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October 11

Topological group counterexample

This may be a little advanced for this page, but I'll give it a shot anyway. Does anyone know of an example of a topological group which is connected but not path-connected? Or do such groups not exist? Of course, such a group cannot be locally path connected and so cannot be anything nice like a Lie group. -- Fropuff 17:59, 11 October 2006 (UTC)

Consider in RxR, with the usual topology, X={(x,sin(1/x)) | x is in [1,0)}, and Y={(0,y) | y in [-1.1]}, now, XUY is connected and not path-connected. By the way, it is not an advanced topic.

This example is a topological space which is connected but not path-wise connected (if [1,0) is interpreted to mean (0,1]). However, the question concerned a topological group; or, in other words, a set which simultaneously has a group structure and a topology, such that the group actions 'respect' the topology in a natural manner. That is indeed a more advanced topic, and a harder question to answer. JoergenB 21:08, 11 October 2006 (UTC)

Right :P, sorry my wrong

But there still remains the interesting question as to whether a topological group which is connected has to be path-connected. I can't believe that the topologists sine curve could be a topological group - but I can't see any obvious reason why it is impossible to define a group structure on the points of the curve ... Madmath789 21:47, 11 October 2006 (UTC)

It's not possible. A topological group has to be homogeneous to a certain extent; given any two points you can define a (topological) autohomeomorphism of the whole space that takes one of the points to the other one. (Just multiply everything by the quotient of the two points.) The space above does not have this property; there's a point that has no path-connected neighborhood, whereas every other point does have a path-connected neighborhood. --Trovatore 21:51, 11 October 2006 (UTC)

Spot on! I think I should have known that ... Madmath789 22:06, 11 October 2006 (UTC)

---

I do not believe such groups exist. Google doesn't seem to be coming up with either a proof or a counterexample, so I'd try Bourbaki next.

RandomP 18:47, 11 October 2006 (UTC)

Morava states here that the Pontryagin dual of $\mathbb {Q}$ should be a counter-example.--gwaihir 21:14, 11 October 2006 (UTC)

Thanks for the suggestion. I've never seen Pontryagin duality defined for non-locally compact groups. I assume the definition is the same. I'll have to try and think about why the dual of Q is not path-connected (if that is indeed the case; Morava doesn't sound entirely sure). -- Fropuff 05:10, 12 October 2006 (UTC)

• Hopefully this is a less advanced question--What nontriv. spaces are everywhere connected but nowhere pathconnected? I'm trying to piggyback on Trovatore's reasoning above.Rich 18:57, 12 October 2006 (UTC)

Okay, I have a nice answer compliments of Daniel Asimov (who responded to a question posted on sci.math.research by Rich). The p-adic solenoid is a compact, connected abelian topological group which is neither path-connected nor locally-connected. I have yet to work out what the path component of the identity is, or the group of path components, but at least I have something to play with. -- Fropuff 06:36, 18 October 2006 (UTC)

TeX Supporting non-Wikimedia sites/servers for editing/previewing?

Given the uselessly slow loading of images (including TeX) the last few days (at least sporadically), it seems practical to find other wiki sites that are fully TeX supported (and with open preview privileges) to work on articles (some, like http://uncyclopedia.org, seem to support most TeX, but not the most recent commands/effects——e.g., "{\color{white}.}V=Q", to add a hidden character for space value at the beginning). Anyone know of any alternate sites to use for such purposes?  ~Kaimbridge~19:14, 11 October 2006 (UTC)

Be grateful if you could be more specific than http://uncyclopedia.org where I could not find anything related to TEX. Thank you in advance. Twma 01:06, 12 October 2006 (UTC)

Just hit the "Random page" link, then the "edit": Blank out the text that is there, cut and paste whatever you are working on, work on it, then cut it back out and save it wherever you had it before (you are just "previewing" on uncyclopedia, not saving anything).

Add a space and a negative space if you want to force TeX. \colour (IIRC) is not TeX, it most probably is a LaTeX addition, and it's not the best solution anyway. Dysprosia 02:07, 12 October 2006 (UTC)

After hitting the "Random page" link on the left column menu, the first time, I got a cooking page. For the second time, I got Editing Top Ten Numbers, etc. I do not understand what is a nagative space. I typed in space and delete. I typed in \frac{1}{2} but it did not show the fraction. Tried again with $\frac{1}{2}$ but failed again. I am interested in HOW IT WORKS but not whether it works. Clearly, it was not working probably due to my misunderstanding of NEGATIVE SPACE.Twma 12:20, 13 October 2006 (UTC)

Okay, after you landed on the "cooking page", you should have hit the "edit" link at the top, deleted the text, then work on whatever. The wiki TeX endmarkers are "<math>" and "</math>" (not "$"s) and the negative space is "\!", so a small space ("\,") and a negative space cancel out: Using your example, the proper way to format a fraction is "<math>V=\frac{N}{D}\,\!</math>", which should come out as $V={\frac {N}{D}}\,\!$.
There is a good "cheat sheet" page, Help:Formula (though I prefer an older version, as the format layout is better——besides using the "printable yes" link). P=) ~Kaimbridge~14:13, 13 October 2006 (UTC)

It's not a matter of forcing TeX to activate, but recognizing the initial character/command as spacing (it ignores it):

$\ V=Q+3\,\!$ (<math>\ V=Q+3\,\!</math>)

$\;V=Q+3\,\!$ (<math>\; V=Q+3\,\!</math>)

$\quad V=Q+3\,\!$ (<math>\quad V=Q+3\,\!</math>)

${}_{\color {white}.}V=Q+3\,\!$ (<math>{}_{\color{white}.}V=Q+3\,\!</math>)

${}_{\color {white}8}V=Q+3\,\!$ (<math>{}_{\color{white}8}V=Q+3\,\!</math>)

${}_{\color {white}.}\quad V=Q+3\,\!$ (<math>{}_{\color{white}.}\quad V=Q+3\,\!</math>)

Is there a better way? ~Kaimbridge~10:08, 12 October 2006 (UTC)

October 12

PROPOSED CHANGES TO THE REFERENCE DESK

If you haven't been paying attention to Wikipedia talk:Reference desk, you may not know that a few users are close to finishing a proposal (with a bot, now in testing and very close to completion) which, if approved by consensus, will be a major change for the Reference Desk.

Please read the preamble here, and I would appreciate if you signed your name after the preamble outlining how you feel about what we are thinking.

This notice has been temporarily announced on all of the current desks.  freshofftheufoΓΛĿЌ  06:58, 12 October 2006 (UTC)

For convenience, I propose any reactions to this anouncement be limited to Wikipedia:Reference_desk/Miscellaneous#PROPOSED_CHANGES_TO_THE_REFERENCE_DESK. DirkvdM 07:59, 12 October 2006 (UTC)

Bounded polynomials

I'm looking for some polynomials $p_{n}(x)$ such that $p_{n}(0)=p_{n}(1)=0$ and $|p_{n}(x)|\leq 1$ for $0\leq x\leq 1$ and most importantly, for some fixed, real $\theta$, have $|p_{n}(e^{i\theta })|>1$ when n is large enough. In a sense, I want $|p_{n}(e^{i\theta })|$ to be "as large as possible". Such polynomials are easy to find for $\theta =-\pi$ but I'm having trouble dreaming something up for small values of $\theta$. linas 15:28, 12 October 2006 (UTC)

Idea: take ${\frac {x(1-x)^{2}}{1-2\cos \theta x+x^{2}}}$ and approximate by using the geometric series.--gwaihir 15:47, 12 October 2006 (UTC)

Excellent! That's exactly what I needed! And its so ...obvious, yet I would have scratched my head for days :-( Thanks! linas 02:39, 13 October 2006 (UTC)

Let's see $p(x)=3x(1-x)\;$ — it is zero for x being 0 or 1, it does not exceed 3/4<1 for $x\in [0,\ 1]\subset \mathbb {R}$, and for $z=e^{i{\frac {\pi }{2}}}=i$ it is $3i(1-i)=3(i+1)\;,$ which gives an absolute value equal $3{\sqrt {2}}.$
Is $\theta ={\frac {\pi }{2}}$ small enough and $3{\sqrt {2}}$ big enough for your needs? --CiaPan 16:46, 12 October 2006 (UTC)

Well, I needed it to work for any $\theta$, no matter how small, and to be unbounded as n got large. gwaihir's answer was spot-on. linas 02:39, 13 October 2006 (UTC)

Weird question

If x + 1/x = y then what is x^3 + 1/x^3. Can anyone help? even my math teacher can't figure it out. (This is NOT a HW problem.)John 16:52, 12 October 2006 (UTC)

You'll need to do something to the first equation to make it look like the second. Try the most obvious thing; it won't work perfectly, but it'll get you halfway there. Melchoir 17:00, 12 October 2006 (UTC)

You can also just solve the corresponding quadratic equation for x (multiply by x to get rid of 1/x), no real problems there. But Melchoirs solution is definitely shorter.--gwaihir 17:53, 12 October 2006 (UTC)

• • It is not a weird question. You are within a millimeter of reinventing Chebyshev polynomials.Rich 19:18, 12 October 2006 (UTC)

y³ ? --RedStaR 20:53, 12 October 2006 (UTC)

hm well if you cube x + 1/x, i.e. (x + 1/x)^3 = x^3 + 1/x^3, so you have timesed both sides by x + 1/x)^2, but since x + 1/x = y i guess you could substitute y in for it on the other side giving x^3 + 1/x^3 = y(y^2) = y^3. So yeh, I'd go with y³. Philc _T^E_C^I 22:28, 12 October 2006 (UTC)

(x + 1/x)³ is not x³ + 1/x³!!!! When you exponentiate a binomial (ie a + b), you use the binomial theorem. See my answer below. Oskar 22:30, 12 October 2006 (UTC)

(edit conflict) Yeah, that's not how math works.

