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[edit] December 24

[edit] Green's second identity and Green's functions for the Laplacian

Hi all,

I'm trying to prove that the for Green's function for the Laplacian, G(r;r₀) in any arbitrary 3D domain, symmetry holds between r and r₀; i.e. G(r;r₀)=G(r₀,r). My friend suggested I should try using the Second Green's identity (I sometimes wonder if my life would be a more interesting place if Green were never born!), but I can't seem to get anything out, perhaps I'm being slow this time of night.

Does anyone else have any luck using Green's 2nd identity? Thanks very much, Delaypoems101 (talk) 02:30, 24 December 2009 (UTC)

[edit] sequence space

Resolved.

I'm trying to prove that the sequence space of all complex sequences is a metric space with the metric $d(x,y)=\sum_{j=1}^\infty \frac{1}{2^j}\frac{|x_j-y_j|}{1+|x_j-y_j|}$ . My questions are how can I show that this series is always convergent and why does d(x,y)=0 imply x=y. Thanks-Shahab (talk) 06:36, 24 December 2009 (UTC)

Doesn't really matter, d fails to satisfy the triangle inequality.--RDBury (talk) 07:18, 24 December 2009 (UTC)

No it satisfies the triangle inequality. I can reproduce the proof for that given in my book.-Shahab (talk) 07:26, 24 December 2009 (UTC)

My apologies, I got mixed up when I was checking it.--RDBury (talk) 12:13, 24 December 2009 (UTC)

(ec) Second question is easy: all fractions $f_j = \frac{|x_j-y_j|}{1+|x_j-y_j|}$ are non-negative, so d() is a sum of non-negative terms, and thus can only be zero if all terms are zero, which implies all numerators are zero, so x=y. Now the first question gets easy: as all terms are non-negative AND less than 1 (because for $f_j \neq 0$ we have $f_j = \frac 1 {\frac 1 {|x_j-y_j|} + 1}$ which is a reciprocal of something greater than 1), the sum $d(x,y)=\sum_{j=1}^\infty \frac {f_j} {2^j} \leq \sum \frac 1{2^j} = 1$ --CiaPan (talk) 07:25, 24 December 2009 (UTC)

Thank you, it's clear. Instead of saying d() is a sum of non-negative terms, and thus can only be zero if all terms are zero isn't it more appropriate to say that d(x,y)=0 is a limit of a monotonic increasing sequence of non-negative terms which is only possible if all terms are zero. I tend to think of series as sequences only.-Shahab (talk) 07:39, 24 December 2009 (UTC)

Both are valid arguments. CiaPan's argument is rooted on the assertion that if $\sum_{i=1}^\infty a_i$ is a convergent sum, with each term in the sum non-negative, $a_j < \sum_{i=1}^\infty a_i$ for all j. The argument you have suggested is rooted on the assertion that if $S_j = \sum_{i=1}^j a_i$ is the jth partial sum of a convergent series, $0\leq S_k\leq S_{k+1}$ for all k. Essentially, both arguments are correct (and similar in nature). However, you are correct to note that in a situation where basic intuition does not apply, it is often more appropriate to employ a formal argument. --PS T 09:13, 24 December 2009 (UTC)

By the way, which book are you studying?

--PS T 09:14, 24 December 2009 (UTC)

Kreysig's. I found an online copy.-Shahab (talk) 09:40, 24 December 2009 (UTC)

Note that you can use other functions $\scriptstyle\phi:[0,\infty)\to[0,\infty)$ in place of t/(1+t) in the definition of d(): precisely, any bounded continuous subadditive increasing function φ such that φ(0)=0 and φ(t)>0 if t>0 produces a distance $\scriptstyle d(x,y)=\sum_{j=1}^\infty 2^{-j}\phi(|x_j-y_j|)$ on the space of sequences. These are topologically equivalent, and induce the product topology. For instance,

φ(t): = min(t,1)

is often used. --pm a (talk) 12:24, 24 December 2009 (UTC)

Thanks everyone and merry christmas-Shahab (talk) 04:34, 25 December 2009 (UTC)

[edit] Rolling sphere

An unconstrained sphere resting on the top of a fixed one is in unstable equilibrium. Suppose a minute disturbance (e.g. it's given an initial velocity of one millionth of the fixed sphere's radius per second) starts it rolling under gravity. Assuming no slipping, are there any circumstances which will make it leave the surface of the fixed one before the 90° point has been reached?→→86.155.184.27 (talk) 17:21, 24 December 2009 (UTC)

I suspect the answer might depend on whether "rolling" means it's not "slipping".

