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June 11

Rendering symbol in matrices using MathJax

I expect that someone has picked this problem up somwhere already, but I wonder whether a solution has been found to the problem of the & symbol not translating correctly in the renbdering of matrices in Wikipedia articles using MathJax. From what I have seen since MathJax has been made available to use for mathematical equations in Wikipedia, it works really well, other than for matrices. With matrices it keeps coming up as amp; instead of playing its role as a tab key equivalent to line up the elements inside a matrix. Can someone help on this problem, please? --Chewings72 (talk) 06:31, 11 June 2012 (UTC)[reply]

I added a title to your Q. StuRat (talk) 14:49, 11 June 2012 (UTC) [reply]

I worked it out - I just needed to add some extra code to my "skin". --Chewings72 (talk) 09:17, 14 June 2012 (UTC)[reply]

Automorphisms of direct products of groups

Our article on the direct product of groups states the following: "When G and H are indecomposable, centerless groups, then the automorphism group is relatively straightforward, being Aut(G) × Aut(H) if G and H are not isomorphic, and Aut(G) wr 2 if G ≅ H, wr denotes the wreath product. This is part of the Krull–Schmidt theorem, and holds more generally for finite direct products." However, when I click through to Krull-Schmidt, the conditions are stated in terms of ascending and descending chain conditions, rather than the groups being centerless (and decomposable). Can someone explain how the component groups being centerless is sufficient to ensure ACC and DCC are satisfied? Also, am I to understand that the generalisation is that the automorphism group of a direct product of n groups is a semi-direct product of the direct product of the n automorphism groups, with an appropriate permutation group acting? Thanks, Icthyos (talk) 14:26, 11 June 2012 (UTC)[reply]

Fence panels on curves

Hi.

This may sound like a textbook question, but can someone suggest a hint on where I can find information on a soloution to the following..

The centere line of a path is an arc is defined on AB.
The length of the chord AB is known as is the radius of the curve.

At some distance from the centre line I'd like to put a fence.

The Fence consists of fixed length

How do I work out how many fence panels I need, and where I put the fence posts?

Sfan00 IMG (talk) 15:21, 11 June 2012 (UTC)[reply]

Is the "curve" a circular arc (as the mention of a single radius implies) ? StuRat (talk) 18:47, 11 June 2012 (UTC)[reply]

Let me restate the question as I understand it, then you can tell me if my interpretation is correct:

Take a circle of radius R. Draw a chord on that circle, also of length R. Mark the endpoints of the chord as A and B. Take the circular arc, AB, and offset it both inside and outside the circle, normal to the circle, by distance X. What is the arc length of these offset circular arcs ?

If that's correct, then we still have the additional issue that you seem to be describing the fences as a series of line segments, which will only approximate a circular arc. So, then, the question comes up as to whether the fences should lie completely inside the circular offsets, completely outside, one inside and one outside, or straddle them ? Also, do you want the same number of fence segments on both the inside and outside edges ? If so, you will need to cut the inside edge panels to shorten them. If not, then the width of your path will vary. StuRat (talk) 18:51, 11 June 2012 (UTC)[reply]

If you are doing this is real life, I suggest using a piece of graph paper, a compass, and a ruler, to create a scale model drawing. This should be more than accurate enough for your needs. (If you don't have a compass, you can get a cheap child's school compass, made out of pressed sheet metal, for maybe $1.) StuRat (talk) 19:00, 11 June 2012 (UTC)[reply]

(i) Curve is a circular arc.

(ii) The 'fence' should lie as close to the ideal offset arc as practical.

NB What I think I am wanting to do is describe (or inscribe) 'part' of a regular polygon (whose sides are of a fixed length)

on a given arc. Sfan00 IMG (talk) 19:09, 11 June 2012 (UTC)[reply]

NB At the moment I am considering the curve to be a 'fixed' circular arc.. I know in a lot of cases 'curves' are anything BUT circular :) Sfan00 IMG (talk) 19:09, 11 June 2012 (UTC)[reply]

OK:

1) Do you just want the polygon inscribed on the single, outward offset from circular arc AB ? That is, there is no inside fence ?

Yes, although presumably the technique is similar for an 'inside' fence. Sfan00 IMG (talk) 19:18, 11 June 2012 (UTC)[reply]

But, as I noted, having a different number of segments in the inside and outside fences would both look bad and give a variable width path. You need to abandon the fixed length fence segments approach, if you want to do this. StuRat (talk) 19:28, 11 June 2012 (UTC)[reply]

2) Is this a real life fence you are building ? If so, I still suggest the graph paper method, as that allow you to plot exactly where each post belongs on the graph paper, from which you can find the spots to dig on the ground. StuRat (talk) 19:13, 11 June 2012 (UTC)[reply]

It's not a 'real' fence, I was thinking about how to make a 'fencing' generator script for some computer graphics code. Sfan00 IMG (talk) 19:18, 11 June 2012 (UTC)[reply]

This is how I would do the final step (inscribing a fence inside a circular arc):

A) Define the circular arc in polar coordinates.

B) Create a loop, say I = 1 to 1000.

