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March 14

This is Pi Day. But when, in history, was the best approximation to pi equal to 3,14 ?

Have you checked out approximations of pi? 3.14 was apparently suggested by the Chinese as "good enough" (though they had what they believed a more correct value) in AD 263. — Lomn Talk 16:18, 14 March 2006 (UTC)

This is also Steak and Blowjob Day. Happy S&BJ Day! Black Carrot 19:52, 14 March 2006 (UTC)

How cool that Pi Day just happens to be Albert Einstein's birthday! JackofOz 05:52, 15 March 2006 (UTC)

Hmm. I'm pretty sure AD 263 is wrong. I think its AD 1592. About 6 in the morning, 53 minutes and 58 seconds and 979 milliseconds after the hour. linas 06:18, 19 March 2006 (UTC)

SA of a sphere

How was the forumla for the surface area of a sphere derived? Are there any good sites with information about this? 64.198.112.210 16:41, 14 March 2006 (UTC)

It's a simple calculus problem. I'll give my own intuitive explanation, which you may or may not like: The volume of a sphere is $V={4 \over 3}\pi r^{3}$, and when the radius changes by dr the volume changes by dr times the surface area. (You can visualize this as adding or removing an infinitesimally thin spherical shell.) The surface area is then ${dV \over dr}=4\pi r^{2}$. —Keenan Pepper 18:13, 14 March 2006 (UTC)

The question is based on a faulty premise; the surface area of a sphere has been calculated many times in many ways. A proper question is "how did so-and-so do it at such-and-such a time", or "how might we do it". One of the more ancient and intriguing methods first demonstrates that we can project the sphere out onto a cylinder wrapped tightly around it at the equator without affecting the area. Now we need only calculate the area of a cylindrical band with circumference 2πr and height 2r; and this, of course, is simply 4πr². Still, this depends on being able to show that the projection leaves area unchanged. A simple argument for area preservation is that tilt of the sphere surface towards the central axis gives the projection onto the cylinder less area (by foreshortening), but expansion of the surface out to the cylinder gives more area, and these two factors exactly cancel everywhere on the sphere. Archimedes found this relationship of sphere to cylinder so appealing that he is said to have asked for it to be drawn on his tombstone. --KSmrq^T 19:43, 14 March 2006 (UTC)

March 15

A number theory problem

Find all n such that (20n+2) divides evenly into (2003n+2002)

L33th4x0r 05:46, 15 March 2006 (UTC)

Shouldn't you be doing your own homework? The way to do it is to define a variable x as the quotient of the division of (20n+2) into (2003n+2002). Then derive n as a function of x, so that every whole number value of x = 1,2,3... will give the values of n you need. --Canley 07:46, 15 March 2006 (UTC)

Found it. It's not The Answer to Life, the Universe, and Everything - quite the opposite, in fact. Gandalf61 17:23, 15 March 2006 (UTC)

About 2 D creatures traveling on a 2 D plane seen from a 3 D point of view

I've got this strange theory that I conjured out of thin air som 15 years ago, have there been anybody else than me thinking about this stuff.

I postulate hereby that C a 2D creature's fastest way (like in shortest/less resource expenditure) of traveling (in an XY plane) from point A to point B would be to travel in one direction (X or Y) (not distance AB) until it would be at the distance AB - (A - (X or Y)) = B and then the distance B.

This way the creature C would only have to change direction of travel (90 degree change) once.

Try to think about this and and tell me how much further would 2 D creature have to travel compared to a 3D (capable) creature, my answer is squareroot 2 (appr. : 1.4) times more.

Reason: while the 2D creature could change direction of travel (X or Y) for every second or mm (or whatnot) that it went, this would take even more time/resources ..stop/change direction etc.... than it would to just travl the distance in the X direction and then after that in the Y direction....one start/stop/start/stop

Relate this to a 3D creature like us in a 4D world....and the answer would be 0.5 times the value of pi.....and bring in the double slith photon/electron experiments and other relativistic points... anybody got any thoughts about this...?

Could this be elaborated a bit more? [Azalin1999]

Uh. Perhaps you should phrase your question better. Are A and B points or lines? You use them as both. What's X and Y? And if it travels from A to B in a straight line, it doesn't ever need to change its direction. --BluePlatypus 18:05, 15 March 2006 (UTC)

I think maybe I see what he's getting at, but it doesn't really have anything to do with 2 vs. 3 dimensions. It sounds to me like he's describing the Manhattan distance. At least in the first part where he's talking about 2 dimensions. If you compare the Manhattan distance (let's call it m) with the ordinary, Euclidean distance (d), between any two points in a Euclidean plane, it's not too difficult to show that $1\leq {\frac {m}{d}}\leq {\sqrt {2}}$. But I have no idea where he's getting the ${\frac {\pi }{2}}$ factor from for three dimensions. In three dimensions, you have $1\leq {\frac {m}{d}}\leq {\sqrt {3}}$. More generally, in n dimensions, $1\leq {\frac {m}{d}}\leq {\sqrt {n}}$. Nor do I see what it has to do with quantum mechanics or relativity. Chuck 18:40, 15 March 2006 (UTC)

Hi again I'm the one who phrased the question. A little note about myself. Never did score high in maths and equations but I'm really curious about this 'theory'. Much thanks for the link to the 'manhattan distance' article, I think I remeber reading about it some years ago. My native tounge is Danish so that's my excuse for the inaccurate phrasing.

I originally made some drawings on a XY crossed paper to illustrate the theory. And I had a really hard time to explain my arguments for those I presented it to (mostly indiferrent friends). And I still have difficulties atm. because the thing I would like to verify is that it would be possible to apply something like this 'manhattan distance' to distances in a 3D world seen from the perspective of a fourth dimension. Which should be possible. Basically like --BluePlatypus went about distances in an euclidian plane...

The distance 1/2 Pi was derived from projecting a imaginary path (distance traveled) through 4D space compared to what a 3D creature would have go by not using the 4th dimenson...like a 2D creature can't (see, thus) travel the path that a 3D creature can. In regards to the 'Manhattan Distance' this would be akin to a race between a cab an a helicopter going over the cityblocks.

Too bad your friends are indifferent; this stuff is fun! As Chuck wrote above, yes, you can define Manhattan distance in 3D, and you don't even need a 4D perspective to do it. I'm not sure where the pi is coming from.

As for bringing in modern physics like quantum mechanics and relativity, let's just say that they're not really compatible with a Manhattan metric. If you lived in a world where the physics was affected by a Manhattan metric, you would easily be able to experimentally determine the coordinate axes you lived on, whether you're a 2D or a 3D creature. But we've never observed any such special directions in space. Melchoir 19:30, 15 March 2006 (UTC)

That's an understatement! :) If all directions weren't equivalent, you wouldn't have conservation of angular momentum, and that'd really yank out the carpet from under physics-as-we-know-it. --BluePlatypus 20:08, 15 March 2006 (UTC)

OK, I kind of see where you're going here. First of all, it's important to remember that the ${\sqrt {2}}$ factor in 2 dimensions is a maximum; it doesn't hold for all pairs of points. For example, if you're travelling from (0,0) to (5,5), the person constrained to taxicab geometry has to travel 10 units, but the person who can travel in any direction only has to travel $5{\sqrt {2}}$ units, and the person limited to taxicab geometry has to travel ${\sqrt {2}}$ times farther in that particular case. However, if you're traveling from (0,0) to (7,24), the Euclidean traveler has to travel 25 units, but the taxicab traveler only needs to travel 31 units--just 1.24 times further than the Euclidean traveler, which is considerably less than ${\sqrt {2}}$. And of course, if you're going from (0,0) to (13,0), both people travel the same distance.

