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August 29

Why use Poisson distribution as an approximation to binomial?

I understand that the Poisson distribution can be used as an approximation to a binomial distribution (if n is sufficiently large and p sufficiently small). I've seen it demonstrated, and I think I get the math involved. But why would you want to use this approximation instead of just using the binomial? What is the advantage of using the Poisson? —Preceding unsigned comment added by 118.241.57.8 (talk) 02:03, 29 August 2009 (UTC)

One could just consider it an out-of-date notion, relevant for a time when calculation was more difficult.Julzes (talk) 02:16, 29 August 2009 (UTC)

One reason is you might have no clue how big n is, or p, but you may still have a good estimate of λ = np. Michael Hardy (talk) 02:32, 29 August 2009 (UTC) ...and besides, obviously it's simpler. Michael Hardy (talk) 02:33, 29 August 2009 (UTC)

The last answer is better than mine.Julzes (talk) 03:56, 29 August 2009 (UTC)

• As Michael said, the Poisson distribution is much simpler. Why should you (the mathematician, not the computer) do calculations involving two parameters, and think about the problem in terms of two parameters, when all the relevant information is contained in one?
• The faster computers become, the more difficult the calculations people use them for. Calculating just one value of a distribution is of course easy, but it's not unthinkable a problem would require billions of such calculations. In such cases there may be a significant difference between evaluating a binomial distribution, which requires 3 factorials, and Poisson which only requires one.

-- Meni Rosenfeld (talk) 21:43, 29 August 2009 (UTC)

The binomial distribution is (a limiting case of) the hypergeometric distribution for large populations, and the poisson distribution is (a limiting case of) the binomial distribution for large samples. You may like to study Cumulant#Cumulants_of_some_discrete_probability_distributions to see that the relationship between binomial and poisson distributions are like the relationship between ellipses and parabolas. The parabola can be used as an approximation to an ellipse. But why would you want to use this approximation instead of just using the ellipse? Bo Jacoby (talk) 22:47, 29 August 2009 (UTC).

The prospective user of a quick approximation to something where an exact calculation is possible but may take more time needs to keep in mind that the trade-off may not be a good one because of the need to keep track of errors. If the approximation really is good enough on its face, that's one thing; but where there is uncertainty it may be necessary in very complicated cases to do some sort of prior study of the nature of the trade-off before one decides whether to use the quick approximation.Julzes (talk) 04:00, 30 August 2009 (UTC)

Some folk here might be interested to check out Stein's method, which is one way of generalizing the binomial -> poisson limit. HTH, Robinh (talk) 08:29, 30 August 2009 (UTC)

The Tychonoff theorem

There's every reason to believe that the product space X is (in terms of sets) different from all the spaces X(alpha) in the family. Now take the product space X of the collection of all compact spaces. X is compact hence it belongs to the collection, and so X is different from X (in terms of sets). Is this where proper class come in? Standard Oil (talk) 12:35, 29 August 2009 (UTC)

Right. Within the category of (compact) topological spaces you are only allowed to do products of families of spaces indicized by sets. Indeed what you proposed is just a topological version of a paradox that you may repeat in all versions: the sup of all ordinals is not an ordinal; the union of all sets is not a set.. &c. This does not mean that you can't endow a proper category with a structure that mimes the usual "small" structures (a group structure, a topological structure and so on).--pma (talk) 13:10, 29 August 2009 (UTC)

Question about sets

Hey, I've got a logic question relating to sets that I havn't been able to quite figure out. Maybe someone here could help me. OK, here goes:

There are 6 hypothetical sets (some of them are real sets, but some of them are non-standard). These are N, W, Z, Q, I, and R.

${\sqrt {17}}$ belongs to sets R and I,

$0.86$ belongs to sets R and Q,

${\sqrt {64}}$ belongs to sets R, Q, Z, W, and N,

$-10/2$ belongs to sets R, Q, and Z.

I've gathered that set R are all real numbers, set I are irrational numbers, set Q are rational numbers, and set N are natural numbers. I am have some difficulty with sets W and Z, however. Can anyone help? Thanks in advance, --Anonymous —Preceding unsigned comment added by 做工器 (talk • contribs) 19:59, 29 August 2009 (UTC)

$\mathbb {Z}$ is standard notation for the integers (it comes from German). I've never come across "W" as a standard name for a set before and can't guess what it would be from the information provided. --Tango (talk) 20:07, 29 August 2009 (UTC)

Ah, Blackboard bold says it could be "whole numbers". I guess whoever is using this notation uses W to refer to whichever of {0,1,2,...} or {1,2,3,...} isn't being called the natural numbers (definitions vary, you see - some people consider 0 natural, some don't). --Tango (talk) 20:09, 29 August 2009 (UTC)

The sets in the order presented are contained within the next one - so N would be {1,2,...} and W is {0,1,2,...}. Acb314 (talk) 10:55, 30 August 2009 (UTC)

... maybe, but Q (whatever it is) cannot be contained in I (whatever it is) since Q contains at least three given members that are not in I. Logically, there is no way to infer the entire contents of these hypothetical sets from a few members. We can, however, say that a minimal solution is R={√17, 0.86, 8, -5}; I={√17}; Q={0.86, 8, -5}; Z={8, -5}; W=N={8}. Gandalf61 (talk) 12:06, 30 August 2009 (UTC)

I think "the order presented" means the order in the statement "${\sqrt {64}}$ belongs to sets R, Q, Z, W, and N". That makes Acb314's argument as good as any, but it's of course true that we can't know much about these sets. —JAO • T • C 12:17, 30 August 2009 (UTC)

That was lack of reading comprehension on my part -indeed Q will not be contained in I! Assuming N, Z, Q, R have the usual meanings, which they seem to, N is inside Z inside Q inside R, so my logic was that N would also be inside W from the order they were written in - not mathematically rigorous I know. I throws a spanner in the works unfortunately, although I is "bigger" than Q and contained within R. I agree that there isn't enough information in the examples to tell what N and W are just from those. Acb314 (talk) 14:31, 30 August 2009 (UTC)

Average Value of a Function

Hello, someone asked me a question about this and I have been thinking about it ever since. Working only in the reals, the question is to find the average value of a function over a given set. Obviously, if the set has a finite number of points, then using the definition of average, I can just evaluate at each point and then take the average. If the set is a finite interval, then we just integrate over the interval and then divide by the length of the interval. Extending this, if the set is a bounded region in R^n, then I just integrate over the set and then divide by the volume of that region. Here are the questions:

1.To find the average, it is okay to divide by the Lebesgue measure of the underlying set over which we integrate, right? Is it correct to say, for a bounded set, that to find the average value of a function over that set, just integrate over that set and divide by its Lebesgue measure, assuming that its Lebesgue measure is nonzero?

2.If this is correct, then how do we deal with sets that have measure zero? If the set is finite, we are good. But what if the set has an infinite number of points with measure zero, like the Cantor set or the rationals in [0,1]? I don't want to be dividing by zero but I can't simply evaluate at each point and take the average.

3.In addition, how do we deal with sets with infinite Lebesgue measure? I am sure I can't also "divide" by infinity after the integration. For example, what is the average value of $f(x)=e^{(-x^{2})}$ over $[0,\infty ]$?

4.Furthermore, what is meant when we are asked to find the average value of a function without specifying the region? Is it safe to assume that the entire domain of the function is meant or is there some other standard? Thanks!-Looking for Wisdom and Insight! (talk) 20:52, 29 August 2009 (UTC)

#1 is correct not only for bounded sets, but for any set whose measure is finite. In #3, the way it's usually done involves conditional convergence, so you might get different answers depending on how the region of infinite volume is approached. The idea is: take an average of a subset of finite measure, then take a limit as as that subset in some way approaches the whole domain. There's no non-uniqueness problem if the function is everywhere nonnegative, nor if the positvie and negative parts both have finite integrals; it's only when both are infinite that there is such a problem. Simple example:

$\lim _{a\to \infty ,\ b\to -\infty }\int _{b}^{a}{\frac {x\,dx}{1+x^{2}}}.$

If b = −2a, then the limit is (1/2)log(1/4), but if b = −a, then the limit is 0. See also Cauchy principal value. Michael Hardy (talk) 21:19, 29 August 2009 (UTC)

[ec] I don't think the term "average value of a function" is generalizable to the kind of cases you are thinking about. For these cases it may be more useful to consider the expected value of a function of a random variable, where the latter follows a given distribution.

1. This seems correct.
2. There is probably no generic solution. For a countable set you may choose to assign a weight for each point, or choose an enumeration and take $\lim _{n\to \infty }{\frac {1}{n}}\sum _{i=1}^{n}f(q_{i})$. For a non-countable set you may want to choose a countable subset of "key points" (for the cantor set, a natural choice is the set of interval endpoints).
3. You can choose a weighting function with a finite integral. In some cases (e.g. periodic functions) it is appropriate to consider $\lim _{t\to \infty }{\frac {1}{t}}\int _{0}^{t}f(t)\;dt$ or $\lim _{t\to \infty }{\frac {1}{2t}}\int _{-t}^{t}f(t)\;dt$.
4. It seems safe to assume the entire domain.

-- Meni Rosenfeld (talk) 21:25, 29 August 2009 (UTC)

With periodic functions, I'd just take the average value over one period. Michael Hardy (talk) 02:20, 30 August 2009 (UTC)

Fun little problem

Here's a fun little problem I saw yesterday. Which is bigger $e^{\pi }\,$ or $\pi ^{e}\,$? No calculator, computer, or search engine allowed; just your brain and some paper. Try to find a proof, i.e. no estimations allowed. It took me a little while to think of a proof, but in the end it's not that difficult. I would like to see if anyone can come up with a proof that uses a different method to mine. If this is too easy for you, then after you've posted a proof (which must use a different method to any others that yours follows) then please turn your attention to my serious problem above. ~~ Dr Dec (Talk) ~~ 23:13, 29 August 2009 (UTC)

Consider the inequality $\scriptstyle e^{x}>x^{e}$, that is $\scriptstyle x>e\ln x$. So we're interested in the sign of $\scriptstyle f(x):=x-e\ln x$, which is nicely continuous and differentiable for $\scriptstyle x>0$. $\scriptstyle f'(x)=1-e/x$ so the only solution to $\scriptstyle f'(x)=0$ is $\scriptstyle x=e$. $\scriptstyle f(e)=0$, of course, and it's readily seen that this is a minimum. Hence, $\scriptstyle e^{x}>x^{e}$ for all positive $\scriptstyle x$ (except $\scriptstyle x=e$). —JAO • T • C 23:40, 29 August 2009 (UTC)

Please. It's $\scriptstyle e\ln x$, not $\scriptstyle e\,lnx$. The backslash BOTH prevents italicization and provides proper spacing. Michael Hardy (talk) 06:24, 30 August 2009 (UTC)

Oops, of course. I thought my spacing problems were somehow due to \scriptstyle, so I just thought it weird instead of thinking myself stupid. Thanks for setting me straight on that; I fixed those uglies now. —JAO • T • C 08:59, 30 August 2009 (UTC)

Essentially the same as the above. Taking logs we see we are comparing $\pi$ to $e\log \pi$. Now, differentiate $x/\log x$ to compare it to $e$. Phils 23:46, 29 August 2009 (UTC)

But I'd like to see a non-log proof, if at all possible. ~~ Dr Dec (Talk) ~~ 23:59, 29 August 2009 (UTC)

It is easy to re-write Jao's proof to not use logs. Let $f(x)=e^{x}x^{-e}$ for x > 0. We have $f'(x)=e^{x}x^{-e}(1-e/x)$. This is zero exactly when x = e, so f has a unique minimum at e. Proceed as before. Eric. 98.207.86.2 (talk) 01:18, 30 August 2009 (UTC)

Declan, you've been posting some interesting topics, but they really aren't Reference Desk questions. Have you thought about starting a blog? 67.122.211.205 (talk) 06:16, 30 August 2009 (UTC)

Really? I know that this one is boarderline, but all of my other threads have been perfectly valid reference desk question. My latest questions have been The real numbers and subsets of the real numbers, Commutivity of addition, Associativity of composition and Line bundles and cohomology. I've just read them all again, and they seem perfectly valid reference desk question. I don't know, what do other people think of these five questions? ~~ Dr Dec (Talk) ~~ 09:54, 30 August 2009 (UTC)

I think Dr Dec's questions are entirely appropriate for the Mathematics Reference Desk, and I welcome him as a new regular contributor here. But this is off-topic so if any further discussion is required. let's take it to Wikipedia talk:Reference desk. Gandalf61 (talk) 11:55, 30 August 2009 (UTC)

I agree. I'll admit that this thread is a very unusual RD thread, posing a problem the OP has already solved. But as the goal was not to see if others can solve it, but rather to see if there are other ways to solve it, I took it as a sincere fragment of a pursuit of deeper understanding. Which, after all, is what the RD is intended for, isn't it? (Dec's questions have also certainly been more interesting than the typical homework questions, but I don't think that is disputed.) —JAO • T • C 12:04, 30 August 2009 (UTC)

Fun little macro with a (math) issue...

I posted this here because this is more a math question than a computer question, though it involves both...

Im making a little macro for fun with AutoIt v3 which opens up MsPaint and draws. There are currently values called xMomentum and yMomentum which guide the pen. Basically the program starts with xMomentum = 0, yMomentum = 0, then each step it generates a random number between -2 and 2 and adds that to xMomentum , and does it again seperately to yMomentum, then the mouse is moved that many pixels.

What i am seeming to get is momentums that build up, so the pen starts to fly across the screen in a (non straight) line. Although a few straight bits are cool, id like more changes to happen. My thought is this:

I could have variables xTop and xBottom, where the random number is no longer between -2 and 2 its between xTop and xBottom. I would adjust these based on xMomentum so that the "faster" the pen goes in either X direction, the more likely it is to change direction slightly. (of course, i would do the same for corresponding y values.)

I dont know how to do this mathematically though, any ideas? All thoughts are welcome! Thanks! :)

97.112.117.236 (talk) 23:38, 29 August 2009 (UTC)

There are a lot of ways you could do this. One way would be to subtract k times the momentum from the momentum at each step where k is a small positive constant. This is basically like introducing a frictional force proportional to the velocity.

Another way would be to just cap the momentum. If at any step it exceeds the cap, reduce it to the cap. Rckrone (talk) 00:19, 30 August 2009 (UTC)

I spoke shortly with my father, and he unearthed a problem. Lets say i add a random integer in the set [-2, 2] as i did before. this is added to a momentum, so there are 3 integers which do not decrease the speed, and 2 that do. Thus 0 needs to be removed from my random integers.

I sort of like the frictional thing.... but can i use that and still get the possibility of both increase or decrease? Maybe i should add my random number (-2, -1, 1, or 2) and then do this subtraction? this would have the possibility of speeding up or slowing down, but add the friction as well. am i correct?

97.112.117.236 (talk) 00:26, 30 August 2009 (UTC)

Yeah I meant add the friction part in addition to what you're already doing. With the friction alone it would be pretty dull: just slowing to a stop and then sitting still. Using the set {-2, -1, 0, 1, 2} doesn't cause any problems, although {-2, -1, 1, 2} works fine too. As long as the average is 0 the momentum won't favor one direction or the other. Rckrone (talk) 00:43, 30 August 2009 (UTC)

Perhaps so. I have opted to eliminate 0 just in case :) 97.112.117.236 (talk) 00:49, 30 August 2009 (UTC)

Resolved

August 30

Principle of mathematical induction

Please solve this problem of mathematical induction

sinx + sin3x + ... +sin(2n-1)x =sin^2nx/sinx —Preceding unsigned comment added by Tipusultan11 (talk • contribs) 07:08, 30 August 2009 (UTC)

If you can tell us where you are getting stuck in solving this problem maybe we can help you with it. What have you tried to do so far? Do you understand induction? Eric. 98.207.86.2 (talk) 07:47, 30 August 2009 (UTC)

Have you to use trig identities? You can do it easily using Euler's formula‎‎ Dmcq (talk) 11:37, 30 August 2009 (UTC)

...and powers thereof. ~~ Dr Dec (Talk) ~~ 11:43, 30 August 2009 (UTC)

As an induction problem it seems simpler than taking powers of the Euler formula, expanding, then comparing terms in the binomial expansion. The base case would be when n = 1, i.e.

