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April 28

WATER'S DEPTH

I HAVE A RECTANGULAR TANK MEASURING 55 METRES BY 48 METRES CONTAINING 33 LITERS OF WATER. HELP ME FIND THE DEPTH OF THE WATER. — Preceding unsigned comment added by 208.131.187.81 (talk • contribs) 02:22, 28 April 2006 (UTC)

First, don't write in all captical letter, this is considered shouting and rude. Second, it is obvious that this is a homework problem, please read the instruction on the top of this page. Simply put, Volume of rectangular prism is length times width times height. Just plug in the numbers, and divide to get the answer. Besure to use consistent unit. --Lemontea 02:57, 28 April 2006 (UTC)

Convert the volume from units in litres to units in metre cube and the answer is as plain as daylight. Ohanian 09:15, 28 April 2006 (UTC)

I thought yards & gallons were still in use in some parts of the world. Do they like plain daylight ? --DLL 17:45, 28 April 2006 (UTC)

1 gallon is 231 cubic inches. The rest is easy. Chuck 18:55, 28 April 2006 (UTC)

Bloody little. I'd call that "moisture" rather than "depth of water". —Ilmari Karonen (talk) 19:37, 30 April 2006 (UTC)

Do you perhaps mean centimeters (cm)? You could always convert to chains and hogsheads.... Skittle 23:24, 1 May 2006 (UTC)

Trig Identities

How would you go about doing the question below...

${\frac {2\sin ^{2}\theta -1}{\sin \theta \cos \theta }}={\tan \theta -\cot \theta }$

All I could think of was changing 2sin²θ-1 into -2cos²θ + 1. I can't change sinθcosθ (at least I don't think so). I tried the right hand side next, and I got:

${\frac {\sin ^{2}\theta -\cos ^{2}\theta }{\cos \theta \sin \theta }}$

By the way, I keep on getting Failed to parse (lexing error) whenever I try to properly write a fraction, I just copied and pasted one from this page and changed it to this question but it still doesn't work, sorry about that. Thanks

C-c-c-c 03:34, 28 April 2006 (UTC)

It parses now. Also, you're close. Compare the numerators of your two fractions --Deville (Talk) 03:43, 28 April 2006 (UTC)

Mhm, well I've actually located a little thing on the formula sheet, and noticed that 2cos²θ -1 = cos²θ - sin²θ(yes I should have looked earlier, but it had it as Cos2θ = 2cos²θ -1, Cos2θ = cos²θ - sin²θ and I didn't link cos²θ - sin²θ and 2cos²θ-1 together until then.) I have -2cos²θ +1 so that means it should be -(cos²θ - sin²θ) so it should be sin²θ - cos²θ. Yep it works, thanks a lot. --- C-c-c-c 04:02, 28 April 2006 (UTC)

Hey, just remember that all you ever need to know is $sin^2 + cos^2 = 1$, and the angle-addition formulas... everything else comes from these. --Deville (Talk) 15:18, 29 April 2006 (UTC)

Try using the definitions of tan and cot. Does that help? Confusing Manifestation 04:09, 28 April 2006 (UTC)

Retire

How much money do I need to retire? And what % return can I expect to get from investing?

If I can achive X% Average return, does that mean I can always take out X% of the princaple each year? 12.183.203.184 03:52, 28 April 2006 (UTC)

How to calculate various interest rate problem
----------------------------------------------

extremely long post cut by me. For full version, see this diff. -lethe ^{talk +} 09:25, 28 April 2006 (UTC)

Note that taxes and inflation also need to be subtracted from the interest rate each year to find your true gain (or loss). StuRat 17:28, 29 April 2006 (UTC)

Duplex Numbers

What are duplex numbers? Ohanian 09:06, 28 April 2006 (UTC)

The first page or so of this document has some background of duplex numbers. It looks like they're a parallel to the complex numbers or quaternions. Confusing Manifestation 12:42, 28 April 2006 (UTC)

The same concept is covered by split-complex number.--gwaihir 12:19, 3 May 2006 (UTC)

Hex, square and triangle grids

Some board games or maps use squares (4 exits from every cell, or 8 if you include diagonals, but the latter are not the same distance away as the former), some use hexes (6 equidistant exits), some use triangles. Is it possible for a map to use other shapes, and allow other numbers of possible equal exists besides 4, 6 or 3? Is it possible to have with 8, for instance? Or why is it impossible? The shape doesn't have to be symmetrical, but ideally the exit points should be as equidistant as possible from the centre of one to the centre of another.--Sonjaaa 10:28, 28 April 2006 (UTC)

You are looking for List of uniform planar tilings (the main article is tessellation, the current Mathematics Collaboration of the Week). There are only the three regular tilings that you mention, but there are some semiregular ones, too. Rasmus (talk) 11:25, 28 April 2006 (UTC)

If you start putting together congruent regular polygons in a repeating pattern, the overall fate of the surface is determined by the value of 1/(number of edges per polygon) + 1/(number of polygons meeting at each corner). If that's equal to 1/2, you get a tiling of the plane. Only three pairs of unit fractions fractions add up to 1/2: 1/6 + 1/3, 1/4 + 1/4, and 1/3 + 1/6. These correspond to tilings of the plane by hexagons, squares, and triangles. If the value is less than 1/2, you get a tiling of the sphere (a.k.a. a Platonic solid). If the value is greater than 1/2, you get a tiling of the hyperbolic plane. —Keenan Pepper 13:05, 28 April 2006 (UTC)

The life game believes that a square has 8 neighbours, including diagonally opposites. Hexes cant' choose (the diagonal is just between two other hexes) ; and triangles are back to hexes, but with more density. So we have 3, 4, 6 and 8 on a plane (according to OEIS : very round numbers ?!) --82.227.17.30 17:37, 28 April 2006 (UTC) (DLL)

Displaying 1-10 of 1281 results found. Wow, useful. —Keenan Pepper 17:58, 28 April 2006 (UTC)

It's in the 10 first. P.S. I feel dizzy now I found a life game played on a semiregular pentagonal tessellation. --DLL 19:52, 28 April 2006 (UTC)

Geometry question

Dear Wikipedians. I'll be very pleased if I'll the solution for the following geometrical rider: ABC is a triangle in which angle B = 2 angle C. D is a point on BC such that AD bisects angle BAC and AB=CD. Prove that angle BAC=72 degrees.

Thank you very much. User Akshitapatel

---

This sounds like a homework question, but it's an interesting one--took me quite a while to work out--so I'll give an outline which I hope will lead you in the right direction.

• Given that $\angle ABC=2\angle BCA$, and $\angle ABC+\angle BCA+\angle CAB=180^{\circ }$ (assuming Euclidean geometry!) we can define everything in terms of a single angle. Let $\angle BCA=2\theta$ (using $2\theta$ rather than $\theta$ avoids some ugly fractions later on, but is not strictly necessary).

• Determine $\angle ABC$, $\angle ADB$, $\angle BCA$, and $\angle CAD$ in terms of $\theta$.

• Apply the sine rule to determine a relationship between $AD$ and $CD$, and another between $AD$ and $AB$.

• Substitute $CD=AB$, and rearrange both equations from the previous step to generate two different functions of $\theta$, both equal to ${\frac {AD}{AB}}$. Set them equal to each other. (It may also help to apply the fact, at this point, that $\sin(90^{\circ }-x)=\cos x$, but it is not strictly necessary.)

• At this point there are two approaches, depending on how rigorous you need to be:

• Easy, but non-rigorous approach: See if you can find a value of $\theta$ which makes the equation true. It's not too hard to do by inspection with the right insight. (And it's even easier since you know what $\theta$ is supposed to work out to.)
• Hard, but rigorous approach: substitute the trigonometric functions you have in your equation of $2\theta$, $3\theta$, and $4\theta$ with the equivalent values in terms of trigonometric functions of $\theta$. Solve for $\sin \theta$, and then for $\theta$. This involves a fifth-order equation in $\sin \theta$, but it has a quadratic factor.

It's possible there's a geometric approach to this proof which doesn't involve trigonometric functions, but I wasn't able to find it. Chuck 20:43, 28 April 2006 (UTC)

Found it! Actually it's the famous classical Pentagon or Pentagram. Anyway, the geometric approach goes below:

For simplicity, call the angle BAD as b, and call angle ACD as a. Now the first and most important step, Find a point on AC such that BE bisect angle ABC. Join BE as well as DE.

Because angle EBC = angle ECB = a, triangle BEC is isosceles and so BE = EC.

We also know that AB = CD, and that EB = EC, and that angle ABE = angle DCE. Hence, by SAS, triangle ABE and DCE are congruent.

Consequently, AE = DE, so angle ADE = angle EAD = b. Also angle EDC = angle EAB = 2b. So in summary, angle ADC = angle ADE + angle EDC = b+2b=3b.

But by exterior angle of triangle, angle ADC = angle BAD + angle ABD = b+2a. So 3b=b+2a, and therefore a=b.

The rest should be obvious enough for you to see. --Lemontea 01:32, 29 April 2006 (UTC)

The data are : (1) B = 2C ; (2) AD bissects BAC ; (3) AB = CD ; (4) A = 72° (to be proven).

Let (4) be true. (4) => (5) B+C = 108° ; (5) and (1) => (6) C = B/2 = 36° ; (2) and (6) => (7) now we have thoses triangle measures : ABD (36°, 72°, 36°) and ADC ( 36°, 108°, 36°). Then both are isosceles (golden triangles) and remind of pentagram angle measures.

(7) isoscele means that => (8) AB = AD and AD = DC ; since (3) it is true. --DLL 07:50, 29 April 2006 (UTC)

Let me clarify the above comment's purpose: it demostrates why the figure is part of the pentagram, and shows that there is no contradiction for the assertion(that A=72 degrees). But it isn't a solution, just an, err, exploration, if you ask me. It doesn't show (4) is the only possible solution, in particular, see Affirming the consequent and Logical fallacy. Thanks. --Lemontea 09:01, 29 April 2006 (UTC)

Sin x into Sec x

Hello fellow Wikipedians. Today I had the unfortunate task of writing a trig test (but I don't think it went too bad for that matter), and the question below stumped me.

