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February 2

pentiminoes

Can someone show me how to fit 12 diffrent penteminoes into a six by ten rectangle in 5 diffrent ways?

You might want to try and look at the article on pentaminoes. The question sounds like homework, so I'm not going to help you further. Oskar 07:16, 2 February 2007 (UTC)[reply]

Well, the proper article would be pentomino. — Kieff | Talk 15:51, 2 February 2007 (UTC)[reply]

Here's one possiblity as a starter:

   ***@@@@@**
   *%::::%%:*
   *%%%:%%::*
   @@*%@%*:@*
   @***@**:@@
   @@*@@@**@@

– b_jonas 19:04, 2 February 2007 (UTC)[reply]

Poincaré conjecture

The Poincaré conjecture says that if a 3space is enough like a 3sphere, then it is like a 3sphere. Um, so what does that mean? If a Poincaré-violating manifold (PVM) exists, what properties might it have that distinguish it from a 3sphere? —Tamfang 08:14, 2 February 2007 (UTC)[reply]

The conditions that make a 3-dimensional space "close enough" to a 3-sphere for the Poincaré conjecture to apply are:

The space must be a manifold - if you look at a small enough neighbourhood of any point, it will resemble Euclidean 3-space.
The space must be simply connected - given two points in the space, any path between these points can be continuously transformed within the space into any other path between the same two points.
The space must be a compact space - roughly speaking, it has a finite size and does not extend to infinity in any direction.
The space must be without boundary - roughly speaking, it doesn't have any edges where you can suddenly fall out of the space.

A Poincaré-violating manifold would have to posess all of these properties, yet still fail to be topologically equivalent to a 3-sphere. I'm not actually sure what topological property might allow you to easily distinguish such a manifold from a 3-sphere. Gandalf61 11:33, 2 February 2007 (UTC)[reply]

seeking hyperbolic cookbook

It appears that the algebraically easiest way to do stuff in hyperbolic geometry is with the hyperboloid model, in which each isometry apparently can be expressed with a matrix. But which matrix? I've read a couple of books on h.g. and plenty of websites, but found nothing that goes beyond "it can be done." Where do I find a book (or website) that will tell me:

Given a line, what matrix expresses a unit translation along that line?
Given a plane, what matrix expresses a reflexion in that plane?

...and like that. I might be able to work it out for myself, given free time and a better night's sleep than I've had in recent years! —Tamfang 08:47, 2 February 2007 (UTC)[reply]

If you want to experiment with the hyperbolic plane, a nice place to start is

Stillwell, John (1992). Geometry of Surfaces. Springer-Verlag. ISBN 978-0-387-97743-0.

One reason we use so many different models is because each has its virtues. Also remember that all isometries can be composed using reflections (reflexions) only. In the hyperboloid model, a hyperbolic line corresponds to a plane of the ambient space, and any isometry must map the surface to itself. --KSmrq^T 18:23, 3 February 2007 (UTC)[reply]

Does the Stillwell book contain answers to the specific questions I ask, or only principles that I already know as in your paragraph? —Tamfang 00:41, 4 February 2007 (UTC)[reply]

I did find the knowledge I needed (though no matrices) in Euclidean and Non-Euclidean Geometry: An Analytic Approach by Patrick J. Ryan, published circa 1986 iirc. —Tamfang (talk) 00:20, 1 August 2023 (UTC)[reply]

triangle angle bisectors2

From Wikipedia:Reference_desk/Mathematics#triangle angle bisectors

Somebody asked "is there an expression to get each side of the triangle from given values of the angle bisectors?…86.132.237.140 21:07, 29 January 2007 (UTC)"

I've got this equation

(C²-ab)(a+b)²/c = (B²-ac)(a+c)²/b = (A²-bc)(c+b)²/a

Where capitals are bisector lengths, lower case are side lengths..

