Timeline for Do formulas involving fewer repetitions of variables give higher numerical precision?
Current License: CC BY-SA 3.0
9 events
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| Sep 30, 2018 at 22:20 | vote | accept | nalzok | ||
| Jan 19, 2017 at 7:14 | comment | added | nalzok | I'm really sorry to say that, but it seems that our discussion has gone beyond the boarder of what my little brain can understand. I didn't expect things to become complicated like this... While it is true that, technically speaking, every program is comprise of a set of expressions, I think this exercise concerns simple algebraic expressions only. And your Interval Newton Method example is just too unconventional and complicated for a layman like me to understand. Anyway, I've learnt a lot about how numerical things work from your great answer and patient comments. Thank you again! | |
| Jan 19, 2017 at 6:49 | comment | added | Derek Elkins left SE | The way I'd approach the case I gave, with +,-,* is by noting that such expressions are polynomials and can be put into Horner form. Then attempt to show that every other equivalent expression can be reached from it by transformation that preserve or widen the interval. Either way, I believe, though I haven't thought too hard about it that the only ways duplicate variables can arise are: distributivity, additive inverse (X-X=0), and affine combinations of copies of a variable with non-interval coefficients (arguably, a case of distributivity). | |
| Jan 19, 2017 at 6:39 | comment | added | Derek Elkins left SE | $\log$ is monotonic so it's not going to cause problems. I didn't state that it was false, though largely because it's not clear what "uses" means in that context nor what the range of allowed operations would be. As a simple example though, doing Newton's method with intervals diverges but if you "use" a variable an extra time to intersect, it works correctly: www2.math.uni-wuppertal.de/~xsc/xsc/node12.html | |
| Jan 19, 2017 at 6:29 | comment | added | nalzok | I've tried many expressions with more "advanced" functions such as $\log_a b =\frac{\log b }{\log a }$, but all of them show me that "the more inter-interval operations (i.e. operations between two intervals) are performed, the more information on precision is lost, and thus the less accurate the result would be". Could you please give me an example of an "arbitrary program" for which Eva Lu Ator's statement is false? | |
| Jan 19, 2017 at 6:03 | comment | added | nalzok | To divide two intervals, we can simply multiplies the first by the reciprocal of the second, so I guess the statement holds true for all expressions involving only addition, subtraction, multiplication, and division. I'm glad to perform the proof on my own, but could you give me some hints on how to prove it? You know, even elementary arithmetic can produce a broad class of expressions, and I don't know where to get started. Also, by "arbitrary programs", do you mean weird things like the Dirichlet function? | |
| Jan 19, 2017 at 5:38 | comment | added | Derek Elkins left SE | As a generally true fact about interval arithmetic, I don't know. I suspect that you can prove that it holds for broad classes of expressions. For example, I'm fairly confident that it's true for expressions involving only addition, subtraction, and multiplication. In general "expressions" can be arbitrary programs for which this statement is almost certainly false. | |
| Jan 19, 2017 at 3:22 | comment | added | nalzok |
This answer is great! However, I think compensated summation mainly deals with the error brought by floating-point arithmetic, on which this exercise doesn't lay much emphasis. Let's assume floating-point arithmetic always gives accuracy result. (Alternatively, let's assume rational numbers are stored as numerator / denominator in the memory, so the representation is exact.) In this case, considering only the error caused by interval arithmetic itself, is Eva Lu Ator right?
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| Jan 18, 2017 at 17:56 | history | answered | Derek Elkins left SE | CC BY-SA 3.0 |