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June 28

trignometry

show that tan 15+cot 15=4

Except it isn't. tan 15 + cot 15 = -2.024. --Zemyla^t 12:09, 28 June 2006 (UTC)[reply]

It is if you work in degrees not radians. Madmath789 12:15, 28 June 2006 (UTC)[reply]

Punch it into your calculator, write down the answer, hand in your homework, get a bad grade, go home and brood, stop using Wikipedia to try and do your homework for you. — QuantumEleven 12:41, 28 June 2006 (UTC)[reply]

\tan 15^{\circ }+{\frac {1}{\tan 15^{\circ }}}={\frac {\tan ^{2}15^{\circ }}{\tan 15^{\circ }}}+{\frac {1}{\tan 15^{\circ }}}={\frac {\tan ^{2}15^{\circ }+1}{\tan 15^{\circ }}}={\frac {8-4{\sqrt {3}}}{-{\sqrt {3}}+2}}=4

—Mets501 (talk) 12:57, 28 June 2006 (UTC)[reply]

... or prove general relationship

\tan {A}+\cot {A}={\frac {2}{\sin {2A}}}

then plug in 15 degrees for A. Also think about why you should expect the general solution to be symmetrical about A=45 degrees (as indeed it is).Gandalf61 14:00, 28 June 2006 (UTC)[reply]

Real-time Remote Mac Software

Is there (Mac) software to allow the remote use of a computer over a fast local network at fast enough speeds to make no visual difference? For example, I want to use a desktop Mac via a (Mac) laptop, but at full refresh, large resolution and 32 bit colour with very little lag. This would be for general real-time use rather than admin purposes. Finally, it would not be X, but a streamed display.

In fact, is this even possible?

Something like Timbuktu perhaps? I doubt that you could have a network that is as fast as a hardware bus, you'd have to decide if the lag is tolerable. --LarryMac 15:58, 28 June 2006 (UTC)[reply]

You don't need to match the speed of the hardware bus, just the speed of the video connector. A 100Mbit ethernet connection is fast enough for small displays; for larger displays, you need gigabit ethernet. You'll have a latency of about two screen refreshes no matter what (one for the screen to be composed on the host, and one for it to be displayed on the client), and content such as video from capture cards or 3D graphics might not display, as those are drawn directly to the video card's buffer, not to a buffer in main memory. --Serie 23:55, 28 June 2006 (UTC)[reply]

Why not use Chicken of the VNC? It is very full-featured, free, and open-source. -- 24.75.133.178 18:28, 10 July 2006 (UTC)[reply]

Programming in LISP

I am looking for an online LISP environment to do things in, I would like to be able to save programs as well, can anybody give me a link?

There's a Scheme implementation for your web browser here but it isn't really useful for writing code because it uses a simple text box for editing, and thus has no parenthesis matching. You can find links to more useful Lisp implementations here. 84.239.128.9 16:07, 28 June 2006 (UTC)[reply]

Gestalt Value .

wat is the principle of GESTALT VALUE ? --59.161.8.150 15:09, 28 June 2006 (UTC)Roman Nagpur.[reply]

I'm not sure if it is mathematical either. But mathematicians are so much at loss of proper words for fuzzy concepts ... --DLL 20:53, 30 June 2006 (UTC)[reply]

this may have to do with something that is 'holistic' and has to be viewed as a hole instead it will not be comprehended. it can also refer to completing an experience...but I don't see how can it be applied to math because it's a psychological concept.--Cosmic girl 19:51, 10 July 2006 (UTC)[reply]

Sum

how do we calculate the product of ((r^2)-1)/(r^2), where r = prime numbers starting from 2.

{\frac {r^{2}-1}{r^{2}}}=1-r^{-2}

, then use Riemann zeta function#Relationship to prime numbers.--gwaihir 17:51, 28 June 2006 (UTC)[reply]

Um... the best non-giveaway hint I can think of is to simplify the fraction, invert the whole thing, and think about geometric series. Melchoir 18:25, 28 June 2006 (UTC)[reply]

(Never mind, I should refresh before responding to these things!) Melchoir 18:26, 28 June 2006 (UTC)[reply]

June 29

Windows XP Volume Muting Question

Hello,

I am in need of an automated process (registry file, script, etc.) that I can run on several Windows XP computers that will perform two actions:

1) Add the volume control speaker icon to the taskbar system tray.

and

2) Mute the master volume.

Repeated searches of Wikipedia have yielded no assistance in solving this for me. I've also done some Google searching but have yet to find a solution. Your help would be greatly appreciated!

Thank you.

A Google search for the terms visual basic muting volume brought me to this. If you download it and view the class, you can see that it uses API calls to winmm.dll. You would have to make calls to that DLL through whatever programming language you are using. Changes to the Windows Registry would probably not be reflected immediately in the interface and would be overwritten at shutdown. I found another program that will mute the volume called Wizmo if you aren't up to the required programming task. —Bradley 19:58, 29 June 2006 (UTC)[reply]

An answer to your first question might be on this site, specifically #320, but I make no guarantees. My Google search was registry setting volume control system tray. —Bradley 20:19, 29 June 2006 (UTC)[reply]

about percentage

Hi Guys and Gals! Wikipedia defines percentage as:

"...a way of expressing a proportion, a ratio or a fraction as a whole number,.

what if a percentage is expressed in decimal form? (e.g. 2.45%)

Is there an exact term to call it? Thanks for your help.

See Wikipedia:Reference desk/Language#question on percentage. Melchoir 02:50, 29 June 2006 (UTC)[reply]

iTunes/outdoor speaker set up

In August I'm having a large outdoor gathering, a party of sorts, at my house. I'd like to have iTunes playing music for the roughly 100 guests who will be just outside the house (i.e. not very spread out). What I'm looking for is ways to have good quality sound played outside with equipment that I can use for some other purpose once the gathering is over. Like using it for a stereo set up that I don't yet have or using it for an improvement over the stock speakers on my television or playing music from the computer throughout the house etc.

What I'm working with is a dual 1 Gig G4 tower. If there's a need for wireless, I'm using a Sonnet internal card for wireless internet with a Linksys wireless router. I also have an iPod (and iTrip) which will be able to hold the desired playlist if need be.

What's the best (while not going into thousands of dollars into debt) way to do this? Dismas|^(talk) 07:52, 29 June 2006 (UTC)[reply]

Doesn't Apple have AirPort Express which permits easy music streaming? Dysprosia 12:00, 29 June 2006 (UTC)[reply]

If you have the music on your computer, and you already have a stereo, it's just a matter of taking the output from the computer to input into the stereo. All you need is the right cable. Note, though, that "all in one" music players do not necessarily have any inputs. If you do have a CD player that is suitable, almost certainly the most cost effective thing to do is to just cut CDs of your music choice. If you have a stereo player supporting MP3 playback, a single CD can easily contain the whole night's music. If you do want to go down the wireless route, be sure to test it thorought, and make sure the signal is strong; filling an area with people will block off a lot of the signal, and you don't want your oh-so-high-tech wireless system to stop just as it gets crowded. Also not what you asked for, but bear in mind that for a one-off party it may make more sense to hire a good quality/loud system that is louder than you will ever be able to use it for again. Oh, and if you are using your computer to drive it, make sure people understand this, so they don't reboot it, or start playing games that kill iTunes. Notinasnaid 12:07, 29 June 2006 (UTC)[reply]

The computer does have all the music on it and will be inside where it should be safe from the curious guests who think that they can pick and choose their own selections. The last thing I need is one of the younger guests to start playing Nine Inch Nails' "Closer" while Grandma is telling Auntie how nice the weather has been.

I knew a bit about AirPort Express but wanted other ideas as well.

I don't currently have a good stereo (just a little bookshelf system) but that may be what I end up getting to meet my needs for this event as well as just because I want an actual stereo.

Thanks for the ideas! Dismas|^(talk) 12:34, 29 June 2006 (UTC)[reply]

I play CDs through the DTS receiver I have attached to my television and DVD player. Mine has additional input, so I imagine I could hook-up the line-out from the computer to the line-in of the receiver. Get some speaker cable and drag the speakers outside (and hope it doesn't rain). Otherwise, they do make outdoor speakers that are water-proof. Perhaps this the time for you to upgrade your home theatre system. Personally, I'd skip any wireless options as they would cost more for a solution I don't neeed—running cheap speaker wire outside is not an issue for me. —Bradley 20:43, 29 June 2006 (UTC)[reply]

June 30

Concepts of Non-linear optimization

Hi, I have looked at the optimization, lagrange multiplier and quadratic programming. I have some confusions here. Suppose we are consider optimization in Euclidean space. Since the use of lagrange multiplier can be reduce our problem from constraint to unconstraint ones, and also can be used for all non-linear problem, why do we have to invent some other techniques such as linear/quadratic programming, or some heuristic such as simulated annealing ? -- 131.111.164.110 11:05, 30 June 2006 (UTC)[reply]

From what I can tell, your question is a general why are there different approaches to different problems question. The simple fact of the matter is that mathematical solutions to problems require several stages: First, the problem must be well formed; next, the problem must be translated into a mathematical formalism suitable to capture all the necessary paramters; and finally, the solution that the mathematical formalism yields must be re-translated into the original problem's setup (e.g., 'plain English'). Of course, different problems have different best approaches. Do you have more specific questions about particular fields? Nimur 20:34, 30 June 2006 (UTC)[reply]

General non-linear optimization is difficult, uncertain, and expensive. A mathematician may use a general result to assert the existence of a solution, but that is not the same as a practical algorithm to find a specific solution.

Let's take a simple example that most people can follow. We can write a closed form solution in surds for the roots of any univariate polynomial of degree four with complex coefficients. This form is horrendously complicated. If we have a linear polynomial, the general method applies but we would be crazy to use it. For higher degree polynomials, we cannot write a closed form but we do have an extremely sophisticated and expensive method called cylindrical algebraic decomposition that would allow us to isolate roots to some extent. To use this for a routine quadratic polynomial, rather than one of the two quadratic formulae would again be madness.

One of the methods used to find the solution of some complicated non-linear optimizations is sequential quadratic programming, locally approximating a difficult objective function by a quadratic objective subject to safety bounds. An analogous procedure for polynomial roots is Newton's method, which solves a series of linear approximations.

To optimize a quadratic objective function subject to linear equality constraints, we have an option that is much simpler and more efficient that using Lagrange multipliers: we can project the problem onto a space of lower dimension, with each constraint effectively removing one variable. For example, to minimize x²+y² subject to x−y = 1, we need merely consider 2x²−2x. Equating the derivative to zero, we immediately obtain the solution x =1⁄2, y = −1⁄2.

Thus as a practical matter we seek the most restrictive classification of our optimization problem, not the most general. --KSmrq^T 23:35, 30 June 2006 (UTC)[reply]

Additionally, the method of Lagrange multipliers does not work if the problem has inequality constraints (more precisely, it does work but it does not get rid of the constraints). Even if the problem has only equality constraints so that the method of Lagrange multipliers yields an unconstrained problem, we do not have a good method for solving unconstrained nonlinear problems. I think that the people working in the field have found that the unconstrained problem is harder to solve in practice than the original constrained problem. It is an interesting question, though, and I'm not sure of my last point; it may be just that problems arising in practice always have inequality constraints. -- Jitse Niesen (talk) 04:23, 1 July 2006 (UTC)[reply]

Inequality constraints have been accommodated by treating them as temporary equality constraints (the "active set" idea), and by penalty/barrier functions, just to name two options. In the latter case, the constraint is replaced by a term added to the objective function that gives a prohibitive increase upon approach to the constraint wall. This sounds promising on paper, but is difficult to get working well; it can easily turn a nice problem ugly.

One level up, it is possible to express a variety of problems as optimizations; and again, it is often not wise to do so.

The literature of optimization takes time to penetrate. Because so much of the work has grown up as applications in industry, the field evolved its own language, its own terminology and world view. It also takes time to build a mental model of the geometry of non-linear optimization complications, especially since the typical problem lives in a space of many dimensions (many variables). As an introductory example, try to understand why conjugate gradient is more efficient than steepest descent in a long narrow valley. --KSmrq^T 19:06, 1 July 2006 (UTC)[reply]

Thank you for all answers; Then can I just conclude that we have many methods to attack many specific problems, mainly for the reason of efficiency? 131.111.164.226 13:50, 3 July 2006 (UTC)[reply]

Yes, mainly. An overly-general algorithm may also give less accurate results, or even fail to find a solution. --KSmrq^T 04:37, 4 July 2006 (UTC)[reply]

Parabola?

I was thinking of a pattern in which you start with two numbers, say 1 & 10, and add one to the first number and subtract one from the second number, and take the product of every pair. For example:

1*10=10

2*9=18

3*8=24

4*7=28

and so on.

I decided to graph this, except using a larger range (-10*21 to 21*-10) and a smaller interval (one tenth). The result is the graph you see on the right.

I am guessing this is a parabola, simply by what it looks like, but is there any way to tell if it is one?

Thanks for any help.

--Tuvwxyz 21:53, 30 June 2006 (UTC)[reply]

The second differences are constant so it is a quadratic equation and hence a parabola. 128.197.81.181 22:06, 30 June 2006 (UTC)[reply]

Oh, to clarify second differences: 18-10 = 8. 24-18 = 6. 28-24 = 4. Taking the differences of those: 8-6 = 2. 6-4 = 2. The number of times you have to do this until they all come out the same tells you the order of the polynomial. 128.197.81.181 22:07, 30 June 2006 (UTC)[reply]

Thanks for the response, I think I've heard of the second differences before. --Tuvwxyz 22:13, 30 June 2006 (UTC)[reply]

You can also put it into an equation of a parabola, namely

y=30.25-(x-5.5)^{2}

. StuRat 22:16, 30 June 2006 (UTC)[reply]

Good eye! Shall we try a little algebra? Call the starting numbers p and q. After n steps these have become p+n and q−n. Their product is (p+n)(q−n) = pq+(q−p)n−n². Since the graph depicts the product versus n, it is the graph of a quadratic polynomial in n. Letting n be fractional makes no essential difference.

The graph of a quadratic polynomial, y = ax²+bx+c, is always a parabola. If a is positive the "arms" go up; if negative, down. The constant term, c, is the height of the lowest point (arms up) or highest point (arms down). The remaining term, b, shifts that center point left or right, to −b⁄2a.

To see the proposed graph in simpler terms, take the starting values p = 0, q = 0. Then q takes negative values, but the polynomial reduces to −n², the quintessential parabola inverted. --KSmrq^T 00:12, 1 July 2006 (UTC)[reply]

July 1

Mathcad to Wiki math markup

Is there an engine to covert mathcad format to wiki math markup format. I have a bunch of derived equations for solved variables that will take about two weeks to convert if done by hand? ...IMHO (Talk) 10:44, 1 July 2006 (UTC)[reply]

Wikipedia uses TeX, so that might help you in your search (i.e. don't search for "mathcad to wiki", search for "mathcad to tex"). —Keenan Pepper 18:38, 1 July 2006 (UTC)[reply]

Number speak

I realize this question is elementary to mathematicians but how do you speak the number, 1,072,915,200,000,000,000.

Thank You

See Names of large numbers. -- Meni Rosenfeld (talk) 16:51, 1 July 2006 (UTC)[reply]

I'd say "one point zero seven two nine one five two times ten to the eighteenth", but if you wanted to, you could say "one quintillion, seventy-two quadrillion, nine hundred fifteen trillion, two hundred billion". —Keenan Pepper 18:37, 1 July 2006 (UTC)[reply]

The honest answer is that we don't. For a number with that many digits it is not particularly helpful to say it. If the precise digits are meaningful, it is more helpful to see the number written down. In the common case where we don't care about those details, it is more helpful to use rounded scientific notation ("one point oh seven times ten to the eighteenth", 1.07×10¹⁸). Use of scientific notation is one effective way to combat the "size stun" effect of large numbers. ("A billion here, a billion there, and pretty soon you're talking real money!" — attributed to US Senator Everett Dirksen) --KSmrq^T 19:24, 1 July 2006 (UTC)[reply]

Number System

Why everything to the power of zero is equal to one ?

Because it follows from the rules of exponents. x^n/x^n = x^(n-n) = x^0 = 1. Splintercellguy 17:20, 1 July 2006 (UTC)[reply]

The question is almost true, but not quite. Zero to the power of zero is commonly considered undefined, not one.

The core of the answer is consistency, to make a system where everything works nicely together. Consider a sequence of powers of 10.

…	10⁻²	10⁻¹	10⁰	10¹	10²	10³	…
	1⁄100	1⁄10	?	10	100	1000

As we step right through the powers, each number is 10 times more than the one before. As we step left, each is 1⁄10 of its neighbor. The only way to fill in the entry for the power zero and be consistent is to use 1. This is true whether we use powers of 10 or of any other number besides zero. (Try it for zero!) --KSmrq^T 19:44, 1 July 2006 (UTC)[reply]

Empty product provides a good overview, and a nice quote to boot. RandomP 20:15, 1 July 2006 (UTC)[reply]

Much as I endorse the choice of Knuth and colleagues to define 0⁰ as 1, it is important to understand that they are advocating a position, not stating a commonly agreed convention. The argument for that choice is again consistency, in fact a consistency beyond that shown in my previous post. However, every positive power of zero is zero, and every negative power is undefined, so this is a qualitatively different sort of choice, not so easily made. --KSmrq^T 02:21, 2 July 2006 (UTC)[reply]

I'm not sure what your definition of "common" is, but I sincerely doubt there is even one reasonably recent textbook of analysis which hasn't adopted that convention when talking about power series.