But then on another couple of seconds' thought (I haven't thought this one through) I would think it would have to leave the surface before reaching the 90° point, because its motion has a horizontal component and there's inertia. Reaching the 90° point would mean going straight down with no horizontal component to its motion. Michael Hardy (talk) 19:53, 24 December 2009 (UTC)

OK, now I see there's an explicit statement that it's not slipping. I don't know whether that actually matters. Michael Hardy (talk) 06:43, 25 December 2009 (UTC)

This problem is suited for Lagrangian mechanics. Bo Jacoby (talk) 23:09, 24 December 2009 (UTC).

I think in the general case of an object rolling off the fixed sphere (assuming the size of the rolling sphere is much smaller than that of the fixed one), it will depart at $\cos \, \theta = \frac{2}{3 + \frac{I r^2}{m}}$ (where

I

is the moment of inertia of the sphere,

m

its mass, and

r

its radius). Note that this reduces to a constant in the special case of a particle sliding down the sphere (

I = 0

), giving $\theta \approx 48.2^{\circ}$ . Michael is completely right---since the object acquires some horizontal velocity, it has to leave before the 90° point. You can analyse this by working out the velocity as a function of angle (by conserving energy), then working out the angle at which the gravitational pull on the object (directed towards the sphere) is no longer enough to keep it in circular motion around it. — Zazou 00:40, 25 December 2009 (UTC)

[edit] December 25

[edit] Easy way of deciding if two lines cross?

I am writing a computer program where many lines are stored as a pair of x.y coordinates. I would like to be able to decide if two lines cross. What would be the easiest way to program this please? I can think of changing the lines into y=mx+c format, doing a simultaneous equation (I think) to find the point of intersection, and then checking that this intersection point is within each line segment. But is there any easier way please? (I am not fluent with matrices and the language I am using has no matrix commands). Maybe something regarding the angles between the four points - I'm guessing. A simple way to find the x.y coordinate of the point of intersection would also be useful. Thanks 92.24.44.4 (talk) 14:52, 25 December 2009 (UTC)

Just determine whether or not they are parallel. If they're parallel, then either they never intersect or they're the same line. If they're not parallel, then they must intersect at exactly one point. No need to determine the point of intersection. I'll leave it as an exercise to you to figure out how to determine if they're parallel, and to explain why this technique doesn't work in 3 dimensions. --COVIZAPIBETEFOKY (talk) 16:58, 25 December 2009 (UTC)

Sorry, I should have made clearer than the lines are not infinate. They can be non-parallel and still not cross. 78.146.194.118 (talk) 17:07, 25 December 2009 (UTC)

Your method sounds best to me. You should first check they aren't the same line (if they are you just need to check the order of the endpoints to see if they overlap) then that they aren't parallel (if they are and they aren't the same line, they won't intersect) and then you can find the point of intersection and see if it is in both lines. To get the intersection point from two equations, y=mx+c and y=nx+d, you can just do x=(d-c)/(m-n) (to derive that just put mx+c=nx+d and rearrange) and then plug that into y=mx+c to get y. --Tango (talk) 17:37, 25 December 2009 (UTC)

I'm wondering if the four end points of the two lines would always make a polygon with a concave part in it if they do not cross? 78.146.194.118 (talk) 17:49, 25 December 2009 (UTC)

One way you could do it is that if the segment from A to B and from C to D don't cross then either (B-A)×(C-A) and (B-A)×(D-A) will have the same sign or (D-C)×(A-C) and (D-C)×(B-C) will have the same sign. Here × is the cross product, A×B = x_Ay_B - x_By_A. There might be some more efficient way to get to that though.

