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September 7

Latex align* multiple points of alignment.

Resolved

I am trying to type up a displayed equation for which I want to have multiple points of alignment, one at equals and one at the point where I give reasons for each line. I have tried doing it like this:

  \begin{align*}
     \frac{x}{y} + \frac{a}{b} &= xy^{-1} + ab^{-1}  & ( \text{by definition})\\
        &= xy^{-1}1 + ab^{-1}1  & (\text{where 1 is the multiplicative identity})\\
        &= xy^{-1}bb^{-1} + ab^{-1}yy^{-1}  & (\text{inverse property of mulitplication } bb^{-1} = yy^{-1} = 1) \\
  \end{align*}

But the problem is that the reasons get right aligned whereas I want them to be left aligned, preferable such that there is at least a \quad spacing between the equation and the reason. Is there any way to do this? AMorris (talk)●(contribs) 04:41, 7 September 2009 (UTC)[reply]

The default alignment for the align environment is left right left right... so it alternates left and right. You can just put two & before the text instead of one, making it align to the left. --Xedi^Talk 05:56, 7 September 2009 (UTC)[reply]

To expand on Xedi's answer:

  \begin{align*}
     \frac{x}{y} + \frac{a}{b} &= xy^{-1} + ab^{-1}  && ( \text{by definition})\\
        &= xy^{-1}1 + ab^{-1}1  && (\text{where 1 is the multiplicative identity})\\
        &= xy^{-1}bb^{-1} + ab^{-1}yy^{-1}  && (\text{inverse property of mulitplication } bb^{-1} = yy^{-1} = 1) \\
  \end{align*}

The idea is that the &s alternate between being /alignment markers/ and /column separators/. So the first & means "align the first column at the = sign". The second & delimits the end of the first column and the start of the second column. The third & means "align the second column at the (". And so on. Eric. 67.169.125.37 (talk) 07:57, 7 September 2009 (UTC)[reply]

Perfect - that sounds like it will do exactly what I wanted. It was frustrating because I couldn't find this documentation anywhere when I looked up how align works. Thanks guys! AMorris (talk)●(contribs) 12:05, 7 September 2009 (UTC)[reply]

Yeah, Latex is a pain. Sometimes you just have to know the right magic incantation, or know about some obscure behavior that you'd never think to look up. Eric. 67.169.125.37 (talk) 19:46, 7 September 2009 (UTC)[reply]

Oh, now that I looked at the align again, I seem to recall that having a \\ on the last line gives you some extra whitespace after the equations that you probably don't want. You can see for yourself whether you like it. Eric. 216.27.191.178 (talk) 23:06, 7 September 2009 (UTC)[reply]

Reduction to Canonical Form

Hello, I have been trying to reduce $2u_{xx}+u_{xy}+yu_{yy}=0$ in the region y>1 for the last couple of hours but not getting anywhere. In this region, the PDE is elliptic. Letting a=2, b=1, and c=y, the discriminant is negative. So I set

${\frac {dy}{dx}}={\frac {b+{\sqrt {b^{2}-4ac}}}{2a}}={\frac {1+{\sqrt {1-8y}}}{4}}={\frac {1+i{\sqrt {8y-1}}}{4}}$

and (I think) that I am supposed to solve this differential equation and then let $\phi (x,y)$ be the real part of the solution and let $\psi (x,y)$ be the imaginary part and then I will have my change of coordinates which will reduce this PDE to the canonical form (the Laplacian equals some function). But the problem is that when I try to do this integral I get

$-i{\sqrt {8y-1}}+\ln(-i+{\sqrt {8y-1}})$

which I can't separate into the real and imaginary part. And then when I tried to integrate the real and the imaginary part separately (which is correct I hope) and use the results as the transformation, it doesn't work. I let ${\frac {dy}{dx}}=1/4$ giving me $\phi (x,y)=4y-x$ and then ${\frac {dy}{dx}}={\frac {\sqrt {8y-1}}{4}}$ giving me $\psi (x,y)={\sqrt {8y-1}}-x$ . The mixed partials don't go away and also it is a nightmare when I try to invert it. What am I doing wrong? Is there an easier way of doing this or am I making some stupid mistake? Thanks! 97.118.56.41 (talk) 07:24, 7 September 2009 (UTC)[reply]