Use only sec x:

${\frac {\sin x}{\sin 2x}}$

I didn't know what to do, and I looked at all the formulas on the formula sheet to see how I could convert sin x into anything, but alas, nothing. The question below has to be in terms of sec x, that's the challenging part. I spent a good deal towards the end trying to think this through, and all I could come up was converting sin x into 1/cscx, but that didn't help either. Also, I converted sin 2x into 2sinxcosx (edited) I believe or something similar, I don't know it off the top of my head. Any thoughts on this would be greatly appreciated. --- C-c-c-c 22:52, 28 April 2006 (UTC)

See Trigonometric identities#Double-angle formulae. You appear to have confused sin(2x) with cos(2x). Melchoir 22:58, 28 April 2006 (UTC)

I have an OT question—I've just recently administered my first final exam as an instructor in Canada, and have come across this usage of "write an exam", which confuses me a little. If the student writes the exam, just what is it the instructor does, when creating it? Isn't that also "writing", or is there a different verb used? --Trovatore 23:55, 28 April 2006 (UTC)

Right, never mind, I did confuse it, but I think it was 2sinxcosx I had on the bottom, whatever it was, I got it from the identity sin 2x on my formula sheet. --- C-c-c-c 02:44, 29 April 2006 (UTC)

Sorry, I may not have expressed myself clearly. My confusion has nothing to do with the mathematics, but with this usage of the phrase "write an exam". To me, an American, writing an exam means devising questions, TeXing them up, printing them out, and copying them for the hapless victims^W^W eager students. But in Canada, I find that it's the students who "write the exam" (what in the States we would call "taking an exam"). So if it's the students who wrote my exam, then what exactly did I do to it? --Trovatore 02:48, 29 April 2006 (UTC)

Mhm true, the people often around me say "write an exam", (here in Canada) or take an exam, both are interchangable in the high school setting at least. I would assume that actually creating an exam would be referred to "making up the exam", though now as I reread that it sound goofy, but maybe adding more detail to the end of the phrase, "I'm writing an exam...for my students". I can't think of a different verb you'd use though. --- C-c-c-c 02:57, 29 April 2006 (UTC)

How about "set an exam"? Mon4 11:27, 29 April 2006 (UTC)

Is that what's used? To my ear that sounds like deciding when an exam will be held, how much it's worth, maybe what it will cover. But this could be a difference between Canadian and American usage, I guess. That would bring my count of such differences up to, I don't know, five or so. Canadian speech, customs, way of life, are a lot more similar to the American ones than Canadians like to think, but that sometimes makes it all the more disconcerting when you run into one. --Trovatore 14:41, 29 April 2006 (UTC)

Oh, you wacky Canadians!!!!  :) Seriously, my experience is that in American English usage, the instructor "writes", or "writes up" the exam, and the students "take" it. In British English usage, the instructor more commonly "sets" or "designs" the exam, and the students "write" it. Actually it makes a little more sense, as I don't "write" exams, I "TeX" them, and the students are the only ones doing actual writing... --Deville (Talk) 15:17, 29 April 2006 (UTC)

In New Zealand, an instructor sets an exam, and students sit it. I don't think I've ever heard the term "write an exam" before seeing this.-gadfium 20:29, 29 April 2006 (UTC)

That would be true of Australia as well. JackofOz 06:47, 2 May 2006 (UTC)

The students sit it because they can't stand it. --Lambiam^Talk 00:17, 5 May 2006 (UTC)

So, you got as far as:

${\frac {\sin x}{\sin 2x}}={\frac {\sin x}{2\sin x\cos x}}$

What did you do next? Melchoir 04:23, 29 April 2006 (UTC)

Sinx in the numerator cancels out with the sinx at the bottom which leaves you with 1/(2cosx). Since 1/cosx = secx, 1/(2cosx)=0.5secx. Hope that helped.

Oh, so that's how you would do it. Thanks C-c-c-c 05:00, 30 April 2006 (UTC)

Sine, cosine, tangent

Today we started learning trig in my geometry class (our teacher calls it plain-vanilla trig, as it does get harder) and we learned the simplest of rules. We took notes, started on homework with extra class time, etc. When I got home to finish, I found some one had taken my notes and homework (while it is entirely possible I just lost them on the floor at school somewhere, I will stick with them being stolen). What I need to do is just make sure I'm sure. Sine=Opposite side/Hypotenuse; Cosine=Adjacent side/Hypotenuse; Tangent=opposite side/Adjacent side; Area= (1/2)(angle a)(angle b)(sinC). I pretty sure about all but the last one. If any corrections are needed please help me. Thanks. schyler 00:05, 29 April 2006 (UTC)

I don't understand what the last one is even saying, but the rest look fine! There's plenty of information at Trigonometry and related articles if you want to verify any other formulas. Melchoir 00:53, 29 April 2006 (UTC)

Actually... you probably mean the formula at Triangle#Using trigonometry, where a and b are side lengths, not angles. Melchoir 00:57, 29 April 2006 (UTC)

Have you been taught the popular trigonometry mnemonic "SOHCAHTOA" for the basic formulae? --Bth 08:23, 29 April 2006 (UTC)

I've always preferred "sex on holidays can advance happiness to outrageous amplitudes" - more memorable, and less chance of mixing up sohcahtoa and sahcohtoa. Confusing Manifestation 14:59, 29 April 2006 (UTC)

I prefer "Some Old Hippy, Caught Another Hippy, Tripping On Acid". StuRat 17:24, 29 April 2006 (UTC)

We always used "The Old Arab Sat On His Camel And Humped", which takes a different order. I think the main reason we used it was because our teacher was embarrassed by it, since it was clearly not quite ~right, and she tried to make it a little better by using 'hoped' instead of 'humped', which clearly missed the point of why it wasn't ~right and spoiled the pun at the same time. Bet she wished she never made that poster. (PS, humped as in sulked) Skittle 23:38, 1 May 2006 (UTC)

I think you must mean area = side a * side b * sin gamma / 2. – b_jonas 18:10, 29 April 2006 (UTC)

Thank you. Now that you mention it, my teacher did tell us this story about an indian princess named Sohcahtoa. I forgot. Also, that is the formula I was thinking of. Again, thanks. schyler 01:32, 30 April 2006 (UTC)

When my physics teacher (Mr. Rippetoe) related that story, he claimed that Sohcahtoa was related to his own Indian ancestors, the Rippetoas. Black Carrot 22:09, 4 May 2006 (UTC)

April 29

The Basic rules of Differential Calculus

I have been looking through the proofs of the rules of differential calculus, and I understand how we find d/dx(x^n)=nx^(n-1) (something I was taught to take for granted), as well as the quotient rule and d/dx(sin x) and d/dx(cos x). However, I reached a stumbling block with the proof of the Chain rule. From the first line:

$g(x+\delta )-g(x)=\delta g'(x)+\epsilon (\delta )\,$ where ${\frac {\epsilon (\delta )}{\delta }}\to 0\,$ as $\delta \to 0.$

I assume this was reached by:

$g'(x)={\frac {g(x+\delta )-g(x)}{\delta }}$

$\delta g'(x)=g(x+\delta )-g(x)$

Where did the epsilon(delta) come from?

Also, in the proof of the product rule (which is more clearly laid out), it is said that:

$g(x+\Delta x)h(x+\Delta x)-g(x)h(x)=g(x)(h(x+\Delta x)-h(x))+h(x+\Delta x)(g(x+\Delta x)-g(x)),\,$

I assume this is done with some sort of factorisation, but nothing I can see. The two terms on the left have no common factor. What am I missing?

Thanks in advance. --Alexs letterbox 05:00, 29 April 2006 (UTC)

For your first question, the epsilon comes from replacing an equation that only holds in the limit with a straightup equation. If lim f(x) = lim g(x) as x → a, I cannot conclude that f(x) = g(x), but I can say that f and g are approximately equal. That is, that f(x) = g(x) + ε, with lim ε =0 as x → a. Proof: let ε = f(x) – g(x). In the equation you cite, on one side of the equation you have the fraction [g(x+δ) – g(x)]/δ and on the other you have g'(x). Your mistake is assuming these are equal. They're not equal, they're only approximately equal, up to a small error which vanishes in the limit that δ → 0.

For your second question, I think the easiest way to see that the two sides are equal is to distribute out the right-hand side. Once you do that, you'll realize that the way to procede from the left-hand side to the right-hand side was not to just factorize, but to first add zero in the form of 0=g(x)h(x+Δx)–g(x)h(x+Δx). Then you factorize, then you get the equation. -lethe ^{talk +} 09:15, 29 April 2006 (UTC)

Thanks. --Alexs letterbox 23:15, 29 April 2006 (UTC)

Just looking through the working again (I'm nearly there!), once again in the Chain rule, how does one get from f(g(x)+δg'(x)+ε(δ))-f(g(x)) to (δg'(x)+ε(δ))f'(g(x))+ɳ(δg'(x)+ε(δ))? What justifies taking the δg'(x)+ε(δ) out of f? --Alexs letterbox 11:00, 30 April 2006 (UTC)

For function f we have a similar equation as for g. Using all different variables to avoid mixups, it is: f(z+ω) – f(z) = ωf'(z) + η(ω). Now simply plug in g(x) for z and δg'(x)+ε(δ) for ω. --Lambiam^Talk 14:08, 30 April 2006 (UTC)

Edit conflict: The equation h(x+ζ) = h(x) + h'(x)ζ does not hold exactly, but only in the limit ζ→0. In other words, h(x+ζ) = h(x) + h'(x)ζ + η, for some η for which η/ζ → 0 as ζ→0. Substitute fg for h, and δg'(x)+ε(δ) for ζ, and denote the error η for this choice as α_δ, and you have your equation. -lethe ^{talk +} 14:21, 30 April 2006 (UTC)

Difference between Absolute Graph and Relative Graph

Hi,

I would like to know the difference between Absolute Graph and Relative Graph? I have searched all over the internet and i havent found any information. Its not even in my college notes. Any help would be appreciated as i have no idea what absolute and relative graphs are!

Please help me out. Thanks!

I've never heard of them, and Wikipedia has nothing. (See Graph for common usages.) If I were you, I would be really suspicious of whoever told me about these things! Melchoir 08:16, 29 April 2006 (UTC)

Its for my assignment for Quantitative Maths. I am stumped at it too!

The two most likely looking results I dug up from Google are:

Something to do with electronic engineering, and measures of components' response, or something.

