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January 22[edit]

about infinity[edit]

why do we define something like infinity when you dont reakky know it's value —Preceding unsigned comment added by Joy.vivin (talk • contribs) 06:45, 22 January 2009 (UTC)[reply]

"Infinity" in mathematics is completely another thing, that has nothing or very little to do with the meanings it has in various philosophies or in poetry etc. In mathematics both the substantive "infinity" and the adjective "infinite" are used in various situations, in a technical and precise meaning, usually very simple. You can find something in the article here, but maybe it is not the "infinity" you are interested to --131.114.72.215 (talk) 09:23, 22 January 2009 (UTC)[reply]

Just ask yourself "Why do we define the number -1, -2, -3, etc, when you don't really know its value": it's true! When you say -5 you "really" only know the value 5, and the negative sign tells you more information about how to use the value you "really" know. So the answer to why we DEFINE something like infinity, or -5 (one of which mathematics considers a number, the other not -- neither is in the intuitive sense, and for tens of thousands of years negative numbers didn't have a value in anyone's mind) is because it is useful. —Preceding unsigned comment added by 82.120.227.136 (talk) 10:27, 22 January 2009 (UTC)[reply]

Maths tries to conquer it. As Sun Tzu says "All warfare is based on deception". Dmcq (talk) 15:14, 22 January 2009 (UTC)[reply]

Physics: power of water pump[edit]

A pump, taking water from a large reservoir, is used to spray a jet of water with speed 20 m/s and radius 0.05 metres, from a nozzle level with athe surface of the reservoir. Calculate the power of the pump.

Answer: 31.4KW

Need the procedure or solution derivation work please. —Preceding unsigned comment added by 202.72.235.202 (talk) 13:25, 22 January 2009 (UTC)[reply]

So, every second the pump sprays a cylinder of water 20 m long and 0.05 m in radius. What is the volume of this cylinder ? What is its mass (you need to know that the mass of 1 m³ of water is 1000 kg) ? If this mass of water is moving at 20 m/s, what is its kinetic energy ? How much kinetic energy does this water have in the reservoir ? So how much kinetic energy does the pump add to this mass of water ? Does the pump give the water any potential energy when it removes it from the reservoir ? So, how much work is the pump doing on the water per second ? That gives you the pump's power. Gandalf61 (talk) 13:39, 22 January 2009 (UTC)[reply]

YOU ARE THE MAN! I have been trying to do this damn problem for hours, drawing the problem diagrametically but saw that we needed mass to find power (kgm²)/(s³). But you figured out that the pump must be cylindrical. Found volume, mass, KE of water at 20 m/s. Power found at 31.4 KW. Thanks man! APPRECIATE IT! —Preceding unsigned comment added by 202.72.235.202 (talk) 15:31, 22 January 2009 (UTC)[reply]

Resolved

It's nice that you want to thank someone for helping you at the reference desk but please stay calm. :) --PS T 13:31, 26 January 2009 (UTC)[reply]

Commuting elements in SO(3)[edit]

How would one best find out/show which elements in SO(3) commute? Obviously 2 rotations around the same axis commute, but I know that only from intuition - do any others commute?

I thought I might be able to use eigenvectors to explore the problem, but I couldn't see any obvious way of doing so.

Many thanks,

131.111.8.97 (talk) 14:25, 22 January 2009 (UTC)Zant[reply]

Every non-trivial rotation has exactly one (up to scaling) eigenvector with value 1, its axis. Let A, B be nontrivial rotations with B having eigenvector v. What is the axis of ABA^-1? Algebraist 14:33, 22 January 2009 (UTC)[reply]

