Wikipedia:Reference desk/Archives/Mathematics/2011 September 13

= September 13 =

Name for variable update procedure
When a variable is used to describe some attribute of a sequence, and it is desired for that variable to weight the values of recent observations more greatly than earlier ones it is common to update it as so:

var = var * (1 - a) + observation * a

where a is some real number between 0 and 1

It's quite a natural way to do it, and I'm sure people do it all the time without really considering it. Is there a name for this? I'm sure I read it in an article here fairly recently but now I've not got a clue which... Thanks! --Iae (talk) 01:07, 13 September 2011 (UTC)


 * Sorry, came to me as soon as I posted: Exponential moving average --Iae (talk) 01:12, 13 September 2011 (UTC)

Relationship between Computer Science and Mathematics
How is computer science related to pure and/or applied mathematics.I see that computer science is interwined with discrete maths,combinatorics.By the way is computer programming a a mathematical faculty.In some universities Maths Honours students have to clear a C programming paper.Numeric analysis may require to do programming.Scientific visualization may also require programming as also computational electromagnetics but these are applications.My question is- why and when does a student of CSE dont understand algorithms and programming and find it all very dull and disgusting.Ichgab (talk) 07:50, 13 September 2011 (UTC)
 * That sounds all very specific to the requirements of particular universities in particular places, along with a specific complaint. (You don't like programming?) It might help if you gave us your location. But even then, I'm not sure exactly what your question is. HiLo48 (talk) 08:01, 13 September 2011 (UTC)
 * Hmm. If a student on a computer science course finds algorithms and programming "very dull and disgusting", I suspect he or she might just be studying the wrong subject. At some universities computer science is part of the mathematics department but this does not necessarily mean that either discipline is taught as part of the other. As HiLo says, it just depends. Many aspects of mathematics are quite unlike computer science. However, both academic disciplines require careful, logical thinking (ideally with some inspiration too!) and the ability to carry this forward in a sustained way. Many mathematicians are not that much interested in computing and much computer development does not require mathematics. However, I suspect that someone who found algorithms uninteresting might well find the same for many areas of mathematics. Thincat (talk) 08:29, 13 September 2011 (UTC)
 * Computers are pretty bad at thinking up fresh, new ideas. However, computers are often a great help to visualize these ideas, some of which are ideas that humans are not made to grasp, such as higher dimensions. Aacehm (talk) 02:36, 17 September 2011 (UTC)

universe where pi is 3
you know how people study newtonian physics even though it doesn't actually describe our universe? could a teacher teach a math universe where pi is 3 (for simplified calculations) even though it doesn't describe our particular universe? 82.234.207.120 (talk) 14:47, 13 September 2011 (UTC)


 * That could happen if space was warped enough. The circumference of the circle may stay the same, but while the center bumps upward due to warping in space, the diameter gets a little longer. So, the value of pi shrinks a tiny bit. We live in space that is pretty much as flat as possible (as far as any experiment can tell), so we don't see that happen. -- k a i n a w &trade; 15:09, 13 September 2011 (UTC)


 * That's not actually the case -- even warped spaces are locally flat (unless they are extremely weird). In any case, $$\pi$$ shows up in much more than geometry, and it would be easy to prove that $$0 = 1$$ starting from the assumption that $$\pi = 3$$. Looie496 (talk) 16:12, 13 September 2011 (UTC)


 * Yes. A space curved enough to produce pi=3 would also produce varying values such that 0=1. Space that curved wouldn't have a consistent value for pi. It would change as the diameter changed. All in all, Euclidean geometry simply won't work in that universe, so attempts to apply our Euclidean-based geometry is a bit pointless. -- k a i n a w &trade; 16:17, 13 September 2011 (UTC)


 * The elephant in the room is: how do you define a circle in a curved space? There are many different properties that hold for a circle in Euclidean space that may not be equivalent in non-Euclidean space; let alone curved space. You could define them as the curves with constant, non-zero curvature − but how do you define curvature of a curve in a curved space? — Fly by Night  ( talk )  18:47, 13 September 2011 (UTC)


 * ... so if the ratio would be variable, and a "circle" would be undefined in this space, the logical conclusion is that there does not exist a universe in which pi can be consistently defined as equal to 3. Of course, in the days when all measurements were approximate, it was reasonable to use 3 as an approximation for pi, as in the building of Solomon's Temple (1 Kings 7 verse 23, though this might be explained by the difference between inside and outside diameter).    D b f i r s   19:10, 13 September 2011 (UTC)


 * I think we can define a circle as long as we have a metric - a cirle with radius r and centre P is simply the set of points at a distance r from P. However, the ratio of circumference to radius will, in general, not be a constant but will depend on the size of the circle (i.e. circumference will not be a simple multiple of radius, but will be some more complicated function of radius). Gandalf61 (talk) 08:04, 14 September 2011 (UTC)


