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May 29

Roots

recently my maths teacher gave me a problem to solve, but i have had trouble with it. i need to find out what z is, in the equation z^3=1, the obvious answer is 1, but he said that it was more difficult than the previous question he gave me, solving the square root of i. He also asked me to find n in the equation, z^n = 1, and he then said there were n answers. any help would be much appreciated, sorry if this is too simple.

This has to do with the beautiful roots of unity. Essentially, the full solution involves complex numbers. If you plot these solutions out in the complex plane, they always end up on a unit circle centered around the origin. --HappyCamper 03:38, 29 May 2006 (UTC)[reply]

They are also equally spaced around this circle. —Bkell (talk) 03:39, 29 May 2006 (UTC)[reply]

THANKS GUYS

No problem! Actually, I think we can add more stuff to the article, like a diagram of a circle along with the position of the roots. Feel free to come back if you have more questions. --HappyCamper 04:32, 29 May 2006 (UTC)[reply]

ok i've had some problems. To solve z^3=1, do you just replace z with (a+bi)^3=1? Because then i get 1 = a^3 - b^3i - 3ab^2 + 3a^2bi and i dont know where to go from here. thanks.

The trick is not to use the representation z = a + bi, but rather z = r * exp(i θ). Now, I'll just give a colloquial sketch of what is going on. Because you know it's a root of unity, r = 1. So, you know that z = exp(i θ).

Now, to go around a circle, you need to traverse 2π radians. Since you are trying to find z^3=1, you look at the exponent and take a note that there is a 3. What angles (in radians) give multiples of 2π when multiplied by 3?

One of them is obviously 2π/3. There's also 4π/3. There's also plain old zero. Note how these can all be written like

\theta =k{\frac {2\pi }{3}}

, where k = 0, 1, 2.

So, your solutions are of the form

z=\exp(i{\frac {k2\pi }{3}})

Now, let's just substitute in the k...these are your values for z

z=\exp(0),\exp(i{\frac {2\pi }{3}}),\exp(i{\frac {4\pi }{3}})

Finally, use Euler's formula,

e^{i\theta }=\cos \theta +i\sin \theta \,\;

and you get your answers

z=1,-{1 \over 2}+i{\frac {\sqrt {3}}{2}},-{\frac {1}{2}}-i{\frac {\sqrt {3}}{2}},

Does this help? --HappyCamper 05:02, 29 May 2006 (UTC)[reply]

yeah, thanks a lot. Do u have a maths degree?

On second thought, I think your question can be answered much much simpler, although I haven't worked it out entirely yet.

Take the approach you had before, set z = a + bi, and just expand. You get a³ - ib³ - 3ab² + 3ia²b = 1. Now, you invoke this insight: The real and imaginary parts on boths sides must be equal! So, you get two simultaneous equations for a and b:

a³ - 3ab² = a(a² - 3 b²) = 1

3a²b - b³ = b(3a² - b²) = 0

From the second equation, you get that either b = 0, or

b=\pm {\sqrt {3}}a

. You see where this is going? Substitute these into the first and solve for a, and you get the same solutions as above. Much much easier than having to wave magic wands with complex exponentials and such. However, there is a lot of insight you can get from understanding the complex exponentials. I'd encourage you to revisit this problem again when you get the chance to. You might also wonder about...what is the quantity

i^{i}

, where i is the imaginary unit? Hm.... --HappyCamper 05:22, 29 May 2006 (UTC)[reply]

Right, that's a very important princple: any equality of complex numbers corresponds to two independent equalities of real numbers. —Keenan Pepper 05:46, 29 May 2006 (UTC)[reply]

Is there an article with a formal proof of this on Wikipedia? --HappyCamper 05:47, 29 May 2006 (UTC)[reply]

That's the definition of equality of complex numbers: Two complex numbers are equal if and only if their imaginary parts are equal and their real parts are equal. —Bkell (talk) 05:51, 29 May 2006 (UTC)[reply]

This was missing from complex number! [1] --HappyCamper 05:54, 29 May 2006 (UTC)[reply]

The problem as posed makes a hidden assumption. Any time we look for zeros (roots) of a polynomial, we must first know what kinds of values we are allowed to substitute for the unknown. For example, the polynomial

x+1\,\!

has a zero when x is −1, but that may not be admissible if we require a positive whole number. If we allow a negative whole number then we have exactly one root. Similarly, the polynomial

2x-1\,\!

has a zero when x is ¹⁄₂, but that is not a whole number and so may not be acceptable. Or consider

x^{2}-2\,\!

with a zero when x is √2, which is not a fraction! (This is the length of the diagonal of a square by the Pythagorean theorem, and the discovery that it was not a "proper" number deeply disturbed ancient Greek mathematicians.) Notice that all these polynomials are extremely simple, and all have integer coefficients.

Now look at the polynomial

x^{2}+1\,\!

which might seem to have no roots at all. For, the numbers we are accustomed to using in arithmetic have the property that any non-zero number squares to a positive number, yet we require a number whose square is −1. Is there no end to the need to enlarge our concept of number, from natural numbers to integers to rational numbers to real numbers to these new beasts? To our great relief, when we expand from real numbers to complex numbers by including a value that squares to −1, and commonly denoted by ⅈ, along with all multiples and sums with it, we find that no polynomial can ever again stop us. Hurray! Complex numbers make our life simple.

The usual picture is that real numbers extend smoothly left and right along a linear continuum, and complex numbers extend above and below that line as well. To turn 1 into −1 requires a 180° reversal; to do that in two steps (squaring) requires a 90° turn, which is what ⅈ accomplishes. Notice that in this case the magnitude of the number — its distance from zero — remains one before, during, and after the turn. Observe that 1 itself has two numbers that square to it: +1, which involves either a 0° or 360° turn, and −1, which involves two 180° turns. Also observe that −1 has two square roots: ⅈ, which involves two 90° turns, and −ⅈ, which involves turns of 270° or −90°.

With a little imagination we can picture three options for a turn that, when repeated three times, gives (an integer multiple of) 360°. These would be "cube roots of unity". We note that paired with every "positive" turn there will be a matching "negative" turn; this corresponds to replacing ⅈ by −ⅈ, an operation called complex conjugation. --KSmrq^T 05:56, 29 May 2006 (UTC)[reply]

It should be pointed out that the reason that squaring numbers like 1, −1, i, and −i does not change the number's magnitude is that the magnitude of all these numbers is 1, and 1² = 1. In general, when you square a complex number z, the magnitude of the resulting number will be the square of the magnitude of z. —Bkell (talk) 06:11, 29 May 2006 (UTC)[reply]

well. that was a pretty hard question considering the last thing we did in maths was inverse functions and we havent even been taught complex numbers yet.

ok i get up to where you said, but when trying to solve a, how does a^3 + 9a^2 - 1 give you -1/2 for a? and also with a^3 - 9a^2 - 1 = -1/2?

HappyCamper's formula

b=\pm {\sqrt {3}}a

was presented in a somewhat non-conventional way. More conventional is

b=\pm a{\sqrt {3}}

, which makes clear that the occurrence of a is not under the square root's vinculum, which would have been

b=\pm {\sqrt {3a}}

. So b² = 3a² (which is where we came from), and not 3a. --Lambiam ^Talk 10:19, 29 May 2006 (UTC)[reply]

Thank you to everyone who contributed in answering this question.

Math Question 1

If there is a ticket raffle, and the prize is worth $1,000, with 100 tickets outstanding I should clearly purchase tickets at $1 each, as they are worth more than that, but as I do, I increase the # of tickets oustanding, and at some point (1000 tickets I think) it is no longer advantagous to purchase tickets. (is this correct?)

But lets say I own 5 of the 100 oustanding tickets. These tickets would be valued at $10 each ( I think). What if instead of ONE $1000 price, there are 5 $200 prizes. How would this effect the "intrinsic value" of each ticket if:

-after a ticket is drawn it is put back into the pool and can be drawn again for one of the other prizes

-After a ticket is drawn it isnt put back but the ticketholder can win on the other tickets.

-After a ticket is drawn the ticketholder is ineligable to win other prizes regardless of his remaning tickets?

Anyone? 12.183.203.184

If there are 100 tickets outstanding, and you buy one, then there is a 1/101 chance you will win $1000, and a 100/101 chance you will win nothing. So the expected value of your winnings is

{1 \over 101}\times \$1000+{100 \over 101}\times \$0\approx \$9.90.

So for that first ticket, you should be willing to pay up to $9.90. Suppose you buy it. Now you consider whether to buy a second ticket. Since there are now 101 tickets outstanding, the chance that your second ticket will win $1000 is 1/102. So now the expected value of your winnings from the second ticket is

{1 \over 102}\times \$1000\approx \$9.80.

So it seems that you should pay up to $9.80 for that second ticket.

The problem, though, is that buying the second ticket changes the odds, and so the calculation for the first ticket is no longer valid. If you buy two tickets, then you have two chances in 102 of winning the thousand dollars, so the expected value of your winnings is

{2 \over 102}\times \$1000\approx \$19.61,

so if you are planning to buy two tickets the total amount you should pay should be no greater than $19.61. (This of course assumes that no one else will be buying any other tickets; if you buy your two tickets for $19.61, and then someone else buys a hundred tickets, then you will have overpaid.)

You can continue this line of thought to figure out how many tickets you should buy for a dollar each. If you buy n tickets, then the expected value of your winnings is

{n \over 100+n}\times \$1000.

You're going to have to spend n dollars to buy n tickets, so if this expected value is less than n, you shouldn't buy so many. —Bkell (talk) 02:34, 30 May 2006 (UTC)[reply]

Now suppose that you have five of the 100 outstanding tickets and there are five $200 prizes. If tickets are put back into the pool after being drawn, then in each drawing you will have five chances in 100 of winning $200. This means that the expected value of your winnings is

{5 \over 100}\times \$200+{5 \over 100}\times \$200+{5 \over 100}\times \$200+{5 \over 100}\times \$200+{5 \over 100}\times \$200=\$50,

which means that your five tickets together are worth $50 (that is, $10 each). In fact, as long as the tickets are returned to the pool after each drawing, it doesn't matter how the $1000 is divided up. There could be a hundred thousand drawings for a penny each, and your expected winnings would still be $50. (You will find, however, that as the $1000 is divided ever more finely, you have a greater chance of winning closer to $50; that is, it becomes less and less likely that you will win $0 or $1000, and more and more likely that you will win $50, or $45, or $55.) —Bkell (talk) 02:53, 30 May 2006 (UTC)[reply]

Math probibilty question 2

Say a casino sells the "oppertunity" to roll a 6-sided dice. You win the number rolled x 100. e.g. 2 = $200. The value of this appears to be $350. If the casino allowed you to reroll at your choice, what would the value be? If you could reroll twice (3 rolls total) what would the value be, and what number should you take the reroll option if the dice number turned up is less than? I think this should be higher on the 1st reroll than on the 2nd option. Is that right? How would one calcualate that? Thanks! 12.183.203.184 08:17, 29 May 2006 (UTC)[reply]

With one reroll, you would reroll if the expected value of the reroll is more than the amount you have already won: that is, if the die currently shows a 1, 2 or 3 ($100, $200 or $300). Thus the expected value of that situation is

{\begin{matrix}{\frac {1}{6}}\end{matrix}}\left(\$350+\$350+\$350+\$400+\$500+\$600\right)=4{\begin{matrix}{\frac {1}{4}}\end{matrix}}\times \$100

.

From this point on, we can remove the $100 factor, because it just serves to confuse things. This means that after the first roll of three, the remaining two rolls are worth

4{\begin{matrix}{\frac {1}{4}}\end{matrix}}

if exercised; thus they will be exercised if the current value of the die is less than

4{\begin{matrix}{\frac {1}{4}}\end{matrix}}

i.e. on a 1, 2, 3 or 4. So the expected value of 3 rolls is

{\begin{matrix}{\frac {1}{6}}\end{matrix}}\left(4\times 4{\begin{matrix}{\frac {1}{4}}\end{matrix}}+5+6\right)=4{\begin{matrix}{\frac {2}{3}}\end{matrix}}

. EdC 14:13, 29 May 2006 (UTC)[reply]

In other words, denote the value of n rolls by

v_{n}

. What Edc meant was that this value satisfies a recurrent equation (see Conditional expectation)

v_{n+1}=E(x|x\geq v_{n})P(x\geq v_{n})+v_{n}P(x<v_{n})

and forms a sequence 3.5,4.25,4.66,4.94,5.13,5.27,...(Igny 16:52, 29 May 2006 (UTC))[reply]

Or, equivalently but conceptually simpler:

v_{n+1}={\frac {1}{6}}\sum _{i=1}^{6}\max(i,v_{n}).

For n ≥ 5 a closed form expression is given by v_n = 6 − (6768/3125)(5/6)ⁿ. --Lambiam ^Talk 17:19, 29 May 2006 (UTC)[reply]

While we just provided a correct answer, I wonder if the proof is needed that

v_{n}

is indeed the maximal value of n rolls, and our strategy is optimal. (Igny 18:05, 29 May 2006 (UTC))[reply]

I think it's obvious. But on demand a proof can be supplied. --Lambiam ^Talk 23:04, 29 May 2006 (UTC)[reply]

Salary Statement

How to prepare a SALARY STATEMENT of any type in Ms Excel?--86.62.239.145 10:55, 29 May 2006 (UTC)[reply]

You will probably like Microsoft InfoPath (even although it is used to make XML forms, it can be used for receipts; anyway, if you dont have that you should just lay it out as it appears in most wage slips. Kilo-Lima|^(talk) 16:50, 29 May 2006 (UTC)[reply]

Windows Media Audio

Hi, I went and ripped some files from a bought music CD and it was extracted as .wma format. I then put it into WMP library. I then wanted to burn them to a CD, so I put in a blank disk and I keep getting the same error that it is not blank—I can assure you, it is blank! Does anybody know what's happening here? Thanks, Kilo-Lima|^(talk) 16:46, 29 May 2006 (UTC)[reply]

Ugggh...now you know why I like iTunes (no WMA crap). But anyways...maybe your CD is messed up (defective, etc.)--could somebody have "borrowed" when you weren't looking? --M1ss1ontomars2k4 ^{(T | C | @)} 21:41, 29 May 2006 (UTC)[reply]

Yeah, iTunes is totally free of DRM and everything, too. Dysprosia 22:20, 29 May 2006 (UTC)[reply]

May 30

Cell CPU

Hey ppl, i hope this is the right place to ask about the Cell CPU thats gonna be used in the Playstation 3. I'v asked SOOO many ppl and they aren't quite sure how it works either. Please, can someone explain to me how all the Synergistic Processing Elements (SPE's) work together to process the info and why game developers developing games for the PS3 say its so hard to program for 7 cores? Even the main wikipedia article on the Cell says its too complicated. Thx! KittenKiller

It's mostly because it's a very uncommon architecture, so almost nobody has any experience with it.