If one did not have any geometric idea of what SO(3) was, then one could still solve the problem using eigenvectors. Elements of SO(3,R) are normal matrices, and so if they commute they are simultaneously unitarily diagonalizable. Since their determinants are 1, the possibilities are any {1,z,1/z}, where |z|=1 (use that if the eigenvalues are real, |z|=1 means one either has the identity matrix or eigenvalues 1,-1,-1, and if there are complex eigenvalues, they come in complex pairs, so |z|=1 forces the real eigenvalue to be 1). Now the eigenvector associated to 1 is probably that thing you guys call an axis. In the complex eigenvalue case you need the axes to be the same, but in the real eigenvalues case there are probably some others. I think the symmetry group of an expensive geometry textbook has some examples. JackSchmidt (talk) 15:35, 22 January 2009 (UTC)[reply]

Algebraist's approach is a good one, but you need to be wary of special cases. Apart from rotations about the same axis, there's one other type of pair that works: two 180° rotations about perpendicular axes. Joeldl (talk) 12:07, 23 January 2009 (UTC)[reply]

How do you know that those are the only rotations which work? Rotations around the same axis is obvious - i've derived that ABA^-1 must have the same rotational axis as B, but how do you show that this is also/only otherwise true for 2 180° rotations about perpendicular axes? Thanks again,

131.111.8.97 (talk) 16:13, 23 January 2009 (UTC)Zant[reply]

If you assume that A and B commute, then yes, ABA^-1 and B have the same axis, since they're equal. But Algebraist's point is that the axis of ABA^-1 can be described in another way (even without the assumption that A and B commute). To do this, try and think of a vector that's fixed by ABA^-1. Now compare your two descriptions of the axis of ABA^-1 and try and draw conclusions from that.

As for showing that the other kind of pair A, B, commutes, one idea would be to pick a basis in which the matrices corresponding to A and B will become simple.Joeldl (talk) 16:25, 23 January 2009 (UTC)[reply]

If it helps, this is exactly what I was describing. Choose a simultaneous eigen-basis over the complexes to get a pair of commuting, diagonal, unitary matrices. Amongst all such pairs, the ones that come from SO(3,R) are the ones where the diagonals "match up" in terms of being real or not (and assuming the eigenvalues are either all three real, or one real and a pair of complex conjugates). The all real eigenvalues describe the 180° rotations that leave a rectangular prism fixed. The complex ones are rotations that share an "axis". JackSchmidt (talk) 17:01, 23 January 2009 (UTC)[reply]

I'm sorry but I can't honestly say I follow re. the description of the conjugate matrix' axis, it's a slow day for my brain apparently. If a non-trivial rotation fixes a vector it must be a vector along the axis no? This would be the eigenvalue corresponding to the eigenvalue '1' of ABA^-1, but does knowing that get me any further? I understand the relation between the rotational axis of B and its eigenvalues, but not when conjugated with an arbitrary matrix A. A hint perhaps? Thanks ;) 131.111.8.97 (talk) 17:01, 23 January 2009 (UTC)Zant[reply]

Okay, here are two approaches which can be viewed as amounting to the same thing:

If v is the name of a nonzero vector such that Bv = v, you might be able to identify a nonzero vector w satisfying ABA^-1w = w by looking at this and guessing. A hint would be that the A^-1 ought to be made to cancel out with something.
Given linear mappings φ and χ represented, in a basis e₁,...,e_n, by matrices A and B, with A invertible, how can the linear mapping ψ = φχφ^-1, represented in the same basis by ABA^-1, be described? Answer: ψ has the matrix B in the basis φ(e₁),...,φ(e_n). Thus ψ, viewed relative to the second basis, is entirely like χ relative to the first. Joeldl (talk) 05:03, 24 January 2009 (UTC)[reply]

Physics: Work from force on inclined plane[edit]

A crate of mass 5kg is pulled directly up a rough slope, of inclination 10 degrees, by a constant force of magnitude 20 N, acting at an angle of 30 degres above the horizontal. Find the work done by the force as the crate moves a distance of 3 m up the slope.

Answer: 56.4 J Need the derivation work though.