A page about investment which talks about "absolute" and "relative" graphs in the context (I think) of looking simply at a stock's performance (absolute) or its value divided by its value on some chosen reference date ("relative").

The second one seems like it might possibly be the general idea, but it seems too simple really. --Bth 08:57, 29 April 2006 (UTC)

Hey thanks alot. Atleast that gave me some idea. Maybe i will be able to expand on it but any new info is always welcome. thanks!

Er, maybe you should ask your instructor instead? Melchoir 09:07, 29 April 2006 (UTC)

If you gave us some context, we might be able to make good educated guesses. What course is this for? -lethe ^{talk +} 09:19, 29 April 2006 (UTC)

Quantitative Maths. I have my asked my instructor and hopefully she will respond quickly. Its an online course.

I suppose any graph of an index, like the Dow Jones Industrial Index, would be a relative graph. That is, the actual values each day are meaningless, it's only the change from day to day that's significant. In a normal (absolute) graph, the individual values do have some meaning, like the price of an individual stock each day. StuRat 17:08, 29 April 2006 (UTC)

Newton

Hi is the Newton who created the Newton-Rapson Method (for finding roots of an equation) Sirr Isac Newton. (the apple one)

Yes. See Newton's method#History. Melchoir 21:15, 29 April 2006 (UTC)

April 30

Computer Audio Recording

Hello, everyone. I'm interested in making my own "book on tape" type of project, except on a CD which is made with my computer. Does anyone know what I would need to do this and what software I could use, preferably downloadable for free from the internet from a site such as tucows? Thanks. --Think Fast 01:13, 30 April 2006 (UTC)

So you are looking for a software which can record your voice right? You can get Audacity. Its free. You can download it from here: http://www.download.com/Audacity/3000-2170_4-10514927.html?tag=lst-0-1

—The preceding unsigned comment was added by 213.42.2.28 (talk • contribs) 08:26, April 30, 2006 (UTC).

Negative exponential distribution

Say shoppers enter a shop at a mean rate of 4 per minute. This can be modeled as an exponential distribution parameter 0.25. How do you show that the difference in times between successive shoppers entering is distributed exponentially as well? Is this to do with Exponential distribution#Memorylessness? x42bn6 Talk 12:53, 30 April 2006 (UTC)

Yes, that is what the "memorylessness" gives you. Just after a shopper entered at time t is no different from any other moment, including time 0. Lambiam^Talk 13:24, 30 April 2006 (UTC)

Er, I don't think that's strictly speaking true. If your model is "an exponential distribution with parameter 0.25", then by definition you're only modelling the time till the first customer. So you haven't said anything about any subsequent customer. However, if you also assume that shoppers occur independently, then you can use the memoryless property (which tells you how much longer to expect to wait for a shopper if one hasn't occurred yet—read the article—not what you want) together with that assumption (which tells you that the time you expect to wait is the same whether a shopper has occurred or not) to prove that the shoppers collectively form a Poisson process, and hence in particular you can use the strong Markov property to prove the required result. —Blotwell 22:07, 30 April 2006 (UTC)

I agree, that is entirely correct. What I was trying to say is that if you assume the process has no memory (which I took to mean that whatever happened or did not happen in the past was irrelevant for the expected value of future events) the maths needed to show that the time differences have negative exponential distribution proceeds essentially as sketched under Exponential distribution#Memorylessness (even though this covers a different question, namely the expected time to a single event, and not an ongoing process). I thought that that was what was being asked. This was somewhat cryptically expressed as 'that is what the "memorylessness" gives you'. I should perhaps have said: 'that is what the assumption of "memorylessness" gives you'. Lambiam^Talk 23:51, 30 April 2006 (UTC)

Ugh, I misread the question. It is modelled by X~Po(4) where X is the number of shoppers per minute. Can you then assume that Y~exp(0.25) from X?

And then, how on Earth do you prove that the difference in times is distributed as a negative exponential distribution?

I'm pretty sure I don't have to learn about this for my exam, but seeing this makes me annoyed... x42bn6 Talk 03:23, 1 May 2006 (UTC)

The solution is implied by the previous answers. Did you look up the Poisson process article? The basic observation that may help you make the bridge is that if it takes time τ for the first event to happen after time t, then there were zero events in the interval [t,t + τ). So if x is the random variable standing for how long it takes for the first event after time t to happen, then Prob(x < τ) = Prob(#{events in the interval [t,t + τ)} = 0). I hope this helps. Lambiam^Talk 23:56, 1 May 2006 (UTC)

Yeah, figured it out, though by not that article, though that would've helped a lot. x42bn6 Talk 07:30, 2 May 2006 (UTC)

Ellipse-like shapes

Are there any formal names for 3-focus, 4-focus, etc. ellipse-like shapes?

Tuvwxyz 18:53, 30 April 2006 (UTC)

Perhaps if we had a definition, or even an illustration, we could be of more help. --KSmrq^T 19:02, 1 May 2006 (UTC)

I can provide a fully-rigorous definition, despite not knowing the answer. As an ellipse with the usual 2 foci is the locus of all points such that the sum of the distances between the respective foci and the point is some constant, the natural analog is simply to increase the number of distances to sum to that constant. For example, with foci (0,1), (0,-1) and (1000,0) with a total distance of 2010, the shape would be very narrow near the third focus, but around the origin would more nearly resemble a 500-radius circle. (A contour plot of $\sum _{i}d({\vec {f_{i}}},{\vec {x}})$ shows how these look -- in particular, mine looks like a circle mated to a square-ish corner. I can't seem to get MediaWiki to like mine, though.) Note that it's meaningful for the constant to be so small that none of the foci are even within the curve! --Tardis 22:37, 1 May 2006 (UTC)

daffy definitions from gfeometry

i need help find it some definitions for word like PI,ratio ect where i can look?

Well, Wikipedia's a good place to start. Try Pi and Ratio. —Keenan Pepper 22:05, 30 April 2006 (UTC)

May 1

Pair of equations

How can you solve the system

$a+b=10\,\!$

$ab=10\,\!$

without graphing? (i.e., algebraically)

I know the answers approximately, having graphed as a last resort.

I tried substitution, but to no avail.

Guidance?

139.55.19.50 11:02, 1 May 2006 (UTC)

It's a quadratic equation. The first line gives an expression for b in terms of a; rewrite the second line in those terms and resolve to get your quadratic. Notinasnaid 11:06, 1 May 2006 (UTC)

Yes, the canonical method is the Viète's formulas. Let's see:

$x=a$ or $x=b$ is equivalent to $0=(x-a)(x-b)=x^{2}-(a+b)x+ab=x^{2}-10x+10$ from which you get $x=5\pm {\sqrt {5^{2}-10}}=5\pm {\sqrt {15}}$, thus $a=5+{\sqrt {15}}$ and $b=5-{\sqrt {15}}$ or the same two values swapped. – b_jonas 11:19, 1 May 2006 (UTC)

Algebra solves this easily, but we can also approach numerically. Since the product of a and b is positive, either both are positive or both are negative; and since their sum is also positive, we conclude that 0 < a,b < 10.

Actually, we can do better, because if one factor is 10 the other is 1, and if one summand is 1 the other is 9. Repeat! If one factor is 9 the other is ¹⁰⁄₉, and if one summand is ¹⁰⁄₉ the other is ⁸⁰⁄₉. Then we get ⁹⁄₈ and ⁷¹⁄₈; then ⁸⁰⁄₇₁ and ⁶³⁰⁄₇₁; then ⁷¹⁄₆₃ and ⁵⁵⁹⁄₆₃. Numerically we now have 1.127 and 8.873, which are correct to the number of decimal places given. In fact, the last two steps agree to this precision, so we seem to have converged to an answer. The correct answers are approximately 1.1270166537926 and 8.8729833462074, so we have done well.

As requested, we solved the system without graphing. In this particular case we could get an exact expression with algebra, but in more complicated cases it's good to have more tools. A similar computation led to the intriguing discovery of the arithmetic-geometric mean. --KSmrq^T 20:13, 1 May 2006 (UTC)

This is a nice solution. It appears that this method works for any quadratic equation which has two distinct real roots except for the ones with zero linear term.

Let's see. We have the equation $0=x^{2}-2bx+c$, where $0<b^{2}-c$ and $0\neq b$. The solutions are $x_{1}=b-{\sqrt {b^{2}-c}}$ and $x_{2}=b+{\sqrt {b^{2}-c}}$, for which $x_{1}+x_{2}=2b$, and $x_{1}x_{2}=c$. Now if $f(x)=2b-c/x$, then $x_{1}$ and $x_{2}$ are fixpoints of $f(x)$. The derivative is $f'(x)=c/x^{2}$. Now it's easy to see that one of ${\sqrt {c}}$ and $-{\sqrt {c}}$ is between the two roots, so at the larger absolute valued root $|f'(x)|<1$. Thus, the iteration $y_{n+1}=f(y_{n})$ will converge to that root if started from a sufficently small environment of it.

This way we can indeed find the roots of any quadratic equation where $0\neq b$: we find the larger absolute value root with iteration and the other root in the usual way by dividing $c$ with the first one.

The only problem with this iteration is that it converges only slowly. Let me show an example.

Suppose we want to determine the numeric value of ${\sqrt {5}}$. We can use the method shown above: start with $x_{0}=3$, iterate $x_{n+1}=2+4/x_{n}$, and finally ${\sqrt {5}}=x_{\inf }-1$. The iteration gives 3.00000 3.33333 3.20000 3.25000 3.23077 3.23810 3.23529 3.23636 3.23596 3.23611 3.23605 3.23607 3.23607 so the approximate value is 3.23607 - 1 = 2.23607. Compare to this the fast Newton-iteration: $x_{0}=2$, $x_{n+1}=(x_{n}+5/x_{n})/2$. This gives 2.00000 2.25000 2.23611 2.23607 2.23607. It's obviously worth to use the latter one, even though the iteration rule is a bit more complicated.

– b_jonas 12:50, 2 May 2006 (UTC)

                a + b = 10           (1)
                a * b = 10           (2)
(1)             a = 10 - b           (3)
(3) & (2)       (10 - b) * b = 10    (4)
(4)             10b - b^2 = 10       (5)
(5)             b^2 - 10b + 10 = 0   (6)

After that the answer is as plain as daylight.

