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August 9[edit]

Constructing Triangles[edit]

Can we construct a triangle whose perimeter is given and two of its angles are given. If yes, please give the steps of construction and its justification too. -- $\xi ri\pi i\vartheta \alpha \xi$ 04:15, 9 August 2009 (UTC)[reply]

Such a triangle always exists, yes. What do you mean by "construct", though? Do you just want to know the side lengths, or do you have something more like Compass and straightedge constructions in mind? Algebraist 04:50, 9 August 2009 (UTC)[reply]

Yes, I have compass and straightedge constructions in mind. --ξΓιΠιγαξ 04:58, 9 August 2009 (UTC)[reply]

Not only can one say that it exists (provided of course the sum of the two given angles is not too big and the perimeter is positive) but also that it is unique. But I think if you want a compass-and-straightedge construction, that might depend on what the angles are. Michael Hardy (talk) 05:36, 9 August 2009 (UTC)[reply]

Assuming you have a way to construct the angles, you can make a similar triangle to the one you want, and find the perimeter of that. By using the ratio of the perimeter you want and the perimeter of your similar triangle, you can find one of your side lengths from the side length of the similar triangle. Rckrone (talk) 05:43, 9 August 2009 (UTC)[reply]

Law of sines is relevent here.83.100.250.79 (talk) 12:42, 9 August 2009 (UTC)[reply]

You'll have two big issues with your compass and straightedge constructions. First is that it's a compass and not a protractor. It is exceedingly difficult to construct arbitrary angles. Small, whole number fractions of pi radians are doable, but trying to construct an angle like 0.923876 radians is tedious at best. Secondly, it's a straightedge, not a ruler. Compass and straightedge constructions are usually taken to be scale invariant. That is, the internal scaling is self-consistent, but the absolute size is arbitrary. You'd need to have some sort of external scale bar to reference off of for the size of your perimeter. That said, one way to do the construction is to draw a baseline, construct the two angles facing each other, sharing the common baseline. Extend the sides until they meet at the third point of the triangle. Next, you'll have to to scale the triangle appropriately. How you do that will probably depend on how you're measuring the perimeter, and what you're comparing it to. Note that scaling bit can be avoided if you pre-compute the length of the base, and construct the angles that distance apart. -- 76.201.158.47 (talk) 17:44, 9 August 2009 (UTC)[reply]

There are different things that can be meant by saying that a length or angle is "given." It's true that there's no way to establish an overall objective scale with compass and straight edge, so if you "have a length" in the sense that you have a number measurement there's no way to produce that. But you can also "have a length" in the sense there are other elements in the construction that need to have the same length. If you're drawing a triangle in a vacuum, the scale is arbitrary, but if your triangle is part of a larger construction it is meaningful to require for instance that the perimeter be the same as the diameter of some circle you've drawn. A similar story holds for angles. If you "have an angle" in the sense that you have a measurement like π/5 radians, it may be difficult or impossible to construct the angle on paper, but if you "have an angle" in the sense that the angle has already been constructed somewhere, then its easy enough to copy it somewhere else. Presumably if you've already constructed it, then it's an angle that can be constructed, but having knowledge of how to construct it isn't necessary in order to copy it. In that sense, if you are given a paper with two angles and a length drawn on it, whether they be part of another compass and straightedge construction, or just elements floating on the page, you can use compass and straightedge construction to build a triangle with those angles and that perimeter. If you are given a paper with "26°, 75°, 14cm" written on it, the length can't be found without a ruler and the angles may be problematic. Rckrone (talk) 19:32, 9 August 2009 (UTC)[reply]

I'd have thought it goes without saying that in a straightedge-and-compass problem, saying that the length is given means that there's a segment of that length and you can set the center at one endpoint and take the other endpoint to be on the circumference, so you can draw that circle. Michael Hardy (talk) 09:10, 10 August 2009 (UTC)[reply]

the 24 cells of the 24-cell and their intersections[edit]

Hello, I studied 24-cell but I am still not sure about this. As I understand [1], the 24 vertices can be explicitly described in a fourdimensional affine space over the reals. If I'm not mistaken, the 24 cells correspond with the 24 affine three-dimensional subspaces, with equations $+/-x_{i}+/-x_{j}=1$ . That seems to make sense: each of the 24 vertices described in that article are one exactly 6 such 3-spaces, and every such 3-space contains exactly 6 of those vertices. But I have a problem with the intersection. It seems that if I fix one of these affine 3-spaces, there are exactly 8 others meeting it in 3 vertices, 6 meeting it in 1 vertex, and nine in no vertices at all. This surprises me as I had expected, based on more combinatorial texts, that there would be 8 meeting it in a plane (3 points), 6 meeting it in a line (2 points), 8 meeting it in a point and finally one meeting it in no points at all.

I must have a mistake somewhere, but I found it hard to visualize such a four-dimensional structure. I would be very happy if someone could tell be a bit mor about this.

