Wikipedia:Reference desk/Archives/Mathematics/2008 March 24

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March 24

Checking number of citations

Newbie question: how can I verify how many citations a math paper has? For example:

http://www.math.princeton.edu/~annals/issues/2004/Sept2004/Agrawal.pdf

Thanks in advance. Mdob | Talk 00:13, 24 March 2008 (UTC)

Google Scholar ([1]) says it has 352 cites. I'm not sure how accurate that is. --Tango (talk) 00:18, 24 March 2008 (UTC)

Try MathSciNet -mattbuck (Talk) 00:24, 24 March 2008 (UTC)

MathSciNet finds 27 (+5 reviews) - that's a lot less than 352... --Tango (talk) 00:36, 24 March 2008 (UTC)

Thanks, Tango (and mattbuck). I already knew of MathSciNet, but somehow missed this feature in Google Scholar. Indeed, 300+ is the expected number for a paper in Annals, wich has a very high impact factor.

Lebesgue measure of a subset of [0, 1) invariant under translation by Q

Let $\Omega =[0,1)=\mathbb {R} /\mathbb {Z}$; that is, addition is taken modulo Z (so that 0.7 + 0.6 = 0.3). We say that $S\subset \Omega$ is Q-invariant if it is invariant under translation by Q, i.e., for every $q\in \mathbb {Q} ,S+q=S$. My question is, if S is Q-invariant, does that imply that the Lebesgue measure of S is either 0 or 1? I've proven a few basic lemmas (for example, that $m(S\cap [a,b))=(b-a)\cdot m(S)$), and I've gained a better understanding of Lebesgue measure for it, but neither I nor the number theory professor I asked this to got anywhere (the question arose from proving the Thue-Siegel-Roth theorem, if I remember). Eric. 86.152.32.69 (talk) 02:29, 24 March 2008 (UTC)

What if you start with a non-measurable set, and take the union of that set with all Q translations of it? Would that set still be non-measurable? -GTBacchus^(talk) 03:12, 24 March 2008 (UTC)

I thought about that too. Starting with a Vitali set would give you the entire real line, so that's no good. Maybe there's another non-measurable set which works, though. —Bkell (talk) 05:19, 24 March 2008 (UTC)

If S is Lebesgue measurable, then yes. Look at the Lebesgue density theorem. If S has positive measure, then for any ε>0, there's a point x and an arbitrarily small open interval containing x such that the measure of S on the interval is at least 1−ε times the measure of the interval. Now shift the interval around by rationals to show that the measure of S is at least, say, 1-2ε. Since ε was arbitrary, the measure of S is one. This general phenomenon is called ergodicity. --Trovatore (talk) 06:16, 24 March 2008 (UTC)

Thanks for your help, everyone. I see now that I had conjectured the Lebesgue density theorem... it's reassuring to know that it's true, although that puts me no closer to finding a proof of it (the Lebesgue density theorem). Hmmm... is the Lebesgue density theorem easy enough that I should try to figure out a proof for myself, or does it take, say, 20 pages of obscure higher math to prove? Eric. 81.157.254.39 (talk) 15:15, 24 March 2008 (UTC)

Honestly I'm not sure I've ever been through the proof of the full result, but I doubt it's hard. I came across a slightly weaker version of it on my own once before I recall hearing about the standard result. Think about this: Suppose there were a set so "evenly smeared out" on the real line that it picked up, say, 75% of the measure of every interval. Its measure on [0,1) is 0.75, its measure on [0.3,0.7) is 0.3, etc. Well, apply the definition of measure, and cover this set with a countable collection of disjoint open intervals, such that the sum of the lengths of the intervals is, say, 0.8. But now for each of the intervals in the cover, the measure of the set in that interval, by assumption, is 0.75 times the length of the interval, so what's the total measure of the set?

Fiddle with this sort of idea long enough and I expect you can prove the full theorem, though I personally have never bothered. --Trovatore (talk) 18:21, 24 March 2008 (UTC)

The question of whether there's a non-measurable set closed under rational translation is also interesting. We want a set A, closed under translation by rationals, such that A has neither measure zero nor measure one. Any measure zero set is a subset of a measure zero G-delta set; any measure one set contains a measure one F-sigma set. So enumerate the measure-zero G-deltas and measure-one F-sigmas together in the smallest possible order type. We'll build up two sets A and B by transfinite recursion. If the next set to be considered is a measure-zero G-delta, we want to refute the claim that A is contained in it. So pick a real number x that's not in the measure-zero G-delta, nor in A or B as constructed so far, and throw x and all its rational translates into A. If the next set is a measure-one F-sigma, we want to refute the claim that A contains this set. So pick a real number x that is in the F-sigma set, but not in A or B so far, and throw it and all its rational translates into B (thus forever preventing them from getting into A). How do we know we can always find such an x? Well, at any point, we've done less than $2^{\aleph _{0}}$ iterations, and at each iteration we've thrown only countably many points into A or B, so both A and B have less than full cardinality. However each measure-one set has full cardinality, so there's a point in it that's not in A or B, and each measure-zero set has a complement with full cardinality, so there's a point not in it and also not in A or B. When we're done, we've guaranteed that the final version of A has neither measure zero nor measure one, and by construction it's closed under translation by rationals --Trovatore (talk) 07:04, 24 March 2008 (UTC)

Interesting. So, picking the numbers x at each step requires the axiom of choice, right? (As I understand it, it is not possible to prove the existence of a non-measurable set from within ZF.) I'll admit to not really understanding transfinite recursion or ordinals, but I think I've managed to wrap my head around your argument. Actually, this is the first time I've seen transfinite-anything make any sense. Eric. 81.157.254.39 (talk) 15:15, 24 March 2008 (UTC)

Right, picking the numbers requires AC, and so does picking the wellordered enumeration of the measure-one F-sigmas and measure-zero G-deltas. --Trovatore (talk) 22:26, 24 March 2008 (UTC)

Non-roots of unity having modulus 1

While refreshing some elementary complex number theory, I realise that I have a problem in proving that a complex number is *not* a root of unity. For instance, it is quite obvious that ${\frac {1}{4}}(\pi +i{\sqrt {16-\pi ^{2}}})$ is not a root of unity, even if it has modulus 1: too much irrational stuff around. But how may I prove it formally? Thanks!