Quadratic_equation Ohanian 06:06, 2 May 2006 (UTC)

Yes, this is a slightly more explicit version of the first response, by Notinasnaid. It does assume that we know how to solve a quadratic equation. And if we require a numerical answer, we eventually still use numeric iteration on a computer to compute the square root of 15.

A more sophisticated response would be to point at the article on resultants. Although it could use some work, such as mentioning the existence of an assortment of resultants (Sylvester, Bézout, Dixon, Macaulay, Kapur-Saxena-Yang), it explains enough to free us from custom manipulation of the equations. Better still, it will work with a much broader range of equation systems than this simple pair.

Related ideas are the Gröbner basis and Hilbert's Nullstellensatz. We even have Cylindrical Algebraic Decomposition, despite the apparent lack of a Wikipedia article, which can handle inequalities.

The deeper point of mentioning alternatives is that when we are confronted with a mathematical challenge in real life, we do not know in advance how best to confront it, or whether it is intractable. In elementary classes we are lulled into thinking there is always an answer, and a right way to find it. This is not a helpful mindset, and the sooner we escape it the better. --KSmrq^T 23:36, 2 May 2006 (UTC)

Matrix problem

Suppose I have an unknown square matrix A. I know that there exists a regular matrix Q such that:

$B_{1}=Q^{-1}AQ$

$B_{2}=QAQ^{-1}$

Only B₁ and B₂ are known to me. Is it possible to reconstruct A with this information? Under what assumptions? How can this be done in practice? -- Meni Rosenfeld (talk) 16:48, 1 May 2006 (UTC)

You'll need stronger assumptions than those. If B1 = B2 = diag (1,0), then you might have A = diag (1,0), Q = I or A = diag (0,1), Q = antidiag (1,1). Melchoir 19:58, 1 May 2006 (UTC)

This looks somewhat like an eigenvalue decomposition, if the B_i are diagonal. Typically the Q matrix would be orthogonal, but that's not essential. Do you know more than you have said? Otherwise, reconstruction is not possible. --KSmrq^T 21:30, 1 May 2006 (UTC)

$Q^{2}B_{1}-B_{2}Q^{2}=QAQ-QAQ=0$

This a linear system of equations for $Q^{2}$, so if it is possible to determine $Q^{2}$ uniquely, then $Q^{2}=0$. That is $Q$ could be any nilpotent matrix. In some cases, $Q^{2}$ does not exist or not unique, see the Fredholm alternative. (Igny 21:48, 1 May 2006 (UTC))

Thanks. The main problem I am interested in is as stated, which I see is unsolvable. I'll work on it some more to see if a solution to a weaker problem can be useful to me. Any ideas which additional information can solve the problem? Or limitations on the solution (is every matrix similar to A a solution to the problem, or must the solution satisfy additional restrictions?) Thanks. -- Meni Rosenfeld (talk) 11:52, 2 May 2006 (UTC)

$B_{1}$ and $B_{2}$ have to be similar, because $B_{1}=Q^{-1}AQ=Q^{-2}B_{2}Q^{2}$. In that case, the equation $Q^{2}B_{1}-B_{2}Q^{2}=0$ mentioned by Igny does not have a unique solution for $Q^{2}$, which is just as well, because Q cannot be nilpotent if it is regular. -- Jitse Niesen (talk) 08:16, 3 May 2006 (UTC)

Programming Languages (Moved from 24 April)

Hello. I'm a beginning programmer. Two years ago, I tried to learn Visual C++ .NET. That was really, really hard. I wasn't able to do anything besides the tutorial in the book, so I quit. About four months ago, I picked up Liberty BASIC. I am able to do a lot and understand the language. I really enjoy it and am doing a lot of fun stuff. I wondered if anyone had any suggestions on other languages to try next or any suggestions about a good progression of languages for a learning programmer. Any stories about what you did, what you wish you did, or just simply any advice you have would be great. Thanks for your help. --Think Fast 01:23, 24 April 2006 (UTC)

Replies should be in the archive. Basically

• Compiled object oriented languages: C++, Java, Pascal, C#
• Compiled not very object oriented languages: Visual Basic, C
• Interpreted languages: Python, Perl, PHP, Ruby, ...
• Other stuff work knowing: javascript, MySQL, Ajax

see Programming language for more. I doubt if the reference desk will be able to help much more than this. --Salix alba (talk) 10:43, 2 May 2006 (UTC)

Hello i used ABC language a lot and i liked it a lot. It is much more simpler than C++ and very easy to learning and alsxo to programmin g. But is nor ported to macos X and i cant use it any more. Now i using python language a lot which is also fun for programming but not so easy to learning. If you have pc you can start with ABC language. 88.233.136.138 19:09, 2 May 2006 (UTC)

May 2

Cooper-Scharlemann's graph

Can anybody please give me advice about how to obtain the properties of Cooper-Scharlemann's graph using Thom's Transversality Theorem? Cthulhu.mythos 09:29, 2 May 2006 (UTC)

I mean, I think Thom's theorem is to be used, but if it can be done some other way it is ok, of course. Cthulhu.mythos 09:31, 2 May 2006 (UTC)

Introductory advanced math websites

I've all but given up on Wikipedia for help with math, and most other comparable sites (e.g. MathWorld) aren't that helpful either. This wouldn't be a problem, except that I have a test in a month that I'm frantically studying for, and I still haven't gotten my head wound around all the material yet. My matrices and vectors are very shaky, and everything I know about three-dimensional geometry is stuff I reasoned out by myself (and reinventing the wheel is obviously a bad way to go about this). I also have a difficult time with trigonometry (I rely on my calculator a lot), but I can struggle through. What I'm mainly concerned with are the topics I just mentioned -- matrices, vectors, and three-dimensional geometry (planes and all that crap). Oh, and permutations/combinations as well. I'm currently learning them in statistics class, but any complementary materials would be a great help. I don't need the fancy stuff Wikipedia has -- a simple introductory webpage (or three) for the topics I mentioned will suffice. (Yes, I know I could just Google it, but filtering the wheat from the chaff is a bit hard when it comes to math.) Johnleemk | Talk 15:23, 2 May 2006 (UTC)

The best we can offer on matricies is Matrix (mathematics) and Vector (spatial), otherwise I would sugest a bookshop/libary, you are likely to much better introductory texts in print than anywhere on the web. Wikibooks might be worth a look. (Oh and reinventing the wheel can be a great way to learn for some people, practice and making lots of mistakes is an esential part in the learning process). --Salix alba (talk) 23:54, 2 May 2006 (UTC)

I looked at those before -- they aren't much help (as I said). Johnleemk | Talk 15:56, 3 May 2006 (UTC)

Wikipedia isn't supposed to help you with math. That's never been it's purpose. Wikipedia aims to be an encyclopedia. As far as I understand the concept, encyclopedias are reference works -- a repository of facts. Wikipedia aims to explain the facts and concepts in a lucid and illustrative manner, but it never aims to fully teach you anything. Wikibooks however is more for this purpose, but it is far less complete. Dysprosia 03:47, 3 May 2006 (UTC)

Uh...exactly. Johnleemk | Talk 15:56, 3 May 2006 (UTC)

Sorry that wasn't so helpful, but I felt that needed to be said. Dysprosia 07:10, 4 May 2006 (UTC)

As far as the matrix goes, why not try http://joshua.smcvt.edu/linearalgebra/ ? It's a complete book on matrix, though of course you still need to pour in a lot of efforts- matrix is quite a large area. And tips on the trignometry: if you know a little about complex number, then you only need to remember one formula... (drumroll) Euler's formula. With some simple substitution you can derive compound angle formula from that, and this formula is much more inituitive then the trig formula. (Oh, no, but you still have to remember sine law and cosine law) However, maybe the exam paper will print out all those formula anyway, at least this is the case in my exam. --Lemontea 10:02, 3 May 2006 (UTC)

I don't need to know everything -- just the basics (multiplication, etc.). I'll have a look, though. I've been taught Euler's formula before, but I never really learnt how to apply it. Thanks anyway!

You might try asking for clarification on those topics you don't understand on the wikipedia math reference desk. Not ideal, but an available option. -lethe ^{talk +} 00:02, 4 May 2006 (UTC)

Euler's formula say $e^{i\theta }=\cos {\theta }+i\sin {\theta }$. Now substitute a+b into θ, you get $\cos(a+b)+i\sin(a+b)=e^{i(a+b)}=e^{ia}e^{ib}=(\cos {a}+i\sin {a})(\cos {b}+i\sin {b})=\cos {a}\cos {b}+i\cos {a}\sin {b}+i\sin {a}\cos {b}+i^{2}\sin {a}\sin {b}$ Knowing that i²=-1, and matching coefficients gives $\cos(a+b)=\cos {a}\cos {b}-\sin {a}\sin {b}$ and $\sin(a+b)=\cos {a}\sin {b}+\sin {a}\cos {b}$. By substituting a-b, you will get the other two formula. As for the product to sum and sum to product formula, your best bet is to start from these compound angle formula, or substitute θ and -θ into euler's formula, add or subtract them to express the sin and cos function with exponential function. Then just add them or multiply them. --Lemontea 13:23, 4 May 2006 (UTC)

Benefit vs Cost of using MLE calculation to measure central tendency of a data set

Dear Sir or Ma'am,

We are in a research and analysis organization here at Fort Gordon that works with computer simulations and insights into their output. We have recently run a study that was made up of multiple simulation runs that developed different sets of data at the end of the run, that are of particular interest to us. These data sets range from large (>300 data values), to small (300<n>30), and statistically insignificant (n<30). Our customer required 1xinput value for each of these data sets as a representation of the measurement of central tendency (i.e. arithmetic mean, median, and mode). Analysis of our data set ditributions showed that many were severely skeweed (postive skew) due to extreme outliers, and therefore we also calculated (harmonic mean and geometric mean) as a method to mitigate the extreme outliers in a large data set and still take into account all the data values in set while calculating its measurement of central tendency.

Our question, is what are the benefits and costs/limitations of using the Maximum Likelihood Estimator (MLE) calculation to provide a measurement of a value (i.e. central tendency value) to represent these data sets vs the method of calculation we used previously. Also is there certain criteria of a data set (e.g. normal distribution, large sample set, etc.) that should be met in order to confidently use the MLE for this type of estimation? -----16:16, 2 May 2006 (UTC)

Thanks for any help you can provide!

v/r,

Klaus G. Sanford

[phone number removed]

Possibly you need robust statistics which can cope with outliers. --Salix alba (talk) 00:05, 3 May 2006 (UTC)

Squaring the circle differently?