The analagous situation in three dimensions, it seems to me, is someone who can travel in any one of six orthogonal directions (which we might call "north," "south," "east," "west," "up," and "down," but it's important to remember here that these are idealized directions where all north-south lines are parallel with each other and do not meet, as are the east-west lines, and also the up-down lines; and not the directions as defined on the earth, where the "north" lines all meet at the north pole, the "south" lines all meet at the south pole, and the "down" lines meet at the center of the earth).

In this case the maximum factor between the distance of the taxicab traveler and the distance of the Euclidean traveler would be ${\sqrt {3}}$, for example going from (0,0,0) to (5,5,5) the taxicab traveler has to move 15 units, as opposed to the Euclidean traveler's $5{\sqrt {3}}$ units. But as before, that's a maximum: in going from (0,0,0) to (3,4,12) the taxicab traveler goes 19 units to the Euclidean traveler's 13, but ${\frac {19}{13}}<{\sqrt {3}}$. It's possible to choose a pair of points where the taxicab traveler's distance is exactly ${\frac {\pi }{2}}$ times as long as the Euclidean traveler's, but that's only because such a pair of points can be found for any number x such that $1\leq x\leq {\sqrt {3}}$, which ${\frac {\pi }{2}}$ happens to satisfy.

The "car vs. helicopter" analogy is an interesting way to think about it, but at the same time it's misleading because you don't actually have to move in the 3rd dimension for this to happen. You can have one traveler who can move in any direction, and one traveler who is limited to moving in a set of orthogonal directions, and still have both of them confined to a plane, so it's not really an issue of one person being constrained to two dimensions while another can move in three (or three vs. four, or four vs. five, etc.) Chuck 20:33, 15 March 2006 (UTC)

You're Chuck making some good points. But what if we assume that in our (unlimited) reference 3D space we are looking down on the 2 plane (which could be from any arbitary reference point/perspective) let's say head on, straight down.

Let's asume some more - Our little 2D creature can only observe (look) through either the X or the Y, and thus only observe it's destination when its in line with either the x axis of the destination or th Y axis, whichever comes first, depending on which route it took.

From the 3D POV this would satisfy my squareroot 2 clause about a 3D being would be able to go the shorter distance right? Because it can sort of observe the direct line of vision...X & Y combined.

Now if the 2D creature were able to in some way observe the destination it could of course just travel the direct route. But this would only happen after it had moved all the way along either the X or Y axis to be in line with the destination. If this 2D thing were sentinent the percieved distance the 2D being traveled would be? the squared X+Y which is less than the distance it really took if a 3D being measured...also there are 2 routes that are the same distance but starting with etiher going out of the X or the Y axis...

Relate this to a 3D world...(damn crunching)hmm non-euclidian geometry..........Spherical trigonometry

That's all true--but the "2D" and "3D" specifications are red herrings. What is important is that you have one creature which can only travel parallel to the X and Y axes, and another that can travel any direction within the XY-plane. The first creature, who can only travel parallel to the X and Y axes, will have to travel at least 10 units to go from (0,0) to (5,5). It doesn't matter if he's two dimensional, or if he can perceive three, or four, or eighteen dimensions, as long as his movement is limited to directions parallel to the X and Y axes, he has to travel at least 10 units. Similarly, for the creature that can travel any direction in the XY-plane, he can get from (0,0) to (5,5) by moving only $5{\sqrt {2}}$ units, regardless of whether he can travel, or even see, in any additional dimensions or not. (For this second creature, imagine not a helicopter, nor a car confined to a grid of city streets, but instead a car on a wide open dry lakebed.)

In other words, it's true that in your example with a 2D-creature confined to moving parallel to the X and Y axes, and a 3D-creature which can move in any direction, everything you say is true. But your choice of "2D" and "3D" are arbitrary, and there is nothing special about 2-dimensional or 3-dimensional creatures which makes your statements true; I could just as easily talk about a dog which could only move parallel to the X and Y axes, and a cat which could move in any direction, and the statements would all still be true. Chuck 22:44, 15 March 2006 (UTC)

Graph Gradients

Hi, I'm working with gradients at the moment but can't for the life of me rember if you count the squares (i.e. measure) or use the unit values for the graph. To see an example of what I mean: Click here is the gradient 3 over 6 (as attained by counting squares) or 15 over 12 (as attained by using the unit values). If it helps I'm dealing with "real life" data here. Thanks! --Christopher Denman 17:50, 15 March 2006 (UTC)

I guess that would depend on the context, but since the scales of the x- and y- axes are different, it's more plausible to discuss the units and not the squares. So the slope is 15/12 = 5/4. -- Meni Rosenfeld (talk) 18:01, 15 March 2006 (UTC)

You need to use the units, otherwise you'll be off by a factor 5/2, since you presumably want the slope given in "y units per x unit", and not in "5 y units per 2 x units" the latter being the equivalent to "y lines per x line". --BluePlatypus 18:11, 15 March 2006 (UTC)

Thanks very much both of you for your answers!--Christopher Denman 18:34, 15 March 2006 (UTC)

When we talk about a road having a steep gradient, the term "gradient" has a different meaning than is usual in mathematics, where the common term would be "slope". To compute slope, a dimensionless number, we divide the vertical displacement by the horizontal displacement, being careful to use the same measurement units for both. Your graph is therefore an invitation for trouble, misleading both the eye and the calculation by displaying as a square what is actually a rectangle 5 high and 2 wide. See the slope article for a more extended discusssion. --KSmrq^T 21:18, 15 March 2006 (UTC)

Making things appear 3D without axes

Two ropes

On right is a plot of x-, y-, and z- coordinates of two ropes on which some forces are acting. I made this using the plot3 command in matlab. Now I want to remove the axis labels, and marks. In that case, this will look like 2d. How can I make it look 3d? Some kind of shading/lighting/perspective tricks? I need to do this in MATLAB. Thanks! deeptrivia (talk) 21:39, 15 March 2006 (UTC)

It depends on what you want to emphasize, but I guess that's not too helpful... how about conical arrowheads? I have no idea about MATLAB. Melchoir 23:21, 15 March 2006 (UTC)

Do a little reading on depth cues. Two of the easiest for you might be motion and shadows/occlusion. If the curves were rendered as ropes, the variations in texture size would be a natural cue. For simplicity, let MATLAB provide data points along the curve (or spline coefficients), and use a free rendering package like POV-Ray or Blender to produce the fancy images. This would allow fog (aerial perspective), for example. --KSmrq^T 23:48, 15 March 2006 (UTC)

Thanks for your answers. I tried shadows and it helps a bit. I have no choice but to use matlab because this will be a part of a GUI I'm making in Matlab. I think I can manage if I get a solution to the problem regarding cylindrical surfaces below. Cheers, and I appreciate it highly. deeptrivia (talk) 01:30, 16 March 2006 (UTC)