For the intersection point, I think it should be A + (((C-A)×(D-C))/((B-A)×(D-C)))(B-A) if I didn't screw anything up. You could also use that intersection point to decide if the segments cross, although I think this way is more computationally intensive unless you need the intersection point anyway. Rckrone (talk) 21:23, 25 December 2009 (UTC)

(Answering 78.*)... No, they do not make a concave polygon. This is a common homework or quiz question in algorithms programming. Nothing in the question assume that the direction of the lines is from the Y axis towards infinity. One may be right-to-left. The other may be left-to-right. This creeps in again in processor/ALU design. Division is a very nasty time consumer. Comparison is not. So, using less-than/greater than, you can sort the points to form a concave polygon. Then, if you go around the four points in a clockwise manner, you can detect that each turn is a right turn by only using subtraction (which is actually a very cheap addition process inside the computer). -- kainaw ™ 21:47, 25 December 2009 (UTC)

I forgot to mention that some student always comes up with the idea of just comparing the endpoints. It is a bit trivial to come up with an example that nullifies anything that depends solely on comparing endpoints. -- kainaw ™ 21:50, 25 December 2009 (UTC)

See also Wikipedia:Reference desk/Archives/Mathematics/2009 October 4#Best way to calculate if a line crosses another line, or a polygon.. Is there an article to point to on this?--RDBury (talk) 21:59, 25 December 2009 (UTC)

To the OP: please do look at the link that RDBury gives, and the explanation that RDBury and BenRG give there. Intuitively, if we wish to check whether AB and CD cross, we check whether A and B are on opposite sides of the line CD, and whether C and D are on opposite sides of AB. To check which side of a line that a point is on, we compute the appropriate cross product (or equivalently, the signed area of the triangle the three points form). BenRG provides code in C++; while I haven't checked it myself, it looks correct.

Don't use methods that involve computing intersection points, because these are generally more difficult to code correctly (with special cases like infinite slope, etc.), not numerically stable, and slower (although speed is unlikely to be a concern either way). Although there is nothing mathematically wrong with this approach (and this is a mathematics reference desk, after all), from a programming perspective it is not preferred. Eric. 131.215.159.171 (talk) 23:54, 25 December 2009 (UTC)

The formula: (x₂-x₁)(y₃-y₁)-(y₂-y₁)(x₃-x₁) :is commonly referred to as "turn" in computer programming - mainly in graphics. Going from point 1, to point 2, to point 3, if the value is positive, you made a left turn. If it is negative, you made a right turn. If it is zero, the three points are on a line (note: it could be a 180 degree turn). Calculating turn comes in handy in a lot of graphics programming. -- kainaw ™ 02:25, 26 December 2009 (UTC)

I didn't know that... in the context of computational geometry I've only heard it referred to as the "signed area". Eric. 131.215.159.171 (talk) 08:25, 26 December 2009 (UTC)

To answer the OP and my own question, we have an article, Line segment intersection in this topic but it's in dire need of expansion. I get the impression that mathematicians look at the problem and see the main issue as determining whether two line segments intersect; multiple line segments are just a matter of applying the solution multiple times. While to people in computer science, the problem of whether two line segments intersect is simple algebra and the real issue is to organize the problem so you don't have to test all possible pairs of segments. It seems to me that the article should cover both points of view and right now it just gives an outline of the second. I found lectures notes [1]] which give a pretty good introduction to the subject except they are not self-contained. (For example pseudocode calls a function CCW whose implementation is not given.)--RDBury (talk) 12:16, 26 December 2009 (UTC)

Thanks, as the OP is there any consensus on what the easiest way to check if two lines cross or not, given that I am only fluent in an old version of BASIC and that my maths education stopped when I was 16 years old? 89.240.110.255 (talk) 16:28, 26 December 2009 (UTC)

[edit] December 26

[edit] dual spaces of Sobolev spaces

The Rellich-Kondrachev theorem gives compact embeddings of $W 1, p$ into $L^{p^*}$ , but what can we say about, say, $L q$ and $W - 1, q$ ? I remember it was straightforward, but I'm having trouble finding it in the references, and am rather embarassed that it's not working out easily. Many thanks. 96.235.177.218 (talk) 03:47, 26 December 2009 (UTC)

(To be precise, there is compactness only when q is strictly below the critical exponent: q<p*). I'm not sure of what's exactly your question though. One thing is that dualizing the RK embedding you still get a dense, injective, compact map of the dual of

L q

into the dual of

W 1, p .