I know about how squaring the circle in the manner described is impossible, but I remember reading about a mathematical result in which someone proved that a square could be divided into a finite number of pieces and the pieces could be recombined into a circle. Anyone know when and where this was? --Zemyla 17:39, 2 May 2006 (UTC)

Well, dividing a circle into a square is Tarski's circle-squaring problem, but I guess it's equivalent to the reverse. Melchoir 18:04, 2 May 2006 (UTC)

Thank you, you wonderful person of indeterminate gender. --Zemyla 20:34, 2 May 2006 (UTC)

But of course! Pronoun trouble, eh? I'm male; I guess I'll put that in my user page. Melchoir 20:36, 2 May 2006 (UTC)

Drawing a circle around a bunch of circles

I have a bunch of circles and want to draw a circle around them which is the tightest possible. Is there simple way of finding it? Now I use an iterative method which seems to work but I can't prove it. Also, same question for a circle around three other circles. There I use not iteration but geometry and trig formulas which are ugly. Is there a beautiful method for solving this? This is not a homework but for a drawing program I made for fun. 88.233.136.138 19:02, 2 May 2006 (UTC)

See Descartes' theorem, especially the part at the bottom about complex numbers. —Keenan Pepper 19:47, 2 May 2006 (UTC)

Right, if you start with three circles that all touch each other, then there is a beautiful method: Descartes' theorem. Or, if the circles all have the same radius, then you could find the minimum bounding circle of the triangle formed by their centers (see Circumcircle for hints) and then enlarge it by the common radius.

For three general circles, the question is harder. It seems that most authors are more interested in finding a tangent circle that excludes three circles, rather than including them; see, for example,

• http://www.ijcc.org/on-line(pdf)/2(1)45-54.pdf
• http://citeseer.ifi.unizh.ch/gavrilova00another.html
• Or just Google for Apollonius Tenth Problem.

Possibly you can co-opt such a solution to answer your own problem. Anyway, if you can deal with three circles, you can deal with any number. Melchoir 20:00, 2 May 2006 (UTC)

You don't say what your iterative method is. If it's to replace two circles with the tightest circle containing them, then it's wrong. For example, let us have three very small circles with centers {(-4, 0), (4, 0), (0, 7)}. The smallest circle containing all three is the one with center (3, 0) (0, 3) and radius 5. If, however, you first take the tightest circle containing the first two circles, that's the one of center (0, 0) and radius 3; and then if you take the smallest circle containing both that circle and the small one at (0, 7), you get the circle with center (2, 0) (0, 2) and radius 6, which is suboptimal. – b_jonas 21:41, 2 May 2006 (UTC) (Corrected some numbers – b_jonas 22:34, 2 May 2006 (UTC))

You're right, I think, that this problem doesn't have the greedy property like that. But your numbers are still wrong -- (0, 3) is only 4 from (0, 7), so the first circle is suboptimal. Moreover, I think the idea was to iterate with a solution to the three-circle problem, not the two-circle problem. Still no idea if there's a better extension to $n>3$ than the obvious one I give below, though. --Tardis 14:54, 3 May 2006 (UTC)

Start with two circles. The answer is obvious (a circle centered between them, though not at the midpoint of the segment connecting the centers unless the circles are the same size), but we can go farther and get some insight into the three circle problem by considering non-optimal solutions. We have the centers ${\vec {c_{1}}}$ and ${\vec {c_{2}}}$, and radii $r_{1}$ and $r_{2}$. Call the new circle $({\vec {C}},R)$, and the points of tangency ${\vec {t_{1}}}$ and ${\vec {t_{2}}}$. We know that the segments ${\vec {C}}\rightarrow {\vec {t_{1}}}$ and ${\vec {C}}\rightarrow {\vec {t_{2}}}$ have the same length $R$, and that they are respectively collinear with ${\vec {c_{1}}}\rightarrow {\vec {t_{1}}}$ (length $r_{1}$) and ${\vec {c_{2}}}\rightarrow {\vec {t_{2}}}$ (length $r_{2}$) because radii to a point of tangency are collinear. So (using $d({\vec {x}},{\vec {y}})$ for distance), $d({\vec {C}},{\vec {c_{1}}})=R-r_{1}$ and $d({\vec {C}},{\vec {c_{1}}})=R-r_{2}$. So $d({\vec {C}},{\vec {c_{1}}})-d({\vec {C}},{\vec {c_{2}}})=r_{2}-r_{1}$.

But this is precisely the definition of a hyperbola; all circles tangent to two given circles lie on (one of the two arms of) a hyperbola with the centers of the circle as the foci. To do three circles, then, we just need to find the intersection of any two of the three hyperbolae defined by the three handshakes. (This is an extension of using the intersection of the perpendicular bisectors to find the circumcenter of a triangle.) Basically, characterize each hyperbola with a center ${\vec {c}}$ (just the midpoint of the segment connecting the circles' centers), a "focal length" $f$ (half the length of that segment), and a $d=\Delta r$ that is the constant difference in distance. Some coordinate geometry gives you that the usual parameters for the hyperbola are then $a=d/2,b={\sqrt {f^{2}-a^{2}}}$. Parameterize with sinh and cosh: ${\vec {x}}(t)={\vec {c}}+a{\hat {m}}\cosh t+b{\hat {n}}\sinh t$, where ${\hat {m}}$ points from the center to the bigger circle and ${\hat {n}}\cdot {\hat {m}}=0$. Then set up ${\vec {x_{1}}}(t_{1})={\vec {x_{2}}}(t_{2})$ (two possibly-different parameter values, since this is an intersection and not a collision), and solve. You'll need to write sinh and cosh in their exponential forms and use the substitution $\tau _{i}=e^{t_{i}}=1/e^{-t_{i}}$. Solve for (say) $\tau _{1}$, and then evaluate ${\vec {x_{1}}}(\tau _{1})$ to get your center; the radius is then $d({\vec {x_{1}}}(\tau _{1}),{\vec {c_{i}}})+r_{i}$ (for any i). This is analytically solvable, but the algebra is insane. (If I get a simple result, I'll post it; I notice that the link at ijcc.org mentions the hyperbola intersection, but they seem to not be using the analytic solution for some reason...?)

(Yeah, apparently it's inefficient to intersect them either numerically or symbolically, and of course it's less sexy than applying Möbius transformations. It still might be easier to code and/or a good exercise, though.) Melchoir 22:05, 2 May 2006 (UTC)

There's more! It's possible for the tangent circle to contain but not touch one of the three circles. In this case, the hyperbolae do not intersect. It's also possible that the tangent circle for all three is not the smallest circle possible (consider the points (circles of vanishing radius) (-1,0), (0,ε), and (1,0)). But this is trivial to overcome: construct the obvious containing circle for each pair of circles and check if it contains the third. If so, use that and never set up the equations (which would yield impossible values of $\tau _{i}<0$ in the first case and valid but poor choices in the second).

Finally, to deal with $n>3$ circles, as far as I can tell you'll just have to consider all ${n \choose 3}={\frac {n(n-1)(n-2)}{6}}$ triplets and construct the circle for each, then check that it contains all the other circles. While you're at it, for each triplet check to see if any other circle is contained within the circumcircle of the triangle formed by the three circle-centers and discard any that are. This sounds like extra work but it's only $O(n)$ and will (hopefully) help reduce the effective $n$ in the main $O(n^{4})$ algorithm. Hope this helps! --Tardis 21:54, 2 May 2006 (UTC)

Actually, it occurs to me that in the optimization I mentioned you must only discard circles contained in the triangle formed by the centers of other circles; a circle can be contained by the circumcircle and still be relevant to the overall problem. (Consider the circumcircle of the three points I mentioned earlier, and how much could fit in it!) --Tardis 22:42, 2 May 2006 (UTC)

There's a simpler problem I've heared that's similar to this: given a list of points, find the tightest circle contaning all of them. Clearly this is a special case of the original problem. To do this, you can check all the circumcircles of three points and all Thales-circles of two points, but there's probably a more efficent way to do this. I don't know how fast the fastest solution to this is. – b_jonas 22:41, 2 May 2006 (UTC)

Thank you so much all of you. It is very interesting and i would never have known to google for apollonius tenth problem, did he have more problems? Maybe the wikipedia can tell more about apollonius problems. Thank you again. 88.233.136.138 06:12, 3 May 2006 (UTC)

Hmm... there are some problems at Apollonius of Perga, but they're not numbered. Melchoir 06:25, 3 May 2006 (UTC)

CIH author

I've read that the CIH virus was also written by Sergei Kozachkov, who was recently condemned. He is a co-author? Brand 21:32, 2 May 2006 (UTC)

Sergei Kozachkov? Never heard of him, I get no Google results for "Sergei Kozachkov"... SCHZMO ✍ 22:48, 2 May 2006 (UTC)

The closest thing I found to this subject was a guy named Sergey Antimonov whose DialogueScience antivirus software helped overcome the CIH virus. —Mets501^talk 23:29, 2 May 2006 (UTC)

Recently condemned? Sounds ominous. --Deville (Talk) 00:41, 3 May 2006 (UTC)

Rolutte

If I put $1 on red, and the Roluette table comes up black, my net loss would be $1. I could then bet $2 on red and if it lands black, my net loss would be $3. I could then wager $4 on red. Should I lose again, my net losses are $7. if I were to then bet $8 on red, & win, I would win $1 (my origional bet). Why then can I not do this until the table turns up red? Eventually it will. Of course I woudl need a substantial amount of money to make sure I can keep betting until it hits red....but why couldnt I if I had say $250,000? Lord Westfall 23:28, 2 May 2006 (UTC)

You wouldn't do that because you wouldn't win anything. It's just like when the people buy a scratch-off lotto ticket for $5, and then they win $5 and they're so happy about it. But if you have the money, and you enjoy just playing roulette, then be my guest and play that way. :) —Mets501^talk 23:35, 2 May 2006 (UTC)

Wouldnt you slowly accumulate $1s from your winnings? Lord Westfall 23:36, 2 May 2006 (UTC)