You could also use color, with, say, red increasing in the X direction, blue in the Y direction, and green in the Z direction. This method is particularly useful for surface plots. Showing the graph from different views would be another option. StuRat 03:48, 16 March 2006 (UTC)

Or use perspective, can u make thickness of the lines varying? (Igny 18:04, 20 March 2006 (UTC))

definition of manipulatives

I need the definition in mathmatical terms for manipulatives

Aren't those those little wooden blocks that come in sticks of ten, plates of a hundred, and cubes of a thousand? Those things were great. —Keenan Pepper 01:01, 16 March 2006 (UTC)

Cuisenaire rods are one of the early tools for "hands-on" mathematics teaching. Today there are many others. --KSmrq^T 03:28, 16 March 2006 (UTC)

March 16

Equation of cylindrical surface

This is a follow up question from above (3D effects without axes). Suppose we have a spatial curve defined by x = f(s), y = g(s) and z = h(s). What equations will describe a cylindrical surface of radius r which has this curve as its axis? I can do real neat 3D things in matlab if I plot this surface. Will post the results here, of course. deeptrivia (talk) 01:28, 16 March 2006 (UTC)

I'm not sure what you mean. Are you talking about an envelope of spheres centered on the curve? —Keenan Pepper

Well, I'm not 100% sure what would an envelope of spheres exactly mean, but I think that's what I mean. Think of a tube/pipe (eg a water hose) whose axis is twisted to coincide this curve. I need the equation of the surface of the tube. deeptrivia (talk) 02:39, 16 March 2006 (UTC)

Assuming the curvature never vanishes, you can use

$r'(s,\theta )=r(s)+N(s)\cos \theta +B(s)\sin \theta ,$

where N and B are defined at the article on the Frenet-Serret formulas. Melchoir 02:52, 16 March 2006 (UTC)

Thanks for the answer. Since this is a 3D curve, we need r, $\theta$ and $\phi$ to describe the surface. deeptrivia (talk) 03:01, 16 March 2006 (UTC)

Oh...I think I better understand your answer now. Let me try. Thanks ! deeptrivia (talk) 03:04, 16 March 2006 (UTC)

For everyone's benefit. I got a readymade solution based on Melchoir's answer. See [1] deeptrivia (talk) 03:08, 16 March 2006 (UTC)

Yeah, I should have figured that it's been done. Well, it doesn't hurt to rederive the wheel every now and then, eh? Melchoir 03:11, 16 March 2006 (UTC)

Surface Area of a Rotated Curve

I am writing this as an AP Calculus BC student. Assume f(x) describes a continuous, differentiable curve, such as

$f(x)={\sqrt {x}}$

I want to find the surface area of the solid that results from a rotation of part of the curve, say 0 to 4. For this example, let's assume I'm rotating about the x axis. I know the correct method to solve this problem (surface areas of cone frustrums). However, why can't it be found as a summation of the surface areas of right circular cylinders (excluding the bases)? Thus, my presumed integral would be

$\int _{0}^{4}2\pi r\,dx=$

$\int _{0}^{4}2\pi {\sqrt {x}}\,dx$

Obviously, it is wrong. However, why is it wrong? My textbook helpfully includes the formula above (though I found it myself before seeing this). It then says it can not be used because it has "no predictive value and almost never gives results consistent with other calculations." However, I find this somewhat disappointing, as there is no explanation of why the formula is flawed. Please do not simply show me the right way to do it, and explain that this is therefore the wrong way; that is unhelpful Superm401 - Talk 01:58, 16 March 2006 (UTC)

Try visualizing the geometric interpretations of the two formulas really close to the origin, where the slope of f is infinite. The correct formula accounts for the "part" of the surface area facing perpendicular to the x-axis; the incorrect formula cannot. Whatever you do, don't try to visualize the difference where the slope of f is small. Melchoir 02:21, 16 March 2006 (UTC)

Thanks. That's a helpful distinction. I can begin to see the problem now, though a more formal explanation would help too. Superm401 - Talk 02:46, 16 March 2006 (UTC)

When calculating the area under a curve, one can approximate either by rectangles, or by trapezoids bounded by some "diagonal segments" on top. I think the second approximation is called the trapezoid method, and it's better than the rectangle method, but both work in the limit. Both converge to the Riemann integral.

On the other hand, when calculating the length of that same curve, you have to approximate the curve by diagonal chords. If you approximate the curve with rectangles, the sum of the rectangles will always add up to b–a (where the domain of the curve is the interval [a,b]).

The technical reason behind this is that the form which gives you the length is not a linear differential form. In other words, the Pythagorean equation is not linear, and that's needed to calculate length. The form that gives you the area is linear, and therefore any sum will do. Stated even more simply, the triangle inequality tells you that the hypotenuse of a triangle is greater than the leg. Rectangles get you the leg of the triangle, and never approximate the length, though they do approximate the area. Calculating surface area is just calculating length and throwing in a 2πr, so the same issues apply to volume versus surface area. -lethe ^{talk +} 03:31, 16 March 2006 Th(UTC)

Thank you. That's very helpful. Superm401 - Talk 04:18, 16 March 2006 (UTC)

To illustrate the problem somewhat differently, consider a cone obtained by rotation of "y=ax" around the x-axis. The area and the volume are given by integrals

$S=\int _{0}^{h}2\pi r(x){\frac {dl}{dx}}dx=\int _{0}^{h}2\pi ax{\sqrt {1+a^{2}}}dx=\pi h^{2}a{\sqrt {1+a^{2}}}$

$V=\int _{0}^{h}\pi r^{2}(x)dx=\int _{0}^{h}\pi (ax)^{2}dx={\frac {1}{3}}\pi h^{3}a^{2}={\frac {1}{3}}\pi hr^{2}(h)$

You can see that omitting dl/dx will lead to a wrong result. For the curve $y={\sqrt {x}}$, dl/dx varies with x. Another example is the surface area of a unit circle

$S=\int _{-1}^{1}2\pi r(x){\frac {dl}{dx}}dx=\int _{-1}^{1}2\pi r(x){\sqrt {1+\left({\frac {dr}{dx}}\right)^{2}}}dx=\int _{-1}^{1}2\pi {\sqrt {1-x^{2}}}{\frac {1}{\sqrt {1-x^{2}}}}dx=4\pi$

(Igny 13:04, 16 March 2006 (UTC))

And how do you know that result is wrong? Because it has "no predictive value and almost never gives results consistent with other calculations"? -lethe ^{talk +} 17:13, 16 March 2006 (UTC)

Volume of cone

can anyone give the derivation for this?

A cone is a special type of pyramid, and the derivation is of the volume formula is therefore the same. It involves finding the area of a "step" pyramid, and then making the steps finer and finer. You can take a look at this site. Sjakkalle (Check!) 13:59, 16 March 2006 (UTC)

Or alternatively:

$V=\int _{0}^{H}\pi \left({\frac {hR}{H}}\right)^{2}dh={\frac {\pi R^{2}H}{3}}$

-- Meni Rosenfeld (talk) 18:11, 16 March 2006 (UTC)

It's very interesting that the volume of a pyramid (including a cone) is equal to one-third of the volume of a prism with the same base area and height. :P —OneofThem 19:10, 16 March 2006 (UTC)

I guess so, but it's just a special case of the fact that the volume of an nth dimensional pyramid is 1/n the volume of an nth dimensional prism - which itself is a consequence of the equality

$\int _{0}^{1}t^{n-1}dt={\frac {1}{n}}$

-- Meni Rosenfeld (talk) 19:35, 16 March 2006 (UTC)

determinat of a parabola

how do you find the determinant of a parabola?? please help

Did you mean, "discriminant"? The discriminant of the parabola $y=ax^{2}+bx+c$ is $\Delta =b^{2}-4ac$. -- Meni Rosenfeld (talk) 18:13, 16 March 2006 (UTC)

March 17

discontinuous linear functional on test spaces

On page 74, L. Schwartz, Mathematics for the Physical Sciences,Hermann

Quote: The existence of linear functionals which are discontinuous on $\script D$ may be demonstrated mathematically using the axiom of choice. UN-Quote.

