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Hi, I have a question on the "convergence linearly" . Since the absolute error of bisection convergence is |b-a|/2^n, so the rate of convergence is O(1/2^n) which means the growth of error is exponential. I think it is more precise to use this information instead of the phrase "convergence linearly"! <span style="font-size: smaller;" class="autosigned">—Preceding [[Wikipedia:Signatures|unsigned]] comment added by [[Special:Contributions/132.235.37.166|132.235.37.166]] ([[User talk:132.235.37.166|talk]]) 13:24, 3 October 2008 (UTC)</span><!-- Template:UnsignedIP --> <!--Autosigned by SineBot-->
: This is called linear convergence in numerical analysis because the number of digits in the answer grows at a linear rate. Equivalently, the running time is linear in log(eps). [[User:McKay|McKay]] ([[User talk:McKay|talk]]) 06:13, 1 April 2009 (UTC)
I don’t agree with the statement that the bisection method has a linear convergence or, in other words, a 1st order convergence. The classification of a 1st order method has to do with the error along the sequence (distance between the successive values and the root) and not with the estimation of the absolute maximal error, incidentally quite imprecise in the bisection method.
In the 1st order methods the observation of the sequence of values converging to the root reveals a steady increase in the number of correct figures. On the contrary, through the bisection method it is possible for the distance to the root to increase in two consecutive iterations. It follows that the bisection method is just a method that narrows the interval that contains the root, halving it on each iteration. It doesn’t guarantee the same happens with the distance to the root, therefore it is not a 1st order method. Ana Maria Faustino (FEUP)
== Computation of the midpoint ==
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