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April 18

Conditional probability in LaTeX

Does anyone know how to do the "bar" or "given" in LaTeX, besides "|"?

I mean:

$P(B|A)={\dfrac {P(A|B)P(B)}{P(A)}}$

looks fine, but when we start putting bigger text (say, unions):

$P\left(\bigcap _{i=1}^{k}E_{i}\right)=P(E_{1})P(E_{2}|E_{1})P(E_{3}|E_{1}\cap E_{2})\dots P\left(E_{k}|\bigcap _{j=1}^{k-1}E_{j}\right)$

looks poor, so I have to kludge things:

$P\left(\bigcap _{i=1}^{k}E_{i}\right)=P(E_{1})P(E_{2}|E_{1})P(E_{3}|E_{1}\cap E_{2})\dots P\left(E_{k}\left|\bigcap _{j=1}^{k-1}E_{j}\right.\right)$

However, the spacing seems a bit off. I was wondering if there was anything like \colon in LaTeX, which has better spacing:

$\Omega _{1}:\mathbb {R} \to \mathbb {R} \qquad {\mbox{vs.}}\qquad \Omega _{2}\colon \mathbb {R} \to \mathbb {R}$

Thanks in advance. x42bn6 Talk Mess 00:43, 18 April 2009 (UTC)

If you use the amsmath package, you can get different sizes with appropriate spacing use \bigm|, \Bigm|, \biggm|, \Biggm| . Unfortunately they don't work in Wikipedia markup. Once you get used to \bigm, also use \bigl and \bigr (and bigg, Big, Bigg) and leave \left and \right for cases like huge matrices where their bad choices (look how the extra -1 in your example caused the size to jump up) and gratuitous spacing (look at the huge space after your Ps) don't matter. McKay (talk) 04:02, 18 April 2009 (UTC)

Other useful commands are \vert and \vrule; experiment to see how they look for you. I wrote a custom command as follows:

\newcommand {\set} [2] {\left\{#1\ \vrule\ #2\right\}}

to get the appearance I want, and just invoke the command like \set{x \in \Q}{x^2 < 2} to get the set of rationals with square less than 2. Another useful trick is the \vphantom command:

\newcommand {\set} [2] {\left\{#1 \left| \vphantom {#1} #2 \right. \right\}}

The \vphantom command places an invisible box of zero width and height equal to the height of its argument. Eric. 131.215.159.99 (talk) 09:02, 18 April 2009 (UTC)

How about this?

$$P\left(E_k\ \left|\quad \sum_{j=1}^{k-1}E_j\right.\right)$$

You may replace \quad by \,\, or other spacing control sequences. twma 05:35, 19 April 2009 (UTC)

Uniform population sampling noise

Hi Wikipedians:

In the statistics textbook I am reading right now it says that a sample of size 5000 taken from a uniform population with a = 0 and b = 10000 would be subject to more sampling noise than a sample of size 5 taken from a uniform population with a = 0 and b = 10.

I am wondering why that would be the case? I googled "sampling noise" but could not find any definition of "sampling noise" in a statistical context.

Thanks,

76.65.13.187 (talk) 02:25, 18 April 2009 (UTC)

I assume that sampling noise refers to the variability of an estimate of a population parameter. For a uniform continuous distribution between a and b, the SD of the sample mean (for example) is (b-a)/root(12n), giving 40.8 and 1.29 respectively for your two cases, i.e. the first has more uncertainty and hence "noise". I'm not surprised that you could find nothing in a simple statistical context, as noise is usually taken to be something unwanted which is added to a signal, causing trouble in its reading or processing. The innate variability of an estimate, which I think that you are asking about, isn't the same thing at all.Special:Contributions/86.148.186.106|86.148.186.106]] (talk) 09:39, 18 April 2009 (UTC)

Thanks! That is exactly it! And indeed, everything I could google about "sampling noise" is in terms of digital signal processing (DSP) and such. 76.65.13.187 (talk) 14:05, 18 April 2009 (UTC)

See sampling error, shot noise etc. 66.127.52.118 (talk) 09:02, 19 April 2009 (UTC)

(n-1)-size linearly independent subsets of n vectors

If every (n-1)-size subset of the vectors $a_{1},...,a_{n}$ is linearly independent, is it also the case that the full size n set of vectors is necessarily linearly independent? I'm leaning towards yes, but I'm not sure how to prove it - I don't imagine it has a very complicated proof, it just seems to be going over my head right now. Thanks, Otherlobby17 (talk) 05:55, 18 April 2009 (UTC)

n=2 provides a counterexample. Bo Jacoby (talk) 06:25, 18 April 2009 (UTC).

And if you want it for a given n: take n-1 linearly independent vectors, $a_{1},...,a_{n-1}$, and define $a_{n}:=a_{1}+...+a_{n-1}$. --84.221.69.233 (talk) 07:49, 18 April 2009 (UTC)

I'm just wondering here ... a more amusing question would be, given n vectors, any m of which are linearly independent, what is the minimum possible dimension of the space spanned by the n vectors? Ray^Talk 07:55, 18 April 2009 (UTC)

m —Preceding unsigned comment added by 84.221.69.233 (talk) 09:25, 18 April 2009 (UTC)

For infinite fields, the answer is m. For finite fields, the formula will be more complicated: for example, consider $n=q^{m}+1$ vectors over the field of q elements, any m of which are linearly independent. Certainly those n vectors must span a space of dimension strictly greater than m. Eric. 131.215.159.99 (talk) 10:38, 18 April 2009 (UTC)

For infinite fields of nonzero characteristic, is the answer always m?

Yes. For any infinite field F, one can find n vectors in F^m such that any m of them are independent. Algebraist 19:48, 22 April 2009 (UTC)

Is a given set of group elements a set of coset representatives?

If G is a group and if g1, ..., gn are elements of G, is there a criterion or an algorithm (or a function in some dedicated program, like GAP) to determine whether there is a subgroup of G such that those elements form a set of representatives for the cosets of the subgroup? (We may assume that G is a finite permutation group, and probably even the full symmetric group.)

(There are of course several algorithms to find the cosets of a given subgroups, like Todd-Coxeter algorithm; this is a kind of inverse question.)

Thanks! Goochelaar (talk) 12:41, 18 April 2009 (UTC) (Apparently I had not signed. Sorry!)