$P(1):\ \sin(x)={\frac {\sin ^{2}(x)}{\sin(x)}}\ ,$

and this is clearly true. Next, assume that P(k) is true and prove that this implies that P(k+1) is also true. Well for the case P(k) we have

$P(k):\ \sin(x)+\sin(3x)+\cdots +\sin((2k-1)x)={\frac {\sin ^{2}(kx)}{\sin(x)}}\ .$

We assume that this equality holds. To prove that P(k) implies P(k+1) we simply need to show that

${\frac {\sin ^{2}(kx)}{\sin(x)}}+\sin((2k+1)x)={\frac {\sin ^{2}((k+1)x)}{\sin(x)}}\ ,$

which I won't do because I'm not going to do all of your homework for you ;oP ~~ Dr Dec (Talk) ~~ 12:30, 30 August 2009 (UTC)

...and this last step isn't too hard if you use the formula $\sin(x+y)=\sin(x)\cos(y)+\sin(y)\cos(x).$ ~~ Dr Dec (Talk) ~~ 12:42, 30 August 2009 (UTC)

Maybe I'm a stickler for detail, but the conjecture, as written, is false for x=kπ, k integer. --Stephan Schulz (talk) 12:44, 30 August 2009 (UTC)

It's not false, it's true for all x. I assume your problem is that the RHS has a zero denominator for x = mπ (m an integer). Well there's no problem: it's perfectly well defined. The limit of the RHS exists, and is equal to zero. See the article on L'Hôpital's rule. (In fact you don't even need that: just completing the inductive steps that I have outlined above show you that the singularities at x = mπ are removable.) ~~ Dr Dec (Talk) ~~ 12:50, 30 August 2009 (UTC)

Sure, the function can be steadily continued at the undefined points, as limits from both sides agree. But as written, it is not so continued. Even your base case is, strictly, wrong. As I wrote, stickler for details ;-) --Stephan Schulz (talk) 13:15, 30 August 2009 (UTC)

Why is the base case wrong? It has a removable singularity - there is no problem! The base case doesn't require steadily continuing anything. We are canceling functions. I've not tried to evaluate the numerator and denominator before canceling. Would you argue that x/x = 1? I hope not! Well, you're trying to say that there's a problem with the statment that x/x = 1 by saying that the base case isn't well defined. In fact the expression x/x is perfectly well defined; its value at x = 0 is indeterminate, but can be calculated as a limit. Where did I evaluate anything above? ~~ Dr Dec (Talk) ~~ 13:18, 30 August 2009 (UTC)

To make Stephan happy, let's rephrase the problem. Show, using mathematical induction, that for any positive integer n, and any real number y:

$\lim _{x\to y}(\sin(x)+\sin(3x)+\cdots +\sin((2n-1)x))=\lim _{x\to y}{\frac {\sin ^{2}(nx)}{\sin(x)}}\ .$

I always tacitly assume the idea of a limit when I evaluate a function. I'm sure most other mathematicians do too. ~~ Dr Dec (Talk) ~~ 13:30, 30 August 2009 (UTC)

I suspect you're missing a negation somewhere above. But yes, for me (the function defined over the reals by the formula) x/x is different from (the function defined over the reals by the constant formula) 1. In set notation, 1 is $(R\times \{1\})$, and x/x is $(R\backslash \{0\}\times {1})$. The one is a proper subset of the other (and so can of course be extended), and hence they are not the same. --Stephan Schulz (talk) 13:54, 30 August 2009 (UTC)

...and this helps to answer Tipusultan11's question, how? (p.s. So what if I haven't not missed no negatives?) ~~ Dr Dec (Talk) ~~ 14:02, 30 August 2009 (UTC)

Stephan, let's stop this now. You can carry on worrying about the validity of expressions like x/x, so that I can try to help answer people's questions here on the RD. Maybe Tipusultan11 should just hand his homework in with the solution: Question not well defined? ~~ Dr Dec (Talk) ~~ 14:08, 30 August 2009 (UTC)

For OP's interest, it will probably help him get full marks if he puts sin x != 0 somewhere. --91.145.89.58 (talk) 16:28, 30 August 2009 (UTC)

...meaning? ~~ Dr Dec (Talk) ~~ 16:39, 30 August 2009 (UTC)

In the case OP has to compose math test answers, it will steer him away from all trouble if he mentions that there's no equality when sin x = 0; x = pi * n. Forgive if my english is bad. --91.145.89.58 (talk) 16:52, 30 August 2009 (UTC)

Dr Dec. You write: "I always tacitly assume the idea of a limit when I evaluate a function. I'm sure most other mathematicians do too". How can you be that sure after meeting a mathematician who do not tacitly assume the idea of a limit? Your tacit assumption may lead you into trouble. Stephan's objection is perfectly valid. Bo Jacoby (talk) 17:29, 30 August 2009 (UTC).

Even if Stephan were a mathematician, the statement "most other" does not mean "all". Besides that, Stephan is a computer scientist and not a mathematician and as such does not count when it comes to qualifying the valididity of statements involving mathematicians (exhaustive or not). One is not going to go very far with one's mathematical thinking if one stops at the first problem. If one were to be presented with a problem that involved indeterminate forms, then one may stop if one chose. However, I might prefer to proceed with common sense. ~~ Dr Dec (Talk) ~~ 19:01, 30 August 2009 (UTC)

When composing functions with restricted domains, the domain of the composition is the largest set on which all the intermediate functions are defined. For example, working over the real numbers, the function given by f(x) = (√x)² is only defined for x≥0, even though it can be extended to a function defined for all x. (Such an extension must of course be non-unique, which is why we cannot ask what f(-2) is.) If a student claimed that (sin x)/x was defined at 0, I would mark it as incorrect. And I don't mean this pedantically; I don't believe I've ever said "And since (sin x)/x is 1 at 0..."; instead I would only say "And since (sin x)/x goes to 1 at 0..." The distinction between a function and its continuation, completion, or closure is an important one which becomes more important the deeper you go in mathematics; in my experience a practicing mathematician would be unlikely to blur this line, even in informal discussions. Tesseran (talk) 05:53, 31 August 2009 (UTC) [Edit: fixed typo, thanks Pt.] Tesseran (talk) 18:51, 31 August 2009 (UTC)

As functions on R, sin²x/sinx is definitely not equal to sinx. The OP's teacher may or may not care if he/she attends to that kind of detail, but if for instance this were an analysis class that would not fly. 67.100.146.151 (talk) 08:16, 31 August 2009 (UTC)

By the way, (sin x)/x goes to 1 and not 0, as x→0. — Pt _(T) 10:35, 31 August 2009 (UTC)

Program the function f by f(x) = x/x. Then f(x) evaluates to one when x is not equal to zero, and gives either an "zero divide" error message, or the value 0 (because of the zero in the numerator) when x = 0. Which programming languages return f(0) = 1 by evaluating lim_x→0f(x) ? Bo Jacoby (talk) 22:07, 1 September 2009 (UTC).

Big O of sin x

$\sin(x)={\mathcal {O}}(f(x)){\mbox{ as }}x\to \infty$

Can anything meaningful be said about f(x)? SpinningSpark 13:13, 30 August 2009 (UTC)

Could you please qualify your notation. I know ${\mathcal {O}}$ as a function ring. Do you mean O(f)? ~~ Dr Dec (Talk) ~~ 13:19, 30 August 2009 (UTC)

Sorry, I meant Big O notation. I copied the style from sorting algorithm so someone should take a look at that article if it's wrong. SpinningSpark 13:34, 30 August 2009 (UTC)

Well the notation means that, for sufficiently large x, $|\sin(x)|\leq k|f(x)|$ for some fixed positive constant k. This isn't very illuminating. The Big O notation is usually used to describe growth, but since the sine function doesn't grow it's a bit of a non-starter. For example, if $g(x)=O(f(x))$ then we know that the infimum of $|f(x)|$, for very large x, grows like the infimum of $|g(x)|$. But the infimum of $|\sin(x)|$ is zero: as small as it gets; so this doesn't tell us anything. Why did you ask the question? Can you give us more information and context? ~~ Dr Dec (Talk) ~~ 13:58, 30 August 2009 (UTC)

(ec) Saying sin(x) = O(f(x)) means that there is some k such that for x sufficiently large we have |sin(x)| ≤ k|f(x)|. This will always be satisfied provided that 1 ≤ k|f(x)|, which is the same as 1/k ≤ |f(x)|. So sin(x) = O(f(x)) means that the values of f(x) will eventually stay outside of some interval around 0. That is, there is some ε such that ε ≤ |f(x)| for x sufficiently large. It's not hard to see that this condition is equivalent to sin(x) = O(f(x)). Another way to say this is that the lim inf of f(x) as x goes to infinity is not zero. Staecker (talk) 14:05, 30 August 2009 (UTC)

Is this true? $|\sin(x)|$ will become zero infinitly many times, not matter how large x is. For example, try $f(x)=2\sin(x).\,$ Clearly $|\sin(x)|\leq |2\sin(x)|$ for all x (where k = 1), but f is zero infinitly many times and does not have a well-defined limit. I guess that the statement might be that If a limit exists then it will not be zero. ~~ Dr Dec (Talk) ~~ 14:18, 30 August 2009 (UTC)

While $\scriptstyle \lim _{x\rightarrow \infty }2\sin x$ does not exist, $\scriptstyle \liminf _{x\rightarrow \infty }2\sin x$ certainly does; it is –2. But shouldn't it be $\scriptstyle \limsup _{x\rightarrow \infty }|f(x)|$ (which is 2 in your example) that should not be 0? —JAO • T • C 14:30, 30 August 2009 (UTC)

Good point! I missed the inf. In that case then

$\lim _{x\to \infty }\inf |f(x)|\neq 0$

as Staecker quite rightly says. ~~ Dr Dec (Talk) ~~ 14:42, 30 August 2009 (UTC)

I think Dec's objection is valid- |f(x)| = |2sinx| does have lim inf equal to zero, but it is certainly the case that sin(x) = O(f(x)). So there are more possibilities for f. Perhaps the correct weaker condition is that: $\liminf _{x\to \infty }|f(x)|\neq 0$ or $\liminf _{x\to a}|f(x)|=0$ for all a being an integer multiple of π. (Or something like that.) Staecker (talk) 17:38, 30 August 2009 (UTC)

Sorry Dr. Dec, there is no context, it just arose out of some idle scribblings on scrap paper. SpinningSpark 14:38, 30 August 2009 (UTC)

Just to check that I am following this, would it be right to say,

$\sin(x)=\mathrm {O} (1)){\mbox{ for }}k\geq 1$

$\because |\sin(x)|\leq k|1|$

SpinningSpark 15:29, 30 August 2009 (UTC)

That's correct. To say that a function f is such that f(x) = O(1) means that, for sufficiently large x, f is bounded, i.e. |f(x)| ≤ k for all sufficiently large x. Like I said: the big O notation is used to describe the growth of a function. Say that f(x) = O(g(x)) then what does that mean? Well consider the cone given by y = ±g(x). Then f(x) will, for sufficiently large x, always stay inside a cone of that shape. So, e.g., if f(x) = O(x) then, for sufficiently large x, f(x) will always lie inside a cone given by the two straight lines y = ±kx, for some k (i.e. f(x) ≤ k|x|). ~~ Dr Dec (Talk) ~~ 15:50, 30 August 2009 (UTC)

The usefulness of sine and cosine

So, let's assume that sine and cosine haven't been invented. We want to try to express, in big O notation, the statement that the limit of the infimum of some smooth function f is non-zero (whilst also, theoretically, being able to take any other real value). How could that be done if we didn't know about sine or cosine? (If we don't know about sine or cosine then we couldn't use a power series for them either!) ~~ Dr Dec (Talk) ~~ 14:51, 30 August 2009 (UTC)

That's $\liminf f>0$. The big-Oh notation says something about absolute size. You cannot separate statements about the liminf and limsup. Besides, $O(\cdot )$ only implies an upper bound. There is a corresponding notation for lower bounds ($\Omega$, see Big Oh notation), although it's not as commonly seen. The definition also has absolute values. I don't think it is worthwhile trying to express non-quantitative statements in big-Oh notation. Phils 16:16, 30 August 2009 (UTC)

After reading the above more carefully, I realize that you are referring to the fact that $\sin x=O(f(x))$ implies $\liminf |f|>0$, i.e. the function to tested is in the $O(\cdot )$. In that case $1=O(f(x))$ works just as well. My remark about absolute value still holds. Phils 16:26, 30 August 2009 (UTC)

In computer science when we want to be precise, for a given function f, we define O(f) as the set of functions h for which there exists a constant K such that for sufficiently large x, $h(x)/f(x)<K$. So we would then say $\sin \in O(1)$ where we are abusing notation and writing "1" to denote the constant function $x\mapsto 1$. Re Declan: big O notation in the treatment I'm accustomed to is insensitive to additive constants. I don't think it would be used in the way you are asking. 67.122.211.205 (talk) 01:52, 31 August 2009 (UTC)

What mathematics things are vs. what mathematical things do

Is only the latter important? It seems to me like defining 'dog' as something that barks, bites, etc. But not telling what a 'dog' is. On the other hand, the only way of defining 'division' I can thing of is over its output.Quest09 (talk) 16:50, 30 August 2009 (UTC)

...so are you suggesting we define a dog by its output? YUK! :oP ~~ Dr Dec (Talk) ~~ 17:01, 30 August 2009 (UTC)

But surely all objects are defined by their interaction with obsevers anyway? The "are" is only assessed by interaction with the world. --Leon (talk) 19:18, 30 August 2009 (UTC)

Here's an interesting example of this problem (borrowed from a book, but I've forgotten which one EDIT maybe the one in the OUP very short intro series by Gowers). What "is" the king in chess? It's not a particular piece in your chess set, because if you've lost your king you can use any old bit of junk lying around. Is a king anything more than "something which can move only a single square in one direction except for castling, and...."? 87.194.213.98 (talk) 19:25, 30 August 2009 (UTC)

If you can find a difference between the two then perhaps you can find a scientific explanation of Transubstantiation ;-) Dmcq (talk) 19:53, 30 August 2009 (UTC)

Mathematicians are concerned with the fact that the reals form a complete ordered field (what they do), and often impatient with things like the fact that the reals can be regarded as equivalence classes of Cauchy sequences or as Dedekind cuts, etc. (what they (supposedly?) are).

The analogy to the king in chess is on the mark.

But this might be argued about philosophically. Michael Hardy (talk) 01:35, 31 August 2009 (UTC)

The fact that you have mentioned two completely different constructions of the reals that are both very popular shows how unimportant constructions are to the vast majority of mathematics. "The reals are the completion of the field of fractions of a set with at least one element and a successor function." is the most useful definition (albeit lacking a few key details for the sake of conciseness). --Tango (talk) 01:43, 31 August 2009 (UTC)

That's exactly my point in mentioning the two examples. Michael Hardy (talk) 21:05, 31 August 2009 (UTC)

The issue you're up against is the endless quest for syntactic proofs of theorems, since who can really say anything about the semantics of uncountable entities like real numbers, to say nothing of (e.g.) Hilbert spaces (the substrate of quantum theory)? Jean-Yves Girard has some pretty good rants along these lines, e.g. [1] 67.122.211.205 (talk) 03:11, 31 August 2009 (UTC)

PUZZLING PUZZLE

Guys I have been trying to solve this for a long time but I am not able to..please help me with this puzzle.(Thot it would fit in here better rather than the misc. desk)

The diagram shows a 4 x 5 grid with some filled cells. Find the numbers in the remaining cells according to the following rules: a) Each number can only take values from 1 to 5. There are 4 such full sets(1-5). b) Sum of each row is same. c) Sum of each column is same. d) Adjacent numbers cannot be the same.

