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June 2

Mercator versus cylindrical

Quote from "An Application of Geography to Mathematics: History of the Integral of the Secant", by V. Frederick Rickey and Philip M. Tuchinsky, Mathematics Magazine, v. 53, No. 3, May 1980, pp. 162–166:

“ Consider a cylinder tangent to the earth's equator and imagine the earth to "swal [swell] like a bladder." Then identify points on the earth with the points on the cylinder that they come into contact with. Finally, unroll the cylinder; it will be a Mercator map. This model has often been misinterpreted as the cylindrical projection (where a light source at the earth's center projects the unswollen sphere onto its tangent cylinder. ”

What exactly is the difference between these two things? Michael Hardy (talk) 01:54, 2 June 2011 (UTC)

It sounds like the distance of polar regions from the equator is the difference. In the cylindrical projection, the height of the north pole would be the (foreshortened) radius of the Earth, while with the Mercator projection, the height would be the (correct) 1/4th the circumference of the Earth. That should be about 1.57 times more. StuRat (talk) 02:23, 2 June 2011 (UTC)

Here's a diagram of one quarter of the globe and it's two projections:

Cylindrical projection:

   | -> |
 _/  -> |

Mercator projection> (imagine it "unwrapping"):

   | -> |
 _/  -> |
     -> |

Corrected Mercator projection (with polar region stretched out to match East-West stretching):

   | -> |
 _/  -> |
     -> |
     -> |

StuRat (talk) 02:29, 2 June 2011 (UTC)

Well, if that's what it means, it's wrong. The key special property of the Mercator projection is that it's a conformal mapping. In this case, what that means is, however much you have to stretch the horizontal scale at a given latitude, to make the parallel as long around as the cylinder is, that's how much you also have to stretch the vertical scale.

This means that small features at any latitude (except the pole itself) will appear undistorted, at least to first order. (The value of "small" depends on how much distortion you can tolerate, and on the latitude.) --Trovatore (talk) 02:45, 2 June 2011 (UTC)

Saying "The key special property of the Mercator projection is that it's a conformal mapping" seems to neglect the fact that many quite different map projections are conformal without having what should really be considered the key special property of the Mercator projection: that bearings on the map correspond to bearings on the earth. For example, the stereographic projection is conformal, but it doesn't preserve bearings at all. Michael Hardy (talk) 18:21, 4 June 2011 (UTC)

Oh, I think I see maybe what Rickey and Tuchinsky are getting at. The idea is, draw your globe on a rubber balloon. Put it inside an infinite rigid cylinder, radius equal to the original radius of the globe. Now start inflating it. You can see that the poles never do touch the side of the cylinder, no matter how much you blow up the balloon, but (assuming an infinitely inflatable balloon) any point other than the poles will eventually touch the cylinder and be mapped.

I suppose R and T are assuming that, for any given level of inflation, any point not yet touching the cylinder is equally stretched in all directions. I somewhat doubt that that is actually true for a physical balloon, because my intuition suggests that nearby points that are touching the cylinder are "pushing" the non-touching parts a little in the vertical direction, but I could be wrong. --Trovatore (talk) 02:57, 2 June 2011 (UTC)

"Trovatore", I don't see any explanation in your comments of the difference between the projection involving the balloon and the cylindrical projection. That the north and south poles are not part of the domain of the function is of course what I expect of both the cylindrical projection and the Mercator projection; I'm not sure why it's worth mentioning here. I think the latter is characterized by the conjunction of two things: (1) it's conformal (angle-preserving) and (2) curves of constant bearing appear as straight lines on the map; e.g. the curve that goes in a direction 71 degrees east of north, which in fact spirals infinitely many times around the north pole, although the arc length of the spiral is finite, is a straight line on the map. Michael Hardy (talk) 13:14, 2 June 2011 (UTC)

Actually, only the Mercator projection has a problem with showing the poles. They could be shown, as a line at the top and bottom of the map, under both the cylindrical projection and the middle projection I described above. StuRat (talk) 19:38, 2 June 2011 (UTC)

Well, but the cylindrical projection you described is not the same as the one Rickey and Tuchinsky described. They said the light source was at the center. What you've described is more like a light source along the central axis. Drill a hole between the poles and put a light stick through the middle, and you get your projection; put a light bulb in the actual center and you get the one R and T are talking about. --Trovatore (talk) 19:47, 2 June 2011 (UTC)

Yes, but they don't seem to be using the term properly, IMHO. Isn't a projection assumed to be normal to the target surface, unless otherwise specified ? StuRat (talk) 20:31, 2 June 2011 (UTC)

That's "projection" in the sense of a vector space (more precisely, an inner product space). Cartography uses the term a little differently. Any way of mapping the globe onto a flat surface is called a "projection". --Trovatore (talk) 22:05, 2 June 2011 (UTC)

Right, the cylindrical projection they're talking about, with the light source at the center, is not conformal. A point of latitude θ gets projected by the light source to a point of height tan(θ). That means the vertical stretching is proportional to d/dθ tan(θ) equals sec²(θ), whereas the horizontal stretching is only sec(θ). --Trovatore (talk) 19:15, 2 June 2011 (UTC)