If g1 and g2 are both in the same right coset of a subgroup H, then g1 g2^-1 is in H. And vice versa. So let H be the subgroup generated by all products gi gj^-1 — if that doesn't work nothing does. To make it more efficient, note that (gi gj^-1)(gj gk^-1) = gi gk^-1, so it suffices to use any set of pairs gi gj^-1 for which the digraph with edges i->j is strongly connected (such as a directed cycle). McKay (talk) 12:24, 18 April 2009 (UTC)

Reading your question more carefully, you want to know if g1,...,gn is an entire coset. That is true iff {g1 gi^-1 | i=1..n} is a subgroup. My more general procedure above finds the smallest subgroup H such that g1,...,gn are contained in the same coset of H. McKay (talk) 12:33, 18 April 2009 (UTC)

Thanks a lot, but I am afraid I was not clear. I am interested in understanding whether the elements g_i form a set of coset representatives, i.e. a transversal, for some subgroup, not a single coset. Goochelaar (talk)

Ooops, your words were clear enough, it was my eyes that were the problem. :) McKay (talk) 13:57, 18 April 2009 (UTC)

I don't see a much better way than to enumerate the subgroups of index n and check for each. The Todd-Coxeter algorithm takes a finite index subgroup given by generators of a finitely presented group and gives the multiplication amongst coset representations. The inverse problem to my way of thinking is starting with the multiplication amongst coset representatives, try to get a finitely presented group with finite index subgroup given by generators. This is well studied and GAP has fairly efficient functions for it. Of course, it will only return a quotient of the original group by the core of the finite index subgroup. JackSchmidt (talk) 17:48, 18 April 2009 (UTC)

I haven't found anything for the general problem (which seems like a reasonable combinatorics type question) of: (1) determining if a subset of a permutation group is a transversal of a subgroup, and if so (2) determining the subgroup from its transversal. I'd be interested to hear a solution. Again, with more information you might have better luck. For instance if the subgroup is known to be a point stabilizer, then you just look at the images of that point under the proposed transversal. JackSchmidt (talk) 20:05, 18 April 2009 (UTC)

Another problem that seems similar is to determine the subgroups that don't include any of some given set of elements. In the original problem we know that the products gi gj^-1 (i ≠ j) all lie outside the subgroup. I think that any subgroup of order n that satisfies that property is a solution, and I don't see why it should be unique. McKay (talk) 07:07, 19 April 2009 (UTC)

That's what I was thinking too. If we consider e.g. the set

$H^{*}:=\{h\in G\,:\,h^{k}\neq g_{i}g_{j}^{-1}\,\forall k,i,j\,;\,i\neq j\}$,

the problem can be restated as: find $\scriptstyle H\subset H^{*}$ which is a subgroup of G of index n. This gives a first necessary condition like $\scriptstyle \mathrm {card} (H^{*})\geq \mathrm {card} (G)/n$. Then, my experience in algebra is too poor to understand if one can go any further in this direction. --pma (talk) 12:25, 19 April 2009 (UTC)

A friend of mine and I were discussing the OP's question, and eventually reached the same conclusion as you (that any subgroup of the right index excluding those elements works). It is also easy to see that if you can solve them problem for a group G, then you can solve it for any subgroup of G; thus solving the problem for the full symmetric group yields a solution for any finite group (and is thus the hardest case). After calculating the number of subgroups of the n-th symmetric group to be $e^{\Theta (n^{2})}$ he decided that the problem was probably intractable. I am not fully convinced that it is intractable but don't see any reasonable way to approach the problem.

One idea is to start with H being the trivial group, and starting adding elements of $H^{*}$ to the generators of H until H cannot be made any larger; then we have a maximal subgroup of G subject to the condition of being a subset of $H^{*}$. The trouble is it might be too small. If somehow in this manner we could enumerate all maximal subgroups of G that are subsets of $H^{*}$, then the desired H must be one of them, but there's seems no obvious non-exponential way to do that.

For G cyclic the problem is trivial. For $G=(\mathbb {Z} /p\mathbb {Z} )^{2}$ we were unable to make much progress. Eric. 131.215.159.99 (talk) 19:33, 19 April 2009 (UTC)

Yes, the combinatorics is so rich that what you say seems the most reasonable procedure. Or checking all subgroups of index n as JS said. Such a set of representatives looks like a very cryptic object to me (a naive observer, as I told). Here's just an analogy, with an infinite group: if G is $\scriptstyle \mathbb {R}$ and the unknown subgroup H were $\scriptstyle \mathbb {Q}$, any set of representatives for the cosets of H would be a Vitali set: thus not even measurable; so a complicated object that we can only state its existence as a direct consequence of the axiom of choice. This somehow suggests that in general the complexity of such sets of representatives, for finite, even abelian groups, should become untractable for large n.--pma (talk) 08:29, 20 April 2009 (UTC)

Here is the OP again. First of all, thanks to everybody for the food for thought. I am glad I had not missed something terribly obvious. Then, here is how I posed myself such a question. I am investigating (structureless, a priori) sets of permutations with certain properties, and the small degree cases gave me a hunch that those sets might be transversals of suitable subgroups (among other things, they may wlog contain the identity but it is not necessary, have a size dividing n! and more), so I wanted to check this in the next few known examples. So I wondered whether this check was easy to do, before launching myself in a brute-searchish computer search or, as a last resort, some serious thinking. :) Goochelaar (talk) 17:10, 20 April 2009 (UTC)

It would help to have more specific numbers. In the next few examples about how many permutations are there and how many moved points? How large is the subgroup generated by the permutations? JackSchmidt (talk) 19:54, 20 April 2009 (UTC)

Critical graph

Hello. I wish to prove that a 3-critical graph is an odd cycle. All I can think of is that since biparitite graphs have chromatic number 2 so the graph cannot have even cycles (that it has cycles is assured for otherwise it would be a tree, which also has chromatic number 2) and hence definitely contains an odd cycle. But I can't think any further. Any help please.--Shahab (talk) 20:02, 18 April 2009 (UTC)

I think you are done? A 3-critical graph is a graph that is not bipartite, but removing any edge produces a bipartite graph? So every odd cycle is 3-critical, and if a 3-critical graph has an odd cycle, then it *is* an odd-cycle (since removing some other edge leaves an odd cycle, so a non-bipartite graph). If the graph is an even cycle, then it is bipartite, so it is not 3-critical. If the graph has an even cycle and removing an edge from the cycle results in a bipartite graph, then restoring the edge remains a bipartite graph. Hence a 3-critical graph has no even cycles. As you say, a graph without cycles is a tree, so bipartite, so not 3-critical. Hence every 3-critical graph is an odd cycle. JackSchmidt (talk) 20:13, 18 April 2009 (UTC)

Thanks. I can't believe what a big idiot I was.--Shahab (talk) 20:18, 18 April 2009 (UTC)

April 19

Splitting lemma

Hi! Could anyone show me a reference concerning the Splitting lemma? I think about a handbook where the lemma and its proof are explicitly included (a book on homological algebra, ...). Thanks! Mozó (talk) 18:34, 19 April 2009 (UTC)

If I am not wrong, you can find it in the first pages of Homology by Mac Lane. (As to the proof, if you do it thinking to vector spaces, you have it in general). --pma (talk) 19:30, 19 April 2009 (UTC)

