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August 14

Infinity--Question/comment/request for comments

Gauss quote from Actual infinity: "I protest against the use of infinite magnitude as something completed, which is never permissible in mathematics. Infinity is merely a way of speaking, the true meaning being a limit which certain ratios approach indefinitely close, while others are permitted to increase without restriction." In Kenneth Kunen's book on forcing, I recall he said roughly proper classes are not defined in the most popular theory ZFC but are frequently talked about in an informal way and used as an aid to intuition. That is pretty akin to being not permissible, to being a manner of speaking, as Gauss described actual infinity. That makes me suspect that although mathematicians have genuinely tackled infinity and have some really big infinite sets like the reals and more, Aren't we confronting the same issue Gauss had? That is, we've tried to deal with his issue by grappling with infinite sets but ran into it again in proper classes?Rich (talk) 06:47, 14 August 2009 (UTC)

I'm not sure what the question is. What are you hoping for? Some comment that's better than what's already at that article? The reference desk isn't really a discussion board though sometimes it might seem otherwise. Dmcq (talk) 13:19, 14 August 2009 (UTC)

ZFC doesn't axiomatize classes but other set theories like NBG do. ZFC itself postulates the existence of certain collections of objects (called "sets") that obey a certain bunch of axioms (the ZFC axioms). It turns out that any model of ZFC must necessarily also contain some collections of objects (e.g. the ordinals in the model) that can't follow the axioms without leading to contradiction. This was considered paradoxical for a while (Burali-Forti paradox) but it just means that not every collection is a set. 70.90.174.101 (talk) 01:24, 15 August 2009 (UTC)

Well, actually, from a realist point of view, it means a bit more than that. The class of all ordinals is not a completed totality at all (if it were, it would have to be a set). So when we speak of "the class of all ordinals", what we're actually referring to, via a metalingustic circumlocution, is the predicate "is an ordinal", rather than a collection of any sort. --Trovatore (talk) 01:43, 15 August 2009 (UTC)

Responding to the original question: It seems to me that what you are talking about is the notion of the absolute infinite. As I understand it (and this is sort of a reconstruction from a modern point of view, one for which I don't exactly have clear references to point you to), you can look at it like this: Gauss says, "you can't treat infinite collections as being actual". Cantor says "you're right, but many of the things that were formerly thought of as being infinite, say the set of all natural numbers, are not "infinite" in that sense. They are transfinite; that is, beyond some limit, but not infinite or entirely without limit.".

See also limitation of size. Our article on that could use some (or a lot of) work. I started it but never got around to doing the literature search to really bring it up to snuff. --Trovatore (talk) 01:51, 15 August 2009 (UTC)

• Maybe Gauss had a notion of absolute infinite, since he was very smart and didn't publish many of his brilliant ideas? If he did maybe that means his quote was a denial of what he really thought, to avoid controversy, sort of like what some historian claimed was Gauss's lack of courage for not publishing his thoughts on noneuclidean geometry, though I've never heard that he actually REJECTED noneuclidean geometry.(I don't know if the claim of lack of courage is fair, since Gauss was a perfectionist and didn't publish things he hadn't had time to polish.) But if Gauss didn't have a notion of absolute infinity, then I think his quote must mean he rejected all quantities that weren't finite, that no infinity could be actual, really exist, like a zfcer would say some classes could not exist as sets, and only as collections"in a manner of speaking." Also, do we know if Gauss thought in terms of "cardinality of sets" for measuring quantity? Thanks to both of you for your thoughtful answers.Rich (talk) 15:21, 16 August 2009 (UTC)`

Regifting Robin

http://www.regiftable.com/regiftingrobinpopup.html

What's the secret behind this little game? --Halcatalyst (talk) 13:54, 14 August 2009 (UTC)

The same as it was last time. Algebraist 13:57, 14 August 2009 (UTC)

Sometimes, though, it comes up with something other than "board game." --Halcatalyst (talk) 16:29, 14 August 2009 (UTC)

When you take a 2-digit number and subtract both digits, you always get a number divisible by 9. So all that game has to do is label all the numbers divisible by 9 with the same item. --COVIZAPIBETEFOKY (talk) 17:58, 14 August 2009 (UTC)

Take a number x, it is made up of digits a, and b. in other words x = a * 10 + b.

now, take x and subtract a and b to get y: y = x - a - b.

Substitute values for x: y = (a * 10 + b) - a - b. y = a * 10 - a + b - b. y = a * 9.

if the possible values for a are (0-9) then the possible values for y are 0,9,18,27,36,45,54,63,72,81. If you look, you'll see that all of those positions have the same item. 148.134.37.3 (talk) 18:42, 20 August 2009 (UTC)

Showcase Showdown strategy

When should contestants spin again or stay to have the best chance of winning? I guess it would be different depending on whether you go first or second, and whoever goes third doesn't have to worry about it since they are just trying to beat the best score. Recury (talk) 17:29, 14 August 2009 (UTC)

What's the difference?

My statistics text book says this:

Suppose we calculate from one sample in our battery example the following confidence interval and confidence level: "We are 95 percent confident that the mean battery life of the population lies within 30 and 42 months' This statement does not mean that the chance is 0.95 that the mean life of all our batteries falls within the interval established from this one sample. Instead, it means that if we select many random samples of the same size and calculate a confidence interval for each of these samples, then in about 95 percent of these cases, the population mean will within that interval.

My question is - don't the two statements that are bolded mean the same thing? That is, doesn't one imply the other? What's the difference between the two? I have been scratching my head for a long time over this but can't figure it out and I'm feeling extremely stupid now =/ --ReluctantPhilosopher (talk) 20:47, 14 August 2009 (UTC)

I think that "We are 95 percent confident that the mean battery life of the population lies within 30 and 42 months" is talking about the accuracy of the calculation of the mean value. The statement "...the chance is 0.95 that the mean life of all our batteries falls within the interval established" is talking about the distrabution of the data. For example, let's say you tested one million batteries, and you found that 500,000 batteries had lives of 1 month and that 500,000 had lives of 71 months. In this case the mean battery life would be exactly 36, but none of the batteries would have a battery life within 30 and 42 months. ~~ Dr Dec (Talk) ~~ 22:19, 14 August 2009 (UTC)

---

95% of the time the confidence interval will contain the population mean. That means if you have independent repetitions of the experiment with the mean remaining the same throughtout, in 95% of cases that will happen. But you're looking at just one case, where you got the interval from 30 to 42. The conclusion that they're saying is not justified is that you can be 95% sure in that one case. The difference is that being 95% sure in that one case—that one repetition of the experiment—is not a statement that says in 95% of all repetitions of the experiment, a specific thing happens.

In fact, sometimes you may even find something in your data that tells you that the one specific repetition of the experiment is one of the other 5% of repetitions, where the specified method of finding an interval gives you an interval that fails to include the population mean. And sometimes you might find information in your data that doesn't tell you for sure that you've got one of the other 5%, but makes it probable. That only happens when you've got a badly designed method of finding confidence intervals, but nonetheless the 95% confidence level is correctly calculated. Ronald Fisher's technique of "conditioning on an ancillary statistic" was intended to remedy that problem.

The statement that in one particular repetition of the experiment, which gave you the interval from 30 to 42, the population mean has a 95% chance of being in that interval, is a statement about 95% of possible values of the population mean, not about 95% of repetitions of the experiment.

The argument that one should be 95% sure, conditional on the outcome of that one particular repetition of the experiment may actually be reasonable in cases where all of the information in the data was taken into account in forming the interval, but it's not actually backed up by the math involved. Something other than mathematics, not as well understood, is involved. Michael Hardy (talk) 22:33, 14 August 2009 (UTC)

Summary: One statement is about 95% of all independent repetions of the experiment. The other is about 95% of all equally (epistemically) probable values of the population mean, given the outcome of one particular repetition of the experiment. "95% of repetitions" is a relative frequency, not an epistemic probability. Michael Hardy (talk) 22:36, 14 August 2009 (UTC)

---

So, Michael, please tell us: what's the Wikipedia convention that made you put a line (like the one above) before your answer? ~~ Dr Dec (Talk) ~~ 22:53, 14 August 2009 (UTC)

You should not be feeling stupid. Your professor should. The concept of confidence interval is low quality science. The dispute between frequentist and bayesian statistics is behind this sad state of affairs. Bo Jacoby (talk) 11:56, 15 August 2009 (UTC).

I didn't want to indent at a different level from the previous comment but I want the boundary between the previous comment and mine to be clear. Michael Hardy (talk) 19:29, 15 August 2009 (UTC)

One of the problems you're facing is that the population mean is not a random variable. The mean lifetime of your batteries is a fixed number. We don't know what it is, but in repeated experiments it will never change. This raises issues with the the frequency view of probability. You can't really assign a (frequency based) probability to the population mean, as it's always exactly the same, no matter how you conduct your sample. Any discussion of probability with respect to the population mean would refer, rather, to our state of knowledge (or lack thereof) about the mean - a Bayesian or epistemic probability. The catch is that the standard confidence intervals were derived from frequency-based statistics. Bayesian statistics has it's own related measure, the credible interval, but the two are not necessarily equivalent. So formally, the "95%" has to refer to a frequency-based probability for a random variable: "If you carry out random sampling multiple times, 95% of the time the calculated confidence interval (the endpoints are random variables) will enclose the population mean." You can't say: "If you carry out random sampling multiple times, 95% of the time the population mean (NOT a random variable) will be within 30 and 42 months", because the population mean either is in that interval or it isn't. It doesn't jump around with repeated sampling. -- 128.104.112.102 (talk) 19:30, 15 August 2009 (UTC)

Lipschitz, absolute continuity

I'm working through problems in Royden today. Probablem 5.20b says:

Show that an absolutely continuous function f satisfies a Lipschitz condition if and only if |f'| is bounded.