I tried it, and actually it's very simple (I'm surprised that your teacher couldn't solve it). We begin, as expected, with cubing (note: wikipedias TeX engine seems very flaky, I'll use ASCII):

x + 1/x = y

(x + 1/x)³ = y³

Thanks to Messrs. Pascal and Newton, this kind of exponentiation is easily expanded (using the third row of pascals triangle and the binomial theorem, we get(a + b)³ = a³ + 3a²b + 3ab² + b³)

x³ + 3x²*1/x + 3x*1/x² + 1/x³ = y³

x³ + 3x + 3*1/x + 1/x³ = y³

So then we really have our answer don't we? We can just say

x³ + 1/x³ = y³ - 3x - 3*1/x

and that'd be correct. Kind of an anti-climax? Maybe there's something we're missing... Hmm, that 3 looks factorable!

x³ + 1/x³ = y³ - (3x + 3*1/x)

x³ + 1/x³ = y³ - 3(x + 1/x)

Wait, wait! I recognize that expression! It's the very first equation we had! Remember, x + 1/x = y. Let's substitute our little hearts out!

x³ + 1/x³ = y³ - 3y

Voila! We have an answer. I believe that is how you were supposed to solve it, the trick was remembering that y = x + 1/x and realising that you could substitute it. Mini-exercise: to confirm our answer, expand y³ - 3y in terms of x, and make sure that we do, infact, end up with x³ + 1/x³. Cheers! Oskar 22:28, 12 October 2006 (UTC)

dy by dx

Why is ${\frac {dy}{dx}}$ often pronounced "dy by dx"? — Matt Crypto 21:19, 12 October 2006 (UTC)

I suppose that it has to to with the basic idea of differentiation. You take y = f(x), you increase x by dx, and you see what increase you get in y. Ie y + dy = f(x+dx), dy = f(x+dx) - f(x) and finally dy/dx = (f(x+dx) - f(x))/dx. What I'm saying is that dx is the "controlled" variable, the one that you limit down, and dy is the "result". You get dy by changing dx.

In reality, I have no idea, I'm just guessing. It's just something people say, just accept it :) Oskar 22:39, 12 October 2006 (UTC)

On second thought, maybe it's a historical brevity thing. Leibniz, who invented the notation, used infinitismals, not limits. To get the derivative he divided dy by dx. Maybe we today just dropped the "divided", since that's no longer the way we think about calculus. Oskar 22:42, 12 October 2006 (UTC)

I often pronounce it just "dy dx"... when no confusion will result®. Melchoir 22:47, 12 October 2006 (UTC)

I say "the derivative of y with respect to x". StuRat 00:59, 13 October 2006 (UTC)

For the same reason why 3/5 is pronounced "3 by 5", I guess. deeptrivia (talk) 02:14, 13 October 2006 (UTC)

I pronounce that "three fifths". I also pronounce .6 "three fifths" though for less-obvious decimals I just use the powers place instead of simplifying. Oh and I say "dy dx". --froth^{T C} 18:58, 17 October 2006 (UTC)

Perhaps as an origination for the symbol for small changes, the Greek letter $\delta$, which eventually became "d"? x42bn6 Talk 18:55, 13 October 2006 (UTC)

Thanks for your answers! — Matt Crypto 19:59, 14 October 2006 (UTC)

de Bruijn sequence

Hey ya'll!

I'm after a very specific number or equation, used to crack numbered locks. It is used by repeatedly entering the number or equation - and, after continually entering the number, the door, lock, or other password opens.

I know that this number exists from an X-Files fan book, which covered series 1 and 2. In one of the episodes in series one and two, Scully and Mulder used this number/equation/code to get through a number-lock door. They had to repeatedly enter the number, and, eventually, they opened the door.

I can't say more than that, unfortunatly. However, if anyone has any other useful number/equation that does something similar to that, I'd love to hear about it. Scalene 22:03, 12 October 2006 (UTC)

Um, The X-Files is fiction. If you don't have any information about the combination, you have to try every possibility. There's no way around that. The "number/equation/code" you're looking for is called a brute force attack, also known as counting. Am I missing something? —Keenan Pepper 22:56, 12 October 2006 (UTC)

I found it, eventually. The de Bruijn sequence - searched for an hour on the net and found it by luck. SO, that leads me to the second part of my question:

What is the basic algothrism for a de Bruijn sequence, with base 2 and length 2. (As in the length of the code you're trying to crack) And how can this be changed to find the value with different bases and lengths of the required result?

Essentially, I need an equation for base 10, with a length of 4.

Scalene 23:39, 12 October 2006 (UTC)

See the de Bruijn sequence page again - it describes the way to find your sequence in details.

Probably there is no compact equation that would give what you need. You'll have to build a proper graph and traverse it with a Hamiltonian cycle. The graph will have 10⁴ nodes (one for each posible 4–digit code), connected with edges (an oriented edge leads from node A to node B when you can derive B code from A code by removing the first digit and appending one at the end). That graph has a Hamiltonian cycle, which defines the de Bruijn sequence for your problem. --CiaPan 06:35, 13 October 2006 (UTC)

Oh, I see. I assumed the lock had something like an enter key and would accept the correct sequence only in isolation. Still, it only speeds up the process by a constant factor. The sequence produced by a maximal linear feedback shift register is quite close to a de Bruijn sequence, but not exactly, and I'm not sure how to convert from one to the other. Anyone have a reference? —Keenan Pepper 07:02, 13 October 2006 (UTC)

I must have misunterstood then – I assumed the lock accepts any input stream (key sequence), and opens immediately when you enter proper digits, ignoring possible wrong input before. And yes, the de Bruijn sequence reduces time by constant factor only. It is the best you can get, though. It's obvious that every two different codes must begin on different positions, so you can not pack 10⁴ codes in less than 10⁴-digit string.

On the other hand, I'm affraid the only way to break a lock with enter key is trying all 10⁴ possible codes. A lock does not give any Mastermind-like feedback, which you could use to refine your guess in next attempts. --CiaPan 09:29, 13 October 2006 (UTC)

Some combination locks actually do – a slight but perceptible difference in how much you can move things. I've occasionally succeeded in opening a lock whose combination was lost using just that feedback, and some luck. Also, a decimal Gray code sequence may help to reduce the physical effort, but the gain is marginal and typically not worth the complication.  --Lambiam^Talk 12:20, 13 October 2006 (UTC)

The Hamiltonian path problem is, unfortunately, NP-Complete, making the de Bruijn sequence a really tricky thing to calculate. - Rainwarrior 17:08, 15 October 2006 (UTC)

You can do with an Eulerian cycle over the directed graph whose vertices are (n–1)-length words over the alphabet (here digits), with an edge from uσ to σv for every pair of symbols u and v and every (n–2)-length word σ. This is an easy problem. Since every vertex has two incoming and two outgoing edges, such a cycle exists. This link gives an extensive account.  --Lambiam^Talk 18:32, 15 October 2006 (UTC)

I'm afraid you can't. The Eulerian cycle visits each edge in the graph. To do that you would have to visit each node N times for N-digit alphabet (N>1), whilst we need each node to be visited (i.e. each code be generated) exactly once. Also the last part of your comment is false: in general we have more than two edges starting and two edges ending in every node (unless we consider binary codes). It's true, however, that finding the Eulerian cycle is easy, because in each vertex there is same number of incoming on outgoing edges (so you can leave every vertex you visit, until you come back to the starting node). --CiaPan 05:49, 17 October 2006 (UTC)

Read what I wrote carefully again. How many edges are there? What is the length of a de Bruijn sequence? Do you see a similarity? The elements of the sequence are not represented by the vertices but by the edges. Look at the last diagram in our article.  --Lambiam^Talk 06:29, 17 October 2006 (UTC)

Ahh, so can the Euler path be constructed in a greedy fasion? (or at least in polynomial time?) I can't think of a good way offhand. - Rainwarrior 07:01, 17 October 2006 (UTC)

I should have just looked at Euler path. So apparently Fleury's algorithm is what Scalene should be using to find this sequence. - Rainwarrior 07:22, 17 October 2006 (UTC)

Testing whether removal of an edge results in loss of connectivity is expensive, and no incremental method is known. A faster way is to represent the path as a doubly-linked list, take at each vertex just any not yet traversed edge, and when coming to a premature dead end, backtrack – without destroying the tail of the path – to a vertex that still has outgoing edges, cut the path open there, and continue. When you come again to that vertex, connect the loose tail and if necessary backtrack again, etc.  --Lambiam^Talk 08:40, 17 October 2006 (UTC)

That'd do it nicely. - Rainwarrior 04:03, 18 October 2006 (UTC)

Though, I'm wondering how exactly this "Fleury's Algorithm" is actually implemented... I can't seem to find an adequate description. All of the pages I've found say to avoid bridges while possible, basically, but neglect to explain how to maintain the list of bridges. What you suggest sort of does the bridge test, functionally, but while also finishing up the cycle adjacent to the bridge. Hmm... Perhaps instead of your suggestion, we could first build the unconnected cycles, keeping a list of the edges that could not be added, then at the end break these cycles to add the "bridge" edges in this list, which should occur in pairs (or else there would be no Euler path), thus avoiding backtracking entirely? (Both methods should be linear time, anyway, I think, so this would be a minor difference.) Perhaps this is what is meant by Fleury's algorithm?: identify the bridges by forming cycles, then add the bridges? -- Rainwarrior 04:17, 18 October 2006 (UTC)

October 13

Penrose tiling, and its Fourier transform

What does the 2D fourier transform of a penrose tiling look like? --HappyCamper 12:27, 13 October 2006 (UTC)

I'd expect that instead of nice spikes at the relevant frequency vectors, there would be wild oscillation. Surely there would be something. Melchoir 19:02, 13 October 2006 (UTC)

Probably, you need to be more specific about what you actually mean with a Penrose tiling, that is, how you intend to draw it on the plane. The article itself shows several different ways to do so. I think that if you draw them with infinitely thin lines, they would not have any impact at all in the Fourier transform, which would be just zero. And how about different colours? —Bromskloss 22:21, 13 October 2006 (UTC)

This is an excellent question that deserves more than cop-outs. Here's an image (which, unfortunately, you might not be able to see without a subscription), and these articles also have some nice diagrams. —Keenan Pepper 06:37, 14 October 2006 (UTC)

I'm sure it's a good question, I just don't know how the image is defined! —Bromskloss 20:16, 15 October 2006 (UTC)

October 14

I love my Turing Machine!

I have a task to complete which is to design a Turing Machine which will halt on all inputs from a certain language and halt and accept only on inputs from a particular subset of this language. I should mention that in the course I'm taking we've taken the option of considering a TM as a set of rules of the form {READ STATE, READ SYMBOL, CHANGE TO STATE, CHANGE TO SYMBOL, GO LEFT OR RIGHT OR PAUSE} with a "start" state and an "accept" state, the machine starting on the "left-most" letter of the input word, and the machine halting without accepting if it encounters a symbol in a certain state for which there does not exist a rule.

Anyway, I've completed the task, and designed a TM. I'm happy with it - I think it's quite cute in fact. And I should think anyone reading my rules with the accompanying commentary would be happy too. What I would like to do is make a systematic and complete justification of my TM - i.e. prove that it does exactly what it's meant to. The trouble is, when I start writing about it I end up with paragraph upon paragraph of flowery prose and going round in circles. This is silly. It's a simple machine really and is dealing with a simply defined subset of a simple language.

So, my question is this: does there exist a standard approach to justifying a machine of this type which would be concise and rigorous? It's bugging me since I love my machine and it clearly (to me) does its job but I end up going round in circles trying to justify it.

