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

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August 6

Rational function?

The definition of a rational function is a polynomial divided with a polynomial. So I've been wondering- is $f(x)={\frac {P_{m}(x)}{Q_{n}(x)}}$ a polynomial if $a_{0}=3$ and $b_{0}=2$ (so that $f(x)={\frac {3}{2}}$)? Is $f(x)={\frac {1}{1}}$ a rational function- and if so, is $f(x)=1$ a rational function even thought it isn't a ratio (because if ${\frac {1}{1}}$ is a rational function and since ${\frac {1}{1}}=1$...). And as a final question, is ${\frac {0}{0}}$ a rational function even though it's undefined? 0 is a polynomial (the zero polynomial) so it's technically the ratio between two polynomials- does as function have to be defined? (sorry for asking so many questions, this has been piling up in my head for some time and I'm too tired to stylize my question) --BiT (talk) 01:48, 6 August 2009 (UTC)

Oh never mind the ${\frac {0}{0}}$. A quick read through the "where $P\,$ and, $Q\,$ are polynomial functions in $x\,$ and $Q\,$ is not the zero polynomial. The domain of $f\,$ is the set of all points $x\,$ for which the denominator $Q(x)\,$ is not zero." answered my question --BiT (talk) 01:49, 6 August 2009 (UTC)

Yes, functions which are constantly a rational number (which includes integers) are (constant) rational functions. As are polynomials (just take Q=1). A function does have to be defined everywhere on its domain - that is the definition of a domain, really - so 0/0 would be, at best, a function defined on the empty set, ie. the empty function. --Tango (talk) 02:05, 6 August 2009 (UTC)

Is "The domain of $f\,$ is the set of all points $x\,$ for which the denominator $Q(x)\,$ is not zero." (a quote from Rational function) really true? This implies that f : [0,infty) -> R defined by f(x) = 1/(x² + 1) is not a rational function. Is this common usage? (Depending on the context, it might even be taken to imply that g : R -> R defined by g(x) = 1/(x² + 1) is not a rational function either, because its domain leaves out a whole lot of points (all non-real complex numbers except ±i) where the denominator is not zero.) —JAO • T • C 05:40, 6 August 2009 (UTC)

That section is talking specificially about real functions. (There's another section about the complex version). It should probably say that explicitly. I'll change it. Rckrone (talk) 06:37, 6 August 2009 (UTC)

Actually, after thinking about it, the article is fine as is. It's important to keep in mind the field that you're dealing with when talking about rational functions. What looks like the same function can have different meanings depending on what field you're dealing with. x² + 1 as a polynomial over the reals is nowhere zero, so the domain of 1/(x² + 1) as a rational function over the reals is all of R. x² + 1 as a polynomial over the complex numbers has zeros at i and -i, so the domain of 1/(x² + 1) as a rational function over the complex numbers is C\{i,-i}. —Preceding unsigned comment added by Rckrone (talk • contribs) 06:53, 6 August 2009 (UTC)

To answer the other part of the question, it's generally not ok to restrict the domain and still call the function a rational function without qualifying it in some way. When someone says "rational function" or similarly "polynomial", there are certain properties we expect, which may not hold on some smaller domain. For example the fundamental theorem of algebra doesn't hold if we allow "polynomials" to include functions with restricted domains. Rckrone (talk) 07:03, 6 August 2009 (UTC)

I'd say there is a slight abuse of language due to obvious historical reasons. Strictly speaking a rational function (say with real coefficients) is not a function, its primary meaning being that of an element of the field of fractions R(x) of R[x]. Correspondingly, the term "domain" of a rational function f denotes primarily a certain subset of the coefficient ring, attached to the algebraic object f (and of course it turns out to be the set-theoretic domain of a certain set-theoretic function). Note that "domain" is used with special meanings in other contexts: e.g the "domain of a convex function" f:E →RU{+∞} is the subset of the normed space E where f(x)<+∞, although strictly speaking the domain of f is the whole E.--pma (talk) 07:34, 6 August 2009 (UTC)

I find myself using the term "rational functions" to refer literally to functions more often than anything else. At least this suggests that this usage is fairly common.