Is there any reason it couldn't be an irrational root of unity? I mean, we define the square roots of unity as 1^(1/2). Surely there's no reason we couldn't define the pi-th root of unity as 1^(1/pi)? -mattbuck (Talk) 15:40, 24 March 2008 (UTC)

Yes, sorry, I meant "is not an n-th root of unity for some natural n".

Irrational roots aren't very useful, since they aren't going to be cyclic under multiplication. You can define them, but pretty much whenever you want to use a root of unity, it's necessary that it be an integer root, otherwise you end up with an infinite number of them. --Tango (talk) 16:31, 24 March 2008 (UTC)

Let's say you wanted to prove it's not an nth root of unity for some given n. Using standard trig identities, in particular the angle sum formulas and the pythagorean theorem, you can get an algebraic formula for the sine of a whole angle in terms of the sine of one nth that angle, and likewise for cosine. Since the sine and cosine of 2pi are whole numbers, the sine and cosine of 2pi/n must be algebraic. Their quotient is then algebraic, but that would have to equal the tangent of 2pi/n, which equals by assumption sqrt(16-pi^2)/pi. There's no algebraic formula for pi, so that's impossible. Black Carrot (talk) 17:53, 24 March 2008 (UTC)

If it were a root of unity, it would be algebraic. Then so would its real part be algebraic. However, pi is transcendental. The proof can be found in any introductory book on transcendental number theory, though finding a motivated proof is harder. Try "Making Transcendence Transparent: An intuitive approach to classical transcendental number theory" by Edward B. Burger and Robert Tubbs. Boris Bukh (talk) 17:12, 28 March 2008 (UTC)

Polarity associated with a conic, defined by Steiner construction?

Hello,

Steiner defined a conic in a projective plane as "the locus of the points of intersection of corresponding lines of two homographic pencils with distinct vertices". This is a very nice definition... but in most cases conics come with (orthogonal) polarities (mapping points to the polar line). Is there a nice way to visualize this polarity?, just using the two homographic pencils? I've been thinking about this for a while now, but I haven't found an answer yet. Does anyone know more? ThanksEvilbu (talk) 21:16, 24 March 2008 (UTC)

Parabola vs Hyperbola

what is difference b/w a parabola and hyperbola? I know the geometric difference. But can someone explain the difference in terms of mathematics?[2]

Have you read parabola and hyperbola? Those articles should help. I'd say the biggest difference is that a hyperbola has asymptotes, a parabola doesn't. --Tango (talk) 16:51, 24 March 2008 (UTC)

Parabolas have an eccentricity of 1; whereas hyperbolas have eccentricities greater than 1. (Ellipses have eccentricity less than 1.) In the equation for a conic section in Cartesian coordinates, parabolas have $B^{2}-4AC=0$; whereas hyperbolas have $B^{2}-4AC>0$. (Ellipses have $B^{2}-4AC<0$.) --Spoon! (talk) 22:43, 24 March 2008 (UTC)

For one thing, a parabola has no asymptotes. But the question is phrased in a vague way, so beyond that maybe the best thing is to suggest you read the standard articles and then see if you have more specific questions. The "geometric difference" is a "difference in terms of mathematics. Michael Hardy (talk) 23:07, 24 March 2008 (UTC)

The parabola closes in the infinity (it touches the ideal line of the plane), the hyperbola continues smoothly on the other side in the infinity (it intersects the ideal line). – b_jonas 19:23, 25 March 2008 (UTC)

rational expressions

Hello, Take any number (except for 1). Square that number and then subtract one. Divide by one less than the original number. Now subtract your original number. Did you reach 1 for a answer? How does the number game work? (hint Redo the number game using a varible instead of an actual number and rewiten the problem as one rational expression). How did the number game use the skill of simplifying ration expression? state whether you number game uses the skill of simplfying ration expression. Please help I don't get what it is saying!

Go through the process using "x" instead of a number and you'll get a rational expression, you can then simplify that (hint: use difference of 2 squares), and you'll find it equals 1. --Tango (talk) 17:52, 24 March 2008 (UTC)

I'm still not getting it. Could you please give me some example so I can understand a little better thank you

(Please reply in this section, rather than creating a new section for each message, thanks!) I'll start you off: You start with x, you then square it and get x². You then subtract one and get x² - 1 and so on. Once you get to the end of the process, simplify the expression and you'll find it simplifies to simply "1". --Tango (talk) 19:23, 24 March 2008 (UTC)

rational expression

How is doing operations (adding, subtracting, multiplying, and dividing) with rational expressions similar to or different from doing operations with fractions? Can understanding how to work one kind of problem help understand how to work another type. When might you use this skill in real life?

Fractions are just a special case of rational expressions, so the operations are basically the same. Fractions are easier, since there is less to do, but the basic process is identical. I'm not sure you'll use rational expressions directly in real life, but they come up a lot in various areas of maths, and those areas have real world applications. --Tango (talk) 17:55, 24 March 2008 (UTC)