Thanks --87.194.21.177 01:04, 14 October 2006 (UTC)

There's no systematic way to prove a Turing machine does what it's supposed to do. As you know, there's no general method for even proving that a Turing machine terminates. And writing proofs is hard.

Can you do a structural induction proof to show your machine works? That is, show it works for simple cases, and then any expansion of those simple cases must also work?

In any case, you might have to do two proofs - first to show that the machine accepts everything it's supposed to, and a second proof to show that it rejects everything it's not supposed to. --Robert Merkel 01:19, 14 October 2006 (UTC)

Assuming that the TM design is indeed simple, I expect the structure of a nice proof to depend mostly on the nature of the language and its subset. Does the head move to and fro? Is auxiliary data written to the tape? A possible approach may be to come up with an invariant: a predicate that is true for the initial configuration and is preserved by each step, and that implies for a final configuration that the machine is halted in the right acceptance condition. This has to be supplemented with a variant function, a mapping from the space of configurations to some domain with a well order, such that the function value decreases on each step. If a proof can be given at all, it can be put in this form.  --Lambiam^Talk 01:33, 14 October 2006 (UTC)

As a matter of the fact, I would have felt fairly sure that the langie and its subset are regular, if it were not for the phrase 'halts on all input', which seems to indicate that no word in the language has a proper left subword. Now, it might still be regular, of a very simple structure; e.g., there could be one letter that stands at the end of each word, but only there. If you may strip down the verification to the recognition of a regular language, you may construct a finite automaton doing this, and then verify that the Turing machine accepts the same language.

Of course, this is not very helpful, unless you have learned how to relate a finite automaton with a regular language. JoergenB 04:08, 14 October 2006 (UTC)

Thanks guys. I don't know about regular languages - maybe I'll dip a toe. Meanwhile I think I'll start thinking along the other lines suggested. I think that, as Robert says above, "writing proofs is hard" is the essential truth I am struggling with as per usual. Thanks again. --87.194.21.177 15:37, 14 October 2006 (UTC)

Elasticity of substitution

I understand that the Elasticity of substitution is the % change in x/y over the % change in the rate of technical substitution, or MRS, but then I become confused when it is simplified to: d ln x/y over d ln f(x)/f(y).

Could someone state this in terms of operations, or what I should calculate if all I have is their production function and conditional input demand functions? Thanks, ChowderInopa 02:03, 14 October 2006 (UTC)

Hmm...let's try this: take some variable x, and some function of that variable y(x).

${\frac {d\ln y(x)}{dx}}={\frac {d\ln y}{dx}}={\frac {d}{dx}}\ln y={\frac {1}{y}}{\frac {dy}{dx}}$. Does this help? --HappyCamper 04:30, 14 October 2006 (UTC)

If it helps you: interpret "d ln x/y over d ln f(x)/f(y)" as: (d (ln x/y) / dx) over (d (ln f(x)/f(y)) / dx), that is, as the quotient of two ordinary d/dx forms.  --Lambiam^Talk 11:10, 14 October 2006 (UTC)

Further, the notation "f(x)/f(y)" is misleading. This should be something like "U_x/U_y", the ratio of two partial derivatives of the utility function U in x and y.  --Lambiam^Talk 11:24, 14 October 2006 (UTC)

Lambiam, you are right, the bottom is the partial derivatives of the utility function, my bad. I'm still not understanding though...

I am taking the derivative of the ln(x/y) with respect to what? and then, I am dividing this by the derivative of the ln of the ratio of the partial derivatives of the Utility function??? Help! ChowderInopa 23:57, 14 October 2006 (UTC)

I don't know the meaning of the variables here, but in any case, for the formula to make sense, there has to be something coupling x and y, making them dependent on a common variable. It could be that y is a function of x, or x and y both are functions of t (say time). If x and y both depend on a common variable (like x or t), just take the derivatives w.r.t. that variable.  --Lambiam^Talk 18:59, 15 October 2006 (UTC)

Permutations

hello.can you pls give me instant tricks of solving permutations...

Hello, our permutation article tells : It is easy to count the number of permutations "P" of size r when chosen from a set of size n (with r ≤ n).

The formula is P(n,r) = n! / (n-r)! where "!" means factorial.

Our factorial article tells that n! = Π k=1, n, k) where Π "capital Pi" means multiplication, giving n! = 1*2*...(n-1)*n.

One easy trick I'm thinking of is just solving with your brain when, say, n < 10, else using a spreadsheet or any computer algebra system . Does that help ? -- DLL ^{.. T} 18:12, 14 October 2006 (UTC)

Can you explain what you mean by "solving permutations"? Could you give an example of how to solve one (simple) permutation?  --Lambiam^Talk 20:15, 14 October 2006 (UTC)

Linear vector spaces

I am learning about linear vector spaces at university at the moment, and was wondering about slightly odd bases, such as {sin(t), cos(t)}. What are the coordinates of the function f(t) = 3 sin(t) + 5 cos(t) with respect to these basis {sin(t), cos(t)}, for example. Is the answer simply (3,5)? Batmanand | Talk 13:27, 14 October 2006 (UTC)

Yes. The coordinates are just the amounts you need to multiply every vector in the base and add together to get your desired vector. Note that you need an ordered base, that is, (sin, cos) and not {sin, cos}. -- Meni Rosenfeld (talk) 13:37, 14 October 2006 (UTC)

Your last statement is merely a matter of taste.--gwaihir 13:59, 14 October 2006 (UTC)

Taste? How can you tell if (3, 5) means 3 sin(t) + 5 cos(t) or 3 cos(t) + 5 sin(t) if the base is {sin(t), cos(t)}, which is of course equal to {cos(t), sin(t)}? -- Meni Rosenfeld (talk) 14:02, 14 October 2006 (UTC)

If you use unordered bases, sets of co-ordinates are functions on the basis, so it's not (3, 5) but rather (sin → 3, cos → 5) (in either order).--gwaihir 15:27, 14 October 2006 (UTC)

Very well, but he was specifying his expectation to have (3, 5) as the coordinates vector, which can only work if the base is ordered. So, in general it is a matter of taste, but not within the context of this question. -- Meni Rosenfeld (talk) 18:34, 14 October 2006 (UTC)

Batmanand, just to make sure you made a typo and didn't commit one of the beginner's most common mistakes, when you wrote these basis: {sin(t), cos(t)} is one basis. The basis consists of two basis vectors (which in this case may be called basis functions). You'll soon meet situations where you consider several bases simultaneously, but then you should notice that that means several collections of basis vectors. JoergenB 14:42, 14 October 2006 (UTC)

Lol you do me too much of a service; I have only just started to course and did indeed make the elementary error you mention. Thanks for pointing it out. Batmanand | Talk 16:20, 14 October 2006 (UTC)

And it's bases in plural, isn't it? Of course, when unsure of the number, you could write these basis and say that one or the other was just a typo. ;-) In any case, I'm not very fond of writing vectors as a sequence of coordinates (like "$(1,5,3)$"), because I seem to never know which basis I (or someone else) was referring to anyway. If my example was implying the basis vectors $\mathbf {\hat {x}}$, $\mathbf {\hat {y}}$ and $\mathbf {\hat {z}}$ (in that order) I prefer $\mathbf {\hat {x}} +5\mathbf {\hat {y}} +3\mathbf {\hat {z}}$. This way you can also mix basis vectors from different bases in the same expression. —Bromskloss 19:59, 15 October 2006 (UTC)

Not to mention what happens if the vector space isn't of finite dimension. You'd have a hard time writing out the whole coordinate sequence then, whereas by using the other way you can use a summation symbol. —Bromskloss 20:04, 15 October 2006 (UTC)

I don't think that's really relevant; if you can write $\sum _{i=0}^{\infty }f(i){\hat {e}}_{i}$, surely you could also write the sequence (not the set) $\{f(i)\}_{i=0}^{\infty }$ of coordinates? Moreover, for an uncountable basis, there's no way to write the sum (although in some cases an integral can be substituted), but you can still use a coordinate-like function $f(t)$ of some continuous variable t (see, for example, Fourier transform and Laplace transform, although I'm not sure they can precisely be called bases). --Tardis 15:20, 16 October 2006 (UTC)

d2/dt2 = H d2 / dX2

d²/dt² * ß = H * d/dx * ß, Solve for H205.188.116.136 14:36, 14 October 2006 (UTC)

What is ß and what is H? Functions? Constants? --HappyCamper 15:32, 14 October 2006 (UTC)

And what's with the asterisk? Some kind of convolution operation?  --Lambiam^Talk 15:45, 14 October 2006 (UTC)

Assuming you mean

${\partial ^{2} \over \partial t^{2}}\beta =H{\partial \over \partial x}\beta \,,$

it might help to apply the method of separation of variables; compare how the wave equation is solved for the case of one space dimension.  --Lambiam^Talk 16:06, 14 October 2006 (UTC)

Well, if they're functions there's simply no general solution. Separation of variables can't be applied if the solution is inseparable, and there's no reason to automatically assume it is. To solve it, you'll need some conditions. --BluePlatypus 01:33, 18 October 2006 (UTC)

alpha level increase power?

Hi:

I am sitting on my desk, trying to figure out the following statistics problem: if the alpha level increases power also does. Now, I wonder whether power can be less than the alpha level for a hypothesis test. I think it cannot be less as they are somewhat related and depend on each other. But I am not 100% sure. Does anyone know anything that would illuminate my mind and refresh my brain cells a little? I would be thrilled. Thanks much. Hersheysextra 16:15, 14 October 2006 (UTC)

For the moment I can only say that if that were to turn out possible for some test, then it is a strange test. I tried to see if the impossibility follows from the usual requirement of the test statistic being a sufficient statistic, but the (presumably trivial) proof eluded me.  --Lambiam^Talk 23:41, 14 October 2006 (UTC)

As I understand it, power is 1 - beta, where beta is the risk of a Type II error, while alpha is the risk of a Type I error. Thus power and alpha aren't directly related. However, alpha would normally be less than 1 - beta, as both alpha and beta are usually small.--86.136.126.188 17:36, 17 October 2006 (UTC)

The "alpha" is the probability of type I error, which is the probability of rejecting your null hypothesis when it is in fact true. The "power" is (1-P(type II error)), or the probability of rejecting your null hypothesis when it is false. So:

$\alpha >\beta \,$ $\Rightarrow P_{H_{0}}(acceptH_{1})>P_{H_{1}}(acceptH_{1})$

means you are more likely to reject your null hypothesis when it is true than when it is false – which is obviously a pretty stupid thing to do. It is entirely possible (it might take me a few minutes to come up with a specific case, but it just depends on the critical value you choose for your test) but if you come across test that has this characteristic, it would be a smart move not to use it ;-) – AlbinoMonkey ^(Talk) 07:43, 18 October 2006 (UTC)

significance of e

There is a definition for e in wikipedia. Assume the following.

y=f(x); y=e^x; y'=x(e^(x-1));

here, acceleration of y is always equal to velocity of x. Does this point that - acceleration of y is equal to velocity of x - hold good for only e or does that hold good for all integers?