What you possibly had in mind is that if a bounded linear operator $\scriptstyle J:X\to Y$ between Banach spaces is compact/injective/dense, then the transpose operator $\scriptstyle J^*:Y^*\to X^*$ is respectively compact/w*-dense/injective. If 1<p<n the space

W 1, p (Ω)

is reflexive, so that "w*-dense" above is the same as just "dense". Was this your question?--pm a (talk) 16:55, 26 December 2009 (UTC)

[edit] Is the word "induce" used technically or non-technically?

Suppose A and B are groups, and N is a normal subgroup of A. Suppose we have an isomorphism $f : A \to B$ ; then we would say that f naturally induces an isomorphism $A/N \to B/f(N)$ .

I would like to know the limits of the word induce. I see two alternatives:

(1) Is the phrase "the map induced by f" rigorously defined to refer to that map which results from passing to the quotient spaces, as in my example? In this case, the phrase "induced map" would have a formal, unambiguous meaning, just as "the pullback of f" has a formal, unambiguous meaning.

(2) Is the phrase "the map induced by f" used informally to refer to any map that results from some kind of a canonical process? For example, would it be correct to say the restriction map $N \to f(N)$ is induced by f? Could we also say the pullback of f (by some other map) is "induced" by f? Maybe the lift of f is also "induced" by f? In this case, the phrase "induced map" would have a subjective meaning, depending on context to establish what particular process we mean.

Of course the actual usage of the word "induce" could differ from both of the two above descriptions, and can vary from one mathematician to another; but I am most interested in the distinction between a formal, technical meaning for "induce" and an informal, non-technical meaning. Thanks. Eric. 131.215.159.171 (talk) 08:43, 26 December 2009 (UTC)

Let A, B, C and D be objects in a category, and let $f:A\to B$ be an element of

H o m (A, B)

. Formally, I would say that f induces an element $g:C\to D$ of

H o m (C, D)

, if there exist morphisms $h_1:A\to C$ and $h_2:B\to D$ such that the following diagram commutes:

$\begin{array}[c]{ccc} A&\stackrel{f}{\rightarrow}&B\\ \downarrow\scriptstyle{h_1}&&\downarrow\scriptstyle{h_2}\\ C&\stackrel{g}{\rightarrow}&D \end{array}$

We shall now restrict our attention to abelian categories. The above diagram includes the two cases you mentioned, as is demonstrated by the following special commutative diagrams (let $h_1:A\to A/N$ and $h_2:B\to B/f(N)$ be the respective canonical homomorphisms):

$\begin{array}[c]{ccc} A&\stackrel{f}{\rightarrow}&B\\ \downarrow\scriptstyle{h_1}&&\downarrow\scriptstyle{h_2}\\ A/N&\stackrel{g}{\rightarrow}&B/f(N) \end{array}$

The above diagram commutes, as you can check. Similarly, consider the following commutative diagram (in this case, let $h_1:N\to A$ and $h_2:f(N)\to B$ be the respective inclusion maps):

$\begin{array}[c]{ccc} A&\stackrel{f}{\rightarrow}&B\\ \uparrow\scriptstyle{h_1}&&\uparrow\scriptstyle{h_2}\\ N&\stackrel{g}{\rightarrow}&f(N) \end{array}$

Note that the vertical arrows in the above commutative diagram point up, as opposed to those in the other commutative diagrams, which point down. I have merely given you a formal definition (in my view) of "induce" in the mentioned situations. I do not quite understand what you would call an "informal definition"; could you please clarify? Hope this helps (and try not to notice the ugly commutative diagrams...

). --PS T 11:52, 26 December 2009 (UTC)

Thank you for your reply; it was helpful. Perhaps I should be more clear. I am not so interested in what the definition, whether technical or non-technical, of "induce" is, per se, as whether mathematicians view the word "induce" as a technical term (like the terms "pullback", "lift", "inverse limit"), or as a non-technical term (like the terms "trivial", "characterization", "equivalent", "canonical", "natural"). I gave examples of what a technical definition for "induce" (the result when passing to the quotient space) and a non-technical definition for "induce" (the canonical result of some natural process) to clarify my meaning of a technical term vs. a non-technical term, but I am not necessarily convinced that either one of those is what most mathematicians use the word to mean. Eric. 131.215.159.171 (talk) 12:55, 26 December 2009 (UTC)