 * ... but wouldn't such a "circle" have a variable diameter in many curved spaces?   D b f i r s   09:15, 14 September 2011 (UTC)


 * How are you defining "diameter" ? If take a point Q on the circle and define the "diameter at Q" as "the maximum distance between Q and any other point on the circle as measured along a geodesic" then I think the diameter of a circle must always be twice its radius. We know there is at least one other point R on the circle with |QR| = 2r; go along a geodesic from Q to the centre of the circle P then continue along the same geodesic until it intersects the circle again at R. And we know that no point S on the circle can have |QS| > 2r because this would means that |QS| > |QP| + |PS| i.e. it would break the triangle inequality, which is part of the definition of a metric. So "diameter at Q" is always 2r for any circle and any choice of Q. Gandalf61 (talk) 10:13, 14 September 2011 (UTC)


 * Yes, I hadn't read your reply carefully enough, and the weird curved spaces I was imagining wouldn't have a metric.   D b f i r s   20:17, 14 September 2011 (UTC)


 * I think people have been responding in good faith to what the question "should have been", so to speak, but it is worth pointing out definitively here that &pi; cannot be 3. Never never never never.  No matter what physical conditions might obtain.
 * &pi; is a mathematical concept and does not depend upon physical law.
 * As people have pointed out, yes, in very warped regions of space (or less warped ones for very big circles), you can have a circle with a circumference equal to 6 times its radius (for suitable definition of "radius"; in general relativity this is a somewhat problematic concept). But that does not make &pi; equal to 3!  &pi; is never anything but &pi;. --Trovatore (talk) 20:46, 14 September 2011 (UTC)
 * But in Trovatore's warped universe the real &pi; is not equal to the complex &pi;, and the real 3 is not equal to the integer 3. Bo Jacoby (talk) 21:26, 14 September 2011 (UTC).


 * Trovatore has got it exactly right. Thought experiments that explore the results of adjusting physical constants or laws can be frutiful. Likewise, tons of the coolest math is accomplished by adjusting axioms. But adjusting the value of a mathematical constant is tantamount to (and logically equivalent to) identifying 1 with 0: at that point everything goes to pot, by which I mean that one could logically derive propositions and their negations, a useless state of affairs.—PaulTanenbaum (talk) 00:25, 15 September 2011 (UTC)

Variant of Entscheidungsproblem
In 1936 and 1937 Alonzo Church and Alan Turing respectively, published independent papers showing that it is impossible to decide algorithmically whether statements in arithmetic are true or false, and thus a general solution to the Entscheidungsproblem is impossible.

What happens if you amend Entscheidungsproblem to only accept inputs that are possible to decide algorithmically? For instance, a solution to this variation of Entscheidungsproblem could accept "$$\begin{bmatrix}1 & 3\\2 & 6 \end{bmatrix}$$ is invertible" as an input and return "false", and it could accept "11 is prime" and return "true", because there exist algorithms for determining the truth of these statements. However, it would not need to decide if Diophantine equations have solutions, because that cannot be decided algorithmically. Has it also been shown that this variation of Entscheidungsproblem has no solution? Widener (talk) 23:24, 13 September 2011 (UTC)
 * You need to formalize the question further. It doesn't make sense to say a particular input is or is not decidable: for any Diophantine equation, there is an algorithm which, on input that equation, outputs "yes". There's another which outputs "no". In any completion of arithmetic, one of these algorithms has correctly told you if that equation has solutions.
 * Decidability only makes sense on classes of problems, but "class of problems" isn't a formal concept. Some classes of Diophantine equations can be decided quite easily.  If you ask me if a particular equation has solutions, should I interpret it as a member of the full class, so the question can't be decided, or as a member of some subclass that can be?
 * Maybe you mean to restrict to statements which are provable or disprovable from PA? In that case, there's a simple algorithm: simultaneously search for a proof or a disproof. When you find one, halt and give the appropriate answer. Of course, on statements that are independent of PA, this algorithm will never terminate, and since there's no effective way to identify such statements, you can't get away from that problem.--Antendren (talk) 05:37, 14 September 2011 (UTC)


 * You could revise the Entscheidungsproblem, or another undecidable problem like the halting problem, so that the allowable responses from the algorithm are "yes", "no", and "I don't know". It is possible to write an algorithm to solve this modified Entscheidungsproblem—trivially, in fact, by always saying "I don't know". The interesting problem is then to design an algorithm that answers "I don't know" as rarely as possible, and otherwise always answers correctly. Of course, to make that problem rigorous, you'll need to precisely define what you mean by "rarely".
 * Algorithms that answer "yes", "no", or "I don't know" are actually in use for applications that need to solve undecidable problems. For example, the other day I asked Mathematica to solve a Diophantine equation, and the response was something like, "The methods available to Mathematica are insufficient to either find a solution or prove that one does not exist," essentially an "I don't know". —Bkell (talk) 15:18, 14 September 2011 (UTC)