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January 10

Matrix subgroup

I'm interested in identifying the subgroup of GL(n,Z) generated by the squares of transvection matrices (i.e. squares of matrices that differ from the identity matrix by replacing a single zero with a one) -- in particular, I'd like to know a presentation. It's not too difficult to see (or maybe it is) that any matrix in this subgroup must have odd entries on the diagonal and even entries elsewhere, and so it is contained in the level 2 principal congruence subgroup (it might perhaps be the whole congruence group?). Any pointers towards a presentation, or just an identification of this group would be appreciated! Thanks, Icthyos (talk) 10:11, 10 September 2012 (UTC)

Related to your earlier question - Wikipedia:Reference_desk/Archives/Mathematics/2012_March_27#Homomorphisms_from_GL.28n.2CZ.29_into_a_symmetric_group.3F ? The wikilinks there, especially the 4th & 5th section of Special linear group & Steinberg group (K-theory) should help - there's a presentation of SL (n, Z) there. Has to be some modification of the Steinberg relations, and the group you are looking at should (nearly?) generate the congruence subgroup, as the elementary matrices / transvections generate SL. A look at the first chapters of Milnor's book on should help (even if not directly, because it is Milnor).John Z (talk) 09:14, 11 September 2012 (UTC)

It's not completely relevant to that question, I just always seem to end up mapping into GL(n,Z)! I've realised that I can throw in inversion matrices too if I want, as well as squares of transvections, and I'm *pretty* sure that together they generate the level 2 congruence group (the transvection squares alone, however, do not). So, it comes down to finding a presentation of the congruence subgroup...but surely someone must have done this before? I mean, it's a finite index subgroup, so I can definitely churn out a finite presentation using Reidemeister-Schreier, but...really? Must I? Thanks for the suggestions, Icthyos (talk) 09:51, 11 September 2012 (UTC)

Standard metrics for quality of multi-objective optimization?

Tried looking through relevant articles on optimization but couldn't find any notice. Is there any "standard" metric to judge the quality (i.e. how close to the Pareto front are the results found) for a heuristic optimization algorithm? This is in the setting of discrete input variables (hence finite state space). A concrete case study is available for direct comparison of the solutions found, but it would be nice to learn if such metrics are (generally) used.

For example, say the Pareto front lies for two variables to be maximized lies on (0,10), (8,8), and (10,5), but the algorithm identifies (0,10), (7,7) and (10,5), thereby missing one solution on the front (and none in between), out of a state space of size 128? Of course such a metric can be "made up" (e.g. share of missed solutions better than the identified Pareto border, area of the space between identified and real border), but I want to know if there is any such metric which is used as a "standard". 130.237.57.87 (talk) 14:53, 10 September 2012 (UTC)

I don't know if there is a standard for this, but if there is one I would predict that it would take this form: the average, over all points identified by the algorithm, of the squared Euclidean distance to the nearest point on the true Pareto front.

One could simulate the heuristic algorithm, and its resulting average of squared Euclidean distances, over a large number of hypothesized problems for which the entire Pareto front is known, then take an average (of the averaged Euclidean distances) over the set of all hypothesized problems. It seems to me, however, that this approach will be sensitive to the choice of hypothesized problems, so one would have to choose that set carefully based on what sort of problems, with what frequencies, one plans to use the heuristic for.

A possible flaw in the above suggested metric is that it fails to penalize heuristics that don't offer many candidate points Duoduoduo (talk) 17:35, 11 September 2012 (UTC)

Since in this case we have concrete case studies, it would be relatively easy to get such metrics (throwing many problems at the algorithm will be done slightly later...). There is a problem that it penalizes algorithms that have "gaping holes" (i.e. misses obvious candidates which are "on their own" so don't have close borders on the Pareto front...), but at least it's a start. If there was no standard I was thinking about something along those lines too... 130.237.57.87 (talk) 09:09, 12 September 2012 (UTC)

Another approach, which I have not seen, would be this: Make the Pareto front continuous by interpolating between points, perhaps linearly. Do the same with the heuristically generated front. Then take the area between the two continuous fronts. That should penalize a paucity of generated points (since the linear interpolation between two points with a big gap between them will generate a lot of area), and it should also penalize an algorithm which fails to find any point nearest to some point on the front. (That is, if point P on the Pareto front is not the nearest point to any of the generated points, that failure would be penalized.) Duoduoduo (talk) 14:14, 12 September 2012 (UTC)

Thanks, good idea. I actually had it too... 85.228.20.17 (talk) 19:32, 12 September 2012 (UTC)

What is a cross-over point in a rational funciton?