That establishes easily that the cylindrical projection is not conformal. But how does "swelling like a bladder" result in anything other than the cylindrical projection? Michael Hardy (talk) 20:40, 2 June 2011 (UTC)

Well, I'm shakier on that part of it. R and T's method works if you assume that (i) once the rubber touches the walls of the cylinder, it sticks and doesn't move and (ii) for the part of the rubber that hasn't yet touched the cylinder walls, at any point, the tension is the same in all directions. I can easily throw (i) into my intuitive picture (I'm already allowing infinitely stretchable rubber) but I'm less clear on (ii). (However I certainly don't see any reason the method would give you R and T's "cylindrical projection"; why would you think it would?) --Trovatore (talk) 21:17, 2 June 2011 (UTC)

It hadn't occurred to me that the cylinder was a solid thing that would stop the spherical balloon from expanding. I pictured the sphere continuing to expand, so that progressively less of it is inside the cylinder and most of it is outside, but retaining its spherical shape. Each point on the sphere would pass through the cylinder once, and the point on the cylinder where it passes through would be the points it's mapped to. If R & T had something else in mind, they were certainly cryptic about it. Michael Hardy (talk) 22:00, 2 June 2011 (UTC)

Wouldn't that mean that, since the width at the poles is infinitely stretched, that the height at the poles is also infinitely stretched ? Do they handle this by not going all the way to the poles ? StuRat (talk) 02:53, 2 June 2011 (UTC)

Yes, to make a Mercator map finite in the vertical direction, you do have to cut out a ring around each pole. --Trovatore (talk) 02:58, 2 June 2011 (UTC)

Mercator distortion

Right. I'd like to see a Mercator map with very small neighborhoods of the poles removed. I think it would be a good illustration of the shortcomings of that projection. Anyone know where we can see something like this? (i.e. even less polar neighborhoods removed than this example from mercator projection). SemanticMantis (talk) 16:00, 2 June 2011 (UTC)

Well, all flat projections have issues. It's not possible to make a flat projection that preserves all distances, say, because from those you can recover the metric tensor and thereby the curvature tensor. The Mercator projection has some excellent properties if you understand how to read it. It really does show correctly how things are shaped, as long as those "things" are not too large based on their latitude.

In the Seventies or so there were people trying to make a political point out of this, claiming that the Mercator projection was designed to inflate the importance of "northern" peoples. I think that was basically crankery. I read Arno Peters's little handbook on the subject, and it was plain that he just didn't understand the mathematics. --Trovatore (talk) 19:21, 2 June 2011 (UTC)

Must be a conspiracy by those all-powerful natives of Greenland and Antarctica. :-) StuRat (talk) 19:29, 2 June 2011 (UTC)

There aren't many features to map very near to the poles, so such a map would be kind of boring... just endless Arctic sea at the top and endless Antarctic plateau at the bottom, both at an ever huger scale. 86.160.216.165 (talk) 23:26, 2 June 2011 (UTC)

Antarctica has lots of interesting features, including mountains, glaciers, underground (or under-ice ?) lakes, colonies of penguins, research stations, routes taken by explorers (and their associated remains), magnetic south pole, etc. StuRat (talk) 23:06, 3 June 2011 (UTC)

The standard Mercator projection is quite an inappropriate way of mapping Antarctica. Michael Hardy (talk) 00:06, 4 June 2011 (UTC)

Agreed. StuRat (talk) 00:53, 4 June 2011 (UTC)

conical pendulum

Wikepedia calculates by formula the force T exerted by a swinging pendulum. what is the unit of measurement for this force T ?

given a length of 25 inches to the bob , has the vector T force of this magnitude ever been put to any useful purposes?- eg battery recharging/electrical generation? (disregard and force required to start the pendulum movement)

this enquiry is not homework but of general interest

Reference 7300 — Preceding unsigned comment added by Kevin Sefton (talk • contribs) 02:05, 2 June 2011 (UTC)

1. See Force#Units_of_measurement. 2. No. When the force is at a right angle to the velocity, then no power is produced. Bo Jacoby (talk) 04:23, 2 June 2011 (UTC).