You said that you can't seperate into the real and the imaginary part. Well, if

z:=-i{\sqrt {8y-1}}+\ln(-i+{\sqrt {8y-1}})

as you say then we can split this into it's real and imaginary parts:

\Re (z)=\ln \left(\left|-i+{\sqrt {8y-1}}\right|\right)+\Im \left({\sqrt {8y-1}}\right)\ ,

\Im (z)={\mbox{Arg}}\left(-i+{\sqrt {8y-1}}\right)-\Re \left({\sqrt {8y-1}}\right)\ .

Now, if we assume that y > 1 (in fact y > ⅛ would do) then we get

\Re (z)=\ln \left(2{\sqrt {2y}}\right)\ ,

\Im (z)=-{\sqrt {8y-1}}-\arctan \left({\frac {1}{\sqrt {8y-1}}}\right).

(I've ignored the multi-valuedness and just worked with principal values) ~~ Dr Dec (Talk) ~~ 09:59, 7 September 2009 (UTC)[reply]

FRACTALS

210.212.239.181 (talk) 11:24, 7 September 2009 (UTC)harshagg[reply]

I want to know about fractals and their application and how to design them.I am just a beginner in this field even in wikipedia itself i can know about only their theory not use and i don't know can be that i couldn't understand what they say or mean to say.Pls help me.If its not against rule i can give my e-mail id so that u can contact me at for longer period of time.

There are many different types of fractals. Here at Wikipedia we have a general article on fractals, articles on certain types of fractals such as Julia sets, De Rham curves and iterated function systems, and even articles on individual notable fractals, such as the Mandelbrot set and the Koch snowflake. Our list of fractals by Hausdorff dimension gives a list of about 70 different individual fractals. The index of fractal-related articles gives a longer list of our articles about fractals. Gandalf61 (talk) 11:50, 7 September 2009 (UTC)[reply]

One nice fractal with a fairly elementary construction and plenty of (mathematical) applications is the Cantor set. -GTBacchus^(talk) 12:11, 7 September 2009 (UTC)[reply]

If you have MS Windows you might also like to try out the Fractint program. Dmcq (talk) 14:06, 7 September 2009 (UTC)[reply]

Irreducible representations of signed permutations (double young diagrams?)

Hello,

Young tableau describes how one can use Young tableaux to get all irreducible representations of the symmetric group. Another good text on that can be found here [1]

The symmetric group on $n+1$ symbols is precisely the finite Coxeter group $A_{n}$ .

The Coxeter group $B_{n}$ consists of the $2^{n}n!$ signed permutations on the set $\{1,....n\}\cup \{-1,...,-n\}$ I heard that that group has a similar construction for its irreducible representations, using something like "double young diagrams". I can find very little on that on the internet. Can anyone tell me a bit more about this (even a few names that might help my search could be useful).

I am also interested in which irreducible characters appear in the decomposition of the characters of $B_{n}$ , induced by parabolic subgroups. For $A_{n}$ this is completely known, and the coefficients are known as Kostka numbers.

Many thanks, Evilbu (Talk) unsigned by Evilbu

You seem to have made a series of statements; without posing a question. Are we to assume that your question is "Which are the irreducible characters appearing in the decomposition of the characters of $B_{n}$ , induced by parabolic subgroups?" ~~ Dr Dec (Talk) ~~ 12:27, 7 September 2009 (UTC)[reply]

My sincere apologies. My main question is: how to describe the irreducible representations of that group

B_{n}

? My secondary question is: which of these appear (and if possible: how often) in the decomposition of the characters of

B_{n}

, induced by parabolic subgroups? Evilbu (Talk)

I don't know of any particularly special treatment of Bn, but the chapters on wreath products in Kerber's Representations of Permutation Groups is where I learned how to deal with Bn.