Anyway, I agree with Rckrone that a function whose domain is artificially restricted (as opposed to implicitly restricted to our current structure of interest) should not be called rational.

Regarding Tango's comment, any constant function is rational - the constant need not be a rational number. -- Meni Rosenfeld (talk) 16:59, 6 August 2009 (UTC)

Thanks for all the enlightening comments. Especially the fact that "rational function" is often used in English to denote what really is a "rational expression" set me straight from being fooled by my own language, which upholds that distinction. —JAO • T • C 05:01, 7 August 2009 (UTC)

Thank you for your answers and now I just want to verify what Tango said- so you're saying that rational numbers (like 4⁄5), integers (like 5 or 3) and polynomials (like $3x^{2}+4x+16$) are all rational functions? Just to be completely clear. If that's so, then why isn't the introduction to the 'rational function article' formulated like the rational number article? That is to say: "In mathematics, a rational number is any number that can be expressed as the quotient a/b of two integers, with the denominator b not equal to zero. Since b may be equal to 1, every integer corresponds to a rational number." If I understand Tango correctly a better way of putting it would be something like:

“ In mathematics, a rational function is any function that can be expressed as the ratio ${\frac {P_{m}(x)}{Q_{n}(x)}}$ of two polynomial functions, where the denominator $Q_{n}(x)$ is not equal to zero. Since $Q_{n}(x)$ may be equal to 1, every integer corresponds to a rational function. ”

I know I basically copied the introduction from the article about rational numbers- but I think the revision would help other people understand what I was having trouble with. What do you think? --BiT (talk) 23:06, 7 August 2009 (UTC)

As Meni points out, it's not just rational numbers, any real number will do, I made a mistake there. I'm not sure I understand your proposal - the rational function article does say essentially that. --Tango (talk) 00:26, 8 August 2009 (UTC)

1. I actually didn't fully understand Meni- what do you mean by "any real number will do"? (refresh my challenged-at-learning-definitions memory, real numbers include both rational and irrational numbers right?) Does that mean rational functions can be equal to ${\sqrt {x}}$... I feel like I've made some mistake here?
2. Ah it seems I've recalled the article on rational functions incorrectly. I didn't know it said "a rational function is any function which can be written as the ratio of two polynomial functions" and thought it said "a rational function is any function written as the ratio of two polynomial functions", my bad. I do however think that some parts of the introduction I proposed should be included. The "where the denominator $Q_{n}(x)$ is not equal to zero" (I realise it appears in the definition section but I think it belongs in the introduction) and the part about "every integer/real number can be represented as a rational function" --BiT (talk) 00:55, 8 August 2009 (UTC)

As for point one, the numerator and denominator still have to be polynomials in x, which ${\sqrt {x}}$ clearly is not. It's just that the coefficients of these polynomials may be any number (for the meaning of "number" we are currently interested in). So ${\frac {{\sqrt {2}}x^{2}-\pi x+e}{{\sqrt[{7}]{\frac {18}{13}}}x-e^{-\pi ^{2}}}}$, although involving a lot of irrational numbers, is a perfectly rational function (over the real numbers). A constant irrational, like f(x) = e, is just a special case of this. —JAO • T • C 08:42, 8 August 2009 (UTC)

Ah of course, I didn't think of irrational numbers being the coefficients. Back in my head I thought the coefficients had to be integers for some reason. Thanks, I'm learning a lot of new interesting things here. --BiT (talk) 10:17, 8 August 2009 (UTC)

I'm glad that you have clarified this point, but I'm not quite sure that other people will have trouble with the same point as you. Apart the case of truly ambiguous sentences, everybody meets his/her personal point to clarify, and the reason why a certain point is obscure is linked to the reader's personal story (I'm thinking to my personal experience too of course). In most cases this psychological reason is something really doomed to remain obscure, and not so important after all. A rational function is a quotient of two polynomials: why isn't this clear enough? I fear that if you make explicit all possible clarifications the resulting very long explanation will appear going-to-nowhere, and not clear at all. (By the way: what's the need of subscripts m and n ? That's obscure to me). Last remark: if one wants to specify the nature of the coefficients, one says, for instance: "a rational function with coefficients in the field k". Most often, one refers to real or complex coefficients, so in particular, as Meni Rosenfeld pointed out, any constant (rational or irrational) is a rational function; moreover of course P(x)/0 is not a rational function, because it is not even a quotient: it's devil-knows-what). --pma (talk) 00:58, 8 August 2009 (UTC) PS: But maybe the case of polynomials and constants may be added in the subsequent section of examples, that sounds nice.