I think that in specific cases, many mathematicians view "induce" as a technical term; one example being "pullback" (or "pushforward") as you mentioned. However, in general, I do not think that all mathematicians have a specific view as to what induce should mean. If I was talking about the pushforward measure in measure theory, the tangent bundle in differential topology, or even quotient spaces in algebra, I would employ specific aspects of the term "induce"; I would not use it in its full generality. I think that this is the case in most of mathematics - often we would generalize a term if we feel that the generalization sheds new light on concrete (or even abstract) cases. For instance, the snake lemma in homological algebra, amidst all this "abstract nonsense" about generalized abelian categories, actually allows one to construct long exact sequences in homology (Zig-zag lemma); a particularly basic tool in homology. Although it seems unnaturally general at first to the beginning student, it actually does shed light on basic tools in singular homology theory (as an example). To summarize, I do not think that mathematicians have found a "specific purpose" of viewing "induce" as a formal term, as people have done in many other branches of mathematics such as point-set topology or abstract algebra. Rather they have formalized the term in specific situations such as the ones I have mentioned ("pushforward", "pullback" etc) and this has been particularly useful. Does this answer your question? --PS T 13:30, 26 December 2009 (UTC)

I personally use "X induces Y (via Z)" for any object or situation Y whose existence and (essential) unicity is guaranteed by X (as a consequence of the theorem or the construction Z, to be specified unless it is clear). It seems to me that this generic use is the most common. In some cases "deduce" or "produce" may be valid alternatives (although probably nobody cares about the etymology). --pm a (talk) 17:26, 26 December 2009 (UTC)

Thanks, you have answered my question thoroughly. Eric. 131.215.159.171 (talk) 21:44, 27 December 2009 (UTC)

[edit] Does U(1) = SU(1)

Does the circle group equal the special unitary group of one dimension? -Craig Pemberton 08:44, 26 December 2009 (UTC)

No. SU(1) is the trivial group {1} - see special unitary group. Gandalf61 (talk) 09:12, 26 December 2009 (UTC)

An element of U(1) is of the form $\begin{bmatrix} a+bi \\ \end{bmatrix}$ where $a+bi\in \mathbb{C}$ , and $\begin{bmatrix} a+bi \\ \end{bmatrix}\cdot \begin{bmatrix} a-bi \\ \end{bmatrix} = \begin{bmatrix} 1 \\ \end{bmatrix}$ (since $\begin{bmatrix} a-bi \\ \end{bmatrix}$ is the conjugate transpose of $\begin{bmatrix} a+bi \\ \end{bmatrix}$ ). This fact allows one to conclude that which you note; U(1) is isomorphic to the circle group. Now, SU(1) is the set of all elements of U(1) having determinant 1. However, a matrix $\begin{bmatrix} a+bi \\ \end{bmatrix}$ has determinant 1 iff $a + b i = 1$ ; equivalently, iff $\begin{bmatrix} a+bi \\ \end{bmatrix} = \begin{bmatrix} 1 \\ \end{bmatrix}$ . Thus, SU(1) is the trivial group, whereas U(1) is the circle group (perhaps I have over-explained a little...). --PS T 12:05, 26 December 2009 (UTC)

[edit] Linear combination of matrices

Is there a reference book or wiki article or something to direct me to research of eigenvalue problem of matrices like $A + x B$ , in particular, properties of the roots of the characteristic polynomial $det(A + x B - λ I)$ with the parameter $x\in\mathbb{C}$ ? (Igny (talk) 17:29, 26 December 2009 (UTC))

Have you tried this article? If not, I recommend it. If so, do you have a specific question?--Leon (talk) 17:50, 26 December 2009 (UTC)

Well I know the general theory of eigenproblem, and I know implicit differentiation well enough to figure out, for example,

λ'(x)

. However I thought that there was some more obscure research of the roots

λ j (x)

from the point of view of the Galois theory, for example. (Igny (talk) 18:22, 26 December 2009 (UTC))

Tosio Kato, Perturbation theory for linear operators. --pm a (talk) 20:05, 26 December 2009 (UTC)

Thank you, I think I read it quite a while ago, I will look it up again. (Igny (talk) 02:16, 27 December 2009 (UTC))

[edit] Algebra over a ring that is a field?

