Wikipedia:Reference desk/Archives/Mathematics/2012 February 15

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.

February 15

A Quick Proof

Just a quick question here to all, my algebraic topology is a bit rusty. I know that a 2-torus is not homeomorphic to a 3-torus. And my argument is that because their fundamental groups are not isomorphic there is no way to continuously deform one onto another. Is this right? And I am pretty sure there is a theorem like this (which I am using) but I can't remember its name. Oh and just for curiosity, what are their fundamental groups? And yes, the context stems from Valentine's day, having an argument whether human male and females bodies are topologically equivalent or not. I say they aren't.  :-) - Looking for Wisdom and Insight! (talk) 03:48, 15 February 2012 (UTC)

The fundamental group is a homeomorphism invariant; that means that if two spaces are homeomorphic, they have the same fundamental group. Topologists would never have bothered studying the fundamental group if it weren't homeomorphism invariant, just as algebraists are only interested in properties of groups which are isomorphism invariant. So yes, your reasoning is correct. I don't think this fact has a name; it's a straightforward check.

The fundamental group of a product is the product of the fundamental groups. Since the 2-torus is the product of two circles, while the 3-torus is the product of 3, that makes their fundamental groups $\mathbb {Z} ^{2}$ and $\mathbb {Z} ^{3}$, respectively.--121.74.100.56 (talk) 05:05, 15 February 2012 (UTC)

Wait, now I am confused by something in your response. I thought that 1-torus is the ordinary torus (a doughnut with one hole). A 2-torus is a double torus and it has two holes, from attaching two 1-tori together. And a 3-torus (triple torus) has three holes. So I thought that the fundamental group of an ordinary 1-torus is just $\mathbb {Z} ^{2}$. So wouldn't the fundamental group of 2-torus be $\mathbb {Z} ^{4}$ and the fundamental group of 3-torus be $\mathbb {Z} ^{6}$? - Looking for Wisdom and Insight! (talk) 08:14, 15 February 2012 (UTC)

Sorry, I interpreted 2-torus and 3-torus to refer to the number of dimensions, not the number of holes. In retrospect, that doesn't make much sense when discussing humans.

The fundamental groups of the double and triple torus aren't that simple; you need to use the Seifert–van Kampen theorem to compute them (the computation is on that page). In particular, they're not abelian. For the n-torus, the fundamental group is $\langle A_{1},B_{1},\ldots ,A_{n},B_{n}|A_{1}B_{1}A_{1}^{-1}B_{1}^{-1}\ldots A_{n}B_{n}A_{n}^{-1}B_{n}^{-1}\rangle .$--121.74.100.56 (talk) 08:35, 15 February 2012 (UTC)

Maybe I am missing something here, but rather than calculating fundamental groups, wouldn't it be simpler to argue that the 2-torus cannot be homeomorphic to the 3-torus because they have different Euler characteristics ? Gandalf61 (talk) 09:04, 15 February 2012 (UTC)

Well for me the simplest argument for me would be the intuitive one. If two tori have a different number of holes, then one must have more holes than the other which means that the one with the fewer holes must be torn/cut to create more holes which is not a continuous deformation. But for a "formal" proof, I just thought of fundamental groups because I don't know much about algebraic topology and fundamental groups is all I remember from the introductory class I took. - Looking for Wisdom and Insight! (talk) 09:42, 15 February 2012 (UTC)

Looking for Wisdom: that's the same problem as you have with distinguishing a sphere from a torus. It might seem intuitively clear that the torus has a hole and a sphere does not, but it's not easy to give a precise proof that they're not homeomorphic, because it's not easy to define precisely what a "hole" of this kind means. That's why we need fundamental groups or some other algebraic invariant. – b_jonas 10:04, 15 February 2012 (UTC)

The dimension is homeomorphism invariant. 74.98.35.216 (talk) 11:35, 15 February 2012 (UTC)

(An aside on motivation): Why do you think that human males and females have different topological genera? I'd say they both have genus 1, due to the connection between the mouth and anus. Other bits of plumbing (male or female) don't really connect in the same direct manner. There are all sorts of finer scale connections, but if you start counting that way, you'd have to include tear ducts and sweat glands, etc., etc. and end up with a very large genus. I don't want to debate finer points of human physiology here; just curious how you're thinking of the problem. SemanticMantis (talk) 14:10, 15 February 2012 (UTC)