My daughter's math class is working with rational functions and graphing them. What is a "cross-over point" of a function such as y = 3x/(x+4)? IS it where the horizontal and vertical asymptotes cross, (-4,3) in this case? Bubba73 ^{You talkin' to me?} 20:53, 10 September 2012 (UTC)

That sounds reasonable. The only other possibility I can think of is if it refers to the X- and Y-intercepts (both (0,0), in this case). StuRat (talk) 21:12, 10 September 2012 (UTC)

Her textbook doesn't seem to say and she can't remember what the teacher said. But as I thought about it, it was all I could think of that made sense. They are working on asymptotes. Bubba73 ^{You talkin' to me?} 22:44, 10 September 2012 (UTC)

While I've never heard that terminology myself, I believe you have given the most reasonable interpretation. One other thought though, it might be a trick question, with the answer being "this curve never crosses over either asymptote" (some curves do). StuRat (talk) 23:53, 10 September 2012 (UTC)

In a case like this, I just give every bit of info I have:

• The 2 asymptotes cross at (-4,3).
• The curve does not cross either asymptote.
• The X-intercept is at (0,0).
• The Y-intercept is at (0,0).

StuRat (talk) 00:01, 11 September 2012 (UTC)

There were three of these on a test she got back today, so I doubt it was a trick question. (She missed all three.) Bubba73 ^{You talkin' to me?} 00:22, 11 September 2012 (UTC)

Sorry to hear that. Did any of those curves cross the asymptotes ?

In any case, since this terminology was apparently made up by the teacher, she better ask the teacher what it means, so she will know for the finals. StuRat (talk) 00:24, 11 September 2012 (UTC)

Do you mean did they cross the X- or Y-axis? Bubba73 ^{You talkin' to me?} 01:43, 11 September 2012 (UTC)

There were three rational functions. For each they were to graph it, give the x- and y-intercepts, the horizontal and vertical asymptotes, the cross-over point, and the hole. The hole is a point that is not on the graph in the original function, but is after reducing the function by cancelling. She missed cross-over point on all three, and neither of us know what it is. Bubba73 ^{You talkin' to me?} 04:19, 11 September 2012 (UTC)

No, I meant a curve which crosses it's asymptote(s). Here's an example: [1]. That one crosses many times, but it's also possible to have a curve which crosses it's asymptote just once. StuRat (talk) 05:00, 11 September 2012 (UTC)

Two more examples of rational functions, whose graphs cross theis asyptotes just once, can be found in Rational function#Examples. --CiaPan (talk) 05:16, 11 September 2012 (UTC)

Resolved

I got email back from the teacher, and the cross-over point is where the function crosses the horizontal asymptote. Bubba73 ^{You talkin' to me?} 14:42, 11 September 2012 (UTC)

Interesting that it's just the horizontal asymptote. Why have a special term for this special case, and exclude vertical asymptotes and all others ? This is sounding less and less like a standard mathematical term and more like one invented by the teacher. It's up to you, but you could lodge a complaint that the teacher is using non-standard terms not included in the text, and potentially get those questions excluded from the test. (I was once placed in a remedial math class because I answered all the division problems on a test correctly, but in fractional form, when they wanted them in the much inferior remainder form.) StuRat (talk) 18:05, 11 September 2012 (UTC)

It was a function like this one File:RationalDegree2byXedi.svg. I don't see the significance of the point where it crosses the horizontal asymptote. The term isn't in the index of the textbook and I read the entire section on rational functions that they had just covered, and it isn't in there. Bubba73 ^{You talkin' to me?} 18:58, 11 September 2012 (UTC)

Yes, and I didn't find it by Googling either. StuRat (talk) 19:01, 11 September 2012 (UTC)

(responding to "exclude vertical asymptotes" above) A function can't cross a vertical asymptote. I reread the section and now I see where it comes from. The book gives guidelines for sketching rational functions. One step is to see if it crosses the horizontal asymptote and (if it does) plot that point. But it does not use the "cross-over point" terminology. Bubba73 ^{You talkin' to me?}

Well, a function in the form y = f(x) won't typically cross a vertical asymptote, but one in the form x = f(y) often will, as will those in polar coordinates. Also, there can be asymptotes which are neither horizontal nor vertical. StuRat (talk) 19:51, 11 September 2012 (UTC)

Any function f:R → R can be "graphed" which means to plot the value of the function along the y-direction for each argument x. Obviously you can make other 2D representations, but this particular one is standard and in this context we can say a rational function can never cross a vertical asymptote. Rckrone (talk) 00:19, 12 September 2012 (UTC)

