Wikipedia:Reference desk/Archives/Mathematics/2009 March 6

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March 6

About Lagrange's Theorem

Lagrange's four-square theorem states that every positive integer can be written as the sum of four squares of integers.

If S is the set of all non-negative integers except 7, is it still the case that every positive integer is the sum of the squares of four elements of S?

I feel sure that this sort of question has been considered before. Does anyone know where to find it in the literature?Partitioin (talk) 00:43, 6 March 2009 (UTC)TC)

I fooled around with a computer program for a few minutes and didn't find any counterexamples in a few thousand trials, but that means nothing. I wikilinked the theorem for you, hope that's ok. 207.241.239.70 (talk) 05:37, 7 March 2009 (UTC)

I've confirmed solutions (sans 7s) for integers up through 300,000,000. The average number of solutions per value appears to have a slightly less than linear relationship with numbers. In the 1,000,000 range the average number of solutions per integer is on the order of 25,000 or a ratio of about 0.025. However, I've found some numbers having significantly fewer solutions, such as 10,485,760 with only 2 solutions – that's quite a statistical bump. Of course, none of this has anything to do with a proof. -- Tcncv (talk) 05:47, 8 March 2009 (UTC)

I am not surprised that the number of solutions does not behave smoothly - there are an infinite number of integers which only have one representation (up to order) as the sum of four squares - see OEIS A006431. If you omit 5 from S instead of 7 then empirical evidence suggests that every integer has a representation as a sum of four squares of integers in S apart from 79. And, interestingly, 79 has 3 distinct representations as the sum of four squares:

79 = 7² + 5² +2² + 1²

79 = 5² + 5² + 5² + 2²

79 = 6² + 5² + 3² + 3²

... but they all happen to include 5². However, I don't see an approach to a proof. Gandalf61 (talk) 13:17, 8 March 2009 (UTC)

Physics question

Please see Wikipedia:Reference_desk/Science#Force_meter_question, (somebody sensible), for some reason the usually sentient editor "Gandalf61", appears to have lost their marbles, or I'm going blind...

ie Third opinion213.249.232.187 (talk) 14:34, 6 March 2009 (UTC)

Nope, he's right. "Tango" proposes a good analogy near the end of the thread. yandman 16:26, 6 March 2009 (UTC)

Finding $y(x)$ from ${\frac {d}{dx}}$ Finding ${\frac {dy}{dx}}$ from ${\frac {d}{dx}}$

How do you find $y(x)$ from ${\frac {d}{dx}}$?The Successor of Physics 15:19, 6 March 2009 (UTC) How do you find ${\frac {dy}{dx}}$ from ${\frac {d}{dx}}$?The Successor of Physics 13:15, 8 March 2009 (UTC)

It's a little difficult to understand what you mean, but if you mean that knowing ${\frac {dy}{dx}}$ as a function of $x$, you want to express $y$ as a function of $x$, then Antiderivative is your article. —JAO • T • C 17:29, 6 March 2009 (UTC)

Perhaps I should restate my question as "How do you find ${\frac {dy}{dx}}$ from ${\frac {d}{dx}}$?". Thanks anyway.The Successor of Physics 13:15, 8 March 2009 (UTC)

${\frac {d}{dx}}$ is an operator. Applying it to $y$ is how you find ${\frac {dy}{dx}}$. Your question sounds like "How do you find $x+y$ from $+$?" and has the same answer: knowing what operator to apply (${\frac {d}{dx}}$ or $+$, for instance) is not useful if you don't know what to apply it to. —JAO • T • C 17:37, 8 March 2009 (UTC)

I want to know it so in for example ${\frac {dy}{dx}}={\frac {d}{dx}}y=ty,t={\frac {d}{dx}}$ because you can express derivatives more easily, and if ${\frac {dy}{dx}}=ty,{\frac {d}{dx}}=t$, and sometimes I want know y and knows t but not y so that is why I asked the question.The Successor of Physics 10:04, 9 March 2009 (UTC)

Your expression $t={\frac {d}{dx}}$ is complete nonsense. Algebraist 10:58, 9 March 2009 (UTC)

If you do know that ${\frac {d}{dx}}y=t(x)y$ for some known function $t$ which you can integrate, then you can use separation of variables to find $y$. But as Algebraist says, don't try to divide both sides by $y$, because the left-hand side isn't a multiplication to begin with. —JAO • T • C 14:21, 9 March 2009 (UTC)