Fair enough, I just somehow found something wrong with $5$ being a rational function, so I decided to go here for confirmation. "A rational function is a quotient of two polynomials: why isn't this clear enough?"- because the integer $5$ is not a quotient of two polynomials, it's not a quotient at all.

As for the subscripts, my calculus book sometimes used these subscripts to denote the power of the polynomial function. I actually find it handy to use, is it wrong? --BiT (talk) 10:17, 8 August 2009 (UTC)

It's important to distinguish between an object (e.g., a number) and its representation(s). The digit "5" does not represent the number 5 as a ratio. However, the number 5 itself is a ratio - it is 5/1. Similarly, using the symbol "5" to represent the function $f:\mathbb {R} \to \mathbb {R} ,x\mapsto 5$ is not a representation of the function as a quotient of polynomials - but the function is the quotient of the polynomials 5 and 1.

This confusion is also manifested in you thinking that "a rational function is any function written as the ratio of two polynomial functions" could make sense literally. Writing a function differently does not change the function itself. The properties of a function can depend on how it is possible to write it, but not on the specific way you choose to write it.

Regarding subscripts - I wouldn't call it wrong, but it can be unnecessary and confusing when the degree of the polynomial is not of key importance. -- Meni Rosenfeld (talk) 18:29, 8 August 2009 (UTC)

Matrix function

I was reading this article, and began to wonder: is there a type of function in use that describes a matrix based on every entry's position within the matrix itself? For example, consider a matrix in the form

$\mathbf {A} ={\begin{bmatrix}a_{1,1}&\cdots &a_{1,n}\\\vdots &\ddots &\vdots \\a_{m,1}&\cdots &a_{m,n}\end{bmatrix}},$

and a function g of real integers m,n such that

$g:a_{m,n}\mapsto m^{n}\,\!$

and such that we can call say, for example, that for $2\leq m\leq 5$ and $2\leq n\leq 5$,

$g(a_{m,n})={\begin{bmatrix}4&8&16&32\\9&27&81&243\\16&64&256&1,024\\25&125&625&3,125\end{bmatrix}}.$

Do such functions as g have a name, are they used, and which article should I read? —Anonymous Dissident^Talk 08:51, 6 August 2009 (UTC)

$g(a_{m,n})\,$ does not depend on $a\,$ so you would write $g_{m,n}=m^{n}\,\!$ Bo Jacoby (talk) 09:00, 6 August 2009 (UTC)

I had a feeling about that. I was initially going to use g(m,n), but opted with g(a_m,n). Never thought of g_m,n. —Anonymous Dissident^Talk 10:11, 6 August 2009 (UTC)

Let me see if I understand the question properly. You have an m × n matrix $A:=(a_{i,j})\,$ where $1\leq i\leq m$ and $1\leq j\leq n$. You want a function f so that $f(a_{i,j}):=i^{j}?\,$ We could record the behaviour of f by a function $F:{\mbox{Mat}}_{m,n}(\mathbb {R} )\to \mathbb {Z} ^{m\times n}$, where

$F(A)=(1^{1},\ldots ,1^{n},2^{1},\ldots ,2^{n},\ldots ,m^{1},\ldots ,m^{n})\ .$

Well, if this is the case then it sounds like F would be a pretty dull function. The infinite dimensional space of all m × n matrices would get sent to a single point. For example, consider the space of 2 × 2 matrices with real entries (denoted by ${\mbox{Mat}}_{2,2}(\mathbb {R} )$). We have

$F\left({\mbox{Mat}}_{2,2}(\mathbb {R} )\right)=(1,1,2,4)\sim \left({\begin{array}{cc}1&1\\2&4\end{array}}\right)$. ~~ Dr Dec (Talk) ~~ 16:51, 9 August 2009 (UTC)

I think you made a mistake when you said that the space of all m × n matrices is infinite dimensional. It has dimension mn.--Shahab (talk) 08:05, 10 August 2009 (UTC)

Easy contour integral

Hey all, I did this contour integral and a friend did it and we got answers that differ by a sign. I can't find mistakes in either! So, here is my solution and hopefully you can tell me if I made a dumb mistake that I can't see.