This isn't true is it? I mean, |x| is Lipschitz but the derivative does not exist at x = 0 so it is not bounded. Something that doesn't exist can not be bounded. But, whenever it does exist, it is bounded. Would it be correct if it were changed to "if and only if |f'| is bounded whenever it exists."??? StatisticsMan (talk) 23:45, 14 August 2009 (UTC)

Yes; more precisely, recall that an absolutely continuous function f on an interval I is differentiable a.e. in I, so that f' is defined a.e.; then, f is Lipschitz if and only if f' is essentially bounded. If I is a bounded interval youu may also rephrase it saying that Lip(I) coincides with the Sobolev space W^1,∞(I). --pma (talk) 07:54, 15 August 2009 (UTC)

Okay, thanks. So, say f is absolutely continuous and Lipschitz. Then, if the derivative exists at a point a, it is $\lim _{x\to a}{\frac {f(x)-f(a)}{x-a}}$. By the Lipschitz condition, $-M\leq {\frac {f(x)-f(a)}{x-a}}\leq M$, so the limit is also within these bounds, when it exists. Now, assume f is absolutely continuous and the derivative is bounded, whenever it exists. I'll have to think about this one. Thanks! StatisticsMan (talk) 13:01, 15 August 2009 (UTC)

And, as to the first implication, remember that a Lipschitz function is in particular absolutely continuous. For the other implication, remember the generalization of the fundamental theorem of calculus in the setting of absolutely continuous functions. --pma (talk) 13:57, 15 August 2009 (UTC)

Okay, so this is pretty simple too. Assume |f'| <= M whenever it exists, which is almost everywhere since f is absolutely continuous on [a, b]. Also, the derivative is measurable and f is equal to the antiderivative of its derivative, $f(x)=f(a)+\int _{a}^{x}f'(t)\,dt$. Then, for any x < y in [a, b], we have

$|f(y)-f(x)|\leq \int _{x}^{y}|f'(t)|\,dt\leq \int _{x}^{y}M\,dt=M|y-x|$. StatisticsMan (talk) 15:08, 15 August 2009 (UTC)

Measurable functions/Derivatives

Let f : [0, 1] to R be a measurable function and E a subset of {x : f'(x) exists}. If m(E) = 0, show that m(f(E)) = 0.

This is a qual problem from the past that I have not seen a solution for. Any ideas? Thanks StatisticsMan (talk) 23:46, 14 August 2009 (UTC)

Some hints for a completely elementary proof.

For any positive integer k consider

$\scriptstyle E_{k}:=\{x\in E\ :\ |f'(x)|<k\}$.

Since by assumption $\scriptstyle \mathrm {m} (E_{k})=0$, there exists a relatively open nbd $\scriptstyle U\subset [0,1]$ of $\scriptstyle E_{k}$, such that $\scriptstyle \mathrm {m} (U)<1/k^{2}$.

Consider the collection $\scriptstyle {\mathcal {J}}$ of all those (relatively) open intervals $\scriptstyle J\subset U$ such that $\scriptstyle \mathrm {diam} (f(J))<k\mathrm {diam} (J)$, that is, such that $\scriptstyle f(J)$ is contained in an interval of length less than k times the length of $\scriptstyle J$.

Consider the union $\scriptstyle V$ of these intervals. The relevant facts that you can check are: $\scriptstyle V$ is a relatively open neighborhood of $\scriptstyle E_{k}$, and $\scriptstyle V\subset U$; moreover (key point) each connected component of $\scriptstyle V$ is an interval $\scriptstyle J$ that belongs to the class $\scriptstyle {\mathcal {J}}$.

This implies that $\scriptstyle \mathrm {m} (f(E_{k}))<1/k$, whence $\scriptstyle \mathrm {m} (f(E))=0$ since $\scriptstyle E$ is the increasing countable union of the $\scriptstyle E_{k}$. --pma (talk) 09:53, 15 August 2009 (UTC)

August 15

L^p(R^n) limit

I am working through a solution a friend presented at our study group for a problem and I am not sure his first step is even true. All we are given is $f\in L^{q}(\mathbb {R} ^{N})$ for some $q<\infty$. He says for p > q,

$\int |f|^{p}=\int |f|^{p-q}|f|^{q}\leq \|f\|_{\infty }^{p-q}\int |f|^{q}=\|f\|_{\infty }^{p-q}\cdot \|f\|_{q}^{q}<\infty$.

Well, a function can be essentially unbounded but still be integrable (${\frac {1}{2{\sqrt {x}}}}$ on [0, 1]) but he is claiming the essential supremum is finite. Is this true with $\mathbb {R} ^{N}$? Thanks! Sorry so many questions, but I have just a few days before my qual. And, if it bothers you, I am fine with you not answering them. StatisticsMan (talk) 02:01, 15 August 2009 (UTC)

I guess I should mention the point of the problem is to show $\lim _{p\to \infty }\|f\|_{p}=\|f\|_{\infty }$ and the first step is just to show that this makes sense by showing from a certain point on $\|f\|_{p}$ is finite. After that, I have the rest of the solution. StatisticsMan (talk) 02:11, 15 August 2009 (UTC)

Umm, isn't it true that $\|1\|_{\infty }=1$ and $\|1\|_{p}=\infty$ on $R^{N}$?(Igny (talk) 02:34, 15 August 2009 (UTC))

Yes, that's why we have the assumption that $f\in L^{q}(\mathbb {R} ^{N})$ for some q. The second part of the question asks for a counterexample if we do not have that assumption and the one you gave is the one. StatisticsMan (talk) 03:09, 15 August 2009 (UTC)

In fact I do not understand what statement you want to prove. In general, a given function on R^N belongs to L^p for a set of exponents p that is an interval in [1,∞]. Conversely, for any interval J of [1,∞] there is a measurable function f on R^N such that f is in L^p if and only if p is in J. --pma (talk) 07:32, 15 August 2009 (UTC)

Okay, well let me give you the exact question, just to be sure.

If $f\in L^{q}(\mathbb {R} ^{N})$ for some $q<\infty$, show that $\lim _{p\to \infty }\|f\|_{L^{p}}=\|f\|_{L^{\infty }}$. Also, show by example that the conclusion may be false without the assumption that $f\in L^{q}(\mathbb {R} ^{N})$. StatisticsMan (talk) 12:52, 15 August 2009 (UTC)

I think this can still be true. Say f is in L^q for q in [100,10000]. Then, it is measurable so this means that the L^q norm for q > 10000 is infinity. In that case, this is simply saying that the infinity norm is also infinity. And, I think to prove this, we just do 2 cases. One is where the infinity norm is infinity and the other is where it is finite. The finite one is the one I put up there. Thus, in that case, we are assuming it is finite so in that case f is in L^p for every $p\geq q$, though as you said, this is not true in general. StatisticsMan (talk) 13:31, 15 August 2009 (UTC)

Yes, it's a well-known property of Lp norms; unfortunately the article here does not have the proof but it's a simple thing that you can find in almost all textbooks on the subject. (PS: how could one imagine that the question was that one?) --pma (talk) 13:39, 15 August 2009 (UTC)

My second post, right under my first one, says that the first post is the first step in the proof of showing the p-norm goes to the infinity norm. StatisticsMan (talk) 13:58, 15 August 2009 (UTC)

uh yeah I missed it --pma (talk) 14:06, 15 August 2009 (UTC)

Yea, I should have put it there in the first place so it would be less likely to be missed! StatisticsMan (talk) 14:26, 15 August 2009 (UTC)

N disjoint solutions to N-queens puzzle

What attention, if any, has been given to the problem of dividing an N-by-N board into N solutions to the N-queens puzzle? NeonMerlin 02:05, 15 August 2009 (UTC)

We can generalise the solution given in the article to any n × n board where n is divisable by 4, say n = 4m for some positive integer m ≥ 1. Assume that the coordinates are given by $(x_{i},y_{j})$ where $1\leq i\leq j\leq 4m.$ We have three families of queens:

$Q_{1}=\{(x_{i},y_{2i}):1\leq i\leq 2m\}\ ,$

$Q_{2}=\{(x_{2i+3},y_{4i-1}):1\leq i\leq m\}\ ,$

$Q_{3}=\{x_{2i+4},y_{4i-3}):1\leq i\leq m\}\ .$

~~ Dr Dec (Talk) ~~ 11:53, 15 August 2009 (UTC)

What is summation of r! from r=1 to r=n?

I've tried adding manually to find a pattern, but can't find any from 1, 3, 9, 33, 153, 873, ...

The original question is to find the summation of r(r!) from r=1 to r=n, so I split up the summation into the summation of r (I know it's n(n+1)/2) and the summation of r! which I can't find a conjecture for. Doing the summation of r(r!) manually gives 1, 5, 23, 119, 719, 5039, ... which I can't find a pattern for either. But since we are not expected to know the formula for summation of r!, I think it's more likely I need to find a pattern from one of the two manual summations. Any hints?