Hmmm ... "My dog has genus 1" ... "How does he smell ?" ...Gandalf61 (talk) 16:25, 15 February 2012 (UTC)

Ha--Ok, so the genus of a mammal is rather ill-defined. There's all kinds of flaps and switches inside that change the communicativity, I was thinking of the body in the position where the flaps lead air to the lungs, and not the stomach. If we consider that nostrils can be connected to the stomach, then I guess a the dog is topologically equivalent to a block with a Y-shaped hole in it. Doesn't that have genus 2? I guess my point was that, if you can agree on a genus for a representation of a human male, it should be the same as that of a human female. It's not as though there's two different ways to get to open air from inside a uterus. SemanticMantis (talk) 18:22, 15 February 2012 (UTC)

Central tendency and usefulness of measurements

if measures of central tendency and dispersion are given,then how these measurements can be useful? — Preceding unsigned comment added by 180.211.216.154 (talk) 14:13, 15 February 2012 (UTC)

I created a section header for you. This will happen automatically if you hit the "ask a new question" button at the top of the page. SemanticMantis (talk) 15:01, 15 February 2012 (UTC)

I don't understand your question. If you are asking why measures of central tendency are used, they are used to indicate the "middle" value of a set of data. If you are asking when one type or another should be used, here's some ideas:

1) An arithmetic mean (arithmetic average) is perhaps the most common. It's useful when data is symmetrically distributed, like in a bell curve. For example, you might find the arithmetic mean of class grades, to determine how well the students are getting the material. If the mean is 95%, then they are getting it. If it's 50%, then they aren't.

2) A weighted arithmetic mean is useful when some data points need to be weighted more than others. For example, in the US Presidential election, each state gets a different number of electoral college votes. If you were doing polling to try to determine who would win the election, you might want to weight the state results by the number of electoral votes in that state.

3) For data which is not symmetrically distributed, like incomes and wealth, an arithmetic mean is meaningless. Imagine Bill Gates in a room with 1000 minimum wage workers. The average wealth of each person in the room would be millions of dollars, which would make you think the room is full of rich people. Dealing with percentiles is more useful here. You could say 99.9% of the people earn minimum wage, and 0.1% earn billions.

4) The mode is just the most common number in a set. Let's say you run a restaurant and want to add one item to your menu. If you keep track of the things people request, the most often requested item, which you don't already have, might be a good choice to add.

5) Geometric means, logarithmic means, etc., are useful when data varies in those specific ways. See central tendency for others. StuRat (talk) 19:45, 15 February 2012 (UTC)

Re #3, the arithmetic mean is definitely not meaningless for asymmetric distributions. It's just not always what we want to look at. It's also useful to explicitly mention the median which is the 50% percentile, and useful as a central tendency measure in some situations. -- Meni Rosenfeld (talk) 05:45, 16 February 2012 (UTC)

Do you have an example of when the arithmetic mean for an asymmetrical distribution is meaningful ? StuRat (talk) 16:55, 18 February 2012 (UTC)

To be honest your suggestion that the arithmetic mean (or its counterpart for probability distributions, expectation) can be in some situations meaningless is so alien to me that I'm not sure what kind of example you're looking for. Unlike median, mode, geometric mean and many other measures, the mean is additive, and as addition is such a basic concept, so is arithmetic mean. Similarly, variance is additive for independent variables. This is why expectation and variance are the key attributes of distributions that are at the heart of all probabilistic and statistical analysis. When you want to learn about a distribution, first learn its expectation and variance. Anything else is a bonus which only in specific cases is of much use (one such case is your income example, where the median better captures our intuitive notion of what a "typical" person earns).

All that said, here's an example from a domain I'm involved in. For Bitcoin mining, the expected time to find a block (for the whole network or for any individual miner) follows the exponential distribution, which is pretty much asymmetrical (skewness 2). But the mean of this distribution translates directly to the average rate of finding blocks, and hence, the profitability of mining. As anyone who has invested thousands of dollars into mining hardware will tell you, this is pretty important.