-	-	-	1	5
-	-	-	-	1
2	4	-	-	-
4	-	-	-	-

thanks —Preceding unsigned comment added by 117.193.136.45 (talk) 16:53, 30 August 2009 (UTC)

An impractical but possible way to solve any such problem is to formulate and solve it as an Integer Programming Problem. There are softwares which automatically solve IPP's and you could use those.--Shahab (talk) 17:46, 30 August 2009 (UTC)

But, I would hope, 117.193.136.45 wants some kind of mathematical algorithm to solve the problem. I know that computers use algorithms, but that's just no fun! I was trying to find a connexion with sudoku. ~~ Dr Dec (Talk) ~~ 18:28, 30 August 2009 (UTC)

1	3	5	1	5
5	2	4	3	1
2	4	2	5	2
4	3	1	3	4

Just used good old fashioned trial and error. It's clear that the row and column sums must be 15 and 12 respectively.--RDBury (talk) 19:33, 30 August 2009 (UTC)

My solution was

2	4	3	1	5
4	2	3	5	1
2	4	3	1	5
4	2	3	5	1

which has a more obvious pattern. I can't see how it could have taken any length of time at all. Dmcq (talk) 19:41, 30 August 2009 (UTC)

Umm Dmcq, what about the constraint: d) Adjacent numbers cannot be the same? -hydnjo (talk) 19:49, 30 August 2009 (UTC)

Oops sorry, I should have read through to the end. Silly me Dmcq (talk) 20:09, 30 August 2009 (UTC)

I found the same solution as RDBury, also by trial and error. Note that since the column sums must be 12, there are only three ways to complete column 1 and three ways to complete column 5. That gives you 9 starting points, most of which quickly lead to dead ends. Gandalf61 (talk) 19:58, 30 August 2009 (UTC)

Damn. I followed this from Science to Miscellaneous, but didn't realize (didn't read) that is had also been moved here. However, under the premise that it is better the teach a man to fish than to give him a fish, I'll copy my answer here. -- Tcncv (talk) 23:59, 30 August 2009 (UTC)

There is a solution (and only one), but it would spoil the fun for me to give it outright. One way to find it is to use a backtracking approach, which is a way to solve many logic problems like this. (See also the Ariadne's thread article.) First, make a choice and see where that takes you. Fill in other cells that can be derived from that choice. When you cannot fill in any more, make another choice. If you come to a dead end, backtrack to your most recent choice and select another. If you run out of options at that point back up further and select another option for the previous choice.

This puzzle has a couple of good places to start. For ease of reference, I'll refer to the cells by row and column – (1,1) through (4,5). First you need to figure out the needed row and column sums. Since the sum of all of the available numbers is 4 × (1 + 2 + 3 + 4 + 5) = 60, the rows must each total 15 and the columns 12. Take a look at column 1. You have two numbers already whose sum is 6, so the remaining two numbers in cells (1,1) and (2,1) must also equal 6 (giving a column total of 12). Your options are 1 & 5, 2 & 4, and 5 & 1. The combination 3 & 3 would be an immediate adjacency violation; so would 4 & 2, given the existing value in cell (3,1). Tentatively select one pair and fill them in, keeping track of the other options in case you have to backtrack.

Now take a look at row 1. You already have three of five values, so you can apply the same logic. After that, look at column 2, then row 2, then column 3 etc. As you consider numbers, you can immediately eliminate any that cause adjacency violations. You can also watch for cases where the same number is used more than the allotted four times. That would also prompt you to back track.

Good luck. -- Tcncv (talk) 23:20, 30 August 2009 (UTC)

Sudoku boards and groups with 9 elements

Think about a completed sudoku board: it's a 9 × 9 board where we must fill the board with the numbers {1,…,9} in such a way that the same number can't appear in the same colomn, the same row, or in any of the 3 × 3 sub-squares that make up the 9 × 9 total board. Now, for me, the first two conditions (i.e. the same number can't appear in the same colomn or the same row) remind me of a group table for a group, say G, with nine elements. Is there a correspondence between completed Sudoku boards and groups with nine elements? It's been about five years since I did any finite group theory or number theory, so please forgive my ignorance. But I seem to rememeber subgroups forming little blocks in the table like the 3 × 3 blocks we have on sudoku boards. Is there a 1-1 correspondence between group tables and sudoku boards (I doubt it, but I'm so rusty that I forget). Although, I seem to remember some enumeration results from representation theory of finite groups, but again; it's so long ago that they're nothing more than quiet little bells ringing ~~ Dr Dec (Talk) ~~ 18:51, 30 August 2009 (UTC)

There are only two groups of order 9: the cyclic group and the direct product of two groups of order 3, so no such correspondence exists. If you draw the Cayley table of a group, that is, a grid whose rows and cols correspond to the group elements and whose ij entry is the product of the corresponding elements, a subgroup doesn't look like a "sub-square" in which every element (of the big group) occurs once, so this is also not like the small squares in a sudoku grid. You might be interested in Latin square, a correct sudoku grid is a special kind of Latin square. As for rep thy, you might have character tables in mind perhaps.87.194.213.98 (talk) 19:21, 30 August 2009 (UTC)

I already know about Latin squares, but I thought a group theoretical approach might be interesting. I know, from Lagrange's theorem, that the order of any subgroup must divide the order of the group, so subgroups of order 1, 3 and 9 are all possible. Could you please say a few words as to why there aren't any order 9 groups with subgroups of order 3? ~~ Dr Dec (Talk) ~~ 19:29, 30 August 2009 (UTC)

There are, all groups of order 9 have subgroups of order 3. If <g> is cyclic of order 9 then g^3 generates a subgroup of order 3, and if a group G is the direct product of two groups of order 3 then it has several subgroups of order 3. Have a look at the Mathematics_of_sudoku article, they mention that not-Burnside's lemma can be used to enumerate solutions up to a notion of equivalence. That article also talks about Cayley tables and sudoku grids.87.194.213.98 (talk) 19:40, 30 August 2009 (UTC)

Excellent! I'd seen the maths of sudoku section in the sudoku article, but had some how managed to miss the fact that there was an entire article. I know what I'll be doing for the next half hour... ~~ Dr Dec (Talk) ~~ 21:15, 30 August 2009 (UTC)

There are 2 non-isomorphic groups of order 9. The actual number of Cayley tables that form groups is a lot more than that. The problem is that groups require associativity which puts way more structure on the array than is applicable for Sudoko squares.--RDBury (talk) 19:51, 30 August 2009 (UTC)

If we restrict ourselves to the set {1,...,9} then the only isomorphisms are permutations of the symbols and you can permute the symbols in a completed Sudoko board without breaking it, so I don't think that makes much difference (although rigour requires we mention it). What if we consider quasigroups instead? According to the article there is (modulo permutation of symbols) a 1-1 correspondence between quasigroups and Latin squares. Is there anything interesting to say about those quasigroups that correspond to a Suduko board? --Tango (talk) 01:35, 31 August 2009 (UTC)

As an aside note that all groups of order p² (p prime) are abelian and the converse of Lagrange's theorem holds for all abelian groups.--Shahab (talk) 19:55, 30 August 2009 (UTC)

If G is a group of order p² where p is prime, it must equal to its center. To see this, observe that the center of G, Z(G), is either trivial or equals G, since G/Z(G) cannot be cyclic (here we are using the hypothesis that p is prime). Now note the fact that the center of a p-group (a group of prime-power order) can never be trivial, and therefore G is abelian (the fact that the center of a p-group is non-trivial is often proven using a simple argument with orbits and group actions). --PST 12:13, 31 August 2009 (UTC)

Random trig question

I know that it is possible to perform inverse trigonometric functions with a calculator. I have been told that it is not possible by hand. However, trigonometry existed centuries before calculators were invented. So, is it possible to perform these function by hand? If not, how did the ancient mathematicians do it? Intelligentsium 23:21, 30 August 2009 (UTC)

It is possible, it just depends how much time you have on your hands. For example, in the olden days people used log tables to evaluate logarithms. These were basic grids that once you knew the base of the log (say n) and the value of the variable (say x), would give you the answer (say log_n(x)). Well, in that case, finding a number in the middle of the table that was close to the number you wanted to find the inverse of would tell you to which base, and of which value, it was the inverse of. I assume that trig' functions were the same. It's just like a group table: find the number in the group table and you know what's the inverse of what. ~~ Dr Dec (Talk) ~~ 00:13, 31 August 2009 (UTC)

Well, obviously, when you see a table of values of sine and cosine in a book published in 1908, or 1650, that tells you something. Maybe what was meant was that the methods needed in order to do it efficiently by hand cannot be taught at the very elementary level at which instruction in trigonometry takes place today.

As for log tables, where you looked up log_n(x) if you knew n and x, that's not how it was done because it's not an efficient way to do it. Base-10 log tables were commonplace. If you wanted log₂(3), you used a base-10 log table and found (log₁₀(3))/log₁₀(2). Tables of base-e logarithms also existed, but were not in an appendix to every book. Michael Hardy (talk) 01:28, 31 August 2009 (UTC)

Of course you can compute inverse trig functions by hand, for example with Padé approximants. Stupendous amounts of research went into developing methods to minimize the amount of labor required by those computations. In practice you would look your number up in a table and use linear or quadratic interpolation on the nearest values you could find to the one you wanted. But somebody had to prepare the tables, and they did it without computers... 67.122.211.205 (talk) 01:42, 31 August 2009 (UTC)

And the books of tables all had little errors in them which is why Charles Babbage designed his Difference engine. Dmcq (talk) 07:47, 31 August 2009 (UTC)

Maybe we should add that if it were impossible to compute values of inverse trigonometric functions by hand, then it would be impossible to do so via computers as well. Michael Hardy (talk) 21:07, 31 August 2009 (UTC)

I wouldn't say that. I mean, sure, they are Turing equivalent, but some amounts of hand calculation are just impractical. You can calculate pi to 100's of digits by hand, maybe even 1000's of digits, but not millions of digits. But it can be done with computers. 70.90.174.101 (talk) 00:59, 1 September 2009 (UTC)

August 31

Empty set and phi

So, in my topology class last year, a student referred to the empty set as phi. Then, this past week, one of my professors did the same thing. It seems to me that $\emptyset$ is more like a 0 with a line through it than $\phi$, and the 0 would match up symbolically as well. Is it related at all to $\phi$ or are these two people incorrect in calling it $\phi$? StatisticsMan (talk) 01:04, 31 August 2009 (UTC)

I've not heard "phi" for the empty set, and calling it that is incorrect or at least not standard. However $\varnothing \$ is a common variant on $\emptyset \$ that looks a good bit like $\phi \$. Eric. 98.207.86.2 (talk) 01:16, 31 August 2009 (UTC)

Nor have I. People mis-reading the symbol me thinks. ~~ Dr Dec (Talk) ~~ 01:21, 31 August 2009 (UTC)

Ditto. It is an easy mistake to make, but I've always heard it pronounced as "the empty set" or just "empty". --Tango (talk) 01:26, 31 August 2009 (UTC)

This is my thought when it was just a student in the class saying it last year. When a professor says it, then it makes me wonder, though it still seems wrong. Thanks for your thoughts. StatisticsMan (talk) 02:42, 31 August 2009 (UTC)

From the page Earliest Uses of Various Mathmatical Symbols:

The null set symbol (Ø) first appeared in N. Bourbaki Éléments de mathématique Fasc.1: Les structures fondamentales de l'analyse; Liv.1: Theorie de ensembles. (Fascicule de resultants) (1939): "certaines propriétés... ne sont vraies pour aucun élément de E... la partie qu’elles définissent est appelée la partie vide de E, et designée par la notation Ø." (p. 4.)

André Weil (1906-1998) says in his autobiography that he was responsible for the symbol:

Wisely, we had decided to publish an installment establishing the system of notation for set theory, rather than wait for the detailed treatment that was to follow: it was high time to fix these notations once and for all, and indeed the ones we proposed, which introduced a number of modifications to the notations previously in use, met with general approval. Much later, my own part in these discussions earned me the respect of my daughter Nicolette, when she learned the symbol Ø for the empty set at school and I told her that I had been personally responsible for its adoption. The symbol came from the Norwegian alphabet, with which I alone among the Bourbaki group was familiar.

So no relation to phi. Tesseran (talk) 03:35, 31 August 2009 (UTC)

Indeed. The symbol originated as an Ø and is completely unrelated to phi. Gandalf61 (talk) 07:50, 31 August 2009 (UTC)

Wow. This is the first time I've heard of the symbol for the empty set not being phi. For the past 10, maybe 15 years I was sure it was. Our article really should state explicitly that it's not. -- Meni Rosenfeld (talk) 20:37, 31 August 2009 (UTC)

Done - it now states that explicitly. Gandalf61 (talk) 21:56, 31 August 2009 (UTC)

So, how about the way some computer scientists write zero, that is again Ø? Why do they need it? Did it follow the notation of the empty set ?(via von Neumann construction?) or where did it originate? And is it still used? How they distinguish zero from the empty set? :( --pma (talk) 22:22, 31 August 2009 (UTC)

The slash through a zero is to distinguish it from the letter O. You don't usually need to distinguish between zero and the empty set - it should be clear from context. --Tango (talk) 22:43, 31 August 2009 (UTC)

Method of Characteristics to solve PDEs

First, I do admit that this is a homework problem. The question is to solve
$2tu_{t}+xu_{x}=0$
using the method of characteristics. I have done that and the solution is
$u(x,t)=g\left({\frac {x}{\sqrt {2t}}}\right)$. Then the initial condition $u(x,0)=f(x)$ is given but how does this work? I can't just plug in t=0 in my solution and figure out what g(z) is. How can I resolve this? Any ideas/hints? Did I do something wrong? I checked my solution by plugging it into the PDE and it works fine. Thanks!97.118.56.41 (talk) 03:52, 31 August 2009 (UTC)

The method of characteristics requires the characteristics to be transverse to the initial manifold (here it's $\scriptstyle \mathbb {R} \times \{0\}$, the x-axis). On the contrary here the initial manifold is itself a characteristic line (the support of a ch.). As a consequence, the initial values of a solution can't be prescribed on it, and, indeed, the equation already tells you that any solution defined on $\scriptstyle \mathbb {R} \times \{0\}$ has to be constant there. --pma (talk) 06:57, 31 August 2009 (UTC)

Sequence limit

Resolved

--Shahab (talk) 17:22, 31 August 2009 (UTC)

I want to find the limit of the sequence $(an+b)^{1/n}$ where a and b are constants? I know that $n^{1/n}\rightarrow 1$ but how should I use this result. Thanks--Shahab (talk) 10:10, 31 August 2009 (UTC)

Lim x->0 (1+x)^1/x is e. Use this result, and come back if you have any problems. Rkr1991 ^{(Wanna chat?)} 10:24, 31 August 2009 (UTC)

That limit cannot be used, since we are taking a limit as n goes to infinity here.

However, notice that:

$an\leq an+b\leq (a+b)n$

$(an)^{1/n}\leq (an+b)^{1/n}\leq [(a+b)n]^{1/n}$

for n≥1. That should get you your answer. --COVIZAPIBETEFOKY (talk) 13:08, 31 August 2009 (UTC)

(of course, I assumed that b≥0, here; if it's not, simply reverse the signs) --COVIZAPIBETEFOKY (talk) 13:09, 31 August 2009 (UTC)

Thanks. However I can't resolve all cases still. What will happen if $a+b<0$ or $a<0$? Then I can't use the fact that $a^{1/n}\rightarrow 1$. Is the sequence divergent in those cases?--Shahab (talk) 14:49, 31 August 2009 (UTC)

The expression an + b is just a finite scalling and finite translation of n. Since x⁰ = 1 for any real number x, and you already know that n^1/n → 1 as n → ∞. I would say that (an + b)^1/n → 1 as n → ∞. Of course, depending on your choices of a and b you may have complex values for your sequence. This doesn't matter though because you will always have a real limit. For a complex number z we see that |z^1/n| → 1 as n → ∞ and (more importantly) arg(z^1/n) → 0 as n → ∞, i.e. z^1/n tends to a number with modulus 1 and argument 0, i.e. z^1/n → 1. ~~ Dr Dec (Talk) ~~ 16:10, 31 August 2009 (UTC)

I don't understand how you conclude (an + b)^1/n → 1 from n^1/n → 1. COVIZAPIBETEFOKY's approach made sense to me because I used the results related to algebra of sequences and the sandwitch theorem. Perhaps you can quote which result you are using. Thanks--Shahab (talk) 16:24, 31 August 2009 (UTC)

Indeed, COVIZAPIBETEFOKY's response gives a water-tight argument, and will lead you to a solution. I was trying to give some intuative comments as to why the limit is what it is. It is simple to see that the limit is 1, and I have shown that above. But intuition is not a proof, and of course you should use something along the lines of COVIZAPIBETEFOKY's rigorous analytical argument. If you can't see why my comments might lead you to conclude (and then try to prove) that the limit is one then maybe spend some time thinking it over. It's very natural and very intuative. The bottom line is that x⁰ = 1 for all fixed real numbers x. Basically, the only problem you get is from n^1/n. Once you know that it doesn't pose any problems then you're in the clear. ~~ Dr Dec (Talk) ~~ 17:21, 31 August 2009 (UTC)

My first solution didn't have accurate bounds in the general case; this one works (as long as a≥0, which has to be, since you can't always take an n^th root of a negative number), by taking the absolute value of b.