Is it possible for an algebra over a ring-that is, a ring that is not also a field-to be a field? I'm aware that you can describe rational numbers as pairs of integers, but in as much as I understand the term "algebra over a ring", that does not qualify as addition needs to be defined differently to that on a vector space over the ring of integers.--Leon (talk) 17:48, 26 December 2009 (UTC)

What about $\Q$ as an algebra over $\Z$ ..?--pm a (talk) 20:12, 26 December 2009 (UTC)

Didn't I just mention that, and further explain why I figured that it didn't count?--Leon (talk) 21:05, 26 December 2009 (UTC

I don't see why it doesn't count. An algebra over a ring is a module with a suitably-behaved multiplication, that's the only definition I know. $\Q$ can certainly be considered a module over $\Z$ and the usual multiplication is suitably-behaved. --Tango (talk) 21:50, 26 December 2009 (UTC)

Sorry: actually you did, but your explanation was (and I fear, will remain) rather obscure to me. I do not understand your doubts: the definition of algebra is very clear, unambiguous, and standard; and obviously any field is an algebra over any sub-ring of it. --pm a (talk) 23:57, 26 December 2009 (UTC)

I suspect that the motivation for this question comes from the fact that a finite-dimensional algebra (over a field) which is also an integral domain, must necessarily be itself a field; the OP may wish to know whether this can be generalized to an arbitrary ring in place of the field over which the algebra is defined. In general, as pma suggested, the "field of fractions" of an integral domain is an algebra (and the integral domain need not be a field). If a ring has zero divisors, then no algebra over it can be a field. However, if a ring is not necessarily commutative, but has no zero divisors, we can replace the "field of fractions" idea by a "noncommutative division ring" idea via the Ore condition (that is, we can obtain a "noncommutative division ring of fractions" which is an algebra over a given noncommutative ring with no zero divisors satisfying the Ore condition). Hope this helps. --PS T 03:47, 27 December 2009 (UTC)

If K is a field that is also an R-algebra for R some commutative ring with 1, AND such that K is free as an R-module, then I believe R is a field. R is a subring of K, K is a free, divisible module, so R is a divisible R-module, so R is a field. "Free" basically means that the elements of K are ordered pairs, triples, tuples of elements of R with addition defined coordinate-wise, like for vector spaces. If you require the field to be "free", then the coefficient ring R has to have been a field. However, if the elements of K need not have any specific form, but merely need to be able to be multiplied by elements of R, then of course every field is an R-algebra for some R that is not a field (and if you wish, also not the integers). JackSchmidt (talk) 07:53, 27 December 2009 (UTC)

[edit] December 27

[edit] Algorithm to reduce polynary equations to minimum form

I asked this before and maybe the question was ignored due to the holidays...

Is there an algorithm (like for the simplex method in linear programming) to reduce polynary equations to minimum form? 71.100.6.153 (talk) 02:06, 27 December 2009 (UTC)

I believe that the problem is your use of "polynary". That is not a well-defined word. The meaning depends on which group of people is using it. If you just made it up, please define what you mean by it. Otherwise, define what the group of people you got it from intend for it to mean. It literally means "pertaining to many" in the way that binary means "pertaining to two". So, it means "can have many values". Most equations can have many values. Most variables can have many values. Therefore, the usage is very important to make sense of the question. -- kainaw ™ 03:04, 27 December 2009 (UTC)

My intended meaning of polynary is identical to the phrases "multiple state variable" or "many stated variable" or "poly-stated variable" in the same sense that binary pertains to two. My usage covers binary variables and any variable with discrete and finite number of states. My usage does not include fractions directly since probability values can be normalized to percentages and percentages are meaningless beyond a few decimal places which can be rounded or truncated to result in only integer values. 71.100.6.153 (talk) 15:44, 27 December 2009 (UTC)

A polynary equation is a logical equation with variables which may have two states although binary is the specific term used to refer to such an equation as trinary is the specific term used to refer to an equation of variables made up of three states. The algorithm I am looking for, however, should be applicable to all discrete and finite stated equations to include binary, trinary and beyond. 71.100.6.153 (talk) 00:11, 28 December 2009 (UTC)

In short, a polynary variable is any variable having a finite number of discrete states. 71.100.6.153 (talk) 00:20, 28 December 2009 (UTC)

[edit] Measure Theory & Countability

First of all, when I say that a set A is "bigger than" another set B, it means that B is entirely contained in A and that A and B are not equal. So B is a proper subset of A. Working an the interval [0,1] for example, I know that starting from a single point, we can work our way up to the Cantor set which is uncountable and still has measure zero. My question is what is the "biggest" subset of [0,1] that still has measure zero. All of [0,1] has measure 1 obviously.