I'm reading a book on vector calculus, and suddenly remembered that ${\frac {d}{dx}}y=\nabla y$! does this help? —Preceding unsigned comment added by Superwj5 (talk • contribs) 14:53, 9 March 2009 (UTC)

Not really. An actual example of a case where you "know $t$" would help more. (Separation of variables was not helpful?) —JAO • T • C 18:01, 9 March 2009 (UTC)

Look at it this way: if you know that ${\sqrt {}}y=ty$, does that mean that $t={\sqrt {}}$? Of course not. Just because it looks like a fraction instead of a funny symbol doesn't mean that ${\frac {d}{dx}}$ isn't an operator, and to deal with an equation with a differential operator you can't deal with it like it was a normal fraction or took a single value. Confusing Manifestation(Say hi!) 21:23, 9 March 2009 (UTC)

Determinants as Tensor Fields or Matrix Functions

Are Determinants Tensor Fields or Matrix Functions? If yes which and please state the equation and state its inverse.(Please don't be to harsh on me. I'm only new in this field.)The Successor of Physics 15:27, 6 March 2009 (UTC)

The most common use of the word "determinant" is as in "the determinant of a matrix", which is of course just a scalar, not a function. But there's also the determinant function, $\det$, which maps certain matrices to their respective determinants, so that's a matrix function ("matrix function" usually means a function from matrices to matrices, but scalars can be identified with 1×1 matrices so that makes little difference). I don't know enough about tensor fields to answer your first question, and I have no idea what equation you refer to. —JAO • T • C 17:38, 6 March 2009 (UTC)

The determinant is a (rank 0, i.e. somewhat trivial) tensor on the rows (or also the columns) of the matrix, if that's what you're asking. 207.241.239.70 (talk) 05:57, 7 March 2009 (UTC)

Cesaro mean

I am looking into the Cesaro mean right now. My question is, if I take any sequence that is bounded and I do the Cesaro mean over and over, do I eventually get a sequence that converges after a finite number of turns? Thanks for any help. StatisticsMan (talk) 18:10, 6 March 2009 (UTC)

I don't know, but looking at Cesàro summation's iterated section, "The existence of a (C, α) summation implies every higher order summation, and also that a_n=o(nα) if α>-1.", does seem to suggest bounded is a very strong hypothesis. (I think (C,n) summation means do Cesaro mean n times). JackSchmidt (talk) 20:46, 6 March 2009 (UTC)

Ah, unfortunately, it looks like things similar to your question are well studied (so beyond my limited skills). Peyerimhoff's Lectures on Summability doi:10.1007/BFb0060951 MR0463744 appears to have some reasonable discussion of this sort of thing, and in particular describes who and when people studied iterated Cesàro summation (Hölder, Schur) and how to more easily discuss it. I didn't find any specific counterexamples, and couldn't tell if any of the theorems applied to your case, but you might have an easier time. JackSchmidt (talk) 21:00, 6 March 2009 (UTC)

Oooh, you want a Tauberian theorem for iterated Cesàro means, and there is one in that book, Theorem III.2, and I think it says that a bounded sequence is Cesàro summable if and only if it is (C,n) summable for some n ≥ 1. There is an example in the Cesàro means chapter of a bounded sequence that is not Cesàro summable, so i think that means the answer to your question is no. JackSchmidt (talk) 21:07, 6 March 2009 (UTC)

Of course the answer is no. The counterexample has the form of a sequence $c_{k}$ alternating two values on consecutive intervals of larger and larger length. Instead of doing a quantitative computation for you, I'll just try to convince you by this example. You are in your sofa watching TV, say zapping from channel 1 to 9. This way you generate a sequence in {1,2...,9}, each second k, switching from channel $c_{k}$ to channel $c_{k+1}$. The trivial switch is also allowed, so at a certain moment, if you wish, you may start pressing channel 1 button compulsively for 100 times or more, one after the other (as actually happens to addicted TV watchers). The sequence of the Cesaro means, $\scriptstyle {\frac {1}{n}}\sum _{k=1}^{n}c_{k}$, somehow follows the sequence $c_{k}$ (in fact, with a regularization-delay effect). Anyway, after you have pressed channel 1 consecutively for 100 times or more, the means also are quite close to 1, say less than 2, and also the sequence of the means of the means starts being close to 1. Now, you switch to channel 9, pressing it each second, for 10,000 times or more, untill the mean, and the mean of the mean, are both close to 9 (say >8). At that point you go back to channel 1, pressing it till the first 4 iterated Cesaro means are all less than 2 again, even 100,000,000 times if needed. As you see, this way you can make a two-value sequence with no converging iterated Cesaro mean. With a bit of patience you can make the construction quantitative. I think $\scriptstyle c_{k}=0$ for $\scriptstyle (2n-1)!\leq k<(2n)!$ and $\scriptstyle c_{k}=1$ for $\scriptstyle (2n)!\leq k<(2n+1)!$ suffices.--pma (talk) 00:05, 7 March 2009 (UTC)