Find $\int _{\gamma }z^{1/2}dz$ where $\gamma$ is the lower half of the unit circle from +1 to -1.

So, I figure I can parametrize this as $\gamma (t)=e^{it}$ as t ranges from 2pi to pi. Then

$\int _{\gamma }z^{1/2}\,dz=\int _{2\pi }^{\pi }e^{it/2}ie^{it}\,dt=-i\int _{\pi }^{2\pi }e^{i3t/2}\,dt={\frac {-2i}{3i}}\left[e^{i3t/2}\right]_{\pi }^{2\pi }=-{\frac {2}{3}}(e^{i3\pi }-e^{i3\pi /2})=-{\frac {2}{3}}(-1+i)={\frac {2}{3}}(1-i)$

My friend used $\gamma (t)=e^{-it}$ as t ranges from 0 to pi. This gives

$\int _{\gamma }z^{1/2}\,dz=\int _{0}^{\pi }e^{-it/2}(-i)e^{-it}\,dt=-i\int _{0}^{\pi }e^{-i3t/2}\,dt={\frac {2i}{3i}}\left[e^{-i3t/2}\right]_{0}^{\pi }={\frac {2}{3}}(e^{-3i\pi /2}-e^{0})={\frac {2}{3}}(i-1)$

Thanks for any help. StatisticsMan (talk) 14:13, 6 August 2009 (UTC)

I started working out, and then realized something: surely $z^{1/2}$ has TWO solutions for non-zero $z$? Couldn't both answers be right? --Leon (talk) 14:41, 6 August 2009 (UTC)

Leon's right. If we account for the two square roots in your parameterization $z=e^{i(t+2\pi n)}$ for $n=0,-1$. Your calculation just picks up a factor of $e^{i\pi n}$ which toggles between your friend's and your own solution. Martlet1215 (talk) 15:22, 6 August 2009 (UTC)

[ec] In other words, the "mistake" is either $\left(e^{it}\right)^{1/2}=e^{it/2}$ or $\left(e^{-it}\right)^{1/2}=e^{-it/2}$. Which one is the real one depends on the specific branch of $z^{1/2}$ intended. -- Meni Rosenfeld (talk) 15:27, 6 August 2009 (UTC)

What does [ec] mean? I have seen it before and didn't know. As to this specific problem, I get what you are saying as far as the two branches and two possible answers but not exactly how to write it up. Are you saying I should use the parametrization Martlet gave and do the integral? Then my "answer" would be two different values, depending on n = 0, -1? And, this is just part (b) of a question. Part (a) is the same function over the upper half of the unit circle from 1 to -1. So, it would also have 2 answers right? Thanks for all your help! I should have titled this "tricky" contour integral. StatisticsMan (talk) 16:04, 6 August 2009 (UTC)

ec=edit conflict --78.13.139.160 (talk) 16:11, 6 August 2009 (UTC)

In this technical sense. — Emil J. 16:32, 6 August 2009 (UTC)

Indeed, and I have mentioned it primarily to clarify that I left my comment as is, without checking its synergy with Martlet's. -- Meni Rosenfeld (talk) 16:46, 6 August 2009 (UTC)

Yep, it does mean that those are the two solutions, and that both should be included - unless the question specifies one and which one is needed. It's pretty usual to have more than one answer to these kind of things though, so don't worry! --Leon (talk) 16:29, 6 August 2009 (UTC)

I disagree. Depending on the branch, there are infinitely many possible solutions to this integral. The question is pretty meaningless without any clarification of the branch (or the assumption of the standard branch). -- Meni Rosenfeld (talk) 16:43, 6 August 2009 (UTC)