(edit conflict)You're talking about the formulas in Conical pendulum? The units of measurement depend on what units you want to use for all the other variables - if you're working in SI units, then m would be in kilograms, v in metres per second, r in metres, and g in meters per second squared, meaning that T would be in kilograms metres per second squared, or in other words Newtons. But that force just comes from two things - the downwards gravitational force of the Earth on the bob, and the tension from the string. And that force doesn't do any work, since it always acts perpendicular to the direction of motion (see also Work (physics)#Zero work). If you try to extract any energy from the pendulum, you'll just result in the pendulum slowing down, meaning you need to add more energy in to get it spinning at its original speed again. You probably want to read up on Work (physics) to see what you need to have extractable energy from a moving system. Confusing Manifestation(Say hi!) 04:29, 2 June 2011 (UTC)

Indeed, even pendulum clocks (such as grandfather clocks, most tower clocks, and shelftop pendulum clocks) use the pendulum only to regulate the period of the motion of a wheel, with the power being supplied from somewhere else, usually from a heavy weight lowered slowly. – b_jonas 22:02, 4 June 2011 (UTC)

Measure Theory Integral-sum inequality

With regards to a measure theory problem I was discussing with a lecturer recently, it was claimed in the space of seconds that

$\int _{0}^{\infty }{\frac {e^{-({\frac {(y+z)^{2}}{2m}})}}{\sqrt {2\pi m}}}\leq \sum _{k=0}^{\infty }{\frac {e^{-({\frac {(y+k)^{2}}{2m}})}}{\sqrt {2\pi m}}}\leq {\frac {e^{-({\frac {(y)^{2}}{2m}})}}{\sqrt {2\pi m}}}+\int _{0}^{\infty }{\frac {e^{-({\frac {(y+z)^{2}}{2m}})}}{\sqrt {2\pi m}}}$.

Why this is is not obvious to me. I have of course worked with limit/sum inequalities before (the standard one I think is bounding log(x) as the integral of 1/x above and below by sums of 1/n) - however, I can't seem to picture why this one is true. I am sure it is entirely obvious and I am simply out of practice - I have not done anything like this in a few years! - but could anyone quickly explain intuitively to me why the inequality holds? The variable y is meant to take any value, and while the integrand didn't specify what it was with respect to, it looks to be 'z' fairly obviously, and I guess we can lose the square roots, I just added them for the sake of completeness. I have tried to draw a picture of the integral to obtain the inequality but I couldn't quite piece the argument together. Could anyone lend a hand? I'm sure it will seem obvious to me as soon as someone has pointed out why it was true so don't worry too much about the background theory, I just need to see why it is true (hopefully in a way that I could apply to other integrals similar in nature if needs be).

Many thanks all! Typeships17 (talk) 18:54, 2 June 2011 (UTC)

Estimate the integral by its upper and lower Riemann sums wrt the partition consisting of integer points. Sławomir Biały (talk) 19:14, 2 June 2011 (UTC)

Ah smashing, of course. Thanks! :) Typeships17 (talk) 19:21, 2 June 2011 (UTC)

Does $\int _{0}^{\infty }{\frac {e^{-({\frac {(y+z)^{2}}{2m}})}}{\sqrt {2\pi m}}}$ mean $\int _{0}^{\infty }{\frac {e^{-({\frac {(y+z)^{2}}{2m}})}}{\sqrt {2\pi m}}}\,dy$ or $\int _{0}^{\infty }{\frac {e^{-({\frac {(y+z)^{2}}{2m}})}}{\sqrt {2\pi m}}}\,dx$ or something else? Michael Hardy (talk) 00:57, 3 June 2011 (UTC)

It means dz, I believe: but Slawomir solved my problem anyway. Thanks! Typeships17 (talk) 01:05, 3 June 2011 (UTC)

What is the largest consecutive integer of 13 whose sum is 0

w — Preceding unsigned comment added by 116.71.1.172 (talk) 20:38, 2 June 2011 (UTC)

Your question makes no sense. It may help if you give an example of what you mean by "consecutive integer of x" having a sum y. 86.160.216.165 (talk) 21:50, 2 June 2011 (UTC)

Seems pretty straightforward to me: there is a set of 13 consecutive integers that sum to 0, and the OP wants to know what the largest is.

BIG hint: in any arithmetic sequence with an odd number of terms, the sum is equal to the number of terms times the median. -- Bk314159 (Talk to me and find out what I've done) 01:31, 3 June 2011 (UTC)

Oh yeah, sorry, I failed to understand what "consecutive integer of 13" referred to! 86.160.216.165 (talk) 01:56, 3 June 2011 (UTC)

I don't think the phrase "consecutive integer of 13" occurs here. Rather, I'd parse it like this:

largest consecutive integer of {13 whose sum is 0}

Thus, the sum of 13 consecutive integers is 0. What is the largest of them? Michael Hardy (talk) 02:46, 3 June 2011 (UTC)

-6, -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6 is a sequence of 13 consecutive integers whose sum is 0. The largest is 6. Widener (talk) 07:13, 3 June 2011 (UTC)

On a language note, the reason this question is hard to parse is that it doesn't make sense to talk of a consecutive integer. So, instead of how it reads, it should read "What is the largest consecutive integer of 13 consecutive ones whose sum is 0". --86.8.139.65 (talk) 18:59, 5 June 2011 (UTC)

June 3

Strength of a linear relationship

If you have some data with (x,y) coordinates and you plot it on a scatter plot, is the strength of the linear relationship the correlation between x and y, or is it the value of the slope of the best fitted linear regression line? Widener (talk) 07:16, 3 June 2011 (UTC)