Now I could solve this.. But given that I usually fail to manipulate equations through more than two steps without making a mistake could someone check that this is correct.. Or, is there a better way to do this? Thanks87.102.4.6 11:28, 2 February 2007 (UTC)[reply]

I've just read an article in the American Mathematical Monthly of Jan 94 which shows that, given 3 arbitrary values, a unique triangle exists which has these as angle bisectors. As the easily-obtained equations giving the bisectors separately in terms of the sides were not manipulated into a vice versa form, I rather assume that this is not possible, however.86.132.238.155 15:44, 2 February 2007 (UTC)[reply]

Statistics/Normal Distribution

In a mathematics examination the average grade was 82 and the standard deviation was 5.All students with grades from 88 to 94 received a grade of B.If the grades are approximately normally distributed and 8 students received a B grade,how many students took the examination"""" 12:03, 2 February 2007 —Preceding unsigned comment added by 203.128.5.84 (talk • contribs)

For a normal distribution centered on 82 you need to know the fraction of the distribution (a value between 0 and 1 - in this case less than 1) falling between 88 and 94 when the deviation is 5. You can get the equation of the distribution when the average is 82 and the deviation is 5. Then you need to integrate the equation between 88 and 94 (or find the fraction using look up tables..) this gives you the fraction of the students getting a B. Then divide 8 by that fraction to get the total number of students...87.102.4.6 12:44, 2 February 2007 (UTC)[reply]

Note that (from Variance) the "the square root of the variance, called the standard deviation" - therefor your variance is 5x5 = 25.87.102.4.6 12:50, 2 February 2007 (UTC)[reply]

Therfor you need to integrate the below equation with respect to x between 88 and 94, with μ=82, σ=5, σ²=25.(I can't think of a simpler way - but there usually is..)87.102.4.6 12:54, 2 February 2007 (UTC)[reply]

${\frac {1}{\sigma {\sqrt {2\pi }}}}\;\exp \left(-{\frac {\left(x-\mu \right)^{2}}{2\sigma ^{2}}}\right)\!$

The form of the OP hardly suggests that the questioner is going to find integration a meaningful idea, given the essential triviality of the problem. I don't think it unreasonable to tell him/her that the key will be to use Normal tables with z=1.2 and 2.4—86.132.238.155 15:56, 2 February 2007 (UTC)[reply]

Hyperbolic geometry

I too have a simple question about hyperbolic geometry.. What lines are considered to be 'straight' in hyperbolic geometry say of a Hyperboloid surface? It seem to me that for spherical geometry great circles are the equivalent of straight lines - since they represent the longest distance.. For a hyperboloid (of one sheet)

{x^{2} \over a^{2}}+{y^{2} \over b^{2}}-{z^{2} \over c^{2}}=1

(hyperboloid of one sheet),

are lines in the plane x/y=constant condsidered to be the equivalent of straight lines (and also parallel)(think yes)?

What about the circles formed by intersection with z=constant - are two circles formed by z=a,z=b considered to be parallel - and are they considered to be straight lines(thinks not)?

I'm using the term 'straight line' to mean the shortest distance between two points on the surface - I looked at the page and understood I think the parallel concept - but didn't get how to find a line that is the equivalent of a straight line in conventional geometry.

If you can make an answer simple I would appreciate it.87.102.4.6 12:15, 2 February 2007 (UTC)[reply]

First, it's not a hyperboloid of one sheet, it's one sheet (chosen arbitrarily) of a hyperboloid of two sheets.

?? Are you sure about this - this image shows only one sheet - that's the equation I gave? Image:HyperboloidOfOneSheet.PNG 87.102.4.6 16:20, 2 February 2007 (UTC)[reply]

True, that's the equation you gave, and I ought to have looked closer at it, rather than assuming that you had in mind the hyperboloid model of the hyperbolic plane. Since I can't tell you anything about geodesics on the hyperboloid of one sheet (other than to agree with your guess about both questions above), I'd delete my remarks below but that's bad form. —Tamfang 19:49, 2 February 2007 (UTC)[reply]

That's ok. thanks.83.100.183.48 20:05, 2 February 2007 (UTC)[reply]

In this model, hyperbolic straight lines are represented by planes passing through the origin, ax+by+cz=0, or (if you prefer) by the intersections of such planes with the hyperboloid. Note that this also describes great circles on a sphere!

(By the way, great circles are "straight" not because they are the longest possible "lines" on the sphere but because the shortest curve between two points on a sphere is a segment of a great circle.)

