Wikipedia:Reference desk/Mathematics

Reference Desk

Science Mathematics Computing/IT Humanities Language Entertainment Miscellaneous Archives
How to ask a question Search first. It's quicker, because you can find the answer in our online encyclopedia instead of waiting for a volunteer to respond. Search Wikipedia using the searchbox. A web search could help too. Common questions about Wikipedia itself, such as how to cite Wikipedia and who owns Wikipedia, are answered in Wikipedia:FAQ.
Sign your question. Type ~~~~ at its end. Be specific. Explain your question in detail if necessary, addressing exactly what you'd like answered. For information that changes from country to country (or from state to state), such as legal, fiscal or institutional matters, please specify the jurisdiction you're interested in. Include both a title and a question. The title (top box) should specify the topic of your question. The complete details should be in the bottom box. Do your own homework. If you need help with a specific part or concept of your homework, feel free to ask, but please don't post entire homework questions and expect us to give you the answers. Be patient. Questions are answered by other users, and a user who can answer may not be reading the page immediately. A complete answer to your question may be developed over a period of up to seven days.	Do not include your e-mail address. Questions aren't normally answered by e-mail. Be aware that the content on Wikipedia is extensively copied to many websites; making your e-mail address public here may make it very public throughout the Internet. Edit your question for more discussion. Click the [edit] link on right side of its header line. Please do not start multiple sections about the same topic. Archived questions If you cannot find your question on the reference desks, please see the Archives. Unanswered questions If you find that your question has been archived before being answered, you may copy your question from the Archives into a new section on the reference desk. Do not request medical or legal advice. Ask a doctor or lawyer instead.
After reading the above, you may ask a new question by clicking here. Your question will be added at the bottom of the page.
How to answer a question Be thorough. Please provide as much of the answer as you are able to. Be concise, not terse. Please write in a clear and easily understood manner. Keep your answer within the scope of the question as stated.	Link to articles which may have further information relevant to the question. Be polite to users, especially ones new to Wikipedia. A little fun is fine, but don't be rude. The reference desk is not a soapbox. Please avoid debating about politics, religion, or other sensitive issues.

May 9

Cancelling a sent e-mail on Gmail

Well, I know well, that to cancel a sent e-mail is impossible, unless you have some access to the recipient's mailbox. But, now I got into situation when I need it very much. And, maybe, there is a way to do it after all?

I sent a letter form my gmail address to another person's gmail address, and I don't know the other person's password. I know that gmail made a lot of cool things, such as 2.5 G space, internal search, and stuff. Maybe they did implement the ability to cancel sent e-mails too? It's all on their servers after all. (By the way, many message boards like Vbulletin let you delete your unread PMs). I know it is impossible, but, what if. If you know something, your help is very appreciated. I would be ready to pay a reasonable sum of money if you will help me (but note to 'crackers' out there: I will not tell you my password).

Crocodealer 19:49, 16 May 2006 (UTC)

Breaking A Code

• 5
• The 700's
• Riley+Mcmillian=Hill
• 2damsels+1sheep only
• Count the toes+hooves
• King tut the young
• look at the wall subtract 1x100+59
• 69
• Ghost Grant —The preceding unsigned comment was added by 65.183.106.10 (talk • contribs) 02:15, 9 May 2006 (UTC)

Bingo! Melchoir 03:01, 9 May 2006 (UTC)

I reformatted your text to have some resemblance to what I think you meant. Now what the hell are you talking about? Black Carrot 23:32, 9 May 2006 (UTC)

Bloodhound Problem

A prison guard and his bloodhound are chasing an escaped prisoner. The prisoner has a 5 mile head start but the guard is walking 1 mph faster than the prisoner. The bloodhound is trained to run to the prisoner, run back to the guard, and then continue running back and forth between them. If the bloodhound runs 10 mph, how far does the bloodhound run before the guard finally catches up to the prisoner?

How would you go around solving this problem? I want an idea on how to solve this problem and others like it, algebraically preferably. Thanks.

--Proficient 05:33, 9 May 2006 (UTC)

distance travelled is speed×time, keep that in mind, dont get distracted by the implied infinite series. —The preceding unsigned comment was added by Frencheigh (talk • contribs) .

Yes, thank you, but next time sign your post. Proficient, you can either do it by writing an infinite series and then summing it, or you can do it the easy way. The dog always runs at the same speed without stopping, right? —Keenan Pepper 05:49, 9 May 2006 (UTC)

The original story, with the cute John von Neumann twist, involves a fly and two trains. It's a popular old chestnut used to tease new calculus students. --KSmrq^T 08:26, 9 May 2006 (UTC)

There are several ways of doing this, but here is what I think of as the algebraic way.

Variables

d_GP: the distance difference to be overcome by Guard to catch up with Prisoner

v_GP: the speed difference between Guard and Prisoner

t_GP: the time it takes Guard to catch up with Prisoner

d_B:   the distance covered by Bloodhound

v_B:   the speed of Bloodhound

t_B:   the time Bloodhound runs

Equations

d_GP = v_GP×t_GP (by definition of "speed")

d_B = v_B×t_B (same)

d_GP = 5×mile (given)

v_GP = 1×mile/hour (given)

v_B = 10×mile/hour (given)

t_B = t_GP (given)

Solve for

d_B

There are as many equations as there are variables, which looks promising. Let us take miles and hours as units (which just means we leave them out) and start simplifying by plugging in the known values, and using just one variable t for t_GP and t_B. The equations become then:

Simplified equations

5 = 1×t

d_B = 10×t

From here on I think you can do the rest by yourself. --Lambiam^Talk 08:29, 9 May 2006 (UTC)

Well, the guard will get to the prisoner in 5 hours, right? So the entire run will be 5 hours long. Hence the bloodhound will run 50 miles. Cthulhu.mythos 10:00, 9 May 2006 (UTC)

Thank you very much for your oustanding help.--Proficient 05:54, 10 May 2006 (UTC)

'

sums with two summation conditions in latex

Any suggestions on how to improve my poor attempt on Siegel-Walfisz theorem? Mon4 14:48, 9 May 2006 (UTC)

How About

$\sum _{n\leq x \atop n\equiv a\mod q}\Lambda (n)$

? -- Meni Rosenfeld (talk) 15:07, 9 May 2006 (UTC)

Beautiful! Mon4 15:12, 9 May 2006 (UTC)

Recording off computer

If there is a sound played on a website, on a flash webpage, is there any way in which I can get that sound onto an audio file on my desktop? --Chachu207 ::: Talk to me 16:56, 9 May 2006 (UTC)

You may get a hold of software used to decode .swf (normal flash files), which basically will turns the file into its raw materials, like mp3-files, graphics, etc etc. You can ALSO use a mic and standard recording system (an extremely basic version can be found on the Start menu somewhere in most Windows computers), but the result isn't likely to be fantastic.

If you want to actually MIX the flash-sound with an already existing audio file, I have zero idea. Aforementioned Windows standard media-tool might be able to do that, but I wouldn't be able to tell. There are some easy and reliable, perhaps some free sound-mixers out there to be found, I might be able to find one for you if you reply that you want one. 213.161.189.107 21:01, 9 May 2006 (UTC) Henning

You need recording software, try google. Also, sourceforge probably has something of this sort.--Frenchman113 on wheels! 21:03, 9 May 2006 (UTC)

I hear from a friend that .swf-files are most often encrypted, and that opening them with even a flash-editor shouldn't be possible unless there's no encryption. In other words, the Frenchman might be on to something. 213.161.189.107 21:10, 9 May 2006 (UTC)Henning

If you're not concerned about audio quality, the "audio-hole" will work perfectly: obtain a 3.5mm-3.5mm male-male audio cable, plug one end into the speaker socket and the other into the microphone socket. Start an audio recording program in the background, play the SWF -- presto, the audio recording program will pick up the audio from the SWF. Dysprosia 22:36, 9 May 2006 (UTC)

A quick Google search turn up some shareware called Audio Record Expert in the first several links. I'm sure there are others. --LarryMac 14:55, 10 May 2006 (UTC)

Try software that captures (any) sound from Audio Out and saves it into a file. I use TotalRecorder for this purpose, and it works pretty well. --Leapfrog314 04:01, 12 May 2006 (UTC)

caculate displacement

I am a drilling rig supervisor, and am trying to figure out a simple formula that will tell me how many feet from center the bit will be, based on depth and angle of inclination. As we drill a well, we continually monitor how many degrees from vertical the well has moved. So, for example, consider a well that is 10,000' deep, and the angle of inclination is 5 degrees. If you pictured this as a right triangle, the tall vertical leg is 10,000, the angle at the bottom is 90 degrees, and the angle at the top is 5 degrees. I need to know what the length of the short leg is. How do I do this? Les F.

I think you mean that the "angled" leg (how far the bit has penetrated) is 10,000'. Is that correct? hydnjo talk 17:19, 9 May 2006 (UTC)

If the 10,000 feet is the angled length, you need to take the sine of 5 degrees, which is 0.0872. If 10,000 feet is really the vertical depth, take the tangent of 5 degrees to get 0.0875. In either case, multiply the value by 10,000 feet to get the answer, which is either 872 feet or 875 feet. As you can see, for small angles there isn't much difference. Also note that these calcs assume the well is absolutely straight. If the shaft bends, these results are not accurate. StuRat 17:50, 9 May 2006 (UTC)

This sounds like a possible question for a basic trigonometry class, except for a few real-world details. Doesn't it seem peculiar that a company would employ someone to supervising a massive and expensive job like drilling to 10,000 feet (3 km) without adequate training to answer a question like this? --KSmrq^T 21:45, 9 May 2006 (UTC)

Since the oil field is probably quite large, they likely just say it's "close enough" and don't worry about how far off it is. StuRat 20:50, 10 May 2006 (UTC)

First of all, thank you for your insight and suggestions. To Hydnjo..There are two inputs that I have available to me at any time, TVD or true vertical depth, and MD or measured depth. The MD is the convoluted hypotenuse of this triangle, and TVD is the vertical depth or the long leg of the triangle. To KSmrq...I am the person resonsible at the drilling site that makes sure that all the parameters related to drilling are adhered to. On this particular well, I was given a radius from the beginning point of the well that I had to stay within. I was getting the angle of inclination as we drilled, and I was trying to calculate how many degrees from center I could be and still be within the target. It's true, I'm not a petroleum engineer, and it's also true that I should be able to figure this out on my own, but I thought a quick, accurate way to do this would be to post my question her. And although I am not a petroleum engineer, I challenge any engineer to come to the rig, live there 24 hours a day, and make the decisions I make related to drilling fluids systems, pipe stretch, wireline logging decisions, geological changes, cementing procedures, safety management, logistics, and managing 20 roughnecks, a toolpusher, a mud engineer, directional drilling personell, while trying to keep peace with a group of irate landowners, and trying to keep the home office up to date on daily and cumulative costs, and make suggestions related to drill bit selection, fishing tool company selection, access to water, disposal of oilfield waste, road maintenance, etc... And to StuRat...My target on these wells is a 25' radius. Thats after drilling directionally 2,000', and then dropping the bit another 13,000'. Thats right, we drill vertically 300-400', then we kick the well on a predetermined azimuth at a predetermined angle build to 20 degrees, then we drill the predetermined offset, and then steer the well back vertical, then we stay within a 25' radius to 15,000'. Pretty simple, eh? And close enough doesn't cut it. If we don't achieve this objective, we plug the well back, and redrill it until we get it right. If I don't keep this well within the target, I'm out of a job. I've been doing this for 10 years now, so I guess I'm doing something right.

Hey, let me apologize for the unintentional abrasiveness above. It's pretty rare we get genuine posts like this. As for the answer, a diagram would be best...

 |\
 |b\
 |  \ B
A|   \
 |    \
 |a   c\
 -------
    C

Note how I've labeled the triangle. So, from what you're saying, a = 90 degrees, and b is 5 degrees, A is 10000 feet, and you want to find out what C is?

Let's assume you only have A and b. The formula for C is given by C = A tan b. You have to be careful if you are using a calculator. Make sure you know you are working in "degrees" mode and not "radians". --HappyCamper 04:09, 11 May 2006 (UTC)

Great response, Les. Now go back and notice that I worded my "peculiar" remark very carefully; if you had other skills that made you worth hiring, isn't it foolish for the company not to train you with the geometry skills to help you succeed? How expensive is a little education? How expensive is redrilling? The conclusion is simple arithmetic. Tell your boss Wikipedia mathematicians say so!

Meanwhile, let's consider options. First, since drilling holes in the ground is big business, commercial software may be worth investigating, to see if the price and features are right. For example, a quick web search turned up Target and DrillStar.