Mac Lane was a good idea! In Mac Lane, Birkhoff Algebra it is an exercise (p. 328). Resp., Mozó (talk) 21:27, 19 April 2009 (UTC)

On a different note, is the accented "o" in your username part of the Hungarian alphabet? Or is it supposed to be Spanish? --PST 03:26, 20 April 2009 (UTC)

April 20

Topological properties

On the page Normal Property it indicates that the article is mostly concerned with properties that are not topological, and then proceeds to list properties that I always regarded as topological. Is this a typo? I think it should be changed, but I want someone to agree with me first.Aliotra (talk) 02:01, 20 April 2009 (UTC)

I do not know of the article which is referred to, but normality has several distinct meanings in mathematics. Possible distinct interpretations of this term in certain contexts of mathematics are: normal number, normal distribution, normal subgroup, normal space; and it is with no doubt that I believe there are others. --PST 03:19, 20 April 2009 (UTC)

There is no Wikipedia page Normal Property. Are you talking about Uniform property? It lists 2 topological and 4 non-topological properties. Algebraist 10:20, 20 April 2009 (UTC)

Sorry, gentlemen; both of you are correct. I meant uniform, but wrote normal (why I don't know). As well, I read "connected", compact", etc in the list, without investigating further. Even now I haven't done so, but I take it the other four are analytic (or something) properties. I'll try to be more careful in future.Aliotra (talk) 16:37, 22 April 2009 (UTC)

Simplifying logarithms with Maxima

How can I get Maxima to simplify logarithms for me? For example, I want log(4) - 2*log(2) to evaluate to zero. I tried setting logexpand to all or super, but no dice. Is this feature simply missing? —Keenan Pepper 03:46, 20 April 2009 (UTC)

Maxima does not regard this "direction" of transforming as expanding, but contracting. logcontract(log(4) - 2*log(2)) gives the result you want. 79.217.113.11 (talk) 13:25, 26 April 2009 (UTC)

1. Nf3 f5 2. d4 e6 3. Bg5 Be7 4. Bxe7 Qxe7

In chess, what are the name and ECO code (if they exist) for the opening 1. Nf3 f5 2. d4 e6 3. Bg5 Be7 4. Bxe7 Qxe7? Also, are there any free tools online that can answer questions like this automatically (enter the algebraic notation, get the name and ECO code spit out)? NeonMerlin 07:11, 20 April 2009 (UTC)

The opening 1.Nf3 is the Réti Opening; Wikipedia doesn't seem to have anything about the reply 1…f5. Wikibooks has an "Opening Theory in Chess" project that goes as far as 1.Nf3 f5 2.d4 e6. —Bkell (talk) 13:51, 20 April 2009 (UTC)

Chess Opening Explorer seems pretty useful, but you have to get some membership to get unlimited access. —Keenan Pepper 16:34, 20 April 2009 (UTC)

Also jose which is free. (Igny (talk) 16:46, 20 April 2009 (UTC))

Also free is Shredderchess, which has 16 (¿!) games with those initial moves. Pallida  Mors 18:17, 20 April 2009 (UTC)

One can find a list of canonical move orders for all ECO codes in various places on the web, but it is the move order that would be tricky. I would think this is a transposition into a Dutch defense, but cannot vouch for that at this time. Baccyak4H (Yak!) 20:23, 20 April 2009 (UTC)

Numbered system of equations in LaTeX

I am trying to make a numbered system of equations in LaTeX with a brace on the left side. So far I have the following:

\left\{\begin{aligned}
	a^{b+1}b^2&=a,\\
	b^2-1&=b.\\
\end{aligned}\right.

This gives me what I want, but it doesn't have equation numbers. According to More Math into LaTeX, "There is no numbering or \tag-ing allowed in subsidiary math environments [like aligned] because LaTeX does not number or tag what it considers to be a single symbol." Surely what I want can be done; how do I do it? —Bkell (talk) 21:05, 20 April 2009 (UTC)

You see, the book is basically right, TeX equation numbering system really works only for full-width displays. Anything else requires an unsightly kludge, you can try the following:

 \left\{
    \vcenter{\advance\hsize-1em \abovedisplayskip0pt \belowdisplayskip0pt
       \begin{align}
          a^{b+1}b^2&=a,\\
          b^2-1&=b.
       \end{align}}
 \right.

Actually it's not clear what layout exactly you are after. If you want the brace to come closer to the equations for instance, you can play with the \hsize setting, such as

 \left\{
    \vcenter{\hsize8em \abovedisplayskip0pt \belowdisplayskip0pt
       \begin{align}
          a^{b+1}b^2&=a,\\
          b^2-1&=b.
       \end{align}}
 \right.

— Emil J. 12:47, 21 April 2009 (UTC)

The cases package will also do the trick: [1] (read the .sty file in a text editor for documentation). x42bn6 Talk Mess 01:06, 22 April 2009 (UTC)

For numbering each equation in a system you can use numcases environment from cases package.

\begin{numcases}{}
  a^{b+1}b^2&=a,\\
  b^2-1&=b.
\end{numcases}

April 21

A*B*C = B cubed – B, where A, B and C are successive integers?

Hi - my brilliant 15-y-o cousin Tom Schwerkolt-Browne came up with the above rule or observation. Is it original to him? Is it significant or non-trivial or whatever the term is?

Thanks - Adambrowne666 (talk) 12:23, 21 April 2009 (UTC)

It's pretty straightforward if you start with A*B*C = (B-1)*B*(B+1), and multiply it out. -- Coneslayer (talk) 12:28, 21 April 2009 (UTC)

(After edit conflict) Yes, it's a fairly obvious identity. If your cousin is generalising from a few numerical examples, he might want to see if he can prove that it is always true for any three conecutive integers (positive or negative). Difficult way - call your integers x, x+1 and x+2, expand x(x+1)(x+2) and compare it to (x+1)³-x. Simpler way (as per Coneslayer) - call your integers x-1, x and x+1 - then there is less algebra. And notice that there is nothing in the proof that assumes x is an integer ...