Question: Where can I actually find a proof for the existence of a discontinuous linear functional on test spaces?

Be more grateful if you could send your answer directly to me by visiting http://www.maths.uwa.edu.au/~twma/mathematics/index.htm Thank you in advance. twma

See discontinuous linear functional. -lethe ^{talk +} 04:12, 17 March 2006 (UTC)

Dear Lethe, Thank you for your quick response. Because you did me a favor to visit my page above, I was alert immediately without having to wait until tomorrow.

On page 73 of the same book above, a sequence of test functions f_n tends to 0 if

(a) the supports of f_n are contained in the same bounded set and

(b) for each fixed m, the m-th derivative of f_n tends to 0 uniformly.

\sin n^2x/n could not satisfy the above conditions.

I multiplied \sin n^2x/n with a test function g satisfying g(0)=1. Could not get any further. STILL NEED HELP. twma

You have not broken any of our rules, this kind of thing is what this page is here for. Welcome. Now, as for your question, I guess you're talking about the topology of smooth functions of compact support. This is the topology used to define the space of distributions (the space of distributions is the continuous dual space of the space of smooth functions of compact support with the topology you mention: uniform convergence on compact sets of all derivatives). Now what exactly is it that you want to do? The sequence you mention does not converge; the function itself converges uniformly, but none of its derivatives do. But what task are you trying to accomplish? I do not have this text by Schwartz. -lethe ^{talk +} 06:13, 17 March 2006 (UTC)

To find, explicitly by displaying formulas or implicitly by axiom of choice, a counter example that a linear functional T is not continuous, we need to find a sequence of test functions f_n satisfying both conditions (a) and (b) and in addition T(f_n) fails tending to 0 as n tends to infinity. The sequence \sin n^2x/n does not satisfy (a) and not (b).

Distribution theory is well established nowadays. There are many good books on the subject. For example,

Page 222, F. Treves, Topological vector spaces, distributions and Kernels, Academic.

Pages 28, 26, M.A. Al-Gwaiz, Theory of distributions, Monographs and textbooks in pure and applied mathematics, vol 159.

I quote L. Schwartz because the book was written by the creator. I even looked up the French version of the same book with disappointment. My next possible step is to look up the original paper. Twma 18:52, 17 March 2006 (UTC)twma

You people have something like --- 17 March 2006 (UTC) but I do not have it. Am I overlooking some rules and regulations of this site? If so, apologize and please tell me what to do in order to CONFORM.

See top of the page, which explains that you just have to type four tildes to sign (gives IP + date when not logged). --DLL 06:18, 17 March 2006 (UTC)

When editing, just above the window is a row of buttons. Hovering the cursor above them should popup text describing their function. The antepenultimate one, the scrawl between the red no-wiki and the dash, adds a signature with one button click. For those who are typing challenged, this may be easier. Also note that the contents of a signature can be changed by using the "my preferences" link at the top of the page while logged in. This is how some of us include a link to our talk page. --KSmrq^T 16:31, 17 March 2006 (UTC)

I wrongly thought that four tildes are examples of letters of a signature. I am able to conform now. Thank you. My identity is twma in lower case but this site changed it by default to Twma in order to conform.

To save the room, I shall delete part of this section if I have no objection in 24 hours. Twma 18:52, 17 March 2006 (UTC)twma

Which room? What do you want to delete and why? -- Meni Rosenfeld (talk) 19:14, 17 March 2006 (UTC)

The buttons row is java script, does every browser allow it ?

Everything is interesting in our discussions. Errors are amongst the most because everyone may learn from them. There is enough space (room) left on the page, thank you. --DLL 20:50, 17 March 2006 (UTC)

All right! People want the complete record even full of irrelevant junk such as part of mine. Good to consult first and take action later. No action in this case. Twma 22:57, 18 March 2006 (UTC)

OK, twma, now I think I see what you were trying to do. You want a sequence of functions f_n and a linear functional T such that f_n → 0 but T(f_n) does not → 0. But you chose a sequence of functions which does not converge, so clearly that is not going to show anything (your sequence doesn't satisfy your (a) and (b)). Now, the article discontinuous linear functional teaches us that on a complete vector space, all examples are nonconstructive, so you can't just look for a sequence and a functional, you have to invoke the axiom of choice. The vector space you have chosen is complete (though not in the uniform norm). The examples look like the one you find in that article. Let f_n be a sequence of linearly independent functions. Denote by ||g||_m,K the seminorm sup_K D^m|g|. If {K_i}_i is any countable collection of compact sets which cover the real line, then ||*||_Ki,m is a countable sequence of seminorms. Create a directed countable family of seminorms, and denote the kth seminorm by ||*||_k. Then define a linear functional T such that T(f_n) = n||f_n||_n. Then we have for any k, I can choose an f_n so that |T(f_n)| is greater than any ||f_n||_m (or even any finite linear combination thereof). Thus T is unbounded. Now use the axiom of choice to find a basis of the space which contains the f_n, and define T to be 0 on the rest of the vectors. Any unbounded linear map is discontinuous.

Now, this example is the same in essence as the one at discontinuous linear functional, except that I had to get involved with families of seminorms, since the vector space in question is not a Banach space. This makes the details of the proof a little more involved, but doesn't change the idea behind it. Nevertheless, I found it quite confusing dealing with those seminorms, I wonder if you'll find it just as confusing when trying to read it as I did trying to write it. -lethe ^{talk +} 21:39, 17 March 2006 (UTC)

I do NOT know how to produce nice looking mathematical stuff on this page. Be grateful if SOMEBODY could REPLACE this and next paragraph with something equivalent. It compiles well before I pasted here. Thank you in advance. PS: Somebody has done this. Thanks from twma.

twma's example

Here is an example of a discontinuous linear form on the test space D(R) on the real line R. For every compact subset K of R, D(K) is the vector space of all test functions with support contained in K. The topology of D(K) defined by the seminorms $\|\varphi \|_{\alpha }=\sup\{|(d/dx)^{\alpha }\varphi (x)|:x\in K\}$ for all integers α ≥ 0 coincides with the subspace topology induced by the test space D(R) equipped with the locally convex inductive topology induced by D(K) for all compact subsets K of R.