$a\leq an+b\leq (a+|b|)n$

$a^{1/n}\leq (an+b)^{1/n}\leq [(a+|b|)n]^{1/n}$

Where both equations are true for sufficiently large n.

This is pretty easy to generalize to any polynomial of arbitrary degree, by the way. --COVIZAPIBETEFOKY (talk) 17:16, 31 August 2009 (UTC)

Great. Thanks COVIZAPIBETEFOKY and Dr Dec!--Shahab (talk) 17:22, 31 August 2009 (UTC)

COVIZAPIBETEFOKY, there does seem to be one problem this the inequality

$a^{1/n}\leq (an+b)^{1/n}\leq [(a+|b|)n]^{1/n}.$

The sequence can take complex values, for example a = –1 and b = 0. When n is even we get imaginary numbers; so the inequality signs don't make such sense. Besides (–1)^1/4 has four possible values! Surely there should be some modulus signs? ~~ Dr Dec (Talk) ~~ 17:29, 31 August 2009 (UTC)

COVIZAPIBETEFOKY specified that his solution is only valid for a≥0 (although when a=0 b must be positive). In those cases we can assume we're talking about the positive real root. Otherwise the sequence diverges. And if we allow complex values then it's not particularly well defined and definitely doesn't converge in general although I guess you could set up a scheme to choose roots so that it would converge to any value on the unit circle you wanted. However in all cases (except a=b=0) it can be said that |(an + b)^1/n| goes to 1. Rckrone (talk) 18:09, 31 August 2009 (UTC)

I know, I was just prompting him to leave a revised inequality that involved muduli. ~~ Dr Dec (Talk) ~~ 18:13, 31 August 2009 (UTC)

p.s. the sequence does not diverge. It's limit is 1 for all real numbers a and b. As my post above showed: for any complex number z we have that |z^1/n| → 1 and arg(z^1/n) → 0 as n → ∞. This implies that the modulus of the sequence converges to 1 for all real number a and b. Find some a and b so that the sequence diverges. ~~ Dr Dec (Talk) ~~ 18:16, 31 August 2009 (UTC)

arg(z^1/n) doesn't go to 0 in general since z^1/n is ambiguous. The nth roots of z are evenly spaced around the circle. You could pick a value of arg(z^1/n) for each n to make a sequence that went to 0. You could also choose a sequence that goes to any value you like, or one that diverges. Just as an example, if for real a and b with a<0 we take the "obvious" root for each n odd, I would think that would be the real root (which is negative), in which case the odd terms go to -1. Rckrone (talk) 18:32, 31 August 2009 (UTC)

Really? Well, if $z=Re^{i(\theta +2\pi k)}\,$ where k is an integer then $z^{1/n}=R^{1/n}e^{i(\theta +2\pi k)/n}\,$ and clearly, for a fixed choice of k, $i(\theta +2\pi k)/n\to 0\,$ as $n\to \infty \,$. ~~ Dr Dec (Talk) ~~ 19:31, 31 August 2009 (UTC)

k is not fixed for all n. I can choose k_n to be any sequence of integers I want and it'll still produce a sequence of nth roots of z. Going to my example before: Let z = |z|e^iπ and choose k_n = n/2 for even n and k_n = (n-1)/2 for odd n. The nth term of my sequence is |z|^1/ne^iπ(n+1)/n for even n and |z|^1/ne^iπ for odd n. This sequence goes to -1 (the arguements go to π). It's not hard to see how you can create sequences that have lots of different behaviors depending on how you pick the nth root, and in the general complex case there's not an obvious "preferred" scheme for picking roots. Rckrone (talk) 19:58, 31 August 2009 (UTC)

(@dec)I would not worry about complex numbers; that exercise clearly refers to real numbers; it's quite implicit in the notation that a>0 and b is real. One could have asked the same question in a more general setting (complex numbers/matrices/linear operators/Banach algebras or whatever), but since nobody did, COVIZ's is, I think, exactly the right answer to the OP. --pma (talk) 18:18, 31 August 2009 (UTC)

The original question simply says that a and b are constant. It doesn't mention the positivity of a or b. COVIZ has the right solution. With a few moduli signs it would work even when a and b were taken to be complex. ~~ Dr Dec (Talk) ~~ 18:22, 31 August 2009 (UTC)

So, in my opinion the additional information on complex numbers may be of some interest, indeed; but in this and similar situations, it should be better to add it as a further remark or side remark, not as a correction to a totally satisfying answer. Otherwise, the OP will get confused, the first person who answerred will give you an explanation, you'll reply, I'll no help adding my boring remarks, a useless debeate will start, and this desk becames a mess again. It's not to criticize; I'm just looking with great admiration at those fellows here that are able to give short, concise, yet complete and illuminating explanations; let's try to learn how they do. --pma (talk) 21:38, 31 August 2009 (UTC)

Agreed! ~~ Dr Dec (Talk) ~~ 16:54, 1 September 2009 (UTC)

Millimeters

I speak of the probe on the left.

After reading the metre article, I still have a question regarding measurements -- as a dentist, I measure periodontal pockets (the sulci of gum tissue surrounding teeth in patients with periodontal disease) with a periodontal probe to the nearest mm. Because we (dentists in general) round to the nearest mm, and because there is some inherent intra- and interexaminer discrepancy due to probing force, velocity, angulation and the degree of inflammation present at the base of the sulci, any measurement is expected to be within 1 mm of what another competent dentist would record. My question -- since a mm is 1/1000 of a meter and a meter is arbitrary (in that there is no reason to have chosen the half-meridian for the original calculation), had a mm been 0.9 or 0.8 of our actual mm, wouldn't our measurements be more accurate? DRosenbach ^{(Talk | Contribs)} 12:53, 31 August 2009 (UTC)

With the smaller units, if you were still able to reliably measure within one "short" mm, then your measure would be more accurate. But I assume there would be a tradeoff between the this gain in accuracy and the loss in accuracy due to the examiner discrepancies. If the millimeter were smaller, it would be even more difficult for examiners to measure accurately. Staecker (talk) 13:06, 31 August 2009 (UTC)

True, but it's perfectly possible that the optimal trade-off between the stated precision, and the ability to reliably make measurements with that precision, is at an increment less than a mm. However, in other applications the optimum may be higher, so it's not universally preferable to choose a smaller meter. If this is significant you (where "you" is a guild of some sort) can always define your own optimal unit of measurement for your application. -- Meni Rosenfeld (talk) 20:25, 31 August 2009 (UTC)

There are two sources of error: the various sorts of individual variation that you mentioned, and the rounding to the nearest mm. If the definition of mm were changed, then the second source of error would be reduced (because you're rounding less) but the first would not be changed. If the first source of error is the dominant source of error (e.g., dentists can't be more accurate than 2 or 3mm because of measurement error), then changing the definition of mm will have negligible effect, whereas if the second source of error is the dominant source of error (e.g., dentists can measure down to 0.01 mm but this precision is lost because of rounding to the nearest 1mm), then rounding to a smaller unit would significantly reduce error. Eric. 216.27.191.178 (talk) 20:34, 31 August 2009 (UTC)

Octahemioctahedron

This is the octahemioctahedron, with tetrahedral symmetry... 3/2 3 | 3	...but what about this? Can it have octahedral symmetry?
Cantellated tetrahedron 3 3 | 2	Cuboctahedron 2 | 3 4

Does the octahemioctahedron have tetrahedral or octahedral symmetry? Since it's a facetting of the cuboctahedron with octahedral symmetry. Thanks. Professor M. Fiendish, Esq. 12:58, 31 August 2009 (UTC)

The article you linked - infobox - symmetry group. The answer is in there. It seems correct.83.100.250.79 (talk) 15:06, 31 August 2009 (UTC)

If you consider the eight triangular faces as being identical and indistinguishable (right hand diagrams) then it has octahedral symmetry. If you consider the triangular faces as separated into two families of four, distinguished by whether you are seeing the "top" of "bottom" surface of the triangle (red and blue triangles in the left hand diagrams) then the octahedral symmetry is broken and what remains is tetrahedral symmetry. Gandalf61 (talk) 18:21, 31 August 2009 (UTC)

Usually by "symmetries of a polyedron", or more generally "symmetries of a metric space", people mean its group of isometries... so I'd say the former you said is the natural choice. I suddenly had a suspect, and googoled "colored polyhedra" and "colored symmetries": they are there. Everything that may be thought by a human mind is already there. --pma (talk) 21:59, 31 August 2009 (UTC)

I contributed to the article and images. The Wythoff symbol 3/2 3 | 3 has tetrahedral symmetry, two triangle colors. There may be a secondary reflective construction with octahedral symmetryLooking at Schwarz triangle options, I don't see a simple form, maybe as a composite form?! I'm asking around by email too. Tom Ruen (talk) 22:33, 31 August 2009 (UTC)

OK, what about the tetrahemihexahedron? It's a facetting of the octahedron, which has octahedral symmetry.

Tetrahemihexahedron.

Especially since the dual, the tetrahemihexacron, is made up of 3 intersecting infinite square prisms (octahedral symmetry! There's a cube in the central intersection.)

Tetrahemihexacron.

Professor M. Fiendish, Esq. 04:42, 2 September 2009 (UTC)

Oh, maybe we should check the symmetries of the octahemioctacron.

Octahemioctacron.

Professor M. Fiendish, Esq. 04:42, 2 September 2009 (UTC)

Thank you, math reference desk

Hi. I'm just posting here to say that I passed my Complex Analysis qualifying exam, and you guys helped me a lot while I was studying for that. This page is one heck of a resource, and that wouldn't be true if not for the patient and friendly mathematicians who answer questions here. You guys rock, individually and as a group, and I wish a thousand blessings upon all your houses. :) -GTBacchus^(talk) 19:19, 31 August 2009 (UTC)

I second the above totally.--Shahab (talk) 05:09, 1 September 2009 (UTC)

Weighted sum of distance from a curvy line

Consider a busy road which is a curvy line on a flat plane. The sound intensity at every point on the curvy line is the same - call it a sound intensity level of 1, say. The sound intensity at an X,Y point on the plane is I think the sum of the noise from each point on the curvy line, except that the noise from each point is weighted by the inverse of its distance squared, since sound energy decreases with distance.

My question is - what would be the best practical way to calculate the sound intensity at an X,Y point on the plane?

All I have is a paper map which includes the road, and my maths is sub-calculus. Ideally I would like to have contour lines of the sound intensity over the plane, but just being able to calculate the sound intensity at two different points will do. I realise that this is an abstraction, but its a start. The calculations could be done manually or is there any suitable mapping software that could be used? Thanks 92.27.79.62 (talk) 20:30, 31 August 2009 (UTC)

Well, if $\textstyle \gamma :I\to \mathbb {R} ^{2};$ (or $\textstyle \mathbb {R} ^{3}$ if you like) is the arc-length parametrization of your road, the intensity at the point x it is proportional to $\textstyle \int _{I}|x-\gamma (s)|^{-2}ds$. here |v| denote the euclidean norm or v. Then, depending on the curve, it may be possible to compute the integral, or just evaluate it numerically. In any case I hope it is just an ideal calculus, and you do not have to face a real traffic noise! --pma (talk) 21:08, 31 August 2009 (UTC)Warning: In fact, the above integral was a mathematical representation of a quantity induced with a inverse square law by a source uniformly disributed on the curve γ, provided one assume a superposition principle. But I do not see why it should be true here. --pma (talk) 11:51, 2 September 2009 (UTC)

If it's a real road map, one way you could get an estimate manually for a specific point is to draw concentric circles around your point on the map with a compass. Then for each ring between consecutive circles estimate the length of road in the ring and take the distance away for all that road to be the average of the radii of the two circles. So if you have length d of road between circles of radius r and s, then the contribution of that part of the road is d/((r+s)/2)². Then sum up the contributions of all the rings. This is going to be least accurate for road that's very close to your point, so you would probably want to draw your circles closer together as you get closer to your point. Depending on how accurate you want to be, this could be a lot of work. Rckrone (talk) 21:11, 31 August 2009 (UTC)

Yes, but the sample points should be equally spaced on the road, to avoid having to weight each individually, (in what is already a time consuming process..) 83.100.250.79 (talk) 22:09, 31 August 2009 (UTC) Ignore this - had a better idea..:

I have a better idea - if you can get the x,y coordinates of a set of points along the road (choosing points so that the 'zig-zag' straight line is fairly close to the original road) - then enter the points into a spreadsheet (or specially prepared program) - the program could then calculate the contour lines for the sound levels - using the distance between adjacent points as a weighting factor, and the inverse square rule as well.

This way you only need to measure points once - but you can get an entire contour map made.

Would you be able to make a program to do that (any programming experience?) - it would be quite easy to write - someone would probably volunteer to do that on the computer desk, as it is not too difficult (I could write that if necessary).83.100.250.79 (talk) 22:30, 31 August 2009 (UTC)

Yeah, this method is much better than mine since you can get a computer to do most of the work without having to feed it too much data, and it's more accurate. Rckrone (talk) 22:41, 31 August 2009 (UTC)

Assuming there is no equation for the road - one solution would be to scan the map - and then use software to convert the road into a line of points (tricky bit - but possible) - then the sums of intensities to give contours would be actually trivial for the computer.

If you want to follow this route (pun!) then the hardest step is finding software which will take an image (assume you have scanned or photographed the map, and removed any stuff that is not the road) and convert it to a set of points describing the line/curve of the road. You should ask for this if this sounds useful to you. I would guess the maths desk would be better for the algorthym, but the computer desk would be more likely to know of software.

If you don't do this I think the solution will be a lot of hand measuring, and calculations.83.100.250.79 (talk) 22:06, 31 August 2009 (UTC)

Thanks. Remember as I said my maths is beneath calculus, so the first responce means nothing to me. I remember I have a computer program somewhere that is designed to digitise lines from images of graphs, so that would probably help. Or I could do it manually. This is a practical real problem, by the way, not an exercise. You can assume as a starting point that I have coordinates for points along the road. I have not done any amateur computer programming for a long time - the only language I was fluent with was GWBasic. I will try and find the digitisation program and get back to you later. 78.147.28.17 (talk) 10:20, 1 September 2009 (UTC)

There's a side-issue which has occurred to me - since there are an infinite number of points in a line, won't the sum of the sound from these points along the road be in theory infinate? Or at least, wont the calculated intensity of the sound at X,Y compared with that at the road be heavily dependant on how many points along the road I decide to use? If I use for example twice the number of road-points, then the total sound intensity is approximately doubled. 78.144.246.12 (talk) 14:44, 1 September 2009 (UTC)

If you sample the curve with 2x points, you divide the result with 2. You need to sample relatively evenly in space, eg with a rectangular grid. --91.145.89.58 (talk) 15:21, 1 September 2009 (UTC)

Yes using N points divide the total by N, also bear in mind to weight each point by the approximate length it represents. - using the weights - you then divide by the sum of the weighting lengths at the end, instead of N.83.100.250.79 (talk) 18:38, 1 September 2009 (UTC)

I'm not sure about the two comments above - if for example the road formed a hairpin bend, and you went and stood so that you had road noise on either side of you, then you would expect the intensity of the noise to add and be greater where you stood than on the road itself. Dividing the sum of the noise by the number of points, or the total of the inverse squared distance would not give this effect I think. 78.146.3.82 (talk) 20:38, 3 September 2009 (UTC)

It will. Noise intensity is directly proportional to the inverse square of distance from the source because it spreads in 3d space. In this case, it is also directly proportional to the length of the road you have, especially when the road is short. However if you have a long and curvy road the idea of distance is a bit vague as the road isn't exactly point-like and its length isn't always trivial to measure either. So you model the road as n shorter roads, ie you divide your original road to n pieces. The larger you make the n, the more point-like your pieces of road get and you get more accurate results. You compute the final sound intensity as the sum of the intensities of the n pieces. The intensity at a point by a piece is C * x / r**2, where C is some road-specific constant ("base intensity"), x the length of the piece, and r**2 distance from the road squared. If you use 2n pieces instead, x will be two times smaller on average and the result won't double. So ideally the whole sum is not divided, but in practise you might not want to bother find out the length of every piece and would take a shortcut approach instead. --194.197.235.240 (talk) 00:10, 4 September 2009 (UTC)

But I really I have some doubts about the validity of a superposition principle here. What is your definition of "sound intensity"? --pma (talk) 04:59, 4 September 2009 (UTC)

Sound intensity. I don't see why superposition wouldn't apply unless there are weird phase differences and resonance, but my physics are quite basic. --194.197.235.240 (talk) 05:58, 4 September 2009 (UTC)

