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July 11

Math problem concerning discographies

I'm terrible at math, so I was wondering if I could get some help on a math-related question having to do with discography articles. Basically, I've proposed at MOS:DISCOG that chart data should only be included for the top 10 charts where the artist has been most successful. However, I'm not sure how to objectively do that, in a mathematical sense. One could do an average of the chart positions, but that would give undue weight to a chart where they hit #1 once but never charted again over another chart where they hit #2 any number of times. I suppose the problem is that "not charting" usually means the release would have been 100+, but it's not specified exactly which chart number above 100 it's charted at. For a good working example, see Garbage discography, which I'm currently trying to whittle down to 10 chart columns. Any ideas? Did I even make any sense? Drewcifer (talk) 01:08, 11 July 2008 (UTC)[reply]

I'm not a pro in statistics, but here's my take on the problem. Essentially, when you take an average, be it mean, median, mode, or other, you lose information. So, no matter how you choose to filter your charts, (even if you don't choose to average), you're stuck with the fact that you'll be giving weight to certain types of charts (e.g., one week at #1) over others (one year at #2). It's up to you to decide what is the most important criteria for your selection.

Perhaps there is a type of statistic that is frequently used to best describe these sorts of problems; if so, I don't know what that would be. It seems likely, though, that the music industry itself may have some way of saying whithin which genres an artist/band has been most successful. Maybe that's where you should start looking. — gogobera (talk) 03:34, 11 July 2008 (UTC)[reply]

My knowledge of charts and discographies is fairly limited but if you assume that the ranking of any given new album/single follows a distribution you could use the sufficient statistic for that distribution and retain all the information. For example, if you were to assume that the chart locations were normal for some group, you can take their mean / variance and you could chart based on those and you would be able to preserve the distribution (though inevitably you lose some information). One way to do this would be to plot all noteworthy (mean > n), all outliers (n > variance *n) or such. 134.160.173.1 (talk) 05:04 11 July 2008 (UTC)

Linearity and derivations of Kurtosis

The wiki page describes one condition in which kurtosis of a sum is the sum of a kurtosis. I am trying to find the kurtosis of a uniform random variable (with mean m and width 2s) convolved with a Gaussian with mean 0 and variance sigma. In this case the kurtosis should depend on the parameters (ie if s -> 0 then excess kurtosis should be 0) but linearity would suggest otherwise. Is it possible to expand Kurt(Unif + Gauss) or must i work through the integral. 134.160.173.1 (talk) 05:09, 11 July 2008 (UTC)[reply]

Mathematical rule

What is this rule in mathematics: "quae in eodem tertio conveniunt, et inter se conveniunt"? Thanks. --Omidinist (talk) 13:18, 11 July 2008 (UTC)[reply]

If online latin translators aren't failing me, it's the transitivity of equality (up to symmetry of equality, anyway). Algebraist 13:25, 11 July 2008 (UTC)[reply]

I would guess it is the latin translation of the first Common Notion from Euclid's Elements [1]. Aenar (talk) 15:31, 11 July 2008 (UTC)[reply]

Continuing Mathematical Education

Hi everybody. I love math, but am stuck in a rut. I have completed courses in Multivariable Calculus, Linear Algebra, and Actuarial Probability, but have no idea what to do next. I am mostly interested in pure math, but applied is fine as well. I thought aboutAbstract Algebra, Complex Analysis, or Topology, but I would like to able to self-teach it. I've always been able to learn math by myself, but now am stuck because I don't know where to go and what books to use. Thanks a lot for you help! Xen (talk) 18:44, 11 July 2008 (UTC)[reply]

excellent! someone who wants to do homework rather than me do theirs! How about reading Sums#Identities and checking you can prove them all? Maybe you've already done that?87.102.86.73 (talk) 19:28, 11 July 2008 (UTC) Alternatively you could look at this desks archives and see if you can find something you want to know but don't already Wikipedia:Reference_desk/Archives 87.102.86.73 (talk) 19:30, 11 July 2008 (UTC) How about space filling solids and tessalation?87.102.86.73 (talk) 19:33, 11 July 2008 (UTC)[reply]

Locus (mathematics) is something you can easliy self teach and is effectively endless; which could lead you to look at 'famous curves' Category:Curves 87.102.86.73 (talk) 19:36, 11 July 2008 (UTC)[reply]

If you're motivated enough and have enough time, you can probably teach yourself any branch of mathematics. Just find a good textbook called "Introduction to X" or "A first course in X" or similar and go for it. If you pick a topic, people here may well be able to recommend a good textbook. --Tango (talk) 23:07, 11 July 2008 (UTC)[reply]

Needham's Visual Complex Analysis is a reasonable place to learn complex analysis. Another decent topic is discrete mathematics; look for any sophomore level CS or math course's textbook. Another decent topic is non-euclidean geometry; both Greenberg's E and N-E geometry and Sibley's Geometric Viewpoint are reasonable. A nice book to get an idea about topology is Weeks's Shape of Space. JackSchmidt (talk) 07:02, 12 July 2008 (UTC)[reply]