Um, you seem to have differentiated y with respect to e, not x; I don't think that's a meaningful thing to do. Melchoir 17:02, 14 October 2006 (UTC)

$e^{x}={\frac {x^{0}}{0!}}+{\frac {x^{1}}{1!}}+{\frac {x^{2}}{2!}}+...+{\frac {x^{n}}{n!}}+...\,$

${\frac {d(e^{x})}{dx}}=0+{\frac {1x^{0}}{1!}}+{\frac {2x^{1}}{2!}}+...+{\frac {nx^{n-1}}{n!}}+...\,$

${\frac {d(e^{x})}{dx}}={\frac {x^{0}}{0!}}+{\frac {x^{1}}{1!}}+...+{\frac {x^{n-1}}{(n-1)!}}+...\,$

${\frac {d(e^{x})}{dx}}=e^{x}\,$ Richard B 18:06, 14 October 2006 (UTC)

Visual Analog Linear Scale

When it was first invented ? 124.109.18.18 20:17, 14 October 2006 (UTC)

There is an article: Sriwatanakul K, Kelvie W, Lasagna L, Calimlim JF, Weis OF, Mehta G., "Studies with different types of visual analog scales for measurement of pain." Clinical Pharmacology and Therapy 1983; 34(2):234-9. An abstract is here. I can't tell whether this presents the actual invention, but I think this article was instrumental in making them popular, at least for eliciting pain intensity judgements from patients.  --Lambiam^Talk 23:35, 14 October 2006 (UTC)

October 15

The Fourth Axis

Where is the forth dimension's axis? I'm refering to the spatial dimension, not time. THL 14:56, 15 October 2006 (UTC)

In three dimensions, the standard basis includes the unit vectors (1,0,0), (0,1,0), and (0,0,1). In four dimensions, it is similar: (1,0,0,0), (0,1,0,0), (0,0,1,0), (0,0,0,1). If you are about physical space, there is no real fourth dimension, so I don't think we have the proper words to describe it in that sense. - Rainwarrior 16:54, 15 October 2006 (UTC)

There is no visual way to represent full 4 dimensions similarly to how we do with solids in three dimensions. You can project anything in higher dimensions to three dimensions, and you can take several 3-D slices of the 4-D object and then order them in time, but that's still just having your way around to represent the 4th dimension. ☢ Ҡi∊ff⌇↯ 17:30, 15 October 2006 (UTC)

I gathered that from the article, I was just wondering if they had found the axis. Apparently not. THL 18:28, 15 October 2006 (UTC)

I'd just like to point out that modern theories say that physical space has as many as 26 dimensions, though only 3 of them are meaningful in large scales. What do you mean by "if they had found the axis"? They have found complicated explanations for more than 3 dimensions, and they have found that there is not simple explanation for them. It's not like there is any "fourth axis" hiding somewhere in the familiar euclidean 3-dimensional space. -- Meni Rosenfeld (talk) 19:17, 15 October 2006 (UTC)

I meant that I was wondering if they had found the equation of the line passing through the origin of a spatial plot at a 90 degree angle to the X, Y, and Z axes; A.K.A. the axis representing the fourth dimension. THL 19:31, 15 October 2006 (UTC)

Well, that's easy. In parametric form, (x, y, z, w) = (0, 0, 0, t). Melchoir 19:37, 15 October 2006 (UTC)

Yeah, and it was also implicit in Rainwarrior's reply. You can also have it in equations form, which is just as simple - in 4D it's (x = 0, y = 0, z = 0), or, if you like having a single equation, |x| + |y| + |z| = 0. -- Meni Rosenfeld (talk) 20:06, 15 October 2006 (UTC)

When it comes to math, I am very blue collar. I haven't taken anything higher than Algebra 3/4, and I got a C in that. I'm not a genius. THL 20:36, 15 October 2006 (UTC)

In that case, I suggest you read about Flatworld [1], where there is a 2D world full of 2D creatures, and the author describes how 3D objects would appear to them. By analogy, we 3D creatures can't fully comprehend the 4th physical dimension, much less the 26th. We can, however, use that dimension in science and math calcs. StuRat 22:02, 15 October 2006 (UTC)

Much thanks. THL 04:59, 16 October 2006 (UTC)

Let me say that neither Melchoir nor I meant to offend you. I was only trying to suggest that before giving up, you should be as specific as possible about your question, and try to make sure we understand each other. Your first question made it unclear whether you are referring to a mathematical or a physical concept (and there's a world of difference!), and your second was just as vague. I get the impression you didn't fully understand Rainwarrior's reply, in which case the best thing to have done is to ask for an explanation. -- Meni Rosenfeld (talk) 13:45, 16 October 2006 (UTC)

Well, when I suggested a set of four vectors to define an axis, if you want to create the equation of a line that passes through them, there was the parametric form above, or perhaps the more familiar equation "L = A + tB" (t can be any/every real number), where "A" is some point it passes through, and "B" is a direction. In this case, "A" is (0,0,0,0), the origin (where all four axes intersect), and "B" would be (0,0,0,1) for the fourth axis. So, L = (0,0,0,0) + t (0,0,0,1). - Rainwarrior 15:08, 16 October 2006 (UTC)

I wasn't offended. I have a tendency to sound impersonal on the internet, my apologies. My vagueness was a result of my lack of understanding. I get it now, thank you both. THL 05:28, 17 October 2006 (UTC)

Geometry - theorems and axioms

Hello, is there anywhere I could find a quite complete list of theorems in geometry (well, euclidean geometry would suffice). I looked at geometry, euclidean geometry and on pages about Hilbert's and other's sets of axioms, but couldn't find a list of theorems. I'm not really looking for very complicated theorems, just things like that the diagonals of a square intersect perpendicularily (well not just THAT simple actually :p). I would also appreciate if someone could point out where I could find a page that would compare analytic geometry and "traditional" geometry, by saying how one would state theorems and axioms in one or the other. Thanks --Xedi 19:45, 15 October 2006 (UTC)

There are thousands upon thousands of theorems in Euclidean geometry. You don't want a "quite complete" list, I hope. But in any case, I don't know of a list of even "notable" geometrical theorems. As to the comparison between analytic geometry and "traditional" (axiomatic) geometry: the difference is not so much in how you state the theorems but in how you go about proving them. For traditional geometry you have Euclid's axioms (Euclidean geometry#Axiomatic approach). For analytical geometry you could use any axiomatization of the real numbers, but no working mathematician thinks in terms of axioms when proving geometrical results analytically.  --Lambiam^Talk 23:05, 15 October 2006 (UTC)

HOW DO I ADD AUTO TIME.

HOW CAN I ADD AUTO TIME TO MY CALC. I WOULD LIKE TO SIGN PEOPLE IN AND OUT AND KNOW HOW LONG THE WAIT WAS BY CLICKING A CLOCK ICON. PEGGY [email removed] — Preceding unsigned comment added by 68.88.141.66 (talk • contribs)

• Please don't write in ALL CAPITAL LETTERS.
• Do not include your e-mail address.
• Sign your question.

And most importantly, we can't answer your question unless you:

• Be specific.

--Ķĩřβȳ♥ŤįɱéØ 21:30, 15 October 2006 (UTC)

October 16

What calculations were aided by John Napier's tables of logarithms?

From what I've been able to gather, these tables were very incomplete, offering only a small percentage of what would have made up a complete and adequate set (given the base, etc.). Yet histories state that they were instantly very useful for all kinds of important calculations. So, what is an actual example of calculations for which Napier's tables were complete and adequate and used? I'm guessing that either

• I've misunderstood what I've read, either mathematically or historically, and in fact Napier's tables were complete enough to do the whole range of calculations for which people used Briggsian tables before calculators.
• There was a specialized kind of navigational (etc.) problem that determined the range of numbers Napier offered.
• In fact the histories are wrong to say they were useful, and Napier produced his tables merely as a demonstration of what logarithms could accomplish, should others continue with much more industry to make much more complete tables.

Wareh 02:43, 16 October 2006 (UTC)

Napier's table is organized in such a way that it specifically aided in the calculations required for celestial navigation using the formulas of spherical trigonometry. For more details, see our article Prosthaphaeresis. His 1616 Admirable table presents in columnar form, for α going from 0° by increments of 1' to 45°, the values of α, sin α, Log(sin α), Log(tan α), Log(cos α), cos α, 90°–α (so effectively his table goes to 90°). Here Log(_) stands for Napier's Logarithme, which is to the base of exp(–10^–7). At the time, computations for celestial navigation presented the most pressing need for easier multiplication, and Napier's table fulfilled this need. His particular organization was definitely easier for this specific application, although less convenient for general, arbitrary multiplication. It is clear from the accompanying text that Napier was aware of this.  --Lambiam^Talk 11:00, 16 October 2006 (UTC)

Oh, how cool it would have been if celestial navigation was actually to navigate from star to star! :-) —Bromskloss 12:42, 16 October 2006 (UTC)

You want astrogation. But, then again, who doesn't? --Tardis 15:22, 16 October 2006 (UTC)

Thanks for another fine answer, Lambiam. I would love to see someone as informed as you add a couple of sentences at Logarithm#History pointing to Prosthaphaeresis, and an example at Prosthaphaeresis of a calculation using the combined resources of that method and Napier's tables. But of course I am content just to have my curiosity satisfied on this point. Wareh 15:33, 16 October 2006 (UTC)

a?

So in my math class, we're doing factorials. One of the homework problems was "a?" My friend says it means a wants to be n. Nobody knows what it actually means. Therefore, I ask you; WTF does that mean? I'm not asking for the problem to be solved, just for someone to tell me what the question mark is supposed to be.

Thank you. 64.198.112.210 15:34, 16 October 2006 (UTC)

The Minkowski's question mark function uses the question mark symbol, but not like that. I doubt that subject would show up in your math class, anyway. My guess is that it's just a typo. ☢ Ҡi∊ff⌇↯ 15:52, 16 October 2006 (UTC)

Perhaps if you can provide more context as to where and how it appears, we might figure out the intention. -- Meni Rosenfeld (talk) 21:18, 16 October 2006 (UTC)

If it was supposed to be a factorial, that is, a!, that can be expressed as a! = 1×2×...×(a-2)×(a-1)×(a), which is the definition of the factorial. --Ķĩřβȳ♥ŤįɱéØ 07:23, 17 October 2006 (UTC)

My guess is that it was typed by someone not so knowledgeable or not paying much attention. For example, the phrase "how much is a!?" may sound inappropriate for a test if the person is ignoring the fact the "!" is part of the notation of the exercise. So the typist could have just removed the "!" thinking it would make the text more "formal" (since "!?" makes it look like an overly enthusiastic question) ☢ Ҡi∊ff⌇↯ 07:38, 17 October 2006 (UTC)

Combinatorics -- the "obnoxious geek" problem

This was motivated by an annoying guy in my class I hope I never have to work with again. Here's the situation, and it could be generalized, but I'll just wonder about my specific case:

In my class of 16, we've been put into 4 groups of 4 for team projects. We'll be put into different groups of 4 a few more times throughout the semester. My question is, how many partitions of the class into groups of 4 can be made such that any 2 individuals are in the same group at most once?

