Talk:Safe addition: Difference between revisions

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(Some interval arithmetic issues)
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Wouldn't the sum of the two ranges be (''a''+<small>&darr;</small>''b'', ''a''+<small>&uarr;</small>''b'')? Modulo whether they are closed or open bounds, of course (my argument doesn't depend on that). I can't see any logical reason for assuming otherwise, as the “real” values of ''a'' and ''b'' could be anywhere from the lower bound to the upper bound; that's got to be what the range means, with appropriate infinitesimals in the open bound case, of course. (Yes, this does mean that the error will increase as more operations are performed; this is a normal feature of calculations with physical quantities if I'm remembering my Physics-101 class right.) —[[User:Dkf|Donal Fellows]] 22:52, 18 August 2009 (UTC)
 
: I am not sure, if this answers your question. Addition of intervals is defined as [''a'', ''b''] + [''c'', ''d''] = [''a'' + ''c'', ''b'' + ''d'']. With +&darr; and +&uarr; operations its machine implementation becomes [''a'' +&darr; ''c'', ''b'' +&uarr; ''d'']. As for error (rather uncertainlyuncertainty) of the result obtained using interval arithmetic, it is not a simple issue. The result's width grows under the assumption of the worst case scenario. That is, assuming all intervals involved in computations independent. The intermediate results are often dependent, so the inputs, etc. Therefore in many cases the result can be improved in terms of interval width. A simple illustration of the case. Consider interval expression [1, 2] x [1, 2]. According to the interval arithmetic, the result is [1, 4]. But that is when [1, 2] and [1, 2] are fully independent. If they were ''same'' (correlated), we could multiply [1, 2] by 2, and get a two-fold narrower result [1, 2] * 2 = [2, 4]. --[[User:Dmitry-kazakov|Dmitry-kazakov]] 08:46, 19 August 2009 (UTC)