(Of course, 0×∞ is now commonly defined to be 0, too. Measure theory just wouldn't work without it).

Another thing to keep in mind is that when 0 is considered a set, 0⁰ very plainly does have cardinality 1: there is exactly one empty function (by contrast, 0¹ is empty.) At the very least, if you think 0⁰ is undefined, you must go to the trouble of saying which 0 you use.

RandomP 12:20, 2 July 2006 (UTC)[reply]

Three points:

On "common": I have not done a survey of contemporary literature in research and teaching, but I'll wager the "undefined" version is still to be found, and surely literature predating Concrete Mathematics (which is a lot) will be full of it.
The qualitative difference of a nonzero base is clear, both in the arguments given and in the fact that essentially all the literature, current or past, agrees that a nonzero value raised to a zero power should equal 1.
Assuming we're still interested in helping the original poster understand, it's important to highlight the difference between the two cases.

It is indisputable that when Knuth et al. argued for 0⁰ = 1 the convention was not commonly agreed; so if you like, change "they are advocating" to "they were advocating". Either way, we should not ignore the issue. --KSmrq^T 19:20, 2 July 2006 (UTC)[reply]

My understanding is this is currently a high-school vs real mathematics issue: in high school classes, 0⁰ is left undefined (just as

{\sqrt {-1}}

is, commonly), but for someone with more mathematical knowledge, 0⁰ is defined at least when 0 is considered as a cardinal number; it might yet be undefined when 0 is considered a complex number, and I have seen people go so far as to claim that a^b is undefined (or multi-valued, at least) for b a non-integer real number and a a positive real number, unless it is explicitly mentioned that real exponentiation, not (multi-valued) complex multiplication, is meant.

Sorry if that got lost. The important points are:

it's a convention. conventions are chosen for convenience, not for truth.
overwhelmingly, people consider 0⁰ = 1 to be the implied convention when the expression is used by someone else, though some may choose not to use the expression at all, leaving it undefined.
there are situations where the expression 0⁰ is defined, and is 1.

RandomP 20:33, 2 July 2006 (UTC)[reply]

Of course, 0×∞ is now commonly defined to be 0, too. Measure theory just wouldn't work without it.

Can you explain why?--72.78.101.61 12:32, 2 July 2006 (UTC)[reply]

Well, I might have been a bit .. informal there. Of course you could reword all references to that definition, but it just so happens that when you define 0×∞ to be 0, things like the product measure Just Work. The 2-dimensional volume of the x axis, for example, is 0; however, it should also be 0×∞, because it's the cartesian product of a single point (measure 0) and the entire real line (measure ∞).

I believe that it used to be perfectly acceptable to write f(∞) = 7, for example, where today we would write

\lim _{x\rightarrow \infty }f(x)=7

. And there certainly are real functions f, g such that, using the sloppy notation, f(∞) = ∞, g(∞) = 0, fg(∞) = 17, or whatever number you want. However, now that we're standing on the shoulders of those giants (in patience, at least) who've eradicated the sloppy notation, we can easily define 0×∞, and be done with it (though we must then remember that considered as a map from [0,∞] × [0,∞] to [0,∞] (with the usual topologies), multiplication is no longer continuous).

Put yet another way, {0,∞} is the simplest semiring. If you so wish, you can think of that as a model for binary logic, and the only way for that to work is to define 0∞=0.

RandomP 13:31, 2 July 2006 (UTC)[reply]

There is yet another reason to adopt 0⁰=1 — you simply need this for the binomial theorem to work. Put y=0 and try to calculate (x+0)¹. You'll get: (x+0)¹ = x⁰·0¹ + x¹·0⁰. This will give the desired result of x only if 0⁰=1.
:-) CiaPan 17:46, 7 July 2006 (UTC)[reply]

I should point out that not everything to the power of zero equals zero.

0^{0}

, which is the only exception in the entire number set C, has two definite possible values, and one undefined. Here's the proof:

0^{0}=e^{0\log 0}=

undefined

Since

0^{n}=0

therefore

0^{0}=0

Since

n^{0}=1

therefore

0^{0}=1

--JB Adder | Talk 14:12, 2 July 2006 (UTC)[reply]

That

0^{n}=0

only holds for natural numbers n > 0 (with xⁿ interpreted as x multiplied by itself n times), so this isn't a "proof" at all, just incorrect appeal to a definition. Fredrik Johansson 22:43, 2 July 2006 (UTC)[reply]

No, it holds for every real n>0. See Exponentiation article and read how Exponentiation#Real powers of positive real numbers are defined from rational powers by continuity. In the same manner (i.e. by continuity) Pⁿ for positive P and positive n extends to positive powers of zero. Check Image:Root_graphs.png to see how Pⁿ approaches zero when P approaches zero, for different n. This definition is correct. --CiaPan 20:12, 11 July 2006 (UTC)[reply]

what is this function called?

What do you call the inverse to the real function x |-> xⁿ + x + 1. (Maybe only for odd n?) For some reason, I thought this was called the Eisenstein function, but now I can't find that name in any of my references, so I'm wondering if I just dreamed that up. -lethe ^{talk +} 17:11, 1 July 2006 (UTC)[reply]

It isn't one-to-one for even n—f(0) = f(-1) = 1, for example—so you'd better restrict to odd n. Tesseran 07:12, 2 July 2006 (UTC)[reply]

This is not an insurmountable barrier; for example, x² suffers from that defect as well, and that hasn't stopped us from giving it an inverse, the square root. But I think I mucked up my question. What I really want is the function that gives you the root of the polynomial xⁿ + x + a. I guess this function is actually the inverse of f(x) = xⁿ + x, rather than what I said above, whoops. It's the function that you need to write the solution of the quintic in closed form (does it work for polynomials of any degree?) -lethe ^{talk +} 07:33, 2 July 2006 (UTC)[reply]

A somewhat related article is Artin-Schreier theory.--gwaihir 08:24, 2 July 2006 (UTC)[reply]

Simplification

If

${\left({\frac {P+D}{P}}\right)}$

can be simplified as

${\left(1+{\frac {D}{P}}\right)}$

then what is the simplification of

${\left({\frac {P}{P+D}}\right)}$

...IMHO (Talk) 22:50, 1 July 2006 (UTC)[reply]

It can't be simplified further. The top equation only works because P/P = 1, but you can't split a denominator this way. --ColourBurst 23:07, 1 July 2006 (UTC)[reply]

I'd say that

{\frac {1}{1+D/P}}=\left(1+{\frac {D}{P}}\right)^{-1}

is a simpler form, whichever way you prefer to write it. Melchoir 23:59, 1 July 2006 (UTC)[reply]

Oh yes it can!

{\left(1-{\frac {D}{P+D}}\right)}

-- Jokes Free4Me 13:08, 4 July 2006 (UTC)[reply]

ColourBurst is correct. Melchoir is less correct, if not strictly wrong. Computer algebra systems explain. To some extent the criteria for which of two equivalent expressions is "simpler" than the other are in the eye of the beholder. When a computer program is asked to simplify an expression, it may not choose the form the user would prefer. That said, for most purposes, and with most computer algebra systems, the original expression is as simple as possible (assuming P does not divide D). The alternative expression has a fraction in the denominator, which is usually something we prefer to eliminate, not create. However, on some occasions and for some purposes the double division may nevertheless be what we want. It may be easier to appreciate the issues with a different example: Which is simpler, 1−x² or (1−x)(1+x)? The first form has fewer "pieces", but the second reveals the factors. Still, I think the original question is less ambiguous; if it comes up on an mathematics test you are likely to get a better grade with "no simplification". --KSmrq^T 01:48, 2 July 2006 (UTC)[reply]

Going a bit further without creating another section now let me ask whether in either case P can equal zero without resulting in a divide by zero error? ...IMHO (Talk) 03:40, 2 July 2006 (UTC)[reply]

In the first case, no, and in the second case, yes. The second case doesn't have a nonzero denominator (assuming D is nonzero), and in the first case either form has a zero denominator so it would be undefined. If you wanted to ask whether what the limit approaches as P approaches zero in the first case (which different from what you asked), the answer would be that it approaches positive infinity. --ColourBurst 05:18, 2 July 2006 (UTC)[reply]

Without jumping to any conclusion and avoiding the assumption that the answer is "no" allow me to ask then if in a situation where the value of P represents the number of grains of sand in the top portion of an hour glass and the value of D represents the number of grains of sand in the bottom portion of the same hour glass is it possible for the all of the states of the hour glass (upper chamber filled with sand and lower chamber empty and lower chamber filled with sand and upper chamber empty, plus any and all states in between) to be represented by the ratio of D to P or by the ratio of P to D corresponding to all of the states in terms of the location of each grain of sand in the upper chamber or the lower chamber of the hour glass specifically including the state of P where P is equal to zero? If not what equation would allow all states of the hour glass to be represented without incurring a divide by zero error for P. ...IMHO (Talk) 04:50, 2 July 2006 (UTC)[reply]

Hmmm, I think you have the wrong sense of ratio here. A "zero" ratio for D:P means that D is zero, and you couldn't infer anything about P from D. "Infinite ratio" means the opposite (0 for P). But ratios never infer anything about how many particles are really in the glass - just the proportion. --ColourBurst 05:18, 2 July 2006 (UTC)[reply]

Okay then what you are saying appears to be that the proportion of P to D or D to P can not be represented by the division of P by D or of D by P when the denominator is zero and therefore that the relationship expressed by such division is invalid when applied to the situation represented by an hour glass where P or D may have a value of zero? ...IMHO (Talk) 05:25, 2 July 2006 (UTC)[reply]

Well, the ratios in those two cases have a "meaning" - one of those elements is zero, depending on case. It's not invalid - you just can't infer anything about the other quantity without additional information. See elasticity (economics) for an example of what a zero ratio and an infinite ratio means. --ColourBurst 05:41, 2 July 2006 (UTC)[reply]

I'm not quite sure of your objective, but I think you had the right idea with P/(P+D). Given the extra information you have now provided, we know that P+D gives the total number of grains of sand, which should be constant. So this ratio will equal 1 when the top is full (bottom empty) and 0 when the top is empty (bottom full). It's a "time remaining" meter, or more accurately, a "proportion of grains remaining" meter. Assuming there is sand in the hourglass, this expression will never cause a divide by zero. --KSmrq^T 06:03, 2 July 2006 (UTC)[reply]

Such that we might make a direct substitution in the equation below? ~~(Forgive me in the interest of time for not first working the answer out for myself.)~~

$t={\frac {1}{\lambda }}{\times }{\ln \left(1+{\frac {D}{P}}\right)}$

...IMHO (Talk) 06:11, 2 July 2006 (UTC)[reply]

$t={\frac {1}{\lambda }}{\times }{\ln \left({\frac {1}{\frac {P}{P+D}}}\right)}$

...IMHO (Talk) 06:36, 2 July 2006 (UTC)[reply]

~~What I am looking for is an equation that will render the same proportional values for t yet tolerate a value of zero for P.~~ ...IMHO (Talk) 07:05, 2 July 2006 (UTC)[reply]

table

**Hour Glass**
Grains of Sand
Chambers		Proportion			t
Upper	Lower	form	inverse		form	inverse
P	D	1+(D/P)	P/(P+D)	λ	(1/λ)*ln(1+(D/P))	(1/λ)*ln(P/(P+D))
20	0	1	1	0.0121	0	0
19	1	1.05263157894737	0.95	0.0121	4.23911523864054	-4.23911523864054
18	2	1.11111111111111	0.9	0.0121	8.70748063287821	-8.70748063287821
17	3	1.17647058823529	0.85	0.0121	13.4313164874194	-13.4313164874194
16	4	1.25	0.8	0.0121	18.4416158110917	-18.4416158110917
15	5	1.33333333333333	0.75	0.0121	23.7753778885769	-23.7753778885769
14	6	1.42857142857143	0.7	0.0121	29.4772680941101	-29.4772680941101
13	7	1.53846153846154	0.65	0.0121	35.6018938919384	-35.6018938919384
12	8	1.66666666666667	0.6	0.0121	42.2169936996686	-42.2169936996687
11	9	1.81818181818182	0.55	0.0121	49.4080165913736	-49.4080165913736
10	10	2	0.5	0.0121	57.2848909553674	-57.2848909553674
9	11	2.22222222222222	0.45	0.0121	65.9923715882456	-65.9923715882456
8	12	2.5	0.4	0.0121	75.7265067664591	-75.7265067664591
7	13	2.85714285714286	0.35	0.0121	86.7621590494775	-86.7621590494775
6	14	3.33333333333333	0.3	0.0121	99.5018846550361	-99.5018846550361
5	15	4	0.25	0.0121	114.569781910735	-114.569781910735
4	16	5	0.2	0.0121	133.011397721826	-133.011397721826
3	17	6.66666666666667	0.15	0.0121	156.786775610403	-156.786775610403
2	18	10	0.1	0.0121	190.296288677194	-190.296288677194
1	19	20	0.05	0.0121	247.581179632561	-247.581179632561
0	20	#DIV/0!	0	0.0121	#DIV/0!	#NUM!

Further discussion

When a function diverges, you can't fix it by rearranging the way it's written. Probably the math is trying to tell you something. What is the application for all this? Melchoir 02:42, 3 July 2006 (UTC)[reply]

The inverse column does at least show that the divide by zero error is a number too large (probably infinity) to be calculated which tells me that P has attained a value of zero. (The issue is important because others have disputed this and think that P can never reach zero which is obviously not true for grains of sand or any microscopic, but whole nonetheless objects, that are similarly undivisible). What the math is telling me is it is the value of t that is indeterminate when P equals zero rather than the value of P being indeterminate. ...IMHO (Talk) 04:32, 3 July 2006 (UTC)[reply]

Okay... have you considered the possibility that these equations don't appropriately model the behavior of sand grains; and that nothing should be concluded about sand from examining them, anyway? Melchoir 05:01, 3 July 2006 (UTC)[reply]

Absolutely and in fact my reason for bring this matter to the attention of the mathematics desk. I am looking for a mathematical model of an hour glass. ...IMHO (Talk) 05:24, 3 July 2006 (UTC)[reply]

Oh! Well, the underlying microscopic dynamics aren't well-understood,[1] but on a large scale, the flow of sand through an hourglass proceeds at a constant rate, regardless of the amount of sand in the upper bulb.[2] Melchoir 06:07, 3 July 2006 (UTC)[reply]

That is true in terms of the rate of flow as the equations and value of t would suggest. In the formula I am looking for the proportional difference between the upper and the lower bulbs would have significance and be exponential but not related to or based upon a rate of flow but rather only on the proportional difference in the number of grains of sand in each bulb even if the hour glass fell on its side for a couple of hundred years. ...IMHO (Talk) 08:32, 3 July 2006 (UTC)[reply]

What are you talking about? Here's a formula:

D=rt.\,\!

Here's another formula for the uptight:

D(t)={\mbox{mid}}(0,rt,P(0)).\,\!

Given a sufficient reserve of creativity and time, I'm sure you could also work some proportional differences and exponents in there, but why would you even want to? Melchoir 08:49, 3 July 2006 (UTC)[reply]

Actually having now just plotted the values for t using a line graph I may have found the actual solution to the problem. Instead of needing a new equation what it appears that I need (and which I speculated about in an earlier discussion) is simply the need for integer versus decimal calculations or presentations of the results of the calculations. In other words if you do a line graph then you have a nasty line for the plot of t that is returning to zero from the value of 247.58118. If instead you use a bar graph instead of a line graph then the aesthetics prevent the viewer from having a misconception. ...IMHO (Talk) 10:15, 3 July 2006 (UTC)[reply]

Okay, I've had a look at your contributions to see what you're really up to, and it has nothing to do with sand. In fact, it has little to do with even its stated topic. Just don't put it into a Wikipedia article. Melchoir 18:46, 3 July 2006 (UTC)[reply]

LOLFDAC! Well after another effort to manipulate the equation so that it would not end up with a divide by zero error or an "infinite ratio" (by adding 1 to P and other compensating measures) I again came to the conclusion that the problem is not the equation but rather the misinterpretation of the results that P can never reach zero. In fact I believe now that the occurrence of a divide by zero error or an "infinite ratio" is necessary to provide a reliable indication that the value of P has in fact reached zero, i.e., that there are no grains of sand left in the upper bulb if or so long as these results are not misinterpreted.