Historically, consumer-level computers have had a single general-purpose processor. Programming in such a situation is very well-understood: it's simply a matter of dividing up the available time between tasks. (For example: five milliseconds for AI, one for checking for user input, three for updating the sound effects, one for updating the game logic, twenty-five for updating the screen, repeat)

More recently, computers have started being produced with multiple identical general-purpose processors. Writing programs for this sort of system is fairly well-understood in theory: you divide up the tasks in such a way that the workload can be balanced among the processors. In practice, it's a bit harder, because some tasks can't be divided up, and you need to make sure the tasks don't step on each other's toes, but it's still familiar ground. (For example, the game AI would be one or more tasks, physics would be a task, generating sound effects would be a task, playing the background music would be a task, checking for user input would be a task, updating the graphics would be a task, and so on)

Then, there's the Cell CPU. It's got one general-purpose processor and seven special-purpose processors (the SPEs) that are extremely efficient for certain types of tasks, and very slow for others. Nobody's quite sure how to handle such a situation, or what the seven SPEs should be used for. It's known that they're good for graphics, but the PS3 has a dedicated graphics processor that's even better. They're probably good for sound effects, but that won't take the effort of more than one SPE. They're not good for AI, or for user input, or for background music. They might be good for physics, but nobody knows -- simulating physics is cutting-edge research right now. --Serie 22:48, 30 May 2006 (UTC)[reply]

Series

What is this series of numbers called:

1, 11, 21, 1211, 111221....?

Looks like the see-and-say sequence. Dysprosia 08:11, 30 May 2006 (UTC)[reply]

In mathematical jargon this is not a series. Goes to show that mathematicians are strange people. --Lambiam ^Talk 09:58, 30 May 2006 (UTC)[reply]

There's a difference between series and sequences. OEIS A005150 : Look and Say sequence: describe the previous term! (method A - initial term is 1 - Formerly M4780) --DLL 20:32, 30 May 2006 (UTC)[reply]

Of course, but I think it's quite obvious what the poster above meant, so I don't think there's need for excessive concern about this. Dysprosia 22:19, 30 May 2006 (UTC)[reply]

Noo, I'm more concerned about the mathematicians ;-) Lambiam ^Talk 02:50, 31 May 2006 (UTC)[reply]

Well telling someone once is fine, but twice is getting to be a bit onerous ;) Dysprosia 03:09, 31 May 2006 (UTC)[reply]

Watermellon. Black Carrot 00:23, 1 June 2006 (UTC)[reply]

I don't get it. Dysprosia 02:01, 1 June 2006 (UTC)[reply]

Nothing. It was just that several posts had gone by without anyone saying anything, at all. I was taking that a bit farther. Black Carrot 02:13, 2 June 2006 (UTC)[reply]

Read the following aloud: one, one "one", two "ones", one "two" and one "one", one "one" and one "two" and two "ones"... See it? For even more fun, prove that "four" will never appear in this sequence. PS: I did saw it mentioned at the end of a maths book, but it didn't mention its name. --Lemontea 11:25, 5 June 2006 (UTC)[reply]

Font

for some unknown reason, the font in which i view wikipedia has changed to tahoma, and i would like to revert it. When editing articles my font is times new roman, but after i have edited them, it appears in tahoma. Can anyone shed any light on my problem?

Character encoding ? In Firefox the menu offers this entry with plenty of encoding possibilities. In other brothels it may be worse, brother. --DLL 20:28, 30 May 2006 (UTC)[reply]

Browser settings, font availability, and custom Wikipedia monobook.css can all affect what's displayed. Usually the text typed in the edit window appears in a fixed-width font like Courier, not Times New Roman; and usually the text for the preview or regular display of an article appears in a sans-serif font like Arial. The Tahoma font would be a reasonable substitute for Arial, but are you sure the editing font is proportionally spaced? --KSmrq^T 01:50, 31 May 2006 (UTC)[reply]

quantization

2 questions that keep cropping up in old exam papers, which will hopefully get me a few extra marks:

Define a scalar quantizer with M interval
Define a scalar quantization problem

I'm a bit rusty on these, plus Wikipedia has no articles on them. Any help would be appreciated. --Talented Wikipedian File:Kiss.png 10:42, 30 May 2006 (UTC)[reply]

Could you help us by giving a hint what field the exam is about: signal processing, perhaps more specifically image processing, sound processing, and music, or is it perhaps about physics? --Lambiam ^Talk 13:55, 30 May 2006 (UTC)[reply]

It is part of a data compression module, and is some kind of lossy compression, I can't be sure whether it is to do with sound files or image files. --Talented Wikipedian File:Kiss.png 16:36, 30 May 2006 (UTC)[reply]

I suspect that "M interval" is not standard terminology. Could the meaning be "M intervals" (or "levels"), as in 16 intervals, 256 intervals, 65,536 intervals, ..., M intervals? Here are a few links that may be helpful to you:

--Lambiam ^Talk 18:19, 30 May 2006 (UTC)[reply]

Perhaps this is meant in contrast to vector quantization? --KSmrq^T 22:37, 30 May 2006 (UTC)[reply]

astronomy-orbit of planets

What branch of calculus do I need to understand the position of a planet orbiting the sun at any given time?

Thank you, Denis

Assuming that you are happy enough with Newton's law of universal gravitation and do not require General relativity, what you need is the area of Ordinary differential equations, also known as ODEs, together with a general understanding of Newton's laws of motion and their role in Mechanics. --Lambiam ^Talk 18:29, 30 May 2006 (UTC)[reply]

And if that's a bit much for you, try geometry, specifically conic section geometry. StuRat 22:38, 30 May 2006 (UTC)[reply]

That will tell you where, but you'll need Kepler's laws to know when where. I see now that the Kepler article shows how to derive these laws from the ODE's given by Newton's law. --Lambiam ^Talk 02:58, 31 May 2006 (UTC)[reply]

Rope strength

If a single strand manilla rope has a working load of 100 lbs., does that mean that a triple strand rope of the same material has a working load of 300 lbs.? Is there a formula for figuring this out? Thanks, Pdpdpd1

If the box in Exercise 5.14 weighs 20 kg and has a coeficent of friction of .2, what is the minimum strength rope that can be used to pull it?

A rule of thumb for question 1 would say that there is always one of the rope that bears more weight. So do not triple it. But as the rope itself is made of strings, the standard load might be over . So try to triple it and do not stay under the load. --DLL 20:24, 30 May 2006 (UTC)[reply]

Question 1 is a yes, as long as it still obeys Hooke's law at that point. At least I think so. Question 2 deals with Hooke's Law and the Young modulus, but I don't understand what the "strength" refers to. But you could find out what the frictional force is and find the force required to pull the box. Remember, we don't do your homework. x42bn6 Talk 07:56, 31 May 2006 (UTC)[reply]

why is in locally pathconnected topological space, path connected component same as connected component

http://en.wikipedia.org/wiki/Connected_space#Local_connectedness

defines the concept of a locally path connected topological space. Now I wanna prove that in a locally pathconnected topological space, path connected components and connected components are the same.

Now I do know that this is a relevant fact : in a locally pathconnected space, every open set U has (in its induced topology) open pathconnected components.

But how to use this?

I have gotten this far : let C be the component of x, and P the strictly contained in C pathconnected component. Now I now C will be closed. I cannot say that of P. I need to find contradiction though...

Thanks,

Evilbu 23:08, 30 May 2006 (UTC)[reply]

Well, I hope this is not for homework. What you need to prove is that every connected component is also path connected. To do this, fix a connected component

C

and choose

p\in C

. If you show that the set of all points in

C

that can be connected to

p

by a path is the whole

C

itself, you are done.

To do this, you show that it is both open and closed: as

C

is connected by definition, and nonempty because it trivially contains

p

, you are done.

Openness: suppose

q

can be connected to

p

by a path. Let

U

be a neighbourhood of

q

which is path connected (it exists because the space is locally pathconnected). Then any point in

U

can be connected to

p

by a path simply by going to

q

first and then to

p

. Hence, your set is open.

Closedness: consider the subset of

C

that cannot be connected to

p

by a path, and choose

q

as such. Then choose a neighbourhood

U

as above: if any point of

U

could be connected to

p

by a path, then by following the two paths as before you could also connect

q

to

p

. This means that this set is also open, i.e. the set of points that can be connected to

p

by a path is closed.

Q.E.D. Cthulhu.mythos 09:12, 31 May 2006 (UTC)[reply]

Hmm, yes I see now. I guess one could not from the start that in a locally pathconnected space, all pathconnected components are open, right? Nope, this is not homework, yet it is not unrelated to university work, we did it quickly in class and I wanted to clear things up for myself. Evilbu 10:45, 31 May 2006 (UTC)[reply]

May 31

Integration of sin and cos

What happens when you integrate sin(), -sin(), cos() and -cos()

Well, what happens when you differentiate any of them? Notice something interesting? Then just apply the Fundamental Theorem of Calculus and you're done. If that's not enough, check List of trigonometric identities#Calculus. Confusing Manifestation 12:01, 31 May 2006 (UTC)[reply]

Also check out Table of integrals#Trigonometric functions. -- Meni Rosenfeld (talk) 15:23, 31 May 2006 (UTC)[reply]

Implications

The fact that the differentiation of trigonometric functions (sine and cosine) results in linear combinations of the same two functions is of fundamental importance to many fields of mathematics, including differential equations and fourier transformations. (I added this tidbit at the above article). Nimur 18:06, 31 May 2006 (UTC)[reply]

A sophisticated answer uses Euler's formula, cos ϑ+ⅈsin ϑ = ⅇ^ⅈϑ, and the fact that ⅇ^z is an eigenfunction for the linear operation of integration. For example,

\cos \vartheta ={\frac {1}{2}}e^{i\vartheta }+{\frac {1}{2}}e^{-i\vartheta },\,\!

so

\int \cos \vartheta \,d\vartheta ={\frac {i}{2}}e^{i\vartheta }-{\frac {i}{2}}e^{-i\vartheta }+C.\,\!

A more geometric approach observes a unit circle parameterized by arclength,

(x,y)=(x(\vartheta ),y(\vartheta ))=(\cos \vartheta ,\sin \vartheta ),\,\!

and considers that a tangent to the circle is always a unit vector perpendicular to the radius. Thus starting from the initial condition (x₀,y₀) = (0,−1) and integrating (cos ϑ,sin ϑ) traces out a counterclockwise unit circle parameterized by arclength, (cos(ϑ−^π⁄₂),sin(ϑ−^π⁄₂)). This serves to remind us of the important role of constants of integration, here a consequence of the initial condition.

The best approach is to pay attention in class, read the textbook, and ask the instructor or teaching assistant for homework help. --KSmrq^T 18:42, 31 May 2006 (UTC)[reply]

Thank you for your help, I hope I can get to a stage of complete understanding - there is a few years of work ahead of me yet though.

very exact definition of "orientable topological manifold"?

From http://en.wikipedia.org/wiki/Manifold

" Overlapping charts are not required to agree in their sense of ordering, which gives manifolds an important freedom. For some manifolds, like the sphere, charts can be chosen so that overlapping regions agree on their "handedness"; these are orientable manifolds. For others, this is impossible. "

Okay, but strictly speaking, this is still not really exact. I would like to have a very explicit definition of "agreeing on handedness". I have some very fundamental experience with charts of (topological varieties), and I know that in a vector space of an ordered field, two bases are considered having the same orientation if their matrix has positive determinant.

So how can it make this precise?

Thanks,

Evilbu 13:28, 31 May 2006 (UTC)[reply]

The problem here is whether we want the article to report the formal definition or not. If this is the case, suppose

\phi _{U}:U\rightarrow \mathbb {R} ^{n}

and

\phi _{V}:V\rightarrow \mathbb {R} ^{n}

are two charts. Then

\phi _{V}|_{U\cap V}\cdot \phi _{U}|_{U\cap V}^{-1}:\mathbb {R} ^{n}\rightarrow \mathbb {R} ^{n}

is a homeomorphism: if for every choice of

U

and

V

it preserves orientation, then the manifold is said to be orientable. Cthulhu.mythos 16:44, 31 May 2006 (UTC)[reply]

thanks, but could you explain what you mean by 'it preserves orientation"?

Keep in mind that the manifold article is an overview of all different kinds of manifold, so deliberately does not dive into all the details. The topological manifold article should include a precise definition of orientability; however, these specialized articles have been starved for attention as numerous editors have nitpicked the survey article.

</Mount Soapbox>

The definition of a topological n-manifold states that it is a topological space in which every point has an open neighborhood homeomorphic to an open n-ball, Bⁿ. (Typically, the space must also be Hausdorff, paracompact, and second-countable; but those are technical details that needn't concern us here.)

Now fix B^{n^{, and choose a list of n+1 points (vertices) in general position within it to form a simplex (or choose an ordered basis of n vectors for the Rⁿ in which it lives). The choice imposes one of two possible orientations on Bⁿ, which we may call "right-handed" or "left-handed". If we exchange two vertices in a right-handed simplex we obtain a left-handed simplex, and these are topologically distinct in the sense that we cannot deform one into the other. For example, the triangle ABC in a planar disc cannot be deformed to ACB without violating its topology.}}

By the axioms of topology, if two open neighborhoods have a non-null intersection, then the intersection is an open set. And since a homeomorphism is a continuous map with a continuous inverse, we can create a composite map from Bⁿ to itself by way of the two neighborhood homeomorphisms. By a suitable deformation we can pass the simplex through this map and compare the image to the original. Thus we can decide if the overlapping homeomorphisms imply the same or opposite orientations.

The key observation is that we can cover some manifolds with neighborhoods so that overlapping neighborhoods always agree in orientation; these are orientable manifolds. Significantly, this is not always possible.