The best case scenario would be 5 partitions, in which case everyone would have been in a foursome once with every other classmate. It wasn't too hard to sketch out a way to get 3.

Part of my question is even how to understand this problem or put it into rigorous terms. Does it have another name? My initial naive approach is to phrase is something like how many copies of "4 x the complete graph on 4 vertices" can the complete graph on 16 vertices be broken into, but it seems like there must be something easier.

Most importantly, can I be assured I won't have to work with annoying guy for at least 3 more iterations of these classroom teams?

Thanks! - 142.110.227.162 16:05, 16 October 2006 (UTC)

Which of these do you mean to ask ?

• How many groups of 4, out of 16, can be formed without two SPECIFIC members being in the same group more than once ?

• How many groups of 4, out of 16, can be formed without ANY two members being in the same group more than once ?

In the first case, you would also have to define how "different" each group needs to be. That is, can they have 3 out of 4 members the same as from the last time ? Or 2 out of 4 ? StuRat 16:42, 16 October 2006 (UTC)

I guess I wasn't clear. I want to look at "partitions" (the act of dividing the whole class into 4 groups of 4) as a whole, and consider the largest set of partitions where any group of four contains any two individuals at most once.

i.e. what's the most number of times I can partition the class before somebody, anybody, is in the same group with a person from a previous partition? (as you can see part of my problem is with notation) 142.110.227.162 17:05, 16 October 2006 (UTC)

I think three is probably it. It is easy to see that by relabelling the students, the first two partitions can always be written (1 2 3 4)(5 6 7 8)(9 10 11 12)(13 14 15 16) and (1 5 9 13)(2 6 10 14)(3 7 11 15)(4 8 12 16). There are 16×3×2 labellings that will give these for the first two partitions (16 ways to choose 1, 3 ways to choose 2, 2 ways to choose 3, 1 way to choose 4 through 16). In particular, you can permute your labelling of 1 2 3 4. I think you can show that these 4×3×2 relabellings are sufficient to put the third partition into the form (1 6 11 16)(2 7 12 13)(3 8 9 14)(4 5 10 15). It is easy to then prove that no additional partition is possible. Probably a cuter proof is, though. –Joke 20:40, 16 October 2006 (UTC)

For dividing N people into n m-groups (so $nm=N$), you certainly can't do better than $N-1 \over m-1$ since you drain the pool of possible partners by $m-1$ each time. (In your case, $N=16$, $n=m=4$ so that maximum is $15/3=5$, as you said.) I think Joke is right that you can't get this maximum in your problem, but if $n=m$ are prime, I'm pretty sure you can. Start by writing the objects out in a square matrix: $\left[{\begin{matrix}{\mbox{Andy}}&{\mbox{Sue}}&{\mbox{Roger}}&\dots \\{\mbox{Mark}}&{\mbox{Louis}}&{\mbox{Alice}}&\dots \\\vdots &\vdots &\vdots &\ddots \end{matrix}}\right]$. Then form one partition by rows and n more by considering diagonals with various "slopes": 0 is {{Andy, Mark, ...}, {Sue, Louis, ...}, {Roger, Alice, ...}, ...}, 1 is {{Andy, Louis, ...}, {Sue, Alice, ...}, ...}, and 2 is {{Andy, Alice, ...}, ...}. You won't ever get any repeats until you have a slope of n, since n is prime. So you get $n+1={n^{2}-1 \over n-1}={N-1 \over m-1}$ total partitions, as required. This isn't relevant to your original question, of course, and I don't have a formula in general (it seems to be a harder version of the derangements problem), but I thought it was interesting that the maximum is sometimes obtained and (evidently) sometimes not. As an example, the partitions formed for $n=5$ from the set {0...9,'A'...'O'} is

{01234, 56789, ABCDE, FGHIJ, KLMNO},
{05AFK, 16BGL, 27CHM, 38DIN, 49EJO},
{06CIO, 17DJK, 28EFL, 39AGM, 45BHN},
{07EGN, 18AHO, 29BIK, 35CJL, 46DFM},
{08BJM, 19CFN, 25DGO, 36EHK, 47AIL},
{09DHL, 15EIM, 26AJN, 37BFO, 48CGK}

Hope this helps. --Tardis 23:25, 16 October 2006 (UTC)

Thanks for playing around with this and for your observations! (the obvious questions start popping up ... for a maximal n by n partition system, is n prime a necessary condition? for n m-groups, is the case n > m too easy and n < m too hard as far as achieving the maximum? I guess for n < m that's obvious by the pigeonhole principle.... are all maximal partitionings isomorphic? for that matter, are all partitionings isomorphic?) 68.144.123.4 01:09, 17 October 2006 (UTC)

October 17

Roman numerals

i am a standard FIFTH student. We have been assigned to make a Project Report on ROMAN NUMERALS I have found some answers to the CONTENT LIST of the Project. Please help me find answers to -- Uses of Roman Numerals; advantages & Disadvantages

Syed Bilal, Std V (E), Fr Agnel School (Vashi)

You can start with our article on Roman numerals. Other than that, do your own homework. —Keenan Pepper 02:47, 17 October 2006 (UTC)

Advantages? Good luck on finding that. ☢ Ҡi∊ff⌇↯ 03:47, 17 October 2006 (UTC)

How about counting being easier up to three? Fredrik Johansson 05:00, 17 October 2006 (UTC)

Advantages and disadvantages compared to what? One obvious advantage of writing "XLII" rather than "42" is that it does not require extending our list of symbols beyond the Roman alphabet. However, it also limits the size of the numbers we can write. Incidentally, Google has a fun converter for small numbers. --KSmrq^T 13:18, 17 October 2006 (UTC)

The advantages would mostly be relative to the systems used previously, like drawing "bundles" of 6, 60, 360, etc. StuRat 14:02, 17 October 2006 (UTC)

Addition's rather easy: just concatenate the symbols and regroup. 13 + 26 = XIII + XXVI = XXXVIIII = 39. Dysprosia 01:01, 18 October 2006 (UTC)

But what about IIX (8) + XIII (13) ? Concat the symbols and you get IIXXIII, and changing the order to XXIIIII doesn't help at all. Taking them in that order (IIX;XIII) is exactly the same as adding 8 and 13 in the first place. – AlbinoMonkey ^(Talk) 11:45, 18 October 2006 (UTC)

IIX is not the standard way of writing 8. It's usually VIII. But your point is still valid. VIII + XIII does not make VIIIXIII or any other groupiong of those letters that makes any sense to me. JackofOz 21:01, 18 October 2006 (UTC)

Argh yeh sorry, total mind blank on the 8 thing. XVIIIIII might make sense if you just add them up - 10 + 5 + 6x1 = 21, what I meant was when you get 4's and 9's and thus use the backwards numbering thing. Proper example - IX (9) + VIII (8) = IXVIII (?) or XVIIII (19? wrong). You could interpret IXVIII as (10+5+3)-1 = 17, which is correct, but I don't see that as any easier than adding 9+8 in the first place. – AlbinoMonkey ^(Talk) 21:52, 18 October 2006 (UTC)

Are you serious? Just remove one from each. Given numbers in order, concatenate and regroup. Given numbers not in order, first cancel, then continue. So, IX+XII = X+XI = XXI. The numbers to the left are subtraction, right? And to the right is addition? So they cancel. Given something where none are available on the right to cancel, eg XIV+XX, you can generally just concatenate straight across: XIV+XX = XXXIV. There are probably other rules for special cases, but I don't see that they'd be hard to figure out on the spot. Subtraction isn't bad in Roman numerals either. As a matter of fact, glancing at the Roman arithmetic article... Black Carrot 01:29, 19 October 2006 (UTC)

VIII + XIII = XVIIIIII = XVVI = XXI = 21 = 8 + 13 as expected. Remember (I can't remember if this is fact or not) that the Romans didn't create the "subtraction notation", ie., IV was IIII, it was some later creation. Dysprosia 02:20, 19 October 2006 (UTC)

Ok fair enough. i stand corrected. – AlbinoMonkey ^(Talk) 03:07, 19 October 2006 (UTC)

MarkCC recently wrote a short but interesting article about arithmetic using Roman Numerals: http://scienceblogs.com/goodmath/2006/08/roman_numerals_and_arithmetic.php Rasmus (talk) 10:18, 19 October 2006 (UTC)

The History of a Coin

I buy something at a store, I pay cash, and I am given in my change a 20-cent coin dated 1981. I want to come up with reasonable estimates of (a) how many times this coin has changed hands since it was first issued 25 years ago, and (b) how likely is it that I've previously had it myself during that time. I assume the answer would have to include variables such as: the number of 20-cent coins issued in 1986; the population of my country over a certain age (which itself has varied over that period); the changing relationship people are having with cash vs. credit; and surely other factors that I can't think of. Any suggestions would be welcome. JackofOz 07:53, 17 October 2006 (UTC)

Consider what happens to a coin. It sits in someones pocket / money box for a while and then goes into a till. After that it may go back to someone elses pocket or to a bank - from whence to another shop till and pocket. I guess that larger value coins move around faster than lower denomination ones, so I would expect the 20 cents to have slowed down over its life. On average I would think that 1 movement per workday (5 per week) would be a reasonable guess. No data to back this up though. -- SGBailey 10:13, 17 October 2006 (UTC)

Are you Australian or British? The United States did not produce a 20-cent coin in 1981, and Canada never did. If someone handed me a four-dollar bill, it wouldn't stay in circulation long! Anyway, this is a question that begs for empirical study, not mathematics. It's the kind of thing a mint could probably answer. --KSmrq^T 13:49, 17 October 2006 (UTC)

• My user name is Jack of Oz. If that didn't give the game away about my Australian nationality, a quick check on my user page would have told you immediately what you wanted to know, and involved you in far less work than your international excursion through some of the major first world countries - but with fewer scenic attractions, admittedly.
• How would a mint know the answer to this better than anyone here? Mints don't keep tabs on every coin they ever produce. And how would an empirical study work? You would need 18,000 people (assuming a coin changes hands twice a day as a long-term average) - quite a management challenge. No, I really can't see how you could possibly know this with 100% precision, and even if you could, the answer would be different for every coin. But surely it's possible to come up with a reasonable estimate. Is my twice-a-day assumption reasonable? If so, that's the first part of the question already answered. Or is it? JackofOz 14:44, 17 October 2006 (UTC)

Variables to be considered. Add : the mobility of people may be lesser in a small town than in a big town, the same for their pocket's contents.