Related problem - show that n³-n is always a multiple of 6, for any integer n. Gandalf61 (talk) 12:36, 21 April 2009 (UTC)

The point to note after having read the comments here is that mathematics is broader than what your 15 year old cousin and you think. A reasonably bright 12 year old would be able to verify this identity, and usually an indicator of whether your "discovery" is trivial or non-trivial is to check the length of the proof of your claim. Algebraic identities are almost always trivial (with respect to high-school algebra). The other thing that you should learn is that it is very rare for someone as young as your cousin to discover something new in mathematics, unless they have had training in mathematics at undergraduate level or so. There is nothing to discover within the bounds of high-school mathematics as far as I know the difficulty of high-scool mathematics to be. Lastly, please do not be ignorant. I do not mean to sound rude, but it is very ignorant to think that a mathematical discovery can be achieved within the bounds of a few minutes thinking; not to mention the fact that it is insulting to reasearchers in mathematics. I do not mean to discourage you from asking questions here, but it is important that you understand the intended meaning in my comments. --PST 13:39, 21 April 2009 (UTC)

I am certain that Adam's pride in having a bright and inquisitive younger cousin was not intended as a savage attack on the practitioners of your field. With further encouragement, I am sure that his cousin will make meaningful contributions to mathematics or whatever field he chooses to go into, even if that day is some years off. -- Coneslayer (talk) 14:01, 21 April 2009 (UTC)

Certainly not. My comment was not that his question was unacceptable, but rather that he should understand that mathematics is broader than he thinks. --PST 14:14, 21 April 2009 (UTC)

We are here to answer reasonable questions in the same spirit, not to guess the quality of anyone's thoughts.Cuddlyable3 (talk) 14:34, 21 April 2009 (UTC)

PST: my reasonably bright 12 year old grandson was stumped. -hydnjo (talk) 21:49, 21 April 2009 (UTC)

Thanks, everyone, for the answers, and thanks for the defence, Coneslayer and others; but maybe Point-set is right - I was being ignorant, or at least romantic - had a naive hope that Tom had chanced on something interesting. Adambrowne666 (talk) 05:51, 22 April 2009 (UTC)

In my nursery school they taught me the multiplication table of 5 (1x5=5, 2x5=10,...,10x5=50). Some days later, I realized that one can go further (11x5=55,.....) and I was deeply impressed by this fact... Grown-up people didn't show much interest, but for me it was an important observation and I was quite proud of it, and some of my playmates warmly congratulated. --pma (talk) 10:22, 22 April 2009 (UTC)

Indeed. I felt the same way at 10 or 11 when I discovered how to use forward differences to extrapolate sequences of integer powers. I imagine most amateur and professional mathematicians have had some similar experience in childhood or adolescence that motivated their continuing interest in mathematics. A teenager who is actively exploring the world of mathematics should be encouraged by all means possible, not put down because they have not yet reached the edge of the map. Adambrowne666 - if your cousin is interested in widening their mathematical horizons, I recommend they read Courant and Robbins' classic What Is Mathematics?. Gandalf61 (talk) 11:38, 22 April 2009 (UTC)

Absolutely. Just because a discovery isn't original doesn't mean it isn't impressive or significant that you discovered it. --Tango (talk) 12:22, 22 April 2009 (UTC)

Certainly. I never discouraged his desire to discover more identities in algebra. Nor did I criticize his eagerness when he thought that he made a discovery. All that I said (or what I intended to say), which he will carry on in the future (and make better judgements as to what is a discovery and what is not), is that it is unlikely for something like that to be a new discovery - and that although it is nice that he is interested, he should understand this. --PST 13:31, 22 April 2009 (UTC)

In the context of undergraduate research in mathematics (much less research at the high-school level), a common metric for novelty is that the results are new and interesting to the student who discovers them, not that they are new from the point of view of the research literature. Of course not all results that meet this metric will be publishable, but publication is not usually the goal. There are many interesting problems in elementary combinatorics and number theory that, while they may be easily solved by an expert using high-powered techniques, still provide a great experience for a novice. The same can be said for assembling a car in your garage from a kit; you know ahead of time that the factory could do it faster, but that isn't the point.

However, PST is correct that, since number theory has been heavily studied for centuries, one is unlikely to discover any truly new results at the elementary level.

Another book that a bright high school student might want to pore over (and will take work) is A course in pure mathematics by Hardy. — Carl (CBM · talk) 14:21, 22 April 2009 (UTC)

OK. Let me correct myself - if you find a proof that every even number is the sum of two primes (or an equivalent), publish it. It is not trivial. Similarly, there are many other facts of this nature which are seemingly trivial but in actuality non-trivial. The point is however, algebraic identities tend to be usually trivial or not worth any interest to professional mathematicians. It is perfectly OK (in fact, encouraged) to prove such identities, but what I wish to say is that such identities are unlikely to be of any interest (or non-trivial) to professional mathematicians (but certainly maybe of interest to high-school students). In my view (others may not share the same view), high-school students should understand the purpose of mathematics (or what mathematics really is) before venturing out to make discoveries. Therefore, you should probably read the texts indicated above but in any case, I encourage your cousin to continue mathematics - whatever the nature. --PST 03:16, 23 April 2009 (UTC)

There are other areas in which an amateur mathematician can make a difference. Martin Gardner has many very long lists of original discoveries by modern amateurs, and progress continues. This is generally in areas that professional mathematicians don't bother with, such as board games and origami and so on, but it's still original research and can be a good introduction to the subject. Black Carrot (talk) 22:42, 23 April 2009 (UTC)

Quantum mechanics?

I was bored sitting in work and a colleauge said look up Schrödinger's cat.... While reading it...... it seemed as if the only way to explain Quantum mechanics is "the end product of any situation is allready predetermined"; applying this to the riddle, as far as i see it would explain the answer to the Schrödinger's cat experiment...the cat is either pre destined to be dead or alive this was pre determined right back to the big bang hence all future things are all ready predestined....(god knows what that makes to a time theory?) For example we know the Sun will burn out but it hasnt happened yet' its fule will be used up it will swell then cool ECT....this is all ready predetermined.So my question is if the universe surrounding us is all ready played out and the end states are known (but not by the human race) is there an equation that would fit this model? Ok i dont want you to look at this and think this guy is a nutter 'which i am sure you will' but it makes sense to me....we just need to be able to see the future is all :) Adrian O'Brien —Preceding unsigned comment added by 214.13.113.138 (talk) 12:29, 21 April 2009 (UTC)

I'm not sure but I think there is. Even if there is, the equation has to be really complicated. Also, I recommend you on posting this question on the science reference desk instead. Superwj5 (talk) 13:18, 21 April 2009 (UTC)

It is nice that you are interested in this - I suggest you read the article Quantum mechanics. Especially, it is recommended that you read the "overview" section and the beginning of the introduction. Just read what you understand and ignore what you don't. The other thing I wish to comment upon is your belief that mathematics is described by "equations". This is false. Group theory does indeed apply in quantam mechanics as well as probability theory, and the deep purpose of either subject has nothing to do with equations. Just let me stress that mathematics is not equations. If you are referring to the uncertainty principle, then:

$\Delta X\Delta P\geq {\hbar \over 2}.$

Also, I recommend that you read Introduction to quantum mechanics first as this is (apparently) more accessible. I do not wish to discourage you from asking questions but I just wish to stress the mathematics that applies to theoretical physics is much deeper than you think (for example the application of knot theory (a branch of mathematics) to string theory (a branch of physics)). --Point-set topologist (talk) 14:14, 21 April 2009 (UTC)