The lines that you describe are a valid subset of these, with c=0, but they're not parallel: they all intersect at x=y=0. A parallel set would "intersect" in a point on the circle (a line on the cone)

x^{2}+y^{2}=z^{2}

. —Tamfang 16:06, 2 February 2007 (UTC)[reply]

Given that x=y=0 doesn't have a point on the hyperboloid I described would you reconsider your last point - to be honest you've confused me a bit - are you talking about the same thing - look at the picture..87.102.4.6 16:34, 2 February 2007 (UTC)[reply]

I am not aware of any model of hyperbolic geometry based on the hyperboloid of one sheet. As Tamfang says, the model that is usually called the "hyperboloid model" is based on one sheet of a hyperboloid of two sheets. Other standard models of hyperbolic geometry are the Klein model, the Poincaré disc model, the Poincaré half-plane model and the pseudosphere. To define a model based on the hyperboloid of one sheet, you would have to define a family of curves within the surface that meet the axioms of "straight lines" in hyperbolic geometry. The curves x/y=constant within the surface could be a sub-set of this family, but there are not enough of them - there is only one curve through each point, and most pairs of points within the surface are not connected by one of these curves. The circles z=constant are certainly not candidates for "straight lines" because they are finite whereas all "straight lines" in hyperbolic geometry can be extended to infinity (as in Euclidean geometry). Gandalf61 11:07, 3 February 2007 (UTC)[reply]

Thanks.87.102.9.55 11:44, 3 February 2007 (UTC)[reply]

summation of fractions

Σ(1/n) = ?

Σ(1/n²) = ?

Σ(1/n³) = ?

—The preceding unsigned comment was added by 59.95.71.153 (talk • contribs).

Please, do your own homework. Meanwhile, take a look at summation and infinite series. — Kieff | Talk 15:47, 2 February 2007 (UTC)[reply]

For the sums to infinity see..

Σ(1/n) = ? See Harmonic series (mathematics)

Σ(1/n²) = ? See Basel problem

Σ(1/n³) = ? See Apéry's constant

whether or not this will help is another thing - also see Riemann_zeta_function#Various_properties or just Riemann_zeta_function. Good luck87.102.4.6 16:26, 2 February 2007 (UTC)[reply]

Equations

What is the best way to solve equations in fractions? Cam you please show me ways of solving it? Thank you if you have time

It would help if you could be a little more specific - take a look at the pages linked to by this link http://ww.google.co.uk/search?hl=en&q=algebra+fraction&meta= do any of the first couple of pages have the sort of thing you're asking about. - if so tell us and I'll try to give you a good point to start learning.

Also maybe this http://www.tech.plym.ac.uk/maths/resources/PDFLaTeX/alg_fracs.pdf is the sort of thing you want..

I'm assuming that you already understand basic fractions - if so this link should all amke sense http://www.mathleague.com/help/fractions/fractions.htm

Or just type in an example of the type of thing you want to solve...83.100.183.48 18:23, 2 February 2007 (UTC)[reply]

Twin Primes

The sequence 4, 6, 12, 18, 30, 42, 60, 72, 102, 108 is related to twin primes, that much I know. I just don't know the exact nature of their relation, so any help would be appreciated.

Take a look at Twin_primes#The_first_35_twin_prime_pairs - the sequence should become clear - if you can't see what it is ask for a hint...83.100.183.48 18:26, 2 February 2007 (UTC)[reply]

Just type the word sequences in a search engine search box ... you'll find The On-Line Encyclopedia of Integer Sequences [OEIS] which tells you when you paste your sequence : "A014574 Average of twin prime pairs."

Or just enter your sequence in any search box, to find that the first hit is : "Twin Primes -- from Wolfram MathWorld - The first few twin primes are n+/-1 for n==4 , 6, 12, 18" ... -- DLL ^{.. T} 11:08, 3 February 2007 (UTC)[reply]

OEIS brings up this, which i dont understand either. Can someone explain this in simpler terms? Omnipotence407 03:17, 4 February 2007 (UTC)[reply]

Try writing the twin prime pairs out in a list, and then write your sequence above it. Compare each pair in the list with the corresponding element in the sequence. It should be obvious what the connection is. Maelin (Talk | Contribs) 07:03, 4 February 2007 (UTC)[reply]

I guess its not obvious enough for me. Omnipotence407 03:19, 5 February 2007 (UTC)[reply]

Okay, here are the two lists written out.

{\begin{matrix}{\mbox{Twin prime pairs: }}&(3,5)&(5,7)&(11,13)&(17,19)&(29,31)&(41,43)&(59,61)&(71,73)&(101,103)\\{\mbox{Your sequence: }}&4&6&12&18&30&42&60&72&102\end{matrix}}

Can you see the connection now? Maelin (Talk | Contribs) 04:02, 5 February 2007 (UTC)[reply]

Wow, yeah, my bad. I was thinking too hard. Thanks Omnipotence407 14:36, 5 February 2007 (UTC)[reply]