A basic example of a non-orientable manifold is the projective plane, RP². We can think of this as a sphere with opposite points on the surface identified, or as a disc with opposite points on the circular boundary identified. A 2-simplex is merely a triangle, and a triangle in B² maps to a three-sided region in RP², which might as well be a triangle itself. But because the manifold is non-orientable, there is no essential difference between ABC and ACB. For example, put C on the boundary of the disc, with A and B inside; then the identification of opposite boundary points means we cannot distinguish the two orientations of the triangle. We might see this more directly with the sphere model: Draw the triangle as ABC in the northern hemisphere, then use the identification of opposite points to transfer it to an oppositely oriented triangle in the southern hemisphere.

With an orientable manifold, such as an ordinary sphere, S², this kind of reversal is impossible. We can consistently define a right-handed triangle as distinct from a left-hand triangle, and never the twain shall meet. One point of caution is that the dimension of the simplex must be the same as the dimension of the space. For example, we can easily flip a triangle (a 2-simplex) in 3-space; we must work with a tetrahedron (a 3-simplex) to explore orientability in 3-space. --KSmrq^T 23:36, 31 May 2006 (UTC)[reply]

You mention that a manifold must be paracompact. Isn't this redundant for a second countable space locally homeomorphic to Rⁿ? -lethe ^{talk +} 04:21, 1 June 2006 (UTC)[reply]

I just mentioned some of the additional conditions that often appear, without regard for independence. For example, it is not uncommon for a paracompact space to be required to be Hausdorff. Anyway, the paracompact space article mentions that "Every locally compact second-countable space is paracompact." And the locally compact space article points out that "Topological manifolds share the local properties of Euclidean spaces and are therefore also all locally compact." (There may be a caveat about Hausdorffness in there.) Fortunately, we can discuss orientability without concern for these extra conditions. --KSmrq^T 06:35, 2 June 2006 (UTC)[reply]

Evilbu: it is worth mentioning that manifolds with additional structure admit alternate definitions for orientability, which of course agree with the topological definition. A differentiable manifold if it admits an atlas whose transition functions have positive determinant on the tangent bundle. Equivalently, it has a nowhere zero differential n-form. -lethe ^{talk +} 04:24, 1 June 2006 (UTC)[reply]

And also this one: a differentiable manifold

M

is not orientable if and only if it admits a loop

\gamma

(with base point

p

) such that if you choose a base of

T_{p}M

having

\gamma '(0)

as its first vector and then drag it along

\gamma

using charts, when you came back to

p

at time

1

you have a base of

T_{p}M

whose orientation is reversed w.r.t. the base you started from. Cthulhu.mythos 08:56, 2 June 2006 (UTC)[reply]

I would also remark that orientability is an intrinsec property, and does not depend on how the manifold can be embedded in some euclidean space. For example, it occurs frequently to hear that "the Moebius strip is not orientable because it has only one side and one boundary, whereas an annulus is orientable because it has two sides and two boundaries". But that is not correct. A compact surface properly embedded in a 3-manifold (and the Moebius strip is NOT properly embedded in

\mathbb {R} ^{3}

, because of the boundary) is two-sided if it has a tubular neighbourhood which is just a product of a copy of itself with

[-1,1]

, and it is one-sided if its tubular neighbourhood is a twisted

I

-bundle over the same surface. Example (call

M

the Moebius strip once and for all): in the manifold

M\times S^{1}

the levels

M\times \{p\}

are two-sided, whereas the torus

\gamma \times S^{1}

(where

\gamma

is the core of the Moebius strip) is one-sided. Cthulhu.mythos 10:15, 1 June 2006 (UTC)[reply]

Seems like I'll create one-sided (submanifold) and two-sided (submanifold) soon... Cthulhu.mythos 10:17, 1 June 2006 (UTC)[reply]

I don't understand why the Möbius strip isn't properly embedded in R³. On the contrary, I'm pretty sure the Möbius strip can be embedded in R³. I made one in third grade with scissors and paper, and it wasn't on the day we took the field trip to R⁶. -lethe ^{talk +} 17:37, 1 June 2006 (UTC)[reply]

Not properly embedded. The boundary, y'know. EdC 00:57, 2 June 2006 (UTC)[reply]

I guess I don't know what a proper embedding is. Will you tell me? (And what it has to do with boundaries.) -lethe ^{talk +} 01:04, 2 June 2006 (UTC)[reply]

So the article on embedding didn't mention proper embeddings. It does now, although I'm not sure how to make it clearer. A proper embedding preserves boundaries; for example, embedding

S^{1}\subseteq \mathbb {R}

in

D^{2}\subseteq \mathbb {R} ^{2}

as a chord is proper, but embedding it as a radius is not. So the Môbius strip can't be properly embedded in

\mathbb {R} ^{3}

, because the Môbius strip has a boundary (

S^{1}

) and

\mathbb {R} ^{3}

doesn't. (This raises the question of what subspaces of

\mathbb {R} ^{3}

the Môbius strip can be properly embedded in. I think it's obvious that the boundary has to have positive genus, though I'm not sure how to prove it.) -- EdC 04:16, 2 June 2006 (UTC)[reply]

Oh, right. Assume that M is embeddable in R³ less a contractible subspace A. Then ∂M (= S¹) is contained in the boundary of A, so its contraction S* is D². So M ∪ S* is an embedding of the projective plane in R³, which is a contradiction (there is no such embedding). -- EdC 04:29, 2 June 2006 (UTC)[reply]

The definition of a proper embedding was not precise, I fixed it. Cthulhu.mythos 08:48, 2 June 2006 (UTC)[reply]

Ooh, forgot the transversality condition. Thanks. EdC 11:59, 2 June 2006 (UTC)[reply]

Many thanks to both of you. So a proper embedding is an embedding that takes the boundary to the boundary in a nice way, right? -lethe ^{talk +} 20:03, 2 June 2006 (UTC)[reply]

htaccess for Zeus Web Server

I need this htaccess file written for Apache to be re-writen for Zeus Web Server

      Options FollowSymLinks
      RewriteEngine On
      RewriteCond %{REQUEST_FILENAME} !-f
      RewriteCond %{REQUEST_FILENAME} !-d
      RewriteRule ^(.+)$ /index.php?title=$1 [L,QSA]

Thanks, Gerard Foley 21:49, 31 May 2006 (UTC)[reply]

Nellore Telugu Measurement

I would like to know exactly how many feet are there in an "Ankanam" based on Nellore region, AndhraPradesh, India. Thank You

Maybe the question is flawed. One site listing Nellore housing says "One Ankanam =4 sq.yds.", while an online Telugu-English dictionary translates "ankaNam n. space between two beams or pillars in a house." Whichever we pick, the question makes no sense. --KSmrq^T 23:58, 31 May 2006 (UTC)[reply]

Real Analysis - teaching and learning

What is (or should be) gained by undertaking an introductory course in Real Analysis? What are the implications of this for the way it's taught? Thanks. --The Gold Miner 23:35, 31 May 2006 (UTC)[reply]

I always thought real analysis was just "calculus again, but for serious this time". You go over the definition of limit again, use fancy topology words like compact, and focus on pathological examples like differentiable functions whose derivatives are not continuous. —Keenan Pepper 05:39, 1 June 2006 (UTC)[reply]

The main benefit of a real analysis course is not so much the content, as the mathematical maturity gained by taking the course. Quite often RA courses are the student's first serious introduction to writing detailed proofs and working with high-level (i.e. Rudin) mathematics textbooks. 18.228.1.113 13:34, 1 June 2006 (UTC)js[reply]

Vast numbers of first-year university students are required to learn calculus, for use in a variety of different fields. Therefore the typical calculus textbooks and courses are designed to serve this genuine need for the broad community. However, this is rather different from training mathematicians. Weierstraß and others went to a great deal of trouble to build solid foundations under calculus, which is how real and complex analysis began, but the journeyman calculus applications rarely trouble themselves with such details. Complex analysis, which builds on real analysis, provides an indispensable set of tools for a great deal of modern mathematics. One last observation, admittedly a little biased: I don't recall ever hearing anyone, not even a non-mathematician, complain about knowing too much mathematics. In fact, I'll close with two quotations from Albert Einstein:

In 1943 he answered a letter from a little girl who had difficulties in school with mathematics: "…Do not worry about your difficulties in Mathematics. I can assure you mine are still greater."
"Any man who can drive safely while kissing a pretty girl is simply not giving the kiss the attention it deserves."

The first is well-known; the second shows his true genius. ;-) --KSmrq^T 06:48, 1 June 2006 (UTC)[reply]

Oh, we're listing choice quotes, are we? Here's an interesting anecdote. When Weisskopf was asked "How much mathematics does a theoretical physicist need to know?", he answered simply "more." -lethe ^{talk +} 17:47, 1 June 2006 (UTC)[reply]

Thanks to everyone who responded. My reason for asking is that I recently took such a course and found it a demoralising experience. Students at my University come from a wide variety of backgrounds and their levels of ability vary considerably. Success depends largely on diligent preparation for the end of semester examination. In a perfect world success in exams should, I think, depend on having developed a strong working intuition for the subject, and on being sharp and imaginative on the day. However, because of the way this particular course was taught, the exam positively encouraged cramming and parrot-fashion regurgitation of results and proofs, rather than said intuition and imagination. This rendered the course unfortunately somewhat pointless. I just wonder, given the subject matter and students of mixed ability, if it could ever have been otherwise... --The Gold Miner 15:26, 1 June 2006 (UTC)[reply]

Yes, it could have been different. I was a physics major who had only taken calculus type courses until my senior year, when I took advanced calculus. For the first time, my understanding was limited only by the time I spent thinking about things, and not by the vagueness of definitions, etc. Now, I'm a mathematician, and it is very satisfying (both research and teaching). Here's a quote from a recent student: "I maintain that taking Advanced Calculus was the turning point of my undergraduate career. It had reminded me how beautiful math is, how much it means to me, and how much I was willing to work in order to succeed in it." (Cj67 20:12, 1 June 2006 (UTC))[reply]

It's bad enough that a first-year calculus course might be taught so poorly; it's tragic that this should have been afflicted on aspiring mathematicians. But I must warn you that mathematics is not a field in which participants self-select because of their superior social skills (like teaching ability). Some mathematicians are awesome teachers; others are horrid. The ability to see a subject clearly for oneself is not at all the same as being able to communicate that insight clearly to others. So it has always been. This is one reason for Stigler's law of eponymy. --KSmrq^T 01:57, 2 June 2006 (UTC)[reply]

June 1

Problem with complex numbers

Hello,

You are given that the complex number alpha = 1 + j satisfies the equation z^3 + 3z^2 + pz + q = 0, where pand qare real constants. (i) Find and in the form a + bj. Hence show that and p = -8 and q = 10 [6] (ii) Find the other two roots of the equation. [3] (iii) Represent the three roots on an Argand diagram. [2]

Thanks guys as always. DR Jp.

This sounds like a homework problem. Is there a particular thing you don't understand? Maybe you should take a look at complex number and complex plane. —Bkell (talk) 00:47, 1 June 2006 (UTC)[reply]

Thanks, but I'm currently writing a book and I would like someone to check these problems are doable. If they could put their working so I could see how you are thinking that'd be even better.

How much time is needed to travel around the world for the following modes of transportaion -- the space shuttle, a jet airliner, a cruise ship

Please make sure that your proxy server is either configured correctly, or you refrain from using them, since each time you edit you introduce backslashes before "'"s, breaking the markup. Dysprosia 01:02, 1 June 2006 (UTC)[reply]

Do you take us all for idiots? Do your own homework. (People who genuinely write books are typically careful with language, and have students, colleagues, and paid reviewers — not to mention an editorial staff — to check their work. Also, it is not helpful to see how a trained mathematician approaches a problem if you want to know how a student would think.) --KSmrq^T 03:51, 1 June 2006 (UTC)[reply]

Heh, writing a book, that's a good one. —Keenan Pepper 05:36, 1 June 2006 (UTC)[reply]

Yes, the problem is doable. I would start with part (ii) - find the other two roots of the equation. You know the value of α. Once you know that p and q are real, a second root is immediately obvious. And you can find the third root because you know the co-efficient of z². Once you have all three roots, you can find the values of p and q. Gandalf61 08:50, 1 June 2006 (UTC)[reply]

Starting with (i) is also easy enough, though. --Lambiam ^Talk 14:15, 1 June 2006 (UTC)[reply]

Heh, it looks like an IB Math problem. He put the possible marks next to each part of the question "(ii) Find the other two roots of the equation. [3]". --Codeblue87 19:58, 1 June 2006 (UTC)[reply]

How do I tell if I've chosen an appropriate statistical distribution

I have a group of 12 observations. I'd like to predict what my observations will be in the future. I also need the distribution to apply Bayes Theorem.