I now realize that the basis for misinterpretation is that if P reaches zero then the value of t would be infinite (or at least greater than the time since the Big Bang) which would suggest a lower limit of

P=1.25*10^{-9}\

where

P={\frac {D}{\exp {\left({t*\lambda }\right)}-1}}

such that the value of t in the equation

t={\frac {1}{\lambda }}{\times }{\ln \left(1+{\frac {D}{P}}\right)}

does not exceed the amount of time since the Big Bang or approximately 16 billion years. Otherwise we should prehaps question the validity of all equations that produce a historical date which is or can be farther back in time than the date the universe is determined to have been created.

Rather than serving as the basis for a Wikipedia article this realization will instead allow me to determine how large the grain of salt must be that is taken with the Wikipedia articles I read. ...IMHO (Talk) 22:20, 3 July 2006 (UTC)[reply]

July 2

Summation with additional condition in Maple

I'd like to compute

\sum _{x=1 \atop gcd(x,n)=1}^{n}f(x)

for some positive integer n and some function f in Maple. How do I do this? Mon4 21:25, 2 July 2006 (UTC)[reply]

Could you clarify the sum? Do you mean sum from x=1 to n only for x relatively prime to n? And are you looking for a numerical solution or a closed form solution? I expect that a closed form solution does not exist (could be wrong). I would suggest not trying to use the sum command and just writing a for loop that first checks values from 1 to n, and then keeps a running sum for values where the gcd is 1. 128.197.81.223 23:30, 2 July 2006 (UTC)[reply]

I think the summation is of the values of the function which have, as a parameter, a prime number within the range [1,n) (since n cannot be used, because

gcd(n,n)=n

). I don't know Maple, so what I say may not work, but try creating a list or vector of prime values within that range, and using that as your value set in the sum function. --JB Adder | Talk 00:33, 3 July 2006 (UTC)[reply]

You could try the Iverson bracket trick. Create a function say g(x,n) that is 1 if its arguments are coprime and 0 otherwise. Then calculate the sum from 1 to n of f(x)g(x,n). Dysprosia 00:33, 3 July 2006 (UTC)[reply]

I don't know a really clever way to do it. If n is not very large, use

                        add(f(i),i in select(x->gcd(x,n)=1,{$1..n}));

McKay 02:51, 3 July 2006 (UTC)[reply]

Replacing Pi with another number

What would be the effect of changing Pi to another rational? Would a circle change its appearance?

First of all, pi is irrational, so it doesn't make sense to say "another rational". Second of all, the question as a whole doesn't make sense. It's like asking "What would be the effect of changing 2 to another integer?". However, check out Non-Euclidean geometry for examples of spaces in which the circumference of a circle is no longer pi times its diameter (although pi is still pi). —Keenan Pepper 22:52, 2 July 2006 (UTC)[reply]

The question is meaningless, revealing fundamental confusion. As noted, π is not rational. More to the point, π is defined as the ratio of circumference to diameter of a circle in Euclidean space; its value is fixed by the laws of mathematics.

If we feel creative, we can change what it means to be a circle, but that will not change π. For example:

One way to define a circle is as the locus of points equidistant from a fixed point, the center. The Euclidean definition of distance says that a point with Cartesian coordinates (x,y) is at a distance from the origin of (|x|²+|y|²)^1/2. We can replace each "2" by "p" to give different ways to measure distance, leading to a family of different geometries and different shapes for a "circle".
We can use spherical geometry instead of Euclidean plane geometry. Then the equator is a "circle" around the North Pole. On a unit sphere the "circumference" of such a circle is 2π and the "diameter" is π, so the ratio is 2. With different circles we can make the ratio as small as we like.

These investigations are entertaining, but they do not change the definition nor the value of π. --KSmrq^T 23:22, 2 July 2006 (UTC)[reply]

I believe this person is defining pi as the perimeter/circumference of any shape over 2 times its radius (I suppose the shape would have to be a regular polygon). For example, the "pi" of a square is

2{\sqrt {2}}

. The more sides the polygon has, the closer the "pi" gets to the constant pi. --Tuvwxyz 17:33, 3 July 2006 (UTC)[reply]

It could be that the person was just wondering what the universe would be like if fundamental constants had different values. Whilst pi isn't technically a fundamental physical constant, near enough is good enough, so maybe they are saying "What if the fundamental constants that appear to be responsible for the universe working how it is had values different to those they have now?" To which the answer is, of course, who knows? Your guess is probably as good as anybody's. It's a bit hard to imagine a universe in which circles had circumferences 10 times their radius (π = 5), but good luck.

Another point, a circle's appearance is not defined by pi, pi is difined by the circle. If a circle's appearance changed at all, it would simply no longer be a circle. Also, pi cannot exist in polygons, because it has infinite different radii at different angles. I think you meant a ratio of perimiter to the distance between the center and the point farthest away from the center. -- He Who Is^{[ Talk ]} 04:17, 9 July 2006 (UTC)[reply]

Physics and calculus together, hooray!

I'm trying to find the total amount of energy lost when a block travels completely around a frictional loop-the-loop. What I've been albe to figure out so far is below (illustrated here):

N=m(v^²/r + g cos θ)

E lost to friction = r∫_₀^{^θ} N μ_{_k}

½ m (v_₀)^² = ½ m v^² + mgr(1 - cos θ) + r∫_₀^{^θ} [(v^²/r + g cos θ) m μ_{_k}] dθ

½ (v_₀)^² = ½ v^² + gr(1 - cos θ) + μ_{_k} r∫_₀^{^θ} (v^²/r + g cos θ) dθ

½ v^² = ½ (v_₀)^² - gr(1 - cos θ) - μ_{_k} r∫_₀^{^θ} (v^²/r + g cos θ) dθ

[v(θ)]^² = (v_₀)^² - 2gr(1 - cos θ) - 2μ_{_k} r∫_₀^{^θ} ([v(θ)]^²/r + g cos θ) dθ

...

help? Daniel Olsen 22:59, 2 July 2006 (UTC) [reply]

Hopefully someone will check my math, because I'm honestly surprised with the simplicity of the result, but here we are: start over entirely and notice that $E={\frac {mv^{2}}{2}}$ so $N={\frac {2E}{r}}+mg\cos \theta$ . You're correct that $dE=-N\mu rd\theta$ (although you chose a positive sign for energy lost), so we have $E(\theta )=-\mu r\int _{0}^{\Theta }\left({\frac {2E(\theta )}{r}}+mg\cos \theta \right)d\theta =-2\mu \int _{0}^{\Theta }E(\theta )d\theta$ , since the cosine integrates to 0. Then differentiate both sides with respect to θ: ${\frac {dE}{d\theta }}=-2\mu E$ , so (from basic DEs) $E=E_{0}e^{-2\mu \theta }$ , so (using $\theta =2\pi$ ) $\Delta E=E_{0}(e^{-4\mu \pi }-1)$ (multiply by -1 to get energy lost). Hope this helps. --Tardis 20:42, 3 July 2006 (UTC)[reply]
Actually, let me go back on this a bit. $E={\frac {mv^{2}}{2}}+mgh=m\left({\frac {v^{2}}{2}}+gr(1-\cos \theta )\right)$ , as you implied, taking $h=0$ when $\theta =0$ . Moreover, I can't say that the cosine integrates out for all θ; I was thinking too much about the eventual $\theta =2\pi$ . Done more carefully, we get ${\frac {dE}{d\theta }}=\mu (mgr(2-3\cos \theta )-2E)$ , which is a lot harder to solve. It might be possible to salvage this approach via a clever substitution, possibly involving the solution to the simpler (wrong) equation; alternatively, one might attack the equations of motion more directly, but I find $\theta ''+\mu \theta '^{2}-{\frac {g}{r}}(\mu \cos \theta +\sin \theta )=0$ , which looks no friendlier (being nonlinear and all). Ideas, anyone? --Tardis 22:41, 3 July 2006 (UTC)[reply]

July 3

Semi-factorial functions

Working on the problem about pairing up people, I figured out that the number of possible sets of pairs of 2n people is f(1) = 1, f(n) = (2n - 1) * f(n - 1), or alternately for n people, f(2) = 1, f(n) = (n - 1) * f (n - 2). This is similar to the factoral function, but I haven't found a way to refactor it in terms of factorial. Is there a name for this?--Prosfilaes 06:25, 3 July 2006 (UTC)[reply]
Your function is basically f (n) = (n + 1)!!, where "!!" is a double factorial - see the factorial article. It can also be given using factorials:
$f(2n)=(2n+1)!!={\frac {(2n+1)!}{2^{n}n!}}$

-- Meni Rosenfeld (talk) 06:33, 3 July 2006 (UTC)[reply]

graph coloring and maximal clique

Is there an undirected graph for which the chromatic number exceeds the maximal clique size by more than one?
--Henning
Yes. Mycielski proved in 1955 that for every $k\geq 1$ there is a graph with chromatic number k that contains no triangle subgraphs, that is, whose maximal clique size is just 2. I'll make a drawing of such a graph with chromatic number 4 in a minute. —Bkell (talk) 10:24, 3 July 2006 (UTC)[reply]
File:Fourth Mycielski graph.svg —Bkell (talk) 10:34, 3 July 2006 (UTC)[reply]
Mycielski's proof is actually a constructive proof, so you can use it to make a graph with as large a chromatic number as you like with a maximal clique size of only 2. —Bkell (talk) 10:43, 3 July 2006 (UTC)[reply]
J. Mycielski. Sur le coloriage des graphes. Colloq. Math., 3:161–162, 1955. —Bkell (talk) 10:51, 3 July 2006 (UTC)[reply]
Thanks a bunch! :)

Base

how to calculate the value of 678 to the base of 7?
First make a list of powers of 7 until you get to a number larger than your starting value (678). 7⁰ = 1, 7¹ = 7, 7² = 49, 7³ = 343, 7⁴ = 2401. Take the largest power of 7 less than 678, 7³, and subtract it repeatedly until you get a value less than 7³: 678 - 343 = 335, so 678 = 1*7³ + something else. Continue with each smaller power of 7 until you get to 0. 335 - 6*7² = 41. 41 - 5*7¹ = 6. 6 - 6*7⁰ = 0. So in base 7, 678 = 1656 = 1*7³ + 6*7² + 5*7¹ + 6*7⁰. 128.197.81.223 17:05, 3 July 2006 (UTC)[reply]
We're assuming 678 is written in base 10, yes? The above method is fine; here's another.

(Let's work with different numbers so we don't do your homework for you. We'll convert 2097₁₀ to base 5.)
Divide 2097 by 5, obtaining quotient 419 with remainder 2.

Divide 419 by 5, obtaining quotient 83 with remainder 4.

Divide 83 by 5, obtaining quotient 16 with remainder 3.

Divide 16 by 5, obtaining quotient 3 with remainder 1.

Divide 3 by 5, obtaining quotient 0 with remainder 3.

The quotient is zero; stop dividing.

Assemble the remainders left-to-right from last to first:
2097₁₀ = 31342₅.

For computer implementation with very large numbers, more sophisticated algorithms are available.

When fractional components are involved, scaling and termination become issues; see Burger and Dybvig's paper, "Printing Floating-Point Numbers Quickly and Accurately." [3] --KSmrq^T 18:49, 3 July 2006 (UTC)[reply]

Investor Help

I am doing a project on the different types of investments. I was hoping that you could tell me the average growth and decay rate of stocks over 20-50 years. I also need the equation of this, as well as the future projected growth rate for 10-50 years. Please help me!
Thank You, HS
PS If it's not too much trouble, I would also like the same information on gold, silver, bonds, treasury bills, money market funds, mutual funds and antiques and commodities.
Since there are a lot of stock this is a difficult question to answer: there are stock that haven't seen much movement over the last 20 years while others have fluctuated greatly. But say you're interested in the major stock dealt on the New York Stock Exchange, it might be worth looking at the Dow Jones Industrial Average (DJIA). A chart of it shows that it has mainly grown over the last 20-50 years: Chart of DJIA performance (1928-present). Of course this isn't completely fair and for a real analysis of this you probably have to take into account the inflation that has occured over the same period.
The same applies in more or lesser degree to gold, silver, bonds, treasury bills, money market funds, and mutual funds. With antiques it is even more difficult to say something objective, as the kinds of antiques differ greatly and determining what is antique is a quite subjective business. -- Koffieyahoo 02:46, 4 July 2006 (UTC)[reply]
To get the average growth rate of Dow Jones, you can fit an exponential curve to its graph. That is, you can use linear regression to estimate slope of graph of log(DJ price) versus time. (Igny 04:11, 4 July 2006 (UTC))[reply]

July 4

No new questions were asked on 4 July 2006. Road Wizard 00:45, 5 July 2006 (UTC)[reply]
Wow! --Yanwen 01:57, 5 July 2006 (UTC)[reply]
Please use the search box to see if we already have an article on the subject of "questions not being asked on 4 July 2006". If you have no luck, you're welcome to come back here, and I suggest you rephrase your communication in the form of a question. But if we suspect it's a homework question, you're likely to get short shrift. :--) JackofOz 07:43, 5 July 2006 (UTC)[reply]
I tried searching for an article but couldn't find one. On a more serious point though, could you suggest a better way of noting that no questions were asked on a certain date? Having an entirely blank section either here or in the archives could cause someone to waste time checking if vandalism has occurred. :) Road Wizard 07:48, 5 July 2006 (UTC)[reply]
Perhaps we could create a pastel box for that purpose ☺. --cesarb 18:51, 5 July 2006 (UTC)[reply]

July 5

intravital pressure unit of measurement

We would like to know if the table below concerning the unit of measure are still valild or has been changed:
Japan USA Germany Holland France UK Sweden Australia Partial pressure of oxygen in arterial blood MmHg,Torr mmHg,Torr mmHg(kPa)kPa - kPa kPa mmHg Partial pressure of carbon dioxide in arterial blood MmHg,Torr mmHg,Torr mmHg(kPa) kPa - kPa kPa mmHg Encephalon liquid pressure mmH2O mmH2O mmHg mmH2O - mmHg (mmHg) mmH2O Intracranial pressure mmHg mmHg mmHg mmHg - (mmHg) (mmHg) mmHg Intraocular pressure mmHg mmHg mmHg mmHg - mmHg (mmHg) mmHg Central venous pressure cmH2O cmH2O cmH2O cmH2O - cmH2O (mmHg) cmH2O or mmHg Inner pulse pressure cmH2O cmH2O mmHg cmH2O - (mmHg) (mmHg) cmH2O Pressure of rectum & anus cmH2O cmH2O cmH2O cmH2O - (mmHg) (mmHg) cmH2O Intravesical pressure cmH2O - cmH2O cmH2O - (mmHg) (mmHg) cmH2O or mmHg Urethral pressure cmH2O cmH2O cmH2O cmH2O - (mmHg) (mmHg) cmH2O
You might have better luck posting this question to the Science Desk. StuRat 22:00, 6 July 2006 (UTC)[reply]

Calculating standard deviation

I was wondering if someone could explain to me the reason for using squared numbers in calculating standard deviation (mean distance from the mean).
Taking the example in http://en.wikipedia.org/wiki/Standard_deviation, we have 4 data points (5 6 8 & 9) with a mean of 7. If standard deviation is (as I understand it) the average distance from the mean, why is the calculation not simply 5 = 2 units from the mean 6 = 1 unit from the mean 8 = 1 unit from the mean 9 = 2 units from the mean totalling 6 units from the mean, divided by 4 data points = average 1.5 units from the mean? (rather than standard deviation which is 1.5811)?
Many thanks in advance TS
There are at least 2 reasons for that:
In many practical cases where the dispersion has some importance (quite often, a negative effect), the strength of the effect is greater when there are a few large changes than when there are many small changes. That is, in such a scenario, the data points (10, 10, 8, 12), where there are a few large changes, will have a worse effect than the data (9, 9, 11, 11), where there are many small changes. Squaring the deviations models this phenomenon more accurately, since larger deviations will have larger weights.

Since all the terms we add should be positive (otherwise they will cancel each other out), if we don't use squaring, we will have to use absolute values. The absolute value function is not differentiable, therefore expressions involving it are difficult to develop analytically. This limits our ability to build a statistical theory around this measure of dispersion, and ultimately we will not be really able to use it effectively.

This is why the most common measure of dispersion in usage is the standard deviation - it gives very reasonable results, and is easy to work with. -- Meni Rosenfeld (talk) 15:44, 5 July 2006 (UTC)[reply]
The quantity you mention is called the mean deviation. It is very rarely used because the standard deviation is much better behaved mathematically. One example is that the quantity $\sum (x_{i}-y)^{2}$ is minimised when y is the mean. However, $\sum |x_{i}-y|$ does not have a unique minimum - consider two values to see why. McKay 16:09, 5 July 2006 (UTC)[reply]

0 (Zero)

Me and my friends were sitting in class doing homework a couple months ago (schools out) and there was a question about the number 0. We quickly got through it and went on. One of my friends later presented something rather interesting to the rest of us. What he was saying was that there are infinite positive numbers and infinite negative numbers, therefore, 0 is the middle and therefore,
$\infty /2=0$
That means, following simple multipulcation and division rules, that
$0*2=\infty$
But we all know that any number times 0 equals 0. Is there a name for this paradox. I can't find anything doing a quick Google search that relates to my question. Thanks. schyler 19:39, 5 July 2006 (UTC)[reply]
Before even considering the correctness of "zero is in the middle": if it is in the middle, it doesn't follow that $\infty /2=0$ , only that $(-\infty +\infty )/2=0$ (assuming you could do that; it's an undefined operation, see infinity for more info). --cesarb 19:52, 5 July 2006 (UTC)[reply]
Yeah, that "in the middle" stuff is wrong. Zero is also "in the middle" of −1 and 1, but that doesn't mean 1 / 2 = 0. If you want a real puzzler, check this out: $\infty ={1 \over 0}={1 \over -0}=-{1 \over 0}=-\infty$ —Keenan Pepper 01:50, 6 July 2006 (UTC)[reply]
Exactly. It's a paradox, right? Is there a name for it? schyler 03:33, 6 July 2006 (UTC)[reply]
No, there is no paradox, only a little fun at your expense.