All equations including mistaken ones fit the OP's notion of predetermination. It made me post this. Cuddlyable3 (talk) 14:25, 21 April 2009 (UTC)

I was under the impression that it was classical mechanics that deals with a deterministic world, while quantum mechanics deals with a probabilistic one. What I mean to say is that if the position and momentum of every particle in the world were known at one instant, the state of the universe at any future instant is easily obtained by applying Hamilton's (or Lagrange's) equations. However in quantum mechanics it is intrinsically impossible to know the position and momentum of even one particle to arbitrary precision, let alone all of them. I would disagree with the OP's statement that the end result of any situation is predetermined, as we can only state the probability of obtaining any particular end. mislih 23:08, 21 April 2009 (UTC)

As a matter of fact we are in a position to compute precisely the evolution of the universe. The only thing is that, so many are the data, so high the precision needed, so complicated the equations, that the simplest computer able to do that, is the universe itself. It is reasonably fast: it takes not more than 24 hours to predict what will happen tomorrow ;) pma (talk) 07:44, 22 April 2009 (UTC)

Cutesie, small-texted answers aside the question at hand is whether quantum mechanics and, if QM is a valid description of reality, is deterministic. Many take the stand that the only reason that it is treated as a probabilistic is that we have incomplete knowledge of the universe, and if we knew the exact state of the universe now we could predict it at some future date. My understanding of the uncertainty principle is that complete knowledge is not possible, and that the universe is therefore not deterministic. Maybe this question would get more expert attention if it were at the science desk. mislih 17:56, 22 April 2009 (UTC)

I think even without the uncertainty principle, QM is non-deterministic. Even if you have perfect information about an atomic nucleus, say, you can't predict when it will decay because that is inherently random. --Tango (talk) 18:11, 22 April 2009 (UTC)

QM is nondeterministic; that doesn't mean the universe is. I'm not even sure what it means to say that the universe is deterministic or nondeterministic. Among things that didn't happen, how do you distinguish those that couldn't have happened from those that could have and merely didn't? Newtonian determinism doesn't really buy you much in this regard. Newtonian physics would appear to permit the existence of a universe in which Keanu Reeves is never born, yet we don't live in that universe. Is his existence logically necessary, or is there an element of randomness in the world? Those are the only two possibilities. Even without QM most people would probably guess that he's random.

I wouldn't be at all surprised if there turned out to be a deterministic layer underneath QM. For example, as the original poster said, "everything could be predestined" in the sense that there is some kind of constraint on the final conditions of the universe as well as the initial conditions. When a measurement takes place the outcome that is realized is the one that leads to the necessary final state of the universe; if the final condition is specific enough and the second law of thermodynamics holds, then there can only ever be one such outcome. The outcomes could easily look random, at least at times far from the beginning and end of the universe. We've invented deterministic processes that no one, not even their designers, can figure out how to distinguish from "true" (i.e. quantum) randomness. I'm not saying I believe everything is predestined in this way, just that you shouldn't conclude on the basis of QM that the universe is truly random (any more than you should conclude on the basis of Newton that it truly isn't). -- BenRG (talk) 20:46, 22 April 2009 (UTC)

I would define a deterministic universe as one where someone with complete knowledge of the state of the universe at a given time and complete knowledge of the laws of physics could not accurately and reliably predict the state of the universe at a later time - I think that's a standard definition. If the universe were governed by Newtonian physics, it would be deterministic. Someone with a complete knowledge of Newtonian physics and the state of the universe at any time prior to Keanu Reeves' birth would be able to work out whether he will be born or not. There is no randomness, although there is chaos which can appear random - that's why you need complete knowledge of the state of the universe, not just almost complete knowledge. If there universe were governed by QM, then it would be non-deterministic and no-one, however knowledgeable, could know for certain whether or not Reeves will be born until it happens. Now, we know the universe isn't actually governed by either of those sets of laws, so whether or not it is deterministic is unknown - the evidence certainly suggests that it isn't, though. --Tango (talk) 21:27, 22 April 2009 (UTC)

April 22

The Floor Function

I am trying to evaluate

$\int _{0}^{1}\int _{0}^{1}\cdots \int _{0}^{1}[x_{1}+\cdots +x_{n}]dx_{n}\cdots dx_{1}$

which is basically the floor function in a n-fold integral over the unit cube. I tried to start with a simple case and go step by step.

$\int _{0}^{1}[x]dx=0$ that is easy to see.

$\int _{0}^{1}\int _{0}^{1}[x+y]dydx$ can be rewritten as

$0\cdot R_{0}+1\cdot R_{1}$ where

$R_{1}=\{(x,y)\in \mathbb {R} ^{2}|0<x,y<1,x+y>1\}$

which has an area of 1/2 so the entire integral is 1/2. And similarly for three dimensions, I got

$\int _{0}^{1}\int _{0}^{1}\int _{0}^{1}[x+y+z]dzdydx$ as

$1\cdot R_{1}+2\cdot R_{2}$ where

$R_{1}=\{(x,y,z)\in \mathbb {R} ^{3}|0<x,y,z<1,2>x+y+z>1\}$

$R_{2}=\{(x,y,z)\in \mathbb {R} ^{3}|0<x,y,z<1,3>x+y+z>1\}.$

My questions is, I can't find the volume of these regions correctly (because I can't set up the triple integral correctly). Can someone please shed some light on this on how to find the volume of both of these regions and then how to generalize this to n-dimensions? Thanks!69.224.116.142 (talk) 18:08, 22 April 2009 (UTC)

I guess it is $I_{n}=(n-1)/2$. Change variable in the integral putting: $y_{i}:=1-x_{i}$ and sum the two, so you get twice $I_{n}$ is the integral on the n-cube of $\scriptstyle [x_{1}+\cdots +x_{n}]+[n-(x_{1}+\cdots +x_{n})]$; then use the identities $[n+t]=n+[t]$ and $[t]+[-t]=-1$ (a.e.) --pma (talk) 20:33, 22 April 2009 (UTC)

As to the volume of the sets $R_{n,k}$ here they are [2] pma (talk) 22:22, 22 April 2009 (UTC)

By the way, "a.e" as in pma's post stands for "almost everywhere" in case you were wondering... --PST 03:19, 23 April 2009 (UTC)

This is great my now my question is how can I show that making that change of variables in the integral still equals $I_{n}$? It is pretty easy to show it for the n=1 case but for higher n's, it doesn't seem to work out.130.166.159.98 (talk) 02:09, 24 April 2009 (UTC)

It's just the change of variables formula. But here you need a very particular case of it, for the change of variable map φ(x):=(1,1..1)-x is quite an elementary isometry and the integrand is a simple function. So, if you prefer, write your integral as you did,