Oh, sorry for my mistake. You're right. —Mets501^talk 00:08, 3 May 2006 (UTC)

This is true. You would slowly accumulate $1s from your winnings. However, there is a problem, and that is this: you don't have an infinite bankroll. Presumably, you go to the table with a finite amount of cash that you can bet. So if you ever got into a situation where you couldn't double your bet, then you would have lost all of your money. Also, even if you had an infinite amount of money, the casino wouldn't, and they'd quit covering your bets at a certain point, at which point you'd be out of money as well.--Deville (Talk) 23:47, 2 May 2006 (UTC)

The strategy you have discovered is a martingale. It doesn't work because even with $250,000, if you keep playing you will eventually lose it all; this is known as gambler's ruin. Melchoir 23:43, 2 May 2006 (UTC)

I agree. It would only work if you had much more money than the Casino has, and if they don't throw you out from it. A good book (which doesn't require any prior knowledge) on this topic is Warren Weaver, Szerencse kisasszony, Kairosz kiadó, 1997, orig. "Lady Luck", Doubleday & Company, Inc. Garden City, New York. – b_jonas 20:43, 4 May 2006 (UTC)

May 3

Rotated Conics

I would like to know how a rotated ellipse (or other conic) can be expressed as an equation. More specifically, how can an equation be obtained which describes an ellipse with semi-major and semi-minor axes a and b such that the angle between the x-axis and the ellipse's major axis is θ. I would prefer parametric equations, if possible (by the way, the Ellipse article does not list any - surely they exist?). --72.140.146.246 02:10, 3 May 2006 (UTC)

Matrices make this a breeze. Let's take the simplest parametric equations for an ellipse, an origin-centered circle stretched along the coordinate directions. One form of the equations uses trigonometric functions.

$P(\theta )={\begin{bmatrix}x(\theta )\\y(\theta )\end{bmatrix}}={\begin{bmatrix}a\cos \theta \\b\sin \theta \end{bmatrix}}$

Instead, we will use homogeneous coordinates for our convenience; they will be especially useful in implicit form.

$P(t)={\begin{bmatrix}x(t)\\y(t)\\w(t)\end{bmatrix}}={\begin{bmatrix}a(1-t^{2})\\b(2t)\\1+t^{2}\end{bmatrix}}$

Now a 3×3 matrix can express any linear, Euclidean, affine, or projective transformation of points P = (x:y:z). A rotation by φ has matrix form

$R(\varphi )={\begin{bmatrix}\cos \varphi &-\sin \varphi &0\\\sin \varphi &\cos \varphi &0\\0&0&1\end{bmatrix}}.$

The rotated ellipse is merely R(φ) P(t), namely

$P'(t)={\begin{bmatrix}a(1-t^{2})\cos \varphi -b(2t)\sin \varphi \\a(1-t^{2})\sin \varphi +b(2t)\cos \varphi \\1+t^{2}\end{bmatrix}}.$

This is more machinery than we need simply to rotate an ellipse at the origin, but it allows us to do much more. It works for any rational parametric curve, and for the broad class of transformations mentioned earlier.

Also, it gives us a handle on implicit conics. We can write the implicit form for any conic in bilinear form, using a symmetric 3×3 matrix Q.

$0={\begin{bmatrix}x&y&w\end{bmatrix}}Q{\begin{bmatrix}x\\y\\w\end{bmatrix}}$

For our original ellipse, Q has the form

$Q={\begin{bmatrix}1/a^{2}&0&0\\0&1/b^{2}&0\\0&0&-1\end{bmatrix}}.$

Now if M is some transformation we wish to apply, we merely substitute for Q,

$Q'=(M^{-1})^{\mathrm {T} }QM^{-1}\,\!.$

That's it. The same approach works for quadric surfaces in 3D. --KSmrq^T 04:33, 3 May 2006 (UTC)

...............................................

If all you want is a parametric representation for an ellipse with centre at the origin, put C = cos θ, S = sin θ, and use

x = aC cos t - bS sin t, y = aS cos t + bC sin t.

This was obtained from the unrotated case by replacing (x, y) := (Cx - Sy, Sx + Cy), which amounts to rotating the point (x, y) around the origin by an angle θ. --Lambiam^Talk 07:07, 3 May 2006 (UTC)

Thanks, that was very helpful.--72.140.146.246 20:03, 3 May 2006 (UTC)

pi question

There are some numbers like, 4,77777777777777.... 8,765765765765765765..... 9,0192837465019283746501928374650192837465.. , 912,992992992992992992992...

They haven't found the full number pi, ins't because the pi is a numbers like those other that i said before, but a more complex number???

Pi is irrational and transcendental. As far as I can see, the numbers you listed above are rational numbers. Dysprosia 03:44, 3 May 2006 (UTC)

It is not very clear what the question is. What would it mean for the "full number" to be "found"? Has the full number 10021426384/1111111111 been found? And what about the square root of 2? --Lambiam^Talk 07:20, 3 May 2006 (UTC)

The numbers you give are 43/9, 973/111, 10021426384/1111111111 and 912080/999, respectively. These are all rational numbers - numbers that can be written as the ratio of 2 integers. The rational numbers are also exactly those numbers that can be written as a repeating decimals (where the same bunch of digits repeats over and over). π is irrational: It cannot be written as a ratio of two integeres, and its decimal expansion doesn't repeat itself. So it is essentially different from the numbers you give, and it will never be possible to find all of its digits.

Lambiam: I suppose the fact that the nth digit of this rational number can be found with negligible computational effort (just divide mod 10 and look in an array), is what the questioner essentially means by "the full number has been found". -- Meni Rosenfeld (talk) 08:29, 3 May 2006 (UTC)

Perhaps, but then, presumably, the questioner would think that also the full number √2 hasn't been found. And, if you're happy with the hexadecimal system, you could argue that the remarkable result of Bailey, Borwein and Plouffe means that the full number π has been found. By the way, I've found all the digits of π. Here they are: 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9 ;-) --Lambiam^Talk 10:56, 3 May 2006 (UTC)

On this topic, an article on the computational complexity of various mathematical constants would be an interesting addition to Wikipedia. Fredrik Johansson 11:08, 3 May 2006 (UTC)

He thinks the standard proof of Pi's irrationality is that nobody's found a pattern to it yet. He reasons, however, that that's not absolute proof at all, and indeed it isn't. What's the actual proof? I can't find it in Pi. Black Carrot 21:22, 3 May 2006 (UTC)

You're right, I didn't see the proof of irrationality there. It's a neat little proof using just some integral calculus, we should probably have it somewhere. It does mention the proof that π is transcendental at Lindemann–Weierstrass theorem, which of course implies irrationality. -lethe ^{talk +} 00:00, 4 May 2006 (UTC)

Various proofs are mentioned (but not given) at Irrational number#History, such as Lambert's proof, which is notable by being first but not so simple as Bourbaki's (assuming that is the one you are referring to). It would make sense to copy some of this to the Pi article. Lambiam^Talk 00:12, 4 May 2006 (UTC)

I was talking about this (I will give a example because I forgot the right name used to represent those numbers) if you get the number 10 and / by 3 (so 10/3) the answer will be 3.3333333333333333333333333333333............. there is no end this will be like this forever, there are more complex versions of this type of numbers, like 874.454545454545454545......... or 2435.767657676576765767657676576765....... or maybe even more complex 7777.876483900011133876483900011133876483900011133... ins't the pi number a complex (moooooooooooore complex) number like the last???

In a word, no. There is no pattern. For there to be a pattern, a particular set of numbers of finite length must be repeated, over and over, forever. If the numbers repeated only once, or only a finite number of times, and were then followed by other numbers, that means the existence of a pattern is illusory. You could search through the first 10 billion trillion digits of pi and still never get to the end of the set of numbers that might, supposedly, be repeated. And if you did, hypothetically, eventually find that that humungously long set of numbers did repeat, it would not be sufficient for it to repeat just once, it would have to repeat ad infinitum. If that really were the case, that would make pi an algebraic number, which we know is not the case. The numbers of pi just go on and on in an apparently random manner forever and ever and ever. Looking for all of its digits is like looking for the crock of gold at the end of the rainbow, because it is impossible to write them all down even if you had infinite time available to you. There is no pattern. JackofOz 03:08, 4 May 2006 (UTC)

Additionally, be careful with your wording - while in certain ways pi is "complicated", in mathematics talking about a complex number is a very specific thing, which pi is not (well, ok, to be pedantic all real numbers are complex numbers, but its imaginary part is zero which I suppose you could say means it's not stricty complex). Two useful words used to describe pi are irrational (meaning you can't write it as a fraction - and hence as a recurring decimal) and transcendental (meaning you can't write it as the solution to a polynomial with integer coefficients). Confusing Manifestation 12:52, 4 May 2006 (UTC)

The pattern is that there is none. So, absolutely any text, once encoded (or any figure) may be found somewhere in pi's decimals. Your love letters. Our Ref desk history. Wikipedia. The internet. More : it is a coding of the universe since its beginning. Strange pattern indeed. --DLL 21:42, 4 May 2006 (UTC)

Since nobody else is bothering to, I looked up the proof of Pi's irrationality [1]. It's more complicated than you probably know how to deal with, and involves calculus, but it's at least not very long. It proves that, although we can't look at every digit of Pi, if we could, we would not see a really really long repeating decimal, we'd just see randomness. Therefore, Pi is irrational, and not similar to 912080/999. Black Carrot 22:00, 4 May 2006 (UTC)

If the digits in the decimal expansion eventually repeat forever, the number is rational, that is, it can be written as the fraction P/Q in which P and Q are integers (and the other way around: all rational numbers have a repeating decimal expansion). The German mathematician Johann Heinrich Lambert proved in 1761 that π is irrational (not rational), so its digits will not repeat (at least not forever). As to DLL's remark above (which I also found hiding inside π), the property he refers to is that of being a "normal number". It is generally believed by mathematicians that π is normal. But this has not been proved, and to find a proof of this looks like a very difficult task. --Lambiam^Talk 22:20, 4 May 2006 (UTC)