But as long as we're determined to muddy the waters, we might as well meet some higher mathematics. Before we go any further, we must understand that infinity is not a standard integer, nor rational number, nor real number. We cannot get it as an answer, and we cannot use it in a computation. Since we're about to do both of these things, we're leaving the familiar numbers behind.
In projective geometry, great simplification of theorems and proofs is possible by using "points at infinity". Thus we can define a projective line, where each point is represented by a pair of coordinates, essentially treated as a ratio. Points on the usual (non-projective) line might take the form (x:w), with w ≠ 0, while the "point at infinity" takes the form (x:0). Similarly, the projective plane has, not just a point, but a line at infinity. Points are ratios (x:y:w), and the equation of the line at infinity is w = 0. Computer graphics makes heavy use of projective space, with coordinates (x:y:z:w). As an example of a simplification so obtained, in the projective plane any two distinct lines intersect in a unique point, with parallel lines intersecting at a point at infinity. However, the topology of a projective line is the same as a circle, and that of a projective plane more closely resembles a sphere. On the projective line we can define the reciprocal of every number (x:w) to be (w:x), so that 1/0 = ∞.

In the complex plane, we often use a one-point compactification, adding a single point at infinity. This gives something different from the projective plane, which has multiple points at infinity. This new topology is exactly that of a sphere. One of the fun things we can do in the extended complex plane is a Möbius transformation. For example, we can turn a circle "inside out", which has the curious property that circles and lines remain circles and lines (though a circle may become a line and vice versa), and angles are unchanged. Inversion in a circle exchanges the center of the circle with infinity, so we again have a sense in which the inverse of zero (the complex number 0+i0) can be infinity.

In non-standard analysis, sophisticated theorems from mathematical logic are used to embed the standard real numbers in a larger non-standard model that includes many infinities as well as their reciprocals, infinitesimals. Unlike the projective line, there is no "wrapping around", and we prohibit taking the reciprocal of zero. Although we again have multiple infinities, this is also quite different from the projective plane.

In the IEEE floating-point standard (IEEE 754), numerical algorithms are simplified by the inclusion of distinct values representing positive and negative infinity. It also attaches a sign to zero, and includes different kinds of "not-a-number" (NaN). The standard includes rules for arithmetic with these values. This is arithmetic specifically designed for computer use; it is not the mathematicians' idea of real arithmetic.

These are just a few of the diverse possibilities for giving infinity, or infinities, a formal meaning, and of interpreting 1/0. --KSmrq^T 06:33, 6 July 2006 (UTC)[reply]
You've left out Conway's surreal numbers which contain some infinitely biggies too. -lethe ^{talk +}
I got tired of writing! I left out cardinals, and topos-based reals, and … You might say I was exhausted, rather than exhaustive! :-) --KSmrq^T 08:27, 6 July 2006 (UTC)[reply]
To clarify: Your original argument had no paradox, just a fallacy. There's no reason at all why ∞ / 2 should be 0. About Keenan's ∞ = -∞ argument: No paradox there either. It just shows that division by zero can only be sensibly defined in structures where ∞ = -∞ - For example, the real projective line and Riemann sphere which KSmrq mentioned. -- Meni Rosenfeld (talk) 08:31, 6 July 2006 (UTC)[reply]

Your friend's logic is wrong. --Proficient 09:41, 6 July 2006 (UTC)[reply]
There is another possibility to consider. Since infinitesimal is the inverse of infinite and is "a number that is smaller in absolute value than any positive real number" and ${\begin{matrix}{\frac {1}{10}}&=&.1;\quad {\frac {1}{10+{\frac {1}{10}}}}&=&\!\!\!\!\!.09900990099...;\qquad \qquad \qquad \qquad \qquad \\{\frac {1}{100}}&=&.01;\quad {\frac {1}{100+{\frac {1}{100}}}}&=&\!\!\!.0099990000999900009999...;\qquad \qquad \quad \\{\frac {1}{1000}}&=&.001;\quad {\frac {1}{1000+{\frac {1}{1000}}}}&=&.000999999000000999999000000999999...;\end{matrix}}\,\!$

it follows that ( $\iota \,\!$ is infinitesimality or iota) ${\frac {1}{\infty }}=\iota ;\quad -{\frac {1}{\infty }}=-\iota ;\quad 0={\sqrt {{\frac {1}{\infty +\iota }}\times {\frac {1}{-\infty -\iota }}}}={\frac {i}{\sqrt {\infty +\iota }}};\,\!$

This solution takes into account the +/- dilemma! P=) ~Kaimbridge~16:30, 6 July 2006 (UTC)[reply]

Proof?

How does one prove that if 1) a≠0 2) b≠0 3) a+b=ab, then a or b =2? Are those the only possible values? --Tuvwxyz 23:54, 5 July 2006 (UTC)[reply]
As a first step, one might try subtracting a from equation (3). I would also note that the assertion is only true for a and b integers, and that the reference desk is not a place for homework problems. Tesseran 01:34, 6 July 2006 (UTC)[reply]

There are infinitely many possible values. —Keenan Pepper 01:46, 6 July 2006 (UTC)[reply]
A proof would depend on how much mathematics is familiar, and on whether a and b are supposed to be integers, or real numbers or integers modulo n or whatever. Here's a strategy for the integers: Let a be some value other than 2, such as 3. Consider possible values of b and look for a systematic argument about their success or failure. Then take another value for a and do the same thing, and so on. It should not take too many experiments to begin to understand the possibilities. Now turn that understanding into a proof. A strategy for reals might be different, taking advantage of algebraic manipulation not available with integers. This is such an elementary problem that the most valuable thing we can do for you is encourage you to solve it for yourself. If you get stuck formalizing a proof, write down for us your informal understanding and we can see how you're doing. --KSmrq^T 01:53, 6 July 2006 (UTC)[reply]
Almost forgot: working with integers modulo 9, two solutions are a = 5, b = 8 and a = 3, b = 6. (Challenge: Are there others?) But I suggest you not mention these in a homework solution, as it will be obvious that you didn't do the work yourself. :-) --KSmrq^T 03:31, 6 July 2006 (UTC)[reply]

Here's an approach. First, substitute x=a-1, y=b-1 to get a simpler equation in x and y. It should be obvious that the only integer solutions to this simpler equation are (x,y) = (-1,-1) or (1,1). Convert back to (a,b) values and you are done. Gandalf61 13:17, 6 July 2006 (UTC)[reply]

July 6

What is the "-" Symbol called as?

We use "-" sign to get the negative of a number. i.e 20 can be made as (-)20. What do we call this - sign. I just know that this is not called as "minus sign". Could you please answer my doubt?
I call it a minus sign. So do typographers. It is very similar to a hyphen, and often the same character is used. Notinasnaid 11:58, 6 July 2006 (UTC)[reply]

It is a minus sign. That's its official name in unicode too. McKay 11:59, 6 July 2006 (UTC)[reply]

Typographically speaking, the minus sign (−) is different from the hyphen (-), and both of these are different from the en dash (–) and the em dash (—); see dash. The minus sign is usually the same width as the plus sign (+), because math looks better that way. —Bkell (talk) 14:27, 6 July 2006 (UTC)[reply]

It's also called the unary minus, to distinguish it from the binary minus (we seem to be missing an article or redirect on unary minus and binary minus, any takers?). --cesarb 17:02, 6 July 2006 (UTC)[reply]
Please, when giving examples of characters, also provide the Unicode codepoint or HTML named entity or some such. Why? Because the appearance depends on the typeface, so may communicate nothing. For example, in a font like Courier there is little or no visible difference:

- U+002D HYPHEN-MINUS -

‐ U+2010 HYPHEN ‐

‒ U+2012 FIGURE DASH ‒

– U+2013 EN DASH –

— U+2014 EM DASH —

− U+2212 MINUS SIGN −
This possible confusion would not be obvious to a person using a font like Helvetica or Arial:

- U+002D HYPHEN-MINUS -

‐ U+2010 HYPHEN ‐

‒ U+2012 FIGURE DASH ‒

– U+2013 EN DASH –

— U+2014 EM DASH —

− U+2212 MINUS SIGN −
And depending on the character and the font, the only visible mark might be a missing character symbol. So be kind.

Back to the topic at hand, Unicode distinguishes three characters: the hypen-minus (the old ASCII character which does double duty), the hyphen (which seems rarely used), and the minus (which we prefer in Wikipedia mathematics). --KSmrq^T 20:27, 6 July 2006 (UTC)[reply]
don't forget that some elementary/algebra textbooks (the ones we used) like to put the minus at the top of the line, like the bar in T, like T7, without the serifs or the vertical part, so that it doesn't look at all like a -. Thus 5 + T7 doesn't look like 5 + -7 at all. Again, that's not a T, just my representation, just look at the top of the T and ignore the serifs. Maybe "T minus | sans serif". Gives a whole new meaning to t minus. :) 82.131.188.84 20:30, 6 July 2006 (UTC).[reply]
If you are speaking of the distinction between negation and subtraction, probably the notation you want is "9 + ¯4", as opposed to "9 − 4". (The APL programming language also makes such a distinction, and the character used seems to be U+00AF, MACRON, as shown here.) --KSmrq^T 23:22, 6 July 2006 (UTC)[reply]
I suppose it could be either the macron, or the overline:

¯ U+00AF MACRON ¯

‾ U+203E OVERLINE &oline;
depending on how long you want it to be.-gadfium 03:20, 7 July 2006 (UTC)[reply]
You suppose wrong, at least insofar as APL is concerned. This is not a guessing game based on appearance; the Unicode standard defines the codepoint used for macron as equal to the APL overbar. But if you are not concerned about APL, the standard defines nothing specific for unary minus (that I know of).

Frankly, Unicode is confusing for mathematics, because it associates specific meanings with codepoints, but mathematicians do not always use symbols so consistently. --KSmrq^T 04:26, 7 July 2006 (UTC)[reply]

July 7

Set theory and epistemology

Hi guys... I was wondering why is set theory said to be 'beyond logic' can someone explain this to me please? Is it because logic itself is a set? is it because anything that can be, is a set or is in a set?...but then again, isn't that part of our logic?...unless the subject isn't settled yet...like I think it is.--Cosmic girl 19:43, 7 July 2006 (UTC)[reply]
Where is your reference that set theory is said to be 'beyond logic?' ...IMHO (Talk) 19:47, 7 July 2006 (UTC)[reply]
I would say that set theory is not reducible to logic (in the sense, say, that Frege thought it was). That is, it's synthetic rather than analytic. Is that what you mean, or do you mean something stronger, say that set theory is not amenable to the methods of logic? --Trovatore 20:17, 7 July 2006 (UTC)[reply]
Here is my source: http://www.rbjones.com/rbjpub/philos/maths/faq031.htm I mean that set theory can aply to many kinds of logic... is that possible?...meaning thus, that set theory is way more general than logic.--Cosmic girl 02:23, 8 July 2006 (UTC)[reply]
(Corrected header spelling to "epistemology".) I could not find a statement that set theory is "beyond logic" on that page, only a question about the epistemological implications if that view is adopted. Seeing philosophers around mathematics or physics makes me nervous, like seeing small children playing with sharp knives. Games about foundations have little impact on most working mathematicians. We do know that we have a choice of which rock to stand on and which to pick up, so to speak. It is overly simplistic to claim that set theory is beyond logic, or that logic is beyond set theory. In fact, it is too simplistic to speak of either set theory or logic in the singular, because we have choices of set theories and choices of logics. Further confusing the discussion is the difference between how a philosopher conceives of logic and what a mathematician means.

What every trained mathematician does know is Gödel's incompleteness theorems. An "executive summary" is that there is a difference between truth and provability. (But see the article for details.) Some might thus consider arithmetic "beyond logic". --KSmrq^T 06:40, 8 July 2006 (UTC)[reply]
Do you mean by 'beyond logic' that logic is dependent on arithmetic?.--Cosmic girl 19:53, 10 July 2006 (UTC)[reply]

Number theory question

Anyone know if there's a proven non-trivial limit to the number of consecutive composites each a multiple of some prime <= p(n), expressed in terms of p(n)? This limit is certainly less than the primorial of p(n) and I can show (by a constructive method) that it is at least 2*p(n-1) - 1. I also know for at least some values of n it is somewhat larger than 2*p(n). It seems there might be some way to prove a limit of the form C*p(n) for some ridiculously large C. If anyone has seen anything like this, please let me know. -- Rick Block (talk) 20:07, 7 July 2006 (UTC)[reply]
hey, I probably can't help you, but out of curiostiy does consecutive just mean like 77, 78, 79, etc? How could each such composite number be the multiple of the same prime? Or do you mean "a multiple of some prime" but not the same one, in which case isn't the word "composite" enough to say that?? Thank you. 82.131.184.144 22:20, 7 July 2006 (UTC).[reply]
Yes, consecutive, but a multiple of any of the primes less than p(n). For example, for n=3 the question is how many consecutive numbers are there that are each a multiple of 2,3, or 5 (the answer is 5). There's a pattern to the arrangement of multiples and non-multiples of length primorial of p(n) (which gets large in a hurry). Using Π(p(n)) to mean the primorial, ( x * Π(p(n)) ) + 1 is not a mulitple of any of the primes up to p(n), so the trivial limit is Π(p(n)). -- Rick Block (talk) 22:59, 7 July 2006 (UTC)[reply]
I think it's always the next higher prime minus two, but I don't have a proof. —Keenan Pepper 03:01, 8 July 2006 (UTC)[reply]
Thanks, but I know it's not this. Consider [2,3,5,7,11]. There's a sequence of 13 starting at 114 which are all multiples of one of these primes. Given the set of primes through p(n) I can construct a sequence very much like this one of length 2*p(n-1) - 1. If you're curious, the construction is to pick a multiple of Π(p(n-2)) such that one less is a multiple of p(n-1) and one more is a multiple of p(n) (such an arrangement is guaranteed to exist by the Chinese remainder theorem). The chosen multiple of Π(p(n-2)) is the center of a sequence of length 2*p(n-1) - 1 which are multiples of at least one of the primes through p(n). -- Rick Block (talk) 18:12, 8 July 2006 (UTC)[reply]
Oh whoops, I misinterpreted you to mean all its prime factors are below a certian limit, when you mean just one of them. Right? —Keenan Pepper 21:02, 8 July 2006 (UTC)[reply]
So you're talking about sequence A058989, right? —Keenan Pepper 21:06, 8 July 2006 (UTC)[reply]
Yes, I'm talking about sequence A058989. I'm aware of the Weissman conjecture. I had independently conjectured this, and ultimately found the discussion on yahoo's primenumber group about it (including Phil Carmody's counterexamples). -- Rick Block (talk) 23:55, 8 July 2006 (UTC)[reply]
I don't know whether this counts as non-trivial, but I can shove the upper bound down some. Take p(n) and Π(p(n)) to be the areas I'm trying to find gaps in. max(a,b,c) will indicate the largest of what's in the parentheses, eg max(1,2,3)=3. ZSpan(x) and Span(x) will be as Phil Carmody described them. A "prime candidate" indicates a number that is not divisible by any p(x<=n).

1) First, the most obvious: p(n+1) is guaranteed to be fairly close to p(n). Certainly closer than Π(p(n)), but the tightest bound I've heard of puts at least one prime between any number x and 2x. Anyway, close, which cuts the gap between 2 and Π(p(n)) down a bit. So, Span(n) <= max(ZSpan(n), Π(p(n))-ZSpan(n)).

2) Of course, since Π(p(n)) is divisible by all p(x<=n), Π(p(n))-p(n+1) is not divisible by any, so Span(n) <= max(ZSpan(n), Π(p(n))-2*ZSpan(n)). (I'm ignoring the details of +-1, to get the idea across.)