$\textstyle I_{n}=\sum _{k=0}^{n-1}k|R_{k}|$,

where

$R_{k}:=\{x\in [0,1]^{n}:[x_{1}+..+x_{n}]=k\}$

and observe that $\textstyle R_{k}$ and $\textstyle R_{n-1-k}$ have the same measure, because they are obtained from each other (up to a null set) with a simmetry (the change of sign) and a translation. For instance in your computation above for $n=3$, $\textstyle |R_{0}|=|R_{2}|$ so $\textstyle I_{3}=|R_{1}|+2|R_{2}|=|R_{0}|+|R_{1}|+|R_{2}|=1$. pma (talk) 10:45, 24 April 2009 (UTC)

Factoring a cubic

Does the cubic $20u^{3}-65\pi u^{2}+50\pi ^{2}u-16\pi ^{3}$ factor nicely? If it does, what is the factorization? Lucas Brown 42 (talk) 19:23, 22 April 2009 (UTC)

The first step would be to get rid of the pi in the equation. Introduce a substitution $u=a\pi$ and the cubic reduces to

$20a^{3}-65a^{2}+50a-16=0$. Readro (talk) 19:48, 22 April 2009 (UTC)

Which polynomial is irreducible over the rationals. Algebraist 19:58, 22 April 2009 (UTC)

What about the irrationals? 72.197.202.36 (talk) 04:35, 23 April 2009 (UTC)

$\textstyle 20a^{3}-65a^{2}+50a-16=20(a-a_{1})(a-a_{2})(a-a_{3})\,$ ,

$a_{1}={\frac {R}{60}}+{\frac {245}{12R}}+{\frac {13}{12}}\,$ ,

$a_{2}=-{\frac {R}{120}}-{\frac {245}{24R}}+{\frac {13}{12}}+i{\sqrt {3}}\left({\frac {R}{120}}-{\frac {245}{24R}}\right)\,$ ,

$a_{3}=-{\frac {R}{120}}-{\frac {245}{24R}}+{\frac {13}{12}}-i{\sqrt {3}}\left({\frac {R}{120}}-{\frac {245}{24R}}\right)\,$ ,

$R:={\sqrt[{3}]{68525+300{\sqrt {31749}}}}\,$ .

--78.13.138.117 (talk) 06:53, 23 April 2009 (UTC)

In other words: No, it doesn't factorise nicely! --Tango (talk) 16:50, 23 April 2009 (UTC)

You can see right away that it has at least one positive root. If you find such a root, then u minus that root is a factor of the polynomial. If you divide the polynomial by that factor, you get a quadratic polynomial, and then it's just a matter of solving a quadratic equation. But whether the positive root you find can be expressed "nicely" is another question. If "Tango" has the details right, then it's no nicer than the messiest you could expect under the circumstances. Michael Hardy (talk) 22:07, 23 April 2009 (UTC)

The anon did the calculation (using the cubic formula, by the looks of it), I just concluded that that wasn't "nice" (it clearly doesn't simplify significantly). --Tango (talk) 10:24, 24 April 2009 (UTC)

apparent error in pages on kernel smoothing

Hi, there appears to be an inconsistency in the pages Kernel smoother and Kernel (statistics). The second of these gives the requirement for a K function that $\int _{-\infty }^{+\infty }K(u)du=1\,;$ whereas the Kernel smooth article gives the following as a K function: $D(t)=\left\{{\begin{aligned}&1,\,\,\,if\,\,\left|t\right|\leq 1\\&0,\,\,otherwise\\\end{aligned}}\right.$

I add that the full notation for kernel smoothers, using the K function, is: $K_{h_{\lambda }}(X_{0},X)=D\left({\frac {\left\|X-X_{0}\right\|}{h_{\lambda }(X_{0})}}\right)$

This, however, doesn't seem to make any difference, since I assume the K funtion is to be interpreted as a function of X, not X-nought. In this case, it looks the integral of the D function is 2, which is incompatible with the requirement that it be 1. Am I making a fundamental oversight, and if not, what is the resolution for the inconsistency? Regards, It's been emotional (talk) 23:57, 22 April 2009 (UTC)

I suspect it should have said 1/2 if |t| ≤ 1. Michael Hardy (talk) 04:26, 23 April 2009 (UTC)

April 23

In geometry, is a square with rounded corners still considered a square? If not, what is it called, mathematically?

Normally I can answer any question my daughter asked, but this one stumped me! —Preceding unsigned comment added by 216.61.187.254 (talk) 21:30, 23 April 2009 (UTC)

Might be Squircle. Zain Ebrahim (talk) 21:35, 23 April 2009 (UTC)

In practical use it's more probably a composition of line segments and circular arcs, though, which is simply called a "rounded square" in the same article. But the squircle is obviously more mathematically interesting. —JAO • T • C 21:41, 23 April 2009 (UTC)

Yes, right. Squircle is not what OP is looking for but OP maybe pleasantly surprised to find a closed match of rounded square defined actually by an algebraic equation. Squircle nowhere has straight edges, which OP's question seems to insist on. Though as a side note even square may not have straight edges in Non-Euclidean geometry. But, I hope we are dealing with Euclidean geometry here. - DSachan (talk) 21:52, 23 April 2009 (UTC)

Also rounded square, or smoothed square, or similar. The fact is that not every object in mathematics has a standard aknowledged name; just the most common and used ones (for instance: the trigonometric functions sin(x), sin(x)/cos(x) and 1/sin(x) all have special names, but sin(cos(x)) has none). If in a given context (a book, a paper, a theorem) one needs to use frequently something which is not otherwise so common to deserve a special name, one may just introduce a name to be used there. --pma (talk) 21:53, 23 April 2009 (UTC)

Quotient spaces, and cell complexes

I'm currently reading Hajime Sato's Algebraic Topology: An Intuitive Approach, but I'm having a bit of trouble understanding how to understand what particular quotient spaces look like (despite the book's name). I've taken an introductory course in Algebraic Topology, but was hoping for a bit more rigour regarding quotient spaces, but the book hasn't really helped. It's easy enough to visualise simple examples (such as a closed two-dimensional ball which has its boundary identified to a single point being homeomorphic to S²), but for more complicated examples, I'm just not comfortable stretching these shapes around in my head - I'd like some sort of rigorous method to determine what the quotient space looks like, but haven't been able to find one.

Another (somewhat related) issue I'm having with Sato is the idea of cell-complexes. I've come across these before too, but after describing how to construct general spaces using them, he bounds into examples without really explaining what to do. He, for example, say that the torus, T² can be written as $T^{2}=({\bar {e}}^{0}\cup _{h_{1}}({\bar {e}}_{1}^{1}\cup {\bar {e}}_{2}^{1}))\cup _{h_{2}}{\bar {e}}^{2}$ where the ${\bar {e}}^{i}$ are closed i-balls, and the $h_{i}$ are 'attaching maps', which take the boundary of one part of the complex onto a lower-dimensional part, and which he does not actually give the details of. (For instance, I'm interpreting h₁ to be $h_{1}:\delta ({\bar {e}}_{1}^{1}\cup {\bar {e}}_{2}^{1})\to {\bar {e}}^{0}$). I've covered similar ground to this before, having studied n-simplices, which are each obviously homeomorphic to ${\bar {e}}^{n}$ (and with boundary operators in place of the 'attaching maps', however the boundary operators were never used to glue), but I can't seem to get the two ideas to gel.