September 11

A countanle $\pi -weight$

Hi, Does anyone know what is a (countable) A countanle $\pi -weight$ for a topological space? And also, is there a difference between ${\overline {A}}$ to ${\overline {int({\overline {A}})}}$. In other notation: is there a difference between $Cl(A)$ to $Cl(int(Cl(A)))$ ?. Thanks! Topologia clalit (talk) 11:14, 11 September 2012 (UTC)

Don't know about pi-weights. For the second question, let A in R be a closed interval union a point. Staecker (talk) 11:35, 11 September 2012 (UTC)

The π-weight of a space is a minimal cardinality of its π-base.—Emil J. 12:05, 11 September 2012 (UTC)

Thank you both! Topologia clalit (talk) 13:12, 12 September 2012 (UTC)

R[X]/(X^2-1)

If I have a Field (eg R) and I build the quotient as follows R[X]/(p(x)) with a polynomial p(x), then I get a field extension as usual if p(x) is a minimal polynomial. What happens if p(x) is reducible (ie. not a minimal polynomial)? For example in the cases: R[X]/(X^2-1)

or

R[X]/(X^2)

--helohe (talk) 19:45, 11 September 2012 (UTC)

The result will not be a field, but a ring with zero divisors. —Kusma (t·c) 20:13, 11 September 2012 (UTC)

in the case R[X]/(X^2) I guessed it will be something like a subset of R x Z/4Z . Does that make sense? --helohe (talk) 20:24, 11 September 2012 (UTC)

R[X]/(X²-1) is isomorphic to the group ring R[Z/2Z], but I don't know if you can phrase R[X]/(X²) in a similar way. Rckrone (talk) 00:08, 12 September 2012 (UTC)

R[X]/(X²-1) is also isomorphic to the ring of the Split-complex numbers, and R[X]/(X²) is isomorphic to the ring of dual numbers --84.229.150.202 (talk) 04:40, 12 September 2012 (UTC)

If the polynomial p is square-free, you can write it as a product $p=p_{1}p_{2}\cdots p_{n}$ of irreducible polynomials. Then the Chinese remainder theorem ensures that the ring $R[x]/(p(x))$ is isomorphic to the product of fields $(R[x]/(p_{1}(x)))\times \cdots \times (R[x]/(p_{n}(x)))$. In particular, if R is a field of characteristic other than 2, then $R[x]/(x^{2}-1)$ is just $R\times R$.—Emil J. 09:15, 12 September 2012 (UTC)

September 12

Incompleteness of Arithmetic

If Arithmetic is consistent, can it be "first-order complete" - i.e. such that resolves every first order arithmetical proposition, provided that the Axiom of Induction is formulated in its (original Peano's) second-order version? 77.125.123.63 (talk) 18:27, 12 September 2012 (UTC)

Oh, it's better than that — the second-order version is categorical; it determines the model up to isomorphism. So first order, second order, ninety-seventh order, whatever you want. --Trovatore (talk) 18:35, 12 September 2012 (UTC)

1. "up to isomorphism" is insufficient for me. Is there any (first order) arithmetical proposition that can't be proved in such an arithmetic, nor can its negation be proved in such an arithmetic? 2. If there isn't, so why do mathematicians insist on the first order version (of the Axiom of Induction) and claim that Arithmetic is incomplete, rather than use the second order version (of the Axiom of Induction) - which enables arithmetic to be "complete-up-to-isomorphism"? 77.125.123.63 (talk) 19:06, 12 September 2012 (UTC)

OK, we have to be careful here. I'm assuming that, when you talk about the "original second-order version", you also mean to use second-order logic; is that right? That's necessary to get the categoricity.

The reason mathematicians don't "use" second-order logic is that they don't know how. Second-order logic is extremely powerful (for example it either proves or refutes the continuum hypothesis, assuming, well, basically nothing). But it doesn't have a usable proof system. In fact, no proof system for second-order logic can be both complete (in the sense of Goedel's completeness theorem, sound, and computable.

(I'm leaving to the side the question of whether working mathematicians "use" formal logic at all — I would argue that working mathematics is not really based on formal deduction. But that's another discussion.) --Trovatore (talk) 19:14, 12 September 2012 (UTC)

It's possible that what you wanted to ask about was the second-order version of the Peano axioms, but interpreted in first-order logic (two-sorted first-order logic, with one sort for natural numbers and another sort for sets of natural numbers). In that case you get second-order arithmetic, which despite the name is a first-order theory. Then the answer to your original question is "no" — second-order arithmetic satisfies the hypotheses of Goedel's first incompleteness theorem, so there must be a true (first-order) proposition of arithmetic that second-order arithmetic does not prove. --Trovatore (talk) 19:26, 12 September 2012 (UTC)

Yes, this is what I meant. Thank you.