Another important example is Von Neumann–Morgenstern utility. A rational agent will maximize his expected utility. In general there is no utility function for which maximizing the median (or any other measure) isn't irrational.

And a simpler example - A restaurant receives 100 orders pay day, with amounts distributed in some asymmetric way. Given that, what you need to know to deduce the total revenue of the restaurant is the order amount mean, not some other quantity. -- Meni Rosenfeld (talk) 19:41, 18 February 2012 (UTC)

I don't follow. Surely to get the mean, the total revenue was divided by the number of orders in the first place. Or did you have in mind that the historic mean would be multiplied by the number of orders one night to come up with an estimate of that night's receipts ? I'm not sure how useful that would be, say if the asymmetrical distribution is due to different demographics, like drunken businessmen paying on their company account (spending $$$), versus families trying to save money (spending $). In that case, you would really need to at least weight the averages by the number of each type of order, to get a reasonable estimate. StuRat (talk) 23:47, 18 February 2012 (UTC)

Yes, you could for example base future mean estimates on history. But that doesn't really matter, the point is that the mean is a meaningful thing to say about the order amounts, other measures don't carry any useful information. Of course each amount should be weighted by its probability/frequency. -- Meni Rosenfeld (talk) 06:49, 19 February 2012 (UTC)

Limit conditions

Under what conditions on the functions f and g does the following statement hold:

$\lim _{x\to \ell }\left(f(x)*g(x)\right)=\left(\lim _{x\to \ell }\,f(x)\right)*\left(\lim _{x\to \ell }\,g(x)\right),$

Where * denotes either addition, subtraction, multiplication or division. I understand that there may be different conditions depending upon the operation. Let's assume that the left and right limits of f are the same, and likewise for g. Is the statement always true? I'm really looking for the least possible. — Fly by Night (talk) 15:30, 15 February 2012 (UTC)

For addition, subtraction and multiplication you just need that the limits in the RHS exist; this is for any * that is continuous. For division you need the denominator to be non-zero to get continuity so add that the limit of g is not 0, which you would need for the RHS to be defined anyway. So the short answer is if the RHS is defined then the LHS is defined and they are equal.--RDBury (talk) 15:54, 15 February 2012 (UTC)

... but of course the LHS may be defined when the RHS is not. This seems to relate directly to a subset of the classic indeterminate forms. — Quondum^☏✎ 16:15, 15 February 2012 (UTC)

When you say "are defined", what are the explicit conditions? Do we simply need f and g to be continuous, and for each function's own left and right limit to be equal? If so then what is the proof for this? Which articles contain this information? — Fly by Night (talk) 16:09, 15 February 2012 (UTC)

Continuity at the limit is not required, only the two-sided limit (one-sided for limit at ∞), and that the limits on the right are finite (for them to be defined). — Quondum^☏✎ 16:24, 15 February 2012 (UTC)

Thank you for your reply, but I don't find your reply particularly illumination. Please supply more details and, as we are at a reference desk, links to the appropriate articles. — Fly by Night (talk) 17:15, 15 February 2012 (UTC)

When RDBury says "defined", the limit is being referred to, not the functions f and g. Implied is that the limits are finite. Continuity of a function is not a prerequisite at any point of its domain for the limit to exist; I can provide a simple example of an everywhere-discontinuous function that nevertheless has a limit. The domain can even be discrete. So the problem statement is a little sparse on detail, and the natural thing is to assume that we are dealing with real numbers, that the limit meant is the ε–δ definition, and that the limiting value ℓ is either −∞, finite or +∞. Anyhow, perhaps Limit of a function#Properties will answer much of your question. — Quondum^☏✎ 18:44, 15 February 2012 (UTC)

I'm sorry, but having read your reply and followed your link, I can't really say that it helped. Thank you very much for your efforts, but I would be very grateful if someone with a friendly nature were to answer my question. I am very sorry to have to say this, but I find your replies somewhat condescending. If people are to learn then they should feel welcome and comfortable. Your replies to this post, and the post below, do neither of these. Let me thank you once again, but request that you do not involve yourself any further. — Fly by Night (talk) 22:32, 15 February 2012 (UTC)