C) For each value of I, create that many points, plus one, evenly spaced, along the circular arc. Do this by subdividing the number of degrees (or radians) swept, by I.

D) Find the straight distance between the first two points on the circular arc.

E) If that distance is less than the segment length allowed, you are done. If not, continue, with the next number of points.

This could be done more efficiently, using more complex code and math, but, as is, it's probably under a second of run time, so I don't see the need (and more complex code makes for more bugs). Would you like me to whip up some FORTRAN sample code ? StuRat (talk) 19:41, 11 June 2012 (UTC)[reply]

IF you want to, I think I have enough in thinking to write my own code though :) Sfan00 IMG (talk) 20:03, 11 June 2012 (UTC)[reply]

OK. BTW, if it isn't already obvious; a normal, constant offset outside a circular arc is itself a circular arc, with the same center point and degrees swept as the base circular arc, but with the radius increased by the offset amount. StuRat (talk) 20:23, 11 June 2012 (UTC)[reply]

A related question (which is the inverse of the above), For a given regular polygon which has sides of length x , determine the radius of a circle lying entirely within or without the polygon. As this is a 'textboox' problem, I'm not going to comment further :) Sfan00 IMG (talk) 11:07, 12 June 2012 (UTC)[reply]

Inscribed polygon and circumscribed polygon are the names of those problems, but our articles seem messy and deal with irregular polygons. It seems to be an easy problem if the regular polygon has an even number of sides, where the diameter of the circumscribed circle is the distance from a vertex to the opposite vertex, while the diameter of the inscribed circle is distance from one side to the opposite side. StuRat (talk) 17:46, 12 June 2012 (UTC)[reply]

Do these ambiguities affect curvature?

I am using Ctrax, which records flies' positions and angular orientations per frame (in 2D space) and from the framerate, computes velocities and accelerations. It then spits the output in a MATLAB file that details this data per tracked fly per frame. For some reason, the u and v components of velocity relative to the fly's orientation and doesn't even do this for acceleration. I wish to calculate Curvature K = ||r'(t) x r' '(t)|| / ||(r'(t))||^3 , where r = [ax , by].

Depending on the type of video file that generated the tracks, the y positions will sometimes be flipped (i.e. for 640x480 video, (x,y) = (40,70) will be (40,410)). Also, I cannot figure out from the documentation whether orientation starts from East or North and whether increasing angle proceeds clockwise or anticlockwise.

What I want to know is, do any of these ambiguities impact curvature? Basically I do not know whether v_x = mag(v) * cos(theta) or mag(v) * sin(theta), likewise for v_y. I do know whatever v_x is, v_y has to be the complement. Same for acceleration. Do these ambiguities "cancel out" when calculating curvature, i.e it doesn't actually matter if v_x and v_y are flipped with regards to their formulas (along with acc_x and acc_y)

Looking at the cross product formula (n * |a||b| sin(theta)) , it appears that it doesn't matter if the magnitudes |x| and |y| are flipped, because their angle wouldn't change. But for ||(r'(t))||^3 I am less sure. Mixing up whether v_x is proportional to cos(theta) or sin(theta) wouldn't necessarily be trading places with v_y, because v_x and v_y have different magnitudes. Furthermore, because of the flipping, I imagine that trading of places might not necessarily be the only case, if for example v_x = ||v|| sin(theta) but v_y could be minus or plus cos(theta), if y-flipping was ambiguous. 207.114.92.194 (talk) 18:13, 11 June 2012 (UTC)[reply]

It shouldn't matter if the coordinate system is flipped, provided it doesn't change (between data points) during a calculation. StuRat (talk) 18:45, 11 June 2012 (UTC)[reply]

It shouldn't matter as long as the resolution is the same in the X and Y directions. But any time you have worries like this, you really ought to check explicitly. In this case the way to check is to write an M file that flips the data, then run the flipped data through your analysis to make sure that it produces the same results. Looie496 (talk) 18:51, 11 June 2012 (UTC)[reply]

The resolution is 640x480...I am about to write the code I just wanted to check beforehand. 71.167.234.136 (talk) 08:54, 12 June 2012 (UTC)[reply]

binomial distribution vs. normal distribution

What's the difference between the binomial distribution and the normal distribution? I see in the latter article that they are usually very close ... but why? How do they differ? Comet Tuttle (talk) 21:03, 11 June 2012 (UTC)[reply]

The main difference is that a binomial distribution is a discrete probability distribution and the normal distribution is a continuous probability distribution. The fact that they become close is the central limit theorem. Rckrone (talk) 23:53, 11 June 2012 (UTC)[reply]

A random variable X having a binomial distribution can assume the nonnegative integer values X = 0, 1, 2, ... , N. A random variable having a normal distribution can assume any real value −∞ < X < ∞. So if you know that X is an integer, then X does not have a normal distribution. And if you know that X is nonnegative, then X does not have a normal distribution either. Bo Jacoby (talk) 08:03, 12 June 2012 (UTC).[reply]

Thank you both! Comet Tuttle (talk) 15:49, 12 June 2012 (UTC)[reply]