repeat the process from Iteration #1 to do Iteration #2:

Solution:

Then we check if is positive or negative:

The value of is positive, which means that is greater than .

We therefore keep 1.5 as the left endpoint of our new interval and use as the right endpoint.

### Iteration #3 - Iteration #14

[edit | edit source]Complete Iteration #3 - Iteration #14:

Solution:

or ? | ||||
---|---|---|---|---|

3 | [1.5, 1.75] | 1.625 | -0.35937 | |

4 | [1.625, 1.75] | 1.6875 | -0.15235 | |

5 | [1.6875, 1.75] | 1.74875 | -0.0459 | |

6 | [1.71875, 1.75] | 1.73438 | 0.00807 | |

7 | [1.71875, 1.73438] | 1.72657 | -0.01896 | |

8 | [1.72657, 1.73438] | 1.73048 | -0.0544 | |

9 | [1.73048, 1.73438] | 1.73243 | .00131 | |

10 | [1.73048, 1.73243] | 1.73146 | -0.00205 | |

11 | [1.73146, 1.73243] | 1.73195 | -0.00035 | |

12 | [1.73195, 1.73243] | 1.73219 | 0.00048 | |

13 | [1.73195, 1.73219] | 1.73207 | 0.00007 | |

14 | [1.73195, 1.73207] | 1.73201 | -0.00014 | N/A |

the final answer is follow or describe below

### Conclusion

[edit | edit source]Thus, we have found that

Solution:

is an approximation of correct to within 10^{-4}.