In my mind, the above example starts with two disjoint closed intervals, ${\bar {e}}_{1}^{1}$ and ${\bar {e}}_{2}^{1}$ whose boundaries (end points) are attached/glued by h₁ to a single point, ${\bar {e}}^{0}$, giving a sort of figure of eight shape. This is then glued by h₂ to the boundary of ${\bar {e}}^{2}$, S² (edit: obviously I meant S¹) ...which I again have no idea how to visualise, or see how this could possibly give a torus.

Sorry for this massive outpouring, I just really can't get my head around it. Thanks a lot! Icthyos (talk) 22:27, 23 April 2009 (UTC)

Don't apologize - we are here to help you. Topology is a subject on which there are too few a number of textbooks; at least half of which try to simplify the subject down. I have attempted to understand why this is the case (when I first saw the definition of a topology (a long time ago), I did not see the purpose of having such a complex set-theoretic definition, but after a week I got used to it and saw the definition more intuitively than ever - the definition of a topology is really ingenious and I respect Hausdorff for this). Quotient spaces are rigorously defined as follows: Let X be a topological space and ~ an equivalence relation on X. Let X^* be the set of all equivalence classes under this relation. Define U to be open in X^* iff the union of the equivalence classes that belong to U, is open as a subset of X. Then X^* is a topological space with this topology - it is called the quotient space of X by the equivalence relation ~ (verify that this is a topological space, if you have not done so already). Note that some authors prefer to define the quotient space using the quotient map (which is probably a less confusing way to define it), but I think this definition is alright too. Think about the following - it should improve your intuition of the concept:

a) Consider the natural projection $p:X\to X^{*}\,$. Is this map continuous? Is the inverse of this map continuous? Is this map surjective? Deduce some properties of X^*, given properties of X, using this map.

b) Suppose X is homogeneous (i.e the homeomorphism group acts transitively on X) (equivalently, if x is in X and y is in X, there exists a homeomorphism of X taking x to y). Under what conditions is X^* homogeneous?

c) Consider the circle with ~ defined by - x ~ y iff x is antipodal to y (i.e x = - y). Consider the (two-point) equivalence classes (actually viewed as a single entity)- one equivalence class ((1,0) and (-1,0)) essentially determines the x - axis. As you move clockwise around the circle, the equivalence classes are distinct until you have rotated 180 degrees. So essentially, to consider the quotient space, you have to examine the geometry of the equivalence classes - i.e points on the upper semicircle. Essentially, the equivalence classes "close" to (-1,0) are also close to (0,1) because both points are equivalent. What is the quotient space X^* homeomorphic to?

d) With a similar equivalence relation defined on the sphere, note that the resulting quotient space is a two-manifold. However, it is impossible to embed the resulting quotient space in R³. What is the smallest n, such that this quotient space can be embedded in Rⁿ? What is the smallest n for which the resulting quotient space can be embedded via a diffeomorphism in Rⁿ (define a smooth structure on the quotient space using the natural smooth structure on the sphere, and the projection (i.e quotient) map).

e) Define x ~ y on an arbitrary topological space iff there is a homeomorphism of that topological space carrying x to y. Find a space for which the resulting quotient space is the Sierpinski space. Is there a topological space for which this is homeomorphic to the countable discrete space?

f) Is the quotient space of a manifold, again a manifold? If not, under what conditions is it a manifold? Under what conditions does it have the same dimension (as a manifold) as the original manifold?

g) Let G act on a topological space X. Use this action to define a quotient space of X. Prove that the resulting quotient map to the quotient space, is a covering map, under certain conditions.

I think that these questions are useful to develop your intuition. As I don't know your exact level, I can't say whether these questions will be challenging for you or not. However, most should be relatively easy. If you have any other questions, or any difficulties, feel free to ask again. I will help you on your other question, if no-one else will do so, but I have got to go now. --PST 03:00, 24 April 2009 (UTC)

For the torus example in particular, I think you can consider this by looking at the fundamental polygon for the torus, which is realising the torus by the following construction . You can see that this has only one point ($e^{0}$), two 1-cells and one 2-cell. So you attach these together as the above diagram says, and you get the torus.

I don't think there really can be a general method for determining what the quotient space looks like, you need to consider what the attaching maps involved actually are and see if you can relate that to anything you know; you can realise many spaces by a similar construction as above, for example I remember a nice example of taking a cube and identifying opposite faces with a quarter-twist, you can put a cell structure on this quite easily by counting points, edges, faces and the whole volume, and you end up with some space with fundamental group $Q_{8}$, the quaternion group, which I thought was quite nifty. - Xedi^Talk 02:56, 24 April 2009 (UTC)

April 24

Growth of a bacterial colony

Hi I was wondering if I could obtain some help with the following question: Starting with 1 bacterium that divides every 15 minutes, how many bacteria would there be after 6 hours? Assume none die etc. I thought ok so first of all how many sets of 15 minutes in 6 hours? There are 4 lots of 15 mins in 1 hour therefore 4 x 6 = 24 15 minute intervals in 6 hours. Nothing too difficult there. I thought then to work it out I would do e^{original number x time} making the equation e^1x24 which gives an answer of 2.649 x 10¹⁰. I had a look at the answer and it said 16,777,216 which is obtained by the equation 2²⁴ but I have no idea where this equation 2²⁴ comes from. I'm also why my calculation involving e was incorecct. If anyone could explain these two things to me it would be a great help. Thanks. —Preceding unsigned comment added by 92.22.189.144 (talk) 08:24, 24 April 2009 (UTC)

Dividing every 15 minutes means doubling, so there will be twice as many as before. Hence it is the appropriate power of 2 which is wanted.81.154.108.6 (talk) 08:52, 24 April 2009 (UTC)

Right. After 15 minutes, there will be 2 (2¹) bacteria. They both will divide after 30 minutes (15x2), and there will be 4 (2²) bacteria. After 45 minutes (15x3), these 4 bacteria will double and number of bacteria will be 8 (2³), so you see there is a pattern. After 15xn minutes, there will be 2ⁿ bacteria. In your case n is 24, so there will be 2²⁴ bacteria after 15x24 = 360 minutes = 6 hours. I have no idea why you think there should be a connection with 'e'. - DSachan (talk) 11:21, 24 April 2009 (UTC)

Many thanks for the reply. I am sure there is an equation of exponential growth involving e or would it be log e^kt? Not sure. Thanks anyway! —Preceding unsigned comment added by 92.21.233.141 (talk) 13:08, 24 April 2009 (UTC)

The article on exponential growth is quite helpful in this regard. We could write the number of bacteria as a function of the number of minutes that have passed as

$b(m)=2^{m/15}$

but as the original poster guessed, we could just as well write this formula using base e.

