April 5

What does A^⊕n denote? 76.14.188.230 (talk) 10:27, 5 April 2011 (UTC)[reply]

(just a guess) Maybe ${\displaystyle A\oplus \dots \oplus A}$, n times? Staecker (talk) 11:28, 5 April 2011 (UTC)[reply]

Yes, this is correct. Invrnc (talk) 11:45, 5 April 2011 (UTC)[reply]

Thank you! 76.14.188.230 (talk) 12:12, 5 April 2011 (UTC)[reply]

Resolved

Can anyone recommend a nice BibTex style file, please? I like to use numbers for references, e.g. [1], [2], [3], instead of [Smith99]. I saw a paper where the numbers were in bold: [1] instead of just [1]. I usually put book titles and journal names in italics and in speech marks. To recap, I would like:

• Bolded, number citations.
• Book or journal titles it italics with speech marks.

Also, is there a way to personalise them? I have found one or two that are okay, but they put the journal number as no. 1 instead of No. 1, and I don't like that. How hard is it to write one from scratch? Any suggestions boys and girls? Fly by Night on Tour (talk) 16:15, 5 April 2011 (UTC)[reply]

You will probably find it much easier to tweak existing .bst files than to start from scratch. How easy it is depends wholly on your experience with coding in general, and the idiosyncrasies of *TeX in particular. The file pnas.bst does almost what you want, changing to bold shouldn't be too hard. See file here: [1] SemanticMantis (talk) 17:09, 5 April 2011 (UTC)[reply]

You are right that the coding is quite hard. But I managed to find an automated program that makes one for you. I run TeXnicCentre. I opened CMD (which is a DOS window) and typed latex makebst. It asked me a big long list of multiple choice questions. At the end it made a .dbj file, that it called a "batch file". It gave me the option to run that, which I did, and I now have my own personalised .bst file. However, I made a mistake and told it to put the year after the author. I don't like that and want to change that to putting the date at the end. How would I tweak the .bst file? Fly by Night on Tour (talk) 17:37, 5 April 2011 (UTC)[reply]

Deep in the bst file you will find something like FUNCTION{book}. You will see a list of things inside the curly brackets. That is the order that they are shown. You can move the date to a lower position. Notice that there is a completely different function for articles, manuals, injournals, etc... You may want to change more than just one of them. Personally, I download the bst file for the publication that I am submitting my article to. For example, I'm writing one for IEEE right now, so I'm using the IEEEtran.bst file. -- kainaw™ 18:45, 5 April 2011 (UTC)[reply]

Does anyone know how I might tweak my current .bst file to change the position of the dates? I've run the auto-compiler three times and it won't even create a .dbj output any more. It is obviously unhappy about some of my choices, but can't tell me because it's a DOS window. Fly by Night on Tour (talk) 18:23, 5 April 2011 (UTC)[reply]

Consider functions ${\displaystyle f:X\rightarrow Y}$ on the sets ${\displaystyle X}$ with measure ${\displaystyle \mu _{1))$ and ${\displaystyle Y}$ with measure ${\displaystyle \mu _{2))$. We may say that ${\displaystyle f}$ is well behaved wrt measure if there exists a ${\displaystyle \lambda }$ such that whenever ${\displaystyle S\subset X}$ is a measurable set, ${\displaystyle \mu _{2}(f(S))=\lambda \mu _{1}(S)}$.

Is there a name for such functions? Some special cases are linear maps between ${\displaystyle \mathbb {R} ^{n))$ and ${\displaystyle \mathbb {R} ^{m))$, as well as any function which maps into a set of measure 0. I'm sure there are much more exotic examples. --COVIZAPIBETEFOKY (talk) 16:47, 5 April 2011 (UTC)[reply]

Did you mean a less than or equal instead of equal? Not many linear maps even would satisfy that. Dmcq (talk) 17:13, 5 April 2011 (UTC)[reply]

No, I meant equal. I was under the impression that for a linear map f between ${\displaystyle \mathbb {R} ^{n))$ and ${\displaystyle \mathbb {R} ^{n))$, the absolute value of the determinant ${\displaystyle \lambda =|\det(f)|}$ behaves in exactly this way, and the wikipedia article gives a generalization of that property. Have I missed something? --COVIZAPIBETEFOKY (talk) 19:04, 5 April 2011 (UTC)[reply]

Well, you seem to have started with measures of arbitrary (presumably arbitrary measurable) sets, and retrenched to measures of parallelepipeds. I am not aware of any way to take the determinant of an arbitrary measurable subset of R^n. --Trovatore (talk) 19:09, 5 April 2011 (UTC)[reply]