Re DLL's statement: absolutely any text, once encoded (or any figure) may be found somewhere in pi's decimals. Your love letters. Our Ref desk history. Wikipedia. The internet. More : it is a coding of the universe since its beginning. Can someone convince me this is true, because at the moment I rather doubt it. Just because the digits proceed randomly forever does not necessarily mean that any string of numbers you can possibly devise will eventually occur, does it? If this were true for pi, it would also be true for e, and any irrational number, wouldn't it? JackofOz 03:36, 5 May 2006 (UTC)

No one can convince you, because it hasn't been proven. DLL is referencing a conjecture that pi is a normal number. Melchoir 04:00, 5 May 2006 (UTC)

Although, to be fair to DLL, it's often stated as fact on the internets. Melchoir 04:07, 5 May 2006 (UTC)

And no. Not every irrational number is normal. The transcendental Liouville's constant is certainly not. However, e is also thought to be normal. And if the decimal expansion is truly random, it is easy to show that every given sequence of digits has a probability of 1 of appearing somewhere. -- Meni Rosenfeld (talk) 06:48, 5 May 2006 (UTC)

Thanks Melchoir and Meni. I'd be interested in seeing that proof. JackofOz 07:01, 5 May 2006 (UTC)

Of the normality of Pi? So do we. Hopefully we'll live long enough. -- Meni Rosenfeld (talk) 07:47, 5 May 2006 (UTC)

No, the proof that "if the decimal expansion is truly random, it is easy to show that every given sequence of digits has a probability of 1 of appearing somewhere". JackofOz 09:19, 5 May 2006 (UTC)

Oh, that. As I said, this is remarkably easy - but it involves some technicalities. I'll sketch a proof for a seemingly stronger statement: Given a real number x, the decimal expansion of which is random, and a digit string of length m, there is a probability of 1 that there exists a (natural) number n such that the digits of x from n*m to (n+1)*m-1 is identical to our given string (I'll call a number with this property a "good number"). Equivalently, there is a probability of 0 that there are no good numbers. I'll denote this probability by P. I'll also denote P(n) the probability that every natural number up to n is not good. Clearly P ≤ P(n) for every n. The probabilty for i not to be good is (1-10^-m). This is independent of any other number being good or not, since every number describes a separate set of digits, and because of our assumption of randomness. So the multiplication principle holds, and therefore

$P(n)=\left(1-10^{-m}\right)^{n}$

$\lim _{n\to \infty }P(n)=0$

And it follows that P = 0. I hope this was clear - do tell if it wasn't. -- Meni Rosenfeld (talk) 09:42, 5 May 2006 (UTC)

Somewhat simpler: If your love letter, or the Constitution of the United States, or any other text (encoded as a finite string of digits) fails to appear, then, by definition, the decimal expansion is not (truly) random. This is true for the von Mises definition, which applies to infinite sequences. Strangely enough it is not mentioned in the Randomness article. --Lambiam^Talk 13:17, 5 May 2006 (UTC)

Thanks. This makes sense to me now. JackofOz 14:27, 5 May 2006 (UTC)

Properties

I'm aware of two (connected, as it happens) ways of describing almost any number. One is factors: A*B=C, with few pairs of A and B for each C. The other is difference of squares: A²−B²=C, once again with few pairs of A and B for each C. Does anyone know of similar pairs or small sets of numbers? Black Carrot 21:40, 3 May 2006 (UTC)

You mean something like this: A^B = C, as in 1073741824¹ = 32768² = 1024³ = 64⁵ = 32⁶ = 8¹⁰ = 4¹⁵ = 2³⁰ = 1073741824? Or more like this: A²+B² = C, as in 120² + 115² = 132² + 101² = 141² + 88² = 144² + 83² = 155² + 60² = 160² + 45² = 164² + 27² = 165² + 20² = 27625? The last one is not "almost any number", of course, but A²−B² won't give you any numbers of the form 4N+2. --Lambiam^Talk 00:03, 4 May 2006 (UTC)

Well, I'm not sure what you mean by 4N+2, but given C=A²-B²=(A+B)(A-B)=D*E, it will only give pairs of squares that correspond to pairs of factors that are an even distance away from each other. Otherwise, B is not a whole number. So, it's guaranteed to work on odd numbers, not so much on evens. I don't think those two suggestions are quite what I'm looking for; the first as far as I can see only works for values of C that are powers of whole numbers. The second looks interesting, but I'm not sure how much system there is to it. Please get back to me. Black Carrot 00:56, 4 May 2006 (UTC)

4N+2 is a multiple of 4 plus 2. So a number C is of the form 4N+2 if there is value for N such that C = 4N+2. For example, 14 is of the form 4N+2 since 14 = 4×3+2. Indeed, you cannot solve A²−B² = 14 in integers. C has to be odd, or a multiple of 4, and must not be a multiple of 4 plus 2. The first suggestion always works, since any number is itself raised to the power 1. It is again related to A×B = C. Just let N be any natural number (in the example 2) and define A' = N^A and C' = N^C. Then (A')^B = (N^A)^B = N^A×B = N^C = C'. The second has not such a simple pattern, but I did not see that as a requirement. Is this just for playing with numbers, or is there something else behind the question? --Lambiam^Talk 03:06, 4 May 2006 (UTC)

Every natural number can be expressed as the sum of 4 or fewer squares; 9 or fewer cubes; 19 or fewer fourth powers etc. - see Lagrange's four-square theorem and Waring's problem. Gandalf61 09:14, 4 May 2006 (UTC)

Ooh. Enticing. I'll see what I can do with Waring and Lagranges' work. And I'm sorry Lambian, of course you're right, any number is itself to the first. What I'm doing: for the past year, I've been trying to find shortcuts to the factorization of a number. Finding out that there was another way to express numbers, as the difference of squares, has been very instructive and has, I think, helped me figure some of it out. So, I figure finding even more ways will help even more.

I found another example. Any number can be written as the difference between two triangular numbers, in much the same way as the difference of two squares. The math is a bit different, and it can handle different numbers - The squares can't handle even-odd pairs of factors, and the triangles can't handle even-even. So, the squares represent 16=8*2 as 16=(5+3)(5-3)=5²-3², but the triangles can't, and the triangles represent 12=3*4 as 12=4+4+4=3+4+5=T(5)-T(2) but the squares can't. Anything else you think of, let me know. Black Carrot 00:01, 5 May 2006 (UTC)

16 = T(16)-T(15). --Lambiam^Talk 00:23, 5 May 2006 (UTC)

Indeed. However, that can be proved by way of 16=1*16=16=T(16)-T(15) (following the format above). That's an even-odd pairing of factors, which I said triangular numbers can handle just fine. On the other hand, the squares can't handle that, as it's equivalent to 16=16*1=(17/2+15/2)(17/2-15/2)=(17/2)²-(15/2)², which includes non-whole numbers. Black Carrot 01:22, 5 May 2006 (UTC)

May 4

Pi

Is there any known pattern of pi? Are there any formulas that represent pi without giving an aproximation? If so, what are they?

Pi is fairly helpful. And yes, Pi = Circumference/Diameter is, by definition, exactly accurate, though difficult to use in practice. Black Carrot 00:58, 4 May 2006 (UTC)

Because Pi is irrational, there is definitely no simple pattern to the digits (in terms of having them repeat), and I daresay there's unlikely to be any complicated pattern either (even though it's possible to construct other trancendental numbers that do) although I have no proof for that. Any of the formulas for pi will give you an exact answer, the only problem being that they're mainly infinite sums which means you have to be able to add an infinite number of terms (and need an infinite sheet of paper on which to write the decimal digits), so in that sense the best you can do is to take a partial sum (the first n terms) and get an approximation good to however many decimal places you need. Confusing Manifestation 01:03, 4 May 2006 (UTC)

Keep in mind that a decimal expansion is an infinite sum as well. When you say 1/3 = 0.333..., what you mean is $1/3=\sum _{n=1}^{\infty }{3\cdot 10^{-n}}$. —Keenan Pepper 01:31, 4 May 2006 (UTC)

Pattern? Whether a pattern is visible in how a number is described depends largely on the nature of the description. Is there a pattern in the square root of two, √2 = 1.4142135623…? Written as a decimal expansion, apparently not; written as a regular continued fraction, it exhibits repeating 2s, √2 = [1; 2, 2, 2,…]. That's an algebraic number; surely there can be no pattern with a transcendental number? Wrong; we have e = [1; 0, 1, 1, 2, 1, 1, 4, 1, 1, 6, 1,…], exhibiting a very simple pattern. In fact, π has the simple generalized continued fraction 3+ ¹⁄₆₊ ⁹⁄₆₊ ²⁵⁄₆₊ ⁴⁹⁄_6+…. And note that we can do arithmetic with continued factions directly; we don't need to convert to a decimal expansion.

As for exact formulae, we have the well-known π = 4 tan⁻¹ 1, using the arctangent function; we have π = (Γ(¹⁄₂))², using the standard generalization of the factorial function, the gamma function; and we have many others. --KSmrq^T 02:04, 4 May 2006 (UTC)

Maybe you will also be interested in List of formulae involving π. -- Meni Rosenfeld (talk) 12:19, 4 May 2006 (UTC)

multiplication

Shortcuts or techniques to make multiplication easier?

In your head, or on paper? —Keenan Pepper 03:23, 4 May 2006 (UTC)

It is essential to memorize the single digit multiplication table, but there are tricks to help. For example, multiplication by nine can be done by a rule.