So this is PART of a homework question I have problems with. Later I have to prove the conjecture using mathematical induction. —Preceding unsigned comment added by 59.189.57.133 (talk) 02:07, 15 August 2009 (UTC)

If it's not too big a hint, find out what "OEIS" stands for ;) 70.90.174.101 (talk) 03:01, 15 August 2009 (UTC)

Let's go back to the original question $\scriptstyle \sum _{r=1}^{n}r(r!)$: your first step was not the best thing to do. Just observe that r(r!)=(r+1)!-r!. --pma (talk) 08:47, 15 August 2009 (UTC)

Exactly! summing r·r! is a lot easier than summing r! alone. ~~ Dr Dec (Talk) ~~ 10:38, 15 August 2009 (UTC)

Even if the simpler-looking sum had been easy to compute, there's no reason to believe that knowing $\scriptstyle \sum _{r=1}^{n}r$ and $\scriptstyle \sum _{r=1}^{n}r!$ would help in determining $\scriptstyle \sum _{r=1}^{n}r(r!)$. $\scriptstyle \sum a_{k}b_{k}=\sum a_{k}\cdot \sum b_{k}$, for instance, is very seldom true. —JAO • T • C 11:01, 15 August 2009 (UTC)

Differential definition

An editor tried to change the definition of Differential (infinitesimal) with this edit [1] to have:

${\mathrm {d} f(x)=f'(x)\Delta x}\,$

instead of

$\mathrm {d} y={\frac {\mathrm {d} y}{\mathrm {d} x}}\mathrm {d} x,$

I reverted this and argued against him on the talk page at Differential_(infinitesimal)#The precise definition of a differential. as he had derived it from his own reasoning and dx isn't delta x it was just a possible value. However I have now looked at the Springer Maths dictionary for the subject differential and it does the same sort of thing. Is what he was putting in really right? Dmcq (talk) 14:47, 15 August 2009 (UTC)

I would have agreed with you because you need to take the limit for the editor's expression to work, which was what he apparently did not do. If he had done that then both you're expressions would become equivalent and equally valid. Or in more mathematical terms:

For $y=f(x)$

$dy=d(f(x))={\frac {dy}{dx}}dx=\lim _{x\to 0}f'(x)\Delta x\neq f'x\Delta x$Superwj5 (talk) 15:28, 15 August 2009 (UTC)

If we see where the discussion in the talk page seems to lead, I propouse the article Differential (infinitesimal) to be renamed Infinitesimals in calculus as that seems to be the subject of the article and change where it says "a differential is traditionally an infinitesimally small change in a variable" whith other where there is reference to Leibniz notation about taking dy and dx as infinitesimals, and how that is related with differentials. As I see differentials are one thing (we have two references now on that) and infinitesimals are other. And then create an article about differentials.Usuwiki (talk) 16:08, 15 August 2009 (UTC)

Shouldn't that be $\lim _{\Delta x\to 0}f'(x)\Delta x$? -- 128.104.112.102 (talk) 18:27, 15 August 2009 (UTC)

No. That is an infinitesimal. When you take the limit for $\Delta x$ to approach 0 you are turning the definition of the differential into an infinitesimal.

Unfortunately Leibniz's notation has every one of us confused. dx can be a differential, or an infinitesimal, depends on you taking one notation or the other.Usuwiki (talk) 20:32, 15 August 2009 (UTC)

So is a "differential" only defined at $x=0$, or how does that $f'(x)$ factor into things? (I guess I'm not understanding your notation. You're taking the limit as $x$ goes to zero, but not as $\Delta x$ goes to zero. What is $\Delta x$ doing as $x$ goes to zero? Is it constrained by the limit in some fashion, or is it considered an "independent" variable?) Am I correctly surmising that the "differential $dy$" has only one value - that is, it is not a function of $x$? -- 128.104.112.102 (talk) 14:31, 17 August 2009 (UTC)

Sorry, I think you where trying to correct Superwj5 using some logic. From my point of view, both expressions are wrong. In the case, I will write

$\mathrm {d} y={\frac {dy}{dx}}\mathrm {d} x=[\lim _{h\to 0}{\frac {f(x+h)-f(x)}{h}}]\Delta x$

Usuwiki (talk) 05:06, 18 August 2009 (UTC)

Saying that it's

$dy=f'(x)\,\Delta x\,$

is a bit of nonsense that modern textbook writers have adopted out of squeamishness about infinitesimals, stemming from the fact that you can't present infinitesimals to freshmen in a logically rigorous way. Insisting on logical rigor is clearly a mistake—typical freshmen can't be expected to appreciate that. The absurdity of that convention becomes apparent as soon as you think about expressions like

$\int _{0}^{1}f(x)\,dx.$

Michael Hardy (talk) 19:17, 15 August 2009 (UTC)

Here's the actual edit. Michael Hardy (talk) 19:22, 15 August 2009 (UTC)

People, those of you thinking dx, dy, dt as infinitesimals have to stop calling dx a differential. Either you say dx is a differential or say that dx is an infinitesimal, if you stick to the first interpretation you are on standard calculus, if you stick to the second you are on Leibniz's notation. Do not call dx a differential when you will interpret it as an infinitesimal.

Where did you get that a differential is an infinitesimal?

It seems you got confused because Leibniz used dx, dy, dt as infinitesimals and then you readed that dy is a differential in standard calculus.Usuwiki (talk) 20:02, 15 August 2009 (UTC)

These is why my proposal stands. From the title, the article is wrong. Perhaps changing the title as I sugest, separating the concepts, and clarifying the notation that is used will be helpfull.Usuwiki (talk) 20:32, 15 August 2009 (UTC)

My understanding is that the traditional idea in calculus is that dx etc are indefinitely small quantities and Δx etc refers to finite amounts. More recently they have become viewed as linear maps or in other terms a covariant basis for the tangent space so they have escaped the constraint of being infinitessmals and become more what the word 'differential' means in english. However I can't imagine myself ever mixing Δx and dy and having them linked to each other the way the analysis textbooks seem to do now. I'd simply write f'(x)Δx rather than dy as mixing the two just will cause confusion. f'(x) is dy/dx and is defined as the limit of Δy/Δx at the point, then to mix the two so dy linearly depends on Δx just sounds like it is asking for trouble.

Despite my dislike for the usage I guess it is notable and that's really all that matters. So it will have to be accommodated somehow. This is an encyclopaedia though so both the old and new views and both what happens in topology as well as this analysis view have to be dealt with. In my view the article as it stands is not wrong, just incomplete. Dmcq (talk) 21:49, 15 August 2009 (UTC)

Ok, differential can have another interpretation in english, that's my bad. This way the title of the article stands for something like an infinitesimal differential of something? Disregarding what is mathematically called a differential?

Anyway I have created this article on the differential of a function.Usuwiki (talk) 22:20, 15 August 2009 (UTC)

Yes a differential is normally thought of as an infinitessmal in straightforward calculus. As far as I'm aware there is no difference in the English and Spanish treatment of the word in this context. Differential calculus refers to the limit differential and not a finite version. The use of differential as you have pointed out in analysis is an extension of its redefinition as a linear map. Saying dy is a linear function of Δx using what one might in this context call the infinitessmal differential which defines the tangent space is not an obvious way of proceeding for most people as you can see from the comments above. Dmcq (talk) 00:08, 16 August 2009 (UTC)

We need to get together on this. There needs to be literature about the subject you are pointing, that is, some book that defines the differencial as a limit or something. Other way the only thing I understand is that you are confused.

I'm trying to use the time I have right now to move Wikipedia forward, I want the next edit to be this one. But we need to get together and unify concepts before I continue with this.Usuwiki (talk) 00:36, 16 August 2009 (UTC)

Just go ahead with the edit, it doesn't remove anything and it's obviously a good place to put it. The problem with the other article was changing the leader to a rather confusing business which didn't reflect the contents and citing a book which didn't actually give the equation you wrote down. As to the article you set up a major aim of wikipedia is to explain things for the audience liable to reach the article, it will be edited in strange ways by people who are confused unless it is explained well. Dmcq (talk) 08:11, 16 August 2009 (UTC)

Traditionally, differentials are infinitesimals. As I explained, the recent (probably less than 50 years ago) meme that differentials are finite was invented out of unjustified squeamishness about heuristics, and is seen to be absurd when you apply it to integrals.

