Talk:Bisection method
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Hmm, my textbook give the formula for the absolute error as (b-a)/2^n. Whereas the one here is (b-a)/2^(n+1). I don't know which one is correct... can someone clarify for me?
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- Its (b-a)/2^(n+1). Think about it... Let n=0. then if it was (b-a)/2^n then the error would be b-a, which would extend beyond the limits of b and a, a domain in which we know MUST contain the target. Strange though, that a textbook had it wrong....
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- Also, I would like to tell readers of this article that bisection can break down if f is non-continous or is not monotonic or not a function. It is too much of an unnecessary complication for the article, but worth mentioning.
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- I don't think that f needs to be monotone (why do you think so?), but it indeed needs to be a continuous function. -- Jitse Niesen (talk) 12:11, 13 November 2005 (UTC)
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- Please explain why you think it needs to be differentiable. -- Jitse Niesen (talk) 02:20, 3 March 2006 (UTC)
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- The error is |b-a|/2^n. His text had it right. After 1 subdivision (n=1), the error is at most half the original interval, not 1/4. |b-a| doesn't extend beyond the domain in which we know must contain the target -- it's exactly equal to it. I also added the missing absolute value. 165.189.91.148 20:46, 20 April 2006 (UTC)
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- It depends on what you take to be the approximation to the root. If you use the midpoint of the interval, the error is |b-a|/2^(n+1); if you use either endpoint, the error is |b-a|/2^n. Which formula is used, is not very important, in my opinion. -- Jitse Niesen (talk) 02:26, 21 April 2006 (UTC)
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- Agreed, and I've edited the section to reflect the "controversy". The problem arises because the method doesn't give a single estimate for the root's location, and the concepts of "absolute error" and "length of the interval" are easily confused. The actual formula indeed isn't important, save in Calculus 1 classes where it's fairly often an exam question. Majromax (talk) 22:38, 19 November 2007 (UTC)
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[edit] Pseudo Code completion
The Pseudo Code presented in this article needs to be made more accurate. As of right now, it does not account for the possibility of encoutering the 0 exactily on xL or xR. A theird condition needs to be added.
- No. Bisection method will happily converge to a root even if it occurs at the endpoint. It will just converge to the endpoint with the root. The given pseudocde is proper in that it will consider an interval beginning or ending at 0 to have a sign change. Admittedly, this isn't exactly an efficient approach, but it certainly gives clear code without extraneous conditions. Majromax (talk) 22:38, 19 November 2007 (UTC)
[edit] Visual Basic Code
state which version. Visual Basic.net is not backward compatible with previous versions anymore —Preceding unsigned comment added by 192.197.54.27 (talk) 14:10, 2 October 2007 (UTC)
- Although in general the two are not compatible, this code runs in both VB.net and the older VisualBasic. CRGreathouse (t | c) 21:32, 2 April 2008 (UTC)
[edit] enumeration
- If f has several roots in the interval [a, b], then the bisection method finds the odd-numbered roots with equal, non-zero probability and the even-numbered roots with zero probability (Corliss 1977).
with respect to which enumeration??? --Philtime (talk) 08:28, 11 May 2008 (UTC) PS: what if there are is an infinite and/or uncountable amount of roots? --Philtime (talk) 08:28, 11 May 2008 (UTC)