Let ρ be a test function with ρ(0) = 1 and ρ(x) = 0 for all |x|>1. Choose 0<a_n+1<b_n+1<a_n<b_n ≤ 1 for all n ≥ 0, for example recursively by b₀=1, a_n=b_n/2 and b_n+1=a_n/2. Let c_n=(b_n+a_n)/2 and r_n=(b_n – a_n)/2. For g_n(x)=ρ[(x – c_n)/r_n], we have |g_n|₀ ≥ g_n(c_n) = 1. Then each $f_{n}={\frac {g_{n}}{n\sum _{\beta =0}^{n}\|g_{n}\|_{\beta }}}$ is a test function with support contained in [a_n,b_n] ⊆ K=[-1,1]. For each α and all n > α, we have ||f_n||_α ≤ 1/n → 0. Now f_n → 0 in D(K) and hence also in D(R). Next because all intervals [a_n,b_n] are disjoint, the sequence {f_n} is linearly independent and hence it can be extended to an algebraic basis B of D(R). Define T(φ)=1 for all φ in B and extend T to a required linear form on D(R).

By the way, I am also looking for a native English speaking functional analyst to work on a book. Anybody who is no longer active in research but needs something to work with would be ideal. For further information, please contact me without any commitment from your part. You may find my email address from the find-people-button of the page https://www.maths.uwa.edu.au/index.html Thanks to the enthusiasm of people of this group that provides encouragement to me in sharp contrast to DingoBabyAffair. Twma 08:24, 19 March 2006 (UTC)

I'm trying to read through your example, but your ρ function bothers me. It is discontinuous at x=0, whereas we usually choose test functions to be smooth. For starters, none of the seminorms is defined for this function (except the 0th one). Well, anyway, examples are not hard to come by. Just find any linearly independent sequence and make an unbounded operator on it. PS I'm glad you find this place hospitible; feel free to hang around or come back any time. But who or what is DingoBabyAffair? -lethe ^{talk +} 08:54, 19 March 2006 (UTC)

Let ρ be a test function with ρ(0) = 1 and ρ(x) = 0 for all |x|>1 (typo corrected). Movie: Lost in translation. Hope it should be all right now (waiting for your approval). In order to make life easier for myself, I only look at the positive side but rarely at the ugly part of the world until I HAVE TO face it NOW. I agree that based on one example, we can find many more without any difficulty but we DO need one example for a copycat. ONCE AGAIN, THANKS TO THE HELPING HANDS OF THIS GROUP. DingoBabyAffair appeared on the front page of many Australian newspapers for several years in early seventies. A baby had been severely hurt continuously but nobody wanted to help while the politicians argued whether the surrounding dingoes or the mother should be responsible. In SHARP CONTRAST, this group helps without any argument. After more than 30 years of protracted war, this baby grew up as a lost child of cold war. Now he/she is rising like a fabulous phoenix from the ash, new and fresh and young. Saving the Private Ben happened only in the movie while the politicians argued once again whether it did happen recently in reality. Perhaps I should claim copyright for another movie. Am I violating the rules and regulations of this site by comparison in my thank-you-note and by answering query about something irrelevant to Mathematics? Any way, I do not think it will drag on. Twma 00:51, 20 March 2006 (UTC)

OK, with the typo fixed (and the fix was obvious, I should have been able to figure it out), I think it's a fine example. In the end, you have lim f_n=0, but lim T(f_n) = 1. I chose to make my operator unbounded, while you chose to make the discontinuity explicit. We can also check that your operator is unbounded: for any α, 1/||f_n||_α outgrows any bound. You also explicitly constructed a convergent independent sequence, which was nice. I like it. About the DingoBabyAffair, now I realize you're talking about the Azaria Chamberlain disappearance. I think I saw a made-for-TV movie about that when I was a little kid. And then there was some famous scene in Seinfeld about it. So you're comparing the response you got on other internet sites to the public response to the dingo affair? Well OK, whatever. Anyway, you're not violating our rules. You came here with a good math question. It's OK to chat about other things on the side while you're doing math, I think. -lethe ^{talk +} 15:42, 20 March 2006 (UTC)

PS I've copied your example to talk:discontinuous linear map, for the possibility of inclusion in our article. Hope you don't mind! -lethe ^{talk +} 17:39, 20 March 2006 (UTC)

• I am NOT talking about different internet sites but the mentality of different ENVIRONMENT.
• In the academic world, we do not make money. If we are lucky, we make a name. In this case, your name should be with the example too. Do whatever people want. I do not claim copyright and take no responsibility if any mistake is found. It appears that THIS CASE IS CLOSED. Thanks again. Perhaps I should help the others whenever I have one or two days free. I got this site through some search engine. Twma 23:47, 20 March 2006 (UTC)

Simple Equation

How would one solve the equation 3^x = x^2? Is there a rule which specfies which explains the solvability of equations? --Alexs letterbox 05:35, 17 March 2006 (UTC)

Take the square root of both sides, and then use Newton's method to try to find the fixed point. There might be more than one fixed point for the different signs, so consider positive and negative signs first. --James S. 05:44, 17 March 2006 (UTC)

As a total cheater, I graphed it on my TI-83 and got x = -0.6860276 --DevastatorIIC 07:41, 17 March 2006 (UTC)

I asked something similar a few weeks ago (but was 2^x = x^2). Apparently there's a tool for this, called Lambert's W Function ☢ Ҡi∊ff⌇↯ 08:06, 17 March 2006 (UTC)

Yes, this equation can't be solved using elementary functions. It can be solved numerically using your favorite numerical method, or with Lambert's W function, in terms of which the solutions are:

$-{\frac {2W(\pm {\frac {\ln {3}}{2}})}{\ln {3}}}$

One of them is real, and is roughly -0.686026724536251319. -- Meni Rosenfeld (talk) 08:38, 17 March 2006 (UTC)

Hello: Angle questions

(Title was changed to actually say something useful. StuRat 16:01, 17 March 2006 (UTC))

I know this might be a simple question but can someone please explain to me what Perpendicular mean? And the angle of Depression and Elevation mean ? Thankz

Is there something wrong with the article Perpendicular? —Keenan Pepper 13:09, 17 March 2006 (UTC)

About the angles of depression and elevation, I guess that would depend on the context. -- Meni Rosenfeld (talk) 15:41, 17 March 2006 (UTC)

Perpendicular is used to describe the convergance of two lines so that they form a ninety degree angle. If a line is given, then naturally a perpendicular line is at a ninety degree angle to that line.65.95.44.127 17:12, 17 March 2006 (UTC)

Generally, angles of depression or elevation are the angles you have to look down or up (respectively) to see something. I found a concise explanation for you. Superm401 - Talk 20:32, 18 March 2006 (UTC)

Primorials

I need to know the growth rate of Primorials(for factorials it is X!=O(e^(x*ln(X)) using Big O notation.) I also need to know how fast you can calculate primorials, is it polylogarithmic? For factorials it is not. Ozone 23:12, 17 March 2006 (UTC)

You're wrong about the growth rate of factorials, but you're close. Sterling's approximation is

$n!\sim {\sqrt {2\pi n}}\left({\frac {n}{e}}\right)^{n}$

As for primorials,

$\pi (x)\sim \int _{2}^{x}{\frac {\mathrm {d} t}{\ln(t)}}$

suggests, (using some manipulations I am unable to generate here, because I can't figure out how to get the "#" symbol within a <math></math> equation)

n# =O(e^n)

— Arthur Rubin | (talk) 22:44, 18 March 2006 (UTC)

How about $n\#\sim e^{n}$? —Ilmari Karonen (talk) 00:20, 19 March 2006 (UTC)