Well, I would be a bit puzzled by a scalar physical quantity that obeys to both an inverse square law and to a superposition principle. The reason is that 1/|x|² is not a solution of any physically reasonable linear PDE in 2D or 3D. Said in another way, a superposition principle does hold for sound waves, what reflects their mathematical description as solution of the linear wave equation $\scriptstyle (\partial _{tt}-\Delta )w=0$; but the sound intensity as a scalar quantity, seems related in a nonlinear way to the sound waves, so that the additivity should be destroyed. In fact, your link clarifies this a bit: it defines sound intensity as a vector, which in the case of point source is radially directed and decays as 1/r². If the sound intensity defined that way also admits a (vectorial) superposition principle, what makes sense (but it is still unclear to me) then it would be very analogous to a Newtonian force, and it could be treated in terms of potential theory. --pma (talk) 14:49, 4 September 2009 (UTC)

It looks like from what has been written that I should divide the road up into small segments and take both the segment length and the noise intensity into account. So I could say the noise intensity is 1 per 1 metre of road. This would give a formula of I(x,y)=I(road) x L / (D**2) where I is the noise intensity, L is the length of the segment, and D is the distance between the road and the point. But this formula does not seem to be dimensionally balanced: the L / (D**2) part should be dimensionless for the two sides to balance, but it is not. Where am I going wrong please? 89.240.207.84 (talk) 14:51, 4 September 2009 (UTC)

It does not seem wrong; I(x,y) and I(road) in the formula just are not homogeneous quantities, as e.g. are not homogeneous the force and the mass in the analogous Newton law. --pma (talk) 14:58, 4 September 2009 (UTC)

Sorry I cannot believe that. I'm sure any true physical law would be dimensionally consistant. 78.146.70.127 (talk) 19:13, 4 September 2009 (UTC)

I mean, quoting for instance the above link: "For a spherical sound source, the intensity in the radial direction as a function of distance r from the centre of the source is: I_r=P_ac/4πr²..."; as you see there is no "sound intensity" term on the righ-hand side; correspondingly, the term denoted "I(road)" in your equation is not an intensity sound. But that is quite clear; the main point remains the fact that (I think) there is no superposition principle for this scalar sound intensity. --pma (talk) 20:04, 4 September 2009 (UTC)

No-no-no! The intensity we hear is proportional to the average of the pressure disturbance squared. We may very well take your “vector intensity” and call its module the scalar one I will subsequently be concerned about. Two different points/cars on the road certainly produce sound totally independently, that is, their phases and frequencies are different. Add a wavepacket $\scriptstyle p_{1}=A_{1}\sin(\omega _{1}t)$ and another one, $\scriptstyle p_{2}=A_{2}\sin(\omega _{2}t+\varphi )$. Let's choose such units that the intensity is the time average of the amplitude squared ($\scriptstyle p^{2}$). Then the intensities of the packets are, of course, $\scriptstyle I_{1}=\langle p_{1}^{2}\rangle =\lim \limits _{\tau \to \infty }{\frac {1}{\tau }}\int _{0}^{\tau }p_{1}^{2}={\frac {A_{1}^{2}}{2}}$ and similarly $\scriptstyle I_{2}=\langle p_{2}^{2}\rangle ={\frac {A_{2}^{2}}{2}}$ (just notice that both $\scriptstyle p^{2}$-s are sinusoids). Now find the total intensity you hear:

$I=\langle (p_{1}+p_{2})^{2}\rangle ={\begin{cases}I_{1}+I_{2}&{\text{if }}\omega _{1}\neq \omega _{2}\\I_{1}+I_{2}+2{\sqrt {I_{1}I_{2}}}\cos \varphi &{\text{if }}\omega _{1}=\omega _{2}{\text{.}}\end{cases}}$

The last formula is easily provable imagining the $\scriptstyle p$-s as a projections of phasors and using the cosine law (and almost being wrong like me, when I first misplaced the $\scriptstyle \varphi$ in my mental diagram $\scriptstyle {\ddot {\smile }}$). So the superposition principle does not hold only in an unphysical special case! The wavepackets to be added have, in reality, always different frequencies and thus are incoherent. Even if they had the same frequency $\scriptstyle \omega$, but random phases, superposition of intensities would still hold if we average over the phase shift (and the wavepackets are short if they're not being specifically generated, so the averaging is what our brain does): then

$I=\left\langle {\frac {1}{2\pi }}\int _{0}^{2\pi }\left[A_{1}\sin(\omega t)+A_{2}\sin(\omega t+\varphi )\right]^{2}d\varphi \right\rangle _{t}=$

$=\left\langle I_{1}+I_{2}-I_{1}\cos(2\omega t)\right\rangle _{t}=I_{1}+I_{2}{\text{.}}$

Oh my, if it were otherwise: we would be maddened about all that interference, especially visually, since all the theory applies also in optics… I know that the electric field $\scriptstyle {\vec {E}}$ is a vector there, however, interference is most easily discussed between waves with the same polarization, so they still add up scalarly. In optics, Poynting vector is the “vector intensity” and its time-averaged magnitude the usual intensity AKA irradiance. I am trying to think about the PDE $\scriptstyle I$ follows, but here I am as lost as pma was. What about an integro-differential equation? I don't say I already have one, though. Also, a superposition principle that does not hold in the most general case (just like here) does not imply linearity at all: we have solitons etc, although, in turn, non-interaction of solitons does not imply their linear superposition. — Pt _(T) 20:57, 7 September 2009 (UTC)

On another thought, this conditionality of the superposition is irrelevant here: it concerns only the way intensities arise from pressure waves, not the intensities themselves: you can always change the frequency of any source to make it incoherent with everything else, thus restoring the superposition. Thus, the question now is: if we know that the intensity obeys both superposition law and inverse-square fall-off (because of symmetry and energy conservation), then which PDE (or alike) does it follow? May it be nonlinear? (Sorry, original poster, this has become interesting in its own right! Actually, this could give us a ready-made mathematical machinery just for you.)  — Pt _(T) 21:16, 7 September 2009 (UTC)

Actually, electromagnetism gives us a PDE! Coulomb's law is an inverse-square vector equation that after integration gives rise to a inverse-first-power potential law, a scalar equation written as a Poisson equation as $\scriptstyle \Delta \varphi \propto -\rho$, where $\scriptstyle \rho$ is the density of sources and the coefficient of proportionality can be made 1 by choice of units. Now, this inverse-square dependence comes from the fact that in 3 space dimensions the area of a sphere is proportional to its radius squared, and Gauss' law. In 4 dimensions (not Minkowski spacetime, but just 4 Euclidean dimensions), the sphere's area $\scriptstyle \propto r^{3}$, so the Gauss' law ($\scriptstyle \operatorname {div} {\vec {E}}\propto \rho$) gives now rise to an inverse-cube vector equation and integrating (going from force to work) gives us a inverse-square linear scalar equation! It is simply a 4-dimensional Poisson equation, which we now write for acoustics again ($\scriptstyle \varphi \mapsto I$):

$\sum _{i=1}^{4}\partial _{x_{i}x_{i}}I\propto -\rho {\text{.}}$

Hence we have to put in a known source density $\scriptstyle \rho$ (which can be non-zero only where $\scriptstyle x_{3}=x_{4}=0$, since our original problem is 2D), solve the equation in 4 dimensions while assuming the solution to vanish at infinity (is this boundary condition enough?) and finally restrict it to the 2D plane we are interested in. If the road is infinitely thin (and in the two extra dimensions it has to be!), $\scriptstyle \rho$ is no longer an ordinary function, but a distribution involving the Dirac delta function, but it should not be a problem. Finally, someone $\scriptstyle {\overset {\_\,.}{\smile }}$ should explain a good numerical Poisson equation method to the original poster! — Pt _(T) 00:57, 8 September 2009 (UTC)

September 1

Integration

Earlier this year our lecturer gave us an integral to solve, then realised she had made a mistake, and gave us an easier one.

The Integral was ʃ { 2 / [ 3X Ln ( 2x + 1 )]}dx, which she changed to ʃ { 2 / [ 3X Ln ( 2x)]} dx, which equals ⅓ Ln[ Ln ( 2x ) ] + C

This was relatively easy, but what I would like to know, is how to solve ʃ { 2 / [ 3X Ln ( 2x + 1 )]}dx, and why is it that simply adding a 1 to the term in the parentheses makes it so much harder ? I am only a first year Canterbury ( New Zealand ) University student. Thanks. —Preceding unsigned comment added by 202.36.179.66 (talk) 04:08, 1 September 2009 (UTC)

Probably because if you use the u substitution method (which I'm guessing was the suggested one) to try to evaluate the second ("mistake") integral, it doesn't work out too well, whereas with $\int {\frac {2}{3x\ln {(2x)}}}\,dx$, if you let u = ln(2x) then you end up getting du = 1/x dx which substitutes in nicely. However if you use the same method with $\int {\frac {2}{3x\ln {(2x+1)}}}\,dx$ and set u = ln(2x+1) then du = 2/2x+1 dx, which unlike the first problem, does not substitute in nicely. Ginogrz (talk) 04:44, 1 September 2009 (UTC)

Just to clarify, in the "easier" integral, the solution should be ${\dfrac {2}{3}}\left(\ln {(\ln 2x)}\right)$ (as the u-substitution yields ${\dfrac {2}{3}}\int {\dfrac {du}{u}}$). --Kinu ^t/_c 04:54, 1 September 2009 (UTC)

As far as I can tell, the more difficult integrand doesn't have an indefinite integral in terms of elementary functions. ~~ Dr Dec (Talk) ~~ 07:42, 1 September 2009 (UTC)

How do you determine that? Black Carrot (talk) 22:22, 1 September 2009 (UTC)

Risch algorithm has some details, although only in that superficial hand-wavy kind of way that encyclopedia maths articles always seem to gravitate to. Zunaid 09:57, 2 September 2009 (UTC)

Also found Differential Galois theory, according to which the problem of solving indefinite integrals using elementary functions is analogous to the problem of finding roots of polynomial equations using radicals. Read the article for more details. Zunaid 10:08, 2 September 2009 (UTC)

Hyperbolic Functions

Given the value of, say, $\cos \theta$ one can determine $\sin \theta$ and $\tan \theta$ by drawing a right angled triangle with the appropriate side lengths. Is there an equivalent for the hyperbolic functions? I know you can use relations such as $\cosh ^{2}\theta -\sinh ^{2}\theta =1$ but this isn't what I'm looking for. Thanks. 92.4.122.142 (talk) 17:39, 1 September 2009 (UTC)

If you know $\cos \theta$ then you only know the angle. You would also need to know the length of one of the sides to determine the triangle unically. If we knew the length of the hypotenuse, say h, then we would know that the adjacent side has length $h\cos \theta$ and the opposite side has length $h\sin \theta .$ Knowing one of the non-right-angle angles only gives us the triangle up to a scalling. After all: cosine gives the ratio of the length of two sides, and that's invariant under scalling. ~~ Dr Dec (Talk) ~~ 17:53, 1 September 2009 (UTC)

You have picked up on a minor slip up when writing my initial post that has no bearing on either the statement "Given the value of, say, $\cos \theta$ one can determine $\sin \theta$ and $\tan \theta$" or my question. 92.4.122.142 (talk) 18:12, 1 September 2009 (UTC)

Ay, ay, ay. No need to be so touchy. I was trying to help! You don't need to use any triangles. If you know $\cos \theta$ then you know $\sin \theta$ up to sign: $\sin \theta =\pm {\sqrt {1-\cos ^{2}\theta }}.$ If you know $\cos \theta$ and you know $\sin \theta$ up to sign then you know $\tan \theta$ up to sign:

$\tan \theta =\pm {\frac {\sqrt {1-\cos ^{2}\theta }}{\cos \theta }}\ .$

If you're trying to fit these around a triangle then you'll want 0 < θ < π/2, and so all the ±'s just become +. The same goes for the hyperbolic trig functions: we know that $\cosh ^{2}\theta -\sinh ^{2}\theta =1$, so once we know either $\cosh \theta$ or $\sinh \theta$ we can work out the rest with some algebra. Although there won't be a nice circle picture, you'd get branches of a hyperbola. Please, 92.4.122.142, assume good faith; I was honestly trying to help you. ~~ Dr Dec (Talk) ~~ 18:44, 1 September 2009 (UTC)

My apologies. I thought you were deliberately being pernickety about my question just for the sake of it. I see now that was a wrong assumption to make. Thank you for your help and again, I'm sorry. 92.4.122.142 (talk) 18:47, 1 September 2009 (UTC)

It does not appear that you have to apologize. It does not appear that your initial post was uncorrect. You were talking of determining sin(θ) and tan(θ), not a triangle. --84.221.68.104 (talk) 22:57, 2 September 2009 (UTC)

For trig functions, we can draw the unit circle (x² + y² = 1) and then any point on the circle has coordinates (cosθ, sinθ) for some value of θ. The value of θ corresponds to twice the area of the circle swept to our point, which is also equal to the angle from the x-axis. This let's us draw a right triangle with unit hypotenuse, and one angle θ that has legs cosθ and sinθ. For the hyperbolic functions we can do a similar thing with the unit hyperbola (x² - y² = 1). Any point on the right branch of the hyperbola has coordinates (coshθ, sinhθ) for some value of θ. This θ is also twice the area swept out along the hyperbola to that point, but unlike with the circle that doesn't correspond to any angle that I know of, so I think that's pretty much where the analogy ends. Rckrone (talk) 18:17, 1 September 2009 (UTC)

Actually I guess you could draw a right triangle with one leg equal to 1, and then the hypotenuse would be coshθ and the other leg would be sinhθ for some θ. θ wouldn't correspond to any obvious property of the triangle though. Rckrone (talk) 18:24, 1 September 2009 (UTC)

September 2

Variables confusing me

The "Circular references in computer programming" section of Circular reference gives an example of how spreadsheets can be confused by circular references, using the following wording: "This [the previous example] leads to what is technically known as a livelock. It also appears in spreadsheets when two cells require each others' result. For example, when the value in Cell A1 is equal to the value in Cell B1 plus 5, and the value in Cell B1 is equal to the value in Cell A1 plus 5." Is this possible? I can't imagine a situation in which x = y + 5 AND y = x + 5, unless our spreadsheet putter-togetherer is making more basic errors than circular references. Unless I misunderstanding something here, would someone please revise this statement so that it's far more useful? Nyttend (talk) 02:38, 2 September 2009 (UTC)

It isn't possible that x = y + 5 and y = x + 5 and that's not what the article is implying. It is possible to enter these formulas into a spreadsheet though. The spreadsheet won't be able to compute anything from them and that's why you get an error message. You can enter all sorts of mathematical nonsense into a spreadsheets and a lot of other computer programs, and they give you error messages back because they are nonsense.--RDBury (talk) 06:04, 2 September 2009 (UTC)

That sentence in the article is written very badly. It implies that each cell contains a specific value, and that each of these values is five more than the other. This is of course simply impossible. What is meant is that A1 contains the formula "B1+5" and B1 contains the formula "A1+5", which is possible but leads to an error when the spreadsheet tries to evaluate the formulae. Algebraist 10:46, 3 September 2009 (UTC)

Thanks for the explanations; I wasn't altogether sure that I wasn't missing something. Algebraist, your comment gets at it exactly. Nyttend (talk) 13:31, 3 September 2009 (UTC)

Continuation of the Method of Characteristics to solve PDEs

pma, thanks for the answer! I understand some things but not all so I have a couple of conceptual questions about the method itself if you (or anyone) be kind enough to help me. I state the PDE here just for completeness. $2tu_{t}+xu_{x}=0,\,u(x,0)=f(x)$

• First of all, you say that the PDE tells you that any solution defined on the initial manifold (the x-axis) will be a constant. Is it because the PDE is homogeneous? So that when we compare it to the complete derivative of u(x,t) we obtain

$u'(x,t)=u_{x}x'+u_{t}t'\Rightarrow u'=0,x'=x,t'=2t$

so u must be a constant? Is this how you concluded because u'=0?