3) Now, at the center of the full pattern, at Π(p(n))/2, there is a number divisible by all p(x<=n) except 2. Moving outwards from it a distance a in either direction, you'll find prime candidates fanned out at 2^{a^{for 1<2^a<2*p(n+1). I'll call the largest such a A, which is equal to floor(log₂p(n+1)). There are of course many prime candidates just beyond that, but they're much more difficult to locate. So, Span(n) <= max(ZSpan(n), Π(p(n))/2-ZSpan(n)-2^A, 2^a-1). The last is to take into account the largest gap in the fan. However, since it's defined as being less than p(n+1), I guess it can be left out in this case.}}

4) A similar system will work for any division of Π(p(n)) by one of its factors p(x<=n) (eg Π(p(n))/5, 2*Π(p(n))/5, 3*Π(p(n))/5, 4*Π(p(n))/5 or Π(p(n))/7, 2*Π(p(n))/7...6*Π(p(n))/7). The density of the fan around each decreases substantially, however, as p(x) increases, and for any p(x)>2 each of the p(x)-1 centerpoints given by this system is guaranteed to have at least one prime candidate exactly adjascent to it.

5) Getting into division by two primes at once (Π(p(n))/15, Π(p(n))/21, Π(p(n))/22 etc.) or more makes the fan significantly more dense in general, since a is at not only 1 (sometimes) and powers of each prime, but at combinations of their powers. However, as the number of primes gets higher, I can't quite guarantee there will always be a prime candidate right nearby. However, by following the trend downwards, it's pretty clear that although a multiple of a recent prime may not be the right answer, it's in the ballpark.

6) Now, I'd like to see if you can help with an extension of that. It seems to me that the multiples of p(n) that aren't multiples of p(x<n) tend to have quite a few other prime candidates between them. I'm not certain of that, though, and don't see how to prove it. If someone can, though, then any gap in n is the sum of at most two gaps in n-1, so Span(n)<=2Span(n-1). I tested out the theory that this is the case on the values of Span(x) I can find (2, 4, 6, 10, 14, 22, 26, 34, 40, 46, 58, 66, 74, 90, 100, 106, 118, 132, 152, 174, 190, 200, 216, 234), and the proportion of each to the one before is less than 2 (except for 4/2, which is equal to 2 and demonstrates my point nicely). In fact, the proportions seem to get closer to 1 the farther out I go, ending with 234/216≈1.08. This is consistent with what I would expect, but nowhere near proof. I eagerly await your response. Black Carrot 20:25, 11 July 2006 (UTC)[reply]
I numbered your points for reference.
1) That there is a prime between n and 2*n is Bertrand's postulate. There are tighter bounds that have been proven, I think current is close to between n and n + ~sqrt(n) (it's actually +n to some exponent a little larger than 1/2). And, yes, Span(n) <= max(ZSpan(n), Π(p(n))-ZSpan(n)), but ZSpan(n) is so small compared to Π(p(n)) that this really isn't much different from saying Span(n) <= Π(p(n)).
2) Agreed (but again, Zspan(n) << Π(p(n)) ).
3) I follow the fan out from Π(p(n))/2 with prime candidates at Π(p(n))/2 +/- 2^a (these are really numbers that are relatively prime to p(x<=n) ), but I think there are some notation issues here. Specifically, I think the "prime candidates" are at Π(p(n))/2 +/- 2^a for 1<= a <= A, where A is floor(log₂Π(p(n))/2)). I think this leads to a potential span of length 2^A - 2^(A-1), which I believe simplifies down to Π(p(n))/4. So I think we end up with Span(n) <= max(ZSpan(n), Π(p(n))/4). It's certainly the case that Π(p(n))/4 is far larger than Zspan(n), so I think we're at Span(n) <= Π(p(n))/4.
4) Agreed, and this leads to Span(n) <= Π(p(n))/p(n), i.e. Span(n) <= Π(p(n-1)), and some other more complicated (and smaller) terms which I don't have time at the moment to figure out. They're terms of the form p(i)^A - p(i)^(A-1) where p(i)^A is a piece of Π(p(n))/p(i).
5) In general, I think this means we can't really say anything about these cases. Specifically, if we divide Π(p(n)) by any two primes it might well be the case that the adjacent numbers are divisible by the two primes we've divided by.
6) I think that this boils down to a claim that there's always a "prime candidate" (with respect to primes p(x<=n+1) ) in any span of length 2*p(n+1). I don't know how to prove this (it may well be true). Note that even if this is true and, therefore, Span(n)<=2Span(n-1), we end up with Span(n) <= 2ⁿ. On the other hand, if there's always a "prime candidate" (with respect to primes p(x<=n+1) ) in any span of length 2*p(n+1), because p(n+1) <= 2*p(n) we have Span(n) <= 4*p(n).
If you'd like to continue this conversation (which I would be quite happy to do), I think we're distinctly in the realm of original research which means we should take it out of Wikipedia (my "email this user" link is enabled). The original question was "has anyone seen anything like this", and I'm starting to suspect the answer is no. -- Rick Block (talk) 15:19, 12 July 2006 (UTC)[reply]
BTW - there's a formula for the number of "prime candidates" in one iteration of the full pattern, which is ${\prod (p(i)-1)}$ from 1 <= i <= n. Since no more than two of these can occur in every 6 consecutive integers, and the pattern is symmetric about the midpoint, it's true that $span(n)<=({\prod p(i)})/2-3/2*{\prod (p(i)-1)}$ . -- Rick Block (talk) 16:49, 12 July 2006 (UTC)[reply]
Also note, although it is perhaps not particularly useful for a proof, the "average" span is ${\prod p(i)}/{\prod (p(i)-1)}$ , which grows pretty slowly (less than 9 for p(25) = 97). -- Rick Block (talk) 19:24, 12 July 2006 (UTC)[reply]

July 8

problem of two independent databases

There is a problem that teacher gave us at the university, and he claimed he has a very good solution (see below) for it, but unfortunately I was missing from the class when he explained that, so I don't know this solution, nor the name of the problem to find it on the internet (I tried).
The problem is: There are two identical databases each with n records and cannot communicate. Invent the most efficient way (in terms of number of records you need to read from databases) how to get a certain record from the database in such a way that database owners would get no information about what record did you retrieve.
The simplest way to do that is to download the whole database and look at the record you want (this will take n records to download). There is more efficient way though - if you consider records in database are ordered in the square. You can select a random subset of rows. Then ask the first database to give you the logical xor of all rows in your subset, and ask the second database also to give you the logical xor of all rows in your subset, except (or included) the row that you want. Then you can make logical xor of results and look at the correct column. The databases got no information though, because the subset is random and then the subset with/out one row is random too. So you can manage to do it in O(n^(1/2)) records retrieved.
The point is, our teacher claimed that he has solution with O(n^(1/3)). I thought about it a lot, but I really don't know and would like to know. Has anyone ever heard of this problem and could perhaps point me to this solution? Samohyl Jan 07:53, 8 July 2006 (UTC)[reply]

Is 0=2 ???

Is 0=2?Italic text''Consider x=1
Now (x-1)^2 = (x^2)-(1^2) Simplifying further we get (x-1)^2 = (x-1)*(x+1) [as (a^2)-(b^2) = (a-b)*(a+b)] If we take (x-1) to the LHS then it will cancel out with one (x-1) from the LHS thys we will get the equation x-1 = x+1
Substituting the value of x as 1 in the above equation we get
1-1 = 1+1 hence we get 0=2
See division by zero for an explanation. Isopropyl 08:01, 8 July 2006 (UTC)[reply]
Also, your first line is incorrect. $(x-1)^{2}=x^{2}-2x+1$

which gives 1=2 for your "answer". Isopropyl 08:03, 8 July 2006 (UTC)[reply]
Well, it's true that $(x-1)^{2}=x^{2}-1^{2}$ ... if and only if x=1, in which case we run into the division by zero problem in the last step. Generally though, $(x-1)^{2}\neq x^{2}-1^{2}$ , try some examples where x is any number but 1. -GTBacchus^(talk) 08:10, 8 July 2006 (UTC)[reply]
Thank u very much Isopropyl

DVD won't work!

I'm trying to rip a music DVD to put it on my ipod. I'm using PQ DVD and it comes up with an error.
It says "Invalid Disc Region. This DVD-Video disc cannot by played, because it is not authored to play in the current sysytem region. The region mismatch may be fixed by changing the system region (with DVDRgn.exe)."
I can't get it even to play in anything in my pc. Could you also give recomemdations to a good *free* DVD player program.
Thanks!
~Cathy T.~
Your DVD player must be set to the correct region in order to play your movie. This is done to reduce pirating. For example, if you have a region 3 DVD then it's a DVD that will play on region 3 players. Basically, if you get a region 0 disc then it'll play in any player. Most DVD firmwares will allow you to change your region 3 times before locking your DVD player into a region.
Also, a great free DVD player (and not to mention anything else player) is VLC Player. You can find this fantastic program right here
Hope this helps -Nitrodist 5:12 AM CST, July 8, 06
Sigh; please don't propagate myths. Region codes are a bizarre invention to control distribution, not piracy. For example, if I find a copy of Jacques Tati's delightful Les Vacances de Monsieur Hulot in France, I won't be able to play it on my North American player. Or if I take my perfectly legitimate DVD of W. C. Fields' classic The Bank Dick purchased in North America I won't be able to view it with a player whose region code is for Europe. It's implausible that either of these decades-old films is of interest to pirates. This abuse of digital rights management should be illegal (and is in some places), but purse strings control politicians, and politicians control laws. --KSmrq^T 12:12, 8 July 2006 (UTC)[reply]
So... how do I change it to work in PQ DVD? (I'm so confused! I just don't understand any of this technological computer stuff... hehe)
Thanks for the DVD player sugestion! I'm going to download it now! ~Cathy T.~

Euler-Mascheroni constant approximation

The article on the Euler-Mascheroni constant provides many different formulas equivalent to the constant. Does anyone know what formula is most commonly used by computer algebra systems and other programs in computing it? Or, more accurately, which formula converges fastest with the least computational intensity? -- He Who Is^{[ Talk ]} 15:40, 8 July 2006 (UTC)[reply]
The documentation for Mathematica says it uses the algorithm in
Brent, R. P. and McMillan, E. M. "Some New Algorithms for High-Precision Computation of Euler's Constant." Math. Comput. 34, 305-312, 1980.

For those who do not have ready access to that paper, a description of their method (and others) is available online, but the cited web page uses the deprecated "font face" markup with the Symbol typeface. Here's a wikified transcription of the relevant section:

A better method is based on the modified Bessel functions and leads to the formula
$\gamma ={\frac {A_{n}}{B_{n}}}-\log(n)+O(e^{-4n}),\,\!$

with
$A_{n}=\sum _{k=0}^{\alpha n}\left({\frac {n^{k}}{k!}}\right)^{2}H_{k},\qquad B_{n}=\sum _{k=0}^{\alpha n}\left({\frac {n^{k}}{k!}}\right)^{2},\,\!$

where α = 3.5911… satisfies α(log(α)−1) = 1.
This technique is quite easy, fast and it has a great advantage compared to exponential integral techniques : to obtain d decimal places of γ, the intermediate computations can be done with d decimal places.
A refinement can be obtained from an asymptotic series of the error term. It consists in computing
$C_{n}={\frac {1}{4n}}\sum _{k=0}^{2n}{\frac {[(2k)!]^{3}}{(k!)^{4}(16n)^{2k}}}.\,\!$
Brent and McMillan suggest that
$\gamma ={\frac {A_{n}}{B_{n}}}-{\frac {C_{n}}{B_{n}^{2}}}-\log(n)+O(e^{-8n}).\,\!$
This time, the summations in A_n and B_n should go up to βn where β = 4.970625759… satisfies β(log(β)−1) = 3. The error O(e⁻⁸ⁿ) followed an empirical evidence but the result had not been proved by Brent and McMillan. This formula has been used by Xavier Gourdon with a binary splitting process to obtain more than 100 millions decimal digits of γ in 1999.

Here, as usual, H_k denotes the k-th harmonic number, with H₀ = 0, and for positive k,
$H_{k}=1+{\frac {1}{2}}+{\frac {1}{3}}+\cdots +{\frac {1}{k}}.\,\!$

For example, n = 3 (k = 14) should give an approximation good to more than 10 decimal places, and the computations are relatively easy. One word of advice for those new to numerical methods: for better results sum the terms of a series from smallest to largest. --KSmrq^T 23:32, 8 July 2006 (UTC)[reply]
I wrote a naive implementation in Python of the Brent-McMillan formula a while back. It'll calculate a few thousands of digits or so of γ within reasonable time. I'm not sure how they managed to do it with only d digits internally, though, since there's a huge cancellation in (A/B) - log n. Choosing 2d digits seemed to work for my program; more detailed analysis would be welcome.
For very high precision, you'd need to use FFT multiplication and binary splitting of the series. There's also an iterative version, given in Borwein & Bailey, Mathematics by Experiment - Plausible Reasoning in the 21st Century, page 138: Choose n appropriately (details not given), set
$A_{0}=-\log n,B_{0}=1,U_{0}=A_{0},V_{0}=1,$

and for k = 1, 2, ... iterate
$B_{k}=B_{k-1}n^{2}/k^{2},A_{k}=(A_{k-1}n^{2}/k+B_{k})/k$

$U_{k}=U_{k-1}+A_{k},V_{k}=V_{k-1}+B_{k}$

until U and V don't change anymore. Then γ ≈ U/V. I haven't attempted to implement this yet. Fredrik Johansson 00:11, 9 July 2006 (UTC)[reply]
Thank you all. I feel for my purposes the iterative process from Fredrik will be most useful. -- He Who Is^{[ Talk ]} 02:33, 9 July 2006 (UTC)[reply]

Proof no one (who could achieve it) wants world peace?

1. Assume x can achieve world peace.
therefore (from 1):
1b The only way for x not to achieve world peace, is x must not want it. (Because if he wanted it but couldn't achieve it, 1 would be violated.)
2. X wants world peace.
2b it follows from 1b and 2 that "There is world peace".

However, there isn't world peace. Therefore, there is no x, ie there is no one who can achieve world peace and wants it, and also it follows that anyone who can achieve world peace doesn't want it, and anyone who wants world peace can't achieve it?
I think "there isn't world peace now" is pretty self-evident, but the conclusions in the last paragraph certainly aren't. Is there something wrong with my reasoning? Thank you. 82.131.185.76 16:48, 8 July 2006 (UTC).[reply]
What if x can achieve it, and wants to achieve it, but doesn't know it can achieve it? For example, replace "achieve world peace" with "eat food" and assume x is in a room with food, but the food is hidden. 128.197.81.223 17:32, 8 July 2006 (UTC)[reply]
The statement "No-one who could achieve world peace wants it" could indeed be true. —Ilmari Karonen (talk) 17:45, 8 July 2006 (UTC)[reply]
This makes (at least) one other false assumption: that achieving world peace is one person job, which it almost certainly isn't; imagine if 10 people in the world could each get 10% of the way to world peace, but they never met; although world peace would be achievable, it would never be reached. There are probably thousands of people who want and can help achieve world peace, but there will probably never be enough of them working together. Secondly, 2b does not logically follow from 1b and 2; what does follow is "x is working towards world peace". smurrayinchester^{(User), (Talk)} 17:49, 8 July 2006 (UTC)[reply]
to the first response: You can't achieve it if you don't know how to. You can't achieve it if psychological blocks keep you from doing it. You can't achieve it if whenever you start to someone distracts you. Etc. By "able to achieve it" I mean just that. I don't mean physically able, etc. The second response is exactly what I was thinking too: It's like a vote. 2/3 of congress can change the constitution, but the fact that the constitution isn't being changed means (trivially) that no one who can change it is doing so. So, the fact "no one can achieve world peace" must be read very narrowly as meaning "no one can singly achieve world peace instantly upon resolving to do so."
to the vacuous truth person: a vacuous truth isn't LOGICALLY MANDATED, for example "All elephants inside a loaf of bread are pink." is a probably vacuous truth. But not because it's a tautology.
Is it a tautology (not mere vacuous truth) to say: "In the unlikely case that there is currently someone in a position to instantly achieve world peace upon wishing to do so, it logically follows that this person does not currently wish to do so."
Thanks for your analyses! 82.131.185.76 18:25, 8 July 2006 (UTC)[reply]
Another factor you've left out of your reasoning is time. It is possible that there is a person which wants world peace, and is capable of achieving it all by himself, but this is a process that could take some time. So, there isn't currently world peace, but he's working on it... But I think the truth here is that there is no person who is able to achieve world peace. It seems unrealistically optimistic to believe that such a person exists. So yes, your conclusion is IMHO true - every person is not able to achieve world peace, including those who want it. -- Meni Rosenfeld (talk) 19:50, 8 July 2006 (UTC)[reply]
Another possibility is that those who can achieve world peace do wish to do so, but have decided that the cost is too high. For example, an equitable distribution of the Earth's wealth would bring us a long way toward world peace, but would leave us each with only $9500 GDP per person, which is a huge step down for most people in the developed world. StuRat 20:08, 8 July 2006 (UTC)[reply]
Indeed. It takes the idea of "want" a bit simplistically. People do things, or fail to do them, for many reasons at once. You could say that I don't "want" a donut if I choose not to eat it. But what if I said, "I do want to eat it, but I want to stay healthy more." Such a situation is difficult to fit into your framework. As a matter of fact, I would kill for a donut right now, but actually I wouldn't, because killing is wrong. So, do I want the donut, or not? Black Carrot 16:59, 10 July 2006 (UTC)[reply]

Three-dimensional coordinate Math system

What is the name of the "z" coordinate in a three-dimensional coordinate math system?
The x coordinate is called the abscissa.