If we're going to apply mathematical ideas, it's important for us to approach this as a real-world engineering task, not some simplified homework exercise. For homework we can assume an ideal triangle. In reality, we should be cautious about assuming that the drill path is a straight line, or that it stays in a plane with an ideal vertical. Angle measurements will always be approximate, and if we accumulate them repeatedly the errors can add up. What we would like to do is to track the 3D path as a series of angles and distance increments. And for that, angle from the vertical is not enough; we should be sure of the compass direction as well.

Another thought is that since integrating the drill path may be subject to various errors, it would be nice to have a way to get a true 3D position reading, at least occasionally, perhaps by acoustic triangulation if that's feasible both economically and with different kinds of intervening rock.

The more you can tell us about what information you have available (including its reliability), the more we can help.

Just remember, it's not our job on the line; and while some of us are professional graduates, others are pre-college.

Which reminds me to remind you: please sign your post by using four tildes, "~~~~". It will automagically become a real Wikipedia signature, with user ID and time stamp. Thanks. --KSmrq^T 06:35, 11 May 2006 (UTC)

Hey Les, thanks for getting back to us, not many folks do that. But now that my interest has been piqued by your response, just how do you determine the TVD. The MD is pretty obvious and I can see now that its convoluted length could indeed confound any simple approach to calculating the "off-center" distance. There might even be enough interest here to start the Mathematics of deep well drilling (or some such) article! The reason for my asking is that if the TVD is derived by knowing the MD and the angle from vertical then not much has been gained. If on the other hand, the TVD is determined independently well then it becomes a useful piece of data in determining other values particularly if you have knowledge of either the MD or the TVD for each drilled azimuthal segment (and of course the azimuth of those segments). Hell, you may even invite one of us out to the site to observe such a fascinating operation. I, for one, would love to know how you "kick" the drill off to another aximuthal direction, it boggles my mind, and all the more reason for an article about all of this --hydnjo talk 20:32, 11 May 2006 (UTC)

May 10

Cross Sections of a Cube

Why is it that a cube cannot be sliced by a plane in such a way that the cross section formed will be a regular pentagon?

You can probably answer the question by considering all possible cases that a plane can intersect a cube, and describing the cross-section. By exhaustion of cases, you won't obtain the desired cross-section. Dysprosia 04:43, 10 May 2006 (UTC)

I remember this problem from a Mu Alpha Theta competition. Proving it rigorously would be a real pain. We ended up making some reasonable assumptions (bilateral symmetry was one) and then doing an optimization to convince ourselves it couldn't be done. —Keenan Pepper 04:50, 10 May 2006 (UTC)

Aren't two of the sides of a resulting pentagon necessarily parallel? Because there's only 3 non-parallel sides of a cube...  Grue  07:21, 10 May 2006 (UTC)

Ah, thank you! I knew there had to be a slick answer. Melchoir 08:12, 10 May 2006 (UTC)

In that vein: is there a standard proof or theorem that states something along the lines of: "when two parallel planes are intersected by a third plane, then the points of intersection on the third plane are parallel lines"? or does one have to prove it every time? --Seejyb 21:12, 10 May 2006 (UTC)

It is a very simple lemma. If the lines were not parallel, they would intersect. The point of intersection should lie on both parallel planes, which is impossible.  Grue  21:23, 10 May 2006 (UTC)

Damn. I wish we had thought of that, instead of doing all that hairy calculus (which probably didn't even prove it anyway). —Keenan Pepper 04:07, 12 May 2006 (UTC)

series sum

I'm certain this is incredibly simple but I've had a complete mind blank over the past couple of days. Is there a general expression for 1^2 + 2^2 + 3^2 + ... + n^2 ???? (sorry i don't know how to use the maths markup) 129.78.64.105 04:16, 10 May 2006 (UTC)

From Summation:

• $\sum _{i=0}^{n}i^{2}={\frac {n(n+1)(2n+1)}{6}}$

Melchoir 04:32, 10 May 2006 (UTC)

Argh the title of that article was too obvious for me, haha. THANKS! 129.78.64.105 04:44, 10 May 2006 (UTC)

OEIS names the sequence Square pyramidal numbers. --DLL 20:04, 10 May 2006 (UTC)

i think that you are going to fail trig. and then one day, when your at checkers someone is going to ask you this question. and your not going to know it — Preceding unsigned comment added by 65.9.35.40 (talk • contribs)

paper on data structure

hello all techis! i need to work on some good and new idea on sorting algorithms, plz help...

Try Category:Sort algorithms. Melchoir 10:06, 10 May 2006 (UTC)

Transform a circular image into a square image?

This seems like a basic question, but it's proving to be very hard. Say that you have an rectangular image of a circle (like a picture of the moon), and you wanted to transform that so that the moon stretched to fill the entire image. How would one transform that image mathematically?

You select your favourite map projection according to the properties that you want your resulting image to have, and then you apply that projection to get a rectangular image that's probably less accurate than the circular one you began with. -- AJR | Talk 11:55, 10 May 2006 (UTC)

For the problem as stated, one might apply the Riemann mapping theorem. But be warned; as the article says:

"Even relatively simple Riemann mappings, say a map from the interior of a circle to the interior of a square, have no explicit formula using only elementary functions."

Since the question did not require a conformal map, there are simpler options. From the center of a circle to any point on its boundary we have a line segment of a certain length; to map (invertibly) from rectangle to circle, scale all points on each line segment by the inverse of that length.

A typical map projection is a map from (a part of) the sphere to the plane, which is a different idea. --KSmrq^T 12:34, 10 May 2006 (UTC)

Finding a mapping only solves half the problem. You will also need a resampling algorithm, Bilinear interpolation or Bicubic interpolation may help. EricR 14:14, 10 May 2006 (UTC)

Various questions

Okay, I know this sounds like I'm asking to be spoonfed, but I'm not. Basically I'm self-studying some advanced maths which isn't part of my school curriculum, so getting help is kind of difficult. I'm not asking for the answers; I just need an explanation of the principles which will help me obtain the answers, because I have absolutely no idea how to do these questions. I've never taken a formal course which taught any of these principles, so hopefully you guys can bear with me despite the variety of questions I'm about to ask.

Trigonometry

What is $\sin[\operatorname {arccot}(-3/4)]$? It should be simple, but I've found there are two apparently correct ways to get two different answers. The first gives a value of -0.8. I determine this by first seeing that the cotangent of the unknown is equal to -3/4. In other words, $1/\tan(x)=-3/4$. Using my calculator to find arctangent -4/3 and then finding the sine of this value, I obtain -0.8. My textbook, however, approaches the problem by determining the arccotangent of -3/4 is equal to $\pi /2-\arctan(-3/4)$. The sine of this value is then found to be 0.8. Which approach is correct? Perhaps I've a terrible memory, but I can't even recall being taught the principle behind the textbook's solution.

Given a number a, there are infinitely many numbers x such that cot(x) = a. The function arccot returns just one of them, and this is usually taken to be the one value in the range 0 < x < π. The approach you used generates, indeed, a number whose cotangent is (-3/4), just not the one that is denoted by arccot (-3/4). So the approach in the book is more accurate (though your solution can also be considered correct, depending on the context). -- Meni Rosenfeld (talk) 19:15, 10 May 2006 (UTC)

Periodicity of sine and cosine

Periodicity of tangent

The cotangent of an angle is the ratio of cosine to sine, and ranges from −∞ to +∞. Like sine and cosine (shown left), it is periodic; but like the tangent function (shown right, and noting cot(ϑ) = tan(^π⁄₂−ϑ)), its period is π, not 2π. This implies we have two choices for the value of sin(arccot(−³⁄₄)), one positive and one negative. It may be of interest to note that standard mathematical libraries for programming languages provide a two-argument arctangent, to allow choice of the desired quadrant. --KSmrq^T 23:38, 10 May 2006 (UTC)

3-D geometry

The object $3x+4y+5z=18$ lying in the first octant forms a three-dimensional object with the co-ordinate planes. How do I find its volume? The answer given in the text suggests it's a pyramid, but I can't seem to imagine how this can be inferred. I'm not exactly clear on how I would proceed from this to finding the volume either.

Find the intersection points of the plane with the axes. The resulting object is a pyramid, which you can think of as having for a base a triangle in the x-y plane, and height along the z axis. Use the formula $V={\frac {1}{3}}Ah$. -- Meni Rosenfeld (talk) 20:11, 10 May 2006 (UTC)

The geometric object defined by the zeros of the first degree polynomial

3x + 4y + 5z − 18

is a plane. (In general, a single polynomial equation in 3D defines a surface, and a linear equation defines a plane.) It cuts the x axis where y and z are zero, at x = 6; similarly for the y axis (at y = ⁹⁄₂) and the z axis (at z = ¹⁸⁄₅). This volume is an irregular tetrahedron. It is a pyramid in a rather general sense, not a regular pyramid like the Great Pyramid of Giza. --KSmrq^T 00:04, 11 May 2006 (UTC)

2-D geometry

$9x2-axy+4y^{2}=16$ represents a pair of straight lines for certain values of a. Why is this so? I have absolutely no knowledge of this area, and my book isn't helping with its terse "explanations".

If a is +12 or -12 (the values for which the discriminant is 0), the left hand side becomes:

$9x^{2}\pm 12xy+4y^{2}=(3x\pm 2y)^{2}$

And the equation becomes:

$(3x\pm 2y)^{2}=16$

So $3x\pm 2y=4$ or $3x\pm 2y=-4$, which are equation of straight lines. -- Meni Rosenfeld (talk) 19:20, 10 May 2006 (UTC)

Algebraic (or merely analytic) geometry is involved. In the plane, the zeros of a single polynomial, here a member of the family

9x² − axy + 4y² − 16,

typically describe a "curve", for a suitable understanding of what that word means. If we add the degrees of the variables in a term, we get the "total degree" of the term; the maximum over all the terms is the (total) degree of the polynomial, here 2. Ordinarily a line would intersect a degree-2 curve in two distinct points, per (a sloppy version of) Bézout's theorem. However, some curves are degenerate, in a precise technical sense. For a conic, which is what we have here, we could have a degenerate circle of radius zero,

x² + y²,

or a degenerate hyperbola forming a pair of intersecting lines, such as

xy,

or a degenerate parabola forming two parallel lines (not necessarily distinct), such as

x² − 1.

In geometric terms, a degenerate curve has a point which is "double"; an independent line through that point lacks the other intersection(s) we ordinarily expect. (For parallel lines, that point is at infinity.) In algebraic terms, we can detect degeneracy by asking if the polynomial for the curve has a double root. For a quadratic polynomial, this is easier because we can convert to bilinear form, leading to a matrix form. The matrix for the example is

${\begin{bmatrix}16&0&0\\0&9&-a/2\\0&-a/2&4\end{bmatrix}},$

with determinant 576−4a². We have degeneracy when this expression is zero, at a = ±12. --KSmrq^T 04:02, 11 May 2006 (UTC)

Volume

A circle with a radius of 10 inches is divided into two sectors; one is 240 degrees, the other is 120. Both are bent into cones. How do I find the ratio the smaller cone's volume to the larger cone's? From the answer given, it seems that there's some relation between the slant length of a cone with its height and base, but there's no real explanation or even statement of what principles were applied to derive the formula giving the answer.

If you check Cone (solid), you will find the formula for the slant height, $s={\sqrt {r^{2}+h^{2}}}$. Both of your cones will have a slant height of 10 inches, and you can derive the radius of the base of each from the original problem. You can then solve for the height of each cone to determine their volumes. --LarryMac 17:10, 10 May 2006 (UTC)

A little note. You can find that equation for the slant height if you draw an imaginary right-angled triangle made up of the radius, height and slope of the cone, then apply pythagoras' theorem. That means you don't have to remember the formula, just remember to draw triangles where possible. Skittle 21:28, 10 May 2006 (UTC)

Pardon my ignorance, but how do I derive r? Johnleemk | Talk 17:24, 10 May 2006 (UTC)

You're not ignorant, you're asking questions. I am hoping to give you nudges in the right direction instead of just spelling out the answer. Anyway ... Compute the circumference of the circle you start with. Note that by dividing the circle as specified, the sectors contain arcs whose lengths are 1/3 and 2/3 the total circumference. When you bend your sectors into cones, those arcs become the circles at the base of each cone; thus the lenght of the arc is now the circumference of the base. Solve for r. Ask again if I'm being more obtuse than a 240 degree angle. --LarryMac 18:41, 10 May 2006 (UTC)

Properly speaking, ignorance is not knowing. The correction is to ask questions, which we encourage. --KSmrq^T 04:19, 11 May 2006 (UTC)

This problem has a hidden simplicity. When a circle sector is converted to a cone, the circle radius becomes the cone slant height, the circle radius times the sector radian measure becomes the base circumference, and the cone radius is proportional to the cone base circumference. Both cones will have the same slant height, and one will have twice the radius of the other. --KSmrq^T 04:19, 11 May 2006 (UTC)

1-> s = (r^2 + h^2)^0.5

                        1
#1: s = ((r^2) + (h^2))^-
                        2

1-> h

                         1
#1: h =  ((s^2) - (r^2))^-
                         2

1-> v = pi * r^2 * h/3

        pi*(r^2)*h
#2: v = ----------
            3

2-> eliminate h
Solving equation #1 for (h)...