• Second, what does "transverse" mean? I looked at the article but does it mean that the characteristics have to be at exactly right angles or they just have to be NOT tangent to the initial manifold?
• Third, how do you know that here the initial manifold is itself a characteristic line (the support of a characteristic)?
• Fourth, why does this method fail if the initial manifold is itself a characteristic line? Is there like a test (or something you look at) in the very beginning to determine if this method will work on this PDE or not so that one doesn't do all the work only find out at the very end that the initial condition cannot be satisfied?
• Fifth, what is the resolution in a case like this? Do I just say that the method of characteristics just cannot be used to solve this PDE? Can another method be used or does this (somehow) guarantee that a solution does not exists or if it does exists, it may not be unique? There a lot of confusion (in my head) regarding the vocabulary and what we really are "doing" when we use this method. So thank everyone!97.118.56.41 (talk) 02:56, 2 September 2009 (UTC)

1. Correct. Indeed, the computation you did proves that any solution is constant along any characteristic line. For the x-axis in particular, the argument could be somehow simplified this way: let $u(x,t)$ be a solution of class $C^{1}$ defined on $\textstyle \{(x,0):\,a<x<b\}$. Then the equation in these points becames $u_{x}(x,0)=0$, so $u(x,0)$ is constant.

2. Correct: as a particular case of a more general definition, a vector is transverse to a submanifold iff the submanifold has codimension 1 and the vector is not tangent to it. Notice that (answerring a subsequent question of yours) it is simple to test whether the characteristic lines are transverse to submanifold M. Here, it just means that at any point (x,t) of M, the vector (x,2t) is not tangent to M. You can check e.g. that $\mathbb {R} \times \{1\}$, and also the unit circle $\{(x,t):\;x^{2}+t^{2}=1\}$ enjoy the transversality condition.

3. Is it clear to you how to write the system of ODE for the characteristics lines? It is:

$\xi '(s)=\xi (s)$,

$\tau '(s)=2\tau (s)$.

You can check that these curves (more precisely, their supports) are a family of (half) parabolas with vertices in the origin, and the 4 semiaxes. Any solution $\textstyle u:\Omega \subset \mathbb {R} ^{2}\to \mathbb {R}$ has to be constant along each of them : precisely, along any arc interval of a characteristic line contained in its domain $\scriptstyle \Omega$. Note that since each characteristic springs off from the origin, it also follows that any solution defined on $\mathbb {R} ^{2}$ is necessarily a constant.

Also note that if the equation is not homogeneous, e.g., if you have 1 on the RHS, then for each solution of it you'll have $u(\xi (s),\tau (s))=s+c$ along a characteristic (the constant c depending on the characteristic). More generally, if you consider the semilinear version of your equation, that is

$\textstyle xu_{x}+2tu_{t}=a(x,t,u)$,

again the solutions are determined along any of the characteristics line above, as soon as you know the value of the solution in one of its points. The reason is that now the PDE restricted to the characteristic lines gives you an ODE for $v(s):=u(\xi (s),\tau (s))$, precisely

$v'(s)=a(\xi (s),\tau (s),v(s))$,

which is again enough to determine it completely from the value in one point. (Let's say that $\textstyle a(t,x,u)$ is at least Lipschitz, to have unicity for the Cauchy problem).

4. In particular, clearly, you can't prescribe arbitrarily the value of a solution at more than one point in each characteristic line; or, to be precise: in each arc interval of characteristic contained in the domain of the solution. On the contrary, for instance, $\mathbb {R} \times \{1\}$ instead of $\mathbb {R} \times \{0\}$ is OK, and in fact you can check that you can solve uniquely in $\mathbb {R} \times \mathbb {R} _{+}$ the PDE with the condition u(x,1)=f(x). Also, you can solve uniquely the PDE on $\mathbb {R} ^{2}\setminus \{(0,0)\}$ if you prescribe the values of a solution on the unit circle as initial manifold. Indeed, each arc of parabola in the domain meets (trasversally) this initial manifold exactly once. In general, if the transversality condition holds on the (codimension 1) submanifold M, thanks to a result of general topology, there exists a nbd U of M such that each arc of a characteristic line in U meets M exactly once (transversally). In this situation, each function $f$ of class C¹ on M extends uniquely to a solution of the PDE on U.

5. It should be clear that the original PDE can't be solved prescribing e.g. u(x,0)=f(x) for all x>0 if f is not a constant: it's not a limit of the method, but just a matter of fact. BTW, the method of characteristics can be generalized to fully nonlinear first order equations, that is $F(x,u(x),\nabla u(x))=0$ with $x\in \mathbb {R} ^{n}$.--pma (talk) 19:23, 2 September 2009 (UTC)

PS: I took the liberty of compressing a bit your post, to make it more readable.

Period

Can someone please help me find the period if cos[n²], where n can take only integral values ? I know the answer is 8, but I can't get the method. Rkr1991 ^{(Wanna chat?)} 13:04, 2 September 2009 (UTC)

cos(0) = cos(64) certainly does not hold, so the answer can't be 8... or don't you mean period as in Periodic function? —JAO • T • C 13:13, 2 September 2009 (UTC)

My mistake, the question should read cos[πn²]. Rkr1991 ^{(Wanna chat?)} 13:33, 2 September 2009 (UTC)

Let's see ... cos(0) = 1 ... cos(π) = −1 ... cos(4π) = 1 ... cos(9π) = −1 ... cos(16π) = 1 ... is that beginning to look periodic ? Gandalf61 (talk) 14:07, 2 September 2009 (UTC)

Is that comment supposed to be mocking me ? cos(0)=cos(64π)=1, cos(π)=cos(81π)=-1, and so on... The answer is 8, my question is how ? Rkr1991 ^{(Wanna chat?)} 09:25, 3 September 2009 (UTC)

No, the answer is not 8, as Gandalf's helpful comment shows. (Hmm - just looked at Periodic function - the definition of period there is somewhat defective.) AndrewWTaylor (talk) 09:39, 3 September 2009 (UTC)

No mocking intended - it was meant to be a hint. The correct answer to the question as you posed it is definitely not 8. Let's start again.

You have a function f(n) defined on the integers by f(n) = cos(πn²). A quick check shows that f(0) = f(2) = f(4) = 1, and f(1) = f(3) = f(5) = −1. This can be generalised to f(2m) = 1 and f(2m+1) = −1 - or, more concisely, f(n) = (−1)ⁿ. So we have f(n + 2) = f(n) for all n, and f(n + 1) ≠ f(n). Therefore the period of the function is 2.

You are correct to say that f(n + 8) = f(n), but this only shows that 8 is a multiple of the period. To conclude that the period was 8 you would have to eliminate all possible smaller periods. Gandalf61 (talk) 09:46, 3 September 2009 (UTC)

I'm sorry if I was too hasty, but still, how do you prove that the period is 2, I mean, apart from the method of observation ? Rkr1991 ^{(Wanna chat?)} 09:56, 3 September 2009 (UTC)

Assuming that you can use the fact that the period of cos is 2π, just consider what is the parity of n² for integer n. — Emil J. 11:17, 3 September 2009 (UTC)

Just as a random question, are all non-constant periodic functions essentially trig functions? 92.4.122.142 (talk) 10:15, 3 September 2009 (UTC)

At least pretty much all periodic functions that occur in reality (or engineering) can be written as an infinite sum of trig functions; see Fourier series. So in that sense, pretty much yes (but not strictly all periodic functions). —JAO • T • C 10:34, 3 September 2009 (UTC)

How about all "single valued continuous periodic funtions are essentiall trig functions"83.100.250.79 (talk) 18:50, 3 September 2009 (UTC)

Could you give an example of a periodic function that cannot be expressed as the sum of trig functions? 92.4.150.32 (talk) 22:19, 3 September 2009 (UTC)

f(x)=0 if x is rational, f(x)=1 if x is irrational. So for example, f(x+1)=f(x) for all x, satisfying the definition of a periodic function. 67.122.211.205 (talk) 22:38, 3 September 2009 (UTC)

Or a simpler example is f(x) = 1 if x is an integer and f(x) = 0 everywhere else. The Convergence of Fourier series article goes into some of the questions of when the Fourier series of a function f converges to f and when it doesn't, which is generally not an easy problem. You can usually expect that the Fourier series won't necessarily converge to the value of the function at a point of discontinuity. But for nicely behaved functions such as ones that are differentiable everywhere, the Fourier series converges at every point. Rckrone (talk) 01:16, 4 September 2009 (UTC)

Continuity is not enough. Most (in the sense of Baire category) continuous periodic functions are not sums of trig functions. Algebraist 08:44, 5 September 2009 (UTC)

Non-Archimedean fields

We haven't got much on this topic, I'm noticing.

I'm studying non-Archimedean fields, and I'm stuck on what I think is an embarrassingly simple question. The notes I'm working from begin by defining a non-Archimedean absolute value, which is a map from a commutative ring to the non-negative reals with the following 3 properties:

1. $\left|a\right|=0\Leftrightarrow a=0$
2. $\left|ab\right|=\left|a\right|\cdot \left|b\right|$
3. $\left|a+b\right|\leq \operatorname {max} \left\{\left|a\right|,\left|b\right|\right\}$

Here's my issue. To do some of the very basic first proofs, I find myself wanting to say that $\left|-a\right|=\left|a\right|$. I can prove that quite easily... if the ring has unity. Otherwise, how do I know that a ring element will be the same size as its additive inverse? I think I must be missing something quite obvious. Thanks in advance for any hints. -GTBacchus^(talk) 21:19, 2 September 2009 (UTC)

Yes sir: since $(-a)^{2}=a^{2}$ from 2 it follows $|-a|^{2}=|(-a)^{2}|=|a^{2}|=|a|^{2}$ thus $|-a|=|a|$ because they are both non-negative. --84.221.68.104 (talk) 21:52, 2 September 2009 (UTC)

Ah, duh! Thank you. I knew it had to be something simple. -GTBacchus^(talk) 07:32, 3 September 2009 (UTC)

Actually, thinking about it, I'm not sure why $(-a)^{2}=a^{2}$ if the ring hasn't got unity. On the other hand, I spoke with the professor who prepared the notes I'm studying, and he said that by "ring" he means unital ring. I suppose most people do. -GTBacchus^(talk) 10:06, 3 September 2009 (UTC)

ab=(-a)(-b+b)+ab=((-a)(-b)+(-a)b)+ab=(-a)(-b)+((-a)b+ab)=(-a)(-b)+(-a+a)b=(-a)(-b) Algebraist 10:41, 3 September 2009 (UTC)

Mmm, nice... The first step is really all I had to see; it's clear from there. Thanks, though. -GTBacchus^(talk) 11:19, 3 September 2009 (UTC)

"He said that by "ring" he means unital ring. I suppose most people do." Note however that the theory of non-unital rings is quite different to the theory of unital rings (so that deciding upon whether the ring should have unity is not merely a convention). For instance, by Zorn's lemma, any unital ring has at least one maximal (proper) ideal - simply consider the set of all ideals which do not contain a unit (invertible element); this set is non-empty since (0) is a member. If the ring does not have a 1, however, this argument fails since one cannot define a "unit" and therefore non-unital rings need not possess maximal (proper) ideals. In particular, various constructions relating to maximal ideals have to be altered in the case of non-unital rings; notable is the concept of the Jacobson radical or the notion of quasiregularity.

There are also several interesting theorems which guarantee the existence of a 1 in a (not necessarily unital) ring. For instance, a right artinian "ring without 1" with no non-zero nilpotent ideals must have a 1. This is interesting because it shows how one can "locate" a 1 in a ring through almost "indirect arguments."

Similarly, the theory of non-unital modules also requires special distinctions when translating theorems from the theory of unital modules. For instance, if one defines a simple (not necessarily unital) module, as a module with proper annihilator as well as having no non-trivial proper submodules, one can show certain basic results. One example of this is that such a module is necessarily isomorphic to the quotient of R (the ring over which the module is defined) by a maximal right ideal, M. Additionally, one can show that there is an r in R, with x − r · x an element of M for all x. For unital modules, this is trivial since r = 1 works; for non-unital modules this requires a little more basic work.

I know that I have probably been slightly pedantic in this argument but nevertheless I feel that it is worthwile that one knows of this important distinction. Quite a few authors (such as I. N. Herstein) do not assume the existence of an unity in their rings and study this chosen distinction. Anyway, I think that you get the idea now. --PST 13:12, 3 September 2009 (UTC)

So, I was reading about non-unital rings, and I saw that there seems to be a canonical way to adjoin a multiplicative identity. Would this be another solution to the problem of showing that abs(a) = abs(-1); i.e. adjoin a "1" element, and then proceed as in a unital ring? Somehow that seems too easy... -GTBacchus^(talk) 13:19, 3 September 2009 (UTC)

How do you extend the absolute value from the nonunital ring to the adjunction? — Emil J. 14:01, 3 September 2009 (UTC)

Ah, good point. What I should have asked is whether one could use the extension with unity to more easily show that $(-a)^{2}=a^{2}$. I'm not sure if there's a natural way to extend the absolute value. -GTBacchus^(talk) 14:38, 3 September 2009 (UTC)

Yes, you can use the extension to prove that (−a)² = a². More generally, the same argument shows that the theory of unital rings is conservative over the theory of not-necessarily-unital rings wrt universal formulas in the language $\langle +,0,-,\cdot \rangle$. However, it's not easier than a direct proof: checking that the extension is a ring is a lot more work than Algebraist's proof above (though it is no less elementary, just longer and more tedious); furthermore, you need to show that (−a) = (−1)a in a unital ring, which is more or less the same proof as showing (−a)b = −(ab) in a nonunital ring, and that already implies your identity (it gives also a(−b) = −(ab) by symmetry, hence (−a)(−b) = −(a(−b)) = −−(ab) = ab). — Emil J. 15:28, 3 September 2009 (UTC)

Thank you for that explanation. -GTBacchus^(talk) 16:12, 3 September 2009 (UTC)

September 3

Econometrics

Does anyone have any suggestions for a good undergraduate level econometrics text book?--98.240.70.102 (talk) 00:02, 3 September 2009 (UTC)

I used Gujarati's Basic Econometrics, a very user-friendly introductory book, if somewhat lacking up-to-date treatment of recent topics such as the asymptotic (large-sample) approach.
A more modern choice can be Woolridge's Introductory Econometrics. Pallida  Mors 18:28, 3 September 2009 (UTC)

Binomial Expansions

Resolved

I have to determine the expansion in powers of x up to $x^{4}$ of $(1-x^{3})^{6}(1-x)^{-6}$. Now I can do this by some simple division and then expanding $(1+x+x^{2})^{6}$ but is there any way of reaching the same answer by expanding the two sets of brackets separately? Thanks 92.4.122.142 (talk) 12:47, 3 September 2009 (UTC)

Yes, expand the first factor as $1-6x^{3}+...$ and the second as $1+6x+(-6)(-7)x^{2}/2+...$ Ignore powers of x above 4, and then multiply out the resulting expressions. Tedious, but it gives the same answer as your "simple division" method.Caution, expressions may contain typos. AndrewWTaylor (talk) 13:55, 3 September 2009 (UTC)

Exactly! See Newton's generalised binomial theorem. (r = – 6 is the case of the second bracket) ~~ Dr Dec (Talk) ~~ 17:56, 3 September 2009 (UTC)

Uniform convergence of exp(-x²)sin(x/n) over R

Hi there guys,

could anyone please suggest a test or approach to check whether convergence of $f_{n}(x)\,=\,e^{-x^{2}}\,\sin({\frac {x}{n}})$ to $f(x)\,=\,0$ over $\mathbb {R}$ is uniform? I've tried everything I could think of (not much sadly) such as checking that obviously both $f_{n}(x)$ and $f(x)$ are continuous, and trying to find a value for the maximum of $e^{-x^{2}}\,\sin({\frac {x}{n}})$, from which all I got was an ugly formula in xtan(x/n) (for x, not for f_n(x)), and I don't seem to be making any headway. I don't need to be walked through what to do, but if I could just get myself aimed in the right direction that'd be great (e.g. the name of a test or a property to look at) - thanks!

Spamalert101 (talk) 18:20, 3 September 2009 (UTC)

No special machinery is needed here. To make f_n bounded by epsilon, just choose M large enough that exp(-x²) is less than epsilon outside [-M,M] and then make n large enough that sin(x/n) is smaller than epsilon inside [-M,M]. Algebraist 18:28, 3 September 2009 (UTC)

Also, you may use these elementary inequalities for all $t\in \mathbb {R}$

$\textstyle |\sin(t)|\leq |t|;\;\mathrm {e} ^{t}\geq 1+t;\;2t\leq 1+t^{2}$ and obtain

$\textstyle |f_{n}(x)|\leq {\frac {1}{2n}}$ for all $x\in \mathbb {R}$.

However, the first answer also shows a general principle (uniform convergence on intervals, with domination by a function vanishing at infinity, implies uniform convergence over R).

