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)
[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.