                                  1
        pi*(r^2)* ((s^2) - (r^2))^-
                                  2
#2: v = ----------------------------------
                        3

Note: r = ( r_original * Angle_in_radians ) / ( 2 * pi )

Miscellaneous

Given a right regular hexagonal prism, how can I calculate the number of parallel edges it has? The answer given implies there's a way to mathematically find this through the addition of two different combinations.

Isn't it just 1 set of 6 parallel edges, and 3 sets of 4 parallel edges each? Am I missing something? -- Meni Rosenfeld (talk) 20:16, 10 May 2006 (UTC)

I have trouble visualising it, so maybe that's why I found the problem somewhat difficult to solve. Johnleemk | Talk 10:56, 11 May 2006 (UTC)

Thanks for all the help (again), guys. Johnleemk | Talk 16:44, 10 May 2006 (UTC)

Your questions give me the impression that your textbook may not be very good, at least not for self study. You should consider trying to get books which actually teach the subjects and don't just give exercises. -- Meni Rosenfeld (talk) 20:15, 10 May 2006 (UTC)

I actually rely on it only for questions -- the problem is that my other texts don't focus on some applications of trigonometry which appear in this text. Likewise, they don't cover anything on 3 dimensional geometry, which is why I needed help from the refdesk. Again, thanks so much for the help everyone. :D Johnleemk | Talk 10:56, 11 May 2006 (UTC)

They might be asking about the number of pairs of parallel edges, which is 33, according to a quick calc in my head. StuRat 18:46, 11 May 2006 (UTC)

Ah, I see. So the "addition of 2 combinations" he mentioned referred to ${6 \choose 2}+3{4 \choose 2}$. -- Meni Rosenfeld (talk) 19:29, 11 May 2006 (UTC)

This is a nice set of mathematical questions. I hereby nominate this as the question of the week (or day or whatever template we have available). – b_jonas 19:40, 11 May 2006 (UTC)

Problem with earth radian

Open title, now ain't it. Anyway, I have a standard geometrical issue: One flagpole casts a shadow of 18 meters, next to it is a pole of 2,1m which casts shadow long as 3,4m. Thus the length of the flagpole is in the region of 11,1174m. However, I want to impress my math teacher. Earth's mean diameter is 12 754 591m. I want to use rad to show that the flagpole's shadow is only 18m on a perfectly plane ground, while in truth the earth is curved, and so the flagpole's shadow is longer. The question is just, how do I do this? Any answer would be enormously appreciated! I don't know if my current calculations are right, so I won't bring them up here. :) 213.161.189.107 19:38, 10 May 2006 (UTC) Henning

You should right write down the equations describing a straight line and a circle, and find their intersection. -- Meni Rosenfeld (talk) 20:35, 10 May 2006 (UTC)

I think he means "write down". :--) JackofOz 02:05, 11 May 2006 (UTC)

Now you see why I went to bed shortly after this edit... :-) -- Meni Rosenfeld (talk) 06:34, 11 May 2006 (UTC)

I did this the hard way finding an answer, but it's actually straightforward if you're not dumb like me. Consider the triangle TCS, where T is the tip of the pole (assumed vertical), C is the center of the earth, and S is the tip of the shadow. What we want is the arclength of BS, where B is the base of the pole. Equivalently, we want to know the angle $\phi$ subtended by the segment BS at C. B is on the segment CT, so this is just the angle $\angle TCS$. By the law of sines, we have ${\frac {\sin \theta }{R}}={\frac {\sin(\pi -\theta -\phi )}{h+R}}$, where $\theta$ is the angle from vertical of the sun (and thus the angle $\angle CTS$), h is the height of the pole (length of BT), and R is the radius of the earth (CB, CS). Rearranging and using $\sin(\pi -x)=\sin x$, we have $\sin(\theta +\phi )={\frac {h+R}{R}}\sin \theta$, so $\phi =\arcsin \left({\frac {h+R}{R}}\sin \theta \right)-\theta$. Multiply by R to get the arclength. Hope this helps. --Tardis 04:27, 11 May 2006 (UTC)

Note that local variations in the Earth and the angle of the poles will have a far greater effect than the curvature of the Earth. So, while your answer is a nice mathematical exercise, it has no real application. StuRat 18:41, 11 May 2006 (UTC)

Thank you so very much Tardis! I am having problems understanding the need for arcsin, as I'm not really good at this stuff, but I am sure it will come to me! And yes, StuRat, I realize this. For the sake of good truth I must of course assume the angle is perfectly right and that the earth allows RB and RS to be completely identical. Impossible, but much fun is had trying to grasp a solution. :) So thanks! 213.161.189.107 13:16, 12 May 2006 (UTC) Henning

Multiplying PDF

Suppose I want to find out what is the probability that I will die of a heart attack at 2pm tomorrow on a 747 jet plane.

Answer = Prob(heart attack at 2pm) * Prob(On a 747 at 2pm)

So far so good.

But if I only know the PDF (prob dist func) of

• pdf( heart attack at 2pm )
• pdf( on a 747 at 2pm )

Then how to I multiply those two PDF together to a the PDF of the answer? To make life simple, let's assume I'm completely ignorant of my chances and the PDF of the two events is a uniform distribution (ie constant 1).

Ohanian 21:37, 10 May 2006 (UTC)

Should you die at 2.01 pm, does it still count? And 1 second after 2pm? Is it the local timezone of where the plane happens to be that counts? To have a probability distribution, you need a random variable, and a range of the possible values it may assume. Saying "constant 1" does not give enough information. With very high probability you will die between now and January 1st, 7382236374. It would be a bit silly to assume complete ignorance and claim that you are as likely to die tomorrow as on December 31st, 7382236373. The probability over all days has to add up to 1. --Lambiam^Talk 22:53, 10 May 2006 (UTC)

Godamn it! It does not matter. How about 2pm sharp within a interval of 1 second! Happy now? Ohanian 23:01, 10 May 2006 (UTC)

When asking strangers for help, swearing at them does not typically elicit better answers. Mathematicians are in the habit of demanding clear and precise questions, for very good reasons. If that doesn't appeal to you, ask your mother, not a mathematician (assuming your mother is not one, of course). Beyond that, the top of this page states explicitly, in boldface, Be courteous. Your understanding and cooperation will be appreciated. --KSmrq^T 23:49, 10 May 2006 (UTC)

What are you talking about? It is perfectly legitimate to have a uniform pdf of 1. Ohanian 23:03, 10 May 2006 (UTC)

I think what he is saying is that you have not made it clear what the event space is. Having a heart attack on a plane at 2pm is a single event. You need some unknown (or "random") event space to have the PDF with respect to if you want something more interesting. For example, you might say, given that you're on the plane at 2pm, there is some random number of snakes on that plane. You could think of the number of snakes as a random variable and it would have a PDF. --Deville (Talk) 23:15, 10 May 2006 (UTC)

I'm trying to keep it as simple as possible. A pdf for a specific event occuring at a particular point in time. A pdf for being in a particular location at a particular point in time. So how to get a pdf of (both being in a particular location and having a specific event occuring at a particular point in time)? You may assume that these two pdf are completely independent of each other. Ohanian 23:40, 10 May 2006 (UTC)

Then could you give me an example of a PDF which would correspond to "having a heart attack on a plane at 2pm"? I must say I think there is still some confusion here about what the question really is.--Deville (Talk) 23:49, 10 May 2006 (UTC)

Arrggh! I'm tearing my hair out!

If two events A and B are independent of each other then the probability of (A and B) is

Pr(A and B) = Pr(A) * Pr(B)

But if we do not know the probability of A and B. And instead we are only given the pdf of the probability of A and the pdf of the probability of B.

So we should be able to calculate the pdf of the probability of event (A and B) by some how multiplying pdf(A) with pdf(B)

pdf(A and B) = pdf(A) multiply pdf(B)

My question is "How do you multiply two pdf together?"

Ohanian 02:31, 11 May 2006 (UTC)

That's a good question. I don't know why they're having such trouble with it. Although, you did use pretty difficult examples. Your comment came at the same time as my response, which was going to be

I think I follow you, more or less. Let me repeat what I think the question is: "I know how to take the odds of simple events (5/6, 1/2, etc) and turn them into the odds of those events occuring simultaneously (5/6*1/2=5/12, etc). How do I take two continuous probability distributions (of which the uniform distribution is the simplest) and do the same?" I don't know, but for what it's worth, those are really bad examples. Try perhaps the odds of being more than six feet tall, the odds of earning more than $200,000 a year, and the odds of doing both. Of course, that assumes the two are independent, which they probably aren't.

I'd like to add something. It may have helped confuse them that you kept saying 'multiply' when, most likely, multiplication isn't the answer. 'Combine' might have been clearer. Also, given that most probability distributions are continuous, it isn't meaningful to ask the odds of an exact event (which are zero. what are the odds it'll be right down to the nanosecond?), it's only meaningful to ask the odds of an event within a range (say, of five minutes). Black Carrot 02:39, 11 May 2006 (UTC)

I get it now, I confused a lot of people when i use the term "pdf(event X)".

I meant pdf of the true probability of event X. ie. the random variable in the pdf is "the actual probability".

I should have used " pdf( actual probability of event X ) "

So the problem becomes

pdf(actual prob of event A and B) = pdf(actual prob of event A) combine pdf(actual prob of event B)

Ohanian 02:50, 11 May 2006 (UTC)

Assuming that the events are independent (though their probabilities may not be), then the probability of "A and B" is x iff $P(A)P(B)=x$. So just integrate over the possible values of the individual probabilities that multiply to that: $f_{\mbox{A and B}}(x)=\int _{x}^{1}f_{\mbox{A,B}}(t,x/t)dt$, where $f_{\mbox{A,B}}(x,y)$ is the PDF for A and B's probabilities (which is $f_{A}(x)f_{B}(y)$ if the probabilities themselves are independent). I dunno right now about dependent events, but I'll let you know if I discover something. Hope this helps. --Tardis 04:32, 11 May 2006 (UTC)

Note that Meni's answer to the recast question below is similar to, but more general than, mine and probably more accurate. I wrote this in a rush, and there's some sort of Jacobian-like modification that must be made to the integral; hopefully Meni got it right. --Tardis 18:04, 11 May 2006 (UTC)

Hopefully :-) -- Meni Rosenfeld (talk) 19:57, 11 May 2006 (UTC)

What made the question so unclear is that no event spaces were given, so what were the distributions over? Also, usually PDF stands for probability density function, which is related but not quite the same. For independent continuous random variables the joint event space is the product of the individual spaces, and if both have probability density functions (not all distributions do) the joint probability density function is (by definition of "independent") simply the product of the individual ones; see Probability density function#Independence. So if time and location are independent, just multiply the densities. --Lambiam^Talk 06:12, 11 May 2006 (UTC)

The reason your questions were meaningless is that you talk about a "pdf of a probability". There is no such thing (at least not in the probability theory I know of). There is a probability of an event, and a pdf of a random variable, and a probability of the event that a given random variable will lie in a given range. You only discussed events and their probabilities, so your use of the term "pdf" was completely misplaced. You could describe a random variable X, the number of seconds since midnight in which you die. You can then also assume that this variable is distributed uniformly between 0 and 86400. The pdf of X will be 1/86400 (not 1!) in the range 0≤t≤86400, and 0 elsewhere. Then you can ask about the probability, say, that X wil be in the range 50400≤t<50460, meaning that you will die in the minute 14:00. The result is 1/1440. If you also know the probability that you will die due to a heart attack in a 747 (no continuous random variable can be described for this), and you know that it is independent of your time of death, you can multiply the probabilities and obtain the probability that both will happen. -- Meni Rosenfeld (talk) 06:51, 11 May 2006 (UTC)

Multiplying PDF Redux

Consider a simple rectangle with height (h) and width (w).


******************************************************   |
*                                                    *   |
*                                                    *   |
*                                                    *   |
*                                                    * 
*                                                    *   h
*                                                    *
*                                                    *   |
*                                                    *   |
*                                                    *   |
******************************************************   |

<----------------------  w  ------------------------->


Now the pdf(height) is


pdf(height) =  1       if  0 < h < 1
            =  0       otherwise


Now the pdf(width) is

pdf(width)  =  1       if  0 < w < 1
            =  0       otherwise


What I'm looking for is the pdf of the area of the rectangle


pdf(area)  =  pdf(height)  <multiply>  pdf(width)


So the question is this.