Right now, I'm using the normal distribution but I don't know if that's the right choice. I've calculated the skewness and kurtosis of the data, but I don't have any idea what they're supposed to be! I mean, I know if my observations were truly normally distributed, the skewness would be zero, but I don't know if my skewness of 1.65 is "close enough" or what. Are there rules of thumb for this? moink 05:50, 1 June 2006 (UTC)[reply]

Under the assumption of normal distribution, the probability that a sample of 12 observations has a skewness whose absolute value is at least 1.65 is about 0.002. That is fairly low, and normally ground to reject the null hypothesis of normalcy. What is the source of the observations and how critical is the accuracy of the estimated distribution? Often the physical or other origin of the data suggests a plausible crude model for the distribution that is good enough in practice. --Lambiam ^Talk 06:36, 1 June 2006 (UTC)[reply]

It is not particularly critical. I was actually kinda hoping not to have to share the type of data, but since it's apparently a very poor fit to the normal distribution, I guess I will. It's the length of my menstrual cycle. Now all the boys on the math RD can get all grossed out. :) I like to know if I should carry tampons on me, and the Bayes' theorem thing... well, if you're very bright you may be able to figure it out but I will not provide an explanation. Here's the data: 32, 29, 28, 28, 26, 27, 27, 29, 36, 25, 26, 28. moink 07:41, 1 June 2006 (UTC)[reply]

You know, you could use a neural network for precisely this task. Neural networks can be used to predict the length of menstrual cycles as well as stock market values or other things. Choose an encoding for the lengths, train some sort of recurrent network on the data you have, and then get it to generate predictions. If I get time, and I am sufficiently bored, I might even try this for you. Dysprosia 07:49, 1 June 2006 (UTC)[reply]

Sounds cool but beyond my abilities. Right now, though, I'm less interested in predicting exactly the length of the next cycle, and more in knowing the approximate probability that it is at least some length so I can apply Bayes' theorem. moink 07:56, 1 June 2006 (UTC)[reply]

Using your data and the formula at skewness, I find a skewness of 1.27, which is still significantly different from the null hypothesis but less so. Looking at the data, the problem appears to be the outliers at the high side. If you censor the data by discarding values > 30, you get a good agreement with a normal distribution. Given the application censoring at the high side is acceptable, since you want confidence at the low side. The sample is still a bit small, though, to really confidently assume the low end behaves normally, without outliers. --Lambiam ^Talk 14:38, 1 June 2006 (UTC)[reply]

So much for trusting my spreadsheet software. I thought about dropping the large ones, but it's in the higher range that I'm most interested in the probability, and since it seems that it does occasionally get that large, and not that rarely, I wanted to take that into account. moink 15:28, 1 June 2006 (UTC)[reply]

Well, I'm by no means suggesting that this is what you are trying to calculate, but just for the sake of argument: if A=pregnant, and B=menstruation has not yet occurred, and one were interested in P(A|B), then P(B|A) would of course be very close to unity, but what value should be used for P(A)? Would the age specific fertility rate be correct? --vibo56 15:01, 1 June 2006 (UTC)[reply]

Addendum: P(A) would obviously have to be either zero, or a lot higher... --vibo56 15:11, 1 June 2006 (UTC)[reply]

Why would you say that? I mean, it could be zero, but it could be the small numbers you'd get using the failure rates of certain contraceptives. Even with several instances of unprotected sex in a month, it will generally not go above 25-30%. moink 15:24, 1 June 2006 (UTC)[reply]

Agreed. You are right. --vibo56 16:35, 1 June 2006 (UTC)[reply]

Chi squared might be your answer to whether the data is normal or not. Basically this works by dividing up the domain in to a number of boxes, you then count how many of your data items fall into each box and compare with the number predicted from the normal distribution. Add up the square of differences and compare with the approptite Chi-squared statistic. This should give a confidence interval as to whether the difference is significant or not. I suspect with only twelve points you don't really have enough data to meaningfully talk about skew. --Salix alba (talk) 15:10, 1 June 2006 (UTC)[reply]

Sigh. Ok, so I'm transparent. My prior distribution in this context is from a record of instances of penetrative sexual intercourse along with the underlying numbers used by this site combined with a pdf of the date of ovulation using the pdf above and the possibly quite poor assumption of a constant luteal phase of 14 days. moink 15:14, 1 June 2006 (UTC)[reply]

If the goal is to get pregnant, I'd start off by measuring my body temperature, to get a more precise estimate of the time of ovulation. After one year with no success, I would definitely go see a gynecologist. If, on the other hand, the goal is not to get pregnant, and you want a statistical tool to tell you when to start worrying, I'm afraid your approach won't work. Biological distributions tend to have very heavy tails, and you simply do not have enough data to make a sensible estimate of the distribution. With a limited dataset, however, you could make control charts. Here's a link to a how-to (powerpoint), courtesy of the British NHS Modernisation Agency. --vibo56 17:18, 1 June 2006 (UTC)[reply]

When reading my previous comment: forgetting to mention this was maybe a male freudian slip, but anyway: if the goal is getting pregnant, it would be a good idea to have your partner checked as well. --vibo56 21:33, 1 June 2006 (UTC)[reply]

Well, the goal is complicated. It is one, the other, or both of the above, in addition to saving on costs of Human chorionic gonadotropin tests (which have high false negative rates, especially when used too early) by using them at the right time. For example, applying an additional Bayesian update rule, a negative test with a sensitivity of 25 mIU of hCG would reduce my probability by a factor of nearly three if I used it today, while it would reduce the probability by a factor of eight if used tomorrow. And buying a basal thermometer would negate those cost savings. :) The other main goal is the fun of overanalyzing these things. :) moink 04:12, 2 June 2006 (UTC)[reply]

If you check out the presentation that I linked to, and are able not to get too irritated about the "for dummies" manner in which it is presented, you will see that this might be exactly the tool that you are looking for. It is a tool for decision-making, primarily in the manufacturing industry, but it is now mandatory also in blood banks throughout the EU. As you can see from this article, it has been around for a long time, and has stood the test of time. It is a curious mix of parametric and non-parametric statistics.

Your statement on the goal leaves me with the impression that timing is a rather critical issue. I would definitely invest in that thermometer! I wish you all the best, and hope that you achieve your goal and that it brings you happiness. Best regards, --vibo56 23:39, 2 June 2006 (UTC)[reply]

Newton was right after all

Ok, here is this little idea I had today. It has the potential of solving all problems of theoretical physics in one stroke of genius. Basically, the idea is that the interactions in the world are subject to very simple rules, namely Newtonian mechanics. But hey, I hear you say, wasn't there a guy called Einstein who has proven Newton wrong? Well, he did, but it would be Newton who will be having the last laugh.

The world of Newtonian mechanics is very simple: a Euclidean space and a few differential equations, solutions of which are nicely differentiable curves. There is one problem with this world: it is continuous, which makes its "implementation" extremely hard. There are no continuous things in our world: the space is discrete, the time is discrete, and this brings us to the next point: the world is a finite state machine. The world is, basically, a computer. The bit-twiddling aspect of our world is studied by quantum mechanics.

N-body problem is a classical problem of mechanics. There are a few particles floating around in space under the Newtonian law of gravity. It translates into a system of differential equations, which in general is not solvable by mathematical means. We'll try to simulate this problem on a computer. When n=2 theory states that these two bodies would have elliptic orbits (well, not exactly, the orbits may also be hyperbolic or parabolic, but we assume that they are ellptic). What happens when we simulate this on computer? At first everything seems okay, the smaller planet rotates around the bigger one. Let's modify the program so the path of the smaller planet is visible. Then, after a few rotations we'll see a strange effect: the elliptical orbit is slowly turning! More interesting is that the same thing happens in real life: a result predicted by General Relativity theory. But isn't it strange, that Newton's law of gravity when emulated on a computer produces the same effect?

There are several numerical methods for solving differential equations. They all have the same flaw: when you run the method for a long time, round-off errors and discretization errors build up and the result strays off from the right solution. The same thing happened in our simulation - during the rotation the error builds up as we do little discrete time steps, and it rotates the orbit. The same thing, I suspect, happens in our real world, which faithfully tries to solve Newton's differential equation by discrete means.

Comments are welcome. Grue 12:36, 1 June 2006 (UTC)[reply]

It appears unlikely that this theory can be tweaked to give a quantitative agreement with the observations. Using methods such as Runge-Kutta integration, it is easy enough to get a precision that is able to differentiate between Newtonian and Einsteinian gravitation. And since Wolfram we all know that the universe is a cellular automaton. --Lambiam ^Talk 14:08, 1 June 2006 (UTC)[reply]

I think there's a simple explanation for your observation. Newtonian mechanics predicts precession of planetary orbits due to small pertubations caused by other planets. In your computer models, I suspect that the rounding errors introduce similar pertubations into a 2-body situation, and so give rise to qualitatively similar precession. Both Newtonian mechanics and general relativity predicted a precession of Mercury's orbit - the difference was that the Newtonian prediction of the rate of precession did not agree with the observed value, whereas the GR prediction did agree (within the limits of observational error) - see Tests of general relativity. Gandalf61 15:33, 1 June 2006 (UTC)[reply]

Methods exist for integrating differential equations while respecting conservation laws, and special perturbation methods have been devised for orbital simulations. When JPL computes long-term ephemerides and certain critical spacecraft trajectories, they find it necessary not only to be extraordinarily careful with their numerical methods, but also to include the effects of general relativity. Some tests of general relativity effects require meticulous care to distinguish the sought effect from numerous other sources of perturbation; a notable example is Gravity Probe B. Other effects, like gravitational lenses and black holes, are not subtle at all, and are not only thoroughly observed but also dramatically different from Newtonian predictions. To speak bluntly, it is absurdly arrogant to imagine that your uninformed "little idea" is more insightful and clever than the work of large numbers of trained professional physicists. We are convinced Einstein's theory will eventually be replaced by a more integrated theory, something Einstein himself attempted (unsucessfully) in his later years; but we can never go back to Newton. --KSmrq^T 23:50, 1 June 2006 (UTC)[reply]

You all seem to miss the point. It is true that there are high-precision numerical methods for solving differential equations. It is not true that the world is using these methods to calculate trajectories. It is using the simplest one, that is, Euler integration. With Planck time being a very short interval, the error would be noticeably high. "Continuous" Newton mechanics doesn't take this error into account. Relativity does. Grue 07:10, 2 June 2006 (UTC)[reply]

I said and still maintain that it appears unlikely that this theory can be tweaked to give a quantitative agreement with the observations. How so is that beside the point? The onus of showing that you can get a good agreement is on you. Einstein predicted a difference of 43 arc-seconds per century with Newtonian theory; what does your theory predict? See further Scientific method. For the rest, Euler integration? – now if you could explain away action at a distance, that might make it interesting even if zany. --Lambiam ^Talk 14:35, 2 June 2006 (UTC)[reply]

One way hash/encryption function that is guaranteed unique

Any crypto fans here? I need to generate object IDs from other object IDs to mask the true object with one that is a plausible replacement.

Say I have a unique object ID, is there an encryption algorithm I can use against it to produce a different (cyphertext) object ID that is one way (can't easily get the original plaintext Object ID from the cyphertext Object ID, either not at all, or at least short of a computationally expensive attack), and unique, meaning that no 2 original plaintext object IDs produce the same cyphertext object ID and that I never get different cyphertext object IDs from the same plaintext object ID. The text lengths need not be the same I guess, although obviously the cyphertext one can't be shorter. I have looked at Message_digest and it speaks of cypher/hash functions that are unlikely to have different plaintexts go to the same cyphertext. I need guaranteed... the cyphertext length need not be a hash, it can be as long or longer as the original (exactly the same size would be convenient, though). thanks! ++Lar: t/c 18:34, 1 June 2006 (UTC)[reply]

PS... another way of saying this is that I need Collision_resistance that is not just hard but impossible, or at least practically impossible. ++Lar: t/c 18:36, 1 June 2006 (UTC)[reply]

Well, the whole notion of hash functions is that they're practically good enough, but if you want to reduce the possibility of collisions further, just encrypt the original object. There aren't (AFAIK) formal methods for one-way-only full-length encryption, but you can get the effect by generating a public/private key pair (say, via PGP / GPG). Encrypt normally via the public key, and then hide or destroy the decrypting private key as required. Public-key encryption is preferred here since knowledge of the only in-use key can't expose the data, but there's no reason any other encryption won't work so long as you protect the key. — Lomn Talk 19:13, 1 June 2006 (UTC)[reply]

a further note on "practically good enough": the SHA-1 hash function will have its first random collision, on average, after about 10²⁴ entries. Do you really anticipate needing that sort of collision resistance? — Lomn Talk 19:20, 1 June 2006 (UTC)[reply]

It's a customer requirement, and their answer is yes. Mine would be no though! Thanks for your help on this. I'm thinking public key/private key (with a destroyed private key) is the way to go, the object ID is short enough that encrypting it is fine, should not be too compute intensive ++Lar: t/c 20:02, 1 June 2006 (UTC)[reply]

One other option to consider would be sticking to a standard hash but appending some additional nonce such as date-of-entry alongside the hash output. Of course, if hash function collision rates aren't "practically impossible" enough for the customer, I don't know that this is really any better (unless it simply looks stronger to a corporate perspective). Another consideration, if you have short object IDs, is the possibility that object contents will collide more frequently than the cryptographic process -- for instance, I'm quite certain you'd find the MD5 for "password" far more frequently in a password hash file than mere chance would suggest because the source entries aren't random. This may necessitate a nonce in its own right (and, of course, it could be appended prior to hash/encryption as well). — Lomn Talk 20:27, 1 June 2006 (UTC)[reply]

That's why Unix passwords use salt. -- EdC 00:52, 2 June 2006 (UTC)[reply]

Right. But these aren't passwords and I have to be able to reproducably and repeatably produce the same encrypted OID from the cleartext one without having a salt value stored. Whether THEIR algorithm for making an OID is unique or not is a different matter. ++Lar: t/c 01:37, 2 June 2006 (UTC)[reply]

A method that gives good scramble of an N-bit string is to XOR it with a randomly chosen bit string, apply a randomly chosen permutation to the bits of the string, multiply modulo 2^N with another random but odd bit string, and XOR+permute again (using different choices). Since each step is invertible, you are guaranteed to have no collisions. For additional security the procedure can be repeated. I have no idea how safe this is against various cryptanalytic attacks; for all I know there is a way of breaking the scheme that is obvious to more devious minds than mine. Use at your own risk. --Lambiam ^Talk 14:53, 2 June 2006 (UTC)[reply]

In general, any block cipher should satisfy the conditions you specify, as long as you never reveal the key. If there's a limit on the block size, you can always construct your own ad hoc cipher, which is pretty much what the previous suggestion does. To be safe, however, you probably want to construct your custom cipher around an established cryptographic hash function, using a well-studied construction such as a Feistel cipher. For my own earlier take on a similar problem, see [2]. (Note that the code actually rather overdoes it, using 2*5 = 10 rounds where, per Luby and Rackoff, 4 should suffice.) —Ilmari Karonen (talk) 17:27, 2 June 2006 (UTC)[reply]

Thinking about this a bit more, it occurs to me that, in principle, any (conjectured) one-way function should also satisfy your requirements without requiring a secret key. However, unless the OIDs are randomly chosen from a very large set, any solution without a secret element is vulnerable to an exhaustive search — in which case you're back to the "block cipher with secret key" solution. —Ilmari Karonen (talk) 22:14, 5 June 2006 (UTC)[reply]

notation

What do you call this type of notation?

${\begin{matrix}{}_{n}\\{}*{}\\{}^{k=1}\end{matrix}}k$ or ${{}*{} \atop {}^{k\in S}}{k \atop {}}$

As far as I can remember, ∗ is always been one of Σ, Π, ∩, ∪, or ∐ (coproduct), but theoretically couldn't it be extended to apply to any binary operation? Or maybe even to any function?

I don't understand why this ubiquitous notation does not seem to have a name.

Don't forget \bigwedge (wedge product or join), \bigvee (meet) \bigoplus (direct sum), \bigotimes (tensor product), and I'm sure there are more lesser used examples. It's not binary operators that you do this with, but n-ary operators (especially infinitary operations). Of course any associative binary operation can be extended by induction to an n-ary operation. I think it'd be more risky to try to do this with a nonassociative operation, and of course the notation really earns its lunch for infinitary operations, which do not arise from binary operations by induction. As for what the notation is called, well I don't view it as a problem that the notation doesn't have a name. Lots of notations don't have names. What's the dx symbol that sits next to the integrand called in an integration? But if you have to have a name, I think the AMS-LaTeX guide calls them cumulative operators, which might serve you. -lethe ^{talk +} 19:19, 1 June 2006 (UTC)[reply]

So could a non-binary operation be extended to an n-ary operation with this notation? I'm particularly thinking of the hyper operator.