Many thanks, Evilbu (talk) 13:17, 9 August 2009 (UTC)[reply]

I don't think it would be possible to have 6 other cells meeting a given cell only in a line and still have the whole thing regular. An octahedron has 12 edges, so the number would have to be a multiple of 12. The angle at each edge of a regular octahedron is >90°, which means there has to be fewer than 4 cells around each edge for it to be convex, so there must be exactly 3. That would mean that no two cells share only an edge. 3 cells around each edge requires 6 cells around each vertex: 4 that share a face and 1 that shares only a vertex with a given cell. Since an octahedron has 6 vertices, that makes 6 other cells that share only a vertex. Rckrone (talk) 16:46, 9 August 2009 (UTC)[reply]

Thanks for your answer, it seems to confirm correctness of my approach with coordinates. In any case, I still expect that, "from the point of view of one fixed cell", there are five types of cells, of size 1,8,6,8,1. So that means that there would be two "kinds" of disjointness (parallel and not parallel). When I said that I was also thinking of more combinatorial texts, like: [2] where metasymplectic spaces are considered. One axiom is: "(M1) the intersection of distinct symplecta is empty, a point, a line, or a plane." My intuition made me very convinced that this 24-cell is an example of such a metasymplectic space, with the 24 points, 96 lines or edges, 96 planes or faces, and finally 24 "symplectons" or cells. (Note that the name 24-cell does not appear in that pdf, but the term "thin F4 geometry" does, and F4 is precisely the "Coxeter group" given in the article 24-cell). So somehow I must have a wrong intuition or something like that? Many thanks,Evilbu (talk) 17:22, 9 August 2009 (UTC)[reply]

Functional square root[edit]

What is the functional square root of the exponential function $e^{x}$ and how is it derived? Thanks.--12.48.220.130 (talk) 20:05, 9 August 2009 (UTC)[reply]

Interesting stuff here and here, for instance. —JAO • T • C 20:14, 9 August 2009 (UTC)[reply]

I typed 'Functional square root' into the search box at the top of the page in wikipedia and guess what? it came up with Functional square root rather than a list of possibilities. Amazing thing the search facility. If wikipedia fails (perish the thought) one can always try google. Dmcq (talk) 11:44, 10 August 2009 (UTC)[reply]

Correct but irrelevant to the question the OP asked. Although the article defines what a functional square root is, there is no indication in the question that the questioner was confused on that point. He wanted to know the functional square root of $e^{x}$ . The article contains no mention of

e^{x}

at all. He also wanted to know how to derive a functional square root. Again, the article in no way covers how, given a function, one finds the functional square root of it. - Amazing thing basic reading comprehension skills ;-). - Please save your snark for cases where the obvious articles clearly answers the questions. (Or, better still, don't use it at all.) -- 128.104.112.100 (talk) 18:06, 12 August 2009 (UTC)[reply]

By the way, there is no unique functional square root, it can be built up of chunks. Also if you try turning it into a power series and assume it is going to work everywhere then the derivatives diverge. Not pretty. e^x-1 is a little nicer to work with if you want to give it a go yourself but still nothing very clean. Dmcq (talk) 11:52, 10 August 2009 (UTC)[reply]

I just did a search myself with Hellmuth Kneser and superlogarithm and half-terate and it gave more information. Seemingly what he did was find a fixed point of the complex exponential e^z rather like 0 is a fixed point of e^x-1 and then just expand round there. A bit more than that but that's the general idea. Dmcq (talk) 13:43, 10 August 2009 (UTC)[reply]

The article doesn't mention how to derive the functional square roots or provide examples of them, except for 2x^2. That's why I asked.--12.48.220.130 (talk) 20:16, 10 August 2009 (UTC)[reply]

Ditto my question. Anyone else take a shot at it? --12.48.220.130 (talk) 19:32, 14 August 2009 (UTC)[reply]

A way of going round it that's easy to think about when you have a fixed point like 0 for 1-e^x is to assume you have a series a+bx+cx²... and plug it into itself at that point. For this case we have :a+b(a+bx+cx²..)+c(a+bx+cx²..)² ... = x+x²/2+...

a(1+b+...)=0 so a=0

bb=1 so b=1

bc+cbb=1/2 so c=1/4

etc etc so we get x+x²/4+...

Not the easiest way to generate a series and not easy to analyse and in fact the sequence eventually diverges nastily in this case. Dmcq (talk) 21:35, 14 August 2009 (UTC)[reply]

That was a way of trying to get an analytic function satisfying the condition. If you don't need an analytic function there can be lots of other untions that satisfy the functional equation. For instance for e^x you just need to take a point between 0 an 1 say 0.5 Define f on 0..0.5 as some monotonic function going from 0.5 to 1. Then define f for 0.5 to 1 as e^x for f(x) where x ranges from to 0.5. Then the rest of the line follows firsly straightforwardly. There is quite a bit of arbitariness about such a function though. Dmcq (talk) 21:48, 14 August 2009 (UTC)[reply]

Finding a Permutation Matrix[edit]

Suppose we have two nxn Adjacency matrixes A, and B and we want to find a Permutation matrix P such that

$A=PBP^{T}$ , we also know that since P's Eigenvectors are Orthonormal

$P^{T}=P^{-1}$

We also are given that at least 1 such P exists but not that P is necessarily unique though it might be. In the article on Permutation matrix, there is a section that describes a method of doing this, but as noted the process is apparently flawed.

So my question is what would be a correct method? Also, what makes P unique? A math-wiki (talk) 22:45, 9 August 2009 (UTC)[reply]

This is just the graph isomorphism problem. For practical problems, nauty is a very good solution. This is somewhat of a FAQ. You can see the most recent incarnation at Wikipedia:Reference desk/Archives/Mathematics/2008 November 2#Equivalence_of_0-1_matrices. JackSchmidt (talk) 00:13, 10 August 2009 (UTC)[reply]