Correlation and dependence says "The Pearson correlation coefficient indicates the strength of a linear relationship between two variables". This makes sense, because the correlation coefficient is unaffected by rescaling of the variables, whereas the slope of the regression line does depend on the scale of the variables, so it is somewhat arbitrary. Another way to think of this is to see that the slope of the regression line actually has units (which depend on the units in which you have measured the dependent and independent variables), but a useful mesaure of relationship strength should be a dimensionless quantity. Gandalf61 (talk) 09:25, 3 June 2011 (UTC)

I seem to recall that if x is regressed on y, then y regressed on x, the product of the two slopes is the correlation coefficient. So in this sense, the slope(s) of the best fitted regression line(s) do give the strength of the linear relationship - but it's really the correlation coefficient which does so.→86.132.165.117 (talk) 21:24, 3 June 2011 (UTC)

That is complete nonsense. The product of their slopes is not generally between −1 and 1, and is not dimensionless. The correlation is a dimensionless number between −1 and 1. Michael Hardy (talk) 15:49, 4 June 2011 (UTC)

I think the product of the slopes of the two regression lines must be the square of the correlation coefficient. Gandalf61 (talk) 08:23, 4 June 2011 (UTC)

Gandalf, that is also nonsense. The square of the correlation is obviously dimensionless; the product of the slopes is not. Michael Hardy (talk) 15:50, 4 June 2011 (UTC)

Agreed, my memory was correct but incomplete - for negative correlation, both slopes and the cc would be negative, but the product would be positive. It can be seen from this that it's only if all the points lie on a straight line that the regression lines coincide, and so have slopes reciprocal to each other.→86.132.165.117 (talk) 15:44, 4 June 2011 (UTC)

The slope says nothing about the strength of the relationship. You can have the same slope when the correlation is 0.01 or when it's 0.99. Michael Hardy (talk) 15:47, 4 June 2011 (UTC)

Agreed. But the slopes of the two regression lines have recriprocal dimensions - one is dim(y)/dim(x), the other is dim(x)/dim(y) - so their product is dimensionless. Slope of one regression line is:

$\beta _{1}={\frac {\sum _{i=1}^{n}(x_{i}-{\bar {x}})(y_{i}-{\bar {y}})}{\sum _{i=1}^{n}(x_{i}-{\bar {x}})^{2}}}$

slope of the other regression line is:

$\beta _{2}={\frac {\sum _{i=1}^{n}(x_{i}-{\bar {x}})(y_{i}-{\bar {y}})}{\sum _{i=1}^{n}(y_{i}-{\bar {y}})^{2}}}$

and their product is:

$\beta _{1}\beta _{2}={\frac {\left(\sum _{i=1}^{n}(x_{i}-{\bar {x}})(y_{i}-{\bar {y}})\right)^{2}}{\sum _{i=1}^{n}(x_{i}-{\bar {x}})^{2}\sum _{i=1}^{n}(y_{i}-{\bar {y}})^{2}}}=r^{2}$

Gandalf61 (talk) 18:06, 4 June 2011 (UTC)

If I didn't know something, I'd be more courteous (and wary) than to describe someone else's statement as complete nonsense, in bold for emphasis - these things sometimes come back to bite you. My statement was only slight nonsense, as I'd forgotten the squaring, but Gandalf's were spot on.→86.132.165.117 (talk) 20:25, 4 June 2011 (UTC)

Don't hold back anon! Spit out the truth: Michael Hardy is a cunt. --72.179.51.84 (talk) 21:25, 4 June 2011 (UTC)

The second statement, by Gandalf, was right: the product of the slopes is dimensionless and is the square of the correlation. Haste makes waste: I saw the first attempt and was hasty thereafter. Michael Hardy (talk) 22:00, 4 June 2011 (UTC)

Look up Beta_(finance) and Pearson_product-moment_correlation_coefficient. The regression coefficient beta is the covariance between the two variables divided by the variance of the independent variable. The linear correlation determined by Pearson's r is the covariance divided by the product of the standard deviations. — Preceding unsigned comment added by 99.100.92.26 (talk) 14:06, 5 June 2011 (UTC)

That old earth curvature chestnut again, but with a difference

Turn the page if you've already been bored to death with this. But if you haven't, you may be able to help us non mathematicians (who came up with a variety of answers you wouldn't believe). Two tall people are standing at sea level, facing each other across the water. Their eyes are exactly six feet above sea level. How far apart are they when they can just see each other's eyes, but not the bridge of the noses? OK, it's a fabulously clear day, flat calm sea, and they have fabulously keen eyesight. We're interested in the distance, not the possibility/impossibility of being able to view each other's eyes. Moriori (talk) 21:41, 3 June 2011 (UTC)

The distance between their eyes is twice the distance to the horizon. PrimeHunter (talk) 22:50, 3 June 2011 (UTC)

The same approach answers the question of whether the highest summits in Scotland (Ben Nevis), England (Scafell Pike) and Wales (Snowdon) are visible from each other, assuming perfect visibility and the absence of anything in between - only the last two are.→86.132.165.117 (talk) 16:12, 4 June 2011 (UTC)

June 4

Minor confusion on "if G has faithful complex irreducible representation then Z(G) is cyclic"

Hello all,

I am confronted with the following proof:

Claim: If G has faithful complex irreducible representation then the centre Z(G) is cyclic.