No, because the hyper operators (past multiplication) are nonassociative. The notation could conceivably be extended to nonassociative operators, but order of evaluation would need to be provided. -- EdC 00:50, 2 June 2006 (UTC)[reply]

But could this notation be defined for a function that would, without it, take 3 or more arguments? For example, if you have:

f\left(a,b,c\right)=f\left(a,c,b\right)=f\left(b,a,c\right)=f\left(b,c,a\right)=f\left(c,a,b\right)=f\left(c,b,a\right)

can you define the following?

{\begin{matrix}{}_{n}\\{}f{}\\{}^{k=1}\end{matrix}}{\left(a,b\right)}

Ugh. No. Absolutely not. No way. Look, you have to consider that these cumulative operators are essentially functions defined on multisets (or, at a pinch, sequences); the point is that an associative binary operator (with range a subset of its domain) extends to a function on finite sequences (if it's commutative, on finite multisets). You can't do that with operators of greater arity; for one, the length of the sequence (size of the multiset) is constrained (for ternary operators) to be an odd number ≥ 3. (And where do you grow the evaluation tree?) You'd do better to package the sequence as pairs ((n-1)-tuples) and define a F on sequences of pairs:

{\begin{matrix}F(())&:=&k_{0}\\F(A::(b,c))&:=&f(F(A),b,c)\end{matrix}}

Less convenient, but at least you'll be understood unambiguously. -- EdC 02:59, 2 June 2006 (UTC)[reply]

Then

{{}\gcd {} \atop {}^{k\in S}}{k \atop {}}

and

{{}\circ {} \atop {}^{k\in S}}{k \atop {}}

(function composition) are fine and valid uses of the notation, but

{\begin{matrix}{}_{L}\\{}{\mbox{hyper}}{}\\{}^{k=1}\end{matrix}}{\left(k,n\right)}

(where

L,k,n\in \mathbb {N}

) isn't.

Isn't dx a differential? —Keenan Pepper 21:16, 1 June 2006 (UTC)[reply]

Does differential mean infinitesimally small number? If so, then yes, it was true in Newton's day that the dx symbol in an integration was an infinitesimal number, but it's not true today. Nonstandard analysis puts the infinitesimal on rigorous footing, but even that still does not allow you to interpret the symbol under the integral sign as an infinitesimal. You may interpret the symbol as a differential form in some cases. So I ask you, what does the word "differential" mean to you? The WP page is just disambiguation. -lethe ^{talk +} 21:44, 1 June 2006 (UTC)[reply]

Statistical process control

In statistical process control using control charts, I have noticed that presenters often recommend calculating the standard deviation in a, so to speak, nonstandard way. The recommended procedure is to calculate a mean moving range, i.e. $\sum _{i=1}^{n-1}{\frac {|x_{i+1}-x_{i}|}{n-1}}\!$ , using a relatively small dataset, and then divide the mean moving range by the magic number 1.128. If you google for "(1.128 and calculate)" and are feeling lucky today, you will find a such a presentation. The number 1.128 is often represented by the symbol d₂. Does anybody know the maths behind this non-standard estimator of the standard deviation? --vibo56 19:00, 1 June 2006 (UTC)[reply]

Indices

Can anyone help me with the problem below. I am aware that it is a homework question, but that is why I don't want you to answer it! The question is to simplify it and I assume that you have to multiply out the parenthesises:

a^{\begin{matrix}{\frac {3}{2}}\end{matrix}}(3a^{\begin{matrix}{\frac {7}{2}}\end{matrix}}+a^{\begin{matrix}{\frac {-3}{2}}\end{matrix}})

I obviously tried WP, but couldn't find the right article. Thank you very much. Kilo-Lima|^(talk) 20:05, 1 June 2006 (UTC)[reply]

It looks like your title may be your problem, as those appear to be exponents rather than indices. Given that, you're right that you need to multiply out; now you just need to check the rules for a^x * a^y = a^?. — Lomn Talk 20:18, 1 June 2006 (UTC)[reply]

Ok, I won't answer it. You are correct in assuming that you have to multiply out the parentheses. I am fairly confident that your maths textbook will provide all the information you need to solve it. Just do it step-by-step, slowly and carefully. Write down each line, and be sure to know what rules you are applying in getting from one line to the next. Good luck with your homework. --vibo56 20:24, 1 June 2006 (UTC)[reply]

If you're still having trouble, recall that a^x * a^y = a^x+y --Codeblue87 21:48, 1 June 2006 (UTC)[reply]

Also remember that multiplication is distributive. The first equation of the distributivity article should help you; you can ignore the more complicated stuff after that. moink 22:40, 1 June 2006 (UTC)[reply]

This question is a study in arithmetic properties of exponents. We first meet exponents in the limited form of positive integer powers, such as a³ = a·a·a; but here we are challenged to extend our understanding to fractional powers.

It is a property of all real numbers, however obtained, that multiplication distributes over addition: r(s+t) = rs+rt. (This is also a property of complex numbers.) Therefore, if we follow your hunch and "multiply out", we obtain

a^{\frac {3}{2}}(3a^{\frac {7}{2}})+a^{\frac {3}{2}}a^{-{\frac {3}{2}}}.\,\!

Still being thoughtful and cautious, we observe that multiplication is both associative and commutative, so we may slightly rearrange to get

3(a^{\frac {3}{2}}a^{\frac {7}{2}})+(a^{\frac {3}{2}}a^{-{\frac {3}{2}}}).\,\!

Lacking experience with the fancier forms of exponentiation, it may not be obvious that anything we have done so far is particularly helpful. However, it does bring factors a^x, for various x, next to each other. In fact, this is helpful, for it brings us to the crux of the problem. But to simplify any further we must begin to understand the meaning of fractional exponents, and learn the rules that apply to them. In other words, just as we have used rules of multiplication and addition — such as distributivity, associativity, and commutativity — to get this far, we must use analogous rules involving exponents to proceed.

The big picture will take time to see, but eventually we are able to construct a formal correspondence between expressions like

ax+ay\,\!

on the one hand, and

a^{x}a^{y}\,\!

on the other, so long as a is positive. At that point (which most of us have reached) the last steps of simplification will be obvious. The purpose of this question is to help you get to that point. So, read the article on exponentiation, and enjoy. --KSmrq^T 00:59, 2 June 2006 (UTC)[reply]

A common blunder among people first learning this stuff is that exponentiation distributes over addition. It does not. In other words,

(a+b)^{2}\neq a^{2}+b^{2}

. Don't make that mistake. To figure out the right way to deal with exponentiation of sums, consult binomial formula. -lethe ^{talk +} 01:14, 2 June 2006 (UTC)[reply]

how to run a Monte Carlo simulation on a DEM using ArcGIS

PLEASE I WOULD LIKE TO KNOW HOW THE Monte Carlo simulation techniques IS USED to evaluate the impact of DEM error on viewshed analyses.

I would like to know what the caps lock key does. -lethe ^{talk +} 01:02, 2 June 2006 (UTC)[reply]

June 2

Challenging Integral

$\int {\ln[(e^{x}\ln x)^{\frac {1}{x}}]}dx$

What kind of substitution should I use?Patchouli 06:18, 2 June 2006 (UTC)[reply]

integrand= ${\frac {1}{x}}\ln(e^{x}\ln x)=1+\ln(\ln x)/x\,\!$

integral=

x+\ln[\ln(\ln x)-1]\,\!

Patchouli 07:17, 2 June 2006 (UTC)[reply]

Particle leaving a path

I have a question involving circular motion. Visualise a roller coaster's path. At the start, a straight part of length x, at an angle of 10° above the horizontal exists. A particle rolls down this, and meets an arc of radius 8m sloping downwards, and declines to a straight path 40° below the horizontal. The path is smooth and is only affected by gravity. I have to find the maximum value of x so that the particle stays on the path. I've only ever encountered particles with a mass, and this one has none specified, so I assume it cancels somewhere.

Anyway,

If the particle leaves the path, would it leave as soon as it encounters the curved component? (i.e. it takes off and misses the first part of the arc.) Or is it possible that it leaves the path during the arc?
Does weightlessness have anything to do with it when a particle is on the verge of leaving a path?

I'll figure stuff out from here.

Thanks. x42bn6 Talk 08:57, 2 June 2006 (UTC)[reply]

Once the particle leaves the path, there will be no upward force acting on the particle any more, so the only force acting on the particle will be gravity. What shape of path will this particle take? If you need help, take a look at the article about trajectory. I think how you want to solve this problem is to find out when this "natural" trajectory lies above the path, so that the particle will tend to "take off" instead of following the path. —Bkell (talk) 11:07, 2 June 2006 (UTC)[reply]

A particle follows a path only as long as the resultant force can supply the acceleration needed to keep it on that path. In this case, the resultant force comes from gravity; to stay on the path the centripetal component of the gravitational vector must be at least as great as the centripetal acceleration of a particle taking that path. Now, the centripetal acceleration depends solely on velocity and arc radius; velocity depends solely on height lost (assuming a frictionless path i.e. total conversion of gravitational potential energy into kinetic energy).

This implies that the particle becomes more prone to leave the path as it proceeds round the arc: the velocity increases, and the centripetal component of the gravitational vector decreases. So the maximum x is that which has centripetal acceleration equal to centripetal component of gravitational vector at the end of the arc. Quite beautifully, both mass and gravity cancel out. See here to check your solution.

Finally, what's special about 70.529 degrees? -- EdC 11:46, 2 June 2006 (UTC)[reply]

(1) Yes, if the particle is going to depart at all, it'll depart at the very start of the circularly curved section. Both factors point in the same direction:

If energy is conserved, the particle is moving fastest at the start of the curve. So it requires the most centripetal acceleration to hold to the path there.
At the start of the curve, the slope is the steepest. So the normal component of gravity is the weakest there.

Melchoir 20:42, 2 June 2006 (UTC)[reply]

Are we looking at the same path? I thought it was like this:

        ____    x
        10° ˇˇˇˇ----____
        ˇˇˇˇˇˇˇˇˇˇˇˇ _--ˇˇˇ--_
                   /ˇ         ˇ\
                  |         8m  |\
                  |      ·<---->| ˇ\
                  |             |   ˇ\
                   \_         _/  40° ˇ\
                     ˇ--___--ˇ   ˇˇˇˇˇˇˇˇ

Oh, if it's that then, conversely, the particle might leave at any point, depending on its energy. Melchoir 21:37, 2 June 2006 (UTC)[reply]

Actually, never mind, my exam is over. Luckily it didn't come out. But I did use the argument that if the particle reaches the next straight path, and we want it just to, R must be minimized, so we look for the upper value of

\cos \alpha

or something. Thanks, anyway. x42bn6 Talk 04:51, 5 June 2006 (UTC)[reply]

Age in Chocolate

Why does this work?

1. First of all, pick the number of times a week that you would like to have chocolate
   (more than once but less than 10)
2. Multiply this number by 2 (just to be bold)
3. Add 5
4. Multiply it by 50 -- I'll wait while you get the calculator
5. If you have already had your birthday this year add 1756 .... If you haven't, add 1755.
6. Now subtract the four digit year that you were born.
You should have a three digit number
The first digit of this was your original number (i.e, how many times you want to have chocolate each week).
The next two numbers are:
  YOUR AGE! (Oh YES, it is!!!!!)

I got it from ebaumsworld and it works for me. I find it really wierd. Thanks. schyler 13:48, 2 June 2006 (UTC)[reply]

For an explanation, look here: http://mathforum.org/library/drmath/view/61702.html (the numbers are a bit different, but obviously they must be adjusted each year). --Lambiam ^Talk 14:14, 2 June 2006 (UTC)[reply]

See elementary algebra. If you want chocolate n times a week and you were born in year y, then you get

50 (2n + 5) + 1755 − y if you haven't had your birthday this year.

50 (2n + 5) + 1756 − y if you have had your birthday this year.

This rearranges to:

2005 − y + 100n (if you haven't had your birthday)

2006 − y + 100n (if you have had your birthday)

Hopefully it's now obvious that the (2005 − y) or (2006 − y) bit is your age, and the 100n bit gives n in the hundreds column (given the range you specified for n).

Arbitrary username 14:19, 2 June 2006 (UTC)[reply]

You know, this wouldn't work if your age was at or beyond a century. Black Carrot 18:01, 2 June 2006 (UTC)[reply]

The puzzling part is why anyone would be surprised. If I call a psychic hotline and give them my name, phone number, and credit card number, should I be surprised that they "mystically know" a lot about me? Just so, this song-and-dance routine is telling you no more than you tell it. You provide the chocolate number, you provide your year of birth and an adjustment for whether you've had a birthday this year; the calculations are merely an obscure and entertaining way of producing the stated result: (chocolate)×100 + (age), where (age) = (this year, adjusted)−(birth year). --KSmrq^T 20:00, 2 June 2006 (UTC)[reply]

no, I put in 3 for how many per week and I'm 14 and I got 313,

Check your work, especially step 5. If your chocolate number is 3 and if you are about 14 (born 1992) you should get either

1 →	3	2 →	6	3 →	11	4 →	550	5 →	2306	6 →	314
1 →	3	2 →	6	3 →	11	4 →	550	5 →	2305	6 →	313

depending on whether you have had your birthday this year (first line) or not (second line). --KSmrq^T 08:02, 3 June 2006 (UTC)[reply]

June 3

Laurent Series

I have a complex analysis exam coming soon, and I'm not very confident with Laurent series. Suppose I have a complex function f(z) which has a finite number of poles. Is there a general method for finding its Laurent expansion? Admittedly, my knowledge of Laurent series is somewhat limited. I know what they are, but not really how they work. Can anyone help? Maelin 03:12, 3 June 2006 (UTC)[reply]

Have you read Laurent series? Conscious 06:56, 3 June 2006 (UTC)[reply]

Yes, but it's an infinite series from -infinity to infinity. If I'm asked to find the Laurent series of a function, where do I start? The examples in the article don't show where the expressions come from, it's just "consider this function. Now, abracadabra! Here are some Laurent expansions for it depending on where you're interested!"