How do I multiply two pdf together.


=========================================================

Can you understand why I'm asking for the following.



Pr(A and B)  =   Pr(A)  *  Pr(B)    (given A and B are independent)


so logically


pdf(A and B)  =   pdf(A)  *  pdf(B) (given A and B are independent)

Ohanian 08:10, 11 May 2006 (UTC)

Now you're asking a meaningful question.

Here's the solution:

First you should consider the cdf. Let x be constant. For a given value of w, if x≤w≤1, you have:

P(0≤A≤x) = P(0≤w h≤x) = P(0≤h≤x/w) = x/w

And if 0≤w≤x, you have P(0≤A≤x) = 1.

Therefore:

$P(0\leq A\leq x)=\int _{0}^{x}1dt+\int _{x}^{1}{\frac {x}{t}}dt=x+xln(1/x)$

The derivative is the pdf: ln(1/x). As you can see, the formula you suggested is incorrect. -- Meni Rosenfeld (talk) 09:13, 11 May 2006 (UTC)

More generally, given independent variables X and Y, with pdfs f and g in [0, ∞), the cdf of X*Y is (assuming my calculations are correct):

$H(t)=\int _{0}^{\infty }\int _{0}^{t/x}f(x)g(y)dydx$

And the pdf is:

$h(t)=\int _{0}^{\infty }{\frac {f(x)g(t/x)}{x}}dx$

These can also be presented as several equivalent forms. -- Meni Rosenfeld (talk) 09:29, 11 May 2006 (UTC)

The original question had meaning too -- it was just confusing. If you have incomplete information about the probability P of an event, it makes sense to speak of knowing only the PDF for P, with $f(p)dp$ being the probability that $P\in [p,p+dp]$. It's actually quite interesting to consider observing repeated tests of an event (or set of disjoint events) and determining the PDF for the event's probability (or set of unit-sum probabilities). It's really more of a confidence, since the probability is assumed to be already chosen and constant, but it's mathematically very similar. Also, that $1/x$ factor in your h integral shows that my answer to the previous question was not quite correct; you have to consider some sort of chain-rule-like determination of the "thickness" of the line I was tracing through the space of possible probabilities (or rectangle dimensions). Any thoughts on how to go about it as I was, but accounting for that? --Tardis 18:01, 11 May 2006 (UTC)

Oh, now I get it. Ohanian - what could really have made things clear was if you mentioned that the probability is itself a random variable. My apologies for not understanding this. Tardis - my guess is that any attempt to modify your original calculation would be more complicated than the direct calculation (finding the cdf and differentiating) - I had a try and it didn't quite work out. I think the way to go about it (I'll use my notation here, where t is the area) is to find the displacement of the graph of y=t/x, when t changes by dt, in a direction orthogonal to the graph - but I'm not completely sure. This will give the "thickness" of the line, which, if we're lucky, will amount to the 1/x factor. -- Meni Rosenfeld (talk) 20:10, 11 May 2006 (UTC)

May 11

Mathematics

Is math a closed system? or an open one? or is that unknown? what did Russell,Gödel and all of them say?. (as in closed sets or open sets).--Cosmic girl 16:25, 11 May 2006 (UTC)

Huh? Please suitly emphazi your question. Luigi30 (Ταλκ το mε) 18:22, 11 May 2006 (UTC)

It's hard to understand what you mean by a closed or an open system. I'll give it a try - Since mathematics is everything, it is, by definition, both a closed set and an open set. I guess that's not the answer you wanted to hear - If you clarify, we can be more helpful. Russel and Gödel would say what they always say. -- Meni Rosenfeld (talk) 20:14, 11 May 2006 (UTC)

That's the answer I wanted, but why do you say it's both a closed and an open system? can u explain that?... I guessed it was closed since it's everything like u said... but why open?.--Cosmic girl 22:01, 11 May 2006 (UTC)

Mathematics is a field of study. It is not a set, though sets are one of the things one studies in mathematics. Neither is mathematics a system. Like science, literature, politics, music, architecture, medicine, philosophy, mathematics is a field of study. The question probably has no meaning. -lethe ^{talk +} 22:37, 11 May 2006 (UTC)

I see... but I meant math as a language... it's a system! ( I guess).--Cosmic girl 00:05, 12 May 2006 (UTC)

Mathematics is not a language, though language is used to do mathematics, and mathematicsl languages are one of the subjects that are studied in mathematics. So what is it that you want to ask? Is the language used by mathematicians in their work a closed language? Well, mathematicians often use a form of the English language (with a fair amount of jargon added). Do you consider English to be a closed language? How about Spanish? I'm not sure what you think "closed" means for a language, but perhaps you have some idea. So my answer to you would be: mathematics is closed if and only if English or Spanish is. -lethe ^{talk +} 01:24, 12 May 2006 (UTC)

Meni, what do you mean by "mathematics is everything"? JackofOz 22:44, 11 May 2006 (UTC)

Same thing Pythagoras meant when he said "all is number", I guess. —Keenan Pepper 04:11, 12 May 2006 (UTC)

More or less. Hard to describe exactly here, though. It would suffice to say that AFAIK our physical universe is just one object in the "universe of all possibilities", aka mathematics. About both open & closed: Given a topological space (X, T), the universal set X and the empty set Φ are, by definition, open. X is the complement of Φ, and so it is, by definition, closed. -- Meni Rosenfeld (talk) 08:51, 12 May 2006 (UTC)

No guys, I'm being missunderstood I guess...Lethe, when I say language I don't mean 'english' because it's the one you use to do math... when I say language I mean MATH MATH in any language... like... math has syntactic rules and all, so why not be a language? .--Cosmic girl 16:15, 12 May 2006 (UTC)

Huh, you might be interested to read Godel's incompleteness theorem, Peano axioms, Zermelo-Fraenkel set theory, First order logic, Formal system or Formalism, and Foundations of mathematics. In short, if you're asking whether a mathematical system with axiom and theorem can be both complete and consistent, then godel proved(the first link) that in general the answer is negative, which is somewhat a blow to the attempt of formalism. However, this doesn't mean it is entirely impossible, as some system that are not strong enough to define natural number by itself can be made to do both, such as euclidean geometry or real numbers. Hope these answer your question. --Lemontea 01:50, 13 May 2006 (UTC)

thank you :), but what can euclidean geometry and real numbers do? I didn't understand that part.--Cosmic girl 20:55, 13 May 2006 (UTC)

Be both complete and consistent. "Consistent" means there are no contradictions, that is, you cannot prove both a statement and it's negation. "Complete" means that there are no "unprovable" problems - for every statement, you can always either prove the statement, or its negation. According to Godel's first incompleteness theorm, any system as strong as the Peano axioms can't be both complete and consistent - If it is consistent (which we should hope it is), than it cannot be complete (so there will always be "open" problems which cannot be solved with a proof). However, weaker systems are capable of being both consistent and complete. -- Meni Rosenfeld (talk) 12:32, 14 May 2006 (UTC)

Thank you Meni, great explanation! :) I didn't know real numbers where complete and consistent! I didn't think so either, thnx! :) hey Meni, I have a doubt though... are they both consistent and complete because tey are Axioms? or that doesn't have anything to do. :|.--Cosmic girl 16:07, 15 May 2006 (UTC)

As for the question "what can Euclidean geometry do?", well, it can do a lot. It can prove the Pythagorean theorem, and many other useful theorems. The theory of real numbers can also prove some useful results. -lethe ^{talk +} 14:30, 14 May 2006 (UTC)

For a more principled answer, different views on what mathematics "is" are possible; see Foundations of mathematics and Philosophy of mathematics. One particular school of thought is called Intuitionism; in that view mathematics is very much "open" and always essentially "incomplete". This then necessarily extends to the "language" of mathematics. For a more pragmatic answer, in mathematical practice working mathematicians invent concepts, terminology and notation on the fly, as they go, as long as it (hopefully) serves the purpose of getting an idea across. See Mathematical jargon, Mathematical notation, and, for example, Abuse of notation. In that sense too, mathematics is not closed. --Lambiam^Talk 16:45, 15 May 2006 (UTC)

Great, I see... but the ideas they want to get across themselves, are closed, right?, because they are not 'invented as they go', they are...'discovered'...this is what I mean, not the mathematical jargon, but mathematical escence itself..does intuitionism say anything about this? or no philosophy of math ever touched the subject and just focused on 'how' are ideas expressed rather than, 'what' are this ideas in the first place and how can we 'prove them'.--Cosmic girl 01:53, 16 May 2006 (UTC)

Maths Sum

The cow valued Rs 9600 is sold to three persons. The first person sells it at 10% profit, second person at 12% profit and the third person sells it at the profit of Rs 500. Find the last selling price.

Rs 3 lakh--Deville (Talk) 17:14, 11 May 2006 (UTC)

Assuming 9600 is the initial price and there are no expenses, we get:

Final price = 9600(1.10)(1.12) + 500

StuRat 18:35, 11 May 2006 (UTC)

$ python
Python 2.4.1 (#1, May 27 2005, 18:02:40) 
[GCC 3.3.3 (cygwin special)] on cygwin
Type "help", "copyright", "credits" or "license" for more information.
>>> Final_price = 9600*1.10*1.12 + 500
>>> print Final_price
12327.2

Google calculator. --hydnjo talk 14:13, 14 May 2006 (UTC)

The Age Of Her Life

Problem Sarah started school at the age of five. She spent one quarter of her life being educated, and went straight into work. After working for one half of her life, she lived for fourteen happy years after retiring.

How old was she when she retired?

Plz tell me answer?

We don't do your homework for you, unfortunately. :) Luigi30 (Ταλκ το mε) 18:21, 11 May 2006 (UTC)

Here's the setup, though:

Let X = how many years she lived

5 + X/4 + X/2 + 14 = X

Solve for X

StuRat 18:31, 11 May 2006 (UTC)

And then remember that the question is how old was she at retirement, not when she died. --LarryMac 18:35, 11 May 2006 (UTC)

I wonder why her parents called her "Problem Sarah". A clear cut case of setting up the poor kid for failure, if you ask me. JackofOz 22:43, 11 May 2006 (UTC)

:-) How humiliating when your childhood nickname sticks, even when you achieve textbook status. Skittle 14:59, 12 May 2006 (UTC)

Ah yes, textbook status. For how many centuries has man dreamed of achieving textbook status. But it can have its drawbacks, as we now know. :--) JackofOz 07:20, 13 May 2006 (UTC)

(Removed apparently copyrighted homework answer because it was copyrighted and giving a homework answer.) Skittle 13:04, 12 May 2006 (UTC)

But if you just thought about the problem, you could solve it easily. She lives a quarter of her life at school. She lives half her life at work. That leaves how much of her life? Skittle 13:06, 12 May 2006 (UTC)

I think this question deserves a rewrite:

If Problem Sarah starts contributing to Wikipedia at age 14, spends one quarter of her life making contributions to Wikipedia, and one half of her life in flame wars with various Wikipedians, then spends the last 5 years of her life in a straight jacket in an inane asylum, how old was she when first committed (to the asylum, not Wikipedia) ? StuRat 11:49, 14 May 2006 (UTC)

StuRat 11:49, 14 May 2006 (UTC)

How disappointing to find oneself in an inane asylum, after going to all the trouble of going insane. Skittle 10:38, 15 May 2006 (UTC)

I was wondering if anyone would notice that I slipped that in. Good eyes ! StuRat 01:56, 16 May 2006 (UTC)

Isn't the minimum age of retirement in the United States 67? I think it's a safe wager that 67 is in fact the age at which she retired ;) — Ilyanep (Talk) 01:11, 15 May 2006 (UTC)

While individual employers may set retirement ages, there is no mandatory retirement age in the US. StuRat 01:56, 16 May 2006 (UTC)

How can we see left derived functors even as actual functors?

Hello,

I am trying to understand more of homology.

Suppose you have a functor F (covariant and additive) from R mod (the R modules) to the category of Abelian groups.

If you want to take the nth left derivative, you are going to have to use projective resolutions. You create a projective resolution, apply the functor, you get a complex over Fm, and you take the nth homology module. Morphisms are defined a similar way.

My problem : you created 'a' projective resolution. So it depends on that choice. Now those modules will all be the same up to isomorphism, but... I don't feel too good about it.

This isn't really a functor anymore, or is it? Perhaps one should not change the functor but the categories?

