Talk:Bisection method

message: 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)
replies: text:  Incorrect. The code checks if it is < 0, and if the left, right, or midpoint is exactly zero, an infinite loop would have occurred. Added a simple else - no performance hit compare dto the normal function. It could still be efficientized though.  —Preceding unsigned comment added by 207.161.134.144 (talk) 18:25, 4 February 2010 (UTC)
thread_title:  Pseudo Code completion 

message: 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)
replies: text: 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)
replies: text:  Anyway, Visual Basic isn't free, it would be wiser to write the pseudocode in Python or Java. --PhDP (talk) 01:30, 26 November 2008 (UTC)
replies: text: As long as the section header states 'pseudo-code', I find both Visual Basic, Python and Java inappropriate. Either change the header, or change the code into something that is more or less language-independent. Also keep in mind that source code is not particularly encourared in Wikipedia (WP:NOT). --Berland (talk) 11:35, 26 November 2008 (UTC)
thread_title:  Visual Basic Code 

message: 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"!  —Preceding unsigned comment added by 132.235.37.166 (talk) 13:24, 3 October 2008 (UTC)
replies: text: 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). McKay (talk) 06:13, 1 April 2009 (UTC)
thread_title:  Convergence 

message: In section 'Practical consideration', 2. point, the passage 
:Another method relies on the assumption that (right+left)/2 exactly equals left or right if left and right are adjacent non-equal floating-point values. This is true for most hardware, including IEEE arithmetic in the absence of overflow, but can be violated if intermediate expressions are calculated to greater precision than stored variables (depending on computer optimization).
is totally dubious to me. What does this other method do about? Is that what's implemented in the code below i.e. use left+(right-left)/2 instead of (left+right)/2? And if so, does it take care of the case when the assumption is true (it sounds like that) or when the assumption is violated (makes more sense in my opinion)? Furthermore, when is the assumption violated-any example? Final question: if I know that IEEE arithmethic is used: is there any advantage of preferring left+(right-left)/2 over (left+right)/2? I can't see any, whether left and right are very close to each other or not...  Ezander (talk) 08:48, 26 November 2010 (UTC)
replies: text: Ahh, now I see, it was about the stopping criterion, not the computation of the midpoint. Then it makes sense. However, still the simpler (hi+lo)/2 has been changed to lo+(hi-lo)/2. Any reason for this? Ezander (talk) 08:53, 26 November 2010 (UTC)
thread_title:  Computation of the midpoint 

message: I don't think it is very sensible that both code examples are in different languages. One in VB and the other one (it states in the source tag that it's C, which it clearly isn't) in some hodgepodge of languages (I've seen all that notation in different languages, but never together in one language: assignment (:=) from Pascal-like languages, comments (//) from C++, loop constructs (endwhile, endif) from PHP(?), very strange...
I'm not a friend of VB. However the code looks quite clear, nearly like pseudo-code... Maybe the second example should be changed also to VB? Ezander (talk) 09:03, 26 November 2010 (UTC)
thread_title:  Language of Code examples 

message: Does anyone know when the bisection method was first used? The method seems quite obvious but that doesn't necessarily mean it came before other methods, like secant for example.  — Preceding unsigned comment added by 2.222.44.210 (talk) 12:36, 2 December 2011 (UTC)
thread_title:  History 

message: I think the bisection method mentioned above can be only used for finding only one root case and it is not effective for multiple roots case. If I assume that the function has multiple roots on [a, b] and f(a) and f(b) have opposite signs,then by using the method above, I can find the midpoint c between a and b. If f(c) and f(b) have opposite sign, then I can use the new interval [c, b] and keep going to find a smaller interval until I find one root on that interval,but in this way, I neglect the possible roots on [a, c].Since f(a) and f(c) have the same sign,I can not apply bisection method to find roots on [a, c]. It seems to be difficult to find all the roots of the function on the interval using bisection method because it is hard to know each interval which includes only one root of the function.
So my opinion is to find one of the roots on the interval first by bisection method and then we can apply other methods ,for example Newton's method to find other roots of the function. In fact, we can at first find the monotonicity of the function on the given interval. If the function has monotonicity on interval[a, b] and f(a),f(b) have opposite signs, then we can apply bisection method to find the only one root of that function, otherwise we can not only use bisection method to find all roots of the function unless we know all the local maximal and minimal points of that function by solving the first and second derivative.After that I can split the original interval into some small intervals which include one maximum and one minimum and check if the maximum and the minimum have the opposite sign. If they are not, then skip that interval and continue checking the neighbor interval until you can see the opposite sign of those two values.So we can use bisection method in that interval to find one root and keep going the steps above until we find all the roots of the given function.
YangOu (talk) 01:50, 12 September 2012 (UTC)
thread_title:  The disadvantage of bisection method 

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thread_title:  Mathematical formula