Kerber, Adalbert (1971), Representations of permutation groups. I, Lecture Notes in Mathematics, Vol. 240, Berlin, New York: Springer-Verlag, doi:10.1007/BFb0067943, MR 0325752
Kerber, Adalbert (1975), Representations of permutation groups. II, Lecture Notes in Mathematics, Vol. 495, Berlin, New York: Springer-Verlag, doi:10.1007/BFb0085740, MR 0409624
Young, Alfred (1930), "On Quantitative Substitutional Analysis 5", Proceedings of the London Mathematical Society. Second Series, 31, doi:10.1112/plms/s2-31.1.273, ISSN 0024-6115, JFM 56.0135.02

The chapters in question are (Kerber 1971, Ch. 1 doi:10.1007/BFb0067945, Ch. 2 doi:10.1007/BFb0067946) and (Kerber 1975, Ch. 1 doi:10.1007/BFb0085742). The Coxeter group of type Bn is also known as the hyperoctahedral group, and is originally discussed in (Young 1930). JackSchmidt (talk) 16:35, 7 September 2009 (UTC)[reply]

A^nB^n ≠ (AB)^n for matrices (get around)

In the following series. F=Fo(A^n*B^n*(1-B)^0+Bicof*A^(n-1)*B^(n-1)*(1-B)^1......Bicof*A*B*(1-B)^(n-1)+(1-B)^n)

Bicof =(Binomial coeficient)

I would like to simplify this to Fo(AB+(1-B))^n. Using the binomial theorem

However A and B are both Matrices so A^n*B^n ≠ (A*B)^n. A is a upper triangular matrix and B is a diagonal matrix, with all elements less than 1.

Once that is done the function can be further simplified using the limit form of the exponential function and manipulating B.

Any ideas would be useful. The problem arose from tedious crusher breackage formulas. Startibartfast —Preceding unsigned comment added by 41.240.93.114 (talk) 15:09, 7 September 2009 (UTC)[reply]

This question has me totally blown away. What does Fo mean? What exactly does the binomial coefficient of a matrix-binomial power expansion mean? For example, for square matrices M and N we have: (M + N)² = M² + MN + NM + N². The binomial coefficients don't appear, correct me if I'm wrong, until we have a commutative algebra. ~~ Dr Dec (Talk) ~~ 16:05, 7 September 2009 (UTC)[reply]

I'd guess the OP wants something for binomials like the exponential map formula for Lie groups, but that's fairly hairy. There's an inversion formula for a sum of two matrices also which is probably somewhere on wiki but I don't think it would help either. Dmcq (talk) 16:53, 7 September 2009 (UTC)[reply]

Diagonal matrices commute with everything, don't they? Doesn't that mean that, in this case, A^n*B^n = (A*B)^n? --Tango (talk) 16:58, 7 September 2009 (UTC)[reply]

I thought that for a moment too, but it's not true:

\left({\begin{array}{cc}1&2\\3&4\end{array}}\right)\left({\begin{array}{cc}1&0\\0&2\end{array}}\right)=\left({\begin{array}{cc}1&4\\3&8\end{array}}\right)

\left({\begin{array}{cc}1&0\\0&2\end{array}}\right)\left({\begin{array}{cc}1&2\\3&4\end{array}}\right)=\left({\begin{array}{cc}1&2\\6&8\end{array}}\right)

~~ Dr Dec (Talk) ~~ 17:06, 7 September 2009 (UTC)[reply]

Scalar matrices commute with everything. Algebraist 17:13, 7 September 2009 (UTC)[reply]

Ah, yes. Sorry. --Tango (talk) 17:20, 7 September 2009 (UTC)[reply]

To the OP: if you rewrite your question in a more readable and understandable way, then I will think about it.--pma (talk) 18:45, 7 September 2009 (UTC)[reply]

A topological counterexample

Does there exist a compact T₂ first-countable separable space which is not second-countable? I would especially like it if the example was totally disconnected (i.e., does there exist an uncountable Boolean algebra B such that every ultrafilter in B is countable generated, and B − {0} is the union of countably many ultrafilters?) The Sorgenfrey line almost works, but it's not compact, only Lindelöf. — Emil J. 17:24, 7 September 2009 (UTC)[reply]