--pma (talk) 22:09, 3 September 2009 (UTC)

Thanks guys, that's a great help - didn't realize it could be that simple! Spamalert101 (talk) 14:02, 5 September 2009 (UTC)

September 4

12 people in a family, all with birthdays in different months

Say there's a family of 12 people: 2 parents and 10 children. I'm told all the kids came from single births. Someone claims that none of the 12 have a birthday in the same month as any of the others, i.e. there's exactly one birthday for each of the 12 months. How likely is this claim to be true? -- JackofOz (talk) 21:26, 4 September 2009 (UTC)

As a rough estimate if you took 12 random people and assumed randomly distributed birthdays among the 12 months (which they aren't), the probability of them having one birthday per month is 12!/12¹², which is about 1 in 19000. That's not to say it's a lie. Unusual things do happen some of the time, and when they do there's typically disproportionate attention brought to them, which can make them seem like they happen more often then they rightfully should. I think there's probably a name for that effect but I don't know what it is. Rckrone (talk) 21:52, 4 September 2009 (UTC)

That is "how likely it is to be true" if the family were chosen uniformly from among all families with 12 children.

Rckrone wrote:

assumed randomly distributed birthdays among the 12 months (which they aren't)

Here's another instance of the use of "randomly" to mean "uniformly". This is an oddly persistent meme. I don't know why. Michael Hardy (talk) 22:09, 4 September 2009 (UTC)

What I meant to say was that 1/19000 is a rough estimate of the probability of this happening among 12 random people. I agree that the probability that the claim is true is much much higher. In fact I don't see any reason to think it isn't. As for misusing "randomly" I have no explanation. Rckrone (talk) 22:17, 4 September 2009 (UTC)

Suppose there are n families with 2 parents and exactly 10 children, and the 12 birthdays are uniformly distributed in the 12 months. The probability of at least one occurrence of one birthday for each of the 12 months is 1 - (1-12!/12¹²)ⁿ = 1 - 0.999946276783ⁿ. This passes 50% at n = 12902, and it is 99.999999999999999999999% at n = 1000000. I don't know how many such families there are and how far from uniform distribution the birthdays really are but it seems highly likely that there exists such an occurrence. PrimeHunter (talk) 23:33, 4 September 2009 (UTC)

Oh, I have no doubt it has happened many times, because it's obviously well within the realms of possibility. But if I were a betting man, and someone asserted this particular family had this characteristic, I was wanting to know whether it would be a better bet to lay money they were telling the truth, or they were telling a lie (or, to be charitable, they were mistaken). That may not be exactly what I originally asked, but it's what I meant. -- JackofOz (talk) 23:43, 4 September 2009 (UTC)

It would depend on the reliability of the person making the claim. If they knew the family and I didn't then I would be reluctant to offer a bet to them. If they accept the bet then it may be because they feel fairly confident they are right. Almost everybody knows some things which are rare, and rare things often draw attention and may be told many times. PrimeHunter (talk) 00:13, 5 September 2009 (UTC)

"the use of "randomly" to mean "uniformly". ... I don't know why". The article randomness explains that the word "random" refers to a discrete uniform distribution: "Governed by or involving equal chances for each of the actual or hypothetical members of a population". Mathematicians change the meaning of words to include special cases, such that a coin with two tails still produces a random variable even if the outcome is neither random nor variable, but predictable and constant. Bo Jacoby (talk) 06:43, 5 September 2009 (UTC).

The 12!/12¹² calculation assumes that the birthdays are independent of one another. I'd say that it would be hard work to establish/defend any statement of independence here. Robinh (talk) 09:20, 5 September 2009 (UTC)

That's certainly true (and even the assumption that the distribution is uniform measured by months is weird and not supported by the principle of indifference -- we know, after all, that some months are longer than others). But as a practical matter, to get a first approximation without doing too much work, it's a reasonable guess. --Trovatore (talk) 20:59, 5 September 2009 (UTC)

For a busy family like that, independence is the last thing I'd assume. Mom gets pregnant in month X, kid #1 is born 9 months later, Mom spends 2 months nursing and getting her parts back to normal, and at month X+11 gets pregnant again. Kid #2 is born at X+20, rinse, repeat. A new pregnancy every 11 months, and since 11 (months between pregnancy starts) and 12 (months in a year) are relatively prime, you get all the births in distinct months. See also Cheaper by the Dozen. 67.122.211.205 (talk) 07:49, 6 September 2009 (UTC)

• "Uniformly distributed" does not mean the same as discrete uniform distribution to most readers. Surely "randomly distributed" is less likely to be misinterpreted? Both are unlikely in these circumstances. Dbfirs 09:56, 6 September 2009 (UTC)

Canonical Forms for PDEs

Okay so working with 2nd order linear PDEs, I cam across the fact that there are canonical forms for different kinds of PDEs. For hyperbolic PDEs, the canonical forms are $u_{xx}-u_{tt}=f(x,t,u,u_{x},u_{t})$ or $u_{xt}=g(x,t,u,u_{x},u_{t})$. My question is why are there two canonical forms for hyperbolic PDEs? Is there a significance to this distinction? Thinking with analogies to conics sections, I think that we just "rotate" one form by 45 degrees to get another form. Why do we do that? Is one easier to solve in some cases than the other?--97.118.56.41 (talk) 22:37, 4 September 2009 (UTC)

This is a very nice question. There are some papers by Farid Tari, and some other co-authors, that address the classifications of implicit differential equations.

1. "Bifurcations of binary differential equations." Proc. Roy. Soc. Edinburgh Sect. A 130 (2000), 485–506. (With J. W. Bruce and G. J. Fletcher)
2. "Duality and implicit differential equations", Nonlinearity 13 (2000), 791–812. (With J. W. Bruce)
3. "Two-parameter families of implicit differential equations", Discrete Contin. Dynam. Systems, 13 (2005), 139–262.
4. "Two-parameter families of binary differential equations", Discrete Contin. Dynam. Systems, 22 (2008), 759–789.

Maybe these will help ~~ Dr Dec (Talk) ~~ 09:57, 5 September 2009 (UTC)

It's nothing special. As you know, both form are very common in applications. E.g. as you know, waves and oscillations are described by the first one. As you know, $u_{tx}=0$ is easier to solve, in fact immediate. And, as you know, one can easily pass from one form to the other. So, if you already know the answer, why are you asking? ;) (@Dec: the OP's eq. are semilinear hyperbolic equations, there's nothing implicit!)--pma (talk) 18:18, 5 September 2009 (UTC)

No, but it's a question of classification. Such classifications always follow similar lines. I thought it might have been of some interest to him/her. ~~ Dr Dec (Talk) ~~ 20:24, 5 September 2009 (UTC)

September 5

The Sorgenfrey plane

Ok I know how the usual proof of non-normality goes, but I've been thinking about something that smells fishy to me. Take the line y=-x. The proof goes by showing the set of all rational points and irrationals can't be separated by open sets using Baire's category theorem. But what if we do this: for every rational p/q, cover the point on the line with [p/q,p/q+1/(q^q)) X [-p/q,-p/q+1(q^q)), where X denotes cartesian product. Notice how this half open square is VERY small. It certainly looks plausible that every irrational has a half open square (with the irrational at the bottom left corner) that's disjoint from the union of the rational squares, because every irrational is surrounded by a cloud of rationals with very large denominator. I've been trying to show this is not possible but it seems to have something to do with approximation of irrationals by rationals, continued fractions maybe (which I don't know much about). What am I doing wrong? Breath of the Dying (talk) 09:00, 5 September 2009 (UTC)

Eh, only quadratic irrationals like square roots can be approximated by continued fractions. (At least if you mean the conventional type with only 1's as numerators.) Professor M. Fiendish, Esq. 09:17, 5 September 2009 (UTC)

That is not true. Algebraist 12:16, 5 September 2009 (UTC)

You were probably thinking about the fact that quadratic irrationals have a repeating continued fraction expansion. -- Meni Rosenfeld (talk) 17:15, 5 September 2009 (UTC)

That particular scheme will fail for irrational points that can be well-approximated by rationals. For example, setting $c=\sum _{j=1}^{\infty }10^{-j!}$, any neighbourhood of (c,-c) intersects your neighbourhood of the rational points. Algebraist 12:22, 5 September 2009 (UTC)

As to the first question, as you recalled: in this topology, the closure of any neighborhood of the closed set $\{(r,-r):r\in \mathbb {Q} \}$, meets the anti-diagonal in uncountably many points. If this is OK for you, I really don't see why it is not OK for you that in particular, the nbd you wrote has this property.--pma (talk) 17:58, 5 September 2009 (UTC)

Great retrosnub icosidodecahedron

Great retrosnub icosidodecahedron.

I keep seeing conflicting vertex figures for this one.

1. Some sources say the vertfig is 3.³/₂.3.⁵/₃.3.
2. Other sources say it is ³/₂.³/₂.³/₂.⁵/₃.³/₂.
3. And still others say it is ^{(3.3.3.3.5/₂)}/₂.

Which is right?

On Wikipedia, the vertfig shown is ^{(3.3.3.3.5/₂)}/₂, but application of the Wythoff symbol |³/₂ ⁵/₃ 2 gives 3.³/₂.3.⁵/₃.3. No idea where ³/₂.³/₂.³/₂.⁵/₃.³/₂ came from. Professor M. Fiendish, Esq. 09:04, 5 September 2009 (UTC)

Just going by the vertex figure image there, which I'm assuming is correct, it looks to be a pentagram with the faces being four triangles and one pentagram so I guess the problem is how do you represent that with the notation? Just listing the 5 faces would be ambiguous with the case where they're arranged in a pentagon, so wikipedia handles that with the "/2" at the end to represent 5/2 instead of 5. It looks like another way is to specify the orientation of the faces relative to the vertex. 3/2 is a triangle but connecting the vertices in the opposite order, and similarly 5/3 is an inverted pentagram. I'm not really sure on how the orientations dictate the shape of the vertex figure, but it could be that those other two versions you mentioned both represent the same thing. Vertex configuration seems to explain. Rckrone (talk) 18:04, 5 September 2009 (UTC)

Uniform Continuity of sin(x³)/(x+1)

Hi again guys, sorry about the influx of analysis from me on here, I'm working through a bundle of questions and saving the ones which really stump me for here! I'm trying to work out whether ${\frac {\sin(x^{3})}{x+1}}$ is uniformly continuous over $[0,\infty )$ and I have no idea how to begin. The fact sin(x³) has a derivative tending to infinity leads me to think it might not be, but then the function decreases in magnitude to 0 too which may affect things, and I'm unsure even what answer to expect, let alone how to proceed.

Any help would be greatly appreciated!

Thanks a lot, Spamalert101 (talk) 14:32, 5 September 2009 (UTC)

It's pointwise continuous, and tends to a limit as x tends to infinity, so it's uniformly continuous. This is a variant of the Heine–Cantor theorem. This is very similar to your last question: in both cases we have a function with good local behaviour that is badly behaved at infinity, so multiplying it by a function that decays to 0 at infinity gives good global behaviour. Algebraist 14:54, 5 September 2009 (UTC)

And to make it quantitative, you may try and prove that your function is Hölder continuous of exponent 1/2: for instance, $\scriptstyle |f(x)-f(y)|\leq 5{\sqrt {|x-y|}}$ holds true for all $x$ and $y$ in $\scriptstyle [0,\infty )$. --pma (talk) 20:50, 5 September 2009 (UTC)

The restriction to any bounded interval [0, a] is a continuous function on a compact set and so is uniformly continuous, and the restriction to [a, ∞] only takes values near 0, so it can't vary by more than ε if a is big enough. Michael Hardy (talk) 20:58, 5 September 2009 (UTC)

Properties of an isosceles triangle

Let ABC be triangle with AB=AC=6 cm . If the circumradius of the triangle is 5 cm, find the length of BC . —Preceding unsigned comment added by 117.98.100.98 (talk) 16:44, 5 September 2009 (UTC)

This looks a lot like a homework problem. I will give you this hint, though: if you draw the triangle with its circumscribed circle, you will notice that the line between the vertex A of the triangle and the center of the circle is perpendicular to the side BC of the triangle. When you get your answer and punch it into your calculator (which you shouldn't do until the last step), the 3rd and 4th digits after the decimal should be 6 and 9 respectively, assuming I did my math correctly.

Note that my approach was thrown together from very faint recollections from my geometry class many years ago, and may not be the way your teacher is expecting you to solve it. --COVIZAPIBETEFOKY (talk) 18:12, 5 September 2009 (UTC)

"...the 3rd and 4th digits after the decimal should be 6 and 9 respectively, assuming I did my math correctly." Of course, I found a mistake in my work. The answer after correcting my work still includes both of those dirty digits, though; just not in the same place. --COVIZAPIBETEFOKY (talk) 18:20, 5 September 2009 (UTC)

A hint: Use Heron's formula. Bo Jacoby (talk) 18:22, 5 September 2009 (UTC).

There certainly may be something I'm missing here, but I'm pretty sure this leads nowhere. The area by Heron's formula and the area given by (1/2)base*height are exactly the same expressions, so setting those two equal certainly doesn't really help at all. Was there something else you had in mind? --COVIZAPIBETEFOKY (talk) 18:36, 5 September 2009 (UTC)

You can use law of sines and the area formula Δ=ab*sinC/2 to relate the circum-diameter to the area. Anyway, Circumscribed circle lists a formula that can be applied directly to this problem. Rckrone (talk) 18:54, 5 September 2009 (UTC)

Here is a simple approach to the problem, based on properties of triangles you should have already seen in school.

• Calling the center of the circumcircle O, note that the line AO bisects BC at a right angle, at say point D (see isosceles triangle).
• Note that triangle AOB is also an isosceles triangle, and the angle bisector of angle AOB bisects the line AB at right angles, at point E. Can you determine the lengths of sides AO, OB, AE, EB and OE ?
• Now what can you say about triangles AOE and ABD ? How do you use that to calculate length of side DB and hence, BC ?

Hint: Look up similar triangles and pythagoras theorem. Also, no calulator is required to solve this problem! 98.220.252.228 (talk) 18:46, 5 September 2009 (UTC)

Triangle ABO has sides 5,5,6. s=(5+5+6)/2=8. Area according to Heron is √(8·3·3·2)=12. This is = (1/2)base*height = (1/2)·5·(|BC|/2). Solve to get |BC|=48/5=9.6 . This, COVIZAPIBETEFOKY, is what I had in mind. Bo Jacoby (talk) 20:54, 5 September 2009 (UTC).

I'm guessing that by "isoscale" you mean "isosceles". Remember that if angles α and β are opposite sides a and b respectively, then

${\frac {a}{\sin \alpha }}={\frac {b}{\sin \beta }}={\text{circumdiameter}}$

(see law of sines). So you have

${\frac {6}{\sin \beta }}=10,$

and hence

$\sin \beta ={\frac {3}{5}}.$

Since two angles are β and one is α and they add up to a half-circle, that gives you α. Then you've got

${\frac {a}{\sin \alpha }}=10$

so you can find a. Michael Hardy (talk) 21:11, 5 September 2009 (UTC)

...and I just noticed the arithmetic is really simple: you don't need numerical approximations from a calculator, since sin β = 3/5. Consequently cos β = √(1 − (3/5)²) and as luck would have it, we've got a Pythagorean triple and so cos β = 4/5 exactly. Therefore sin(β + β) = 2 sin β cos β = 2(3/5)(4/5) = 24/25. Since α + β + β = half-circle, we must have sin α = sin(β + β). So that sine is 24/25. SSo we get a = 48/5 exactly. Michael Hardy (talk) 21:23, 5 September 2009 (UTC)

How nice that Michael's and mine seemingly different roads lead us to exactly the same result. Bo Jacoby (talk) 06:09, 6 September 2009 (UTC).