(

I guess you finally understood I'd referred to the Axiom of Induction when formulated as follows:

$\forall P[[P(0)\land \forall k(P(k)\Rightarrow P(k+1))]\Rightarrow \forall n[P(n)]]$.

)

Additionally, I also guess you also confirm (according to Goedel-Rosser Theorem) that there must be a first-order arithmetical proposition - whether true or false - that second-order arithmetic does not prove nor does it refute, right?

77.125.123.63 (talk) 19:43, 12 September 2012 (UTC)

Yes, that's true. As to the "finally understood" part, though — the difference is not in how the axiom is formulated but in how it's interpreted. In the two-sorted first-order-logic version, you have a sort for the P over which you're universally quantifying, and considering it to range over objects in your universe. In the second-order-logic version (which, I believe, is what Peano originally intended), when you say all P, you really mean all P — that is, all predicates (definable or not) that give a yes-or-no answer to an arbitrary element of the universe. --Trovatore (talk) 20:48, 12 September 2012 (UTC)

Oh, I might also point out that there's no mystery about the truth value of the Goedel–Rosser sentence for second-order arithmetic. It is true. It says "given any proof of me in second-order arithmetic, there is a shorter proof of my negation in second-order arithmetic". Because the sentence has no proof in second-order arithmetic, this claim is (vacuously) true. --Trovatore (talk) 21:08, 12 September 2012 (UTC)

Thankxs. I know that Rosser's sentence is true, however please note that I didn't care about its being true - but rather about whether it can be either proved or refuted - in the second order arithmetic. Btw, I'm from the restrahnt (rather than from the restrnt, if you still remember). 77.125.123.63 (talk) 22:20, 12 September 2012 (UTC)

Gibbs phenomenon

Show that $\lim _{N\rightarrow \infty }{\frac {1}{2}}\int _{0}^{\frac {\pi }{N+1}}\left({\frac {\sin((N+1/2)t)}{\sin(t/2)}}-1\right)dt=\int _{0}^{\pi }{\frac {\sin t}{t}}dt$. For instance, how do you get the limits of integration to match up? You would need a substitution $t'=f(t)$ where $f(0)=0$ and $\lim _{N\rightarrow \infty }f(\pi /(N+1))=\pi$ ; such a substitution $f$ if it exists would therefore not be continuous so it's hard to see what it would be. What's the approach to this problem? Widener (talk) 20:47, 12 September 2012 (UTC)

Your substitution can depend on N. Let u = (N+1)t, although probably you really want u = (N+1/2)t. Rckrone (talk) 23:43, 12 September 2012 (UTC)

Thanks for the tip. Here is my working so far with that substitution:

$\lim _{N\rightarrow \infty }{\frac {1}{2}}\int _{0}^{\frac {\pi }{N+1}}\left({\frac {\sin((N+1/2)t)}{\sin(t/2)}}-1\right)dt$

$=\lim _{N\rightarrow \infty }{\frac {1}{2}}\int _{0}^{\frac {\pi }{N+1}}{\frac {\sin((N+1/2)t)}{\sin(t/2)}}dt-{\frac {\pi }{2(N+1)}}$

$=\lim _{N\rightarrow \infty }{\frac {1}{2}}\int _{0}^{\frac {\pi }{N+1}}{\frac {\sin((N+1/2)t)}{\sin(t/2)}}dt$

$=\lim _{N\rightarrow \infty }{\frac {1}{2}}\int _{0}^{\frac {\pi (N+1/2)}{N+1}}{\frac {\sin(u)}{\sin(u/(2N+1))}}(N+1/2)^{-1}du$

$=\lim _{N\rightarrow \infty }{\frac {1}{2}}\int _{0}^{\frac {\pi (N+1/2)}{N+1}}{\frac {\sin(u)}{u/(2N+1)+O(u^{3}/(2N+1)^{3})}}(N+1/2)^{-1}du$

$=\lim _{N\rightarrow \infty }{\frac {1}{2}}\int _{0}^{\frac {\pi (N+1/2)}{N+1}}{\frac {\sin(u)}{u/2+O(u^{3}/(8(N+1/2)^{2}))}}du$

The desired result then follows if you can interchange the limit and the integral, i.e.