For any given z, there are only two values of z^1/2. It follows easily that there are only two choices of the branch which are continuous along the whole integration path, giving the two solutions computed above. — Emil J. 16:51, 6 August 2009 (UTC)

To the best of my knowledge, there is no problem with a (moderately) discontinuous integrand. -- Meni Rosenfeld (talk) 17:01, 6 August 2009 (UTC)

OK, in principle there is no problem even with bestially discontinuous integrands, in this case it is sufficient (and necessary) that the set of points which take a particular one of the continuous branches is measurable. However, in the given context (path integrals in complex analysis) it is tacitly assumed that we are computing integrals of holomorphic functions along continuous paths; there are only two choices how to construe the given example in such way. — Emil J. 17:15, 6 August 2009 (UTC)

You are not bound to use any specific parameterization. You only need to make sure which branch the question is asking about, and find out how this branch applies to the values in your contour. If nothing else is specified, the intended branch is probably $(re^{it})^{1/2}={\sqrt {r}}\ e^{it/2}$ for $t\in (-\pi ,\pi ]$. For this branch, your friend's solution is correct, because $\left(e^{-it}\right)^{1/2}=e^{-it/2}$ for $t\in (0,\pi )$, while $\left(e^{it}\right)^{1/2}=-e^{it/2}$ for $t\in (\pi ,2\pi )$. -- Meni Rosenfeld (talk) 16:39, 6 August 2009 (UTC)

Okay. Thanks. Here no parametrization is given. After all this, my guess is perhaps the whole point of this problem is to recognize there are two branches and get both answers. Otherwise, it's a very easy problem and it is from an old qualifying exam. StatisticsMan (talk) 16:47, 6 August 2009 (UTC)

Again, if you do not require the integrand to be continuous along the path, there are many more than two branches and two answers.

The parameterization is something you choose yourself. What should be given is a branch. -- Meni Rosenfeld (talk) 17:05, 6 August 2009 (UTC)

Maybe a good answer would include the fact that the square root function is multivalued and that an answer including an integer, say n, giving all solutions would be the correct solution. All other attempts are only partial solutions. For example, consider the equation $\sin(x)=0$. What is the solution? Well, we might ask over which interval x lies (this is like the branch problem above), or we could just say

$x\in \{\pi n:n\in \mathbb {Z} \}\ .$

Then there's no doubt: we have every solution! ~~ Dr Dec (Talk) ~~ 17:06, 9 August 2009 (UTC)

Another integral

$f(z)=e^{z+1/z}$. I already calculated the residue at 0 to be $\sum _{n=0}^{\infty }{\frac {1}{n!(n+1)!}}$. Now, I need to calculate the integral around the unit circle with counterclockwise orientation. Of course, I use the residue theorem. It "seems" obvious that 0 is the only singularity but how can I know this for sure? Thanks StatisticsMan (talk) 14:36, 6 August 2009 (UTC)

I did this stuff a long time ago, but if $y$ belongs to the complex numbers, so does $e^{y}$. That's because $e^{y}$ is an Entire function. A simple calculus exercise can demonstrate that everywhere else derivatives exist. --Leon (talk) 14:47, 6 August 2009 (UTC)

Oh yea, good point. For some reason I was thinking of showing the Laurent series has no negative terms. But, it's pretty clear it's differentiable everywhere but 0. Thanks. StatisticsMan (talk) 14:59, 6 August 2009 (UTC)

In other words: the composition of two holomorphic functions is itself holomorphic. Applying this principle in a systematic way is usually enough to find singularities of compound functions like this f. — Emil J. 15:10, 6 August 2009 (UTC)

Thanks! Another thing I knew but didn't think about. Your mentioning it helps my understanding. StatisticsMan (talk) 15:56, 6 August 2009 (UTC)

Assuming you're working over the complex plane, and not the the Riemann sphere, then the function $f(z)=e^{z+1/z}$ has a pole when its value is infinite. Well,

$z+{\frac {1}{z}}={\frac {z^{2}+1}{z}}$

and this is finite for all z ≠ 0. The exponential function is finite for all finite values, so f doesn't have a pole for all z ≠ 0. It's as simple as that! No need to differentiate.