All hints are welcome! Evilbu 22:26, 11 May 2006 (UTC)

I asked a similar question on this reference desk a while ago. Limits like products are only defined up to canonical isomorphism. You can define a functor which assigns products, but it in order to have a specific product object, you have to invoke the axiom of choice. I found a paper by Mikkai where he describes a construction of a generalization of functors which he calls anafunctors. These are like functors which have many different objects in the image of their object function, and satisfies some appropriate functorial condition. You can turn an anafunctor into a proper functor by invoking the axiom of choice to choose a single object from each isomorphism class. I'm not sure if derived functors admit a similar description, but it sounds like it might. That is, I don't know if I can describe a derived functor as a limit, but to each functor, we assign a derived functor, which is only determined up to isomorphism. Sounds about right. -lethe ^{talk +} 23:35, 11 May 2006 (UTC)

You should view them as functors from the derived category, which is pretty much a category whose objects are resolutions: notice that the entire construction is functorial in the resolution. Sadly our article on the subject is all history and no math. —Blotwell 23:21, 12 May 2006 (UTC)

May 12

Brachistochrone

I am interested in actually constructing a brachistochrone track for a physics project. How does one actually construct a brachistochrone (I mean on paper mathematically, not physically)? I know that it is a segment of a cycloid, but how much of the cycloid? I'm not allowed to just use any cycloid curve, right? It says on Wikipedia

Given two points A and B, with A not lower than B, there is just
one upside down cycloid that passes through A with infinite slope
and also passes through B. This is the brachistochrone curve.

So the tangent line of the brachistochrone at A must be vertical By tinkering with this: http://home.ural.ru/~iagsoft/BrachJ2.html I assume -construct an entire "hump" of a cycloid which much start at A, pass through B, and end at a point level with A

So I think that's right, but where do I go from there? If A and B are given, how would I construct such a cycloid? I can't just say that its amplitude (and therefore twice the circle's radius) is the vertical distance from A to B, because the lowest point may be below B. How would I figure its amplitude/width? -JianLi 01:32, 12 May 2006 (UTC)

Well, the obvious approach is to use the parametric equations for the cycloid and try to solve them simultaneously for t and r. Take A to be the origin, and take $B=(x,y),y<0$. Then we have $x=r(t-\sin t),y=r(\cos t-1)$. Isolating the sine and cosine, squaring, and adding gives us $(t-x/r)^{2}+(y/r+1)^{2}=1$, so $t={\frac {x}{r}}\pm {\sqrt {1-\left({\frac {y}{r}}+1\right)^{2}}}$. Then divide the sine and cosine to get $\tan t={\frac {t-x/r}{y/r+1}}={\frac {\pm {\sqrt {1-\left({\frac {y}{r}}+1\right)^{2}}}}{{\frac {y}{r}}+1}}=\pm {\sqrt {\left({\frac {y}{r}}+1\right)^{-2}-1}}$. This gives four possible solutions for t involving r (one times two from the $\pm$ and times two from $\tan x=\tan(x+\pi )$). Plug each of these into, say, the y equation and you can solve for r in theory, but I'm not sure how to do that analytically. Then you'd have to evaluate x and y and see if they matched for each of the four solutions, I guess. ...Hm, this sort of bogged down eventually. Anyone else have a cleaner way to do it? --Tardis 03:11, 12 May 2006 (UTC)

OK, try 2: start by considering the known value $\xi :={\frac {y}{x}}={\frac {\cos t-1}{t-\sin t}}$ (no r dependence). Then $\cos t+\xi \sin t-\xi t=1$. Unfortunately, the best you can do at this point appears to be solving numerically again; you could first replace the trig functions with one shifted, scaled trig function using $A\cos(x-\phi )=A\cos x\cos \phi +A\sin x\sin \phi$, so in this case $A\cos \phi =1,A\sin \phi =\xi$ so $A^{2}=1+\xi ^{2}$ and $\tan \phi =\xi$, so we have ${\sqrt {1+\xi ^{2}}}\cos(t-\arctan \xi )-\xi t=1$ (remember that $\xi <0$). Note that $t=0$ is automatically a solution, but should be discarded. Hope this helps, although I'm still hoping for a magic analytical solution (from my own brain or otherwise). --Tardis 03:30, 12 May 2006 (UTC)

Thanks for your reply thus far. I have another question: What is the effect of rotational kinetic energy? The calculation of the cycloid as the brachistochrone is calculated only accounting for translational KE. I'm assuming that the cycloid still comes in first, assuming that the ball will roll on all surfaces without slipping. Is this right? JianLi 03:56, 12 May 2006 (UTC)

The translational energy is of course $K=mv^{2}/2$, and the rotational kinetic energy is $U=I\omega ^{2}/2$. Since, without slipping, $v=r\omega$, the overall energy is proportional to $v^{2}$ and so acts like some increased mass. Since the mass doesn't matter, you're right, the cycloid wins -- but as the path of the center of gravity! You'll need to displace the track outwards by one radius. Of course, you'd need to do that anyway if the falling object is of non-negligible size compared to the course. --Tardis 04:13, 12 May 2006 (UTC)

Analytic signal and analytic continuation

Suppose we have a real $u(t)$ function. We can obtain a complex analytic signal $z_{a}(t)$ using Hilbert transform. We can also use an analytic continuation to obtain another complex function, say, $z_{c}(t)$. What can we say about $z_{a}(t)$ and $z_{c}(t)$? At least for harmonic $u(t)$ they seem to be equal. And what about the general case?--Ring0 05:33, 12 May 2006 (UTC)

Sorry, I seem to misunderstand your question. The complex analytic signal you get from the Hilbert transform is a complex-valued function whose domain is the reals, i.e.

$z_{a}(t):\mathbb {R} \to \mathbb {C} ,$ but the analytic continuation of a real function is a function whose domain is also complex, namely $z_{c}(t):\mathbb {C} \to \mathbb {C} .$ In this case we're talking two different animals here. Do I misunderstand something? Your last sentence makes me think I did --Deville (Talk) 02:21, 15 May 2006 (UTC)

Thanks, I've forgotten about the domain, but still we can apply $z_{c}(t)$ only to $t\in \mathbb {R}$ and compare it with $z_{a}(t)$ in the real domain.--Ring0 22:01, 15 May 2006 (UTC)

I think it's possible he means something like the analytic continuation of the Hilbert transformation as well, or something along those lines. At the very least there does seem to be a connection between the Hilbert transform and the analytic continuation by following logic along these lines (I haven't done the actual math involved so I'm not 100% sure of the connection):

Assume f(t) has an analytic continuation f(z) in the domain bounded by $-\epsilon <Im(z)<\epsilon$. Then $g_{n}(z)={\frac {f(z)}{(z-\tau )^{n}}}$ has a pole of order n at $z=\tau$. Let C be a contour which starts at $-\infty$ on the real axis, goes along the real axis to $+\infty$ with a small detour over $z=\tau$, then goes back to $-\infty$ with a small detour under $z=\tau$. Then I think there's a fairly close connection between the Hilbert transforms of the nth derivative of f(t), the integral of g_n(z) around the contour C, and the nth coefficient of the Taylor expansion of f(z) around $z=\tau$. As I said, I didn't do the math so it could be completely wrong. Confusing Manifestation 02:40, 15 May 2006 (UTC)

Johnson Solids

I'm wondering about the names of the last 9 Johnson solids. At first glance most of them seem random, but a closer look seems to reveal patterns.

1. First, the snubs (J84 and J85). According to Mathworld, a disphenoid is a tetrahedron, so the snub disphenoid could also be called the snub tetrahedron (Is this right?). My question is, are the snub disphenoid and the snub square antiprism related to the tetrahedron and the square antiprism in the same way that the snub cube and the snub dodecahedron are related to the cube and the dodecahedron? The snub polyhedron article mentions the Johnson snubs in passing, implying that they may be a little different yet not giving sufficient detail as to what the difference actually is.
2. I understand the naming of J87, as the article is very good on this point.
3. All the others (J86 and J88-92) have strange names, but a close look reveals them all to consist of various combinations of the following:
   • spheno
   • corona
   • mega
   • hebe
   • cingulum
   • luna
   • di and bi
   • rotunda (eg pentagonal rotunda)

   According to this site, a luna is a fragment consisting of a square with two triangles attached to opposite sides. It also gives a definition of a corona which makes no sense to me. But that doesn't enlighten me as to the source of the other name fragments. Is there a polyhedron (or polyhedral fragment) called a spheno and another one called a cingulum? Also, rotunda in this case seems to mean something other than pentagonal rotunda. Can anyone help me here, or are these names truly just random?

4. Finally, the article explains that augmented and diminished may mean the addition or removal of either a pyramid or a cupola, but no rules are given to determine when a pyramid is used and when a cupola. With diminishing, I can see that it might be obvious as some shapes only have cupolas to remove and others have only pyramids. But with augmentation, is the choice of pyramid or cupola arbitrary or based on a rule?

Please answer all three questions if possible. Thank you. --72.140.146.246 17:45, 12 May 2006 (UTC)

Ah, those wonderful, wacky, polyhedra and their names! Probably George Hart's site, listed at the bottom of the article, is your best source. The truth is, people invent these names with some attempt at regularity and some attempt at whimsy. So many names are needed and the shapes are so challenging to describe simply (once we get past Archimedean solids), that it is unrealistic to hope for clear meaning and regularity — unless someone comes up with a new insight. --KSmrq^T 03:20, 13 May 2006 (UTC)

I did visit George Hart's site before asking this question, but I could not view the .wrl files, which discouraged me from looking through it much. Revisiting it just now, however, I noticed something which answers my question in point 3 above. Thank you. But what about the snubs (point 1 above) and the augmentation (point 4 above)? I would like to know whether the Johnson snubs are standard snubs (eg the Archimedean snubs) and how the choice of cupola or pyramid is made for an augmentation. --72.140.146.246 14:53, 13 May 2006 (UTC)

Help with finding a VRML browser (for the *.wrl files) is available. As for the remaining questions, try email to George Hart if nothing else works. --KSmrq^T 16:16, 13 May 2006 (UTC)

Area/Volume → Perimeter/Surface Area?

The area of a circle is $A=\pi r^{2}$. Differentiating this formula with respect to r gives ${\frac {\;dA}{\;dr}}=2\pi r$. So ${\frac {\;dA}{\;dr}}=C$, the circumference or perimeter. With a sphere this also works, so ${\frac {\;dV}{\;dr}}=A$. However, it does not work for a cube or a square, and probably not for other polygons or polyhedra. Is the fact that it works for the circle and the sphere simply a coincidence, or is this a special case of something more general which does work for other polgons and polyhedra? (I notice that for square or cube my method would produce exactly half of the correct value.) --72.140.146.246 17:59, 12 May 2006 (UTC)

This is certainly not a coincidence. It represents the fact that dA, an infinitesimal change in the area of the disc, is the area of a thin ring with thickness dr and circumference C. The area of this ring is C times dr (the area of a thin line is its length times its thickness), and therefore ${\frac {dA}{dr}}=C$. The same thing holds for a ball. For it to work for a cube\square, you need to use a different variable: Instead of a, the edge length (which is analogous to the diameter of a circle), consider b = a/2 (which is analogous to the radius). You'll have A = 4b^2 and p = 8b, where ${\frac {dA}{db}}=p$. It doesn't work with a, for the same reason it wouldn't work with the diameter d for a circle. -- Meni Rosenfeld (talk) 18:23, 12 May 2006 (UTC)

Also, this does work for a cube and other Platonic solids too, and for regular polygons, for the same reason as why it works for a sphere, you just have to do the differentiation with the radius of the inscribed sphere (or circle) as the independent variable. For example, a cube with insphere radius $\rho$ has volume $V=16\rho ^{3}$ and surface $A=(d/d\rho )(16\rho ^{3})=24\rho ^{2}$. Really it works for any solid, if you chose a suitable independent variable (the "average inscribed sphere radius" if such a thing exists), because if you scale a solid, the volume always changes cubically, and the surface square proportionally. – b_jonas 22:50, 12 May 2006 (UTC)

I wonder, does it really work for "any" shape? For example, it does not work for an ellipse, at least not if you choose the independent variable to be the semi-major axis and hold the eccentricity fixed. So the symmetry of the figure must somehow be relevant. -lethe ^{talk +} 03:23, 13 May 2006 (UTC)

I guess a sufficient condition for this to work with respect to a variable r is that the distance from the center to the tangent of the shape at any point (where the tangent exists) is r. If we want this to be some intuitive parameter of the shape (such as the half-side of a square, but not the "average inscribed radius" of an ellipse), this is probably necessary as well. -- Meni Rosenfeld (talk) 07:03, 13 May 2006 (UTC)