$\lim _{N\rightarrow \infty }{\frac {1}{2}}\int _{0}^{\frac {\pi (N+1/2)}{N+1}}{\frac {\sin(u)}{u/2+O(u^{3}/(8(N+1/2)^{2}))}}du={\frac {1}{2}}\int _{0}^{\pi }\lim _{N\rightarrow \infty }{\frac {\sin(u)}{u/2+O(u^{3}/(8(N+1/2)^{2}))}}du$

Is there some rationale for this, like uniform convergence or something? Widener (talk) 04:07, 13 September 2012 (UTC)

Are you asking about the limit and integral commuting? I suppose you're looking for the dominated convergence theorem. The lead says that it does not hold for the Riemann integral (so just consider the integral as the Lebesgue integral). BTW, no need to expand sin into a polynomial, just use $\lim _{x\to 0}{\sin x \over x}=1$. — Quondum 05:08, 13 September 2012 (UTC)

Does that theorem imply $\lim _{N\rightarrow \infty }\int _{0}^{g(N)}f(x,N)dx=\int _{0}^{\lim _{N\rightarrow \infty }g(N)}\lim _{N\rightarrow \infty }f(x,N)dx$ which is what I have got here? Plus is there a better theorem which does not require Lebegue integration? Widener (talk) 10:56, 13 September 2012 (UTC)

It does seem to imply it, minus the limit in the integration bound. I should plead ignorance at this point, lest I lead you astray. The variable integration bound is of a nature that I'd normally ignore as well-behaved. I suppose it depends on what level of rigour "Show that..." requires. Why would you want to avoid Lebesgue integration (sounds like it's a more powerful replacement for Riemann integration)? — Quondum 12:00, 13 September 2012 (UTC)

The statement is true using the Riemann integral, so using a theorem which requires Lebesgue integration doesn't actually prove the statement fully. You would have to independently prove that Riemann integrals are equal to Lebesgue integrals when they exist. I would like to know if you can prove it directly from Riemann integration Widener (talk) 12:36, 13 September 2012 (UTC)

I'm going through Stein and Shakarchi's book which does Fourier analysis in a bit of an unorthodox way; it doesn't use Lebesgue integration (although I have read about it from other sources and I know what it is). Widener (talk) 12:48, 13 September 2012 (UTC)

I was going to mention dominated convergence at the start, but held off because I didn't have time to think it through. Anyway, I'll also point out that the Gibbs phenomenon (and convergence of Fourier series in general) demands the Lebesgue integral in order to be treated rigorously. I mean, the Gibbs phenomenon is a classic example of why L^2 convergence can display counter-intuitive behavior compared to pointwise convergence. So (imo) this problem should be handled by appealing to the appropriate limit interchange theorems you have, not by tedious calculus and substitutions. At the end, you can demonstrate that this is one of the cases in which the Lebesgue integral and Riemann integral agree, and then you'll have shown the result for Riemann integrals as well. SemanticMantis (talk) 14:41, 13 September 2012 (UTC)

September 13

Irreducible nontrivial finite dimensional unitary representations of noncompact connected Lie groups operating on a complex vector space

Well, are there any such animals at all for any such group?

Please note the key words: noncompact, connected, irreducible, nontrivial, unitary, and complex.

I don't require the representation to be complex. There are subtleties in terminology here best illustrated by the induced Lie Algebra representation. Such a representation is complex if it is complex linear, (in which case we get a representation of the complexification of the algebra for free). Thus we can easily have a real representation on a complex space, see last paragraph.

In particular (if yes), are there any such representations for the (identity component of) the Lorentz group?

If yes, are there any n-dimensional ones where n > 1. (The determinant is always a 1-dimentional representation, in the case of the full Lorentz group on would get a nontrivial 1-dimensional representation, but not a very interesting one.)

I can think of a representation of R under addition taking t in R to a rotation matrix: t-> {(cos t, sin t),(-sin t, cos t)} in SO(2). This looks irreducible if operating on R x R. But if we allow it to operate on C x C, then it isn't irreducible because the matrix can then be diagonalized over C.