$b(m)=e^{m/T}$

To get the value of T, we set the two formulas equal and just take the log of both sides.

$2^{m/15}=e^{m/T}$

${\frac {\ln {2}}{15}}={\frac {1}{T}}$.

The number of bacteria at a given time can now be written as $b(m)=e^{m{\frac {\ln {2}}{15}}}$. (any suggestions to make my math look better would be appreciated). mislih 13:48, 24 April 2009 (UTC)

a suggestion to make the math look better: try \scriptstyle : $\scriptstyle {\frac {\ln {2}}{15}}={\frac {1}{T}}$. Bo Jacoby (talk) 14:13, 24 April 2009 (UTC).

sci.mathresearch newsreader question

What's an easy newsreader to use for posting there? (They say you have to use a newsreader now, and I don't know a newsreader from Adam.) thanks, Rich (talk) 09:28, 24 April 2009 (UTC)

e lon to nhat la ai nhi lu khon nan nhat oi vai lon chua kia so lon thi so bi aids nhe kinh —Preceding unsigned comment added by 123.18.213.33 (talk) 11:04, 24 April 2009 (UTC)

Just go to groups.google.com and follow instructions. McKay (talk) 12:02, 24 April 2009 (UTC)

Polynomial division

This is actually a homework sum. I tried in as many ways as possible but i didn't get an answer. The question is :

If the polynomial $x^{4}-6x^{3}+16x^{2}-25x+10$ is divided by another polynomial $g(x)=x^{2}-2x+k$, the remainder comes out to be $x+a$. Find k and a.

I tried by the following method:

$x^{4}-6x^{3}+16x^{2}-25x+10=(x^{2}-2x+k)*q(x)+x+a$

putting remainder on L.H.S, $x^{4}-6x^{3}+16x^{2}-26x+10-a=(x^{2}-2x+k)*q(x)$

If i divide L.H.S by g(x), i get $q(x)=x^{2}-4x-8+k$ and $r(x)=(10+8k)x+10-a-8k-k^{2}$. I equate r(x) to zero coz g(x) is a factor of L.H.S. I get $(10+8k)x+10-a-8k-k^{2}=0$. I am unable to continue. Please help me--harish (talk) 13:51, 24 April 2009 (UTC)

The remainder must be the zero polynomial, meaning that both coefficients are zero. $10+8k=10-a-8k-k^{2}=0$. Bo Jacoby (talk) 14:07, 24 April 2009 (UTC).

(e/c) If $(10+8k)x+10-a-8k-k^{2}$ is the zero polynomial, this means that all its coefficients are zero, i.e., $10+8k=0$ and $10-a-8k-k^{2}=0$. You can extract k from the first equation, and then you get a from the second equation. However, I think that you made a numerical error, I got a different result for q and r. — Emil J. 14:09, 24 April 2009 (UTC)

solve(identity(x^4-6*x^3+16*x^2-25*x+10 = (x^2-2*x+k)*(x^2+b*x+c) + (x+a),x),{a,b,c,k}); McKay (talk) 00:40, 25 April 2009 (UTC)

Holmorphic mapping

Hello, I am trying to prove that any one-to-one and onto holomorphic mapping of $\mathbb {C} \setminus \{0\}$ with a removable singularity at 0 satisfies $\lim _{z\to 0}f(z)=0$ and I am stuck.

I assumed it was something else, c. Then, there must exist some other point that also maps to c, say p. I assume I need to do something with an integral to count the number of zeros of the function f(z) - c, which has two (if I just abuse notation and let this f now represent f with the removable singularity removed). But, I am not sure what I can do. Form a circle big enough to contain 0 and p, call it D. Then, consider the function

$\int _{\partial D}{\frac {f'(z)}{f(z)-q}}\,dz$

which is a function of q. At c, the value is 2. I am guessing that I want to prove that for some small neighborhood around c, all numbers in that disc must also have value 2 for this integral, which will contradict the fact that the original map was one-to-one.

So, is this integral function a continuous function of q at c? I am pretty sure I know it is continuous in z but that's not what I want here. I am not sure exactly how this works.

Thanks StatisticsMan (talk) 21:52, 24 April 2009 (UTC)

How to compute statistical significance of correlation

I have a list correlations R[] and the associated number of samples for each correlation N[]. And I am not sure how to calculate a weight for each correlation/amount which is proportional to the chance that the correlation is non-spurious. Right now I have weight set to sqrt(N)*R^2 but this does a lousy job at discriminating one pair of R,N from another.

I believe you mean "statistically insignificant" instead of "non spurious." Spurious (at least in econometrics) refers to a correlation that is statistically significant due to random chance. Determining whether a correlation is spurious is, at best, non-trivial and, depending on your particular circumstances, quite possibly impossible. Wikiant (talk) 23:22, 24 April 2009 (UTC)

April 25

higher derivatives

Hi, I read the following surprising comment in a maths textbook: "Let $f(x)$ be a function such that $f^{(n)}(c)$ exists. Then $f^{(j)}(x)$ exists in an interval around $c$ for $j<n.$" Surely this is false. Take the function $f(x)=0$ for rational x, and $f(x)=x^{n}$ for some integer n > 2, for all irrational x. At x = 0, all derivatives below the nth will exist, but the function isn't even continuous around 0. Have I got this right? It's been emotional (talk) 00:21, 25 April 2009 (UTC)

It looks like the first derivative exists, but do the higher ones? The first derivative is only defined at 0, don't you need your function to be defined on an interval to differentiate it? --Tango (talk) 00:34, 25 April 2009 (UTC)

<sheepish grin> oops - I was only thinking with reference to the squeeze theorem, in which case the idea makes sense, but I think you are quite right, and there is no such thing as the higher derivatives. However, if instead of using $f'(x)$ in the definition of $f^{(2)}(x)$, you use the definition of $f'(x)$ in terms of $f(x)$, it might be a different story. Then I suspect the squeeze theorem would apply, and the result would follow. Am I right now? It's been emotional (talk) 04:24, 25 April 2009 (UTC)

dice

If I roll 64 standard non-weighted 6 sided dice, what are my odds of rolling <= 1, 6?

Could you explain this further, please? Are you referring to your score on each die individually, or your total score, or what? It's been emotional (talk) 04:25, 25 April 2009 (UTC)

I mean what are the odds that 63 of the 64 dice will not have a 6 pips facing up.