Wikipedia:Reference desk/Archives/Mathematics/2009 September 2

Welcome to the Wikipedia Mathematics Reference Desk Archives
The page you are currently viewing is an archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages.

September 2

Variables confusing me

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

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

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

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

I've adjusted the article to explain the problem better. I also discussed the case where there are solutions, but it's still circular; it might be interesting (if it wasn't too complicated) to include a case where there was only one solution to a set of implicit equations (and so it would depend on the spreadsheet's support for such things); OO Calc has an example in its help. --Tardis (talk) 15:22, 9 September 2009 (UTC)

Once upon a time I read something about using a spreadsheet (likely VisiCalc, it was so long ago) to run a cellular automaton. How would you do that without circular references? —Tamfang (talk) 06:26, 10 September 2009 (UTC)

Continuation of the Method of Characteristics to solve PDEs

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Period

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Non-Archimedean fields

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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