ON A CAUTIONARY NOTE: WE AREN'T HERE TO DO PEOPLE'S HOMEWORK ~~ Dr Dec (Talk) ~~ 17:16, 9 August 2009 (UTC)

This isn't homework. Nor are any of my other recent questions. I am not in any classes. And, both questions of mine that you responded to were questions where I had already done most of the work and I was just asking for a bit of help. Even if these were homework questions, these are types that are allowable here.StatisticsMan (talk) 13:54, 10 August 2009 (UTC)

The math behind general relativity

Hi all:

I have recently begun studying Einstein's general theory of relativity on my own. But man oh man did I not quite expect the math that I was plunging myself into! To me all that math seems to be really bad abuse of all the calculus notations I studied in college. And I ended up not really understanding anything in the book other than the grand conclusions, which I already know from reading relevant entries in Wikipedia.

I studied engineering while in college, so I have only studied the basic higher mathematics, such as single-variate calculus, linear algebra 101 and some discrete math. I did take some system control courses in my junior and senior years that delved into higher order differential equations, and I also studied some multi-variate calculus on my own. As for physics, I have studied freshman's year of mechanics and electromagnetics (didn't get to Maxwell's equations despite prof's promise), and in my sophomore year I picked up some waves, simple harmonic motion, and some quantum theory and special relativity (basically repeated high school physics with calculus thrown in) all rolled up into one second-year course.

My question is, given my background above, what should I do in order to ease the pain with which I read my general relativity textbook? I am thinking of brushing up on my math, I saw many new terms introduced in the general relativity book such as p-forms, manifolds, exterior derivatives, Koszul connection, tensors, Hodge's star operator, wedge operator, etc. I am wondering if there is any one or two math textbook that ties all of these concepts together? Or is there some other better way of "opening up" my mind to general relativity?

On a more curious note, did Einstein himself actually knew all this math and used them when he penned his general theory of relativity back in 1915? (I am asking this because many results in my textbook seems to be derived from papers published in the 1960's, well after Einstein had passed away).

Thank you so much for your help!

L33th4x0r (talk) 22:14, 6 August 2009 (UTC)

There are some GR textbooks out there that don't assume much maths and teach you all the Riemannian geometry and tensor calculus you need. If yours doesn't, find a different one - I suggest one aimed at physicists rather than mathematicians (as a mathematician, I don't have one I can recommend). Einstein would have known quite a lot of maths, but he only came up with the basics of GR (I don't mean to belittle the achievement - getting started is always the hardest bit), a lot of other physicists and mathematicians have added to the theory since then (Einstein added to it after 1915 as well). The more recent results will either be easier ways of doing things that Einstein did the hard way or results used in more recent bits of GR. --Tango (talk) 23:17, 6 August 2009 (UTC)

If you are looking for a physicsy approach you could try Relativity: Special, General, and Cosmological by Wolfgang Rindler. It eases you in through Special Relativity introducing some of the mathematics you'll need for the General stuff later. Also, from a physicist's perspective, most of the terms you mentioned aren't taught in a first, or second course in the subject (at least where I'm from). I think you can safely forget p-forms and wedge operators for a while. Martlet1215 (talk) 00:08, 7 August 2009 (UTC)

General relativity is geometry. I you want to enter the necessary maths, I would suggest you Barret O' Neill's book: Semi-Riemannian Geometry with Applications to Relativity. Of course it is well possible that there exists a textbook on GR, with no or little maths: physicists are able to make it possible such unphysical, paradoxical things... pma (talk) 13:27, 7 August 2009 (UTC)

Thank you all for your help! I feel more confident now. L33th4x0r (talk) 17:16, 7 August 2009 (UTC)

Resolved

L33th4x0r, Although the matter has been resolved, I still wish to give you more advise. I have encountered a similar problem before. I suggest you to read something more understandable such as this [1], especially the long course online.

The Successor of Physics 13:38, 11 August 2009 (UTC)