It holds for an ellipse too, only you have to choose a constant multiply of the semi-major axis as the independent variable. The only condition is that the shapes must really be similar to each other (which holds with an ellipse as the excentricity is fixed), and that the independent variable is scaled the same amount as the figures. The ellipse is a bit difficult example because its perimeter cannot be written as a simple formula, but still we know that the perimiter is surely a constant multiply of the semi-major axis $a$, let's say it's $S=k_{e}a$ (I wrote $k_{e}$ because the constant depends on the excentricity). The area is $A=\pi {\sqrt {1-e^{2}}}a^{2}=K_{e}a^{2}$. Now let's choose $t=2(K_{e}/k_{e})a$ as the independent parameter, then $dA/dt=(dA/da)/(dx/da)=(2K_{e}a)/(2K_{e}/k_{e})=S$. – b_jonas 16:01, 13 May 2006 (UTC)

Yes, I understood this idea, and that it works for any shape. I only said that in cases like this, the variable you have to use is a complicated function of the basic parameters of the shape, as opposed to a shape satisfying my condition above, for which the variable is simply the inscribed radius. -- Meni Rosenfeld (talk) 16:49, 13 May 2006 (UTC)

How would this hold for an equilateral triangle? The area of a triangle is $A={1 \over 2}bh$, which with the Pythagorean theorem becomes $A={{\sqrt {3}} \over 2}b^{2}$, where b is the side length. However, with this variable, $P=3b$. Using half the side length as your variable would not work here as it did for the square, so what variable would work and how is that variable related to the side length? Must the triangle be equilateral for this to work? --72.140.146.246 14:33, 13 May 2006 (UTC)

Here, you must chose the radius of the inscribed circle as the independent variable, which is $\varrho =h/3=b/(2{\sqrt {3}})$. You can use any triangle as long as you scale it uniformly, and the independent variable to choose is always the radius of the inscribed circle. – b_jonas 16:08, 13 May 2006 (UTC)

It will also work for a wide range of polygons, including (but not limited to) regular polygons, and also some shapes which are derived by stitching together straight line segments and arcs of a circle - in general, exactly those shapes that have an inscribed circle (touching all sides, and overlapping curved parts). The variable which should be used is, of course, the radius of the inscribed circle. BTW, it's ${\frac {\sqrt {3}}{4}}b^{2}$, not ${\frac {\sqrt {3}}{2}}b^{2}$. -- Meni Rosenfeld (talk) 16:49, 13 May 2006 (UTC)

Integral of the square of the sine

I suddnly found integrals a bit interesting, but there is one thing that bothers me, and that is the integral of the square of the sine. How can this equality be proven, using the usual techniques for integrals?

$\int \sin ^{2}x\,dx={\frac {1}{2}}(x-\sin x\cos x)+C$

Loool 19:09, 12 May 2006 (UTC)

Use the identity

$\cos 2x=1-2\sin ^{2}x.$

Also, use the identity

$\cos 2x=2\cos ^{2}x-1$

for the integral of cosine squared. -lethe ^{talk +} 19:31, 12 May 2006 (UTC)

Thanks for this one. I had forgotten about those formulas. Now the integral looks very easy. Loool 19:35, 12 May 2006 (UTC)

The identities come from

$\cos(x+y)=\cos x\cos y-\sin x\sin y,$

from which you get

$\cos 2x=\cos ^{2}x-\sin ^{2}x.$

To get the required identities, use the famous

$\sin ^{2}x+\cos ^{2}x=1,$

and substitute for the sine squared or cosine squared as appropriate. -lethe ^{talk +} 19:45, 12 May 2006 (UTC)

Yes, using an identity like that is indeed the easy way to compute this integral but it's also possible with partial integration I belive. – b_jonas 22:13, 12 May 2006 (UTC)

Yes, that's another good way. Requires a little less trigonometry. Another way is to use complex numbers, if you're familiar with them. -lethe ^{talk +} 03:13, 13 May 2006 (UTC)

If you desire only to prove the equality and not compute the integral, then it's sufficient to differentiate both sides. Dysprosia 12:32, 13 May 2006 (UTC)

Infinite Sets

Is it possible to prove that one infinite set can be larger than another infinite set using strictly mathematics [no analogies to cavemen]. — Ilyanep (Talk) 22:30, 12 May 2006 (UTC)

Yes. The Cardinal Number article has some nice information and links. What's this 'caveman' analogy? It sounds amusing. Black Carrot 23:05, 12 May 2006 (UTC)

Basically, you have a group of cavemen who can only count up to four, and afterwards they just say 'a lot'. So technically you could have six or eight and one 'a lot' is greater than another 'a lot'. — Ilyanep (Talk) 02:54, 13 May 2006 (UTC)

Yes, it's Cantor's diagonal argument. There are also various humorous illustrations of the same argument, like infinite hotels or the Book of Sets. – b_jonas 22:53, 13 May 2006 (UTC)

Perhaps you're referring to Hilbert's hotel. —Blotwell 21:51, 14 May 2006 (UTC)

Surface area of a sphere

How do you prove that the surface area of a sphere is $4\pi r^{2}$?

Obviously the volume formula comes from that by integrating the surface area formula.

(Be nice, I'm a freshman in high school who's in Geometry but has dabbled around with some calculus).— Ilyanep (Talk) 22:32, 12 May 2006 (UTC)

Some searching on Google gives [1] and [2]. Hope that helps. Black Carrot 23:11, 12 May 2006 (UTC)

I think it's easier to find the volume first, then differentiate to get the surface area. And a simple way to find the volume is to integrate the area of successive circular cross-sections, i.e. $\int _{-r}^{r}\pi (r^{2}-x^{2})\,dx$ or something. —Blotwell 23:15, 12 May 2006 (UTC)

This is a fun question to learn with. Archimedes, an outstanding ancient mathematician, was so proud of the relationship he proved between a sphere and its circumscribed cylinder that he asked for the figure to be engraved on his tombstone. His ingenious methods foreshadowed the development of integral calculus many centuries later.

The area of a cylindrical strip is unchanged by unrolling it to form a rectangle; the height will be 2r (the diameter of the sphere) and the width will be 2πr (the circumference of the sphere at the equator). Thus the cylinder strip area is 4πr², the product of height and width.

Can we transform our awkward problem, the area of a sphere, into this simple, solved, one? Archimedes says yes. Look at a tiny rectangle on the surface at a latitude of ϑ. As we move toward the North or South Pole, the surface tilts more and more perpendicular to the vertical of the cylinder. This causes the projection onto the cylinder to be foreshortened by a factor of cos ϑ. Simultaneously, the sphere surface shrinks away from the cylinder surface by the same factor, implying that the projection onto the cylinder is larger by the reciprocal of that factor. Thus these two effects exactly cancel, so that the area of the cylindrical strip is identical to the area of the sphere.

Incidentally, this implies that for random points uniformly distributed on a unit sphere, the z coordinate is uniformly distributed between +1 and −1. So is the y coordinate and so is the x coordinate; but these are obviously not independently distributed.

Not bad for no (explicit) calculus. Archimedes did more, and he was justifiably proud. If we try it with explicit calculus, we have two obvious choices: rotate a semicircle of longitude (drawn from Pole to Pole), or accumulate circles of latitude. The rotation method uses a valuable tool of broad utility, the description of a surface of revolution. The latitude method should look very much like our Archimedean calculation.

Let's end with a little table, listing interior measure and boundary measure (volume and area, generalized) for unit balls as a function of even dimension, 2n:

2n	2	4	6	8
V	π	π2/2	π3/6	π4/24
A	2π	2π2	π3	π4/3

In each case the product of dimension and volume equals area. The volumes seem to follow the pattern πⁿ/n!, which hints at a general formula. Is this real? If so, can we prove it? And what about the odd dimensions? (The answers to these questions are well-known, of course, and mostly hiding somewhere within Wikipedia.) --KSmrq^T 05:24, 13 May 2006 (UTC)

Deal or No Deal?

I was wondering how they calculate the "Deal" from the banker on the show and on the internet. I have been trying to figure it out for months. Thanks --Zach 23:34, 12 May 2006 (UTC)

Yes, I've been thinking about this a lot too. I guess that it just has to do with the probability that the person is going to win each amount that is left. The "deal" would just be the amount that the person is, according to probability, would probably win. Beyond that, I don't know. (i.e. I haven't thought that far or I can't think that far.) --Think Fast 02:02, 13 May 2006 (UTC)

I'm glad I wasn't the only one thinking about that. Not quite sure how it works though... —Mets501^talk 02:25, 13 May 2006 (UTC)

The banker's tolerance for risk appears to diminish as the prospective loss (and the std dev of the remaining values) becomes larger; whilst early offers are often substantially less than one's expected payout (and thus should be declined by risk-neutral players), late offers (especially in situations in which five or fewer amounts remain, of which at least two are exceedingly large) tend rapidly to approach one's expected payout (I can recall two cases in which the offer exceeded the player's expected payout, and each was, properly, accepted). This is, I think, by design; Howie has explained on, inter al., The Tonight Show with Jay Leno that the "banker" simply conveys an algorithm-generated offer. The show certainly has some appeal inasmuch as it presents one with the opportunity to earn money without debasing him/herself and in a pleasant environment, accompanied by family and friends, but I find myself increasingly unable to watch, if only because of the complete ignorance of game theory or probability exhibited by the players. Worse, though, is the sincere belief that many contestants ostensibly harbor that they make wise case choices; one wonders whether they actually appreciate that the game--at least the case-selection portion--is altogether one of chance (sure, I sometimes manually select numbers for the lottery, but I do that mainly for fun, and I surely don't think that my selections are somehow more lucky to appear than those generated by a computer). Finally, Howie's frequent intimations that a player has made a great deal are ex post accurate (just as one who doesn't switch in the Monty Hall problem but ends up with the car can be said to have made the right decision) but surely reflect a belief that one made the proper decision on a generalized, ex ante level, which belief is not always correct. Joe 03:31, 13 May 2006 (UTC)

It appears to me that the banker's offer is a rounding off of an algorithmic amount, probably to keep the numbers simple for television's sake. If you play the on-line version on the game at nbc.com, the banker's offers are very unrounded amounts. In the game I just played, I was given offers such as $24,834, $48,049, and $55,573. — Michael J 17:39, 13 May 2006 (UTC)

Note that the primary goal of the banker is not to limit payout, but to maximize viewership. Thus, making stingy offers which are constantly refused would be boring and lose viewers, while making generous offers which are accepted is more interesting (although not so generous as to be a "no-brainer"). StuRat 21:31, 14 May 2006 (UTC)

Triangular Sums

Anyone know how to find pairs of triangular numbers whose sums are triangular numbers? Ditto for squares. Black Carrot 23:50, 12 May 2006 (UTC)

Any triangular number can be represented by the expression ${\frac {x(x+1)}{2}}$. This means that for a triangular number to be the sum of two other triangular numbers, it must be the case that $x(x+1)=m(m+1)+n(n+1)$. You can solve this Diophantine equation to find values. As for square numbers, you are really looking for Pythagorean triples, so I would go there for more information. —Mets501^talk 02:46, 13 May 2006 (UTC)

I have a similar problem too, this time is to find (non-trival) integer solution to ${p \choose 3}+{q \choose 3}={r \choose 3}$. BTW, for the squar number problem, move term to get x²=z²-y². Assuming they're in the simpliest form possible(i.e. coprime), we factorize to have x²=(z+y)(z-y).Because they're coprime, so are z+y and z-y coprime. So z+y and z-y are both square number. We set 2m²=z+y and 2n²=z-y. So x=2mn, y=m²-n², z=m²+n² as a parametric solution. --Lemontea 06:53, 15 May 2006 (UTC)

Here is an approach to the original question regarding generating pairs of triangular numbers which sum to another triangular number. Take an arbitrary triangular number T_n and express it as the product of a power of 2 and an odd number i.e.