The y coordinate is called the ordinate.
I just call it the Z-coord along the Z-axis, myself. Those other silly names do nothing but make math harder, as far as I can tell. If everyone called them the X-axis and Y-axis, instead, we would have two less useless bits of information math students need to learn and they would have more time to learn something of actual importance. StuRat 19:58, 8 July 2006 (UTC)[reply]
I agree with StuRat, but either way, there's a good chance no name exists. In polar coordinates there are no such names. The coordinates are merely (t,r) or (Θ,r). Unless you consider "theta" and "radius" fancy names. -- He Who Is^{[ Talk ]} 20:31, 8 July 2006 (UTC)[reply]
Let us first consider a larger question. Mathematics routinely deals with larger dimensions. When we have 20 dimensions, how shall we name numbers 17, 18, and 19? And when we have 200? The point is, the specialized names "abscissa" and "ordinate" are of limited utility. The abscissa is where a perpendicular from a plotted point cuts the horizontal axis (think of "scissors").

If we plot a z value that is a function of x and y, then by rights we should drop a perpendicular to the xy plane and call that point the abscissa. More often, we use names special to the problem. For example, in describing the topography of Madagascar, the directions would be called "latitude", "longitude", and "elevation".

The abscissa-ordinate naming scheme has meaning when plotting y as a function of x, but it does not generalize well to name dimensions. In fact, that is not its purpose. These peculiar names, admittedly somewhat archaic, are best used not to refer to the x and y coordinates, but to their roles. We might instead refer to "independent" and "dependent" variables, if not for the fact that we may need to distinguish between the variables themselves and the coordinates on a graph. That is, we might say
Given a functional dependency y = f(x), we can draw a graph by plotting the independent variable, x, as the abscissa and the dependent variable, y, as the ordinate.

It's hard to say whether these names are retained for utility, blind tradition, or a scholarly fondness for links to the past. Whatever the reason, graphs in higher dimensions need to move beyond them. --KSmrq^T 21:24, 8 July 2006 (UTC)[reply]
I would have said :
Given a functional dependency y = f(x), we can draw a graph by plotting the independent variable, x, along the horizontal axis and the dependent variable, y, along the vertical axis.
This seems to be much clearer and requires no knowledge of archaic mathematical terms. StuRat 01:47, 9 July 2006 (UTC)[reply]

Prime Counting Function

In Prime_counting_function#The_Riemann_hypothesis, it says that proving the Riemann Hypothesis true would prove a particular restriction on the prime counting function true as well. Why? Also, does it go the other way? Would proving this restriction true reflect on Riemann's idea?
I don't know enough analytic number theory to really understand why, but I feel like this applet gives me a glimpse. The Riemann Hypothesis is equivalent to that stronger bound, so if you prove one you also prove the other. —Keenan Pepper 20:57, 8 July 2006 (UTC)[reply]
One other thing: what's the inverse of x/ln(x)? Some things have said it's xln(x), but I can't find a way to make that work. Black Carrot 20:41, 8 July 2006 (UTC)[reply]
Well assuming that's true, we can say that:
${\frac {x\ln(x)}{\ln(x\ln(x))}}=x$
${\frac {x\ln(x)}{\ln(x)+\ln \ln(x)}}=x$
${\frac {x\ln(x)}{\ln \ln(x)}}=x-1$
After putting in a few random inputs, this seems to hold up, but that's not to say that it's actually true -- He Who Is^{[ Talk ]} 04:09, 9 July 2006 (UTC)[reply]
How exactly does this absurd formula hold up? In any case, The inverse function of x / ln(x) cannot be expressed with elementary functions, but it can be expressed with Lambert's W function:
${\frac {y}{\ln {y}}}=x\iff y=-x\cdot W\left(-{\frac {1}{x}}\right)$

x ln(x) is a fair approximation though; and you can obtain successively better elementary approximations by repeatedly replacing ln(x) with ln(x ln(x)). The reason is that what you actually seek is y = x ln(y), only that you don't know y. If, instead of y, you put some approximation for y, you will get a better approximation. -- Meni Rosenfeld (talk) 14:35, 9 July 2006 (UTC)[reply]
Aha. So, what about the second question? Does proving the restriction prove (or help prove) the Riemann Hypothesis? Black Carrot 17:45, 9 July 2006 (UTC)[reply]
Yes, as Keenan mentioned, the article says that the Riemann hypothesis and the tighter bound are equivalent. Proving this bound will immediately also prove the Riemann hypothesis (and vice versa). -- Meni Rosenfeld (talk) 17:58, 9 July 2006 (UTC)[reply]
Sorry. I'm fairly new to mathematical discussion, and the code words (conjecture, equivalent, heuristic, etc.) take some getting used to. One further question:
Since the average density of primes around x is close to 1/ln(x), the average distance between primes in the same area is ln(x). However, the integral of this, x(ln(x)-1), is suspiciously similar to the best simple approximation of the x^th prime, xln(x), which is the inverse of the prime counting function, the derivative of which is 1/ln(x). Is there any particular reason for this circle? Black Carrot 16:50, 10 July 2006 (UTC)[reply]
That would depend on what you mean by a "particular reason". But how about this: If π(n) is the number of primes up to n, p(n) is the nth prime number, and Δ(n) = p(n+1) - p(n), then Δ(n) is roughly equal to ln(p(n)), which is roughly equal to ln(n) (ln grows slowly enough for this approximation). Then:
$p(n)=p(1)+\sum _{k=1}^{n-1}\Delta (k)\approx \int _{1}^{n}\Delta (k)dk\approx \int _{1}^{n}\ln(k)dk$

Which explains the similarity between the integral of ln(n) and p(n). -- Meni Rosenfeld (talk) 07:32, 11 July 2006 (UTC)[reply]

Also, it is mentioned in Von Mangoldt function that von Mangoldt's explicit definition of the summatory von Mangoldt function using a sum over the non-trivial zeroes of the Riemann Zeta function wconstibuted to the first proof of the Prime Number Theorem. If I'm correct, this was the second connection ever found between the prime counting function and the Riemann Zeta function. The first was the sum over the primes p of 1/(1-p^-s) = ζ(s) -- He Who Is^{[ Talk ]} 19:29, 11 July 2006 (UTC)[reply]

July 9

Satisfying Poll Results Involving "The Average"

On a webpoll I recently saw, participants were asked to rank how well they drove, in comparison to everyone else who filled the poll. Options were "Average", "A bit better than average", "Considerably better than average", "Much better than average", "A bit worse than average", "Considerably worse than average", and "Much worse than average". Unsurprisingly, more people supposed that they were better than average than supposed they were worse, though every option had some takers. But some people seemed to interpret this as being "some people clearly think they are better drivers than they are," and it seems to me that is not necessarily true.
After all, consider a simpler quiz of three options "Average," "Better than average", and "Worse than average", and we rank people's driving out of 100 as a way of deciding an average. If 90 people drive at a 75 level, 5 people drive at a 68 level, and 5 people drive at a 10 level, then the average is 68, and 90% of people drive better than this, 5% drive at this, and 5 drive worse. Those could well have been our more-above-average-than-below results.
This got me thinking. Is it possible, given -any- distribution of people's guesses (satisfying certain requirements), to construct such a theoretical sample that would satisfy the result? Clearly, it could not be true if everyone said they were better than average, but it seems to me (I made a half-hearted attempt, but I can't prove it) that as long as you have an open-ended category on the underpopulated side of average that has at least one taker (e.g. a category like "Much worse than average" that covers every really bad driver, and there is no limit on how bad a driver can be, and there's at least one taker), then you can satisfy any possible set of results. This might not even be necessary. Anybody have any thoughts on the matter? What general rule can we construct about such a thing? Maelin 13:20, 9 July 2006 (UTC) (Minor edit for clarity 13:33, 9 July 2006 (UTC))[reply]
If you have a sample of n people, with mean ${\frac {\sum x_{i}}{n}}$ , and you add one more person to the sample, you can get any new mean y that you like by taking $x_{i+1}=(n+1)y-{\sum x_{i}}$ .

Still, while people might be jumping to a logically unsound conclusion from your webpoll, the Lake Wobegon effect does seem like a plausible interpretation. It's also unclear what respondents think the word "average" means. If they think it refers to the median, not the mean, then your logical loophole disappears. -- Avenue 14:50, 9 July 2006 (UTC)[reply]
Yes, I strongly suspect that by "better than average", they mean "better than half of the drivers", which is really the median, not the average. StuRat 17:24, 9 July 2006 (UTC)[reply]

volume of concrete

Hello, My name is matt and I am helping my dad figure out a volume problem with our concrete. We are pouring gypcrete in our house and the contracter said that 1 1/2 inches of crete would do 18 square feet. My dad only wants to pour the thickness to 1 1/8 and hopefully have enough to do our room that is 20 square feet. How do we figure this problem? Thank You so much
One thing that might help is to get them into the same units, instead of using both inches and feet. 1 square foot = 12x12 = 144 square inches. You could work out the volume of the concrete you have, then find out to what depth you can fill 20 square feet (2,880 square inches) of floor with it. If you can fill higher than you need to, you have more than enough. The volume of a rectagular prism (box shape) is depth x area of bottom. Black Carrot 17:55, 9 July 2006 (UTC)[reply]
If you really want to help your dad (and not your grade), you should advise him to rely on the expertise of the contractor. The contractor is being paid to do the job properly, and might be expected to know his job. Using a thinner layer may prove to be false economy. If that too-thin layer causes problems in the future, the cost will be much more than buying more "crete" now. Issues include strength, noise control, and fire protection. There may also be requirements in building codes for minimum thickness.

Q. "How do I stick beans up my nose? A. Don't! --KSmrq^T 20:53, 9 July 2006 (UTC)[reply]

two different group laws

I have two different group laws on a manifold, but it turns out that they're not so different after all. The manifold is R³ and the two group laws are
$(a,b,c)*_{1}(d,e,f)=\left(a+d,b+e,c+f+{\frac {1}{2}}(ae-bd)\right)$
and
$(u,v,w)*_{2}(x,y,z)=(u+x,v+y,w+z+uy).$
It turns out that these two groups have the same Lie algebra, and in fact the two operations can be understood in terms of the exponential map from this Lie algebra. The Lie algebra is of course also homeomorphic to R³. Let Q, P, E be a basis for the Lie algebra, and in terms of the exponential map we have:
$(a,b,c)=e^{aQ+bP+cE}$
and
$(u,v,w)=e^{uQ}e^{vP}e^{wE}$
so we see that the differences in the group law come from the choice to first combine the vectors, then exponentiate, or vice versa. The two group laws are thus really consequences of two different choices of parametrizations for the group. The relationship between the two is given by the Baker-Campbell-Hausdorff formula, which comes to
$e^{uQ}e^{vP}e^{wE}=e^{uQ+vP+(w+{\frac {1}{2}}uv)E}.$
In case you haven't recognized it, the groups described are (both?) the Heisenberg group and their Lie algebra are the canonical commutation relations of quantum mechanics. This group is a particularly nice place to analyze my question due to the simplicity of BCH, but it could be asked of any (nonabelian) Lie group, I suppose.
Now we come to my question. Are the two groups the same? The most obvious map between them doesn't seem to be a homomorphism, unless I'm mistaken. On the other hand, aren't they just two different parametrizations for the same group, and therefore shouldn't they be isomorphic? I have one reference which calls the first group the Heisenberg group and the second group the polarized Heisenberg group and never mentions an isomorphism between them (which seems telling in its ommission). I have several other references which call the second group the Heisenberg group and make no mention of the first, and one which calls the first the Heisenberg group and makes no mention of the second. Right now the Wikipedia article has both groups, but in an inconsistent way, and it is in attempting to fix that article that I have gotten myself stuck on this issue and have come to you folk in suppliance.
I could ask the same question about any Lie group, I think. For example, for the rotation group, I think it would sound like this: write the group law for rotatations in terms of their Euler angles, and then write the group law in terms of the single axis of rotation. Is there an isomorphism between the two groups? -lethe ^{talk +} 21:47, 9 July 2006 (UTC)[reply]
Perhaps it is enough in this case to observe that both groups are finite-dimensional, simply connected, and have the same Lie algebra. Thus each is the same universal covering group (up to isomorphism), yes? Also, the groups are path-connected and not compact, so the exponential map should be your friend in constructing an isomorphism. --KSmrq^T 23:13, 9 July 2006 (UTC)[reply]
Yes, this is of course correct. There is only one connected simply-connected Lie group to any Lie algebra, so we know at the outset that the two groups must be isomorphic. Furthermore, your suggestion that the exponential map should point to this isomorphism also sounds eminently reasonable. Either the "obvious" map between the two groups is not the correct map, or else my calculation about the map contains mistakes. I guess I will repeat the calculation. I'll post back my findings. Thank you for your input. -lethe ^{talk +} 00:28, 10 July 2006 (UTC)[reply]
I've repeated the calculation, and actually the obvious map is a homomorphism. The groups are isomorphic, as they should be. Nothing more to say about this, I guess, except that I should be more careful, triple-check my calculations before I come to RD with silly questions. Thanks again. -lethe ^{talk +} 13:48, 10 July 2006 (UTC)[reply]

Oi, barkeep, wheres my quid

3 guys in a bar, buy three drinks, cost comes to £30.
Bartender refunds £5 for miscalculation, the three men take £1 each, and give the bartender back £2 for honesty.
Each man spent £10 - £1 = £9
the three men spent £9 * 3 = £27
plus the £2 the tender had, comes to £29.
Where is the other pound?
There are several issues with this problem, the whole thing originates from there drinks costing £8⅓ each. Which is the first problem, I figured thats where the extra pound is, by I still can't explain the incosistency with the amount going in and the amount going out. Philc _T^E_C^I 23:28, 9 July 2006 (UTC)[reply]
I believe the issue is you added when subtraction is needed. The three men did indeed spend £27, which includes a £2 gratuity (£25 for drinks, £2 for the barkeep). Thus adding £27 and £2 is meaningless--a more meaningful quantity is £27 - £2, the total spent minus the tip, which yields the cost of the drinks. --TeaDrinker 00:03, 10 July 2006 (UTC)[reply]
There seems to be a large hole in this problem. Is the £30 the price with or without the miscalculation? -- He Who Is^{[ Talk ]} 01:22, 10 July 2006 (UTC)[reply]
[Edit Conflict] :This seems to be one of the most commonly repeated questions on this desk (because of the various ways to word it I'm not sure you could search for it, but I wouldn't be surprised if it was asked at least once a month) - see Missing dollar paradox. Confusing Manifestation 01:19, 10 July 2006 (UTC)[reply]
There is NO missing £1. What you should do is ask "who has each of the original £30?" The answer then is: the cahier has £25, the three men have £1 each (£3) and the bar-tender has £2, so all 30 are accounted for - no 'missing' £1.
Hmm fair enough, makes sense, it all in the wording. Philc _T^E_C^I 17:07, 10 July 2006 (UTC)[reply]

Start from realizing that the three drinks cost £25. Then see the cash flow:

The three guys have The cashier has The bartender has

Initially 10+10+10 0 0

The three guys pay 0 30 0

and the change goes back 5 25 0

the bartender gets his tip 1+1+1 25 2

The money actually spent 1+1+1 25+2 = 27
Note two things:
No penny is missing – the sum in every row is the same £30.