The second formula of the article gives a general formula for calculating the coefficients in a Laurent series, so it's not magic, just calculation. Admittedly, that's not usually the easiest way to calculate. -lethe ^{talk +} 07:40, 3 June 2006 (UTC)[reply]

You can always use the general formula and well-known series for some functions (see the example for

e^{-{1}/{x^{2}}}

in the article). Conscious 07:49, 3 June 2006 (UTC)[reply]

Okay, I've done some more work with them. In our lecture note examples, we generally have a function of the form f(z) = 1 / g(z) where g(z) is some polynomial in z (usually conveniently factorised into linear terms). Then we reduce it to some form of geometric series and find the Laurent series. The problem is that I have no idea how we end up with a Laurent series that is valid for the particular region we're interested in. We just seem to head off reducing it in one particular way and then voila, we have the right one. In the article, we have an example just like this, and then three different Laurent expansions miraculously appear, correct for each region. Clearly, whoever made the example didn't do an infinite number of integrals around γ and then build a series out of them, so what is going on here? Apologies if I sound terse, this is very frustrating. Maelin 05:19, 4 June 2006 (UTC)[reply]

Well, doing infinitely many integrations for the coefficients is not as bad as it sounds, since you can often tell at a glance what a contour integral will be. Anyway, like I said, that's not usually the easiest way. The example in the article goes like this. Split that thing up by partial fractions. When |z| < 1 and 2, you can use the geometric series with the z – 1 and the z – 2i terms. When z is between the two roots, you can use the geometric series with the z –2i term, but for the other term, you have to use like z/(1 – 1/z). And for the last region, you have to invert z for both terms. Basically, you can only invoke the geometric series when the ration has modulus less than 1, and this fixes how you find the Laurent series in each region. -lethe ^{talk +} 20:58, 4 June 2006 (UTC)[reply]

voronoi diagram

I would like the definition and applications of a voronoid diagram (scatter application). Please put a definition on your web site.

Voronoi diagram Melchoir 03:19, 3 June 2006 (UTC)[reply]

qauadratic formula

what is the quadratic formula — Preceding unsigned comment added by 69.231.27.28 (talk • contribs) 04:03, 2006 June 3 (UTC)

Please read and follow the first bullet point at the top of this page. (And sign your posts with four tildes, ~~~~.) Thank you. --KSmrq^T 05:33, 3 June 2006 (UTC)[reply]

See quadratic formula. Conscious 06:54, 3 June 2006 (UTC)[reply]

Firefox: how to search edit areas?

How do I search edit areas in Firefox? For example, when I edit a long article and want to find where it links to a specific category, the text entered into the search field seems to be searched for only outside the wikicode. Is there any way to get it straight? Conscious 06:07, 3 June 2006 (UTC)[reply]

I don't know of a way to do that. It would be very useful. However, if you use the "Show preview" button, you'll be able to search the edited version (though not the edit box itself), and that makes it a lot easier to find the required text, by counting paragraphs.... TheMadBaron 19:28, 3 June 2006 (UTC)[reply]

Mathematica to Wikipedia

I have Mathematica 5.0. What is the best way to transform the *.nb to a format that Wikipedia understands.

You don't. You write text for an article explaining what you're doing. Dysprosia 04:14, 4 June 2006 (UTC)[reply]

The HTMLSave function comes to mind. Realistically, the text will mostly need to be copied as plain text, perhaps with some UTF-8 characters; and each formula will need to be rewritten in either wiki syntax or TeX format. The TeXForm facility may help with the latter, though compatibility with MediaWiki's limited version is not guaranteed. Come BlahTeX, the MathMLForm version may be of interest, though it would make editing obnoxious. Note that it would be inappropriate to import Mathematica style sheets to override Wikipedia's own. --KSmrq^T 04:27, 4 June 2006 (UTC)[reply]

Upper bound on n!

Given an integer n, what are the tightest bounds on factorial n? More specifically, I want to calculate the number of binary digits required to represent n! for a given n. -- Sundar ^{\talk \contribs} 10:37, 3 June 2006 (UTC)[reply]

The article on factorial gives the following approximation based on Stirling's approximation :

\ln(n!)\approx n\ln(n)-n+{\frac {\ln(n)}{2}}+{\frac {\ln(2\pi )}{2}}.

Can this be taken as an upper bound on

\ln(n!)

? If not what would be an upper bound? -- Sundar ^{\talk \contribs} 10:45, 3 June 2006 (UTC)[reply]

The article on Stirling's approximation says that the error is the same sign and size as the first omitted term. The next term in the series after the one you've written is negative, so your series is an upper bound. -lethe ^{talk +} 11:30, 3 June 2006 (UTC)[reply]

Oh whoops. My comment above will be true if you also include the 1/12n term. So do that. -lethe ^{talk +} 11:32, 3 June 2006 (UTC)[reply]

Oh thanks, Lethe. -- Sundar ^{\talk \contribs} 11:44, 3 June 2006 (UTC)[reply]

Upper bound on

\ln(n!)

:

$\ln(n!)\approx n\ln(n)-n+{\frac {\ln(n)}{2}}+{\frac {\ln(2\pi )}{2}}+{\frac {1}{12n}}$

Sposta be a 12. -lethe ^{talk +} 11:48, 3 June 2006 (UTC)[reply]

Thanks, fixed it. By the way, what would be the equivalent one for log to the base 2 (my original question)? (Excuse my laziness.) -- Sundar ^{\talk \contribs} 12:06, 3 June 2006 (UTC)[reply]

Well, log₂ x = log x/log 2. So divide both sides of the equation by ln 2, and you're good to go. -lethe ^{talk +} 12:10, 3 June 2006 (UTC)[reply]

I now realise. It wasn't laziness, but naivety. -- Sundar ^{\talk \contribs} 12:16, 3 June 2006 (UTC)[reply]

Well considering that summation is cheap for computer, for relatively small n you can get a very good estimate by using

\ln n!=\sum _{k=1}^{n}\ln k

Igny 12:48, 3 June 2006 (UTC)[reply]

Thanks. But, summation was not the main concern, logarithm was. -- Sundar ^{\talk \contribs} 06:29, 5 June 2006 (UTC)[reply]

i cut a little piece out of a torus (donut), what is the remaining space

Consider a torus, cut a tiny CLOSED piece out on the side. So the piece I cut out is homeomorphic to $\mathbb {D} ^{2}$ .

What is the remaining space? My professor tells me I can only see it when i start stretching that hole open 'until my fingers touch on the other side'. However I was born with no 3D mind, I simply do not see it.

It should be an 'easy space involving cilinder(s)' I would like to understand this completely for the understanding of homotopy and homology groups.

Evilbu 11:35, 3 June 2006 (UTC)[reply]

Well, a torus is a sphere with a handle, and a sphere with a disc removed is a disc, so a torus with a disk removed is a disc with a handle. I don't know if that's the answer you're looking for though. -lethe ^{talk +} 11:53, 3 June 2006 (UTC)[reply]

Well uhm, what is a handle, my syllabus says every $\mathbb {D} ^{k}\times \mathbb {D} ^{l}$ is a handle? The article http://en.wikipedia.org/wiki/Handle_%28mathematics%29 doesn't really help me out right now. Evilbu 12:16, 3 June 2006 (UTC)[reply]

A handle is a cylinder attached at its two end circles. It is not the product of two balls, which is trivial. -lethe ^{talk +} 12:29, 3 June 2006 (UTC)[reply]

T\setminus \mathbb {D} ^{2}

is

\mathbb {S} ^{1}\vee \mathbb {S} ^{1}

. To see this, represent

T

as

\mathbb {D} ^{2}

with side identifications; remove a chunk around the corner; shrink the remainder. Alternatively consider

\mathbb {S} ^{1}\vee \mathbb {S} ^{1}

as the skeleton of a torus and imagine growing a patch of skin around the torus from that skeleton; the remaining hole is

\mathbb {D} ^{2}

-shaped. EdC 12:39, 3 June 2006 (UTC)[reply]

Is that supposed to be a disjoint union? The torus with a disc removed is certainly a connected space. Perhaps you mean the wedge sum of two circles instead? But that's not right either, as the latter is 1 dimensional, while the former is two dimensional. They are certainly homotopy equivalent though. -lethe ^{talk +} 12:55, 3 June 2006 (UTC)[reply]

Yeah, I meant wedge sum, sorry. As for dimension - if you cross each with a

\partial \mathbb {D} ^{1}

, so that they join at a

\partial \mathbb {D} ^{2}

, then you get something like two cylinders tangent at right angles. EdC 17:09, 3 June 2006 (UTC)[reply]

isn't the boundary of a 1-ball just a disjoint pair of points? If you go crossing anything with that, you'll get a disconnected space again. Surely not what you want. -lethe ^{talk +} 20:48, 3 June 2006 (UTC)[reply]

T\setminus D^{2}

is not the same as

S^{1}\vee S^{1}

.

S^{1}\vee S^{1}

, the deformation of the space in question, is a 1-manifold except at one point, whereas

T\setminus D^{2}

is a 2-manifold with boundary.

T\setminus D^{2}

looks like a thickened copy of

S^{1}\vee S^{1}

---though it's not just a figure-8 drawn with a really fat-tipped marker. (The boundary of a this figure-8 has three components, where the boundary of

T\setminus D^{2}

has only one by the construction.) To visualize

T\setminus D^{2}

, take a big fat blocky plus sign (like the Red Cross logo), then glue the top and bottom ends together in the back, and the left and right ends together in the front. Tesseran 01:43, 6 June 2006 (UTC)[reply]

I am very confused now. What do you mean, a cilinder attached at its two end circle, you mean take a cilinder, then attach the upper and lower circle? Wouldn't that be a torus? Why would the product of two balls be trivial? What is the union of twice $\mathbb {S} ^{1}$ Evilbu 12:44, 3 June 2006 (UTC)[reply]

A coffee cup has a handle. The handle attaches to the mug in two different places, once at each end. -lethe ^{talk +} 12:58, 3 June 2006 (UTC)[reply]

Seems like I know much less than I thought. Is there a precise definition of handle (I am familiar with the language of quotient spaces) In order to make sure we understand each other, I will say a couple of things, and please tell me when you disagree : $\mathbb {D} ^{1}=[0,1]$ $\mathbb {D} ^{2}$ is the closed disk in two dimensions $\mathbb {S} ^{1}$ is the unit circle (it is one dimensional $\$\mathbb {T} ^{2}$ , the (empty,thus 2d)torus, is homeomorphic with $\mathbb {S} ^{1}\times \mathbb {S} ^{1}$ the full,thus 3d torus is homeomorphic with $\mathbb {S} ^{1}\times \mathbb {D} 2$

I disagree that a 3d torus is equal to a circle times a ball. Rather, such a thing is called a genus 1 handlebody. A 3d torus is the cartesian product of 3 circles. If you'd like a technical definition of a handle, you can take it to be a torus with a disk removed, though that won't help you visualize what it is. Hence the coffee cup description. -lethe ^{talk +} 20:46, 3 June 2006 (UTC)[reply]

Maybe it would be relevant to say why I want this. I want to find the torus' homology groups, especially the $H_{2}(\mathbb {T} ^{2})$ group Now my professor told us to do a Mayer Vietoris trick on the torus, by cutting out a little piece (a disk) . The two spaces I get then, have a 'relatively easy' intersection in my Mayer Vietoris sequence, it is a cilinder, homotopic with a circle, and thus completely known. But what is the other space?Evilbu 15:31, 3 June 2006 (UTC)[reply]

It's two cylinders kissing, homotopic to a bouquet of two circles. EdC 19:26, 3 June 2006 (UTC)[reply]

I suppose if you've previously worked out the homology of the more complicated piece this makes sense. (It also hints at a general construction involving a 2n-gon and edge gluing.) But otherwise, wouldn't it be more natural to use two cylinders (retracting to circles), with overlap deformation retracting to two disjoint circles? Also, this space is simple enough that you could take a rectangle split diagonally into two triangles, and glue the outer edges of the triangles together in such a way as to give a simplicial decomposition for an explicit computation. And if you've gotten as far as Mayer-Vietoris sequences you're probably not far from the Künneth theorem, which makes this T² = S¹×S¹ product a trivial computation. You can check your work by these other means, and also by comparing the known Euler characteristic of the torus, namely zero, to that given by its Betti numbers summed with alternating signs.--KSmrq^T 20:58, 3 June 2006 (UTC)[reply]

Also, isn't H_n(X,Z) always isomorphic to Z when X is a connected manifold? Generated by the fundamental class of X. -lethe ^{talk +} 03:22, 4 June 2006 (UTC)[reply]

How best to respond to a learning exercise? I'm assuming that the purpose is to get comfortable with computations, here using the Mayer-Vietoris sequence, rather than to get the answer. It's not as if computing the homology of a torus is much of a burden. Also, Evilbu doesn't tell us if the coefficients are Z, Z₂, R, or something else. (Has the class covered the universal coefficient theorem yet?) Nor do we know if this is singular homology, though that seems a good assumption, and relatively unimportant. In short, I'm trying hard not to say "The answer is …!" :-D --KSmrq^T 05:03, 4 June 2006 (UTC)[reply]

To (sort of) answer lethe's question, consider the homology groups of Rⁿ; these are the same no matter what n may be. For, Rⁿ is homotopy equivalent to a punctured n-sphere, and deformation retracts to a point. So, what is H₂(R²,Z)? Remember, homology measures cycles modulo boundaries under homotopy equivalence; to get a Z there must be a "hole" to catch a cycle, preventing repetitions from collapsing. --KSmrq^T 19:20, 5 June 2006 (UTC)[reply]

Uhm, no I do not know anything of universal coefficient theorem. If it is relevant, we always consider these groups as $\mathbb {Z}$ modules, thus abelian groups. I know I cold do a Mayer Vietoris trick by using two cilinders, who intersection are two disjoint cilinders, then I find everything except $H_{2}(\mathbb {T} ^{2})$ Basically I was hoping by doing this cutting out of a little sphere, I would be able to find it in another way.Evilbu 09:47, 4 June 2006 (UTC)[reply]

June 4

cost management accounting - indirect labor unit cost.

Please help a new small manufacturing company by locating "indirect labor unit cost." We need this for our financial portion of the business plan as required by Small Business Administration. Our leaders, Maasters of Science in Healthcare, for some reason did not include this information. If anyone out there can help, it will greatly appreciated. Our company sews dresses and suits for premature infants and low birth weight infants. The SBA booklet indicates that this "cost" is needed to be included with total production costs.

