Wikipedia:Reference desk/Archives/Mathematics/2009 August 15

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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)