Empirical sudy : done. In 1999, each Euro country issued coins with only one identical side, the other denoting the country. Go[ogle] and find real studies about their dissemination now. -- DLL ^{.. T} 18:48, 17 October 2006 (UTC)

What does Oz have to do with australia? --froth^{T C} 19:06, 17 October 2006 (UTC)

See Oz, 6th meaning.  :) JackofOz 20:41, 17 October 2006 (UTC)

Oz, similar pronunciation to Aus i.e. a shortening of Australia. And interesting that KSmrq mentions Britain - who have never issued any 20 cent coinsRichard B 22:19, 17 October 2006 (UTC)

Technically, the British coin was twenty pence; but I did give a link to the article on that coin. --KSmrq^T 08:01, 19 October 2006 (UTC)

You may have heard of The Boy from Oz, and Oz, the satirical magazine, both so named because of their Australian connections. But this is becoming too tangential, even for a committed tangentialist such as I. The denomination of the coin isn't important. The country of issue isn't important. It's all about: If X is the number of times a coin changes hands in T years, is it possible to define X in terms of T, and if so how? JackofOz 01:21, 18 October 2006 (UTC)

Hmm, well obviously it all hinges upon getting some kind of good estimate on how long a coin stays with its owner. I think a decent model to arrive at that would probably divide it up into two scenarios First, coins that are in circulation, that is, in someone's wallet/pocket, in a till, or somewhere in the bank system. In that case, the coin will be changing owners at a rate which should be possible to estimate. Second scenario are coins that are taken out of circulation for a more indefinite time, that is, coins which go into people's collections or change jars. And then a number of coins that disappear out of circulation completely. You should be able to get hard figures on at least some of those values, since the currency-issuing authority of whatever country should know exactly how many coins they've minted, and have a fairly good idea of how many are in active circulation. --BluePlatypus 01:22, 18 October 2006 (UTC)

... the coin will be changing owners at a rate which should be possible to estimate. Yes, it should be possible to estimate the rate, but it's beyond my feeble powers. That's why I've come to the Ref Desk for assistance. JackofOz 23:52, 18 October 2006 (UTC)

If the coin had a unique identifying mark, you could check every coin you get to see if it has come back. But wait! Check Currency bill tracking where currency which IS uniquely identified, gets tracked. That is a related but more tractable problem. Then there is the saying "To come back like a bad penny," i.e. the perception that one spends a "bad

penny and later it returns in change.Edison 23:33, 18 October 2006 (UTC)

This is slowly creeping closer to what my question is about. I'm not interested in tracking particular coins. I am interested in coming up with a reasonable estimate of how likely it is that a T-year-old coin that has just come into my possession has been through my hands at some other time in the preceding T years. JackofOz 23:52, 18 October 2006 (UTC)

I like the question, and I think currency (design, history, culture) is a fun topic right under our noses — which is why I enjoyed finding the links above.

I stick by my assertion that the answer will depend on empirical data of the sort a mint might provide. That's not to say they would necessarily have on hand the answer to this specific question, but surely they must know a great deal about the statistics of coins in circulation. How many coins of a given denomination are currently in circulation? What is the typical lifetime of that coin? How does it migrate geographically? What physical properties must it have to survive the wear, and how much wear will that be? If a new coin is planned, how many should be struck to supply the needs? Coins spend some of their life in pockets, some in cash registers, some in vending machines, some in slot machines, some in jars, and some in banks. The life of a two-dollar coin likely bears little resemblance to the life of a one-cent coin. And of special interest are commemorative coins and others whose circulation is limited for reasons independent of denomination.

Another factor is place. Surely the statistics for a major centre of commerce like Sydney will hardly resemble those of a small isolated town like Nullagine.

I suspect counterfeit coins are extremely rare, but counterfeit bills are of great interest to the authorities, who would thus have reason to document the life of currency for their purposes.

Sorry I don't have any definite answers for you, but I hope I've given a few useful questions. --KSmrq^T 08:01, 19 October 2006 (UTC)

Thanks, KSmrq. Seems the question will have a more complex answer than I had assumed. I will go away and work on this. In the meantime, I wonder whether your Sydney/Nullagine comparison bears inspection. Sure, within Nullagine there are far fewer places to spend your money than there are in Sydney. But it doesn't take much for a particular coin to get circulated far more widely than where it happens to be right now. Scenario: A Nullagine resident buys petrol/milk/newspaper/whatever, using a coin; the next customer at that shop, who's on their way to Perth, gets that coin in their change; in Perth, the coin is tendered, it gets given to somebody who's going to Cairns. Then it travels to Melbourne, and is put into Granny's change bottle for the grandkids; it stays there for 5 months and they get the bottle at Christmas. They take the coins to the bank, then off it goes again on its merry journey through history. JackofOz 00:58, 20 October 2006 (UTC)

Angle at centre of a tetrahedron

Given a tetrahedron ABCD, with centre at O (i.e. AO = BO = CO = DO), then what is the angle AOB? (O exsists and is unique)

I know it's *roughly* 72 degrees [EDIT: should've been 108] (which is the interior angle of a pentagon) for the following reason: at school, our chemistry department had an organic molecule construction kit. An atom of carbon was reperesented by a four-pronged star (so the ends of the star formed a tetrahdeon) and bonds / other elemnts were represented by little plastic tubes. Having joined five into a planar ring (cyclo-pentane), I then realised you could make up a deodecahedron using entirely carbon molecules... which I duely did. If the straws had been straight, then the above-mentioned angle would be exactly the interior angle of a pentagon. However, the straws were bent slightly outward, meaning angle AOB is actually slightly more than 72 degrees [EDIT: should've been 108]. So, what *is* the angle? Tompw 11:29, 17 October 2006 (UTC)

Could this have been the angle between two faces, which is arccos (1/3) (about 71°)? For the angle between two spikes from the centre to a corner I find arccos (–1/3) (about 109°).  --Lambiam^Talk 13:48, 17 October 2006 (UTC)

The careful description of the molecule, a circle of five, makes it clear the central angle between vertices is really what is wanted. Since a regular tetrahedron is its own dual, the central angle between vertices should be the same as the central angle between face centers, which is 2 tan⁻¹ √2, approximately 109.47°. That's different from the dihedral angle between faces, which is cos^−1 1⁄₃, approximately 70.53°. The mistake is thinking that the interior angle of a regular pentagon is 72° (that is, 360°/5), which is actually the central angle; in fact, the angle between sides is 180° − 360°/5, exactly 108°. This assumes that a ring molecule is planar, which is roughly true here, but not always. --KSmrq^T 14:50, 17 October 2006 (UTC)

Not accidentally, 109.47° + 70.53° = 180°.  --Lambiam^Talk 15:47, 17 October 2006 (UTC)

The mistake was that the interior angle of a pentagon is 108°, not 72° (it's more than a right angle, right?). -- Meni Rosenfeld (talk) 14:15, 17 October 2006 (UTC)

Thanks for that... everyone who has said the interior angle is 108°, not 72° is correct - I gave the central angle not the interior. Anyway, this shows that the vertex-centre-vertex angle is indeed very close to the interior angle of a pentagon. (It also shows why model dodecahedrons will apear to tessalate space). Tompw 14:38, 18 October 2006 (UTC)

Integration rule

Hi!
This doesn't appear anywhere in my text book, but I was wondering, would the integral of e^g(x) equal (e^g(x))/g'(x)+c? Thanks, --Fir0002 11:55, 17 October 2006 (UTC)

No, at least not in general. To test whether the antiderivative of function f is function F, take the derivative of F and see if you get f back. So let's test. The derivative w.r.t. x of (e^g(x))/g'(x)+c is e^g(x) (1 – g"(x)/g'(x)^2), This is the same as e^g(x) precisely when g"(x) is the constant zero function. Only linear functions (that is, first degree polynomials) g(x) = ax+b have that property. So the answer is: Yes if g is a linear function; otherwise no. In fact, for many cases there is no elementary analytical expression for the antiderivative; for example if g(x) = x^2. See also List of integrals of exponential functions.  --Lambiam^Talk 12:49, 17 October 2006 (UTC)

OK thanks for that --Fir0002 07:39, 18 October 2006 (UTC)

Tessaract diagram

I spent 15 minutes trying to find all the cubes in that diagram on the Hypercube page just for fun. I colored each wireframe differently. Did I get them all? I'm concerned about the orange cube and the grey cube.. they don't quite match up and the cubes next to each other seem to make an impossible shape. Also should the top four points and the bottom four points be considered opposite faces of another cube?--froth^{T C} 19:28, 17 October 2006 (UTC)

Did you count the number of colours? There should be precisely 8 cubes there. The top and bottom squares are not squares of the same cube. Corresponding corners are connected by parallel diagonals (not drawn) of other squares, forming a 1 × 1 × √2 block.  --Lambiam^Talk 20:13, 17 October 2006 (UTC)

Yes I do have eight. thanks --froth^{T C} 21:12, 17 October 2006 (UTC)

physics differentiation

I know that in physics the third derivative corresponds to a "jerk", the second derivative corresponds to acceleration, the derivative corresponds to velocity, when the function itself is position. But does the antiderivative of the function mean anything? --froth^{T C} 19:11, 17 October 2006 (UTC)

Just the reverse. The first antiderivative of jerk is acceleration, the second is velocity, and the third is position. StuRat 22:24, 17 October 2006 (UTC)

No I mean just the antiderivative of the position.