A lad form Bangladesh. —Preceding unsigned comment added by 202.72.235.202 (talk) 15:43, 22 January 2009 (UTC)[reply]

Do you know the formula for work done (in terms of vectors)? It will be in your text book somewhere, I'm sure. You just need to plug the numbers in (and remember how to calculate dot products). Drawing a diagram may help you work out the appropriate angles, if it isn't immeadiately obvious. (I'll give you a clue - the question includes more information than you need, so don't worry if you aren't using everything.) --Tango (talk) 16:03, 22 January 2009 (UTC)[reply]

The chater of combining forces in FORCE AS A VECTOR QUANTITY comes much later. I did try solving the problem diagramatically, resolving the forces into components parallet to the slope. Then I plugged in the numbers but didn't match the book's salutes. I think there's much more to the part: A crate of mass 5kg is pulled directly up a rough slope. What do you guys/girls think? PLEASE HELP! —Preceding unsigned comment added by 202.72.235.202 (talk) 16:13, 22 January 2009 (UTC)[reply]

Got the answer myself with no help FROM OTHERS! The rough slope indicates that the resolved component of the weight parallel to the slope down it is counter neutralized by the rough slope's friction. Its quite odd coz normally surfaces have predetermined value for resistance. —Preceding unsigned comment added by 202.72.235.202 (talk) 16:27, 22 January 2009 (UTC)[reply]

You don't actually need to consider the friction (or the mass) at all. All you need to know is that the object is being pulled by a 20N force for 3m in a direction offset from the force by 20 degrees. The relevant formula is that (work done)=(force).(displacement) (for constant force, anyway). To take a dot product, you take the product of the magnitudes of the vectors (20 and 3 respectively) and multiply it by the cosine of the angle between them (20 degrees). 20x3x(cos 20)=56.4! --Tango (talk) 17:03, 22 January 2009 (UTC)[reply]

Compound compound Poisson Distribution[edit]

Is there a term for what I would call the "Compound Compound Poisson distribution"? What I mean is that suppose we have a sum of iid random variables, but the number of such random variables is not Poisson-distributed (as they are in the compound Poisson distribution) but rather follows the compound Poisson distribution itself? If there is a term for the distribution I described, have any of its properties been examined? Thanks. Mickeyg13 (talk) 17:11, 22 January 2009 (UTC)[reply]

Generalised Gauss-Bonnet theorem[edit]

Hi,

I've been reading about the Generalised Gauss Bonnet theorem, and for me it seems that the correct version is $\int _{M}\mathrm {Pf} (\Omega )=(-2\pi )^{n}\chi (M)$ where dim(M)=m=2n. For example, you can have a look at the article at www.unf.edu/~dbell/GBT.pdf which does agree with the sign. I understand the proof quite clearly and it does show that the constant has to be exactly -2π (as n=1) otherwise it doesn't work. Plus it seems to me that that constant is necessary as in the 2-dimensional case, when working with the Gaussian curvature K instead of Pf(Ω), we get that K dA = - Pf(Ω). However the Wikipedia article gives $\int _{M}\mathrm {Pf} (\Omega )=(2\pi )^{n}\chi (M)$ , and I think that a convincing argument was put forward on the talk page (though I do not have a good enough understanding of characteristic classes to understand it). So I wanted to know, where does the disagreement on the sign come from? It seems that all sources I find use the same definition of the Pfaffian (so that $\mathrm {Pf} {\begin{pmatrix}0&a\\-a&0\end{pmatrix}}=a$ ), which is the only place where I can think of a different sign coming from.

Thanks. 137.205.233.132 (talk) 17:42, 22 January 2009 (UTC)[reply]

Does no one have any idea about this? I'm really confused by this difference of sign, and I also think the Wikipedia article should say something about it because many texts in the litterature do give the formula with the minus sign... 86.20.184.156 (talk) 11:45, 25 January 2009 (UTC)[reply]