Thanks all. I can not read that book online. As far as pma's example, I have thought of this. I just alternated 1 and -1. I could say I want the average to go to at least 1/2 and then at least -1/2 each time to make sure it's not converging to something. So, 1, -1, -1, 1, 1, 1, 1, 1, ... . But, if I do the averages, I get 1, 0, -1/3, 0, 1/5, 1/3, 3/7, 1/2, and if I do averages again this time on the 2nd sequence, I get 1, 1/2, 2/9, 1/6, 13/75, 1/5, 57/245, 149/560, ...

The point is, I am barely getting above 1/4 now (I realize I did very few terms) and all values are positive. So, I am closer to convergence in only one step. No matter how weird and wild you make a sequence, if I iterate this process, I will at least get closer and closer to convergence. In your example with the TV, you also get closer to convergence after only one iteration. No matter what sequence you construct, even with factorial number of terms, each iteration you get closer and closer to convergence. So, your example may be correct, but I am not convinced yet.

I originally asked this question because of a homework problem that we would need this in, involving Banach Limits. But, we decided to use the shift operator instead of this operator in order to invoke the Hahn-Banach theorem (actually a slight generalization of it). Then, the problem became easy. But, I thank you all for answer. I am interested in this just in general and may look into it more sometime when I have more time. StatisticsMan (talk) 14:16, 7 March 2009 (UTC)

No problem. You would probably like the book (or other books with Tauberian theorems or Divergent series in the title). It works out PMajer's example, as well as the theorem that says that if one Cesàro mean is not enough, no finite number of iterations will be enough. It should be available in any reasonably sized university library (over 350 libraries according to OCLC 47540). JackSchmidt (talk) 15:42, 7 March 2009 (UTC)

OK, let's see if I convince you, sorry for the TV example. It's very elementary. Start defining $N_{1}=1$ and $c_{1}=-1$; then define $c_{2}=...=c_{N_{2}}$ to be 1, choosing $N_{2}$ so large that the first 2 Cesaro means at $N_{2}$ differ from 1 less than 1/2. Then define $c_{N_{2}+1}=...=c_{N_{3}}$ to be -1 choosing $N_{3}$ so large that the corresponding first 3 iterated Cesaro means at $N_{3}$ differ from -1 less than 1/3. And so on: by induction on k you define $c_{j}$ to be constantly equal to $(-1)^{k}$ for all $N_{k-1}<j\leq N_{k}$, choosing $N_{k}$ so large that the first k iterated Cesaro means differ from $(-1)^{k}$ less than 1/k. As you see, you do not get any closer to convergence; all the iterated Cesaro means have liminf=-1 and limsup=1.--pma (talk) 19:28, 7 March 2009 (UTC)

Alright, now what you are saying seems to make sense. Within the one sequence there are chunks that guarantee the kth Cesaro mean will not converge, for each positive integer k. Easy: that is, now that I have seen it, it is easy. As far as that book, I looked online and my library does not have it. But, that's okay. I have so much to do that I am required to do, I don't have much time to read some other book, even if it would be interesting. Thanks for the suggestion! StatisticsMan (talk) 20:45, 7 March 2009 (UTC)

You are welcome. In fact, the best solution is the interpolation result quoted by JackSchmidt, that is, as I understand, "a bounded sequence with convergent second Cesaro mean has the first Cesaro mean convergent too". --pma (talk) 23:26, 7 March 2009 (UTC)

Yea, that is a very bizarre result. It seems as if each iteration should get you closer to convergence. But, if one does not work, then 1 billion will not either. StatisticsMan (talk) 01:11, 8 March 2009 (UTC)

We can restate it this way, taking the quotient on the space c of converging sequences: the Cesaro mean defines a bounded linear operator M on ${\frac {\ell _{\infty }}{c}}$ (it's a Banach space, in which the norm of [x] is $\scriptstyle 1/2(\limsup x_{i}-\liminf x_{i})$ for all $x\in \ell _{\infty }$). Then we have $\scriptstyle \ker M=\ker M^{2}$, that is 0 is a semisimple eigenvalue of M (of infinite multiplicity). --pma (talk) 10:01, 8 March 2009 (UTC)