Wikipedia:Reference desk/Archives/Mathematics/2012 August 19

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August 19

Dirac

Let me try this again. It may depend on the particular field. Looking at ${\vec {v}}=k(\phi /s){\hat {\phi }}$ in particular (cylindrical coordinates), $\int _{C}k(\phi /R){\hat {\phi }}\cdot dC=\int _{0}^{2\pi }k(\phi /R){\hat {\phi }}\cdot R{\hat {\phi }}d\phi =2\pi ^{2}k$ where $C$ is a circle of radius r.

So is it true that the curl of the vector field is $2\pi ^{2}k\delta (s){\hat {z}}$? Widener (talk) 12:30, 19 August 2012 (UTC)
Or $2\pi ^{2}k\delta (s)\delta (\phi ){\hat {z}}$? Widener (talk) 12:30, 19 August 2012 (UTC)

Please define your symbols. The article on cylindrical coordinates uses (ρ,φ,z). The symbol k is evidently a constant. What is R? — Quondum^☏ 13:53, 19 August 2012 (UTC)

I use a different convention where $s:=\rho$. That should be "where $C$ is a circle of radius R". Widener (talk) 14:11, 19 August 2012 (UTC)

Your function ${\vec {v}}=k(\phi /s){\hat {\phi }}$ appears to be multi-valued (or discontinuous at a plane determined by convention: what is the range of $\phi$?). Is this what you meant? — Quondum^☏ 15:25, 19 August 2012 (UTC)

Yes I think it is multivalued; I think that happens when you use curvilinear coordinates. Widener (talk) 15:35, 19 August 2012 (UTC)

If have a single-valued vector field, putting it into any coordinate system will not make it multivalued. You might have meant ${\vec {v}}=(k/s){\hat {\phi }}$, rather than one that keeps growing each time you trace around the z-axis? The one you have has an essential singularity on the z-axis, making things particularly difficult. In some sense your path of integration (the circle) does not end where it starts, and the integral will be different depending on which point you choose to start your integration. — Quondum^☏ 16:03, 19 August 2012 (UTC)

You make an excellent point. It is definitely $k{\frac {\phi }{s}}{\hat {\phi }}$ though. Widener (talk) 17:00, 19 August 2012 (UTC)

My only guess is that it should really be $k{\frac {\phi \%2\pi }{s}}{\hat {\phi }}$ where % is modulo. Widener (talk) 17:08, 19 August 2012 (UTC)

I'm severely rusty at this, but how does this look?:

$\int _{C}{\vec {v}}\cdot dC=k\int _{0}^{2\pi }\phi d\phi =k\left[{\frac {\phi ^{2}}{2}}\right]_{0}^{2\pi }=2\pi ^{2}k$

To illustrate what I was saying about where the circle starts, if the discontinuity is at $\phi =\pi$ instead, we get

$\int _{C}{\vec {v}}\cdot dC=k\int _{-\pi }^{\pi }\phi d\phi =k\left[{\frac {\phi ^{2}}{2}}\right]_{-\pi }^{\pi }=0$

It seems from Stokes' theorem that away from the z-axis the curl of the field is zero (even on the surface discontinuity where $\phi =0$ for the first case or $\phi =\pi$ for the second case), so the curl is all concentrated at an impulse in the form of $2\pi ^{2}k\delta (x)\delta (y){\hat {z}}$ (which is independent of $\phi$) in the first case, and is zero everywhere (even on the z-axis) for the second case. It remains to convert this central impulse into cylindrical coordinates, but the transformation is ill-behaved on the z-axis. I suspect that you get something like $\infty k\delta (s){\hat {z}}$. I'll expand on this if need be. Please note that I do not have full confidence in my detail workings here. — Quondum^☏ 03:23, 20 August 2012 (UTC)

Irregular Polyhedron

Dear all,
I need 5 examples of irregular polyhedron. I've searched all around the web and wikipedia, and I got 5 examples of regular Polyhedron, but the irregular ones is nowhere to be found.
Can someone please help me? — Preceding unsigned comment added by 197.69.6.1 (talk) 15:24, 19 August 2012 (UTC)

They are not normally called "irregular polyhedron" where they are discussed. Think of various familiar flat-sided figures. Did you look up polyhedron? — Quondum^☏ 15:40, 19 August 2012 (UTC)

Take a regular polyhedron and cut it into two unequal pieces, you'll get two irregular polyhedra. -- SGBailey (talk) 19:04, 19 August 2012 (UTC)

Let me think of some real-world examples:

1) Any box which isn't a cube. That is, not all the edge lengths are the same. You will find many examples of this shape.

2) A geodesic dome. Since the edge lengths aren't all quite the same, it's irregular.

3) An Egyptian pyramid. Since the base is square and the sides triangular, it's not a regular polyhedron.

4) Pretty much any building without curves (because, if it had curves, it wouldn't be a polyhedron at all). The Pentagon is an example. While the pentagon is a regular polygon (neglecting the hole in the center for the atrium), the result of extruding it is not a regular polyhedron, since the sides of the buildings are rectangles, not pentagons. Then there's also the atrium, which would make it an irregular polyhedron, in any case.