The antiderivative of the position of a joystick is the position of whatever it controls. You could think of it as the joystick's accumulated displacement-time. And given a clever setup of cameras, you could use anything as a joystick... Melchoir 23:04, 17 October 2006 (UTC)

Ohhh I see. Very interesting answer thanks --froth^{T C} 02:04, 18 October 2006 (UTC)

Note that all derivatives of all positions and the position of a joystick take a value in a vector space, since they have natural 0s, but more general positions (e.g., of a galaxy) are better thought of as values in an affine space since there is no "natural" origin. A function whose codomain is an affine space can only be integrated with a weight function that integrates to 1 over the domain of the integration, thus generating a weighted average of the function's values, or a sort of centroid. (Note that the weight function in the position case must have units of inverse time, and the result is still a position.) This makes sense, as any value formed from a collection of arbitrary points in space can only be defined relative to those points rather than to some fixed reference on which the points do not depend. So in general, the antiderivative of position is meaningless, although certain definite integrals can be taken of it. --Tardis 04:45, 18 October 2006 (UTC)

Well then velocity is equally meaningless without a reference position.. which doesn't have a natural origin either. But I've never heard anyone talking about a weighted reference in regard to velocity --froth^{T C} 05:22, 18 October 2006 (UTC)

How does velocity require a reference position? It requires a reference frame, certainly, but all of physics requires that. The point is that it is not coordinate-system-dependent, like position is. You and I may disagree on whether a car is to the east or to the west, but we will surely agree that it is going north if it is (even if you happen to term it "left" and I happen to term it "away from Cape Town"). (I suppose that we can go farther and say that acceleration is even more "portable" in that observers in all inertial Galilean reference frames agree on it. And jerk is consistent in all uniformly-accelerating reference frames, and so on. But that's not really the same thing, since the point is that one observer can describe all of these quantities, and they all have obvious 0s except position, whose origin the observer must pick arbitrarily.) And people certainly do talk about weighted averages of velocity — or at the very least speed — with a trivial weight function $\omega (t):={1 \over t_{2}-t_{1}}$ on a regular basis. It's not that you can't take such averages of vectors; it's that they are the only kind of integrals you can take of affine objects. --Tardis 06:05, 18 October 2006 (UTC)

(edit conflict) I agree with Tardis' point; it contains distinctions that get glossed over too often, and I guess I'm a guilty party. And, actually, Froth, velocity isn't meaningless without a reference position. At the heart of the definition of a derivative is the difference between two points, and the difference between two elements of an affine space is itself naturally an element of a vector space. It doesn't require a reference position to define.

Anyway, given a path x(t) in an affine space, you could define a "generalized antiderivative" to be a path n(t) -- let's not think too hard about exactly where n(t) lives -- such that for some point O_n, the derivative of n(t) equals x(t) - O_n. If you like, this family of generalized antiderivatives would be indexed by 2 "integration constants" instead of the usual 1. So it's not such a terrible price to pay. If you're willing to introduce a "unit time", then you can probably even put n(t) in the same affine space as x(t), so that gives some meaning to my original comment about things moving around in physical space. Melchoir 06:14, 18 October 2006 (UTC)

That's not an unreasonable idea, but it seems to me that $x(t)-O_{n}$ is just another function ${\tilde {x}}(t)$ which does live in a vector space, since it is displacement-valued rather than position-valued. Then you can integrate that as usual, although I'm not sure what physical meaning to attribute to the resulting length-time vector. I guess it's most proper to say that a definite integral of, say, acceleration is "an accumulated change in velocity" (or a velocity displacement), and that the definite integral of velocity is a displacement, not a position. The integration constants associated with the indefinite integrals represent the (relative) "affinity" of the next space, as per my discussion of "portability". When all you know is a set of accelerations, we must assume that you are in some inertial frame, but we can't say which one. So in that sense the velocities we get from you are in an affine space as well; in all cases, it is the addition of other information that allows us to turn displacements into position (or velocity) vectors, even though in a different sense the positions have less information. (I think the paradox of more/less information is resolved by the fact that while displacements are vectors, they also identify elements of an affine space: the space of positions represented by displacements from an unknown point. We have to add an origin to turn that into a vector space for one observer, even if that position space is still affine as considered independent of observer.) (The two integration constants you found are in fact the result of two composed integrations, and get you an added $A+Bt$ just as they do in DEs.) --Tardis 17:28, 18 October 2006 (UTC)

Ah yes thank you the first half of that is what I was trying to say when I said that velocities are equally meaningless without a reference position --froth^{T C} 19:12, 18 October 2006 (UTC)

To riff on that last bit, I suppose I can do away with my O_n by simply requiring the vector identity n’’ = x’. In order to recover n(t) from x(t) through that more natural identity, one doesn't integrate once; one differentiates once and then integrates twice. So they really are integration constants. How queer! Melchoir 19:38, 18 October 2006 (UTC)

Flatland

I'm reading Flatland and I'm just curious.. When the narrator moves through the linear space of lineland, the king of lineland sees only a point. When the sphere moves through the planar space of flatland, the narrator sees a line of varying length. What would us three-dimensional beings see if a four-dimensional "circle" moved through our space? A circle of varying diameter? The narrator can tell that the cross-section of the sphere that's in his plane is circular by the shading of the line... how would the shading of the four-dimensional being's cross-section appear to us, and what would it look like if we walked around it? Certainly there's a concrete answer to this question and not just theory and projections on surfaces. --froth^{T C} 21:21, 17 October 2006 (UTC)

When the sphere moves through a 2-d space, the intersection is a disc which appears to a 2-d 'person' as a line. When a 4-d sphere (hypersphere) moves thorugh our 3-d world, the intersection would actually be a sphere of varying size (starts as a point, grows to a max, then shrinks to a point). You might say that it would appear to us as a disc of varying size, except that with binocular vision, we would probably see it as a 3-d sphere. Madmath789 21:36, 17 October 2006 (UTC)

Four-dimensional pyramid passing through three-dimensional space

For everybodys entertainment and education, here is an animation of "four-dimensional balls stacked into a 4-dimensional pyramid (5×5×5×5), animated through successive 3-dimensional cross-sections". —Bromskloss 00:49, 18 October 2006 (UTC)

Wow very cool thanks --froth^{T C} 01:59, 18 October 2006 (UTC)

statistics

Acountof malaria parasites in 100 field with a 2 mm oil immersion lensgave a mean of 35 parasites per field, standard deviation 11.6 (note that, although the counts are quantitive discrete, the count can be assumed to follow a normal distribution because the average is large). On counting one more filed, the pathologist found 52 parasites. Does this number lie outside the 95% reference range? What is the reference range? What is the 95% confidence hnterval for the mean of the population from which this sample count of parasites was drawn? —The preceding unsigned comment was added by 213.6.140.68 (talk • contribs) .

This sounds like a homework question, so I will not give you a direct answer. However is there some part of the question which you find confusing? Do you have an approach to the question? --TeaDrinker 22:55, 17 October 2006 (UTC)

October 18

aargh eigenvectors

I'm not very good with these, so could someone please indicate whether the following statements are true or false (these are not homework questions, but I am hoping they are true in order to complete an assignment proof):

1) A p x p matrix A has eigenvalues $\lambda _{1},\lambda _{2},...\lambda _{p}$, with corresponding orthonormal eigenvectors $e_{1},e_{2},...e_{p}$. Therefore the matrix can be written as:

$A=\lambda _{1}e_{1}{e_{1}}^{T}+\lambda _{2}e_{2}{e_{2}}^{T}+...+\lambda _{p}e_{p}{e_{p}}^{T}$

2) (the converse) A p x p matrix A can be written as:

$A=\lambda _{1}e_{1}{e_{1}}^{T}+\lambda _{2}e_{2}{e_{2}}^{T}+...+\lambda _{p}e_{p}{e_{p}}^{T}$

where the $e_{i}$'s are orthnormal p x 1 vectors. Therefore the matrix has eigenvalues $\lambda _{1},\lambda _{2},...\lambda _{p}$, with corresponding orthonormal eigenvectors $e_{1},e_{2},...e_{p}$.

3) (hmmm - wishful thinking?) A p x p matrix B can be written as:

$B=\lambda _{1}e_{1}{e_{1}}^{T}+\lambda _{2}e_{2}{e_{2}}^{T}+...+\lambda _{n}e_{n}{e_{n}}^{T}$

where $p\neq n$ and the $e_{i}$'s are orthonormal p x 1 vectors. Therefore the matrix B has eigenvalues $\lambda _{1},\lambda _{2},...\lambda _{n}$, with corresponding orthonormal eigenvectors $e_{1},e_{2},...e_{n}$.

4) (more wishful thinking?) A p x p matrix B can be written as:

$B=\lambda _{1}e_{1}{e_{1}}^{T}+\lambda _{2}e_{2}{e_{2}}^{T}+...+\lambda _{n}e_{n}{e_{n}}^{T}$

where $p\neq n$ and the $e_{i}$'s are orthonormal p x 1 vectors. Therefore:

$tr(B)=\sum _{i=1}^{n}\lambda _{i}$

Thanks. – AlbinoMonkey ^(Talk) 05:57, 18 October 2006 (UTC)

For #1 and #2, look at your matrix in the basis given by your p eigenvectors. For #1, note that in general if $v_{k}$ is the column vector with (0,...,1,...,0) in the kth place, then the image $M(v_{k})$ is the kth column of A. So we know that kth column to be $\lambda _{k}v_{k}$, since our change of basis mapped the old $e_{k}$ to the new $v_{k}$. It follows that A is diagonal[izable] with diagonal entries exactly the eigenvalues. If you're worried about the change of basis, just note that if $PAP^{-1}$ is the change-of-basis-ed diagonal matrix above, then $Pe_{k}e_{k}^{T}P^{-1}=v_{k}v_{K}^{T}$, so the linear expansion you wrote down is preserved by changes of basis.

Note that the important part here is the p linearly independent eigenvectors. The Jordan normal form tells us that as long as all our eigenvalues are in the field (as yours are), we can put the matrix in Jordan normal form -- as "close" to diagonal as we can get, in a sense. The JHF will be diagonal if and only if you have p linearly independent eigenvectors. Orthonormal is nice in general but not necessary here.

For #2, we already have a diagonal matrix (if we pass to the basis given by your eigenvectors), and the eigenvalues of a diagonal matrix are just the entries along the diagonal. So #1 and #2 are true.

#3 doesn't make much sense to me. It's impossible that n>p, since you can't find >p orthonormal (or even linearly indepedent) p x 1 vectors, and a p x p matrix has at most p eigenvalues. But if n<p, then #3 is just #1 where you've forgotten that some of the eigenvalues are 0.

#4 is true, though there's no reason to specify this $p\neq n$ business. We know trace is a class function (constant on conjugacy classes, $tr(XAX^{-1})=tr(A)$) so we can pick any basis. So take your n eigenvectors and extend to a basis of p elements (if n<p). Now as above, A will be diagonal, with your n eigenvalues along the diagonal and the rest filled in by 0. The trace of this is clearly what you claim it is. You've rediscovered, in the restricted case of a diagonalizable matrix, the general fact that the trace of any matrix is the sum of its eigenvalues. It is also true that the determinant is the product of the eigenvalues.