Usuwiki, where did you pick up the weird idea that differentials are not infinitesimals? Michael Hardy (talk) 18:33, 16 August 2009 (UTC)

They aren't technically speaking. Note, for instance in Calculus, Vol. 1: One-Variable Calculus with an Introduction to Linear Algebra (Second Edition) by Tom M. Apostol how there never dares to say that differentials are infinitesimals, it only says that Leibniz called <<differentials>> a pair of infinitesimals. Now I ask, is there a book that says that a differential is an infinitesimal? I got a pair saying that differentials can be given specific numerical values. For instance in the very same article Differential (infinitesimal) there is a reference to here, if you look at charapter 2, seccition 2.2 "Differential and tangent lines", you may find the ideas that I wrote in the article but based on infinitesimal changes. But be careful. Note the text says that "the differential dy depends on two independent variables x and dx" and that "is the real function of two variables". So is a real function. Although the logical explanation in this text is based on infinitesimal changes, for the final deffinition given this fact is irrelevant. What I'm saying is that this text gives the deffinition given in Differential of a function where $\Delta x\,$ (an independent variable) is thought as an infinitesimal.Usuwiki (talk) 00:55, 18 August 2009 (UTC)

Sorry if I do not enter the specific topic, but how many articles are there in wikipedia about "differential"? Maybe this is not the right place to note it, but I observed that, sometimes, the articles on the same argument are just too many; they ignore, if not even contradict each other; sometims they are not even linked together. In some case a double version seems really necessary (e.g. in mathematical topics that are also used by other scientists with a language and notations that are irreducible to mathematics). But in other cases, it seems that these multiple version are generated just as a way to escape annoying editing wars (it's the "leave-and-do-it-again" policy). Is there a discussion on this problem? --pma (talk) 15:32, 18 August 2009 (UTC)

Ricci Tensor

I don't understand the article on Ricci Tensor. Can anyone elaborate the method of obtaining the Ricci Tensor from the Riemann Tensor in simpler terms? The Successor of Physics 15:14, 15 August 2009 (UTC)

John Baez gives a very clear geometric explanation here. Gandalf61 (talk) 15:03, 16 August 2009 (UTC)

P(n+1) more likely to be prime if P(n) is?

I have a lot of empirical evidence supporting the notion that primes tend to cluster somewhat among the values of irreducible polynomials over the integers. That is, it seems that given irreducible polynomial P, if it is given that P(n) is prime then this in some way increases the likelihood that P(n+1) is also (i.e., primes appear to arise in pairs, triples, etc. among the values of a given polynomial). Can this be right? Are there any theorems which either confirm or refute this idea? It doesn't seem to make sense to me, but I have quite a lot of specific data. Perhaps I'm looking at still too small a sample to judge.Julzes (talk) 23:09, 15 August 2009 (UTC)

Formula for primes and Ulam spiral have some related information, but nothing that I can see to answer this question. I don't quite know how to word the conjecture well either, because you need to take a probability over all irreducible polynomials. It's obviously untrue for some IPs, like P(n) = n² + 1 (if P(n) is a prime larger than 2, then P(n+1) is even). —JAO • T • C 12:11, 16 August 2009 (UTC)

I haven't really tried to word a precise conjecture or meet the kind of objection raised by P(n)=n^2+1; I was just trying to get the gist across. To see how this question arose, you might want to take a look at an h2g2 Entry I put together on a sequence of related polynomials: http://www.bbc.co.uk/dna/h2g2/brunel/A55643259. It's strange origin is vaguely hinted at in the final part after the long listing. The Entry lists values of the variable giving prime and almost-prime values, and not only does it seem that what I asked is true, but some broader conjecture. The origin of the sample is anomolous, though, and maybe a conjecture would be incorrect with the specific set of cases being far out of the norm.Julzes (talk) 06:14, 17 August 2009 (UTC)

August 16

Bounded linear operator on sigma-finite measure space

Let $(X,\mu )$ be a $\sigma$-finite measure space. Let $1\leq p\leq \infty$ and $g\in L^{\infty }$. Show that the operator $T:L^{p}\to L^{p}$, $Tf=gf$, is bounded, and $\|T\|=\|g\|_{\infty }$.

Proof starts:

T is bounded: If $1\leq p<\infty$, then

$\|Tf\|_{p}=\left(\int _{X}|fg|^{p}\right)^{1/p}\leq \|g\|_{\infty }\left(\int _{X}|f|^{p}\right)^{1/p}=\|g\|_{\infty }\cdot \|f\|_{p}$.

If $p=\infty$ then it's easier, $\|Tf\|_{\infty }=\|fg\|_{\infty }\leq \|f\|_{\infty }\|g\|_{\infty }$. So, for $1\leq p\leq \infty$, we have $\|T\|=\sup _{f\neq 0}{\frac {\|Tf\|_{p}}{\|f\|_{p}}}\leq \|g\|_{\infty }<\infty$. So, T is bounded and this is half the inequality.

I'm not exactly sure where to go from here. I have a friend's solution but they do something which is obviously wrong at this point. Any ideas? Thanks! StatisticsMan (talk) 00:56, 16 August 2009 (UTC)

For simplicity, make $\|g\|_{\infty }=1$. By definition of the $L^{\infty }$ norm, there is a set $E$, with $0<|E|<\infty$ such that $|g|\geq 1-\epsilon$ on $E$. Let us put $f_{\epsilon }=1/|E|^{1/p}$ on $E$, and 0 otherwise. Then $\|f_{\epsilon }g\|_{p}\geq 1-\epsilon$, and $\|f_{\epsilon }\|_{p}=1$. Phils 02:45, 16 August 2009 (UTC)

I was thinking about a similar thing, but if $g\in L^{\infty }$, that just means it's essentially bounded (we can just say bounded). So, g = 1 is such a g, even satisfying $\|g\|_{\infty }=1$. But $mE=\infty$ then, no matter what $\epsilon >0$ is. StatisticsMan (talk) 03:25, 16 August 2009 (UTC)

Oh, I get it. You're not defining E to be the set where $|g|\geq 1-\epsilon$. If that set is real big, just take a part that has finite measure. And, since $X=\cup X_{n}$ with $\mu X_{n}<\infty$, we can look at the set where $|g|\geq 1-\epsilon$ is true and intersect it with $X_{n}$ to get a set where it is true such that the set has finite measure, just as you said. Thanks! StatisticsMan (talk) 03:28, 16 August 2009 (UTC)

Entropy

What is a simple derivation for the formula for calculating entropy? Mo-Al (talk) 08:14, 16 August 2009 (UTC)

In what context? Do you mean in statistical mechanics? In which case, entropy is $S=-k\sum _{i}P_{i}\ln(P_{i})\!$, which sums over the different microstates that correspond to a given macrostate. But I wouldn't call this "derivable", rather, that it's a definition of entropy.--Leon (talk) 14:38, 16 August 2009 (UTC)

Well I suppose my question is really what motivates the definition -- why is this formula a natural thing to use? Mo-Al (talk) 19:03, 16 August 2009 (UTC)

Ah! I was taught that the definition allows one to "correctly" derive ALL the thermodynamics of any particular system; that is, a non-trivially different definition would lead to different, incorrect (in accordance with experiment) thermodynamic predictions. This definition, however, allows one to correctly predict thermodynamic behaviour. However, there is an intuitive "logic" to it, in that the more microstates corresponding to a given macrostate, the lower the information communicated by the macrostate variables. For instance, in a lowest entropy configuration, with one microstate corresponding to the macrostate in question, the macrostate tells us EVERYTHING about the system. For a high entropy configuration, the system contains much more information than the macrostate variables (temperature, pressure etc.) can communicate.

Does that make any sense?--Leon (talk) 19:20, 16 August 2009 (UTC)

And see this.--Leon (talk) 19:25, 16 August 2009 (UTC)

You may also want to take a look at Entropy (information theory). -- Meni Rosenfeld (talk) 20:02, 17 August 2009 (UTC)

In particular Claude Shannon's 1948 paper "A Mathematical Theory of Communication" gives a derivation/plausibility argument for the entropy formula. He lists some properties such a formula should have and it naturally leads to the familiar result. —Preceding unsigned comment added by 87.102.1.156 (talk) 18:36, 19 August 2009 (UTC)

Please fill the gaps in a table

Please help me to fill the gaps in this table: Derivatives and integrals of elementary functions in alternative calculi--MathFacts (talk) 08:25, 16 August 2009 (UTC)

This sort of thing should be filled from looking up a book or journal - and I doubt you'll find much in that way for those systems, they're pretty obscure! Anyway they can mostly be filled in fairly automatically by formulae in the systems once you can do the normal version so overall I'm not certain about the point. Dmcq (talk) 12:50, 16 August 2009 (UTC)

My computer algebra systems do not give expressions for the empty cells.--MathFacts (talk) 13:39, 16 August 2009 (UTC)

I don't think an algebra system producing results from things you feed in is counted as a notable source. And I had to cope with such a result stuck in an article just a day or so ago where the results were not quite right and had ambiguities and besides weren't in simplest terms. Dmcq (talk) 13:45, 16 August 2009 (UTC)

Source can not be notable or unnotable. It's reliable or unreliable. Notable or unnotable may be topic.--MathFacts (talk) 13:56, 16 August 2009 (UTC)

Running expressions you think of through a program and sticking them in a table is counted as original research. The stuff in wikipedia really does have to have some bit of notability and if you can't find some tables giving the expressions or something very similar then the subject really doesn't satisfy notability criteria. Wikipedia isn't for publishing facts you dreamt might be useful and worked out, they have to be notable. Dmcq (talk) 14:25, 16 August 2009 (UTC)

Nobody will publish something that can be derived with a machine - it is simply ridiculous. Only scientific discoveries are published. Using machine or calculator is not a research of course.--MathFacts (talk) 15:16, 16 August 2009 (UTC)

What you put in is original research as far as wikipedia is concerned. Please read the leader of that article. It is quite specific and is a core wikipedia policy. I know maths doesn't always follow that to the letter and it shouldn't either for straightforward things. However you have set up an article that reflects nothing in published literature full of things you thought of yourself and generated the results using a program without any results in sources to check them against. That really is going way beyond the bounds. Interesting articles I would have preferred kept where the person had cited the facts but where the synthesis was not something that had been written about have been removed because of that rule. Dmcq (talk) 16:11, 16 August 2009 (UTC)