Can you think of the last one ? StuRat (talk) 21:42, 19 August 2012 (UTC)

A pleasing one is the result of a cube being cut in half by a plane through a pair of opposite edges: five faces (a non-square rectangle, two squares, two right-angled isosceles triangles).←86.139.64.77 (talk) 10:19, 20 August 2012 (UTC)

Plenty of non-regular polyhedra! —Tamfang (talk) 04:12, 21 August 2012 (UTC)

Actually, any Archimedean solid would work as an example of an irregular polyhedron. Double sharp (talk) 15:07, 23 August 2012 (UTC)

Help me to calculate the amount of possibilities of a audio cd.

How is the amount of possibilities in a Cd.

CD INFO

Max amount of songs=99

Minimun amount seconds a song can have=4

Max lenght of a cd= 79.8 minutes

Each second is made of 44100 samples for the right side and 44100 for the left side, a cd can't be mono.

Each sample is a value and there are 65536 values.201.78.144.152 (talk) 01:38, 20 August 2012 (UTC)

You have the wrong value for the "Minimum amount seconds a song can have=44100". That's 12 hours and 15 minutes. StuRat (talk) 21:17, 19 August 2012 (UTC)

Let's start with the number of possibilities in a second. Using your data, that would be 65536⁴⁴¹⁰⁰. The result is larger than my calculator can handle.

If we then account for the total number of seconds on a CD (60×79.8=4788), we can then get the total number samples on a CD (4788×44100=211150800). This gives a total of 65536^211150800 possible music CDs. So, once you buy that many, you never need buy another ! StuRat (talk) 21:23, 19 August 2012 (UTC)

For reference, 65536⁴⁴¹⁰⁰ = 5.82023479750073295820859739141...*10²¹²⁴⁰⁶ and 65536^211150800 = 2.84293638246956144950010179132...*10^486192684. -- Meni Rosenfeld (talk) 04:11, 20 August 2012 (UTC)

Yes I got the minimum amount of seconds a song can have wrong, its 4 seconds, changed that.201.78.144.152 (talk) 01:38, 20 August 2012 (UTC)

Not quite, Meni & StuRat. Audio CDs are stereo, that is, 44100x2x16 bits per second.

65536^2x211150800. - ¡Ouch! (^{hurt me} / _{more pain}) 08:17, 22 August 2012 (UTC)

Correct. I missed that part. StuRat (talk) 08:47, 22 August 2012 (UTC)

A finite group

A finite group contains an element $x$ of order 10 and an element $y$ of order 6. What can be said about the order of the group?
All I can think of is that the order of the group must be at least 15. Can YOU think of more restrictive restriction on the order of the group? Widener (talk) 20:01, 19 August 2012 (UTC)

It would have to be of order 30 at least because the orders must divide the order of the group. An abelian group generated by $a^{3}=1$ and $x^{10}=1$ works as $(ax^{5})^{6}=1$ but no lower power equals the identity. Dmcq (talk) 20:33, 19 August 2012 (UTC)

In fact, it would have to be a multiple of 30 ! ! Widener (talk) 20:41, 19 August 2012 (UTC)

If you knew that then why did you ask the question in the first place? — Fly by Night (talk) 22:22, 19 August 2012 (UTC)

Sure, I can. But Lagrange already did all the job. Incnis Mrsi (talk) 20:34, 19 August 2012 (UTC)

That abelian group is in fact the same as $x^{30}=1$, and there's no other group of order 30 satisfying the conditions. Dmcq (talk) 00:12, 20 August 2012 (UTC)

How do you prove that there are no non-abelian groups of order 30 with elements of order 6 and 10? With list of small groups one can see that non-abelian groups of order 30 are certainly possible, but I cannot quickly construct an example with such property. Incnis Mrsi (talk) 13:44, 20 August 2012 (UTC)

There's only four groups of order 30, the cyclic one above Z₃₀, plus three non-abelian ones using a dihedral group - D₁₅, D₅xZ₃, D₃xZ₅. To show that takes a little work so I searched on the web and found [1] (the whole thing not just the first answer) - basically you have to show there is a cyclic normal subgroup of order 15 using the Sylow theorems. To get a factor of two in the order of an element you have to use an element of the dihedral group along with the associated cyclic group. You get order 6 elements with D₅xZ₃ but not order 10, and you get order 10 elements with D₃xZ₅ but not order 6. Possibly one could get there more directly without establishing there are just four possible groups. Dmcq (talk) 10:38, 21 August 2012 (UTC)

I think I have a more direct proof. Suppose G is group of order 30 with elements x of order 10 and y of order 6. Let H = <x> and K = <y>. If H∩K was trivial then HK would have 60 elements, so we must have x⁵ = y³ = z say. Hence HK has 30 elements and so G = <x, y> and z is central. G/<z> has order 15 and every group of order 15 is cyclic, so G/<z> is abelian and therefore [x, y] ∈ <z>. If we want G to be non-abelian then we must have [x, y] = z, that is x^-1y^-1xy = x⁵ and so y^-1xy = x⁶. But x⁶ has order 5 so it cannot be a conjugate of x.

60.234.242.206 (talk) 12:07, 21 August 2012 (UTC)

Works for me okay. Dmcq (talk) 13:21, 21 August 2012 (UTC)