The £27 sum is what the three guys have paid – and it already contains the £2 tip.
So it makes no sense to add 27+2. Such sum does not appear anywhere in the balance.
:) --CiaPan 19:24, 10 July 2006 (UTC)[reply]

July 10

Mix CD in Linux

What's a good Linux program to make a mix CD from tracks that exist on other CDs I have, ideally with no loss of quality and playable in a normal CD player? I use gentoo, any common program is probably available. Thanks. -- Pak aran 00:58, 10 July 2006 (UTC)[reply]
Grip and K3b work for me. Yvh11a 23:02, 10 July 2006 (UTC)[reply]

Setting up OS X Server Managed Clients

Sorry in advance if I am posting in the wrong place. I saw the Computer Science bit and figured that this was the closest thing. I'm a lifelong Windows guy...so I'm sorry if I am getting stuff blatantly wrong in this question or testify to my stupidity in this question.
I just set up a server running Mac OS X Server 10.4. I'm trying to set up a group of ~15 computers running Mac OS X 10.4 (and ~5 with OS X 10.3) as "managed clients". They would derive their authentication, preferences, updates, etc. from the server. Individual logons could be created or disabled by the server, and privileges could be remotely managed.
Trouble is, I have absolutely no idea where I should start. I've set up user directories on the server, and am currently able to create user accounts and edit user privileges. I have no idea, however, what I should do with my individual clients. I've played around with networking, logon settings, everything I could think of. I've searched the internet and read several hundred pages of OS X server manuals.
So yeah, help me Wikipedia Reference Desk. Thanks! Alphachimp ^talk 03:04, 10 July 2006 (UTC)[reply]

Hardest Equation

What is the hardest calculus equation? And what is the answer? — Preceding unsigned comment added by Yamabushi334 (talk • contribs)
42. Alphachimp ^talk 04:31, 10 July 2006 (UTC)[reply]
As Alphachimp suggests, the question you pose is difficult to answer exactly. What is a hard problem is subjective of course, so some there will never be a universal answer to your question. In another sense, one answer might be a difficult question on material usually taught in a first year differential and integral calculus class. Alternatively, the natural extensions of a first year calculus class lend themselves to difficult, and often unsolved problems. A few candidates for the latter might be the N-body problem or Navier-Stokes equations (in the area of Differential equations)--see Unsolved problems in mathematics for some fun reading. Problems which are commonly covered in the first year of calculus have different difficulties to different students. I'm not sure a simple list could even be compiled. --TeaDrinker 06:24, 10 July 2006 (UTC)[reply]

Also, if we exit the realm of calculus, there is Matiyasevich's theorem, which proves that there is no algorithm to solve general diophantine equations. 130.119.248.11 13:52, 10 July 2006 (UTC)[reply]
Also, considering that there are many unsolved problems in all branches of mathematics, one would assume that the "hardest" would be one of these. In that case, we cannot give you an answer, either. -- He Who Is^{[ Talk ]} 14:33, 10 July 2006 (UTC)[reply]
NP-hard. Jon513 14:17, 11 July 2006 (UTC)[reply]

Gosh Numbers

Wikimathematicians, if you are interested, please help determine this afd discussion about Gosh Numbers. Thanks! Bwithh 04:36, 10 July 2006 (UTC)[reply]

doubt...

hello sir,
i have a small doubt regarding matrices.
1. if $G={\begin{pmatrix}1&0\end{pmatrix}}$ ; what would be the diagonal matrix diag(G,G,G) be?????
2. if $H={\begin{pmatrix}1\\0\end{pmatrix}}$ i.e it is a column matrix with entries 1 and 0 what would be the diagonal matrix diag(H,H,H) be ????
thank you sir. can u please mail the answer to evani.subrahmanyam (at) gmail.com . thanks
I can't be sure whether you want to know the answer unless you use at least six consecutive question marks. -lethe ^{talk +} 14:54, 10 July 2006 (UTC)[reply]
I can answer with only 4 question marks, but I'm not sure what "diag" means here. Is it a command in a program or are you using it in the sense of the diagonal matrix article? If the latter, it might be a matrix consisting of the single element 1 for both G and H, if one abuses the notation a bit. As used in the article, it is the elements on the diagonal of a square matrix, so since G and H are not square matrices, it isn't quite defined. Notice that in the article, diag(a1,...,an) means an n-by-n matrix whose elements are 0 everywhere except the diagonal, which contains the elements a1, a2, ..., an, going down by rows/columns. So diag(G,G,G) (or H,H,H) suggests a matrix whose diagonal elements are row or column matrices, respectively. Such a matrix is ill-defined. 128.197.81.223 22:09, 10 July 2006 (UTC)[reply]

power law decay versus exponential decay

Are you reading the news?
7 July 2006
If you think you're reading the news, be warned that this story -- and any other on the web -- will be barely read by anyone 36 hours after it was first posted. That's the message from a team of statistical physicists who have analysed how people access information online. Albert-László Barabási of the University of Notre Dame in the US and colleagues in Hungary have calculated that the number of people who read news stories on the web decays with time in a power law, and not exponentially as commonly thought. Most news becomes old hat within a day and a half of being posted -- a finding that could help website designers or people trying to understand how information gets transferred in biological cells and social networks (Phys. Rev. E 73 066132).
Can someone tell me (mathematically) the difference between a power law decay and an exponential decay.
I know that exponential decay is $A_{0}\times A^{-kt}$

Ohanian 22:12, 10 July 2006 (UTC)[reply]
Yes, exponential decay is when the time variable is an exponent, as you demonstrated above. A power law decay is when the time variable is raised to a power (i.e. y = t^.5). The difference is that exponential decay decreases at a fixed rate, regardless of the amount of the decaying substance (e.g. time) whereas power law decay decreases at a changing rate. Compare the derivatives of the two to see what I mean. 128.197.81.223 22:21, 10 July 2006 (UTC)[reply]

Are you saying that
Power Law Decay is $B_{0}\times t^{k}$ where $0<k<1$
But that is not a decay because the amount gets bigger and bigger over time.
Ohanian 22:26, 10 July 2006 (UTC)[reply]
I am indeed. For more information see power law. 128.197.81.223 22:30, 10 July 2006 (UTC)[reply]
Oops, the answers below are more correct, I made a typo then misread your response. Sorry! 128.197.81.223 15:09, 11 July 2006 (UTC)[reply]

I was always under the impression that exponential decay/growth took the form:
$x=x_{0}e^{kt}\quad \equiv \quad x=x_{0}\,{e^{c}}^{{\frac {k}{c}}t}\quad \equiv \quad x=x_{0}\,A^{{\frac {k}{c}}t}\quad \equiv \quad x=x_{0}\,A^{Kt}$

where x is a quantity, x₀ is initial quantity, and k is a constant. If k<0, it is decay, and if k>0, it is growth. However, I think the link to power law is correct, but for decay, k<0. Then we would get something like:
$x={\frac {a}{t^{k}}}$

where in this case k>0. But same thing, really. You will see that the one above will decay.

I might be wrong... x42bn6 Talk 02:56, 11 July 2006 (UTC)[reply]
Yes, x42bn6 has it right (unless I am myself mistaken). An exponential law would be (anything asymptotical to) $x=x_{0}e^{kt}$ , a power law would be $x=At^{k}$ . In both cases, for k>0 it's growth, k=0 it's constant, and k<0 is decay. Examples: The amout of radioactive isotopes in a sample decreases exponentially with time. The gravitational force between a planet and a projectile with positive total energy (assuming these are the only objects in the universe) decreases quadratically (power law with k=-2). Of course, for both growth and decay, an exponential law is much faster (after enough time) than a power law. -- Meni Rosenfeld (talk) 07:44, 11 July 2006 (UTC)[reply]
Suppose k is a positive integer; then 1⁄k is a positive value less than 1, and T^1/k is the k-th root of T. As T increases, so do all its roots, so this would not be decay. However, if T is always positive then T^(−k) is (1⁄T)^k, which does decrease as T increases. Therefore T^(−k) is a power-law decay, just as e^(−T) is an exponential decay.

The curious thing about the news report is that it suggests that a power-law decay drops off faster than an exponential decay. A proper comparison is more subtle, involving constants, initial rates, and asymptotic behavior. What can we really say? --KSmrq^T 12:47, 11 July 2006 (UTC)[reply]
Let's start at $t=1$ , to avoid problems with $t^{k},t=0,k<0$ . Then suppose that $f(t):=Ae^{\alpha t}$ and $g(t):=Bt^{\beta }$ , and that $f(1)=g(1)=1,f(T)=g(T)=C$ with $0<C<1,T>1$ . Then $A=e^{-\alpha }$ and $B=1$ , and $T^{\beta }=C$ so $\beta =\log _{T}C$ . Finally, $e^{\alpha (T-1)}=C$ so $\alpha ={\frac {\ln C}{T-1}}$ . Now note that $f(t)=e^{\ln C{\frac {t-1}{T-1}}}=C^{\frac {t-1}{T-1}}$ and $g(t)=e^{\frac {\ln t\ln C}{\ln T}}=C^{\frac {\ln t}{\ln T}}$ ; as required, these are equal for $t\in \{0,T\}$ because the exponents are equal; for $t\in (0,T)$ , however, $g(t)<f(t)$ because $\ln t$ is concave down and $t-1$ is linear (the denominators being constant in both cases), so the exponent of $C<1$ is always larger for g. In this sense, powers decay more quickly: for any given amount of decay over a given time, the power law does more of the decay sooner. You can arrive at a similar conclusion by considering ${\frac {f'(t)}{f(t)}}=\alpha$ and ${\frac {g'(t)}{g(t)}}={\frac {\beta }{t}}$ , implying that more of the proportional change in g happens for small t, but it's not as rigorous (and is relatively obvious given the exponential nature of f and g and that $x'/x=(\ln x)'$ ). --Tardis 19:22, 11 July 2006 (UTC)[reply]

What does =? mean?

I saw a statement today that used the (I'm assuming) mathmatical operator =? -- what function is this? -Quasipalm 00:01, 11 July 2006 (UTC)[reply]
If the author does not say otherwise, typically this is used to explore whether a proposed equality holds. For example, we might consider
$(x-y)^{2}=?x^{2}-y^{2}\,\!$

or
$x^{3}-1=?(x-1)(x^{2}+x+1).\,\!$

In the first example we could substitute x = 2, y = 1 to produce 1 on the left and 3 on the right, and conclude that the equation is false; while in the second example we could multiply through on the right and simplify, to conclude the equation is true. Thus we resolve the proposed equalities to
$(x-y)^{2}\neq x^{2}-y^{2}\,\!$

and
$x^{3}-1=(x-1)(x^{2}+x+1).\,\!$

The notation is less common in advanced mathematics, but useful for more elementary teaching. Typically the question mark appears over the equality sign, as in the Unicode character "≟" (U+225F, QUESTIONED EQUAL TO). The TeX markup would be "\mathop{=}\limits^?", but Wikipedia's texvc chokes on it. --KSmrq^T 01:20, 11 July 2006 (UTC)[reply]
Awesome thanks! -Quasipalm 04:27, 11 July 2006 (UTC)[reply]

Programming Language

I want to learn a general purpose programming language, either C, C++ or Java; does anyone have any suggestions as to which would be the best one to learn? Many thanks --86.139.216.231 14:07, 11 July 2006 (UTC)[reply]
This is not really a factual question, and is likely to generate more heat than light (i.e., a flame war). But the important thing is that your choice should depend on your prior experience and what you want to do (immediately). If you've never programmed before at all, you would want C if you wanted the simplest language to start with, or Java if you wanted the most structured as a basis. If you have programmed a lot before, then you might want to learn C++ if you had learned functional programming before and wanted to continue using it (somewhat), or you might want C if you had used Smalltalk before and wanted to try something very different. It is possible to write graphical applications, network applications, scientific applications, and games in any of those three; however, for systems programming on Unix C or C++ is clearly a superior choice, whereas for a simple graphical program that will most easily run on multiple platforms Java is the obvious candidate. C and C++ are faster than Java (somewhat), but that's only important for severe amounts of mathematics; C++ and Java support (differently: see garbage collection and RAII) automatic memory management, which is useful in many programs (especially dealing with strings and advanced data structures), but is entirely unnecessary in others. C has the widest compiler support, but that's only important if you plan to be working with many different architectures (and Java wins in terms of support in many "odd" environments such as cellular phones). In other words, "which one is best" is not at all answerable without knowing more about your goals and experience. I personally know all three and find each to be useful at different times. Hope this helps. --Tardis 19:41, 11 July 2006 (UTC)[reply]
I've been considering the same question myself, since I learned BASIC many years ago and think it might be time to also become competant with a more modern language, something that can interact with the internet or offers the possibilty of object-orientation if I want it.
The conclusion that I can to is that Python is easily the best language to choose. See the article on Python programming language. In addition to the above, two other important considerations are 1) it has a clear easily-readable syntax without those curly brackets or other obscure punctuation, and 2) it is the best supported modern language, with many tutorials for beginners, discussion fourms, etc. Its not just a passing fad, but is here to stay. And its all free!
I'd like to know what language you finally choose.

--81.104.12.70 11:12, 12 July 2006 (UTC)[reply]

Geometrically stumped

O ye mathematicians: I am completely stumped by this - I've been crunching the algebra for an hour, and I can't seem to solve it. So, here's a challenge - one wikipat on the back to whoever can figure this out. And I don't need any partial solution - i have lots of those :).
The situation:
In a 2-D rectangular coordinate system: There is a circle of radius R at the location (x,y) = (0, h) - on the y-axis. There is an ellipse with minor and major axes of length 2*a and 2*b, respectively. The center of the ellipse is at (x,y) = (-L, d) (off to the left). d < h, and L > 0. The two figures don't intersect, and a < R and b < R.
The goal:
The thing I'm trying to do is to find a line that goes between them and is tangent to both figures with a positive slope.
Simple, you think? Well, the algebra becomes very unpleasant, then I get stuck. Anyone want to help me out? It would be MUCH appreciated. And I swear this is no homework problem. If it was, I would have just given up long ago. --Bmk 20:04, 11 July 2006 (UTC)[reply]
I think you need to specify the rotation of the ellipse. For example, is the major axis parallel to the X-axis ? StuRat 22:10, 11 July 2006 (UTC)[reply]
Good point - thanks. The major axis is parallel to the y-axis. And in case anyone was wondering, i need this because I'm trying to do some computer-assisted modeling of a structure with a profile that has that description. --Bmk 22:20, 11 July 2006 (UTC)[reply]
This (with the axis specified) seems to me to be relatively straightforward. Allow me to use different notation that I found simpler: consider a circle of radius R at the origin, and your y-major ellipse (same axes, 2a in x and 2b in y) centered at $(h,k)$ in Quadrant III. Now:
Consider the line's y-intercept and label it B, such that $y=mx+B$ .

Label its point of tangency to the circle $(P_{x},P_{y})$ ; since the radius to that point is perpendicular to the tangent, the radius has slope $-1/m$ , so we have $P_{y}=-P_{x}/m$ and $P_{x}^{2}+P_{y}^{2}=R$ , yielding $P={\frac {R}{\sqrt {1+m^{2}}}}(-m,1)$ (since m and B are positive, the tangency must be in Quadrant II).

This is on the line, so ${\frac {R}{\sqrt {1+m^{2}}}}={\frac {-m^{2}R}{\sqrt {1+m^{2}}}}+B$ and $B=R{\sqrt {1+m^{2}}}$ : $y=y(x;m)$ only. Then plug that into the equation for the ellipse: $\left({\frac {x-h}{a}}\right)^{2}+\left({\frac {mx+B-k}{b}}\right)^{2}=1$ . Expand in powers of x: you get a quadratic, of course (it being a conic and all).

Because we want this line to be tangent, we want there to be precisely one solution for x (think about what the two solutions would be for a line through the ellipse, and what happens as the line approaches tangency). This means that the discriminant is 0: this is then a quadratic in m, since (as it turns out; I'm omitting the algebra) the linear coefficient in x's equation is linear in m and the quadratic coefficient is but quadratic.

Solve this quadratic, and you'll get the two values of m that "hug" the ellipse on top and bottom.

From m you get B, and you have the line.

(Note that the points of tangency can be obtained by then solving for x and then y for the ellipse, and by consulting P for the circle.)

Hope this helps. --Tardis 23:10, 11 July 2006 (UTC)[reply]
I'm inclined to believe this is not a homework problem, because it is too much of a challenge, especially without calculus (for the tangents). There is, however, a magically simple (?) approach using the geometric theory of poles and polars. The tangency problem becomes an intersection problem for two conics, and that reduces to finding the four real roots of a quartic equation. Each root gives a point of tangency. Pick the one you want.
1. Write the circle and the ellipse as implicit equations.
$0=x^{2}+(y-h)^{2}-R^{2}\,\!$

$0={\frac {(x+L)^{2}}{a^{2}}}+{\frac {(y-d)^{2}}{b^{2}}}-1\,\!$

2. Rewrite the equations in homogeneous matrix form.
$0={\begin{bmatrix}x&y&w\end{bmatrix}}{\begin{bmatrix}1&0&0\\0&1&-h\\0&-h&h^{2}-R^{2}\end{bmatrix}}{\begin{bmatrix}x\\y\\w\end{bmatrix}}$

$0={\begin{bmatrix}x&y&w\end{bmatrix}}{\begin{bmatrix}{\frac {1}{a^{2}}}&0&{\frac {L}{a^{2}}}\\0&{\frac {1}{b^{2}}}&-{\frac {d}{b^{2}}}\\{\frac {L}{a^{2}}}&-{\frac {d}{b^{2}}}&{\frac {L^{2}}{a^{2}}}+{\frac {d^{2}}{b^{2}}}-1\end{bmatrix}}{\begin{bmatrix}x\\y\\w\end{bmatrix}}$

3. Calling the circle matrix C and the ellipse matrix E, compute the matrix for a new conic Q = CE⁻¹C.
$Q={\begin{bmatrix}a^{2}-L^{2}&(d-h)L&-(R^{2}+(d-h)H)L\\(d-h)L&b^{2}-(d-h)^{2}&(d-h)R^{2}-(b^{2}-(d-h)^{2})h\\-(R^{2}+(d-h)H)L&(d-h)R^{2}-(b^{2}-(d-h)^{2})h&b^{2}h^{2}-((d-h)h+R^{2})^{2}\end{bmatrix}}$

4. Convert Q to an implicit equation (details omitted). Write the circle in rational parametric form.
$x(t)=(1-t^{2})R\,\!$

$y(t)=2tR+(1+t^{2})h\,\!$

$w(t)=1+t^{2}\,\!$

5. Substitute the parametric coordinates of the circle into the implicit equation generated by Q to get a (numeric) quartic equation in t.