I will add, in case you are interested, that this is true even if the matrix "has no eigenvalues", meaning its eigenvalues do not lie in your field. For example, the real matrix $M={\begin{bmatrix}0&1\\-1&0\end{bmatrix}}$ has characteristic polynomial $X^{2}+1$, which has no real roots, so there are no real eigenvalues. Nevertheless, it makes sense to say that tr(M) and det(M) are the sum and product of the eigenvalues---why? Tesseran 08:23, 18 October 2006 (UTC)

Thankyou very much. I was aware of the fact that the sum of the eigenvalues equalled the trace - ie that if #3 was true (I intended n < p, but the thought escaped me to specify it), #4 was also true, but I did not realise the expression gave me a diagonal matrix and therefore the other e'values were zero. Thanks again. – AlbinoMonkey ^(Talk) 08:32, 18 October 2006 (UTC)

Related rates problem

I'm studying related rates right now and am not having trouble with the concept in general, but I'm stuck on one problem. It states: "An airplane flies at an altitude of 5 miles toward a point directly over an observer. The speed of the plane is 600 miles per hour. Find the rate at which the angle of elevation is changing when the angle is 30^o" I realize trigonometry is necessary here, but I still can't figure out a way to solve it without knowing either the plane's distance from the observer or the horizontal distance between the observer and the plane. Any pointers would be much appreciated. -Elmer Clark 08:04, 18 October 2006 (UTC)

(Edit conflict) From the plane drop a perpendicular to the ground. Now we have a right triangle with angle (at observer) of 30°, and opposite side 5 miles. The distance d from observer to plane is the hypotenuse, and we know that d sin 30° equals 5 miles. Lucky us, sin 30° equals ¹⁄₂, so now we know d. But maybe with this insight we should reconsider methods of solution. --KSmrq^T 08:26, 18 October 2006 (UTC)

Imagine a line directly from the plane to the ground, it is 5 miles long. Then imagine a line (which will be parallel to the plane's flight path) from the point where the first line meets the ground to the observer. Call its distance $x$. Join these to form a triangle, with the angle of elevation being $\theta$. You know ${\frac {dx}{dt}}$, and you can use trigonometry to find an expression for $\theta$ in terms of x, and hence ${\frac {d\theta }{dx}}$. From these, you can find ${\frac {d\theta }{dt}}$, and evaluate it at the given angle. – AlbinoMonkey ^(Talk) 08:22, 18 October 2006 (UTC)

Thank you both very much for the helpful (and quick!) responses :) -Elmer Clark 08:49, 18 October 2006 (UTC)

Can you have a consistent but biased estimator?

In statistics, can an estimator be consistent but biased? If so, what would be an example of one? Thanks so much, Mihail

Choose a consistent unbiased estimator and add 1/n, where n is the sample size. You will get a consistent biased estimator.(Igny 17:36, 18 October 2006 (UTC))

What about unbiased but also inconsistent?

Take any unbiased estimator (t_i)_{i > 0}, meaning that for every possible sample size i > 0 we have an unbiased estimator t_i. We turn it into a new unbiased estimator as follows: Given a sample, discard all data except the last element. Then apply t₁ to it. Clearly, this is not consistent.

A possible objection to this counterexample is that a sample should be viewed as a multiset, so that there is no such thing as a "last" element. An even more silly and artificial counterexample, but one that does not suffer from this problem, is the following. Assume that the population parameter to be estimated is the mean of normally distributed data having mean μ = 0 and variance σ² > 0. This is silly because there is a most excellent consistent, efficient and robust estimator: always estimate 0, without even looking at the sample. However, we are not aiming for excellence but instead use the sum of the sample data as estimator. This is unbiased, but not consistent.  --Lambiam^Talk 22:44, 18 October 2006 (UTC)

October 19

Big numbers

Ok, the sequence goes: ones, tens, hundreds, millions, billions, trillions... what's next? --JDitto 06:28, 19 October 2006 (UTC)

Quadrillion, apparently. See Names of large numbers. – AlbinoMonkey ^(Talk) 06:30, 19 October 2006 (UTC)

Whoa! Talk about quick response! Thank you for answering my question. --JDitto 06:36, 19 October 2006 (UTC)

Geometric growth explained

Very good explanation of exponential growth ... another phrase used loosely and i suspect wrongly in the financial community is "geometric growth". I am not seeing that in wiki so far, so i think that would be a positive contribution. Thank you.

Mikeym1 15:53, 19 October 2006 (UTC)

Wouldn't that be the same thing? —Bromskloss 16:08, 19 October 2006 (UTC)

Our article on geometric progression notes that a geometric progression with common ratio $\notin \{0,\pm 1\}$ shows exponential growth. — Lomn 20:26, 19 October 2006 (UTC)

A spiral function

Is there any nice parametrized function that gives a "tight" spiral when applied to integers? That is, I want this:

21 22 23 24 25 26
20 7  8  9  10 27
19 6  1  2  11 28
18 5  4  3  12 29
17 16 15 14 13 30
36 35 34 33 32 31

That is, f(1)=(0,0), f(2)=(1,0), f(3)=(1,-1), f(4)=(0,-1), and so on (obviously it would be two functions, one for x and one for y). I'd appriciate it if someone atleast pointed me in the right direction :) 83.250.208.83 20:36, 19 October 2006 (UTC)

When I was coding an Ulam spiral generator, I too looked for such a thing. I ended up needing a "square sine\cosine" function (that generates a square instead of a circle), which proved to be too difficult. In the end, I just adopted a much simpler walker algorithm (that is, you have a cursor that walks the plane in one of the four directions, and you just make it turn left\right after N moves). I'm not sure what are your purposes with this, but if you can implement the walking, I'd suggest you to do it instead of the parametric. ☢ Ҡi∊ff⌇↯ 20:58, 19 October 2006 (UTC)

Well, looking at the lengths of the "sides" (straight length segments of the spiral), you're getting 1, 1, 2, 2, 3, 3, 4, 4, 5, 5, 6... So, if you want to figure out what layer of the spiral a particular number is in you can make a comparison against the nearest sum of natural numbers (x2). Once you can do that efficiently, it shouldn't be too difficult to solve the rest of the problem. - Rainwarrior 00:21, 20 October 2006 (UTC)

Impossible mathematical shape?

I found this drawing while surfing Wikipedia: http://upload.wikimedia.org/wikipedia/commons/thumb/b/bb/Autocunnilingus.jpg/651px-Autocunnilingus.jpg

Please just ignore the lady in the drawing and look at the curved object on the shelf behind her. I swear that once in a math book I saw an illustration of this shape, where the tube at the bottom becomes the hollow space at the top, and I have forgotten what mathematician was being talked about or what the name of this object is. I think I might have been reading a Keith Devlin book with illustration plates...

Unfortunately the illustration has a solid line where the bottom curved tube meets the hollow space so I'm not sure if it's the same thing but it is very similar.

Can you help me find more information about that object? Is it really physically impossible to make? Please email katybeans@hotmail.com because I am new to Wikipedia and might not find my way back to the right page here to check answers... thanks!

Looks like a Klein bottle. Fredrik Johansson 21:11, 19 October 2006 (UTC)

Here are a couple Calvin Klein bottles: [2] :-) StuRat 22:28, 19 October 2006 (UTC)

It is indeed a Klein bottle (that's a point on the geek-scale right there, seeing a picture of that and noticing the Klein bottle in the background ;). A Klein-bottle is a four-dimensional object, so it's not possible to build a proper one in our lame three-dimensional world, but when you die and go to the after-life (which is, as we all know, four-dimensional) you'll be able to build it. Cheers! Oskar 22:35, 19 October 2006 (UTC)

It's quite a cute pun, given the subject matter. Dysprosia 07:08, 20 October 2006 (UTC)

The central image is obviously an allusion to Ouroboros, the serpent that encircles the World with its tail in its mouth. We may assume that the object in the background is intended to resonant with that concept and excite topologists, who might otherwise find the main image, a mundane torus, uninteresting. And yes, the background object is one of the standard immersions of a Klein bottle — a torus with a twist — in 3-space.

Take a cylindrical strip, snip it apart, give one end a half twist, and glue it back together. This produces a Möbius strip, used by stage magicians in a trick known as the Afghan Bands. The original strip had two sides, but the twist has joined them into one. Likewise, the original strip had two edges, both circles, but the twist has also joined those into one.

If we curl the original strip inward and glue its two edges together, we get a torus. The Möbius strip has only one edge, but it is twice as long, containing both edges of the original strip. We can glue this one edge to itself, connecting the same points as before, to seal the edge and produce a Klein bottle. Konrad Polthier has an article with lovely illustrations.

Like the Möbius strip, the Klein bottle has only one side. That is, it is a “closed” surface (no longer having an edge) for which there is no distinction between inside and out.

However, we cannot physically produce a Klein bottle in 3-space without having it intersect itself somewhere. That said, Cliff Stoll makes nice glass models, and sarah-marie belcastro and others have made softer models — by knitting. --KSmrq^T 08:06, 20 October 2006 (UTC)

October 20

How do I prove this?

How do I prove that d/dx (cosx)=-sinx?

Thank you!

-EggYolk

If you're working from first principles, you need to use the formula $\cos(x+y)=\cos x\cos y-\sin x\sin y$, as well as the small angle approximation (that $\lim _{x\rightarrow 0}{(\sin x)/x}=1$), and then the only tricky part is a term where you have to prove that $lim_{x\rightarrow 0}{\frac {\cos x-1}{x}}=0$, and I can't recall the proof for that offhand. Of course, if you already have your proof that (sin x)' = cos x, you can use the chain rule: ${\frac {d\cos x}{dx}}={\frac {d\sin(\pi /2-x)}{dx}}=-\cos(\pi /2-x)=-\sin x$. Confusing Manifestation 05:15, 20 October 2006 (UTC)

$\lim _{x\rightarrow 0}{\frac {\cos x-1}{x}}=\lim _{x\rightarrow 0}{\frac {\cos x-1}{x}}{\frac {\cos x+1}{\cos x+1}}=\lim _{x\rightarrow 0}{\frac {\cos ^{2}x-1}{x(\cos x+1)}}=\lim _{x\rightarrow 0}{\frac {-\sin ^{2}x}{x(\cos x+1)}}$

I'll let you finish the proof that $lim_{x\rightarrow 0}{\frac {\cos x-1}{x}}=0$ from there. Yes, it does need the small angle approximation. --Ķĩřβȳ♥ŤįɱéØ 05:57, 20 October 2006 (UTC)

Also, you can check out this newly created article, lol. --Ķĩřβȳ♥ŤįɱéØ 08:53, 20 October 2006 (UTC)

A more geometric argument begins with a unit circle. For any parametric curve, the instantaneous movement at any point is a vector tangent to the curve. If the curve is parameterized by arclength, the tangent vector will have unit length. On the circle, that tangent is the radial vector to the point, rotated counterclockwise 90°. For the radial vector (cos ϑ,sin ϑ), the tangent is (−sin ϑ,cos ϑ). This simultaneously gives the derivative of cosine and of sine.

This argument uses concepts and methods not available in first-year calculus, but the insights can still be of use. --KSmrq^T 09:00, 20 October 2006 (UTC)