There are published rules on how to compute such things and anyone can prove them either by himself or using some mathematical software. Regarding integrals anyone can take a derivative to verify.--MathFacts (talk) 16:19, 16 August 2009 (UTC)

I would like to point out that there is a link at the top of each column in this article (except for 1) which takes you to the article on that specific subject. And, those likely have most of the formulas in the table. So, it's not original research, at least mostly. StatisticsMan (talk) 16:29, 16 August 2009 (UTC)

Check them yourself and you'll see they don't. Dmcq (talk) 16:31, 16 August 2009 (UTC)

And ones which have seem to have been filled in by MathFacts presumably the same way as he did this list. Generating content using his computer without looking up things. Dmcq (talk) 16:41, 16 August 2009 (UTC)

(ec) One of the gaps to be filled in your table asks for the "discrete integral" of $\textstyle f(x):=a^{\frac {1}{x}}$. Checking your link, I see you want a solution of the functional equation $\scriptstyle F(x+1)-F(x)=f(x)$. By the way, the definition there is a bit misleading: you should be aware of the fact that the solution is unique up to a 1-periodic function, not just up to an additive constant C (and the analogous remark holds for your "multiplicative discrete integral"). That said, a particular solution is

$F(x):=x+\sum _{k=0}^{\infty }\left(a^{\frac {1}{k+1}}-a^{\frac {1}{k+x}}\right),$

not a particularly relevant information as far as I know.--pma (talk) 16:39, 16 August 2009 (UTC)

Yes. There is a inconsistency in that article. It needs clarification. I know about it. Still not have enough time to clarify. The abovementioned equation is not enough to define the sum. But it is usually defined through Faulhaber's formula or equivalent.--MathFacts (talk) 17:33, 16 August 2009 (UTC)

So, I don't see a real case of original research, but I do not see any reason for the name "alternative calculi", either. --pma (talk) 18:46, 16 August 2009 (UTC)

Any suggestions?--MathFacts (talk) 18:47, 16 August 2009 (UTC)

"A table of formulas"? Maybe it's too generic... ;) --pma (talk) 07:51, 17 August 2009 (UTC)

Learning about multiple regression online

It is many years since I last was conversant with multiple regression, I need a refresher. And I have never used any recent MR software. Could anyone recommend any easy online learning materials to use please? I want to do multiple regression on a number of economic time series with the aim of forecasting the independant variable. Forecasting, not modelling - I think this means that correlations between the variables is not important, as it would be if I was modelling; but I'm not sure. I'm also aware of the different types of MR and unsure which would be best to use. Thanks. 78.144.207.41 (talk) 17:12, 16 August 2009 (UTC)

Uniformly convergent sequence of polynomials

Characterize those sequences $\{p_{n}(x)\}_{n=1}^{\infty }$ of polynomials such that the sequence converges uniformly on the real line.

Here's another qual question for which I have no solution. If you happen to know this is not true or the solution is very complicated, you can just say that. If you know of a somewhat elementary solution, any help would be great. Thanks. StatisticsMan (talk) 19:27, 16 August 2009 (UTC)

Any such sequence must be a sequence of either constant polynomials, or polynomials with identical leading coefficient after finitely many terms, no? Otherwise, for $m,n>N$, $|f_{n}(x)-f_{m}(x)|\sim |ax^{p}-bx^{p}|=|(a-b)x^{p}|$ which for $a\neq b$, $p>0$ is arbitrarily large as $|x|\to \infty$. Fredrik Johansson 19:45, 16 August 2009 (UTC)

By the same reasoning, wouldn't the next term need the same coefficient after finitely many terms? If the first terms of the polynomials had same coefficient, they cancel out and you're left with a polynomial of degree 1 less. So, would the answer be that the sequence of polynomials must differ only in the constant term after finitely many terms of the sequence, and those constant terms must converge to some real number? StatisticsMan (talk) 20:16, 16 August 2009 (UTC)

Yes --pma (talk) 20:21, 16 August 2009 (UTC)

Of course; my blunder. Fredrik Johansson 22:09, 16 August 2009 (UTC)

I don't think the above is correct. The polynomials don't even have to be bounded in degree (see Taylor series example below) as long as the high order terms decrease in size fast enough. 67.117.147.249 (talk) 04:57, 17 August 2009 (UTC)

What pma said (both above and below) is correct. Eric. 216.27.191.178 (talk) 03:05, 18 August 2009 (UTC)

Since you were asking complex analysis questions earlier, this current question may be getting at the idea that the Taylor series of an analytic function is uniformly convergent. Those are sequences of polynomials whose terms look like (x-a)^k/k! so the higher order coefficients if you treat them as polynomials in x don't stay the same as you change a slightly, but those terms get squashed down by the k! divisor as the order gets higher. 67.117.147.249 (talk) 00:18, 17 August 2009 (UTC)

Is a Taylor series of an analytic function uniformly convergent everywhere, rather than just on compact sets? --Tango (talk) 00:29, 17 August 2009 (UTC)

It's only convergent inside the circle of convergence, and (maybe) at some points on the boundary. (Or another possibility is that I've gotten confused and am thinking of something else--I don't remember this stuff very well any more). 67.117.147.249 (talk) 04:51, 17 August 2009 (UTC)

Actually I probably do have it wrong and am mixing several ideas together incorrectly. The article uniform convergence gives exp z as an example of a function whose Taylor series is uniformly convergent, but it doesn't say that's the case for all analytic functions (within the radius of convergence around a given point). Analytic functions are uniformly continuous but I guess that's not enough. Maybe some expert here can straighten this out. 67.117.147.249 (talk) 05:46, 17 August 2009 (UTC)

What they say is correct; the thing is very simple. The only polynomials that are uniformly bounded on R are the constants. Hence two polynomials have a finite uniform distance ||p-q||_∞ if and only if they differ at most by the constant term. So a sequence of polynomials uniformly convergent on R is, up to finitely many polynomials, a sequence of polynomials that differ at most by the constant term. (The exponential series is not uniformly convergent on C, but only uniformly convergent on compact sets of C) --pma (talk) 06:21, 17 August 2009 (UTC)

Most analytic functions are not uniformly continuous either. f(z)=z² is an example. Algebraist 12:40, 17 August 2009 (UTC)

Here's an intuitive way to think about which may or may not be correct. If we're talking nonconstant entire functions, then the derivative is also an entire function. Only if the original function was a linear polynomial is the derivative a constant. Otherwise, the derivative is unbounded. So, it would seem to me that any nonconstant entire function, other than a linear polynomial, is obviously not uniformly convergent on the entire complex plane. Of course, a linear polynomial isn't either as has already been discussed. It is true for sure that if a holomorphic function has a power series which converges inside some open disk that the power series is uniformly convergent on any closed subset of that disk. You can't really say anything more than that without a specific function. StatisticsMan (talk) 12:43, 17 August 2009 (UTC)

Linear polynomials are uniformly continuous, in fact Lipschitz continuous. Conversely, if f is an entire uniformly continuous function, we can find a δ > 0 such that |f(z) − f(w)| ≤ 1 whenever |z − w| ≤ δ, which implies |f(z)| ≤ |f(0)| + |z|/δ + 1, i.e., f is bounded by a linear function, and therefore it is itself linear by (one of the variants of) Liouville's theorem. — Emil J. 12:53, 17 August 2009 (UTC)

Good point. Let $\epsilon >0$ be given. If f(x) = ax + b and $|x-y|<\epsilon /a$, then $|f(x)-f(y)|=|ax+b-ay-b=|a(x-y)|<a\epsilon /a=\epsilon$. I take back what I said. StatisticsMan (talk)

August 17

Tuk-tuks

How many Tuk-Tuk's (three wheeled vehicles) are in India? —Preceding unsigned comment added by Lanceboe (talk • contribs) 01:01, 17 August 2009 (UTC)

Since I doubt that there are accurate figures on the number of tuk-tuks in India, this sounds like a Fermi problem. In which case, how would you set up the problem, and which factors do you have trouble estimating? Confusing Manifestation(Say hi!) 05:12, 18 August 2009 (UTC)

QQ vs. YoY

I’m pulling my hair out trying to calculate quarter-to-quarter annualized real economic growth such that my results match those published by statistical authorities. Singapore, for example, contracted 3.5% in the second quarter (year-on-year), or it grew 20.7% on a quarter-to-quarter annualized basis. In Excel, my formula for year-on-year growth is =sum((Q2year2 –Q2year1)/Q2year1)*100, which give me, say, 4.3 or a 4.3% rise.