Proof: Let $\rho :G\to GL(V)$ be a faithful irreducible complex representation, and $z\in Z(G)$: so $zg=gz\,\forall \,v\in V$. This is a G-endomorphism on V and so is multiplication by some scalar $\tau _{z}$ say, by Schur's lemma. Then the map $Z(G)\to \mathbb {C} ^{\times },\,z\to \tau _{z}$ is a representation of Z and is faithful as $\rho$ is faithful. Thus, Z(G) is isomorphic to a finite subgroup of $\mathbb {C} ^{\times }$ and so is cyclic.

Now my question is this: where do we actually use the fact that we're working with the centre? I'm sure it's in there somewhere, but I can't spot why this only works with elements commuting with everything in G, and why we can't just apply the argument to the whole of G. Could anyone explain? Thanks very much. Mathmos6 (talk) 12:31, 4 June 2011 (UTC)

Schur's lemma only applies for z in the center. More specifically, you're using the fact that the image of z is in the center of End(V).--RDBury (talk) 21:30, 4 June 2011 (UTC)

Rotation of conic sections in polar form

Hi Reference Desk,

I have been given a conic section in polar form and I was asked to find the angle of rotation and then sketch the graph.

r = 1 / (1 - 2 cos(t) + 2 sin(t))

I rewrote r into

r = 1 / (1 - e cos(t + t'))

where e is the eccentricity

I solved for e and t' and this gives me

r = 1 / (1 - 2 cos(t + pi/4))

Does this mean the conic section is rotated anticlockwise by pi/4 or clockwise by pi/4? Does the sign in front of the cos or sin matter?

Are there any general rules for rotation of conic sections in the polar form? I have read a lot articles on rotation of conic sections but they were all written in terms of the standard form and not the polar form.

Thanks a lot! — Preceding unsigned comment added by 169.232.101.13 (talk) 22:39, 4 June 2011 (UTC)

You probably meant r = 1 / (1 - 2 √2 cos(t + pi/4)) so the eccentricity is 2 √2. In general, replacing t by t+α in a polar equation rotates the graph clockwise by α, just as replacing x by x+a in a Cartesian equation shifts left by a.--RDBury (talk) 05:49, 5 June 2011 (UTC)

Problem with Lagrange multipliers

Question:

Find the maximum and minimum values of f(x,y) = x^2 + y^2 subject to the constraint g(x,y) = x^6 + y^6 = 1

This is how I went about solving it:

grad f = t grad g (where t is a constant) 2x = 6t x^5 2y = 6t y^5

1 = 3 t x^4 --> (1/3t)^(1/4)=x 1 = 3 t y^4 --> (1/3t)^(1/4)=y

I substituted x and y into g(x,y)to determine t

(1/3t)^(3/2) + (1/3t)^(3/2) = 1 (1/3t)^(3/2) = 1/2 1/3t = (1/4)^(1/3)

Substituting 1/(3t) back into x and y

x=(1/4)^(1/12) y=(1/4)^(1/12)

To determine whether this is a maximum or a minimum, I need to use the second derivative test

fx=2x fxx=2 fxy=0

fy=2y fyy=2 fyx=0

The Hessian matrix D = 4 > 0, fxx = 2 > 0. This usually means that the point is a minimum, but the problem is that we did not even substitute x = (1/4)^(1/12) into the determinant of the Hessian matrix.

How should this problem be done? — Preceding unsigned comment added by 169.232.101.13 (talk) 22:52, 4 June 2011 (UTC)

First, you're missing a few solutions. What Lagrange really says is that the vectors (2x,2y) and (6x^5,6y^5) are linearly dependent. So solve by setting the determinant (2x)(6y^5)-(2y)(6x^5) equal to 0. This gives x=0, y=0 and x=±y, and putting these back into the original equation gives 8 solutions (0,±1), (±1,0), (±2^−1/6,±2^−1/6). Second, while there probably is a second derivative test for Lagrange multipliers somewhere, I can't find it the books I have. The Hessian only applies for unconstrained problems. In any case, the second derivative test only tells you if a point is a local maximum or minimum and the problem is asking for global. So just evaluate the function at the critical points; the ones where the values is highest are the maxima and the ones where the values is lowest are the maxima. The constraint curve is bounded so you don't have to worry about what happens at infinity.

Another way to solve this is by setting x=cos^1/3t, y=sin^1/3t. This reduces the problem is one dimension and you can use simpler methods on it. (The algebra might get more complicated though.)--RDBury (talk) 06:25, 5 June 2011 (UTC)

Question about 'a convenient way to type symbols?'