$0\,\!$ ${}=[a^{2}-(R+L)^{2}]\,\!$

${}+t[4(d-h)(R+L)]\,\!$

${}+t^{2}[2(2b^{2}-a^{2}-2(d-h)^{2}-(R+L)(R-L))]\,\!$

${}+t^{3}[4(d-h)(R-L)]\,\!$

${}+t^{4}[a^{2}-(R-L)^{2}]\,\!$
6. Solve the quartic in t for its four real roots (using a numerical method).

7. Substitute each t value into the parametric equation for the circle to get the circle contacts.

8. Multiply each of the four circle contacts by C to get four line equations.
$0=(1-t^{2})x+(2t)y-(R+2ht+Rt^{2})\,\!$

9. Choose the desired line. For positive slope, t must be between −1 and 0, or between +1 and ∞.

Simple, yes? ;-D

Let's try an example with specific numbers, to see how simple it really is.
Four mutual tangents to circle and ellipse
Values
$h=9,R=4\,\!$

$L=6,d=1\,\!$

$a=3,b=2\,\!$

Quartic
$0=-91-320t-218t^{2}+64t^{3}+5t^{4}\,\!$

Roots
$t\in \{-15.3707,-0.758734,-0.416599,3.74602\}\,\!$

Circle contacts

t x y w point (x⁄w,y⁄w)

−15.3707 −941.032 2012.36 237.258 (−3.96628, 8.48172)

−0.758734 1.69729 8.11122 1.57568 (1.07718, 5.14777)

−0.416599 3.30578 7.2292 1.17355 (2.81689, 6.16009)

3.74602 −52.1307 165.262 15.0327 (−3.46783, 10.9935)
Lines
0 = −235.258 x − 30.7414 y − 672.36

0 = 0.424323 x − 1.51747 y + 7.3545

0 = 0.826445 x − 0.833198 y + 2.80457

0 = −13.0327 x + 7.49204 y − 127.559

Contact points for the ellipse can be obtained as LE⁻¹, where L is a row of line coefficients. Although a quartic equation has a closed-form solution in principle, the form is so complicated it is rarely useful in practice. Robust numerical root-finding software is available in the GNU Scientific Library, among other places. If a tangent should touch the circle vertically on its far left side, the t value will be infinite; this can be avoided by changing the circle parametric equations, if necessary. Also, be sure to check for typographical errors in all of this! :-) --KSmrq^T 04:54, 12 July 2006 (UTC)[reply]

Thank you so much Tardis and KSmrq!!!! Both of you took very innovative paths through the problem - I hadn't tried either method. It will take me a while to go through this, but thanks so much for the effort. Allow me to extend wikipats on the back to both of you, even though i haven't studied your answers yet. --Bmk 15:08, 12 July 2006 (UTC)[reply]
I hadn't noticed that for some cases (like KSmrq's picture) where the ellipse is relatively close to the circle, tangents with positive slope (like this blue line) can have y-intercepts below the circle. So you need to consider $B=\pm R{\sqrt {1+m^{2}}}$ . However, while writing this I realize that the discriminant may not be a quadratic in m because of the square root in B; it may in fact become a quartic! This makes sense; if we square away the square root in B, we lose information about the sign of B. So my method seems to already be as general as KSmrq's and must yield four solutions for m. It's not too bad, though: since quartics can technically be solved in closed form, both of these solutions are actually exact (if correct at all). --Tardis 18:13, 12 July 2006 (UTC)[reply]

Downloading streaming videos

OK, so I'm trying to find a good free program to download that will download streaming videos... I've tried a few and they all don't work or don't work right. So any suggestion would be great! Thanks!
I highly recommend VLC Player. --Bmk 20:07, 11 July 2006 (UTC)[reply]
That can download streaming videos!?! I've been having that program! hehe... Well at least I know it does that now. But how do you do that?
Hmm - actually I guess I didn't really answer your question. VLC is ok with playing video files that are still streaming from the internet. I don't know about actual streaming video on the internet - sorry. --Bmk 20:29, 11 July 2006 (UTC)[reply]
MPlayer can copy streams to file, the manual explains how. (I have only tested this with audio streams, but I see no reason why it shouldn't work with video too.) -- AJR | Talk 01:26, 12 July 2006 (UTC)[reply]
Yes, mplayer can do it (use -dumpstream and -dumpfile filename). Very useful for watching streaming media when your connection is too slow for that particular video (you can download first and then watch, instead of having the video skipping). --cesarb 06:11, 12 July 2006 (UTC)[reply]
I've downloaded Mplayer, but I have no idea what ya'll are talking about to download the streaming video/audio. I completely don't get it. I've checked out the manual thing, but it's in pc language, so I can't understand it. (I so need to start teaching myself how to understand all this computer talk... even though I've grown up with pc's it's still hard! hehe oh well).
[note to poster of the previous reply: you deleted an entire discussion when you removed the following line: {{User:Actin/Windows 98}}. I have no idea why it was done that way, it just confuses things, but I don't think you meant to remove it. ]
Oh... I don't know how I did that... sorry!

Windows 98

I'm pretty sure upgrading to a new OS won't help much. They simply have too much security holes to plug, but now that Windows 98 is no longer supported by Microsoft (do they still provide the support files they've built up so far?) my computer has become a sitting security hazard. The Windows XP article doesn't give any system requirements, but that's probably the next one to lose support. What is the latest Windows OS one can still run on a 366 MHz system without slowing it down unneccesarily? Can Microsoft legally drop support for a program millions of people use and force them to upgrade (and pay) for a new operating system when they already have one? Is some outside microsoft denouncer planning on providing support so people won't have to switch?
Yes, I probably should switch, but I don't have the cash to buy a good book let alone an overprized operating system. And no, I can't use Linux instead because the programs I use are not compatible with a Linux system. - Actin 07:18, 12 July 2006 (UTC)[reply]
Looks like you may have a problem - while support for Windows XP won't stop for a while yet (Windows ME and Windows 2000 have to go first), I severely doubt it will run well on a 366MHz machine. The official system requirements say it will, but I would be very doubtful as to how well it will run (and not do things like take two minutes from a click to the start menu opening up). Why not go for Windows 2000? The system requirements are more reasonable, it won't run well on a 366MHz, but it ought to at least run. Windows ME is also an option, however, it's a seriously lousy operating system, so I would strongly recommend against it unless you have no other choice. Another advantage of 2000 or ME is that you're likely to be able to get them at a fraction of the price of XP.

In general, however, a 366MHz is so old that it won't run any modern OS, let alone much modern software, well. I would recommend you stick with Windows 98, and accept the occasional holes in your system's security - as long as you're careful, run a firewall, don't surf on too many dodgy websites and use a good browser, you should be okay. IMO that's a much better solution than trying to upgrade your OS. — QuantumEleven 07:57, 12 July 2006 (UTC)[reply]
Switching to Windows ME won't help; it has reached end-of-life at the same time as the rest of the Windows 98 family, and in many ways it's a worse system. Software developers seem to be able do almost anything they like as far as support goes, unless they made some legally binding promise. This is not strictly a software issue. Can you still buy new parts for a Ford Model T? I don't think so!
I have sucsessfully installed XP on a machine slower than 366MHZ. It is a little sluggish, but it is functional. Using the windows powetoy TweakUI (available from the msft site) you can improve performance by diabeling fancy graphics and amnimations. It certainly is a viable option. 48v 20:00, 13 July 2006 (UTC)[reply]
Of course, not factories can't make old parts forever, but you can offer phone support on an operating system if you want. - Actin 12:45, 12 July 2006 (UTC)[reply]

But are you sure your programs cannot be run under a BSD or Linux system using Wine? Or better still, can you use free alternative software like OpenOffice.org or GIMP instead of some of your Windows apps? You might be pleasantly surprised.
Quite sure. Some will work on a Linux system, but some programs are made specifically for windows. And if I need to run a windows emulator on top of Linux I'm positive my system will crack under the pressure. Besides, I don't think I would be able to make a network connection with the other computer in my home that does use Windows XP. - Actin 12:45, 12 July 2006 (UTC)[reply]

So long as you are running Windows 98, it would be wise to avoid Microsoft apps like Word and Excel, and absolutely avoid Internet Explorer and Outlook; these tend to act as an open invitation for computer viruses. Try Opera or Firefox instead of IE, for example; they are much newer, much nicer, and much safer. Be sure to use virus protection, such as the free AVG Anti-Virus; and also use firewall software, such as the free Zone Alarm.
I can't get certain plugins to work in firefox. Movies and Shockwave games still go dead and a firewall gets in conflict with quite an important piece of software. - Actin 12:45, 12 July 2006 (UTC)[reply]

(Warning: POV!) Finally, start saving your pennies to purchase a Mac mini, with which you can run Mac OS X, Windows Vista, FreeBSD, or a Linux. --KSmrq^T 08:54, 12 July 2006 (UTC)[reply]
Anyway, thanks for the feedback. - Actin 12:45, 12 July 2006 (UTC)[reply]
I feel for you, Actin. I can't believe they are actually allowed to do that without providing free alternatives. - Mgm|^(talk) 08:40, 12 July 2006 (UTC)[reply]
I use to run Windows 2000 a slower processor (266, I think)--upgrading from Windows 95 (for some reason). I found it to run at a speed pretty close to Win95. I think there were enough improvements in the virtual memory handling to make up for the extra bulk. —Bradley 16:31, 12 July 2006 (UTC)[reply]
There is an advantage to an old O/S, not many people will bother writing viruses for it, either (except for Microsoft, who will probably write viruses to get you to upgrade). I am running Windows 98 on two computers, and intend to keep on doing so. Just turn your security settings up and avoid questionable downloads (like porn), and you should be fine. StuRat 20:10, 12 July 2006 (UTC)[reply]
As long as you have a decent amount of RAM (read: 192Mb+), you can run Windows 2000 rather well. Just as long as you don't bloat it with programs in the background. Our old school computers used to run Windows 2000 on 233/266Mhz processors with 256Mb of RAM and an 8Mb graphics card. x42bn6 Talk 02:09, 13 July 2006 (UTC)[reply]
I work on computers like yours. I always recomind Windows 2000 for it's superior stability and security.
Do you also recommend the spell checker ? StuRat 15:56, 13 July 2006 (UTC)[reply]

How to calculate variance from geometric and arithmetic averages?

In the textbook Corporate Finance by Brierly (sp?) and Miers (sp?) it mentions in a footnote that, for time series statistics, you can calculate the variance if you know the geometric and arithmetic averages.
I have tried to work out how to do this. I think the relationship might be:
v^2=g^2-a^2
where ^2 indicates the square v is variance g is geometric average a is arithmetic average
Am I right, or is it some other relationship please?
(I've mentioned this plus other relationships between averages in the discussion to the article Average. (I wonder if there are any other relationships bewteen averages etc that I havnt mentioned?))
Thanks
--81.104.12.70 11:00, 12 July 2006 (UTC)[reply]

Well at first I was completely taken by surprise, but then I inspected some more :it is all correct, except that g should not be the geometric average, but the quadratic mean. Evilbu 13:13, 12 July 2006 (UTC)[reply]

Differential Equation

I need to solve a differential equation, yet I am not sure how to go about it. Could you help? it is:
(-2*d2y/dt2) + (4*dy/dt) + y - 2sin(t) - 7 = 0.
I thought about using (y=e^λt), but it does not seem to work -- any help would be greatly appreciated. Thank you --DragonFly31 11:13, 12 July 2006 (UTC)[reply]
Problems like these are not brainteasers, there is a general procedure for this... but don't expect it to be fun.
Some questions first :
are you familiar with complex numbers? are you familiar with the concept of solving linear differential equations with a particular solution and the space of solutions of the homogenous equation?
anyway complexify your equation like this $w=u+yi$ then we have $-2{\frac {d^{2}w}{dt^{2}}}+4{\frac {dw}{dt}}+w=2e^{ti}+7$ as equation an equation like this always has a particular solution of the form $P(x)e^{ti}+C$ where $P(x)$ is a polynomial what degree? well just bet on a constant one here
that should be quite enough to solve this problem completely Evilbu 12:37, 12 July 2006 (UTC)[reply]
Although I am familiar a little with complex numbers and the concept of solving a linear diff equations for a general solution, I cannot understand how to solve them with a particular solution. Can you help me out a little more as to the procedure to completely solve this? Thanks for your help -- what you've shown me so far seems to be correct.
In order to keep notation a bit lighter, let us take an easier example, consider this linear(!) differential equation : ${\frac {d^{2}y}{dt^{2}}}+4y=1$
One easily sees that the simple constant 1 is a solution. This is called a particular solution. Let us call it f. But is this the only solution? Well suppose there is another solution g. Let us, to keep everything as simple as possible say $w=f-g$
well one easily checks that ${\frac {d^{2}w}{dt^{2}}}+4w=0$ this equation is rather nice, it is homogenous and linear , one can multiply any function with a real constant or add two up, and one still obtains a (new) solution. The solutions form a vector space in other words. Well what is this space. As this equation is of second degree, it is a theorem that they are two dimensional. Thus remains the ask of finding a basis : two independent solutions. If $u_{1}$ and $u_{2}$ happen to be such independent solutions, the general solution will look like $f+C_{1}u_{1}+C_{2}u_{2}$
so... what would be the answer?... --DragonFly31 14:53, 12 July 2006 (UTC)[reply]

Well look : solving differential equations is in general not easy, there is no method, most of times luck, ingenuity is the only way to get a real 'formula', usually they just use computers to approximate. However there are a couple of cases (and they do occur quite frequently, for instance in physics) where it is simple routine. But that does mean you have to study to be able to do thel. What I am hinting at, is that you also know how to solve homogenous differential linear equations like this :
$a*{\frac {d^{2}y}{dt^{2}}}+b{\frac {dy}{dt}}+c\ y=0$
with a , b and c constant. Have you ever seen those, and can you do them?
Please, everyone, sign your posts, as stated at the top of this page. Thanks.

DragonFly31, what you are seeing here is a reluctance to do your homework for you. If this is not for homework, you can persuade us by explaining where you encountered this equation and why you need to solve it. In either case we would help more if we could see any evidence that you were trying to do anything for yourself. I have elsewhere told you enough for you to solve this on your own. Saying "so... what would be the answer?" tells us you are not serious about understanding and working on this, you just want a solution served on a silver platter. --KSmrq^T 23:26, 12 July 2006 (UTC)[reply]

Data Migration: Windows 3.1(!) to Windows XP

(Windows 98, indeed. My old machine is perfectly safe from ~~viruses~~ ~~virus~~ virii. But I'd like to bring some data (Netscape mail files, a few old C programs, a lot of random text files, VPN passwords, etc.) from the old machine to the new one. Unfortunately, I haven't found software/hardware that will do it.
Configuration:
Old machine: 486DX2/66, Windows 3.1, 24MB memory (motherboard will hold a maximum of 48MB), a total of about 6GB of EIDE disks, 2 serial ports, 1 parallel port, CD reader, (broken) tape backup unit, 2 floppy drives.

New machine: HP, Windows XP (home), 1GB memory, 80GB (or so) disk space, two CD/DVD drives (one writable), one serial port, one parallel port, one USB port, possibly a FireWire port, and ports for a number of memory sticks.
Things I've tried:
Laplink serial connection (blocked by the XP operating system; I think it might work from the old machine to a Windows 95 or 98 machine, and use the CD-writer on that one)

SparQ parallel port (pseudo-SCSI) external drive. It might work, but the first one I bought on eBay broke while I was trying to download the driver into the old machine. I don't want to try again without some indication it would work.

A Geek Squad Agent was going to download a Linux system into the old machine which would make it a virtual external disk drive for the other machine, but it requires a minimum of 64MB memory.
I'd prefer not to try things that involve taking off the back of the machine (such as loading the two internal disks of the old machine as secondary drives on the new machine), as the frame (of the old machine) is bent, and I might not be able to put them back. It's been suggested that the old machine, once scrubbed of personal data, could be donated to charity. It does have legal copies of IE 3, Netscape 4.72, WordPerfect Office 6 (with WP Office 7 CDs available), and some other software which might be useful to someone. But it may be too old for anyone to want.
Any ideas? — Arthur Rubin | (talk) 01:25, 13 July 2006 (UTC)[reply]
Hmm, perhaps you can hook a CD or DVD writer up to the old machine ? That comp does sound pretty old, but perhaps a kindergarten class somewhere could use it (they wouldn't want a valuable comp in there, as it would be ruined the first time a kid spills a juice box into the vents). StuRat 01:55, 13 July 2006 (UTC)[reply]

Alternative solution of i^243

I tried to help the site http://www.exampleproblems.com/wiki/index.php/CV15 with the following solution
$i^243=(i^4)^60*i^3=1^60*(-i)=-i$
but I was not allowed to edit it on the ground that I was not confirmed with email address.
My question: do you want to help by editing it in your own name?Twma 03:16, 13 July 2006 (UTC)[reply]