What formula should I use for quarter-to-quarter calculations? DOR (HK) (talk) 03:49, 17 August 2009 (UTC)

I don't understand the word "sum". The expression ((Q2year2 − Q2year1)/Q2year1)*100 is just one term. What terms are you adding? Michael Hardy (talk) 04:42, 17 August 2009 (UTC)

=sum is commonly used in Excel equations, although the "sum" may be optional. DOR (HK) (talk) 07:39, 17 August 2009 (UTC)

Similar terms are "=count" and "=average," if that helps. DOR (HK) (talk) 08:24, 17 August 2009 (UTC)

I am not certain what you are trying to do here - it might be clearer if you can show us your underlying data. Cheap and cheerful way to annualise quarterly returns is to multiply by 4, so 4.3% quarterly growth would be 17.2% annual growth. More accurate method is to compound quarterly growth using the formula (1 + r)⁴ - 1. So 4.3% quarterly growth annualises to (1.043)⁴ - 1 = 0.183 = 18.3% annual growth. Gandalf61 (talk) 08:45, 17 August 2009 (UTC)

OK, some real-world data:

US GDP, in chained 2005 $ billion

Qtr/Yr	$ billion	YoY Growth	Qtr-Qtr Growth
Q1 2008	13,367	NA	NA
Q2 2008	13,415	NA	+1.5%
Q3 2008	13,325	NA	-2.7%
Q4 2008	13,142	NA	-5.4%
Q1 2009	12,925	-3.3%	-6.4%
Q2 2009	12,892	-3.9%	-1.0%

In the last line, year-on-year is calculated thus:

=((12,892 – 13,415) / 13,415) = -0.03898 (i.e., -3.9%)

The quarter-to-quarter annualized growth rate is reported as -1.0%. Question: what is the formula for arriving at -1.0% (or, any of the right-hand column numbers) using this data? Thanks. DOR (HK) (talk) 03:09, 18 August 2009 (UTC)

I think the Qtr-Qtr growth figures are calculated by finding the percentage growth since the previous quarter, then annualising this with compounding. For example, in Q2 2009 we have:

• % growth since previous quarter = (12,892 - 12,925) / 12,925 = -0.00255 = -0.255 %
• Annualised % growth = (1 - 0.00255)⁴ - 1 = -0.01017 = -1.02 %

Using this method, I get values of +1.44%, -2.66%, -5.38%, -6.44%, -1.02% as compared to the published figures of +1.5%, -2.7%, -5.4%, -6.4%, -1.0%. Gandalf61 (talk) 11:20, 18 August 2009 (UTC)

I agree. Michael Hardy (talk) 16:15, 18 August 2009 (UTC)

Many thanks for that. What I have not been able to do is to construct an Excel formula that duplicates the results. Any thoughts? DOR (HK) (talk) 01:33, 19 August 2009 (UTC)

Could it be that you're rounding too early? A small rounding error in an intermediate step can in some cases result in a large error in the bottom line. Unless you've got a really good handle on how big the effect of rounding at some intermediate stepp will be on the bottom line, you should not round beyond what the machine forces you do to, until the last step. Michael Hardy (talk) 01:57, 19 August 2009 (UTC)

...or it could be that the published figures are based on rounding too early. Michael Hardy (talk) 01:58, 19 August 2009 (UTC)

Got it! =(((Q2-Q1)/Q1)*4)*100 Been driving me crazy for the longest time. Many thanks! DOR (HK) (talk) 03:45, 19 August 2009 (UTC)

quick maths question

if ab > 0 and b > 0, can you take out b from the inequality, leaving a > 0? —Preceding unsigned comment added by 59.189.62.104 (talk) 05:46, 17 August 2009 (UTC)

Yes. If ab > 0 then either {a > 0 and b > 0} or {a < 0 and b < 0}. If you know that b > 0 then you also have a > 0. Gandalf61 (talk) 08:34, 17 August 2009 (UTC)

Alternative method, you can divide both sides by b. You can't do that without knowing it is positive since when you divide by a negative number you have to flip the inequality, but you do know it is positive, so that's ok. --Tango (talk) 13:23, 17 August 2009 (UTC)

How many eighth degree monic polynomials are such that...

I don't know if any programmer out there wants to tackle this, but my question is: How many monic polynomials of eighth degree have remaining coefficients in the set {0, 1, 2, ..., 41} and are composite when the variable is between 1 and 41 (inclusive), but prime when it is 42? I have one example, and I would like to know how many if any others there are. My example is

$P(n)=n^{8}+15n^{7}+13n^{6}+17n^{5}+7n^{4}+40n^{3}+3n^{2}+34n+25$.Julzes (talk) 15:17, 17 August 2009 (UTC)

How accurate do you want this figure? A Monte Carlo estimate shows that about 0.2% of the polynomials matching the coefficients criterion also match the primality criterion. Since there are $42^{8}$ of those, this turns out around $2\cdot 10^{10}$. One pleasant-looking example I've found is $n^{8}+9n^{7}+41n^{5}+3n^{2}+37$. -- Meni Rosenfeld (talk) 20:29, 17 August 2009 (UTC)

I guess I'd like a slightly better estimate with the range expanded to between 0 and 68 with the exception at 42, as 69 is where the next prime comes in in my particular example. Ideally, I'm curious about just how far up the most extreme case goes (what the polynomial which presents itself as a prime the second time at the highest value is), but I don't suppose that is a practical inquiry. I'm surprised that my initial question got any kind of answer and so quickly, and that the answer is on the order of one in every five hundred.Julzes (talk) 21:15, 17 August 2009 (UTC)

The high proportion should be no surprise at all - most numbers are composite, so it's not very demanding to require that many values of the polynomial will be. Neither should getting an answer so quickly - this is WP:RD/math. :)

About 0.022% of polynomials are prime for 42 and composite elsewhere between 0 and 68, which is about $2.1\cdot 10^{9}$.

If I understood correctly that your "ideal inquiry" is to find a polynomial P such that:

• The coefficients criterion is met.
• $P(42)$ is prime.
• $P(n)$ is composite for $0\leq n\leq N,n\neq 42$.
• N is as high as possible.

Then $n^{8}+17n^{7}+8n^{6}+22n^{5}+14n^{4}+5n^{3}+3n^{2}+37n+25$ satisfies this with $N=1145$. Of course, there may be polynomials with even higher N; only an exhaustive search (which would take quite a while) or some intelligent reasoning can find the absolute best. -- Meni Rosenfeld (talk) 16:14, 18 August 2009 (UTC)

PS: "Quite a while" = 5 years with my current program and hardware. So far I've improved N to 1697. -- Meni Rosenfeld (talk) 16:29, 18 August 2009 (UTC)

Ok. Thanks for all that. How about prime at n=-41 and n=42, but composite in between, if you're interested in continuing? Of course, you'll need either the broad definition of primality or to take absolute values. I'm not at all surprised it would take 5 years to answer the bigger question.Julzes (talk) 22:05, 18 August 2009 (UTC)

I just noticed that the constant term has to be 25 (if P(0) is to be composite but P(42) prime). Also, the condition on negatives to but not including -41 should be more stringent than the condition involving some number greater than or equal to 69. Julzes (talk) 03:17, 19 August 2009 (UTC)

question about probability

The time on digital clock is 10.38 .where here 1 is in column p ,0 is in q ,3 is in r and 8 is in column s .after x hours time is noted. find the probability that the number in

1:- column q is "9"

2:- column q is less than 5.

—Preceding unsigned comment added by True path finder (talk • contribs) 16:58, 17 August 2009 (UTC)

As you described it, the situation is perfectly deterministic, there is no probability involved. The answer is thus either 0 or 1 depending on the value of x. — Emil J. 17:15, 17 August 2009 (UTC)

(ec) Unless there is some constraint on x them it doesn't matter what the time now is. Is the clock a 12 hour or a 24 hour clock? 1:- 1/12 (09) or 2/24 (the same value) (09, 19) 2:- 7/12 (01, 02, 03, 04, 10, 11, 12) or 14/24 (01, 02, 03, 04, 10, 11, 12, 13, 14, 20, 21, 22, 23, 24) -- SGBailey (talk) 17:18, 17 August 2009 (UTC)

SGBailey, you seem to be assuming x is a uniformly distributed random variable. Why are you assuming that. This is a very vaguely stated question where we have to guess what was meant, so it doesn't seem safe to just assume things. And you didn't even state that assumption. Michael Hardy (talk) 17:37, 17 August 2009 (UTC)

AFAIK, time is uniformly distributed. I take that as a given unless something else is stated. I did say "unless there is some constraint on x" which was intended to specify the conditions of my answer. With no x constraint then the time after x is equally likely to be anywhere in a 24 hour cycle. The OP has now constrained x to be from T+1 to T+25 - this happens to be a 24 hour period, so I think my answer still applies. -- SGBailey (talk) 20:24, 19 August 2009 (UTC)

As a minor aside (obviated now that OP has clarified) it is not even possible for x to be uniformly distributed. As I said minor, but relevant to the solution of some apparent paradoxes.--SPhilbrickT 20:11, 18 August 2009 (UTC)

sorry, condition on x is 1 <x <25.and answer on book is 0.1 of first and 0.5 of second. —Preceding unsigned comment added by True path finder (talk • contribs) 19:55, 17 August 2009 (UTC)

Okay, so the time at which we next look at the clock is later than 11:38 today and earlier than 11:38 tomorrow. Let's assume, for the sake of argument, that the time is uniformaly distributed between those limits. Then I don't think I can see how the answer you gave can be reached. Are you sure you have told us the right column - column q is the least significant digit of the hours number, yes ? Gandalf61 (talk) 12:13, 19 August 2009 (UTC)

round robin tournament

I'm wondering about ways of proving that "a round robin tournament is always possible to construct for an even number of players"

The algorthym in section Round-robin_tournament#Scheduling_algorithm shows that it is always possible. However it's not clear how I could obtain that algorthym without having a 'flash of inspiration'. ie to me it seems that the algorhtym is non-obvious... Is there a more workmanlike proof of do-ability.?83.100.250.79 (talk) 19:43, 17 August 2009 (UTC)

Number the players p1,p2,...p(2n). Draw a point for each player and draw lines (edges) between every pair of points (complete graph on 2n nodes). There will be (2n)(2n-1)/2 = n(2n-1) such edges where each edge represents a game between the two players it connects. Then just enumerate the edges in any way you like, and take n of them in every round, giving a 2n-1 round tournament. 70.90.174.101 (talk) 03:14, 19 August 2009 (UTC)

Yet taking the edges off in certain orders can result in having unplayable rounds - ie certain combinations don't work.