I have seen answers to closely related to this in the literature. They range from a plain "No" to "Such representations are very difficult to achieve". I suspect the anwer is "No, with the exception of a few nearly trivial examples". YohanN7 (talk) 10:59, 13 September 2012 (UTC)

plotting relations

I have a collection of items. I have a "weight" between each pair of items. For example, A-B is 10, A-C is 7, B-C is 19, and so on for around 1,000 items. I want to plot them in 2 dimensions, placing each item such that the mean difference between the plotted distance between any two points and the actual distance in 2D between those two points is minimized. Is there a "best" way to do this? — Preceding unsigned comment added by 128.23.113.249 (talk) 12:41, 13 September 2012 (UTC)

This amounts to plotting a "network visualization". In math, we just call them graphs, but that is confusing to outsiders. So other people use network, but that term is also overloaded. Anyway, I haven't used this program, but Graphvis [2] is a free, open source tool that should do what you want. You can use your weights between the nodes to determine spring constants as a start, though that won't generate a solution that truly minimizes the |given-plotted| error in distances. You could also try just setting the edge lengths equal to the weights. An exact solution might not be possible in 2D space, but it's still worth a try. SemanticMantis (talk) 14:53, 13 September 2012 (UTC)

See also Network_visualization#Layout_methods, and the list of software products at the bottom of that page. SemanticMantis (talk) 14:55, 13 September 2012 (UTC)

What you are trying to do is known in the trade as multidimensional scaling. It's an old concept and a number of methods have been developed, but none of them work perfectly, especially in cases where the "distances" are badly inconsistent. If you have access to a statistical analysis package, there is a good chance it will contain an MDS option. Looie496 (talk) 21:07, 13 September 2012 (UTC)

Looie, your answer seems much more relevant to the OP's problem than my shoddy attempt. Thanks! To partially redeem myself, I'll add that R_(programming_language) has MDS packages, and is free/open source. SemanticMantis (talk) 21:35, 13 September 2012 (UTC)

The link multidimensional scaling#Details is very interesting. At one point it says For some particularly chosen cost functions, minimizers can be stated analytically in terms of matrix eigendecompositions. (1) What objective function can this be done for? (2) And, for the quadratic function given in that link, I get interlocking nonlinear first order conditions that look like they would be impossible to solve in polynomial time--is this problem NP-hard? Duoduoduo (talk) 19:12, 15 September 2012 (UTC)

September 14

Suspected fault in article on wikipedia

Hey

I'm not certain if this is the correct place for this, but then again someone will see it and perhaps edit it.

http://en.wikipedia.org/wiki/Ancient_Roman_units_of_measurement#Liquid_measures

In the above link it is stated that a cubic foot is equal to 8 *unitX*, while a half-foot cubed is equal to 4 *unitX*. I do believe a half-foot cubed should be a fourth of a foot cubed, not a half as stated here.

Pleasant day to you, Sirs and Madames — Preceding unsigned comment added by 94.22.125.155 (talk) 08:19, 14 September 2012 (UTC)

My reading of the section that you link to is that it says a "amphora quadrantal" has a volume of one cubic foot, a "congius" is a half-foot cubed (i.e. its volume is equal to a cube that is half a foot in each dimension), and in the table it says that an amphora quadrantal is equal to 8 congii. This seems correct to me. A half-foot squared would be a quarter of a square foot, but a half-foot cubed is one eighth of a cubic foot because (1/2) x (1/2) x (1/2) = 1/8. Gandalf61 (talk) 08:49, 14 September 2012 (UTC)

Growth Dirichlet Kernel

First of all, if you can show $\lim _{n\rightarrow \infty }{\frac {f(n)}{g(n)}}=1$ where $f$ and $g$ are positive and $\forall n\in \mathbb {N} f(n)<g(n)$ is it true that there then exists a constant $c>0$ such that $\forall n\in \mathbb {N} f(n)>cg(n)$?
If so, show that $\lim _{n\rightarrow \infty }{\frac {\int _{-\pi }^{\pi }\min \left(n^{2}\left|{\frac {\sin((n+1/2)y)}{\sin(y/2)}}\right|,1\right)\left|{\frac {\sin((n+1/2)y)}{\sin(y/2)}}\right|dy}{\int _{-\pi }^{\pi }\left|{\frac {\sin((n+1/2)y)}{\sin(y/2)}}\right|dy}}=1$.
I think equivalently $\lim _{n\rightarrow \infty }\int _{-\pi }^{\pi }\left|{\frac {\sin((n+1/2)y)}{\sin(y/2)}}\right|dy-\int _{-\pi }^{\pi }\min \left(n^{2}\left|{\frac {\sin((n+1/2)y)}{\sin(y/2)}}\right|,1\right)\left|{\frac {\sin((n+1/2)y)}{\sin(y/2)}}\right|dy=0$. Widener (talk) 15:06, 14 September 2012 (UTC)
For proof of the first part, I think you can take the $n'\in \mathbb {N}$ which minimizes the ratio between $f$ and $g$ (minimum exists because $\lim _{n\rightarrow \infty }{\frac {f(n)}{g(n)}}=1$) and then choose $c<{\frac {f(n')}{g(n')}}$ Widener (talk) 15:06, 14 September 2012 (UTC)

Regular structures in the prime spiral о_0

о_0

I tried to scale image of the prime spiral (this image). The result:

To repeat,just download aforenamed image, open it and try to reduce the size of the window of your image viewer (slowly).