The determinant of ${\displaystyle f}$, as a matrix! And I explicitly required that ${\displaystyle S}$ be measurable, and by virtue of requiring that ${\displaystyle f(S)}$ have a measure, I implicitly required that to be measurable as well. --COVIZAPIBETEFOKY (talk) 19:14, 5 April 2011 (UTC)[reply]

Right, sorry, I got confused. I thought you meant the determinant itself was the function you were calling "well-behaved". --Trovatore (talk) 19:19, 5 April 2011 (UTC)[reply]

Oh, I see. I was trying to avoid using awkward language. I'll go back and write it out. --COVIZAPIBETEFOKY (talk) 19:24, 5 April 2011 (UTC)[reply]

Reading what has been written it seems that you have a map ƒ : X → Y, where both X and Y have volume forms. A volume form on a k-dimensional manifold is a non-degenerate, skew-symmetric, differential k-form. Assume that X^m is equipped with the volume form ω, while Yⁿ is equipped with the volume form τ. A necessary condition for the map ƒ to be "well behaved" is that the pull back ƒ_*τ = ω; which implies that m = n. After that, it seems that there is a lot of freedom, and it would take some thought. It's a bit late for me. Let me know what you come up with. I'd be interested to hear (read). Fly by Night on Tour (talk) 23:56, 5 April 2011 (UTC)[reply]

Without bothering about manifolds volume is transformed using the Jacobian determinant. Otherwise Measure-preserving dynamical system might be of interest, the constant there is 1 but that probably doesn't matter. But basically as Fly by Night you just have to make sure the measure of the volume form acts like you want. Dmcq (talk) 10:07, 6 April 2011 (UTC)[reply]

It depends how you want to measure volume in the image. If the volume forms are compatible with respect to a flat, torsion-free Koszul connection then you can use the Jacobian. You really have triples ƒ : (X,D,ω) → (Y,∇,τ) where D and ∇ are connections and ω and τ are volume forms. If Dω = ∇τ = 0 and D and ∇ are both flat and torsion free you're okay. If not then you're in trouble. The pull-back condition takes care of that because it expresses the condition in terms of the volume forms themselves, which are then free to vary as they choose. Fly by Night on Tour (talk) 14:25, 6 April 2011 (UTC)[reply]

This is a CS question, but it seems more suited towards the math desk than the computing desk.

I remember reading a story a while back in which some new encryption algorithms were designed that allowed other people to perform operations on your data without being able to see what it actually is. So, for a trivial example, you might encrypt a large number, send it to someone else, have them multiply it by two using this type of algorithm, and send back the encrypted result for you to decrypt, without them ever finding out what the number is. Can anyone dredge up this article for me, or link to more information? Thanks. « Aaron Rotenberg « Talk « 19:16, 5 April 2011 (UTC)[reply]

See Homomorphic encryption, especially the Fully section and its references. Invrnc (talk) 19:27, 5 April 2011 (UTC)[reply]

Looks like what I'm looking for. Thanks. « Aaron Rotenberg « Talk « 19:46, 5 April 2011 (UTC)[reply]

For some situations there are simpler method like blind signatures. 75.57.242.120 (talk) 07:11, 6 April 2011 (UTC)[reply]

Resolved

I wanted to solve the recursive equation ${\displaystyle r_{i}=br_{i-1}-r_{i-2))$, where b is some constant. I got that

${\displaystyle r_{i}={\frac {1}{\sqrt {b^{2}-4))}\left[{\left({\frac {b+{\sqrt {b^{2}-4))}{2))\right)}^{i}-{\left({\frac {b-{\sqrt {b^{2}-4))}{2))\right)}^{i}.\right]}$

This should simplify into a simple polynomial, but all attempts at doing that have failed. I tried representing the term inside the brackets by as a difference of powers, but I still wasn't able to get rid of the square roots. Is there a way to simplify this? 74.15.137.130 (talk) 22:37, 5 April 2011 (UTC)[reply]

What exactly do you mean by a "simple polynomial" in this case? Normally if you speak of a polynomial function of i, you would not have i in the exponent. By "simple", do you mean expressible in terms of integers, so that the radicals wouldn't need to be there? If so, I'm inclined to suspect that such an expression does not exist. Michael Hardy (talk) 23:17, 5 April 2011 (UTC)[reply]

Use the formula

${\displaystyle (x+y)^{i}=\sum _{n=0}^{i}{\binom {i}{n))x^{n}y^{i-n))$

If you use this on both powers, lots of things will cancel out. Looie496 (talk) 23:43, 5 April 2011 (UTC)[reply]

Thanks. 74.15.137.130 (talk) 23:48, 5 April 2011 (UTC)[reply]