Counterexamples in Topology gives the weak parallel line topology as fulfilling all your conditions, including total disconnection. Algebraist 17:36, 7 September 2009 (UTC)[reply]

Excellent, thanks. This is what I was looking for. — Emil J. 09:45, 8 September 2009 (UTC)[reply]

I was about to suggest this book myself. ~~ Dr Dec (Talk) ~~ 17:41, 7 September 2009 (UTC)[reply]

debeat? —Preceding unsigned comment added by 92.8.193.46 (talk) 01:44, 8 September 2009 (UTC) debeat,debate,debeaten no? --pma (talk) 05:48, 8 September 2009 (UTC)[reply]

Base Conversion

I was wondering what happens when you convert to a base beyond 36. From what I understand after base 10 you use letters to fill in for the numbers (i.e. in hexadecimal A is for 11, so on). But after base 36 you run out of letters. It wouldn't make much sense to express the decimal number 100 in base 37 as 2 26 (2*37 + 26*1). Thanks in advance 66.133.196.152 (talk) 18:26, 7 September 2009 (UTC)[reply]

You could make up more symbols, perhaps using another alphabet, or using both lowercase and uppercase letters. Or you could do what you say wouldn't make much sense, and represent each digit in decimal, with spaces (or some other non-numerical character) to separate the digits. Algebraist 18:32, 7 September 2009 (UTC)[reply]

[2] Base 64

Also Ascii85

Both from computer science for human input of big numbers..

They're common in URL's too eg youtube http://www.youtube.com/watch?v=wc8bLHznswA (probably base 64)

83.100.250.79 (talk) —Preceding undated comment added 18:57, 7 September 2009 (UTC).[reply]

(I fixed your base 37 number.) --Tardis (talk) 14:57, 9 September 2009 (UTC)[reply]

Presentation of dihedral group

I took abstract algebra 2 years ago but am sitting in on it this year. On Friday, the Professor said that $\langle r,s|r^{n}=1,sr=r^{-1}s\rangle$ is a presentation for $D_{2n}$ , the dihedral group on an $n$ -gon. A student asked if he needed also $s^{2}=1$ and he said he did not. The book, Dummit and Foote, lists it with $s^{2}=1$ , as does the article Dihedral group, and the professor OFTEN makes mistakes, so I am just wondering if he is right. Thanks StatisticsMan (talk) 22:46, 7 September 2009 (UTC)[reply]

The professor is wrong. s has infinite order in his version. Algebraist 22:48, 7 September 2009 (UTC)[reply]

Indeed. Take, for example, the dihedral group D₆. This Mathworld article shows us that

D_{6}=\langle x,y:x^{6}=y^{2}=1,xy=yx^{-1}\rangle .

(In fact, if you're interested, the article gives you the conjugacy classes, the centre of D₆, the commutator subgroup, the abelianisation, the left cosets, and the character table of D₆) ~~ Dr Dec (Talk) ~~ 00:22, 8 September 2009 (UTC)[reply]

As far as I understand, a presentation is not unique. It is simply a set of generators and relations that imply the group MUST be the given group. So, that article says the given presentation must lead to D_6. But, it does not rule out that some presentation with less relations also must give D_6. I know that presentation leads to D_6 already. My question is, does the one with one less relation, given above, also lead to D_{2n}.

As far as Algebraist's comment, I do not understand that. Again, based on my understanding of presentations, it is not possible that the presentation I first listed gives that s has infinite order. The reasoning is simple, D_{2n} satisfies those generators and relations. So, that first presentation certainly does not imply that s has infinite order. It may be that there does exist some group with s having infinite order that satisfies those generators and relations. This would then prove that presentation does not give D_{2n}. But, that presentation certainly does not imply s has infinite order. StatisticsMan (talk) 00:38, 8 September 2009 (UTC)[reply]

You don't seem to understand what a group presentation means. A group presentation does not describe a family of groups, it describes a specific group.