$T_{n}=2^{p}q.$

Now express the odd number q as the sum of two consecutive numbers r and r+1, so

$q=2r+1.$

Define s and t by

$s=r+1-2^{p}$

$t=r+2^{p}.$

Then

$\sum _{m=s}^{t}m=\sum _{m=r+1}^{t}{(2r+1-m)+m}=\sum _{m=r+1}^{t}{q}=2^{p}q=T_{n}$

so we have expressed T_n as the sum of 2^p+1 consecutive numbers from s to t. Then

$T_{s-1}+T_{n}=\sum _{m=1}^{s-1}m+\sum _{m=s}^{t}m=\sum _{m=1}^{t}m=T_{t}.$

For example, T₁₁ = 66 = 2*33, so we have r = 16, s = 15, t = 18 and

$T_{14}+T_{11}=105+66=171=T_{18}.$

Gandalf61 10:03, 15 May 2006 (UTC)

May 13

"calculations in your head"

What is the English term for "making calculations in your head", if there is one? It's pretty common in Estonia to have competitions in schools for this, and I only just realized I don't actually know the English term. I recently heard a claim a 13-year old Estonian girl is the "world champion", which I find hard to believe, but I don't know where to look to find out. PeepP 07:08, 13 May 2006 (UTC)

I think this is sometimes referred to as mental calculation. -- Meni Rosenfeld (talk) 07:10, 13 May 2006 (UTC)

Thanks, that's obvious enough. I've added it as a see also to calculation, which is where I looked at first. PeepP 07:53, 13 May 2006 (UTC)

Personally, I call it mental math :) — Ilyanep (Talk) 15:09, 13 May 2006 (UTC)

A useful test: [3] (12.2 million results) [4] (25.4 million results) [5] (4.65 million results)

I'm more familiar with "mental math", probably chosen because it's catchier. Black Carrot 15:12, 13 May 2006 (UTC)

We usually called it Mental Arithmetic when we were tested on it. Skittle 16:00, 13 May 2006 (UTC)

I suspect Mental Arithmetic sounds more official, and Mental Math is more of a colloquial term. — Ilyanep (Talk) 22:50, 13 May 2006 (UTC)

In fact, mental math is more of a kids' term. That's why I know it better; I'm 18, and have spent most of my life as a child. Black Carrot 23:56, 13 May 2006 (UTC)

"THE ALL IS MIND; The Universe is Mental." -- The Kybalion. --DLL 19:53, 14 May 2006 (UTC)

finite geometry: how many (m+1)dim spaces on quadric go to m dim space

Hi,

I would really like to know how to answer this question

consider a nonsingular quadric Q in a projective geometry PG(n,q) with q a prime power

let V be a m dimensional space completely on Q how many m+1 dimensional spaces W go through V and are also completely on the quadric Q

now this question is fairly easy if m=0, given that the tangent hyperplane through a given point p cuts the quadric in a cone, with p as top and as basis a nonsingular quadric of two dimensions lower

but how about the general case? The formule should be quite easy I suspect, very closely related to the number of points in smaller quadrics.

Truly, all hints are welcome.

By the way: the more general case of a finite classical polar space is also interesting (thus the same question for nonsingular hermitian varieties and symplectic polarities)

Thanks, Evilbu 12:21, 13 May 2006 (UTC)

Nobody wants this one? We can at least ask about the field; shall we assume ℝ or ℂ or more general? When I'm not so lazy I'll check Hodge and Pedoe (ISBN 0521469015), a likely source. --KSmrq^T 03:59, 15 May 2006 (UTC)

Thank you for your response! What can be said about the field? Well it is definitely finite, thus its order is a prime power. PG(n,q) is 'slang' for the n dimensional projective geometry over the field of q elements (there is only one field of q elements up to isomorphism)

Now however it will matter whether or not q is even, because then you do not always have a polarity... but still the answer to my question (the number of spaces through it) should not be dependent.

Evilbu 16:10, 15 May 2006 (UTC)

May 14

Kernel

What does "kernel" stand for in Dirichlet kernel and Fejér kernel? Mon4 01:53, 14 May 2006 (UTC)

See integral transform. -- Jitse Niesen (talk) 03:41, 14 May 2006 (UTC)

Good catch! The usual algebraic meanings for "kernel" are linked at the disambiguation page, but the kernel of a linear filter defined by convolution is not, nor are any of its relatives. This is a common term in applications, in some instances equivalent to point spread function or impulse response. --KSmrq^T 04:07, 14 May 2006 (UTC)

May 15

Poisson Process

With $\{T_{n}\}_{n=1}^{\infty }$ a sequence of iid exponential variables with parameter $\lambda$. With definition of $\{S_{n}\}_{n=1}^{\infty }$ sequence:

$S_{n}=\sum _{k=1}^{n}T_{k}$

Which of these two process is a poisson process $N=\{N_{t}:t\geq 0\}$ with $\lambda$ parameter?

$A=\{A_{t}:t\geq 0\}\qquad A_{t}:=Sup\{n\in Z:S_{n}\leq t\}$

$B=\{B_{t}:t\geq 0\}\qquad B_{t}:=Inf\{n\in Z:S_{n}\geq t\}$

Hessam 09:22, 15 May 2006 (UTC)

How come this sounds so much like a homework problem? Could the reason be that it is a homework problem? If it is a multiple-choice test in which at least one answer is correct, then there is an easy way of eliminating one incorrect answer by considering P[A_t = 0] and P[B_t = 0]. --Lambiam^Talk 13:12, 15 May 2006 (UTC)

No it's not a multiple choice problem. With considering both on 0 position we can get nothing. There should be another reason in 3 other conditions of poisson process. I have no idea. No other idea?! ‍‍Hessam 22:47, 15 May 2006 (UTC)

:-(( Hessam 08:00, 16 May 2006 (UTC)

I don't know what the "3 other conditions of poisson process" are. I know the condition given under Poisson process. Taking N₀ = 0, we have P[N_t = 0] = exp(–λt).

Let us see if B fits. Assume t > 0. We have P[B_t = 0] = P[inf{n ∈ Z : S_n ≥ t} = 0] = P[S₀ ≥ t] = P[0 ≥ t] = 0. No, this doesn't agree with exp(–λt). So B is out.

Next we try A again with t > 0. Now P[A_t = 0] = P[sup{n ∈ Z : S_n ≤ t} = 0] = P[S₀ ≤ t ∧ S₁ > t] = P[S₁ > t] = P[T₁ > t] = 1 – P[T₁ ≤ t] = 1 – (1 – exp(–λt)) = exp(–λt). Yes, that fits. This isn't a proof that A is a Poisson process; for that you have to consider the more general expression P[N_t+τ – N_t = k]. But it's a start. --Lambiam^Talk 15:09, 16 May 2006 (UTC)

I knew nothing about this subject, but after a quick glance at poisson process I would guess it's A(Not sure if B is correct), as it resemble the "classic" telephone call example. --Lemontea 13:50, 16 May 2006 (UTC)

Can a dot product return a complex number or just reals?

In the Spanish Wikipedia it says it returns "scalars". In the English "reals". Thanks.

What do you mead by "dot product"? If you mean the scalar product in real vector spaces, of course it never returns complex numbers just because the vector space is, indeed, real. In general, if you have a vector space over some field $\mathbb {K}$, any scalar product will return an element of $\mathbb {K}$, as a definition. Have a look at Scalar product for a general definition. Cthulhu.mythos 11:44, 15 May 2006 (UTC)

In short - Yes, if the vectors are over the field of complex numbers. -- Meni Rosenfeld (talk) 12:22, 15 May 2006 (UTC)

Indeed, as the questioner noticed, the article Dot product states: "the dot product [...] returns a real-valued scalar quantity." If this is wrong (or too simplistic) as stated, it should be corrected. The Dot product article starts with "For the abstract scalar product or inner product see inner product space", but it may not be clear to the reader looking up "dot product" whether what they are looking for fits the notion of "abstract scalar product or inner product". --Lambiam^Talk 13:34, 15 May 2006 (UTC)

A dot product is assumed to be the standard Euclidean inner product of a real vector space. We would not usually call an inner product of complex numbers a dot product. (Next time, it would be better to ask on the article talk page so the discussion stays associated with the article.) --KSmrq^T 14:22, 15 May 2006 (UTC)

Gaussian curvature --> radius

Curvature and radius are reciprocals of each other. So, given $M={\frac {1}{\kappa _{1}}}\mathrm {\;and\;} N={\frac {1}{\kappa _{2}}}\,\!$, where M and N are the principal radii of curvature, the Gaussian curvature equals $\kappa _{1}\kappa _{2}\,\!$, not ${\sqrt {\kappa _{1}\kappa _{2}}}\,\!$, so that means the radius of Gaussian curvature must, therefore, equal $M\!N\,\!$, not ${\sqrt {M\!N}}\,\!$—? But how can that be? Is it actually Gaussian curvature²?  ~Kaimbridge~13:47, 15 May 2006 (UTC)

A different choice is the mean curvature, which for a sphere will be the reciprocal of its radius. --KSmrq^T 15:04, 15 May 2006 (UTC)

(edit conflict) "Radius of Gaussian curvature" is not a defined and meaningful notion. "Radius of curvature" is meaningful for curves in 2-space, and "Gaussian curvature" is the curvature of surfaces in 3-space. See Curvature. --Lambiam^Talk 15:06, 15 May 2006 (UTC)

However the radius of the two principal curvatures is defined, and are used to define the focal surface which have been studied in singularity theory and also have aplications in optics. At a point p on a surface with normal N then the two points $p+{1 \over \kappa _{1}}N$, $p+{1 \over \kappa _{2}}N$, define the focal surface. --Salix alba (talk) 16:26, 15 May 2006 (UTC)

I have been going back and forth between curvature, Gaussian curvature, principal curvature (and others), but the discussion is mostly abstract theory. Is the more complete, clarified relationship, "2-space (I presume, meaning 2-D space?) curvature" is the reciprocal of its radius, and "3-space curvature" (such as Gaussian) is the reciprocal of radius²? As for semantics, the radius is the inverse of "whatever curvature", so it would seem the proper order is "radius of Gaussian curvature", not "Gaussian radius of curvature". Consider the two possibilities for "mean" (a, b are the equatorial, polar radii):

• Mean radius of curvature:

${\frac {M+N}{2}}={\frac {{\frac {1}{\kappa _{1}}}+{\frac {1}{\kappa _{2}}}}{2}}={\frac {M}{2}}\!\cdot \!\left(1+{\frac {a^{4}}{(bN)^{2}}}\right)={\frac {N}{2}}\!\cdot \!\left({\frac {(bN)^{2}}{a^{4}}}+1\right);\,\!$

• Radius of mean curvature:

${\frac {2}{{\frac {1}{M}}+{\frac {1}{N}}}}={\frac {2}{\kappa _{1}+\kappa _{2}}}={\frac {1}{H}}={\frac {2M}{1+{\frac {(bN)^{2}}{a^{4}}}}}={\frac {2N}{{\frac {a^{4}}{(bN)^{2}}}+1}}.\,\!$

 ~Kaimbridge~18:12, 15 May 2006 (UTC)

It is correct that the curvature of a curve at a given point is elementary: fit a circle to osculate there, and take the reciprocal of its radius. But as soon as we increase the dimension of our object, whether to a surface embedded in 3D or beyond to a general n-manifold (Riemannian) with no explicit embedding, curvature becomes a much richer phenomenon. No single number, however computed, can capture everything. For example, at a given point on an ellipsoid we can take slices in different directions and get curves with different osculating circle radii. We can prove that these values have a minimum and a maximum (the principal curvatures), and we can combine them into a single number; but how shall we combine them? No matter what choice we make, we lose information.

If we are interested in minimal surfaces, like soap bubbles, we must average the min and max curvatures (obtaining mean curvature). But if we are working with a surface which is not given as an embedding, then we must work intrinsically and use Gaussian curvature. And as we move up in dimension to curved spaces, we encounter options such as sectional curvature, the Riemann curvature tensor, Ricci curvature, and so on.

The broad idea is to have some way to summarize the "shape" of a (well-behaved) geometric object near a given point. (Of course, now that fractal objects have entered the common awareness, we know that curvature may not always be a meaningful measure.) Yes, the choices make learning more of a challenge, but they also make nice pictures. [6][7][8] [9] --KSmrq^T 09:03, 16 May 2006 (UTC)

Right, I realize these things can and do get complicated! P=) For instance, Gaussian may or may not be apparent, depending on an equation's form. E.g., (where $\phi ,\;\alpha \,\!$ is geodetic latitude and its azimuth):

${\frac {1}{\sqrt {\left({\frac {\cos(\alpha )}{M}}\right)^{2}+\left({\frac {\sin(\alpha )}{N}}\right)^{2}}}}={\frac {M\!N}{\sqrt {(M\sin(\alpha ))^{2}+(N\cos(\alpha ))^{2}}}}\,\!$

But, my question, here, isn't so much about its application, but its basic radius-curvature relationship. Maybe it would be better asked another way. How would you define/describe "MN"?: "Radius of Gaussian curvature", "radius of Gaussian curvature²" or "radius² of Gaussian Curvature"? ~Kaimbridge~17:01, 16 May 2006 (UTC)

As follows:

• MN: inverse of Gaussian curvature: meaningless and undefined notion
• radius of Gaussian curvature: meaningless and undefined notion
• radius of Gaussian curvature²: meaningless and undefined notion
• radius² of Gaussian curvature: meaningless and undefined notion

Not all combinations of mathematical terms have a meaning. This thing just happens to be called "Gaussian curvature". That is an accident of history. It might as well have been called "Intrinsic contortion", in which case you probably wouldn't have thought of asking for its radius. --Lambiam^Talk 18:28, 16 May 2006 (UTC)

Matrix exponentiation

I have a matrix:

0 a b
0 c d
0 e f

that I would like to exponentiate. Can this be done analytically, and if so, what is a good approach? --HappyCamper 19:27, 15 May 2006 (UTC)

Find The Jordan form of A: A = Q^-1UQ. Then exp(A) = Q^-1exp(U)Q. So the problem is reduced to exponentiating a Jordan matrix. Since exponentiation of a block diagonal matrix is done block-wise, the problem is reduced to exponentiating a Jordan block. If the eigenvalue of a block D is λ, then exp(D) = exp(λI) exp(D - λI), so we are left with exponentiating a nilpotent Jordan block. This is easy from the power series. Note, however, that if this is a parametric matrix (a, b etc. are unknown), the result will be a monster. -- Meni Rosenfeld (talk) 19:53, 15 May 2006 (UTC)

Unfortunately, yes, the parameters are unknown. Is there a witty method which I can use to exploit the zero eigenvalue of the matrix, or perhaps use the Cayley-Hamilton theorem? --HappyCamper 19:59, 15 May 2006 (UTC)

I doubt it. I've let Mathematica do this matrix, and the result certainly isn't pretty. So my guess is that any approach would force you to do some dirty work. The approach I described should work here as well - you would obviously have to also divide the problem to distinct cases (unless you are only interested in the probable case that all 3 eigenvalues are distinct). Do the calculations symbolically with the parameters, and denote as auxillary variables expressions that occur frequently (such as ${\sqrt {(c-f)^{2}+4de}}$). Perhaps there is some slightly easier way, but I am skeptical about any miraculous instant solutions. -- Meni Rosenfeld (talk) 20:10, 15 May 2006 (UTC)

Oh, Mathematica can handle this? Hmm...too bad I don't have access to it. --HappyCamper 01:38, 16 May 2006 (UTC)

A free alternative for symbolic mathematics is the venerable Maxima, a decendent of the original MIT Macsyma. The list of computer algebra systems is a valuable resource, since each system has its strengths and weaknesses, including price, specialty, capability, and reliability.