I searched on google but can't find good result, so I came here for help. The only way I type symbols like 'sets relation symbol' 'predicate logic symbol' 'very strange operators' is opening the unicode list then copy it to my sheet. Is there some trick to make this operation faster? (Also I found this 'uncommon symbols' on Wikipedia are showed as PNG pictures.)Nilman (talk) 23:00, 4 June 2011 (UTC)

The Edit tools, which appear below the edit window, have various sets including 'Math and logic'. Select that from the popup menu then click on a symbol to insert it as you type. If a symbol you use a lot doesn't appear there you can request that it's added.--JohnBlackburne^words_deeds 23:14, 4 June 2011 (UTC)

More generally there are two ways of getting symbols that don't appear on your keyboard: (1) Use something like <math> mode that allows you to use special descriptive names such as \alpha for the Greek letter α, or (2) find an input method that does what you need. (The Edit tools are an input method supplied by the Wikipedia editor.) Looie496 (talk) 23:47, 4 June 2011 (UTC)

Thank you.Nilman (talk) 11:35, 5 June 2011 (UTC)

You can also use a program like AutoHotkey to set up keyboard shortcuts to specific symbols. This has the advantage of working in (almost) any program. -Elmer Clark (talk) 15:35, 7 June 2011 (UTC)

June 5

a simple ODE

I have a question about solving a simple linear ODE

${\dot {Y}}=\alpha (v(Y)-s(Y))$

where $\alpha >0$ and v(Y) and s(Y) are linear functions of Y that intersect at one point.

Taking the simplest example

${\dot {Y}}=\alpha (v-s)Y$

gives a solution of $Y(t)=e^{\alpha (v-s)t}$ which will go to infinity if $v>s$ and zero (which is the point of intersection) if $v<s$.

What if zero is not the point of intersection, so that $v(Y)=v_{0}+vY$ and $s(Y)=s_{0}+sY$?

Then the point of intersection is at $Y={\frac {s_{0}-v_{0}}{v-s}}$ and the ODE is

${\dot {Y}}=\alpha (v_{0}-s_{0}+(v-s)Y)$

How do I show the solution to the ODE in this case? I want to be able to show that depending on the parameters of the linear functions, the solution will tend to a stable equilibrium at that point of intersection. — Preceding unsigned comment added by 130.102.78.164 (talk) 00:13, 5 June 2011 (UTC)

This is a case of Linear differential equation#Nonhomogeneous equation with constant coefficients but in this case you can find a particular solution by assuming Y is constant. The general solution is then found by adding solutions to the homogeneous case and you've already done that.--RDBury (talk) 06:34, 5 June 2011 (UTC)

what if any statement implied its opposite?

what if there were a proof that didn't require any axioms (was true for all systems), that from any statement, its opposite followed? --86.8.139.65 (talk) 18:56, 5 June 2011 (UTC)

That would mean that the rules of inference you are using are not a very good choice. --Tango (talk) 19:27, 5 June 2011 (UTC)

What do you mean by "opposite"? What's the "opposite" of 2 + 3 = 5? Michael Hardy (talk) 20:17, 5 June 2011 (UTC)

In this case I would say the opposite is "two plus three very much does not equal five". I don't know how to mark that up. 86.8.139.65 (talk) 20:36, 5 June 2011 (UTC)

Is this like the Liar paradox? Grandiose (me, talk, contribs) 20:40, 5 June 2011 (UTC)

Then that would also apply to the statement that says that: "from any statement, its opposite follows". Count Iblis (talk) 22:34, 5 June 2011 (UTC)

Ex falso quodlibet 83.134.160.239 (talk) 09:42, 7 June 2011 (UTC)

I think that this may be different. The OP was asking about the possibility of there existing a $\phi$ such that $\phi \to \neg \phi$. The link you mention is about something different. Ex falso quodlibet is about being able to imply anything from an assumed contradiction, e.g. $(\phi \land \neg \phi )\to \psi$. Although if there exists a φ such that $\phi \to \neg \phi$ then we would have $\phi \land \neg \phi$ which would then imply any statement you could imagine by ex falso quodlibet. — Fly by Night (talk) 15:40, 7 June 2011 (UTC)

For many statements there's no problem with them implying their opposite. If we then showed the opposite implied the first thing again you start having problems. For instance if we showed that ants are three meters tall implied ants are not three meters tall then there's no problem. Dmcq (talk) 20:50, 7 June 2011 (UTC)

Could you give some examples? If (all) ants are three meters tall then then "(all) ants are not three meters tall" is clearly false, so you have $\phi \not \to \neg \phi$. They all cause problems: "if A is true then A is false" is what the OP was asking about. — Fly by Night (talk) 21:13, 7 June 2011 (UTC)

For instance, let $\phi$ be the proposition $(\psi \land \neg \psi )$. By ex falso quodlibet, $\phi \to \neg \phi$.