Pretty! ☢ Ҡi∊ff⌇↯ 03:35, 19 March 2006 (UTC)

Actually, that formula is wrong, because of the variability of prime gaps. How about (at least as a conjecture),

$n\#=O\left(e^{n}\right)$

— Arthur Rubin | (talk) 17:28, 19 March 2006 (UTC)

Just testing my understanding here: What you mean is that the ratio ${\frac {n\#}{e^{n}}}$ is bounded, but does not converge to a limit? -- Meni Rosenfeld (talk) 17:43, 19 March 2006 (UTC)

I think it's bounded, but if it converged to a limit (other than 0), then the prime gaps would be bounded. — Arthur Rubin | (talk) 17:46, 19 March 2006 (UTC)

I'm almost certain that $\ln n\#=n+o\left(n^{\alpha }\right)$ for α = 1, and possibly (assuming the Riemann hypotheses) for α = 0.5. The conjecture that ${\frac {n\#}{e^{n}}}$ is separate. I stand by it as a conjecture, but I don't have an idea of a proof. — Arthur Rubin | (talk) 18:24, 19 March 2006 (UTC)

March 18

covariant hom functor is left exact

http://en.wikipedia.org/wiki/Exact_functor

this article defines left exact functor

Suppose 0→A→B→C is an exact sequence

I am having difficulty to see that the hom(M,-) is a left exact covariant functor. Why is it left exact? Somehow it seems to me that that means proving that if Y is a submodule of X, and alpha is a module morphism of Y to Z, it can be extended to the full X?

Thanks,

Evilbu 16:48, 18 March 2006 (UTC)

No, the statement you have given is the statement of right exactness of contravariant hom (i.e., hom(-,X)), and as you apparently have noted, this does not hold. The corresponding statement in the context you want says that if you can map M into A, and A is a submodule of B, then the composite map M into B uniquely determines the original map. Perhaps that's a little more believable? —Blotwell 17:13, 18 March 2006 (UTC)

Solving an equation

How would you solve this equation:

$L={\sqrt {J^{2}-P^{2}}}+{\sqrt {K^{2}-P^{2}}}$

for P? (So that P is only on one side.)

Thanks for any help.

65.31.80.100 19:12, 18 March 2006 (UTC)

You need to raise both sides of the equation to the second power, move the remaining square root to its own side, and then once more raise to the second power. From there it's quite straightforward - it's just a biquadratic equation in P. -- Meni Rosenfeld (talk) 19:32, 18 March 2006 (UTC)

I don't think that would work. By squaring the right side, you would not remove the roots. The end result would look something like this:

$L^{2}=J^{2}-2P^{2}+2{\sqrt {J^{2}K^{2}-P^{2}J^{2}-P^{2}K^{2}+P^{4}}}+K^{2}$

or something along those lines. The point is that you do not remove the roots. - APower

You didn't read the whole thing. You square both sides, removing one of the roots. Then you isolate the root and square again. Superm401 - Talk 20:37, 18 March 2006 (UTC)

In other words, we produce

$4L^{2}P^{2}=-J^{4}+2K^{2}J^{2}+2L^{2}J^{2}-K^{4}-L^{4}+2K^{2}L^{2}\,\!$

To solve this, we must assume that L is non-zero. --KSmrq^T 06:31, 19 March 2006 (UTC)

And of course, if L=0, then the solution is J=K=P. -- Meni Rosenfeld (talk) 14:36, 19 March 2006 (UTC)

Be careful; that's too restrictive. Although we assume the (non-negative) principal square root, so each square root must separately vanish in the original formula, that still leaves the signs of J, K, and P to vary freely. And to be thorough, we should note that since we have no control over J and K, there may be no solution. --KSmrq^T 19:54, 19 March 2006 (UTC)

You're right, that's what I meant, I guess I didn't write it accurately enough. -- Meni Rosenfeld (talk) 10:44, 20 March 2006 (UTC)

The signs of J, K and P will be arbitrary anyway, since only their squares appear in the equation. So one might as well just restrict the problem to non-negative values and note that other symmetric solutions also exist. —Ilmari Karonen (talk) 16:22, 20 March 2006 (UTC)

Multiplication by the conjugate (${\sqrt {J^{2}-P^{2}}}-{\sqrt {K^{2}-P^{2}}}$)and some algebraic manipulations result in a system of equations

${\sqrt {J^{2}-P^{2}}}+{\sqrt {K^{2}-P^{2}}}=L$

${\sqrt {J^{2}-P^{2}}}-{\sqrt {K^{2}-P^{2}}}={\frac {J^{2}-K^{2}}{L}}$

Which we can solve to get

$P^{2}=J^{2}-\left({\frac {L^{2}+(J^{2}-K^{2})}{2L}}\right)^{2}=K^{2}-\left({\frac {L^{2}-(J^{2}-K^{2})}{2L}}\right)^{2}$

Interestingly enough, unless I made a mistake with signs I get

$P^{2}={\frac {(K+J-L)(K+L-J)(L+J-K)(L+J+K)}{4L^{2}}}$

(Igny 18:42, 20 March 2006 (UTC))

The signs are ok, but it should be L^2 in the denominator and not L^4. In any case, I'm not sure your method is really easier. -- Meni Rosenfeld (talk) 19:34, 20 March 2006 (UTC)

Thanks, I corrected this. The methods (Viète formula and quadratic formula) are essentially equivalent. (Igny 19:43, 20 March 2006 (UTC))

March 19

Integral of a Rational Function

I know all about integration by parts, but is there any one integral formula that can find the integral of (a^nxn + a^(n−1)x(n−1) + ... + a2x^2 + a1x + a0) / (b^nxn + b^(n−1)x(n−1) + ... + b2x^2 + b1x + b0). This is clearly the classic definition of a polynomial divided by a polynomial. Is there an integral formula that uses similar notation to account for all possible terms? --Chris 03:24, 19 March 2006 (UTC)

I doubt there is. Any such formula would have to be capable of spontaneously generating factors of pi; for an example that's recent in my memory see the integral in Proof that 22 over 7 exceeds π. So you can't expect a simple function of the coefficients and endpoints. Melchoir 03:46, 19 March 2006 (UTC)

Recall that transcendental functions can be the antiderivatives of rational functions. -lethe ^{talk +} 05:08, 19 March 2006 (UTC)

Yes, but I'm trying to point out that the problem in question isn't going to be easy. The integral of a polynomial is another polynomial, and the coefficients are simply related. For a rational function, there's no hope of a formula using "similar notation". Melchoir 05:14, 19 March 2006 (UTC)

Well, I think your argument about π doesn't seem relevant. -lethe ^{talk +} 05:31, 19 March 2006 (UTC)

It proves by counterexample that the definite integral of a rational function is not itself a rational function with rational coefficients of the original coefficients and endpoints. I could've made that point with a simpler (and less interesting) example like 1/x, but it's still worthwhile to point out, and I suspect that it speaks to the question. Melchoir 06:31, 19 March 2006 (UTC)

I suppose you mean indefinite integral. The definite integral of any integrable function is just a number. I guess if your only point was that the antiderivative of a rational function need not be a rational function, then of course I agree with you. But that doesn't meaan that we can't write down a formula for the antiderivative of rational functions. On the contrary, we can. -lethe ^{talk +} 07:03, 19 March 2006 (UTC)

