Wikipedia:Reference desk/Archives/Mathematics/2020 November 24

= November 24 =

Bisection method
On the 24th of November, an anonymous editor made this revision to Bisection method. Was this correct? Thanks for your time. Opalzukor (talk) 13:44, 24 November 2020 (UTC)


 * No, it was not correct. --JBL (talk) 14:00, 24 November 2020 (UTC)


 * An easy counterexample, showing the incorrectness of the claim $$|c_n-c|\ge\frac{|b-a|}{2^n}$$, is given by the function $$f(x)=x$$ on the interval $$[a,b]=[-1,1]$$. Then, for $$n=1$$, $$|c_1-c| = 0$$, which is less than $$\frac{|1-(-1)|}{2}=1.$$ --Lambiam 23:27, 24 November 2020 (UTC)


 * First of all one should note the inequality affected serves as a means to estimate the method's convergence. As such we need it to show how fast the approximation error decreases, hence to bound it from above. A bounding from below with greater-or-equal is completely useless in this context, despite being true or false. --CiaPan (talk) 17:45, 29 November 2020 (UTC)
 * One might reasonably (also) wish to have a lower bound on the number of iterations required to achieve a given tolerance. In this case we know that such a lower bound (one that works for all continuous functions) is necessarily trivial, but this may not be clear a priori. The OP's question was whether a specific revision was correct. An IMO easy and convincing way to establish (instead of merely state) that it is not correct, is to show that it introduces a mathematical falsehood. Since the stopping criterion is based on the size of the bracket [a, b], which is halved each iteration regardless of whatever, while maintaining the invariant a < c < b where c is a zero, the introduction of the sequence of approximations (cn)n is actually an irrelevant complication anyway. --Lambiam 20:19, 29 November 2020 (UTC)
 * All above is true. Anyway, a question may arise, how you knew what example to check. When faced to a general question it's often hard for a beginner to decide whether to seek a general proof or rather a specific counterexample. I gave a simple reasoning that shows which way to go in this case. --CiaPan (talk) 21:04, 29 November 2020 (UTC)
 * I don't know how I knew; it was obvious (to me) at first glance that we could choose &fnof;, a and b such that c1 = c, thereby falsifying the altered inequation. What mattered to me most, after that, was to give a concrete counterexample that was easily checkable, which is why I chose &fnof;, a and b the way I did, and not, for example, &fnof;(x) = (1 − 2x)(x − 2x), [a, b] = [0, 1]. --Lambiam 22:39, 29 November 2020 (UTC)