I think RDBury's last sentence nails it. I'm not sure, but I think Spivak's textbook "Calculus" (really an elementary real analysis text) covers this if you want to get into the gritty details. SemanticMantis (talk) 22:54, 15 February 2012 (UTC)

Just to clarify, f and g don't have to be continuous, what I'm saying is the equality follows from operation * being a continuous function R×R→R. Also I wasn't considering infinite limits but the statement would still be true excluding the well-known indeterminate forms ∞−∞, ∞/∞, etc. Again it's nothing really to do with f and g but whether the operation * can be extended continuously to include infinite values. For example × is a continuous function of R×R→R which can be extended continuously to R∪{∞}×R∪{∞}→R∪{∞} if you exclude the points (0,∞) and (∞,0).--RDBury (talk) 01:21, 16 February 2012 (UTC)

Non-Standard Proof

Does anyone know a proof of the statement:

$\lim _{n\to \infty }\left(1+{\frac {r}{n}}\right)^{n}=e^{r}\,$

which does not use the binomial expansion and the fact that

$\lim _{n\to \infty }\left({\frac {n!}{n^{k}\,(n-k)!}}\right)=1\,?$ — Fly by Night (talk) 16:06, 15 February 2012 (UTC)

Substitute n=mr. — Quondum^☏✎ 16:19, 15 February 2012 (UTC)

Thanks for your reply, but I fail to see how this answers my question. Your suggestion leads to much the same proof. I was hoping or a qualitatively different approach. As we find ourselves on a reference desk, links to articles would be greatly appreciated. (Forgive me for not making this explicit; I had assumed it to be a tacit assumption.) — Fly by Night (talk) 17:17, 15 February 2012 (UTC)

Forgive my brevity – perhaps I was making unwarranted assumptions based on your user page, and it is not clear what level of rigour you need. A proof will depend heavily on how you define the exponential function or exponentiation. The statement can simply be a definition of e^r (def #1 of link), rather than a proof. But first, I think you should give the required starting point: the definition of exponentiation and/or of the exponential function you choose to start with, else it is guesswork what you want. — Quondum^☏✎ 18:18, 15 February 2012 (UTC)

The only unwarrented assumptions you made were those about my mathematical understanding. Because I didn't like your reply, and because I asked you to answer a reference desk question in the spirit of the reference desk (please read the preamble at the very top of this page) does not mean I am some mathematical dimwit that does not understand the exponential function. Please re-read my post, and especially its title. I wanted a non-standard, i.e. non-trivial proof. — Fly by Night (talk) 22:24, 15 February 2012 (UTC)

$\lim _{n\to \infty }\left(1+{\frac {r}{n}}\right)^{n}=\lim _{mr\to \infty }\left(1+{\frac {r}{mr}}\right)^{mr}=\left(\lim _{m\to \infty }\left(1+{\frac {1}{m}}\right)^{m}\right)^{r}=e^{r}$ Q.E.D. Bo Jacoby (talk) 19:34, 15 February 2012 (UTC).

Bo Jacoby's "proof" is exactly what I had in mind, but one should note two steps that one should not consider rigorous (which is why I made reference to rigour before): the premises that z^(mr) = (z^m)^r and that lim _m→∞ w^r = (lim _m→∞ w)^r, so it is more a consistency check or plausibility demonstration than a proof. — Quondum^☏✎ 20:11, 15 February 2012 (UTC)

Thanks chaps. But this proof is no different in method to the proof I gave and then asked for another proof. I wanted a proof using other methods from different areas of mathematics. Thank you both for your effort, but it isn't quite what I'm looking for. I'm sure you'll agree that making a single substitution of a variable does not give a qualitatively different proof. Maybe the answer is that none exists. Thanks again folks. — Fly by Night (talk) 22:17, 15 February 2012 (UTC)

In all good faith, I think to get the answers you want, you'll need to specify exactly how/where you want to start. As you are probably aware, there are several different schemes for what is a definition and what is a result in this area. It's nothing personal, just an admission that, even in textbooks, you will find different perspectives on what which parts are definitions and which parts are logical implications of those definitions. SemanticMantis (talk) 22:49, 15 February 2012 (UTC)

You got what you asked for: "a proof which does not use the binomial expansion and the fact that $\lim _{n\to \infty }\left({\frac {n!}{n^{k}\,(n-k)!}}\right)=1$". If that isn't quite what you are looking for, you should ask for what you are looking for. Bo Jacoby (talk) 23:03, 15 February 2012 (UTC).