I meant the definite integral, to be considered as a function of all the numerical inputs to the problem, which I leave indeterminate. Chris is a college student, and he probably has access to the method of partial fractions. He asks for "one integral formula" using "similar notation", and neither of us has such a thing. The review of partial fractions below is valuable, as is the Risch algorithm. But to start with, there is nothing wrong with explaining why some things can't be done. Melchoir 07:27, 19 March 2006 (UTC)

Um, so you refer to indefinite integrals as "definite integrals with indeterminate bounds of integration"? Weird, but I guess that's your prerogative. In that terminology, is there even any distinction between definite and indefinite integrals? Do you even need two terms? Anyway, I'm still not clear on what it is that can't be done. Is it that the notation can't be kept similar? Or is it that the relation between the coefficients can't be kept "simple", for some definition of simple? -lethe ^{talk +} 08:07, 19 March 2006 (UTC)

Ugh, no, I meant something else, but that's not important. As for how simple a formula one might concoct, we can only speculate, and to that end, it is helpful to consider what feats that formula is expected to perform. That's enough. Melchoir 08:25, 19 March 2006 (UTC)

Chebyshev's theorem says that

$x^{p}(a+bx^{r})^{q}$

has an elementary antiderivative if and only if one of (p+1)/r, q, or (p+1)/r + q is an integer. But I guess q is always an integer for rational functions, so this doesn't answer your question. Probably the answer is: perform synthetic division, factor the denumerator, and integrate by partial fractions. I think you can write a general formula this way. -lethe ^{talk +} 04:40, 19 March 2006 (UTC)

So like, assume you're working over the complexes, so that your polynomial can be factored. And assume that the denominator has greater degree than the numerator; you can always make it so by synthetic division. Then you're left with an integral of the form

$\int {\frac {x^{n-1}+a_{n-2}x^{n-2}+\dotsb +a_{0}}{(x-r_{1})^{m_{1}}\dotsb (x-r_{q})^{m_{q}}}}\,dx,$

with

$m_{1}+\dotsb +m_{q}=n,$

so there are q distinct roots, and the i^th root has multiplicity m_i. You find the integral of this bad boy by partial fractions. The partial fraction method gives you a general formula for the coefficients. The article on partial fractions gives you this formula for the case that all the roots are distinct (have multiplicity 1). I don't know about the general case. Then you perform the integration. For the distinct root case, you're left with

$\sum c_{i}\log |x-r_{i}|.$

For the general case, once you have the coefficients, the integrals are equally easy.

Note that you can't write the entire thing in one formula in terms of the coefficients of the polynomials, because by a result of Abel, there is no formula (in terms of finitely many radicals) for the roots of a polynomial. -lethe ^{talk +} 05:05, 19 March 2006 (UTC)

Try reading about the famous Risch algorithm, and its relatives. --KSmrq^T 06:39, 19 March 2006 (UTC)

Probability

There's something basic I don't understand about probability. It isn't the manipulation of the numbers (which has never been a problem for me), it's why the whole thing's possible. What about a natural series of events causes them to fall, given a large sample size, so perfectly into a particular arrangement like 50/50 or 16.7/16.7/16.7/16.7/16.7/16.7? This might be a better question for the science desk, but it seems to fit here too, and most people work both. Black Carrot 22:05, 19 March 2006 (UTC)

Have you read Law of large numbers? —Keenan Pepper 22:28, 19 March 2006 (UTC)

Depending on the level of sophistication you wish for the answer, these explanations may or may not help you. Suppose you throw a perfectly symmetrical coin 1,000,000 times, and calculate the h, the proportion of heads. Then:

1. The probablity that you will get, say, h<0.49 or h>0.51, can be calculated and shown to be extremely small. This is a kind of a circular argument, explaining probability by using probability - But it shows the the concept is internally consistent.

1. Suppose you get h=0.6. That would mean that the coin "prefers" to land on heads. But why would it do that? We assumed it was perfectly symmetrical. The same would happen if you got h=0.3. Since the coin has no preference, the only "logical" result is h=0.5, with equal proportions to heads and tails. Of course, deviations can occur. But if you threw the coin only 10 times, deviations can be attributed to "coincidence". If you throw it a million times, you can no longer attribute a significant deviation to a "coincidence", but only to a real preference of the coin - Which, again, it shouldn't have.

-- Meni Rosenfeld (talk) 10:38, 20 March 2006 (UTC)

If you accept that the chances of a combination of random (equally weighted) events occurring is 1/c, where c is the total possible combinations of events, then you could demonstrate it yourself. This is because the more tosses of a coin, say, you do, the more combinations of all tosses will end up in any given range about the predicted average, say the 0.4 - 0.6 range. However, the chances of getting exactly 0.5 will actually go down (and of course that probability is zero for an odd number of tosses).

As the number of tosses of a coin goes up, the chances of having all heads or all tails goes down. In the case of a single toss, you get either all heads or all tails (H or T) 100% of the time, while with two tosses, you get all heads or all tails (HH or TT) only 50% of the time, and with 3 tosses (HHH or TTT), only 25% of the time. The probability of getting a head/tail ratio somewhere in the middle goes up as a result, particularly as you get closer to half heads and half tails, because a larger percentage of combinations works out to be near half heads and half tails when you have more tosses. You can see this trend with Pascal's triangle (which models coin toss behavior as well as many other things), where the numbers in the center of each row continue to get larger relative to the edges of the row as you move further down the rows. StuRat 16:36, 20 March 2006 (UTC)

March 20

Line integrals and path independence

Let's say you're taking the line integral (path integral) along four straight lines C1, C2, C3, and C4, where C1 is y=-2, C2 is x=2, C3 is y=2, and C4 is x=-2, so you're integrating along a closed curve C, where C is a 4x4 square centered at the origin. (The lines C1-C4 are segmented so that they only go from x=-2 to 2 and y=-2 to 2; the lines are oriented counterclockwise, therefore positive.) Then, let's say you're taking the integral of -ydx/(x^2+y^2)+ xdy/(x^2+y^2) along the curve.

Now, dP/dy equals dQ/dx, so the differential is exact, so the integral is path-independent. Because the curve is closed, this would make the integral equal to zero, because you can choose any path you want, and you could choose a path that remains at the starting point and not go anywhere. All path-independent integrals over closed curves are equal to zero, right? (Right?)

But then my textbook (this is an example from my textbook) shows how one can use Green's theorem here. Generally, one wouldn't be able to use Green's theorem because the denominator in the integral makes the integral undefined at (0,0). But one could draw a circle C', x^2 + y^2 = 1, around the hole, and the textbook shows how the line integral along C (the square) is equivalent to the line integral along C' (the circle). Evaluating this integral quickly gets the answer of 2pi.