Could you please explain me, why there (semi-)regular structures? Is it a property of the prime spiral or some software bug or some defect in interpolation/scale algorithm or what?

--Ewigekrieg (talk) 20:26, 14 September 2012 (UTC)

It's an aliasing effect, resulting from a combination of structure in the prime spiral with a crude method of image scaling. The structure in the prime spiral, by the way, is the topic of the so-called abc conjecture, although our article doesn't discuss it from that perspective. (Note: I have taken the liberty of tweaking the layout of the question -- it was taking up way too much space.) Looie496 (talk) 20:39, 14 September 2012 (UTC)

Could you please explain the link between this structures and the conjecture? I'm not a professional mathematician. Also, why "islands" on the next to last section of image? --Ewigekrieg (talk) 20:53, 14 September 2012 (UTC)

Sorry, there is a story currently making the news that a Japanese mathematician, Shinichi Mochizuki, claims to have solved the abc conjecture, and some of the stories mention that it relates to the prime spiral, but that's the sum total of what I know about the prime spiral. See for example http://www.huffingtonpost.com/2012/09/12/abc-conjecture-shinichi-mochizuki-prime-numbers_n_1877692.html. Looie496 (talk) 21:54, 14 September 2012 (UTC)

That link doesn't actually claim a connection between the abc conjecture and the prime spiral and I haven't seen others do it. The prime spiral just seems like a nice image often used when otherwise non-photogenic prime numbers are discussed. PrimeHunter (talk) 22:21, 14 September 2012 (UTC)

At a first glance, the structures you are observing would require characterization of the aliasing properties of the scaling algorithm before proceeding any further. For example, whether the scaling algorithm is applied iteratively (pixels generated by a previous iteration being used, or the algorithm being applied to the raw data each time), what sampling/anti-aliasing algorithm is used (use of nearest pixel, or some blend of pixels in a region), variation of pixel interpolation at various interpolation positions (the weights assigned to each pixel in a region according to distance from nearest pixel: this can lead to Moiré patterns), and possible rounding errors/nonlinearities (e.g. choice of pixel may alternate between rows along a samping line between them, and when calculation of pixel intensity using weights). Only then can one answer whether there might be a subtle effect from the prime spiral. A few things are certain: there are aliasing effects occurring, and a "perfect" (Nyquist) interpolating filter will not produce these patterns. On the other hand, if the patterns are a genuine result of true aliasing (i.e not an artifact of the scaling implementation), these patterns would be of mathematical interest (as is the effect of the faint 45° patterns in the spiral). A quick test would be to see whether the same scaling produces similar behaviour from simple patterns, including random pixels (Tv snow). — Quondum 05:55, 15 September 2012 (UTC)

September 15

connected graphs

Is there a known pathological set of vertices for which the Urquhart graph, the Gabriel graph or the relative neighborhood graph is not connected? If not, is nonexistence proven? —Tamfang (talk) 07:06, 15 September 2012 (UTC)

Hmm. If the number of points is finite, then you can prove by induction that the relative neighborhood graph is connected. There are only finitely many pairwise distances between the points. Consider these pairwise distances in increasing order to prove that any two points are in the same connected component: Two points that are separated by the minimum pairwise distance (among all pairs of points) must be joined by an edge. Now consider two points A and B that are separated by more than the minimum pairwise distance. Either they are joined by an edge, or else there is a third point C whose distances from A and from B are strictly less than the distance between A and B. By induction, A and C are in the same connected component, and B and C are in the same connected component; so A and B are in the same connected component. —Bkell (talk) 08:31, 15 September 2012 (UTC)

Or how about this for the Urquhart graph, to separate two figures you'd need to have a cycle of triangles which each had two sides removed. One of those sides in the circle must be the smallest so why was it removed? Dmcq (talk) 08:38, 15 September 2012 (UTC)

If you have a graph where all the lengths round the cycle are the same so there is no consistent ordering then you can separate it in two but that's rather straining it. Eg you can do that with a large seven sided regular figure with a smaller one at its centre. Dmcq (talk) 14:53, 15 September 2012 (UTC)

Recurrence relation

I was thinking about what would happen if a player keeps scoring one more than his average. Let's say a cricketer has an average L_n after n matches. Let's say in the next match, he scores L_n+1. His new average will be

L_n+1 = (n * L_n + L_n + 1) / (n + 1)

Hence we have the recurrence,

L_n+1 = L_n + 1 / (n+1)

Now since L_n+1 - L_n tends to zero as n tends to infinity, I know that the series converges. But how do I find out the number it converges to? Can someone please solve this recurrence for me?