Thanks for your help everybody, I'll look into these topics. Xen (talk) 19:52, 12 July 2008 (UTC)[reply]

I learned point set topology on my own from Kelley's book "General topology". I found it very much worthwhile. It will give you background in topology that will be useful for you if you want to go into Analysis, Probability (of the continuous kind) and other areas of topology. I don't know if it is considered outdated by now. You might want to give it a try. Oded (talk) 10:14, 13 July 2008 (UTC)[reply]

I would strongly recommend some abstract algebra if you intend to learn further pure mathematics. Group theory in particular is enormously useful in a very wide range of areas, and with a good grounding in algebra you will be well-placed to learn other mathematical topics. Unfortunately I can't recommend any good introductory texts, but there are many, many available. Perhaps some others can recommend some. Good luck! Maelin (Talk | Contribs) 02:49, 14 July 2008 (UTC)[reply]

Power Series

Is there any nonconstant power series that's zero for a set dense on the unit circle? Black Carrot (talk) 18:57, 11 July 2008 (UTC)[reply]

The zeros of an analytic function are isolated, can you have a discrete set that is dense on the unit circle? I suspect not, but a proof is eluding me. --Tango (talk) 02:36, 12 July 2008 (UTC)[reply]

If the power series does not converge beyond the unit circle, I think it might be able to do something a little weird, but I'm not sure if it can be that weird. If the radius of convergence is greater than 1, then no, by what Tango said (the unit circle is compact). JackSchmidt (talk) 06:55, 12 July 2008 (UTC)[reply]

Compactness, that's the one! I was trying to work out what property of the unit circle is was that made it impossible and somehow didn't think of compactness, thank you! You're right, if you radius of convergence is exactly 1, interesting things could happen, but I have no idea how interesting - we were always taught that it's very difficult to work out what happens at the radius of convergence so not to bother! --Tango (talk) 16:41, 12 July 2008 (UTC)[reply]

Sure, you could have a power series with radius of convergence equal to 1 such that for a dense set of points on the unit circle the analytic function extends continuously to those points and is zero there. Here is how this can be constructed. We start by constructing a real valued function h that is harmonic and non-negative in the open unit disk and tends to infinity on a dense set of points on the unit circle. Suppose that h has been constructed. Then there is an analytic function g in the unit disk whose real part is h. Then we can take $f=\exp(-g)$ . So we only need to construct h. First, given any point $\lambda$ on the unit circle there is a positive harmonic function $h_{\lambda }$ in the unit disk that tends to $\infty$ near $\lambda$ . For example, we can take $h_{\lambda }(z)=a-\log |z-\lambda |$ with an appropriate constant $a$ . Now, take a dense sequence $\lambda _{j}$ on the unit circle. If $\alpha _{j}$ is a sequence of positive numbers tending to zero sufficiently fast, then $h=\sum _{j}\alpha _{j}h_{\lambda _{j}}$ would converge to a harmonic function in the unit disk, and has the required properties. Actually, Harnack's principle implies that if this converges at a single point in the disk, say at zero, then it converges throughout the unit disk and is harmonic. So $\alpha _{j}=2^{-j}/h_{\lambda _{j}}(0)$ would work. All this is in fact very explicit. Oded (talk) 02:14, 13 July 2008 (UTC)[reply]

Cool. Would that also be zero on the rest of the unit circle, or is it confined to the set it was constructed for? Black Carrot (talk) 03:55, 15 July 2008 (UTC)[reply]

If it extends continuously to the entire circle and is zero on the entire circle, then it must be zero everywhere, by the maximum principle. Oded (talk) 13:26, 15 July 2008 (UTC)[reply]

I_n - AB + BA

I’m reviewing linear algebra in preparation for quals, and this problem has me stumped: Fix $A,B\in F^{n\times n}$ , $F$ a field. Prove that $I_{n}-AB+BA$ is not nilpotent. My least fruitless approach has been to try to show that it’s trace is nonzero, but figuring out what the trace looks like seemed quite formidable. I think my problem is that I’m not seeing any properties that the matrix $AB-BA$ has that general matrices do not – other than the fact that it is zero iff $A$ & $B$ commute. —Preceding unsigned comment added by GromXXVII (talk • contribs) 23:07, 11 July 2008 (UTC)[reply]

Working out the trace should be pretty easy - take a look at the properties listed on Trace (linear algebra), one of them will pretty much give you the answer. --Tango (talk) 23:11, 11 July 2008 (UTC)[reply]

A little bit of an odd question, as the trace can be zero, and indeed the matrix can be nilpotent. The trace is equal to n, and so the trace is zero if and only if the characteristic of F divides n. For example, if F is the field with two elements, and A=[0,1;0,0], B=A^t, then I-AB+BA=0. JackSchmidt (talk) 06:51, 12 July 2008 (UTC)[reply]

Hmm. Curious indeed, the statement fails for all fields of finite characteristic. GromXXVII (talk) 11:08, 12 July 2008 (UTC)[reply]

Ahh, nice. I kept focusing too much on the matrix

AB

, when indeed the trace of

AB-BA

is always 0. GromXXVII (talk) 11:08, 12 July 2008 (UTC)[reply]

Error in Manifold?