Alternative approach

Let $f_{n}(r)=\left(1+r/n\right)^{n}$. Show that for n large $f_{n}\leq f_{n+1}$ and $f_{n}'\leq f_{n+1}'$ and the sequence $f_{n}$ and $f_{n}'$ is uniformly bounded on compact sets. Conclude that $f=\lim _{n\to \infty }f_{n}$ exists, is differentiable, and $f'=f$ and $f(0)=1$. Sławomir Biały (talk) 00:19, 16 February 2012 (UTC)

(The above is for $r\geq 0$. For $r<0$, replace $\leq$ by $\geq$. Sławomir Biały (talk) 10:42, 16 February 2012 (UTC))

Sławomir ment to say $f_{n}(r)=\left(1+r/n\right)^{n}$. Bo Jacoby (talk) 07:29, 16 February 2012 (UTC).

Corrected.Sławomir Biały (talk) 10:42, 16 February 2012 (UTC)

${\frac {d}{dr}}\lim _{n}(1+r/n)^{n}=\lim _{n}{\frac {d}{dr}}(1+r/n)^{n}$ (!) $=\lim _{n}(1+r/n)^{n-1}=\lim _{n}(1+r/n)^{n}$. Thus the function we started with satisfied the DE $f'(r)=f(r)$ and f(0)=1. It must therefore be the exponential function. There is some work to be done to justify swapping the limit and the derivative. Tinfoilcat (talk) 10:00, 16 February 2012 (UTC)

Yes, this is the idea. It follows if the function and its derivative converge uniformly on compact sets, which the above argument shows. Sławomir Biały (talk) 10:43, 16 February 2012 (UTC)

I hadn't read your post properly, now I see that my idea was the same as yours. Tinfoilcat (talk) 11:14, 16 February 2012 (UTC)

Spectral radius of a derivative

I ask for help in understanding the following condition for the use of iterative method in numerical solving of a non-linear equation, given in the relevant article:

...a sufficient condition for convergence is that the spectral radius of the derivative is strictly bounded by one in a neighborhood of the fixed point.

What does "spectral radius" mean and how to calculate it? I've found lots of information both on the Wikipedia and on the web related to matrices and the systems of linear equations, but I am not knowledgeable enough to bind these topics together in my mind. --Esmu Igors (talk) 16:23, 15 February 2012 (UTC)

If you type spectral radius into Wikipedia's search bar then you will find the spectral radius article. If you take the modulus of all of the eigenvalues then the S.R. is defined to be the greatest of these. — Fly by Night (talk) 17:24, 15 February 2012 (UTC)

Thank you, Fly by Night, for your help, but I have already found this article previously and understand almost nothing in it. Furthermore, while I know how to use matrices for solving simple linear equations, I have absolutely no idea about their usage for non-linear equations (not systems). I know I don't actually know math, but I really hope somebody would point my attention to where actually information about matrices for non-linear equations can be found...--Esmu Igors (talk) 19:45, 16 February 2012 (UTC)

(I could be wrong here, I'm rather rusty on this topic) I think the article must be talking about the spectral radius of the Jacobian matrix for f, which is a matrix of partial derivatives that captures how a point x_n in phase space will move in a small neighborhood of a point x (where the partial derivatives are evaluated), when subjected to the dynamics described by the function f. In effect, the Jacobian serves to linearize the system at a point. This is similar to finding a tangent line to a curve. The same concept applies: in a small enough neighborhood, everything is well approximated by linear relationships. The radius being bounded by 1 has the effect that there is some small neighborhood around x, in which nearby points will converge to x. This is because the spectral radius is the dominant eigenvalue. Thus, if it is less that 1, then the displacement between x_n+1 and x will be smaller than the displacement between x_n and x, and the sequence x_n will approach x. Conceptually, this is roughly analogous to Newton's method. Does that help at all? If not, you might get better answers by explaining in more detail what you are trying to do, and what types of math you are comfortable with. SemanticMantis (talk) 21:03, 16 February 2012 (UTC)