My question is this: aren't all path-independent integrals over closed curves 0? What's different here? I'm not so concerned with the specific example - it doesn't have to be that initial curve C. When there's a hole in the region bounded by C, and C is path-independent (b/c the differential is exact), why does a discrepancy arise between a) assuming the integral is 0 because, being closed, the start point is the endpoint, and b) encircling the hole and using Green's? zafiroblue05 | Talk 19:31, 20 March 2006 (UTC)

If I remember correctly my multivariate calculus, I think the problem is with the singularity at (0, 0). True, the integral does not depend on the specifics of the path, but it does depend on the number of revolutions around the singularity(ies). Each revolution (in the positive direction) adds a constant term to the integral (in your case, 2π). You can think of your integral as going between 2 different points on a Riemann surface. -- Meni Rosenfeld (talk) 19:46, 20 March 2006 (UTC)

Meni is exactly right. Poincare's lemma only holds in contractible spaces. Your function is defined on a space with a hole in the middle, therefore Poincare's lemma doesn't hold, the differential is not exact, and the integral is not zero. -lethe ^{talk +} 19:58, 20 March 2006 (UTC)

Viewed another way, the source of confusion is with the exactness of the differential form ω = P dx + Q dy = −y/(x²+y²) dx + x/(x²+y²) dy. Naively, it appears that ∂P/∂y |_x and ∂Q/∂x |_y are equal for all x and y, which should ensure that ω = dα for some differential form α. However, there is a problem at x = y = 0. Also note that Wikipedia's statement of Green's theorem insists that P and Q have continuous partial derivatives over the region in question. Any way we look at it, the vanishing of the denominator, x²+y², cannot be ignored. --KSmrq^T 21:00, 20 March 2006 (UTC)

Thank you all, that explains it perfectly. :) zafiroblue05 | Talk 22:32, 20 March 2006 (UTC)

Coplanarity test

This one should be easy for you guys. Suppose I have a set of 100 points defined by P[i] = X[i]i + Y[i]j + Z[i]k , i = 1..100 . How can I test whether these points lie in a plane or not? I can think of choosing any three points and finding the normal vector of the plane defined by them, and then taking the dot product of all P[i] - P[0] with that normal and checking if it is zero. Is this the best way of going about it, or is there something more elegant? How can I find the normal vector of the plane defined by 3 points. Sorry for asking such elementary questions, but I'm kinda retarded on this. deeptrivia (talk) 21:19, 20 March 2006 (UTC)

There are a number of independent ways of determining whether a set of points is coplanar. The simplest that I can think of is that the 100 × 4 matrix described below has rank 3 (or less, if the points are collinear or coincident):

${\begin{bmatrix}X_{1}&Y_{1}&Z_{1}&1\\X_{2}&Y_{2}&Z_{2}&1\\\vdots &\vdots &\vdots &\vdots \\X_{100}&Y_{100}&Z_{100}&1\end{bmatrix}}$

If the rank is 3, the normal vector is the first three components of the (unique, up to scale factor) vector in the (right) Null space of the matrix. — Arthur Rubin | (talk) 21:42, 20 March 2006 (UTC)

Okay, there's a slight problem with this method in my case. Since these arrays X[i] etc are floating point numbers, the points can be very slightly out of plane (deviations such as 1e-10.) I want the algorithm to somehow ignore these small deviations. How do I manage that? The rank of the above matrix might come out to be 4 even if practically the points are coplanar. How do I increase the tolerance value for calculation of rank. Thanks! --deeptrivia (talk)

A floating point calculation is qualitatively different from a precise calculation with exact real numbers. What you likely want is a regression analysis, typically a linear regression, to find a best-fit plane, followed by an examination of the residuals. Then you must make a judgement call, based on your circumstances, about how much deviation from planarity is acceptable. Some of the deviation will be a result of limited floating point precision, some of it will come from measurement error, and some of it may be honest non-planarity. --KSmrq^T 23:25, 20 March 2006 (UTC)

Thanks for your comment. I'm getting these floating point numbers not from a measurement, but from some functions. If I'm allowed to slightly change my question, suppose you have closed form functions X(s), Y(s) and Z(s). You have to find whether the curve defined by X(s)i + Y(s)j + Z(s)k is a plane curve or not. How can you do that? Regards, deeptrivia (talk) 23:41, 20 March 2006 (UTC)

A function can be "closed form" and yet horrible to contemplate. The ideal goal would be a test for linear dependence. Suppose N = ai+bj+ck is a normal to the plane (if it exists), and that P = xi+yj+zk is a point in the plane. Then for any value of s it must be the case that

$0=N\cdot (C(s)-P),\,\!$

where C(s) = X(s)i+Y(s)j+Z(s)k. Or, in components,

$0=a(X(s)-x)+b(Y(s)-y)+c(Z(s)-z).\,\!$

Choose P however is convenient, such as letting x = X(0), y = Y(0), z = Z(0). If such a normal N exists, it can be found exactly from three independent points. And if the closed forms are not too complicated, then it may be possible to decide if the weighted sum is identically zero for all s.

It may be helpful to experiment with degree 2 polynomials; the curves are guaranteed planar. For example, try

$X(s)=1-s^{2}\,\!$

$Y(s)=2s\,\!$

$Z(s)=1+s^{2}\,\!$

which includes the point P = C(0) = (1,0,1). Subtracting this gives (-s²,2s,s²), so we can clearly use N = (1,0,1). --KSmrq^T 00:59, 21 March 2006 (UTC)

What's the name of this little thing

A high-math question:

  a * x + b
 -----------
  z + y * x

This reads "a * x + b over z + y * x". But what is the English name of this little line betveen the numerator and denominator? It is not "over", not "fraction line". Sign of division? Sounds funny. Can't find anything in relevant Wikis, not in Math books. In Math you can go without mentioning it, if you do some typesetting, you have to name it somehow :-)

Miklos Somogyi

A typesetter would call the line a horizontal rule. In fact, HTML markup for the web uses the tag <HR /> for Horizontal Rule. The MathML discussion of fractions calls it a "bar", a "rule", and a "line", all within a few paragraphs. However, what we might use as a fraction bar here can be used with an entirely different meaning in, say, mathematical logic or elementary arithmetic. --KSmrq^T 23:54, 20 March 2006 (UTC)

Thank you, I've got a few good-sounding names. It was not more difficult than Fredholm integral equations with a double-layer Cauchy-Kolmogorov kernel of the Third Kind, after all :-) Thanks, MS

March 21

the kernels' chicken?

If kernel sanders really makes real mathamatically accurate chicken, shouldn't the kernel's chicken have no calories? hence be = to the 0 vector? serious responses only please-Kolin farrellovsky 18:04, 21 March 2006 (UTC)

Well, if the chicken falls into the kernel of the "calories of" operator, then the chicken has no calories, but you can't assume that the chicken is 0, since the kernel can be a large space. This would be equivalent to kernel sanders himself falling into the kernel of the "calories of chicken made by" operator.

On the other hand, if kernel sanders falls into the kernel of the "chicken made by" operator, then his chicken is equal to 0, but you can make that conclusion without examining the calories. Melchoir 18:29, 21 March 2006 (UTC)

Er, to clarify, the second situation implies the first, but the first does not imply the second. So the moral of the story is that the meaning of "kernel" depends on which operator you're talking about. Melchoir 18:34, 21 March 2006 (UTC)

Where am I?

If I travel 10km North, then 10km East and then 10km South and find myself back where I started where am I?

Given that this is such an old and hoary one, a) it's almost certainly your homework, and b) elementary Google skills should give you the answer anyway (try using quote marks to enclose some relevant phrases and excluding the specific distances as they can be changed without affecting the basic characteristics of the problem). However, I'll give you a hint anyway: think about which areas on the Earth's surface are special in relation to the cardinal directions. --Bth 18:14, 21 March 2006 (UTC)