35 should have been 25. -- Meni Rosenfeld (talk) 07:14, 6 September 2009 (UTC)

In my elementary school times I formed the opinion that almost all rectangular triangles have edges in the ratio (3:4:5). --pma (talk) 21:33, 6 September 2009 (UTC)

Dense in C

Resolved

--Shahab (talk) 19:41, 6 September 2009 (UTC)

I want to show that $\mathbb {Z} [{\frac {i}{2}}]$ is dense in $\mathbb {C}$, the complex plane. By $\mathbb {Z} [{\frac {i}{2}}]$ I mean the set of all evaluations of polynomials with integer coefficients at ${\frac {i}{2}}$ (where $i=(0,1)$). How can I show that any open ball around an arbitrary z contains infinitely many such evaluations?--Shahab (talk) 19:48, 5 September 2009 (UTC)

Let r be an arbitrary positive real number. We need to show that you can find a polynomial that evaluates to a number within a distance of r from z. Choose integer n such that ${\frac {1}{2^{n}}}<{\frac {r}{\sqrt {2}}}$ and n=0 (mod 4). Then choose integers p and q such that ${\frac {p}{2^{n}}}+{\frac {iq}{2^{n+1}}}$ is a close to z as possible. That number is $pX^{n}+qX^{n+1}$ evaluated at i/2 and is less than r from z. --Tango (talk) 20:34, 5 September 2009 (UTC)

Thanks. I have a doubt though. How do we choose p and q such that ${\frac {p}{2^{n}}}+{\frac {iq}{2^{n+1}}}$ is as close to z as possible? I understand choosing n by the Archimedian property. Similarly what property do we use when choosing p and q?--Shahab (talk) 20:55, 5 September 2009 (UTC)

To get p, say, we take the integer part of the real part of z, call it a. Then test a2ⁿ+b for b in {0,1,...,2ⁿ-1} and choose the closest. It is a finite number of trials, so you will definitely find a closest. Do the same for q and the job is done. --Tango (talk) 21:01, 5 September 2009 (UTC)

Nice. You may also say: p and q such that 2p+iq is as close as possible to 2ⁿ⁺¹z (a minimum distance that is certainly not greater than $\scriptstyle {\sqrt {5}}/2$, &c) However as soon as you've observed that Z[i/2] contains all numbers p/2ⁿ+iq/2ⁿ it's done.--pma (talk) 21:12, 5 September 2009 (UTC)

Yes, that is pretty much equivalent. --Tango (talk) 21:42, 5 September 2009 (UTC)

Thank you both.--Shahab (talk) 19:41, 6 September 2009 (UTC)

Divisor of a Riemann surface

Can someone please explain what is meant by a divisor on a compact connected Riemann surface? I cannot find the relevant Wikipedia article (rather unusual). 86.210.207.133 (talk) 23:17, 5 September 2009 (UTC)

Divisor (algebraic geometry) is the relevant article, I think. --Tango (talk) 23:44, 5 September 2009 (UTC)

Grrrrr, no mention of Riemann surfaces... 86.210.207.133 (talk) 01:15, 6 September 2009 (UTC)

Any compact Riemann surface is a projective variety. --PST 03:57, 6 September 2009 (UTC)

In this case (because Riemann surfaces have complex dimension 1) a divisor is just a function $D\colon X\to \mathbb {Z}$ with discrete support, ie for a compact Riemann surface it's just a finite (formal) linear combination of points on the Riemann surface. --Xedi^Talk 04:54, 7 September 2009 (UTC)

Function elements

Is there a Wikipedia page concerning function elements?

Mathworld has one... http://mathworld.wolfram.com/FunctionElement.html —Preceding unsigned comment added by 86.210.207.133 (talk) 23:33, 5 September 2009 (UTC)

Have you tried searching Function elements? Intelligentsium 02:27, 6 September 2009 (UTC)

September 6

Evaluating a probability estimator

Hi. I have random variables X and Y, where X is nominal and Y takes values of 0 and 1. I have n observations of the form $(X_{i},Y_{i})\,\!$, where several observations may share the same X. I denote $p=\mathrm {Pr} (Y=1|X)\,\!$ (so p is a random variable which becomes determined when given X), and ${\hat {p}}=f(X)$ is an estimator for p. I want to know how good this estimator is - specifically, I want to approximately know $\mathbb {E} ((p-{\hat {p}})^{2}).$ Is there a standard way of doing this?

I think I have reduced the problem to that of estimating $\mathbb {E} (p^{2})$, in case this is any simpler. Thanks. -- Meni Rosenfeld (talk) 16:42, 6 September 2009 (UTC)

Actually I had an idea for this, but originally I thought it would be computationally intractable. After looking into it I see it's not that bad. The idea is to assume that p follows the beta distribution, find the MLE of the distribution parameters given the data, and calculate the mean and variance for those parameters. Does this sound reasonable? Is there a better way? -- Meni Rosenfeld (talk) 18:06, 6 September 2009 (UTC)

You might be interested in the following article: http://psychology.wikia.com/wiki/Inferential_statistics . Bo Jacoby (talk) 20:08, 6 September 2009 (UTC).

Automorphisms of Z[x]

What are all the ring automorphisms of $\mathbb {Z} [x]$? I believe Z has only one: the identity. What is the general procedure of approaching the problem of determining Aut(R)? Thanks.--Shahab (talk) 22:01, 6 September 2009 (UTC)

Yes, the (unital) ring automorphism is determined by where you send x. x needs to be in the image, so x has to go to something of degree 1. I suspect it can go to ±x+n for any n in Z, and to no other place. If so, this is the infinite dihedral group, represented as the matrices { [ ±1, n ; 0, 1 ] : n in Z } under matrix multiplication. In general finding the automorphism group of fundamental, innocuous rings can be the source of long-standing open problems, such as (the strong) Nagata conjecture and the description of the Cremona group. I wrote a short thing on this a while back (the enwiki article is just a sentence). JackSchmidt (talk) 22:47, 6 September 2009 (UTC)

Well, it's clear the leading coefficient of the image of x must be ±1. If it is ax + n, then a poly in the image of the automorphism can be written as aQ(x) + constant, so it can't be surjective unless a is a unit.John Z (talk) 00:30, 7 September 2009 (UTC)

September 7

Latex align* multiple points of alignment.

Resolved

I am trying to type up a displayed equation for which I want to have multiple points of alignment, one at equals and one at the point where I give reasons for each line. I have tried doing it like this:

  \begin{align*}
     \frac{x}{y} + \frac{a}{b} &= xy^{-1} + ab^{-1}  & ( \text{by definition})\\
        &= xy^{-1}1 + ab^{-1}1  & (\text{where 1 is the multiplicative identity})\\
        &= xy^{-1}bb^{-1} + ab^{-1}yy^{-1}  & (\text{inverse property of mulitplication } bb^{-1} = yy^{-1} = 1) \\
  \end{align*}

But the problem is that the reasons get right aligned whereas I want them to be left aligned, preferable such that there is at least a \quad spacing between the equation and the reason. Is there any way to do this? AMorris (talk)●(contribs) 04:41, 7 September 2009 (UTC)

The default alignment for the align environment is left right left right... so it alternates left and right. You can just put two & before the text instead of one, making it align to the left. --Xedi^Talk 05:56, 7 September 2009 (UTC)

To expand on Xedi's answer:

  \begin{align*}
     \frac{x}{y} + \frac{a}{b} &= xy^{-1} + ab^{-1}  && ( \text{by definition})\\
        &= xy^{-1}1 + ab^{-1}1  && (\text{where 1 is the multiplicative identity})\\
        &= xy^{-1}bb^{-1} + ab^{-1}yy^{-1}  && (\text{inverse property of mulitplication } bb^{-1} = yy^{-1} = 1) \\
  \end{align*}

The idea is that the &s alternate between being /alignment markers/ and /column separators/. So the first & means "align the first column at the = sign". The second & delimits the end of the first column and the start of the second column. The third & means "align the second column at the (". And so on. Eric. 67.169.125.37 (talk) 07:57, 7 September 2009 (UTC)

Perfect - that sounds like it will do exactly what I wanted. It was frustrating because I couldn't find this documentation anywhere when I looked up how align works. Thanks guys! AMorris (talk)●(contribs) 12:05, 7 September 2009 (UTC)

Reduction to Canonical Form

Hello, I have been trying to reduce $2u_{xx}+u_{xy}+yu_{yy}=0$ in the region y>1 for the last couple of hours but not getting anywhere. In this region, the PDE is elliptic. Letting a=2, b=1, and c=y, the discriminant is negative. So I set

${\frac {dy}{dx}}={\frac {b+{\sqrt {b^{2}-4ac}}}{2a}}={\frac {1+{\sqrt {1-8y}}}{4}}={\frac {1+i{\sqrt {8y-1}}}{4}}$

and (I think) that I am supposed to solve this differential equation and then let $\phi (x,y)$ be the real part of the solution and let $\psi (x,y)$ be the imaginary part and then I will have my change of coordinates which will reduce this PDE to the canonical form (the Laplacian equals some function). But the problem is that when I try to do this integral I get

$-i{\sqrt {8y-1}}+\ln(-i+{\sqrt {8y-1}})$

which I can't separate into the real and imaginary part. And then when I tried to integrate the real and the imaginary part separately (which is correct I hope) and use the results as the transformation, it doesn't work. I let ${\frac {dy}{dx}}=1/4$ giving me $\phi (x,y)=4y-x$ and then ${\frac {dy}{dx}}={\frac {\sqrt {8y-1}}{4}}$ giving me $\psi (x,y)={\sqrt {8y-1}}-x$. The mixed partials don't go away and also it is a nightmare when I try to invert it. What am I doing wrong? Is there an easier way of doing this or am I making some stupid mistake? Thanks! 97.118.56.41 (talk) 07:24, 7 September 2009 (UTC)

You said that you can't seperate into the real and the imaginary part. Well, if

$z:=-i{\sqrt {8y-1}}+\ln(-i+{\sqrt {8y-1}})$

as you say then we can split this into it's real and imaginary parts:

$\Re (z)=\ln \left(\left|-i+{\sqrt {8y-1}}\right|\right)+\Im \left({\sqrt {8y-1}}\right)\ ,$

$\Im (z)={\mbox{Arg}}\left(-i+{\sqrt {8y-1}}\right)-\Re \left({\sqrt {8y-1}}\right)\ .$

Now, if we assume that y > 1 (in fact y > ⅛ would do) then we get

$\Re (z)=\ln \left(2{\sqrt {2y}}\right)\ ,$

$\Im (z)=-{\sqrt {8y-1}}-\arctan \left({\frac {1}{\sqrt {8y-1}}}\right).$

(I've ignored the multi-valuedness and just worked with principal values) ~~ Dr Dec (Talk) ~~ 09:59, 7 September 2009 (UTC)

FRACTALS

210.212.239.181 (talk) 11:24, 7 September 2009 (UTC)harshagg

I want to know about fractals and their application and how to design them.I am just a beginner in this field even in wikipedia itself i can know about only their theory not use and i don't know can be that i couldn't understand what they say or mean to say.Pls help me.If its not against rule i can give my e-mail id so that u can contact me at for longer period of time.

There are many different types of fractals. Here at Wikipedia we have a general article on fractals, articles on certain types of fractals such as Julia sets, De Rham curves and iterated function systems, and even articles on individual notable fractals, such as the Mandelbrot set and the Koch snowflake. Our list of fractals by Hausdorff dimension gives a list of about 70 different individual fractals. The index of fractal-related articles gives a longer list of our articles about fractals. Gandalf61 (talk) 11:50, 7 September 2009 (UTC)

One nice fractal with a fairly elementary construction and plenty of (mathematical) applications is the Cantor set. -GTBacchus^(talk) 12:11, 7 September 2009 (UTC)

If you have MS Windows you might also like to try out the Fractint program. Dmcq (talk) 14:06, 7 September 2009 (UTC)

Irreducible representations of signed permutations (double young diagrams?)

Hello,

Young tableau describes how one can use Young tableaux to get all irreducible representations of the symmetric group. Another good text on that can be found here [2]

The symmetric group on $n+1$ symbols is precisely the finite Coxeter group $A_{n}$.

The Coxeter group $B_{n}$ consists of the $2^{n}n!$ signed permutations on the set $\{1,....n\}\cup \{-1,...,-n\}$ I heard that that group has a similar construction for its irreducible representations, using something like "double young diagrams". I can find very little on that on the internet. Can anyone tell me a bit more about this (even a few names that might help my search could be useful).

I am also interested in which irreducible characters appear in the decomposition of the characters of $B_{n}$, induced by parabolic subgroups. For $A_{n}$ this is completely known, and the coefficients are known as Kostka numbers.

Many thanks, Evilbu (Talk) unsigned by Evilbu

You seem to have made a series of statements; without posing a question. Are we to assume that your question is "Which are the irreducible characters appearing in the decomposition of the characters of $B_{n}$, induced by parabolic subgroups?" ~~ Dr Dec (Talk) ~~ 12:27, 7 September 2009 (UTC)

My sincere apologies. My main question is: how to describe the irreducible representations of that group $B_{n}$? My secondary question is: which of these appear (and if possible: how often) in the decomposition of the characters of $B_{n}$, induced by parabolic subgroups? Evilbu (Talk)

I don't know of any particularly special treatment of Bn, but the chapters on wreath products in Kerber's Representations of Permutation Groups is where I learned how to deal with Bn.

• Kerber, Adalbert (1971), Representations of permutation groups. I, Lecture Notes in Mathematics, Vol. 240, Berlin, New York: Springer-Verlag, doi:10.1007/BFb0067943, MR0325752
• Kerber, Adalbert (1975), Representations of permutation groups. II, Lecture Notes in Mathematics, Vol. 495, Berlin, New York: Springer-Verlag, doi:10.1007/BFb0085740, MR0409624
• Young, Alfred (1930), "On Quantitative Substitutional Analysis 5", Proceedings of the London Mathematical Society. Second Series, 31, doi:10.1112/plms/s2-31.1.273, ISSN 0024-6115, JFM 56.0135.02

The chapters in question are (Kerber 1971, Ch. 1 doi:10.1007/BFb0067945, Ch. 2 doi:10.1007/BFb0067946) and (Kerber 1975, Ch. 1 doi:10.1007/BFb0085742). The Coxeter group of type Bn is also known as the hyperoctahedral group, and is originally discussed in (Young 1930). JackSchmidt (talk) 16:35, 7 September 2009 (UTC)

A^n*B^n ≠ (A*B)^n for matrices (get around)

In the following series. F=Fo(A^n*B^n*(1-B)^0+Bicof*A^(n-1)*B^(n-1)*(1-B)^1......Bicof*A*B*(1-B)^(n-1)+(1-B)^n)

Bicof =(Binomial coeficient)

I would like to simplify this to Fo(AB+(1-B))^n. Using the binomial theorem

However A and B are both Matrices so A^n*B^n ≠ (A*B)^n. A is a upper triangular matrix and B is a diagonal matrix, with all elements less than 1.

Once that is done the function can be further simplified using the limit form of the exponential function and manipulating B.

Any ideas would be useful. The problem arose from tedious crusher breackage formulas. Startibartfast —Preceding unsigned comment added by 41.240.93.114 (talk) 15:09, 7 September 2009 (UTC)

This question has me totally blown away. What does Fo mean? What exactly does the binomial coefficient of a matrix-binomial power expansion mean? For example, for square matrices M and N we have: (M + N)² = M² + MN + NM + N². The binomial coefficients don't appear, correct me if I'm wrong, until we have a commutative algebra. ~~ Dr Dec (Talk) ~~ 16:05, 7 September 2009 (UTC)

I'd guess the OP wants something for binomials like the exponential map formula for Lie groups, but that's fairly hairy. There's an inversion formula for a sum of two matrices also which is probably somewhere on wiki but I don't think it would help either. Dmcq (talk) 16:53, 7 September 2009 (UTC)

Diagonal matrices commute with everything, don't they? Doesn't that mean that, in this case, A^n*B^n = (A*B)^n? --Tango (talk) 16:58, 7 September 2009 (UTC)

I thought that for a moment too, but it's not true:

$\left({\begin{array}{cc}1&2\\3&4\end{array}}\right)\left({\begin{array}{cc}1&0\\0&2\end{array}}\right)=\left({\begin{array}{cc}1&4\\3&8\end{array}}\right)$

$\left({\begin{array}{cc}1&0\\0&2\end{array}}\right)\left({\begin{array}{cc}1&2\\3&4\end{array}}\right)=\left({\begin{array}{cc}1&2\\6&8\end{array}}\right)$

~~ Dr Dec (Talk) ~~ 17:06, 7 September 2009 (UTC)

Scalar matrices commute with everything. Algebraist 17:13, 7 September 2009 (UTC)

Ah, yes. Sorry. --Tango (talk) 17:20, 7 September 2009 (UTC)

A topological counterexample

Does there exist a compact T₂ first-countable separable space which is not second-countable? I would especially like it if the example was totally disconnected (i.e., does there exist an uncountable Boolean algebra B such that every ultrafilter in B is countable generated, and B − {0} is the union of countably many ultrafilters?) The Sorgenfrey line almost works, but it's not compact, only Lindelöf. — Emil J. 17:24, 7 September 2009 (UTC)