Conversely, if φ were any proposition such that $\phi \to \neg \phi$ then by conditional exchange, we can conclude $\neg \phi$. Sławomir Biały (talk) 22:10, 7 June 2011 (UTC)

Well actually in that last bit you can't conclude $\neg \phi$ as ψ may be false and false implies false is true. Dmcq (talk) 22:20, 7 June 2011 (UTC)

Sorry, the psi should have been a phi. I've corrected it. Sławomir Biały (talk) 22:33, 7 June 2011 (UTC)

Niels Bohr is quoted for saying: "The negation of a truth is a falsehood, but the negation of a deep truth is another deep truth". I don't know whether this is a truth, a falsehood, or a deep truth. Bo Jacoby (talk) 09:01, 8 June 2011 (UTC).

Tango gave the correct answer as the first post. If you can prove that no consistent system exists then you should restrict the rules of inference. Taemyr (talk) 11:41, 8 June 2011 (UTC)

How to average ranks?

As an example, there are several listings of the best 100 (or whatever) novels of the 20th. century. A particular novel is not going to have the same rank in every list. To get the "average" rank of each novel, is there any better method than merely taking the mean average of its ranks? Thanks 92.28.240.238 (talk) 20:24, 5 June 2011 (UTC)

Average rank is not a very good way of doing it unfortunately. The main problems are that firstly it is very bad at dealing with incomplete orders and secondly in general the rank can vary more in the middle rather than the beginning and end so it is not combining things with similar ranges. I tried doing a quick search for merging rank order and combining rank order but didn't find the one I was looking for which is iterative and assigns a real number value rather than an integer order. Other ways of doing it involve setting up a matrix of wins/losses and getting the overall order from that. Dmcq (talk) 23:59, 5 June 2011 (UTC)

Instead of attempting to order them, you can group them. For example, assume you have 200 books that appear in 10 different top 100 listings. You can place that data in a matrix and use singular value decomposition on the matrix. Then (assuming the books are on the left vertical column of your matrix) the first table will have 5 columns - one per book. Replace all negative values with 0 and all positive values with 1. Each book will have an identifier like 01101. All books with the same identifier are in the same group. With little work, you can sort the groups into the most popular, somewhat popular, and less popular groups. Further, SVD allows you to fill in a rather good estimate for missing data. So, if a book doesn't appear on a particular top 100 list, you can estimate where it is on the list: 101, 110, 250... -- kainaw™ 00:06, 6 June 2011 (UTC)

Just looking again at the type of stuff you want to order I think you can probably get a quick and fairly good order by averaging the logarithm of the order including something like 1+log(100) for all the missing ones. Rather empirical but far better than just averaging the order without getting the log. Dmcq (talk) 00:27, 6 June 2011 (UTC)

If I were doing this, I would probably just use the median of the reported ranks. The main problem with using the mean is that it can be badly affected by outliers -- a single ranking that widely differs from the others can significantly change the result. The median does not suffer from that problem, and is simple to compute. Looie496 (talk) 00:42, 6 June 2011 (UTC)

Isnt there any maths/stats theory which would indicate the best method to use? 2.97.212.124 (talk) 12:17, 6 June 2011 (UTC)

Well there's lots of studies of different algorithms which are based on different assumptions! I found something mentioning what I was looking for above in http://deepblue.lib.umich.edu/bitstream/2027.42/66929/2/10.1177_001316447303300104.pdf , I'm sure there's much better now with people using Monte-Carlo methods since they've oodles of performance to waste on computers. Dmcq (talk) 13:41, 6 June 2011 (UTC)

There is a hell of a lot of information on this topic. You are dealing with three main issues: synonymy, polysemy, and sparsity. In your specific case, synonymy and polysemy are not too bad, but there is the issue of how books are listed. For example, what if one list has both the hardcover and softcover version of a book listed? What if one list places an entire series as a single book while another list breaks each independent book out? Your main issue is sparsity. Many books will not be on all lists. So, assume you use median. A book is, for some reason, on one and only one list at rank 25. It didn't even make the top 100 list for any of the lists but one. You end up ranking it 25 because that is the book's median rank. You need to figure out how to handle sparsity such that it doesn't inhibit the result you want to get. -- kainaw™ 14:16, 6 June 2011 (UTC)

Before you get hat far you've of course got to ask yourself how the various lists of novels or whatever were compiled. Were they done by a wide poll or seeing library returns or by a journalist asking a few friends? Was it done by some teenagers or some mothers group? So you have the problems that some might have much more weight than others and some might be much more compatible with your purposes than others. Dmcq (talk) 15:56, 6 June 2011 (UTC)

This is pretty similar to the questions investigated in social choice theory—you're trying to take several rankings and produce one overall ranking. This is difficult to do well. For example, you would probably like the method you use to have nice properties, such as these:

• If every list ranks The Grapes of Wrath higher than A Clockwork Orange, then the overall ranking should also rank The Grapes of Wrath higher than A Clockwork Orange.
• If the various lists were different, but the relative rankings of The Grapes of Wrath and A Clockwork Orange do not change (in other words, every list that previously ranked The Grapes of Wrath higher than A Clockwork Orange still does so, and vice versa), then the relative rankings of The Grapes of Wrath and A Clockwork Orange should remain unchanged in the overall ranking.
• No one list determines the overall ranking; every list is taken into account.