Is there a way to proof that there will always be a valid combination of games for all rounds for any n?83.100.250.79 (talk) 12:36, 19 August 2009 (UTC)

I've designed such tournaments from time to time, using the following method. Whether it is equivalent to the algorithm you cite I can't say, but it works for any number of players (for an odd number, create an extra dummy player to denote an idle round). To avoid confusing (to me, anyway) generality, I'm taking the case of 8 players from A to H, but I can see that by simple extension it will work for any number. A table will be constructed showing the number of the round in which each pairing will occur, with rows from A to G and columns from B to H - ultimately, only the "northeast" part will apply. From the top, put 1 and 2 in col B, 2, 3 and 4 in col C, ..., 6, 7, 1, 2, 3, 4 and 5 in col G and 7 in col H. Col H is then completed by transferring the "below diagonal" entry in each row, i.e. 2 in row B, 4 in row C, etc, giving col H the successive values 7, 2, 4, 6, 1, 3 and 5 from the top. You'll see that this process guarantees the 28 matches will be played in 7 rounds.

For any number of players, the columns for all but the last one are filled immediately, that for the last one is completed by transferring the numbers below the stepped diagonal. The quick way to fill the last column, once you've seen the pattern, is to put the highest round number (odd) at the top, then to follow it with the even rounds from 2 upwards then the other odd ones from 1 upwards.→217.43.210.186 (talk) 19:19, 19 August 2009 (UTC)

It sounds basically the same in that it gives one solution, but apart from working - doesn't explain how...

I was wondering if anyone had found how to calculate the number of different sets of complete rounds constructable for every n (clearly each round has (n/2)! degenerate permutations, and the sets of rounds assuming order of play is irrelevent is also (n-1) degenerate).

But ignoring the degenerate cases is there ever more than one way to contruct an entire set of rounds of games?83.100.250.79 (talk) 22:30, 19 August 2009 (UTC)

Yes, there are additional ways. For example, with players 1, 2, 3, 4, 5, 6, we could have a set of rounds containing (1, 2)(3, 4)(5, 6) and another set of rounds containing (1, 2)(3, 5)(4, 6), and these two sets are not related by the degeneracies you identified. But that raises the question of whether there are additional ways after accounting for degeneracies from permuting players. Eric. 216.27.191.178 (talk) 02:17, 20 August 2009 (UTC)

Yes thanks, that's obvious though I didn't think of it.83.100.250.79 (talk) 11:36, 20 August 2009 (UTC)

There are distinct ways to create a tournament with 8 players even after identifying tournaments related by permutation of round order, permutation of games within a round, and permuting the identity of the players. Eric. 216.27.191.178 (talk) 02:27, 20 August 2009 (UTC)

August 18

Combinatorics question

I have a list of N categories with Nk members each. (i.e. the kth category has Nk members, and these number of members may be unique, but not necessarily distinct.) I'd like to take a random member from only 3 of the categories. How many ways can I do this? The brute force way to do select 3 groups (and there are obviously N choose 3 ways to do this) and then multiply the cardinality of each of those 3 groups to get the number of ways to choose an item from each. Iterate over all distinct groups of 3 categories. Is there an analytical way to do this? or a combinatorial formula? Any relevant literature?

Here's an example if my above explanation was confusing.

|A| = 4, |B| = 3, |C| = 5, |D| = 1.
A, B, C = 4*3*5 ways = 60
A, B, D = 4*3*1 ways = 12
A, C, D = 4*5*1 ways = 20
B, C, D = 3*5*1 ways = 15
                     
Sum                   107

Also, if anyone knows how to do this in R (I'm sure it's really simple, that would be much appreciated.)

Thanks, --Rajah (talk) 03:32, 18 August 2009 (UTC)

Notice that, for instance, (a+b+c+d)*(a+b+c+d) = aa+bb+cc+dd+2(ab+ac+ad+bc+bd+cd). So, if the sum of a through d is s1, and the sum of the squares of a through d is s2, ((s1^2)-s2)/2 is the sum of all pairs of different letters. If S_n is the sum of all the nth powers of the letters, I think what you want is ${\frac {1}{6}}S_{1}^{3}-{\frac {1}{2}}S_{1}S_{2}+{\frac {1}{3}}S_{3}$. That would be, in this case, (1/6)(a+b+c+d)³-(1/2)(a+b+c+d)(a²+b²+c²+d²)+(1/3)(a³+b³+c³+d³) = (1/6)(13)³-(1/2)(13)(51)+(1/3)(217) = 107. It doesn't look too helpful for a small example, but imagine how much easier it would be for a list of 20 numbers. Black Carrot (talk) 05:25, 18 August 2009 (UTC)

Some related ideas: multiplicative functions (such as the divisor function) and symmetric polynomials. Black Carrot (talk) 05:34, 18 August 2009 (UTC)

Yes, you want the 3rd elementary symmetric polynomial, and as BC shows you can express it in other bases of symmetric polynomials. If you need to deal more heavily with the combinatorics of symmetric polynomials you may find useful R.P.Stanley's book "Enumerative Combinatorics", (chapt 7, vol 2). [2]. PS: what do you mean by how to do this in R? --pma (talk) 06:01, 18 August 2009 (UTC)

Yeah, I meant R, the programming language, as Bo Jacoby pointed out. I was just being lazy, sorry about that. Elementary symmetric polynomials are exactly what I wanted, nice work. --Rajah (talk) 15:12, 18 August 2009 (UTC)

R (programming language) probably. Note that f(x)=(x+a)(x+b)(x+c)(x+d)= x⁴+(a+b+c+d)x³+(ab+ac+ad+bc+bd+cd)x²+(abc+abd+acd+bcd)x+abcd, so the number you want is the coefficient of x in the taylor expansion of f(x). A solution coded in J (programming language) looks like this

  (<-4 3 5 1)&p. t. 1
107

Bo Jacoby (talk) 14:11, 18 August 2009 (UTC).

Awesome, thanks. All of these replies are very helpful! --Rajah (talk) 15:00, 18 August 2009 (UTC)

Limit of integral

Show that if

$\lim _{h\to 0}{\frac {1}{h}}\int _{a}^{b}|f(x+h)-f(x)|\,dx=0$,

then there is some constant c such that $f(x)=c$ a.e.

Here's the thing. I have a proof, but I just read it through again and I have a statement in the middle that I am not sure is true. The first thing to do is, for any $a\leq x_{1}<x_{2}\leq b$, given $\epsilon >0$

$|\int _{x_{1}}^{x_{2}}{\frac {1}{h}}(f(x+h)-f(x))\,dx|\leq \int _{x_{1}}^{x_{2}}{\frac {1}{h}}|f(x+h)-f(x)|\,dx\leq \int _{a}^{b}{\frac {1}{h}}|f(x+h)-f(x)|\,dx<\epsilon$

for $h\in (-\delta ,\delta )$ (where $\delta$ comes from $\epsilon$ since the limit is 0). What this proves is

$\lim _{h\to 0}{\frac {1}{h}}\int _{x_{1}}^{x_{2}}f(x+h)-f(x)\,dx=0$

for any such $a\leq x_{1}<x_{2}\leq b$. From here, what I have down is set $F(x)=\int _{a}^{x}f(t)\,dt$ and that since f is integrable on [a, b] for $h\in (-\delta ,\delta )$, then $F'(x)=f(x)$ a.e. But, I never showed f is integrable and I am not sure if it is. Any ideas?

Just so you know, the next step I have, assuming that is right:

$0=\lim _{h\to 0}{\frac {1}{h}}\int _{x_{1}}^{x_{2}}f(x+h)-f(x)\,dx=\lim _{h\to 0}{\frac {F(x_{2}+h)-F(x_{2})}{h}}-{\frac {F(x_{1}+h)-F(x_{1})}{h}}$

Assuming f is integrable, F is absolutely continuous so that we end up getting the limit to be $0=F'(x_{2})-F'(x_{1})=f(x_{2})-f(x_{1})$ for those $x_{1},x_{2}$ where $F'(x)=f(x)$ and thus it is equal to the same constant almost everywhere. StatisticsMan (talk) 21:37, 18 August 2009 (UTC)

Ok, so we only have $f:\mathbb {R} \to \mathbb {R}$ with no assumptions of integrability nor measurability; for all $a<b$ in $\mathbb {R}$ that limit is 0 and the thesis is f(x)=0 a.e. (Alternatively, one can assume the limit to be 0 of a fixed pair a<b, and then the conclusion is that f(x)=0 a.e. in the interval [a,b]. One can easily prove each of the two statements from the other).