By definition, the group G with presentation

\langle r,s|r^{n}=1,sr=r^{-1}s\rangle

is the quotient of the free group on two generators (r and s) by the (normal subgroup generated by) the given relations. That is, it is the freest group on two generators satisfying those relations. So, if any group H generated by r and s and satisfying those relations has s of infinite order, then s must have infinite order in G, because H is the homomorphic image of G (with r going to r and s to s) and the order of an element can't go up when you take a homomorphic image. H=

\langle r,s|r^{n}=1,sr=r^{-1}s,r=1\rangle

is an obvious example of such a group. Algebraist 00:44, 8 September 2009 (UTC)[reply]

It is possible that I don't understand it exactly :) I will think about this and check out the book for more info on the topic. Thanks. StatisticsMan (talk) 00:51, 8 September 2009 (UTC)[reply]

To put it another way: the group G is the freest group on two generators satisfying those relations. The only relations holding in G are those which must hold in any group in which the given relations hold. Because (for any n), there is a group satisfying the relations given for G with sⁿ≠1, this relation need not hold, and hence does not hold in G. So s has infinite order in G. Algebraist 01:01, 8 September 2009 (UTC)[reply]

You are correct that a given group can (in fact, will) have many different presentations. Proving that a given presentation has the minimal number of generators and relations can be rather difficult, as can proving that two presentations represent the same group. Algebraist's comment about the order of s is sufficient proof that the presentation your professor gave is a different group, though. --Tango (talk) 01:06, 8 September 2009 (UTC)[reply]

Indeed, even the problem of whether a given presentation represents the trivial group is undecidable in general. Algebraist 01:30, 8 September 2009 (UTC)[reply]

I think StatisticsMan's objection to Algebraist's response is fine (Algebraist's reasoning is fine too). SM probably objects to the argument "there is no order relation for s given, so s has infinite order", which is not a very good argument. SM specifically mentioned preferring an explicit group satisfying those relations where s has infinite order; this is a very good attitude to have, as it can be difficult or impossible to say anything about a particular finitely presented group based solely on its presentation. However, it is very easy to give a concrete, explicit example of a group satisfying those relations in which s has infinite order (and presumably so easy that Algebraist didn't see the need to spell it out). The example is just the semi-direct product of an infinite cyclic group ⟨s⟩ with a normal subgroup ⟨r⟩ in which s acts as inversion. Such a group in fact has the presentation ⟨r,s:r^n=1,sr=r^-1s⟩, but it is also quite easy to see it at least satisfies that presentation while s has infinite order. In case semi-direct products have not yet been covered:

To give an even more concrete realization of the group consider the matrix group generated by r = [ z,0,0,0 ; 0,1/z,0,0 ; 0,0,1,0 ; 0,0,0,1 ] and s = [ 0,1,0,0 ; 1,0,0,0 ; 0,0,1,1 ; 0,0,0,1 ], where z is a primitive n'th root of unity in the complex numbers. It is easy to check that r has order exactly n, that sr=r^-1s, and that s has order infinity, since s^(2n) = [ 1,0,0,0 ; 0,1,0,0 ; 0,0,1,2n; 0,0,0,1 ] and s^(2n+1) = [ 0,1,0,0 ; 1,0,0,0 ; 0,0,1,2n+1 ; 0,0,0,1 ]. JackSchmidt (talk) 01:38, 8 September 2009 (UTC)[reply]

The example I gave (Z, with 1=s, 0=r) is even easier. Algebraist 01:46, 8 September 2009 (UTC)[reply]

Ah, too true. I missed it in your second response. Your example is also a good introduction to a much more widely applicable technique called "quotient methods". You find some nice quotient group where you can work, like the abelianization, ⟨r,s:r^n=1,sr=r^-1s,r=1⟩≅Z, and answer the questions there instead of in the unknown finitely presented group.

Well, at least he now has two concrete realizations of the most general group satisfying the presentation his professor described. Both the semi-direct product and the matrix group are actually isomorphic to ⟨r,s:r^n=1,sr=r^-1s⟩. JackSchmidt (talk) 02:05, 8 September 2009 (UTC)[reply]

That's only the abelianization if n is odd. Algebraist 02:14, 8 September 2009 (UTC)[reply]