Oh. Yes, that's right. But the degree of the thing gets bigger as i increases. Not what I think of when I see the word "polynomial". I guess the poster didn't mean a polynomial in the variable i, but rather, for each value of i separately, a polynomial in b. Michael Hardy (talk) 23:59, 5 April 2011 (UTC)[reply]

Generally ${\displaystyle r_{i}=A_{1}\lambda _{1}^{i}+A_{2}\lambda _{2}^{i))$ where ${\displaystyle \lambda _{1}=a+{\sqrt {a^{2}-1))}$ and ${\displaystyle \lambda _{2}=a-{\sqrt {a^{2}-1))}$ and ${\displaystyle a={\frac {b}{2))}$. When using rational initial values, like ${\displaystyle r_{-1}=1}$ and ${\displaystyle r_{-2}=0}$, then the square roots cancel out and all the ${\displaystyle r_{i))$ turns out rational as they should. The special case ${\displaystyle a=1}$ has the solution ${\displaystyle r_{i}=A_{1}+A_{2}i}$ which is indeed a polynomial in ${\displaystyle i}$ . Bo Jacoby (talk) 11:24, 6 April 2011 (UTC).[reply]

The nth term is a polynomial in b of degree n. These polynomials are related to the Fibonacci polynomials ${\displaystyle F_{n}(x)}$ as follows:

${\displaystyle r_{n}(b)={\frac {F_{n}(ib)}{i^{n))))$

(here, i is the square root of -1). Gandalf61 (talk) 11:00, 6 April 2011 (UTC)[reply]

Is there a convex polyhedron whose vertices correspond to the 15 unordered pairs of elements of a 6-element set { a, b, c, d, e, f}, with an edge between two pairs precisely if they are intersecting subsets of { a, b, c, d, e, f} (thus there would be an edge between ab and ac but not between ab and cf)?

(Each vertex would thus have degree 8, so there would be 8 × 15/2 = 60 edges. Euler's formula V − E + F = 2 would then imply

15 − 60 + F = 2,

so F = 47, i.e. there would be 47 faces.) Michael Hardy (talk) 23:55, 5 April 2011 (UTC)[reply]

The answer is no. The graph you describe is a convex polyhedron only if it is a planar graph. But it's fairly straightforward to construct a subgraph that is a subdivision of the complete bipartite graph K_3,3. I can get one from the subgraph spanned by the eight vertices {ab, ac, cd, bd, ad, be, ef, df}. Specifically, take the complete bipartite graph on the pair of vertex sets {ab,cd,ef} and {ac,bd,df}, and then insert the vertex ad into the (ab,df) edge, and the vertex be into the (bd,ef) edge. The resulting graph appears as a subgraph of the graph in question. Sławomir Biały (talk) 01:18, 6 April 2011 (UTC)[reply]

Very nice. Your answer leaves me wondering why I didn't think of that. I must be getting rusty in some things. Michael Hardy (talk) 01:56, 6 April 2011 (UTC)[reply]

Next question: Is there some visually nice way to display the graph I described? Michael Hardy (talk) 02:03, 6 April 2011 (UTC)[reply]

The graph you've described is the complement of the Kneser graph KG_6,2, so it's quite closely related to the Petersen graph, which is KG_5,2. I'm sure someone has made a nice drawing of KG_6,2. —Bkell (talk) 03:04, 6 April 2011 (UTC)[reply]

I see that the complement of KG_6,2 is called the (6,2)-Johnson graph; there's a bit about Johnson graphs in the Kneser graph article (in the #Related graphs section), but I didn't notice that before. So now you have a name for your graph, at least. Wolfram Alpha will draw a picture of it, but there's probably a more beautiful way to draw it. —Bkell (talk) 03:31, 6 April 2011 (UTC)[reply]

Interesting. Probably Wolfram Alpha's picture will be far better than what I could draw by hand. Thank you, Bkell and Sławomir Biały.

(BTW, Sławomir, am I right in suspecting that the first syllable of your first name is pronounced something like "suave"?) Michael Hardy (talk) 04:11, 6 April 2011 (UTC)[reply]

That's a bit strange, probably worth a look, Mathworld has Triangular Graph with T₆ corresponding to this whereas we have triangular graph redirecting to planar graph as composed of triangles. Dmcq (talk) 08:39, 6 April 2011 (UTC)[reply]

So that picture on Mathworld is it! Thank you. Michael Hardy (talk) 12:50, 6 April 2011 (UTC)[reply]

Another representation is given in [2] where they remove the complete graphs of 5 points. Dmcq (talk) 13:38, 6 April 2011 (UTC)[reply]

I've used Sage_(mathematics_software) in the past to draw graphs. StatisticsMan (talk) 15:48, 6 April 2011 (UTC)[reply]