Manifold, in the section on the formal definition, includes this:

Second countable and Hausdorff are point-set conditions; second countable excludes spaces of higher cardinality such as the long line, while Hausdorff excludes spaces such as "the line with two origins" (these generalized manifolds are discussed in non-Hausdorff manifolds).

While the second countable condition does exclude the long line, it has nothing to do with cardinality, does it? I thought the long line had the same cardinality as the real line. --Tango (talk) 23:56, 11 July 2008 (UTC)[reply]

The long line has as many points as the line, yes. I think it has more open sets, though (it has

2^{\aleph _{1}}

, while the line has

2^{\aleph _{0}}

; I'm not sure if these are necessarily different in the absence of the continuum hypothesis), so this might be what is meant. Algebraist 00:23, 12 July 2008 (UTC)[reply]

That would make sense, although it's not a particularly good wording. I would expect the cardinality of a topological space to refer to the underlying set, not the topology. Is that not standard? That said, why does the long line have more open sets? They both have the order topology on a set of cardinality

{\mathfrak {c}}

, from what I can tell that should have cardinality

2^{\mathfrak {c}}

in both cases (you have a basis of open intervals, there is one of those for every pair of elements, with infinity included, of which there are

{\mathfrak {c}}

, and the topology is then formed from unions of those, and you can combine them in

2^{\mathfrak {c}}

ways). Am I missing something? --Tango (talk) 02:24, 12 July 2008 (UTC)[reply]

There are only

2^{\aleph _{0}}

open subsets of the real line; this is by separability. Every open set can be written as the union of open intervals with rational endpoints, and there are only countably many such intervals. On the long line, for every α, the interval

\left(\alpha +{\frac {1}{3}},\alpha +{\frac {2}{3}}\right)

is open, and there are

\aleph _{1}

such intervals, and for any combination you take of them you get a distinct open set, so that gives you

2^{\aleph _{1}}

open sets as a lower bound (upper bound left as an exercise). --Trovatore (talk) 02:42, 12 July 2008 (UTC)[reply]

Ah, yes, I missed the rather obvious possibility that some of the unions could end up with the same open set! Isn't my "proof" at least valid for an upper bound, though, so it solves your exercise? --Tango (talk) 02:54, 12 July 2008 (UTC)[reply]

Well, that does indeed give you an upper bound of

2^{\mathfrak {c}}

, but what we wanted was an upper bound of

2^{\aleph _{1}}

. --Trovatore (talk) 03:36, 12 July 2008 (UTC)[reply]

To answer Algebraist's implied question: It follows from Martin's axiom plus the negation of CH that

2^{\aleph _{0}}=2^{\aleph _{1}}

. There is a recent tendency towards the view that these values are the same, and that they both equal

\aleph _{2}

. --Trovatore (talk) 02:44, 12 July 2008 (UTC)[reply]

Thanks; I was waiting for you to show up, but I got tired. Should this be added to Martin's axiom#Consequences of MA(k)? On the original question: the comment in 'manifold' is true only under a dubious interpretation and a dubious set theory. But certainly the problem with the long line is that it is in some sense too large to be allowed to be a manifold. How should this be phrased in the article? Algebraist 08:37, 12 July 2008 (UTC)[reply]

That looks like a good phrasing to me: "second countable excludes spaces that are in some sense 'too large' such as the long line", I think trying to come up with a rigorous way of saying how it's too large is doomed to failure since the reason it's too large is that it isn't second countable! There is another issue, though. Long line (topology) describes it as a manifold, but mentions some authors require manifolds to be second countable, so don't include it, but manifold takes the opposite approach. There are differing definitions of manifolds, but we probably need to pick one and stick to it and mention the others as alternatives where appropriate. --Tango (talk) 16:34, 12 July 2008 (UTC)[reply]

Right, I've gone with that until someone thinks of something better. On the manifold-definition issue, I'm not sure what should be done, but will comment that even manifold hedges its bets, with footnote 2 and the 'broad definition' section. And then we have to worry about boundaries... Algebraist 00:30, 13 July 2008 (UTC)[reply]

I know pretty much all our articles on manifolds mention all the options, what I meant by taking the opposite approach was that some articles have one definition are standard and the other as an alternative and other articles have them the other way around. We should probably stick to one definition are standard on all our articles, and mention the alternatives where appropriate. As for boundaries, how common is it to allow manifolds to have boundaries? I've always known "manifold" to mean without boundary and "manifold with boundary" to mean either with or without (which is a contender for the worst name in mathematics, but oh well!). Is that not standard? --Tango (talk) 19:22, 13 July 2008 (UTC)[reply]

Pretty standard, in my (limited) experience, but people still speak of boundaryless manifolds occasionally (~5 times on WP). I see what you mean now about definitions; the place for this discussion is probably WT:WPM, ~~but I don't have the energy to take it there~~. Algebraist 21:40, 13 July 2008 (UTC)[reply]

Posted at WT:WPM#Point-set conditions for manifolds? Opinions welcome there. Algebraist 22:07, 13 July 2008 (UTC)[reply]