Also, more generically, if a cricketer scores (k * L_n + c) more than is average instead of just one more, what would his average converge to? Note that if we ave k = c = 1, it reduces to the case we talked about above.

Help will be appreciated ! Rkr1991 ^{(Wanna chat?)} 11:01, 15 September 2012 (UTC)

I was working some more on this, and I find that the general series seems to converge for any value of k less than 2. Is this correct? And if so, how can I find the limit? Rkr1991 ^{(Wanna chat?)} 11:01, 15 September 2012 (UTC)

You cannot conclude that a sequence converges if the change tends to zero. The initial recurrence relation you give is, aside from an initial additive constant L₀, the harmonic series, which diverges (albeit very slowly). — Quondum 11:48, 15 September 2012 (UTC)

September 16

Probability question

Hi, this is for personal interest and is not a homework question.

1. Say I choose n independent random numbers uniformly in the range 0 to 1, and then sort them in ascending order, calling the resulting sequence x₁, x₂, ... x_n.

2. Now suppose I start with x₀ = 0 and then recursively set x_i+1 = x_i + r where r is a random variable from a distribution P. When I get to x_n+1 I stop, and then I divide all the x₁ through x_n by x_n+1. Is there any distribution P that will ensure that the resulting distribution of x₁, x₂, ... x_n is identical to that obtained in method 1? Failing that, is there any sequence of different (i-dependent) distributions* that would achieve that result?

* But not x_i-dependent, since I already know how to do that.

86.179.116.254 (talk) 00:15, 16 September 2012 (UTC)

Space battle with only one shot

Here is a mathematical question. In a space war, there are two ships fighting each other to the death. There is a large ship and a small ship. Both ships have a single weapon which can only be fired once. The weapon is so powerful that one hit from it would result in total destruction. The small ship has a probability of hitting of S_hit(x) = 0.9^x where x is the unit distance. The large ship has a probability of hitting of L_hit(x) = 0.6^x

Assuming that both ship are approaching each other at a straight line. What is the best strategy for the small ship? What is the best strategy for the large ship? It is obvious that the large ship should wait until the small ship approaches as close as possible before firing it's weapon but if it waits too long, the small ship would fire first and destroy it. For the small ship, it wants to get as close as possible before firing it's weapon because if it fire while too far away, then it would miss and giving the large ship a chance to shoot it at a closer distance. 220.239.37.244 (talk) 03:48, 16 September 2012 (UTC)

Here you get into game theory, as you would probably want to fire just before your opponent, who presumably will do the same. StuRat (talk) 03:54, 16 September 2012 (UTC)

You haven't stated what the payoffs are, so I'm going to make some assumptions. Each ship derives utility 1 from surviving, and utility epsilon from destroying the other ship while surviving (small enough to be ignored in our calculations, but positive to ensure that if one ship misses, the other ship will fire at distance 0). If the large ship fires at distance x, then the small ship has two options: fire at distance $x+\delta$, or wait to fire until distance 0 (assuming it survives). One can evaluate the payoffs for each of these. Similarly if the small ship chooses to fire at distance x. Finding equilibrium then comes down to solving the equation $.6^{x}+.9^{x}=1$. Let a be the solution: the equilibrium occur when one ship fires at distance a and the other fires at distance 0 (if it survives). Of course, this all changes if you change the payoff structure.--121.73.35.181 (talk) 06:36, 16 September 2012 (UTC)

It's pretty clear that the objective is to destroy the enemy ship. Therefore the strategy is one where the probability of the destruction of the enemy ship is maximised. However mathematically that occurs when the distance is zero which makes this a paradox. No real Captain would wait until the distance is zero before firing the weapon because it would most certainly means the enemy would fire first.

It's obvious the strategy is as follows.

If the enemy has already fired (and you are still alive) then wait until the distance is zero then fire the weapon.

If the enemy has not fired yet, then wait until optimum distance Xoptimum then fire the weapon.

The distance Xoptimum is one where the probability that the enemy ship is the greatest. But how to calculate Xoptimum.

It seems to be that Xoptimum must a a function of the current distance to the enemy. Obviously at the start the distance is infinity and Xoptimum is a value. Later when the distance is 10, then Xoptimum can be a completely different value because you are still alive at distance 10. 220.239.37.244 (talk) 10:34, 16 September 2012 (UTC)