This question may not be a typical for Wikipedia, but we have really been trying to find this and to date have not been successful.

"Labor unit cost" is the cost of labor per unit produced. So if in a month your labor cost is $100,000.00, and you produce in that time 5000 garments, the labor unit cost is $100,000.00 divided by 5000 equals $20. "Indirect labor cost" is that portion of labor cost that can be ascribed to activities not directly contributing to production (e.g. administration, salespeople, advertising). And the unit cost is as before. So if in that month your indirect labor cost is $40,000.00, then the indirect labor unit cost is $8. See http://www.uwm.edu/Course/IE360-Saxena/two.pdf section 2.5, elements of cost. --Lambiam ^Talk 13:51, 4 June 2006 (UTC)[reply]

Formula used in civil engineering

Can someone give me a simple mathematical forumla sude in civil engineering? It's for a math homework project. Thanks in advance! --Wizrdwarts 01:38, 4 June 2006 (UTC)[reply]

Somewhere not far from where you live is a firm that does civil engineering. Contact them. Chances are excellent that they would be delighted to spend a few minutes talking to you about what they do, and offer a sample calculation or two. It should be much more fun and educational than asking a bunch of mathematicians on the web. (Though we can be fun, too, in our own way.) --KSmrq^T 05:19, 4 June 2006 (UTC)[reply]

Does Hooke's law count? Dysprosia 09:20, 4 June 2006 (UTC)[reply]

Catenary has a cool formula for suspension bridges... (and that is TYPICAL engineering right?) Oh by the way, you should write formula. Evilbu 10:48, 4 June 2006 (UTC)[reply]

Minor pedantic correction - the curve of a free-hanging uniform chain is a catenary, but the curve of the cables of a suspension bridge, in the ideal case where we assume that the weight of the cables is neglible compared to the weight of the horizontal road deck, is a parabola. Gandalf61 08:59, 5 June 2006 (UTC)[reply]

random drawing of numbers 1–100

Suppose you had the values of 1 to 100. Then, you randomly organized the 100 numbers into 10 groups of 10. One group might contain the numbers 7, 13, 17, 38, 41, 52, 59, 71, 90 and 95 for example. Then, by group, the highest number would be given a corresponding value of 10, the second highest a corresponding value of 9, and so on. Then, the process is repeated, and any value earned is added to the last value earned. For example, the number 100 will always be given a value of 10 (because it is always the highest number out of 100), so after 3 random "draws," its corresponding value would be 30. Likewise, the number 1 would have a value of 3 after 3 "draws." My question is: how many random "draws" would it take so that all the numbers were in numerical order based on their values. Likewise, how many random "draws" would it take so that less than 10 numbers were not in correct placement when organized by values. I understand that this question is confusing (and it is quite hard to word), so if you have any questions as to what I mean, then please ask and I will update and clarify accordingly. Thank you in advance for all of your help. - Zepheus 02:55, 4 June 2006 (UTC)[reply]

Fun challenge, though not a practical way to sort. But you cannot use random draws and ask those precise questions. For example, although it is statistically unlikely, fifty random draws could be exactly the same. Or did you know that? Anyway, perhaps you should think about why you expect that the cumulative "10-ranking" scores will converge to a correct 100-ranking order. That will be essential to answering the first question. For simplicity, consider four numbers in groups of two. Write out all possible draws and consider how they combine. Notice that after one draw the values will be ⟨1,1,2,2⟩, with only two distinct quantities; and after two draws the cumulative values will include 1+1 and 2+2, and typically 1+2 as well, but nothing else. So it is impossible to sort properly with only one or two draws. More generally, after n draws the largest cumulative value will be n times the group size and the smallest cumulative value will be n. Randomness aside, the difference of these must be large enough to permit a distinct value for each number in the full set. Since 11×(10−1) = 99 = 100−1, clearly at least 11 draws are necessary for a full sort. Of course, this necessary condition may not be sufficient; nor does it address the likelihood of a correct sort. --KSmrq^T 06:02, 4 June 2006 (UTC)[reply]

11 draws are sufficient:

 (1-10)           (11-20)    (21-30)    ... (91-100)
 (1,11,21,31,...) (2,12,...) (3,13,...) ... (9,19,...)
 ... (10 times) ...
 (1,11,21,31,...) (2,12,...) (3,13,...) ... (9,19,...)

It's fairly obvious that this schema assigns each number n the value n+10. EdC 14:23, 4 June 2006 (UTC)[reply]

Additionally, this (with permutations) is the only way to get a full sort in only 11 draws.

You're going to have to clarify how many random "draws" would it take so that all the numbers were in numerical order based on their values. As KSmrq pointed out, the numbers can stay unsorted through an indefinite number of draws. One possible question is how many draws are needed such that P(sorted | n draws) exceeds some value (say ½). I wouldn't think that P(sorted | n draws) has a nice form, though. — Preceding unsigned comment added by EdC (talk • contribs)

Thank you for all your help so far, and it's been a long time since I've had a math class so it's tricky to write in mathematical ways. I knew originally that every draw could be the same, but it is statistically improbable. Also, I figured that when the number of draws approached infinity, the number of errors (or numbers not in correct placement when sorted by value) would reach zero. I was just wondering how many draws would probably be sufficient. I think my question has pretty much been answered, unless EdC has more to say on the matter. - Zepheus 16:56, 4 June 2006 (UTC)[reply]

I'm currently running a simulation of the problem, and it seems that the average amount of draws required is around 2045 - If that is of any use. Keep in mind, though, that I have not double-checked my program's correctness, and it is very inefficient, so it could take a while until an accurate result is obtained. -- Meni Rosenfeld (talk) 17:28, 4 June 2006 (UTC)[reply]

I've tried such a simulation too, and I've got a similar result: average 2069 draws needed from a sample of 160 tries. The same warnings apply as above. Btw, see birthday paradox as for why you need so many draws. – b_jonas 18:35, 4 June 2006 (UTC)[reply]

Update^2: run on a faster SMP machine (from 13919 iterations) gives average 2058 draws, quartiles of number of draws are 1521, 1911, 2443. – b_jonas 19:19, 4 June 2006 (UTC)[reply]

Some thoughts on the convergence of the order. Consider a random variable $X^{k}$ , the rank of number k in a random draw. Denote $Y^{k}=X^{k+1}-X^{k}$ . With probability 9/10 the numbers k and k+1 are in different groups, and $X^{k}$ has the same distribution as $X^{k+1}$ , and $Y^{k}$ is symmetric with respect to 0 ( $P(Y^{k}=x)=P(Y^{k}=-x)$ for x>0). With probability 1/10 the numbers k and k+1 are in the same group and $X^{k+1}=X^{k}+1$ and $Y^{k}=1$ . That is,

P(X^{k+1}=x)={\frac {9}{10}}P(X^{k}=x)+{\frac {1}{10}}P(X^{k}=x-1)

Besides, we have $E(Y^{k})=1/10$ . After realization of n draws, the numbers 1,...100 have the correct order if

{\bar {Y}}^{k}={\frac {1}{n}}\sum _{i=1}^{n}Y_{i}^{k}>0

for all k=1,...99

It is known than ${\bar {Y}}^{k}\to 1/10$ in distribution so the convergence to the correct order should be expected. The questions arise whether the r.v. $Y^{k}$ are independent of each other, and for which smallest n the event $\inf _{k}{\bar {Y}}^{k}>0$ occurs the first time, and what is the distribution of such n. (Igny 22:46, 4 June 2006 (UTC))[reply]

These mathematical functions are getting crazy. I wish I could decipher them. I'll definitely archive this page. One more question, the first answer I receive was that 11 draws would be sufficient. The next answer was that roughly 2,050 draws would be needed. How are these related? Also, what is the rough estimate for the number of draws needed for less than, say, 10 mistakes. - Zepheus 19:09, 5 June 2006 (UTC)[reply]

No, the first answer was that 11 draws are necessary. That is, with less than 11 draws you have zero chance of getting it right. No finite amount of draws is sufficient (in the sense of having a probability of 1 of winning). Roughly 2047 (obtained after over 55,000 experiments) is the average number (expectation) of draws until a success is obtained. Assuming the distribution is roughly symmetric, this also means that 2047 draws will give you a 50% chance of success. -- Meni Rosenfeld (talk) 19:27, 5 June 2006 (UTC)[reply]

Well, according to the numbers I gave above, you have about 50% success after less draws than that: about 1911 draws. – b_jonas 20:47, 5 June 2006 (UTC)[reply]

Okay. I understand now. Thanks for the update, and all of your hard work. - Zepheus 21:16, 5 June 2006 (UTC)[reply]

Above is a graph of the cdf of the distribution of the number of draws needed I've made from the output of my simulation. – b_jonas 21:26, 5 June 2006 (UTC)[reply]

Loudspeakers

Can anyone recommend me loudspeakers for an integrated sound card (asus a7v8x-x) with a good quality/price ratio? thanks.

The web site for NewEgg lists numerous speakers along with specs and customer ratings. That should help you narrow down your interests to price range, number of channels, optional subwoofer, wattage, and sensitivity. A quick web search suggests that the integrated ADI 1980 sound chip is not exceptionally good, so if you do not intend to some day add a separate card, it may not be worth investing in top-quality speakers. It does provide 6 channels of output, so you may be interested in a 5.1 speaker setup, consisting of front stereo, rear stereo, center, and subwoofer. Finally, listening habits and tastes vary considerably, so it matters whether you are interested in gaming, hip-hop, classical, and so on. Again, a little reading will be quite helpful in narrowing your options. --KSmrq^T 19:22, 4 June 2006 (UTC)[reply]

free group (abelian?) <->free product<->coproduct confusion

Hi,

I have yet again a topology inspired question. First of all though I would like to express my gratitude for the many people who have helped me here.

Right now I mostly receive from Wikipedia being a student in exams, but I have and I will again give to the community myself:)

I am confused about http://en.wikipedia.org/wiki/Free_product_with_amalgamation

Suppose I take a free product of the groups $\mathbb {Z}$ and $\mathbb {Z}$ the article states it should give me a free group on two generators. Now the article on free groups it links to says that free groups and free abelian groups are not the same. There goes my hope that it would be $\mathbb {Z} \oplus \mathbb {Z}$

but wait! , later that article says it is a coproduct of two groups in the categorical sense. But I was thought in my algebra class that for Rmodules, and thus also abelian groups (as they are the same as \mathbb{Z} modules ), simply taking the outer direct sum of two modules should do just fine to give you a categorical coproduct.

So what is going on, can anyone point out the difference between coproduct and free product. What am I doing wrong?

This confusion has led me to believe than 'eight' or an 'infinity symbol' has fundamental (homopoty) group $\mathbb {Z} \oplus \mathbb {Z}$ }.

Thanks, Evilbu 14:53, 4 June 2006 (UTC)[reply]

The coproduct of

\mathbb {Z}

and

\mathbb {Z}

in the category of $\mathbb {Z}$ -modules (i.e. abelian groups) is indeed

\mathbb {Z} \oplus \mathbb {Z}

. The fundamental group of

S^{1}\vee S^{1}

is the coproduct of

\mathbb {Z}

and

\mathbb {Z}

in the category of groups, that is, the free product (which, for example, has uncountably many elements, so is clearly not

\mathbb {Z} \oplus \mathbb {Z}

). —Blotwell 15:04, 4 June 2006 (UTC)[reply]

But that is bad for me! So you are saying : free product of two groups is NOT the same as coproduct?Evilbu 17:05, 4 June 2006 (UTC)[reply]

Coproducts look different in every category. In the category of groups, the coproduct is the free product. In the category of abelian groups and the category of modules, it is the direct sum. In the category of topological spaces, it is the disjoint union (also the category of sets). In the category of pointed spaces, it is the wedge sum. Z is a set, a space, a group, an abelian group, and a module. So there are many different coproducts you can make out of Z. The fundamental group is a functor from the category of pointed spaces into the category of groups, and the coproduct that you have to use is the free product (the coproduct in Grp), not the direct sum (the coproduct in Ab). So π1 takes coproducts of pointed spaces to coproducts of groups. It is a continuous functor. -lethe ^{talk +} 20:39, 4 June 2006 (UTC)[reply]

π₁ isn't a continuous functor (one which preserves limits) because this would contradict the long exact sequence of a fibration. But more importantly, it isn't cocontinuous (preserving colimits) which is what I imagine you meant. For example it doesn't preserve the colimit of the diagram

{\begin{matrix}I&\rightarrow &S^{1}\\\downarrow &&\\S^{1}&&\\\end{matrix}}

where both arrows take the line segment to a circle by identifying the two endpoints. (Hint: the colimit is again S¹.) —Blotwell 01:37, 6 June 2006 (UTC)[reply]

Oh yes, I see, I think I have made a serious mistake in assuming something. I was seeing the abelian groups as a subcategory of the group category. I cannot take some abelian groups, take the free product (defined in the categorical sense) and assume it will still be that in the bigger category right? Evilbu 20:58, 4 June 2006 (UTC)[reply]

There is nothing wrong with thinking of Ab as a subcategory of Grp. But you're right, the coproduct of groups does not restrict to the coproduct of Abelian groups on this subcategory. Stated more explicitly, the coproduct of two groups which are abelian in the category of groups (this coproduct is a free product; always nonabelian) is not the same as the coproduct of two abelian groups in the category of abelian groups (this coproduct is a direct sum; always abelian). -lethe ^{talk +} 21:02, 4 June 2006 (UTC)[reply]

More generally, it is a mistake to think that any operation must restrict to a suboperation on a subset. Just because Ab and Grp both have coproducts does not mean the coproduct in Grp restricts to the coproduct in Ab on that subcategory. Similarly, the Killing form of a Lie algebra need not restrict to the Killing form of a subalgebra (this will happen for the Cartan subalgebra in the semisimple case, but need not in general). The covariant derivative of a vector in a submanifold of a Riemannian manifold need not equal the induced covariant derivative of that vector. Subcategories are, by definition, closed under composition of morphisms. This does not imply that they must be closed under every operation, like coproducts -lethe ^{talk +} 21:28, 4 June 2006 (UTC)[reply]

Bike problem

I know how easy it is to fall off a bike while not moving - I also know that it's harder to turn the wheel when I am in motion. I've had a look at angular momentum and related topics and I can't figure why a rotating bike wheel is harder to turn from its line of motion than a stationery one. Any help please? Anand 18:54, 4 June 2006 (UTC)[reply]