It does appear that the first column of the result is (1,0,0)^T. However, the middle entry of the last row (one of the simpler entries!) causes Mathematica's FullSimplify to produce this mess:

${\frac {2e\exp({\frac {c+f}{2}})\sinh \left({\frac {1}{2}}{\sqrt {(c-f)^{2}+4de}}\,\right)}{\sqrt {(c-f)^{2}+4de}}}$

Recall that the matrix exponential is defined by the same power series as exp, so will involve not only products of the matrix with itself, but also sums. If we can diagonalize the matrix, as A = Q⁻¹DQ, with Q orthogonal and D diagonal, then we get tremendous simplification. First, notice that

A² = (Q⁻¹DQ)(Q⁻¹DQ) = Q⁻¹D²Q.

This generalizes to all powers. Next, notice that linearity of matrix multiplication (on both sides) allows the Q "sandwich" to group all the diagonal matrices inside.

I + Q⁻¹DQ + ¹⁄₂ Q⁻¹D²Q + ⋯ = Q⁻¹(I + D + ¹⁄₂D² + ⋯) Q.

Unfortunately, an algebraic expression for the diagonal form may be messy or impossible (for an n×n matrix the characteristic polynomial has degree n), and not every matrix can be diagonalized (hence the Jordan form). So unless the matrix A has special properties, its exponential won't be pretty. --KSmrq^T 07:56, 16 May 2006 (UTC)

You can use the zero eigenvalue to reduce it to the exponential of a two-by-two matrix, as follows:

$\exp \left({\begin{bmatrix}0&a&b\\0&c&d\\0&e&f\end{bmatrix}}\right)$

${}=\exp \left({\begin{bmatrix}a&b\\c&d\\e&f\end{bmatrix}}{\begin{bmatrix}0&1&0\\0&0&1\end{bmatrix}}\right)$

${}=\sum _{k=0}^{\infty }{\frac {1}{k!}}\left({\begin{bmatrix}a&b\\c&d\\e&f\end{bmatrix}}{\begin{bmatrix}0&1&0\\0&0&1\end{bmatrix}}\right)^{k}$

${}=I+{\begin{bmatrix}a&b\\c&d\\e&f\end{bmatrix}}\sum _{k=1}^{\infty }{\frac {1}{k!}}\left({\begin{bmatrix}0&1&0\\0&0&1\end{bmatrix}}{\begin{bmatrix}a&b\\c&d\\e&f\end{bmatrix}}\right)^{k}{\begin{bmatrix}0&1&0\\0&0&1\end{bmatrix}}$

${}=I+{\begin{bmatrix}a&b\\c&d\\e&f\end{bmatrix}}\left(\sum _{k=1}^{\infty }{\frac {1}{k!}}{\begin{bmatrix}c&d\\e&f\end{bmatrix}}^{k}\right){\begin{bmatrix}0&1&0\\0&0&1\end{bmatrix}}$

${}=I+{\begin{bmatrix}a&b\\c&d\\e&f\end{bmatrix}}{\begin{bmatrix}c&d\\e&f\end{bmatrix}}^{-1}\left(\exp \left({\begin{bmatrix}c&d\\e&f\end{bmatrix}}\right)-I\right){\begin{bmatrix}0&1&0\\0&0&1\end{bmatrix}}.$

This shows that the first column is indeed [1,0,0]^T. If you insist, you can continue and get an analytic expression out of it. But it won't be pretty. -- Jitse Niesen (talk) 13:10, 16 May 2006 (UTC)

Shouldn't the exponents in the fourth and fifth forms be k-1? --Lambiam^Talk 13:46, 16 May 2006 (UTC)

Pi

My question is, has anyone ever thought of using percentage, instead of Pi, in finding the circumference of a perfect Circle?

Thanks.

Pi is the ratio of a circle's circumference to its diameter. Presumably, a "percentage" would be a ratio of approximately 314.159%; that is, pi. Or have I misunderstood your question? — Lomn Talk 21:05, 15 May 2006 (UTC)

Hi! Lomn,

Below is an example in using percentage instead of Pi.

Example: 3 x diameter plus the percentage equals the circumference of a perfect Cirle.

I think this is right. I hope this example gives a more clear understanding of my question.

Thanks.

You could say: to get the circumference, take 3 × the diameter, and then add approximately 4.71975511966 percent. But what would the point of doing that? --Lambiam^Talk 00:09, 16 May 2006 (UTC)

Well, something similar was attempted - see Grad (angle). While it is a bit easier for doing arithmetic with angles, when you start getting into stuff like calculus the $pi/100$ factors get very annoying. Confusing Manifestation 01:14, 16 May 2006 (UTC)

As noted above, there's no real purpose to this. Your percentage is still a ratio, and thus still bound to pi. Your equation has become, in effect, c = 3d + (π-3)d. — Lomn Talk 14:57, 16 May 2006 (UTC)

Hi! Lomn,

Well, instead of c = 3d + (π-3)d, how about something more like this; 33.333333333333333333333333333333% of d + 3d = C

Thanks.

You can write it that way. The percentage would be 100 x π - 300. Approximately 14.15926...% of the original value, continuing with the digits of pi. You can't escape from pi. If you want the percentage to add to three times the diameter, divide that percentage by 3. Notinasnaid 18:19, 16 May 2006 (UTC)

It isn't clear whether you thought this formula is correct (which it is not), or whether you are looking for a correct formula of the same form. If the latter, than as stated previously, you'll get that C is approximately 14.15926535897932384626433832795% of d + 3d. Again, no escaping π. -- Meni Rosenfeld (talk) 19:18, 16 May 2006 (UTC)

Cauchy & d'alambert convergence criteria

Can anyone give me some "general" guidelines in which when to use each convergence criteria? Thanks.

if you have a certain series, you put it into the formulas of the criterias. if one of the formulas you get seems simpler or easier to prove to pass the criteria use it. if it doesn't work try the other one.

also note that some convergent series pass the cauchy criteria and not the d'alambert, so if you can't prove d'alambert for a series you'll have to try using the cauchy, even though the formula may happen to be much bigger or complicated. and then if this doesn't work too maybe you'll have to use other things ... or actualy find out and prove that the series does not converge. hope this helps. --itaj 00:25, 16 May 2006 (UTC)

thanks

Derivatives of complex exponentials

Suppose I'm considering z^c where z is an element of the cut complex plane (Complex plane excluding the negative real axis), and c is some complex constant. This is defined by z^c = exp(c Log(z) ). Now, if I want to take the nth derivative of z^c, am I allowed to just use normal polynomial term differentiation rules (i.e. d/dz(z^c) = cz^c-1), or do I have to go by the definition and use the chain rule? I tried finding the general term by the definition but it started looking like a horrible mess of product rules and so forth. I'm uneasy about just using the normal standard derivative since c is a complex number and not a real, so is there an easy way to calculate the general nth derivative? Thanks. Maelin 23:20, 15 May 2006 (UTC)

${d \over dz}(z^{c})=cz^{c-1}$ is true in any region where z^c is analytic, because ${d \over dz}(\exp(c\cdot \log(z)))=c{d \over dz}(\log z)\exp(c\cdot \log(z))={c \over z}exp(c\cdot \log(z))=c\cdot \exp((c-1)\cdot \log(z))$. —Keenan Pepper 02:52, 16 May 2006 (UTC)

Wow, that's nice. So as long as you stick to a single definition of log, the identity is automatically true. – b_jonas 17:56, 16 May 2006 (UTC)

Kuratowski closure axioms - topology defined by closure

about Kuratowski closure axioms. well, i already put this question in the talk page, but got no response. i think it's good enough to be here:

after reading the definition in the article, i can't see how to prove the following using the axioms:
$\bigcap _{i\in I}cl(A_{i})=cl(\bigcap _{i\in I}cl(A_{i}))$ - i.e. the infinite intersection of closed sets is closed.

and that's one of the needed properties of the closed sets in a topology. --itaj 00:02, 16 May 2006 (UTC)

Well, for any j:

• ∩ cl A ⊆ cl A_j (essential property of intersections)
• cl ∩ cl A ⊆ cl cl A_j (axiom 3 implies monotonic)
• cl ∩ cl A ⊆ cl A_j (axiom 2)
• cl ∩ cl A ⊆ ∩ cl A (essential property of intersections)
• cl ∩ cl A = ∩ cl A (axiom 1)

Melchoir 03:51, 16 May 2006 (UTC)

Maybe I should poke back in here and explain: I got that by working backwards. It's like a proof in algebra: start with the statement you want, then look for a slightly simpler statement that implies it. At every step, there's often only one reasonable thing to do. Melchoir 03:59, 16 May 2006 (UTC)

May 16

Weak confluence in abstract reduction systems

It has been stated (attributed to Hindley, by Kleene, by my reference) that the abstract reduction system defined by reduction rules { b -> a, b -> c, c -> b, c -> d} is weakly confluent but not confluent. It is clear that elements b and c are weakly confluent, but why must elements a and d be weakly confluent? In general, if n is a normal form, n cannot be weakly confluent as for n to be weakly confluent, n must have a one-step reduction to some other element, but that immediately contradicts the supposition that n is a normal form -- if this reasoning is valid, then no abstract reduction system with a normal form can be weakly confluent, as an abstract reduction system is said to be weakly confluent if all of its elements are. The conclusion is highly counterintuitive. Is there a problem here? Dysprosia 10:02, 16 May 2006 (UTC)

On the contrary, normal forms are confluent. Local confluence of a means something like: for all b and c, IF (a → b AND a → c) THEN there is some d such that (b →* d AND c →* d). So if a is normal this becomes: for all b and c, IF true THEN whatever. This is true. --Lambiam^Talk 11:49, 16 May 2006 (UTC)

But if a is normal, no such reduction a -> b exists, simply because a is normal. Dysprosia 12:04, 16 May 2006 (UTC)

Sorry, I made a mistake. I meant to write this: "So if a is normal this becomes: for all b and c, IF false THEN whatever." Unlike my previous attempt, this is true. Precisely because no such reduction exists, the confluence condition cannot be violated. --Lambiam^Talk 12:44, 16 May 2006 (UTC)

Ah, I see. I'd forgotten about the wily nature of implication with false antedecents. Dysprosia 12:50, 16 May 2006 (UTC)

Rolling dice

ok, im working on a c++ program that does stuff with dice probability....

basically, what i need to know is this: if one has 4 4-sided dice, how many possible combinations add up to 7? 1123 and 2311 ARE different.... i got 20, but i dont think thats high enough

any help is appreciated

--spuck

I got 20 as well - 12 for 1123, 4 for 1114 and 4 for 1222. That's correct. -- Meni Rosenfeld (talk) 20:17, 16 May 2006 (UTC)

Sigma notation

I'm doing a homework assignment about sigma notation and I've come across a problem I can't solve such that if you guys show me how to do it, I can then apply that method to other, similar problems. (I've already checked out the sigma-notation page and I didn't find it to be of much help.) How would you express 35 + 48 + 63 + 80 + 99 in sigma notation?