Unfortunately, it is impossible to satisfy all of these criteria simultaneously; see Arrow's impossibility theorem. —Bkell (talk) 16:49, 6 June 2011 (UTC)

Yep there's no sound mathematical method with even very simple assumptions. It's up to people to make choices on what they find works for them. You just have to make an admission like Google who now say their page ranks reflect their judgement of what will be most relevant for a user's query rather than just that it is determined by an algorithm. Dmcq (talk) 10:04, 7 June 2011 (UTC)

Converting between the forms of equations for an ellipse

I know that the standard form for an ellipse is (x-h)^2/a^2+(y-k)^2/b^2=1 or (x-h)^2/b^2+(y-k)^2/a^2=1 depending on the major axis and the general form is ax^2+cy^2+dx+ey=f. I want to be able to convert between these two forms with a simple set of equations. Can anyone help me? --Melab±1 ☎ 21:01, 5 June 2011 (UTC)

It doesn't help that you've used the constant "a" in both forms, making it impossible to get an exact correspondence. If you take the general form as cx^2+dy^2+ex+fy=g, just expand the first form and match coefficients of x^2, y^2, x, y and the constant term.→86.132.165.117 (talk) 21:40, 5 June 2011 (UTC)

To go from the former to latter, just expand as 86.132 suggests. To go the other way, you complete the square for each variable. SemanticMantis (talk) 20:35, 6 June 2011 (UTC)

June 6

Fermat's Principle

Can Fermat's principle be used as a definition of the straight line? 128.232.240.223 (talk) 22:58, 6 June 2011 (UTC)

Plato's definition of a line was basically that which blocks the view between two points, or in a more modern interpretation, the path followed by a ray of light. (See Heath's commentary on Euclid Book I Definition 4.) Since Euclid however this kind of definition is frowned upon because it appeals to physical theory rather than mathematical truth. This is especially true in more recent times, where space itself is considered curved. The modern viewpoint is that a line is an undefined concept whose behavior is determined by the axioms of geometry.--RDBury (talk) 00:22, 7 June 2011 (UTC)

Fermat's principle is just a physics-y way of putting what is more commonly phrased "the shortest distance between two points is a straight line". That holds in Euclidean geometry, due to the triangle inequality. (In some respects, the triangle inequality is the math-y way of putting what is more commonly phrased "the shortest distance between two points is a straight line".) For non-Euclidean geometry, at least for metric spaces, the triangle inequality is a convenient property to have (although not essential). For those metric spaces with the triangle inequality, it's perfectly feasible to use it to define a line-like object, although that might not match the conventional definition of a straight line that others use for that space. -- 174.31.219.218 (talk) 15:57, 7 June 2011 (UTC)

What is the definition of the straight line that does not require discussion of linear equations then? Your article is somewhat unclear. 128.232.240.223 (talk) 19:46, 7 June 2011 (UTC)

A straight line in Euclidean space is an example of of a geodesic. — Fly by Night (talk) 20:01, 7 June 2011 (UTC)

I appreciate that yours is a correct answer but I would ask, is it a helpful one? If pressed for the definition of a real number, one might define it as an example of a tensor. I don't think we have learnt anything from this definition though. Is there a definition, which mentions linear equations neither explicitly nor implicitly (unless you regard any two, apparently, distinct definitions of a mathematical object as implicitly mentioning the other by virtue of the fact that they are equivalent definitions), of a more basic nature that is still unambiguous? Or are we in rather murky murky waters here, since the straight line underpins Euclidean geometry, and, as such, is challenging to rigorously define? 128.232.240.223 (talk) 21:25, 7 June 2011 (UTC)

If you're willing to accept the "distance between two points" as a primitive concept, then the definition of a line as a geodesic is a natural one. This is, by definition, the shortest continuous curve between two points. I don't see what your issue is. Sławomir Biały (talk) 22:16, 7 June 2011 (UTC)

June 7

Example of a graph

Can someone please give me an example of a k-regular connected graph G, (k>2) which has order 4 or more and contains a triangle but no square?-Shahab (talk) 06:14, 8 June 2011 (UTC)

Markström graph

I have an example of order 10, but I'm going to have to draw it. I did a quick search on Commons to see if there was already an image of it there, and I found the Markström graph (see right), which is another example. —Bkell (talk) 07:20, 8 June 2011 (UTC)

The Dürer graph is also an example. —Bkell (talk) 07:22, 8 June 2011 (UTC)

And Tietze's graph. (I'm running through the list at commons:Graphs in graph theory, by the way.) —Bkell (talk) 07:25, 8 June 2011 (UTC)

Three of the truncated Platonic solids, too. The truncated tetrahedron is pretty close to the example I came up with; here's a poor attempt to draw my example (the O's are the vertices):

   O---------O
  / \       / \
 /   \     /   \
O-----O   O-----O
 \     \ /     /
  \     O     /
   \    |    /
    \   O   /
     \ / \ /
      O---O

—Bkell (talk) 07:40, 8 June 2011 (UTC)