Although your second question should be answered at Gyroscope, it is not generally considered the correct reason for bicycle stability. See Bicycle#Balance, including the link to Bicycle physics. Walt 19:48, 4 June 2006 (UTC)[reply]

June 5

Circular number

I just started this article. If anyone believes they can add anything more to it, even one more sentence, go right ahead. — BRIAN0918 • 2006-06-05 05:52

I added my comments on the talk page. --vibo56 09:05, 5 June 2006 (UTC)[reply]

What is the difference with automorphic numbers? Can you give an example of an automorphic number that is not circular? --Lambiam ^Talk 14:51, 5 June 2006 (UTC)[reply]

If the definition in circular number is interpreted as "there exists a power such that...", than it's the other way around - 4 is circular (4^3 = 64) but not automorphic (4^2 = 16). -- Meni Rosenfeld (talk) 15:09, 5 June 2006 (UTC)[reply]

Discussion on the talk page has reached the conclusion that circular number is equivalent to automorphic number. Gandalf61 16:00, 5 June 2006 (UTC)[reply]

Complex Functions & Poles

A question about poles. Consider a function of this form:
$f(z)={\frac {(z-a)(z-b)}{(z-a)(z-c)}}$
Will it have a pole at z = a? A removeable singularity? Or some other form of oddness? Maelin 06:26, 5 June 2006 (UTC)[reply]

Removable discontinuity. No pole. -lethe ^{talk +} 06:41, 5 June 2006 (UTC)[reply]

f is in fact differentiable at z = a -- there's no oddness. Cancel the z-a factor. Dysprosia 06:55, 5 June 2006 (UTC)[reply]

It's a removable singularity, and as one professor said, "At this point we assume all removable singularities are removed." In other words, while f(z) is technically undefined at z = a, since it is identical to

{\frac {z-b}{z-c}}

elsewhere and can be analytically continued through it, you can essentially work with the analytic continuation instead of the function as you defined it. More interesting is a function like

{\frac {\sin z}{z}}

, which has no obvious cancellation, but because of its removable singularity is basically treated as though f(0) was automatically defined as 1. Confusing Manifestation 09:19, 5 June 2006 (UTC)[reply]

Am I missing something (Analysis has never been my strong suit)? Why is f(z) undefined at z = a -- surely f(a) = (a-b)/(a-c)? Dysprosia 09:31, 5 June 2006 (UTC)[reply]

Simply put, you can only cancel out nonzero terms. When z = a, you cannot cancel. The function is undefined there, though as others point out, it can easily be extended. -lethe ^{talk +} 10:07, 5 June 2006 (UTC)[reply]

Of course. Excuse the diversion. Dysprosia 10:59, 5 June 2006 (UTC)[reply]

Having studied them some more, I can now answer. The reason is that at z = a, you get z - a = 0, and to cancel that you must divide by zero and that's not defined. Everywhere else, z - a is some finite nonzero quantity, and the terms will cancel normally, but zero terms do not cancel in that way.

As an example, (3x) / (3y) = x / y everywhere because 3 is never equal to zero, but (0x) / (0y) is indeterminate. -Maelin 10:00, 5 June 2006 (UTC)[reply]

There could still be a pole at z = a, in the special case that c = a and b ≠ a. --Lambiam ^Talk 14:57, 5 June 2006 (UTC)[reply]

Measurement unit on measuring tapes.

Several of my measuring tapes have a measure mark that is a small black diamond shape. There are five of them for every eight feet or 19 1/5 inches for each mark. What is this measure? Sincerely

Albert J. Hoch

Its purpose is to allow carpenters to divide 8 feet exactly in five. See this page and this page (scroll down to diamond) for documentation. There has been a dispute on the Wikipedia where a user claimed it to be an "English cubit", but this is apparently not correct. --vibo56 09:45, 5 June 2006 (UTC)[reply]

BTW, is there a method to divide a line in five equal parts ? (ruler and compass only). Thanks. --DLL 21:29, 5 June 2006 (UTC)[reply]

Yes. You draw two parallel lines extending from the endpoints of the line, then mark off five equal intervals on each of the parallel lines. Lines drawn through corresponding marks will divide the original line into five equal parts. --Serie 00:12, 6 June 2006 (UTC)[reply]

So the black diamond is a carpenter's mark! Very interesting, I'd been supposing it was some foriegn unit of measurement. I was guessing Chinese!

Thanks very much. Albert J. hoch Jr.

Integral of dirac delta over step function

What does the following integral evaluate to:

\int _{0^{-}}^{0^{+}}{\frac {\delta (x)}{H(x)}}dx

where $\delta (x)$ is the Dirac delta function and H(x) is the Heaviside step function.

What about the following:

\int _{0^{-}}^{0^{+}}{\frac {\delta (x)}{1+cH(x)^{2}}}dx

\int _{0^{-}}^{0^{+}}{\frac {\delta (x)^{2}H(x)}{1+cH(x)^{2}}}dx

\int _{0^{-}}^{0^{+}}{\frac {\delta (x)H(x)}{1+cH(x)^{2}}}dx

\int _{0^{-}}^{0^{+}}{\frac {\delta (x)^{2}(1+c_{1}H(x))}{1+c2H(x)}}dx

If these do not exist, what are reasonable approximations to them I can make? deeptrivia (talk) 18:27, 5 June 2006 (UTC)[reply]

As a distribution, the delta function is really only supposed to be integrated against certain smooth functions, which the Heaviside function is not. Strictly speaking, the integral isn't defined. As for whether a meaningful approximation can be made, I do not know. -lethe ^{talk +} 21:32, 5 June 2006 (UTC)[reply]

Thanks lethe. Is it terribly unsafe to assume, for engineering purposes, that H(0) = 0.5, while integrating these? deeptrivia (talk) 21:42, 5 June 2006 (UTC)[reply]

That's probably perfectly safe. For the purposes of integration, we only care about the equivalence class of functions which differ almost everywhere. You can do anything you want on a set of measure zero without affecting the integration. So you can take H(0) to be anything you want, and 1/2 is a sensible value. -lethe ^{talk +} 23:12, 5 June 2006 (UTC)[reply]

So basically, I guess you want to do these integrals by substituting 0 in as an argument for H. I expect that this bit is unsafe, though I'm not sure exactly how unsafe it is. -lethe ^{talk +} 23:14, 5 June 2006 (UTC)[reply]

Okay, if I ask maple to evaluate:

\int _{-\epsilon }^{\epsilon }\!{\frac {{\it {Dirac}}\left(x\right)}{1+c\left({\it {Heaviside}}\left(x\right)\right)^{2}}}{dx}

it gives:

{\it {Heaviside}}\left(\epsilon \right)-{\it {Heaviside}}\left(-\epsilon \right)

The limit of this expression as epsilon --> 0 is undefined. However, as an engineering approximation, we can assume $\epsilon$ to be something small, like 1e-24, and then ${\it {Heaviside}}\left(\epsilon \right)=1$ and ${\it {Heaviside}}\left(-\epsilon \right)=0$ , and so the integral evaluates to 1. Is there any flaw in this reasoning? I'm asking because there's a significant thing happening between -1e-24 and 1e-24, which will be ignored by this assumption. Another thing is, if I were doing this integration from, say -1 to 1, then, say you pointed out I won't have cared about values at finite points. But here, the integration has to be done in a range that encloses 0 and is as small as can be imagined, so the value at 0 might have a significant effect. Regards, deeptrivia (talk) 01:14, 6 June 2006 (UTC)[reply]

You say that the limit of the expression is undefined, but it appears well-defined to me. The right=hand limit of Heaviside is 1, the left-hand limit 0, and so the difference between the two is 1. Your subsequent comments bear out this limit, and by the way, it does not matter how small an interval you integrate over, the result is always the same. What I fail to understand is how Maple arrived at the expression you quote. I don't know how to arrive at the integral that Maple has given you, so I'm not sure how bulletproof it is, but anyway, your reasoning about the value is correct: it is 1, no matter the interval. -lethe ^{talk +} 01:57, 6 June 2006 (UTC)[reply]

But how can this be? Shouldn't the c from the integrand return in the result? Does Maple assume that Heaviside(0) = 0? After all, you expect the answer 1/(1+c(Heaviside(0))²). --Lambiam ^Talk 02:18, 6 June 2006 (UTC)[reply]

two loops on a cylinder , are in same homology class??

Hi,

consider $\mathbb {S} ^{1}\times \mathbb {R} =X$ a cylinder thus

I am still working on that torus, and I thought, this would be handy :

suppose I have a loop ,so let's say a path

p:[0,1]\rightarrow X:t\rightarrow (e^{2\pi ti},a)\,\!

and another , a path

q:[0,1]\rightarrow X:t\rightarrow (e^{2\pi ti},b).\,\!

Are these two loops in the same homology class, I mean, is their difference, in $S_{1}(X)$ an element of the image of $\partial _{2}$

--Evilbu 20:01, 5 June 2006 (UTC)[reply]

If each loop goes around the cylinder the same number of times, then there is a homotopy taking one into the other: they are homotopy equivalent. By definition, they then are in the same cycle class. Consequently, yes, they are also in the same homology class (cycles modulo boundaries). The Z of H₁ comes from that fact that a once-around cycle can be added or subtracted with itself any number of times to give homotopically different cycles (twice-around, once-around-reversed, and so on). Viewed at a slightly higher level, a deformation retraction of the cylinder produces a circle, S¹; therefore these spaces have the same homology. Furthermore, this is true more generally: X×Rⁿ has the same homology as X. --KSmrq^T 21:10, 5 June 2006 (UTC)[reply]

A nit about your notation. What you've written are not paths. Your paths should have domain [0,1] and codomain X. You should better write something like

[0,1]\ni t\mapsto (e^{2\pi ti},a)

to indicate that the number t in the unit interval is mapped to a point in the path on the cylinder. -lethe ^{talk +} 21:36, 5 June 2006 (UTC)[reply]

Thanks, yet I'm sorry but I don't completely get it. Please be very clear in what you mean : homotopy between paths, or homotopy between continuous maps in general. Here was my idea : a 'push up u' of (b-a) is a continuous map from the cilinder to the cilinder, homotopic with the identity this means $H_{1}(id)=H_{1}(u)$ and thus $H_{1}(u)([p])=[q]$ A little weird I think. Why would a homotopy between those two points suffice? And what kind of homotopy do you speak, usually they mean with 'homotopy between two paths' : the homotopy fixes begin and end point all the time, which cannot be the case here as both are even disjoint. Evilbu 21:42, 5 June 2006 (UTC)[reply]

It's quite easy to see that your two paths are homologous: the boundary of the finite cylinder segment bounded by p and q is p – q. Thus they differ by the boundary of a 2 chain, so they are homologous. As for homotopy, you often consider homotopies with fixed endpoints, but you don't have to. The point is that given a homotopy between two curves with fixed endpoints, the two curves are the boundary of the image of the unit square under the homotopy. This also works on the cylinder for a homotopy without fixed endpoint, because the sides of the square are not in the boundary of the image. -lethe ^{talk +} 23:03, 5 June 2006 (UTC)[reply]

To amplify on what lethe has said, in this example we have two options to consider. The definition of homotopy applied to paths says that the ends can move. A path is a map, f, from the unit interval [0,1] to the space X. Given two such maps, f and g, a homotopy continuously deforms one path into the other. Before we start to compute homology groups, we want to take our huge number of cycles and reduce them to classes by homotopy equivalence. A closed loop on the cylinder is a 1-cycle and also a path for which the start and end points coincide. If we have two such loops a homotopy will necessarily deform loop to loop, but it need not leave any point fixed.

This not quite the same as computing the fundamental group, π₁(X,x₀), which requires a relative homotopy leaving point x₀ fixed. (Of course, the homotopy group π₁ is independent of the choice of x₀ if the space X is a path-connected space.)

A formally different option is the definition of homology groups as cycles modulo boundaries. Even without the reduction by homotopy equivalence this can cause two cycles to be identified in a homology group.

The definitions and implications in algebraic topology take time and exercise to grok. It will come; and besides, (modern) algebraic geometry is worse. I had the strange experience that algebraic topology seemed to have more geometric appeal than algebraic geometry! --KSmrq^T 23:58, 5 June 2006 (UTC)[reply]

Sine-rule like formula for radians

There is a highly useful formula which looks like the sine rule, ie something/something = something/something = something/something but I've completely forgot it. I believe terms like arc length, area, theta etc were included in it but I can't remember the other ones, nor can I remember the order. Thanks, Matt. — Preceding unsigned comment added by 80.229.237.12 (talk • contribs) 19:39, 5 June 2006 (UTC)[reply]

Multiplying 2 16-bit numbers with 32-bit registers?

I'm trying to implement IDEA in assembly language on a 386, and I'm having trouble because it was optimized for 16-bit processors. I want to be able to concatenate two 16-bit numbers to multiply mod $2^{16}+1$ and then separate them later, but I'm having trouble with it. I've tried using $\left(2^{16}x_{1}+x_{4}\right)\left(2^{16}k_{1}+k_{4}\right)$ and $\left(\left(2^{16}+1\right)x_{1}+x_{4}\right)\left(\left(2^{16}+1\right)k_{1}+k_{4}\right)$ but neither lets me extract just the multiplications I want and it's all mixed up. --Zemyla^t 21:05, 5 June 2006 (UTC)[reply]

Geometry question -- truncated icosahedron

I assure you this is not homework, just a question from someone who hasn't taken geometry since high school and did almost nothing with three dimensional shapes, at that.

Let's say I have a truncated icosahedron that should fit into a sphere an inner diameter of 150 cm. How long should each of the vertices be? I'm sure this is probably easy for someone to calculate given all of those wonderful symbols on the icosahedron page but I'm totally baffled by them.

Many thanks. --Fastfission 22:48, 5 June 2006 (UTC)[reply]

A vertex is a single point; it has no length. So, do you mean how long should the edges be? Or do you want coordinates for each of the 60 vertices? If the latter, the section Canonical coordinates has all the data, assuming a sphere of radius r, where r² = 9φ + 10. The associated edge length is not given, but can be computed if so desired. --KSmrq^T 23:08, 5 June 2006 (UTC)[reply]

Length of edges, sorry. I probably picked up the habit of calling the lines between vertices as themselves being vertices from computer graphics or sometihng like that. I don't need the coordinates, just the edge lengths. I can't computer them myself, because I don't understand the formulation and am not really interested in learning it from scratch just for this one question. :-) --Fastfission 02:33, 6 June 2006 (UTC)[reply]