Exercises on the bisection method/Solution
- The following is a possible implementation of the bisection method with Octave/MATLAB:
- The solution of the points 1, 2 e 3 can be found in the example of the bisection method.
For point 4 we have
so we would need at least 70 iterations. The chance of convergence with such a small precision depends on the calculatord: in particular, with Octave, the machine precision is roughly . For this reason it does not make sense to choose a smaller precision. The number of iterations, if we don't specify a maximum number, would be infinite.
- To verify the existence of a root we need to show that the hypothesis roots theorem are satisfied. The first hypothesis requires to be continuous. Obviosly this is a continuous function since it is sum of two continuous functions. The second hypotheses requires the function to have oppiste signs at the interval extrema, and in fact we find
- To show the uniqueness of the root we need to prove that the function is monotone and in fact
- The number of iterations need is given by
- and so we have .
- The interval does not contain aany root as the second hypotesis of the roots theorem fails, in fact
- In the plot we show in red the average errorand in blu the actual error. From the graph, it is clear that the actual error is not a monotone function. Moreover, note that the global behavior of both curves is the same, clarifying the term average error for .
For the solution look at the convergence analysis in the bisection method page.