Maybe there are simpler ways, but one possibility is doing your program for the function $\textstyle g_{y}(x):=f(x+y)-f(x)$ for any fixed $\textstyle y\in \mathbb {R}$ and draw the consequences. By hypothesis, each $g_{y}\in L_{\mathrm {loc} }^{1}(\mathbb {R} )$, and since we have

${\frac {g_{y}(x+h)-g_{y}(x)}{h}}={\frac {f(x+y+h)-f(x+y)}{h}}-{\frac {f(x+h)-f(x)}{h}}$

you can do your complete argument and conclude that the function $g_{y}(x)$ is a constant (depending on y). Going back to the hypothesis on f(x), this tells us that the integral in $x$ is an integral of a constant function, thus just multiplication by (b-a), that is, f is a function with $f^{\prime }(x)=0$ in the classical sense for all x, and the conclusion follows. --pma (talk) 07:07, 19 August 2009 (UTC)

But in fact in the above argument I assumed $\scriptstyle x\mapsto g_{h}(x):=f(x+h)-f(x)$ to be locally integrable. Assuming also w.l.o.g. that f is compactly supported, one can write e.g. $\scriptstyle f(x)=-g_{1}(x)+f(x+1)=-\sum _{k=1}^{N}g_{1}(x+k)$ for N large enough, so f is $L^{1}$ too. If we assume the minimal requirement only, that is, that $|f(x+h)-f(x)|$ is (loc) integrable for all h, I am not sure about whether f is measurable. For sure the original statement is already of interest in the assumption of f measurable or integrable; a result under a more general hypothesis wouldn't necessarily be more useful.--pma (talk) 12:06, 19 August 2009 (UTC)

Well, I have seen a "solution" from a professor where he provides almost no detail. It took me a long time to fill them in, perhaps wrongly. But, he seems to be assuming f is measurable, though it is not in the problem. I actually did not notice that I was assuming that without being told, so good point. His entire solution is: "If the limit holds at the end points, then for every a < x1 < x2 < b it is surely true that $\lim _{h\to 0}{\frac {1}{h}}\int _{x_{1}}^{x_{2}}|f(x+h)-f(x)|\,dx\to 0$ as h goes to 0 and hence the limit without the absolute value signs inside also tends to zero with h. However

${\frac {1}{h}}\int _{x_{1}}^{x_{2}}(f(x+h)-f(x))\,dx={\frac {1}{h}}\left(\int _{x_{2}}^{x_{2}+h}f(x)\,dx-\int _{x_{1}}^{x_{1}+h}f(x)\,dx\right)\to f(x_{2})-f(x_{1})$

and hence the result." Perhaps he is also assuming that it is integrable. It seems so, right? StatisticsMan (talk) 12:25, 19 August 2009 (UTC)

Nice; I suppose you mean the last expression to be written without the big parentheses. For the last equality, f should be assumed to be integrable, yes. After all it seems that the point of the exercise is not that. NOte that your proof with F(x) is almost identical to your professor's one (in both the key point is using that (F(x+h)-F(x))/h converges to f(x) as h tends to 0; you are not using anything else about absolute continuity, and you may even avoid mentioning it) --pma (talk) 14:23, 19 August 2009 (UTC)

What you said would make it what I meant, but what I really meant to do was the big parentheses and not have the 1/h on the inside. So, I just took that out. Thanks for the info. I guess if he is assuming those things as well, I am fine with assuming them. I just didn't know for sure. My test starts tomorrow morning! So, I must go study some more. StatisticsMan (talk) 15:19, 19 August 2009 (UTC)

very good; as your coach, I recommend you to stop studying in the afternoon and relax till tomorrow. You have all to do it. and everybody in the RD/M roots for you!  ;) --pma (talk) 16:33, 19 August 2009 (UTC)

Hehe, thanks. Unfortunately, I am not going to take the afternoon off because I am trying to read through every solution I have written up to qual problems. This will end up taking about 2 full days and I still have 30 problems to read. Then, I have some other problems with solutions I want to look through as well. Thanks for the support. StatisticsMan (talk) 18:58, 19 August 2009 (UTC)

Well, I did enough to pass if everything I did was correct, or so I think. And, I think it was but I could be wrong. They aren't real clear about what it takes to pass. But, the first problem on the real analysis side was to show if the integral of a function is 0 on ever interval, then the function is 0 a.e. So, thanks for explaining to me how to do that problem, pma. Without it, I definitely did not pass. I will find out in a week or so if I passed. StatisticsMan (talk) 20:34, 20 August 2009 (UTC)

August 19

Is this polynomial most extreme?

The polynomial

$P(x):=x^{32}+x^{31}+x^{30}+x^{29}+x^{28}+x^{26}+x^{24}+x^{22}+x^{21}+x^{20}+x^{19}+x^{16}+x^{14}+x^{13}+x^{12}+x^{11}+x^{10}+x^{4}+1$

takes on prime or almost-prime (product of two primes) values for x=1*,2*,7*,16,29,30,31,32,33,36,37*,..., where it is prime for the numbers marked (*). Is it likely that there is another polynomial with coefficients in {0,1} such that over half of the values of x for which it is an almost-prime or prime up to some point at least as high as x=33 occur in a string at least five long ending at that point? Julzes (talk) 06:57, 19 August 2009 (UTC)

I took the liberty of formatting your formula with LaTeX --pma (talk) 07:09, 19 August 2009 (UTC)

Note that I suspect that "five" could probably be replaced with "two" and the value 33 reduced also. Julzes (talk) 09:33, 19 August 2009 (UTC)

Vector perpendicular to plane

So given the equation of a plane. Say... 3x+y-2z=10 How would you find the unit vector orthogonal to it? I ask because I'm not very familiar with vector algebra and this would help me finish a few proofs.--Yanwen (talk) 20:44, 19 August 2009 (UTC)

See Surface normal#Calculating a surface normal. —JAO • T • C 21:47, 19 August 2009 (UTC)

(ec) The vector v = (3, 1, -2) is a normal vector for the plane $P=\{(x,y,z)|3x+y-2z=10\}$. This is why: suppose $p_{1}=(a_{1},b_{1},c_{1})$ and $p_{2}=(a_{2},b_{2},c_{2})$ are two points on the plane P; thus we know $3a_{1}+b_{1}-2c_{1}=3a_{2}+b_{2}-2c_{2}=10$. Then the vector connecting $p_{1},p_{2}$ is $p_{2}-p_{1}=(a_{2}-a_{1},b_{2}-b_{1},c_{2}-c_{1})$, and the dot product of $p_{2}-p_{1}$ with v is

$v\cdot (p_{2}-p_{1})=3(a_{2}-a_{1})+1(b_{2}-b_{1})-2(c_{2}-c_{1})$

$=(3a_{2}+b_{2}-2c_{2})-(3a_{1}+b_{1}-2c_{1})=10-10=0\,$,

so v and $p_{2}-p_{1}$ are orthogonal. Therefore v is a normal vector of the plane P. To get a unit normal vector, just divide v by its length. Eric. 216.27.191.178 (talk) 21:52, 19 August 2009 (UTC)

Let's say the plane has equation $ax+by+cz=d$ for some, not all zero, real numbers a, b, c, and d. A normal vector would be (a,b,c), and so the two unit normal vectors are

${\mathbf {N}}^{\pm }:={\frac {\pm (a,b,c)}{\sqrt {a^{2}+b^{2}+c^{2}}}}\ .$

In fact, if you have some surface S given by an equation f = 0, (where f is a smooth function). So

$S=\{(x,y,z)\in \mathbb {R} ^{3}:f(x,y,z)=0\}\ ,$

then a normal vector to S is given by (f_x,f_y,f_z) where f_x, f_y and f_z are the partial derivatives of f with respect to x, y and z respectively. The surface S will be singular at a point $(x_{0},y_{0},z_{0})\in S$ when

$f_{x}(x_{0},y_{0},z_{0})=f_{y}(x_{0},y_{0},z_{0})=f_{z}(x_{0},y_{0},z_{0})=0\ .$

Assuming that not all three partial derivatives are zero we have two unit normals given by

${\mathbf {N}}^{\pm }:={\frac {\pm (f_{x},f_{y},f_{z})}{\sqrt {f_{x}^{2}+f_{y}^{2}+f_{z}^{2}}}}\ .$

In the plane example f(x,y,z) = ax + by + cz - d, and so f_x = a, f_y = b and f_z = c. In your example a = 3, b = 1 and c = -2 so

${\mathbf {N}}^{\pm }=\pm \left({\frac {3}{\sqrt {14}}},{\frac {1}{\sqrt {14}}},{\frac {-2}{\sqrt {14}}}\right).$

~~ Dr Dec (Talk) ~~ 15:56, 20 August 2009 (UTC)

So for a surface $S=\{(x,y,z)\in \mathbb {R} ^{3}:x^{2}-y^{2}+z^{2}=18\}\ ,$

Would the normal vector at (3,4,5) be ${\mathbf {N}}^{\pm }={\frac {\pm (6,-8,10)}{10{\sqrt {2}}}}\ .$--Yanwen (talk) 21:57, 20 August 2009 (UTC)

August 20

Quadratic turd

My friend tell me there is a number called "quadratic turd? Why is mathematicians so vulgar to call a number turd? 67.101.25.201 (talk) 01:28, 21 August 2009 (UTC)