Also there are methods like the Trachtenberg system that organize larger computations for efficiency. One of the valuable tricks to learn is some alternate computation to verify the result; casting out nines is a popular example, related to the nine rule. --KSmrq^T 09:56, 4 May 2006 (UTC)

A variety of methods are discussed in Multiplication algorithm. JoshuaZ 03:52, 5 May 2006 (UTC)

Double digit Multiplying

2000 minus 72 times 26 equals

Thanks — Preceding unsigned comment added by 203.213.7.130 (talk • contribs)

Either you can work this out easily on paper, or if you just want the answer, people usually have calculators for these sort of things. Dysprosia 07:12, 4 May 2006 (UTC)

This is, of course, a trick question, depending on grouping. Google gives an answer, a power of two; but is it the one intended?? --KSmrq^T 10:07, 4 May 2006 (UTC)

Not quite. If you assume operator precedence (substituting "times" etc. for the necessary operators) there is only one answer. Vocally however the statement is ambiguous. Dysprosia 11:08, 4 May 2006 (UTC)

Shortcut (see preceding section) : 72 * 26 = 72 * ( 100/4 + 1) = 7200/4 + 72 = 1872. --DLL 21:26, 4 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.
>>> print 2000 - 72 * 26
128

Can someone tell me what is the point of this question???? - anon

Only the original (anonymous) poster can say. Take it as a reminder that not everyone who reads Wikipedia is a university graduate; some are still learning the basics. Perhaps we have a very young student who is bright enough to use the Web, but not bright enough (or too lazy) to do assigned problems in arithmetic. The latter assumption explains the form of answers given, typically requiring the student to do more individual work. Obviously it is not a smart question. --KSmrq^T 01:40, 5 May 2006 (UTC)

The answer is 50128 if you do the subtracting first; 128 if you do the multiplication first. Jonathan ^talk File:Canada flag 300.png 02:21, 5 May 2006 (UTC)

Generalized Euler limit

The limit for Euler's constant can be generalized to

$\lim _{n\to \infty }\left[\sum _{k=1}^{n}f(k)-\int _{1}^{n}f(x)\,dx\right]$

where f is any positive decreasing function. Euler's constant is of course given by f(x) = 1/x, and e.g. f(x) = (log x)ⁿ/x gives the Stieltjes constants. Taking f(x) = x^-a gives the zeta function plus a simple ratio (with the nice

$\gamma =\lim _{x\to 1}\left[\zeta (x)-{\frac {1}{x-1}}\right]$

as a special case). In addition, taking f(x) = 1/Γ(x-1) gives e minus the Fransén-Robinson constant.

My questions are: has this limit been studied for other f than those mentioned, and does it have a name? I think it'd be worthy of a Wikipedia article of its own. (It could be mentioned in the article on Euler's constant, but it does seem to be of a more general character.) Julian Havil mentions it in Gamma (page 117), and at least refers to the case of x^-a as "Euler's generalized constants" but I'm having difficulties locating anything relevant based on that name. Fredrik Johansson 12:06, 4 May 2006 (UTC)

celestial mechanics : how to derive formula for angle since periapsis

Hello,

this question may be considered science, but for me the main difficulty is in finding the correct manipulations on more well known equations in order to obtain this result.

This is not homework! :) It is just a personal interest of mine to understand spaceflight as good as possible.

Consider an object describing a (let us say) elliptic orbit, around a sun (which you may consider a point) of mass M , and assume the object has neglectible mass.

The periapis is the point where the object came closest to the sun. Suppose the object has at a certain time : distance r from the sun an angle t of its speed vector with the line connecting it with the sun a speed (in modulus)v

What is the angle $\theta$, viewed from the sun, that the object has travelled from the periapsis

it should be :

$\tan(\theta )={\frac {{\frac {rv^{2}}{GM}}\sin(t)\cos(t)}{{\frac {rv^{2}}{GM}}\sin(t)^{2}-1}}$

G is just the universal gravity constant.

All replies are welcome.

Try starting with the article on orbital elements, and also follow its links (especially orbital state vectors). --KSmrq^T 23:21, 4 May 2006 (UTC)

May 5

What's the significance of e?

"e" is approx. 2.7182818285. You use it in advanced algebra. My question is: Why does e have to be that particular number? Jonathan ^talk File:Canada flag 300.png 02:09, 5 May 2006 (UTC)

Not algebra, but calculus. The number e is the exponential e¹, where e^x is the unique function that is its own derivative, modulo a constant factor. For more, see Exponential function. Melchoir 03:42, 5 May 2006 (UTC)

Oh, and Exponential growth for applications. Melchoir 03:44, 5 May 2006 (UTC)

A good definition is $e=\sum _{n=0}^{\infty }{1 \over {n!}}=1+1+{1 \over 2}+{1 \over 6}+{1 \over 24}...$. —Keenan Pepper 04:01, 5 May 2006 (UTC)

Because a smaller number would be too small, and a larger number would be too large. Fredrik Johansson 04:41, 5 May 2006 (UTC)

This constant appears naturally in different mathematical situations, more so than most of the other named constants (of which there are many). For example, consider the calculation of compound interest. If we have an initial investment A and accrue interest once each year at an annual rate of r = 18% = 0.18, then after a year we will have A(1+r). If the interest accrues every month, the rate is divided by 12 for each increment as we accumulate 12 times, for a total of A(1+^r⁄₁₂)¹². This is more generous. We could go further, choosing weekly or daily or hourly increments. If we choose N times per year, the formula is A(1+^r⁄_N)^N. Now, although it may not be clear without study, we get a definite (and finite!) rate of return even if N goes to infinity (so that we are compounding continuously). Our constant magically appears, as the formula becomes

$Ae^{r}\,\!.$

Another example involves the reciprocal function, f(x) = ¹⁄_x. If we draw a graph for this function, we see that it rises steeply (towards infinity) as x decreases from one towards zero, and it sinks slowly (towards zero) as x increases towards infinity. We are interested in the area bounded by the x axis, the curve, a cut at one, and a cut at c (with c positive). We count areas to the right of one as positive, and areas to the left as negative. Once again our constant appears, for the area is s, where e^s = c. We call s the natural logarithm of c.

A final, famous, and seemingly unrelated, example is in the description of circular movement. If we measure angles by distance traveled along the circumference of a unit circle, so that a full circle is 2π (radian measure), and if 0 is perfectly horizontal and ^π⁄₂ is perfectly vertical, then for a given angle, ϑ, the horizontal position, x, is

$x={\frac {e^{\mathbf {i} \vartheta }+e^{-\mathbf {i} \vartheta }}{2}},$

where i² = −1. The article on Euler's formula discusses this further.

The deeper reason for the frequent appearance of e, and especially exp(x) = e^x (the exponential function) is that we are seeing an eigenfunction with eigenvalue one of the continuous linear operator for the calculus notion of taking the derivative of a function.

${\frac {d}{dx}}\exp(x)=\exp(x)\,\!$

Simply put, the constant b = e gives the unique function of the form b^x unchanged by taking its derivative. That's a Good Thing. --KSmrq^T 06:04, 5 May 2006 (UTC)

What are Groebner bases?

I'm researching current work on Frobenius' linear Diophantine equations. You know, the "Postage Stamp Problem". I've found references to Graph Theory and Geometric work being done on this classic number theory problem, yet can't find any explanation when "toric Groebner base" is mentioned. Perhaps we should start with just what a Groebner base is . . .

Thanks.

JAsbrand

I haven't a clue, but is Gröbner basis what you're looking for? Melchoir 04:03, 5 May 2006 (UTC)

Try combining the idea of a Gröbner basis with the circumstances of toric geometry; that is, compute a Gröbner basis for a toric ideal. This paper may help more than Wikipedia. --KSmrq^T 06:17, 5 May 2006 (UTC)

inner product - wavelets

I'd be right in saying that the inner product of f(t) and g(t), where:

• f(t) = 1, between t=0 and t=1, and
• g(t) = t, between t=0 and t=1, 2-t between t=1 and t=2

(i.e. a square wavelet and a triangle wavelet) ...is 0.5 right? --131.251.0.7 11:03, 5 May 2006 (UTC)

• Or 1, actually. Well, its 1 or 0.5 anyway. --131.251.0.7 11:11, 5 May 2006 (UTC)

Looks like 0.5 to me. Of course, this is only if by "the inner product" you mean the "natural" inner product: the integral of the pointwise product over all space. Sometimes it's useful to use a different inner product; see Orthogonal polynomials. Melchoir 11:18, 5 May 2006 (UTC)

I'm talkin about the inner product in the "integral of f(t)g(t)dt" sense. So, the data compression/ wavelets sense of the term. --PrifysgolCaerdydd (The anon of the first posting) 11:37, 5 May 2006 (UTC)

Righto, then 0.5 it is. Melchoir 11:41, 5 May 2006 (UTC)

Yeah, that was pretty easy actually. Cheers anyway. I just had to make sure I was right (thats a few points in the bag for the exam then). I may well come back to this page later to ask about LFSRs and hash functions and linear congruential generators (stuff that I need for finals). --PrifysgolCaerdydd 11:47, 5 May 2006 (UTC)

hash functions

Yeah, I may need your help again. The question is: Let h: Σ^{* --> Σn be a cryptographic hash function. What is the total number of collisions for h? .... I'm sure this is a trick question, cos there's no more to the question, and you need an extra bit of information surely. Any pointers? --PrifysgolCaerdydd 14:52, 5 May 2006 (UTC)}

If I'm interpreting the question correctly, then Σ* is an infinite set and Σ^{n is a finite set. —Blotwell 15:24, 5 May 2006 (UTC)}

Without any further information, the correct answer would appear to be "infinite". — Timwi 15:51, 5 May 2006 (UTC)

Well, that's one of the multiple choice answers, I'll click on that one then. --PrifysgolCaerdydd 15:53, 5 May 2006 (UTC)

Gödel's(?) logic

When is 10 + 10 not equal to 20? -- PCE

When it's equal to 100? --Bth 16:35, 5 May 2006 (UTC)

[Edit conflict] In binary, 10 + 10 = 100. If + is interpreted as bitwise or, then 10 + 10 = 10. Can't see the connection to Gödel though. -- Meni Rosenfeld (talk) 16:36, 5 May 2006 (UTC)

Correct. Someone else pointed out to me that the base has to be designated as decimal before 10 + 10 = 100. As for Gödel someone else has said that while his Incompleteness Theorem applies to arithmetic it does not apply to binary (or for that matter multiple state) logic, which seems a bit odd. -- PCE 16:56, 5 May 2006 (UTC)

What do you mean "has to be designated as decimal"? Is it a typo for "binary"? (Obviously binary is the only counting system where 10+10 isn't 20 because it's the only that doesn't have a 2). --Bth 17:44, 5 May 2006 (UTC)

Perhaps you mean Propositional calculus? In that case, I don't remember the details but I think it's true - This theory is too simple for the incompleteness theorem to apply to it. -- Meni Rosenfeld (talk) 17:12, 5 May 2006 (UTC)

Are you familiar with Rucker's proof? -- PCE 17:29, 5 May 2006 (UTC)

No, sorry. -- Meni Rosenfeld (talk) 17:30, 5 May 2006 (UTC)

Is it the one about half way down this page? --Bth 17:46, 5 May 2006 (UTC)