I have the same question regarding countable sets (countable means finite or countably infinite) in [0,1] for example. Cantor showed (using his usual ingenious arguments) that starting from a single point, we can work our way up to the algebraic numbers which are countable. Is there any set "bigger" than the set of all algebraic numbers in [0,1] which is still countable? Thanks! -Looking for Wisdom and Insight! (talk) 02:56, 27 December 2009 (UTC)

Adding a single point (or any countable set) to a set which is countable/measure zero yields a new set which is again countable/measure zero. You'll need some better questions to get more interesting answers. Algebraist 03:07, 27 December 2009 (UTC)

I think that the OP is looking for the existence of a subset of [0,1] containing the algebraic numbers within [0,1], maximal with respect to the property that it is countable and has measure zero (you might as well omit the "measure zero" from the OP's question since any such countable set has measure zero). In this case, however, there is no such maximal set, for the reason given by algebraist (if M is any such countable set, $M\cup \{x\}$ for $x\in [0,1]\setminus M$ contains M and is countable, so M cannot be maximal with respect to the property of having measure zero). --PS T 04:08, 27 December 2009 (UTC)

You know what I meant and I was afraid of this answer. I thought about that too but that is not what I was looking for. So what is the largest subset of [0,1] which is countable/has measure zero?-Looking for Wisdom and Insight! (talk) 03:37, 27 December 2009 (UTC)

There is no such countable set for the reason given by algebraist above. There is no such set having measure zero for exactly the same reason (if M is any such set having measure zero, $M\cup \{x\}$ for $x\in [0,1]\setminus M$ contains M and has measure zero, so M cannot be maximal with respect to the property of having measure zero). --PS T 04:11, 27 December 2009 (UTC)

Basically, although many aspects of mathematics are firmed on intuition, which formalizm consolidates (and is thus often tacit), in this case formalizm is important to obtain some interesting intuition. --PS T 04:14, 27 December 2009 (UTC)

The question is analogous to asking the existence of a "largest number"; unless you add extra assumptions to your definition of "number", you cannot obtain a meaningful answer. In this case, you will need to add additional assumptions to your definition of "set". --PS T 04:17, 27 December 2009 (UTC)

Thanks for the explanation, everyone!-Looking for Wisdom and Insight! (talk) 04:29, 27 December 2009 (UTC)

Maybe more interesting answers can be found if we also consider description complexity. "Rational numbers" fails because a much bigger set (algebraics) can be obtained with an equivalently short description. "Algebraic numbers and

1 / π

" fails because the added complexity of specifying

1 / π

is not justified by the increase of one element. In other words, that extra element does not "belong" in this set. So, is there a countable set which is significantly larger then algebraics and yet similarly simple? Is there some quantification of these notions, under which an optimal set can be found? -- Meni Rosenfeld (talk) 06:00, 27 December 2009 (UTC)

Perfect, like if all elements in a set share a common property. That is definitely a better wording of my question.-Looking for Wisdom and Insight! (talk) 07:02, 27 December 2009 (UTC)

The Computable numbers are countable and the only way you'll write a number that's not amongst them is by doing something like throwing a dice for each digit. Dmcq (talk) 22:55, 27 December 2009 (UTC)

Well, it depends on what you mean by the way, will I be man enough to ignore the barbarism a dice??? Probably not. At least I can rant about it in small text. by write such a number. Of course you can't really "write" it (because it would take too long) but you can certainly specify one, with no ambiguity whatsoever. For example, consider the number in binary representation that has a zero at position n if the Turing machine with Goedel number n halts, and a 1 otherwise. --Trovatore (talk) 23:01, 27 December 2009 (UTC)

The wikipedia article dice allows it for the singular form but yes die is better. A set larger than the algebraic numbers that includes practically everything before the 20th century can be got by using the hypergeometric series to generate extra numbers. Dmcq (talk) 23:12, 27 December 2009 (UTC)

We could extend that idea to the set of all numbers for which there exists a formula (in the first-order language of set theory) which your favourite set theory proves to uniquely define the number in question. That set's countable, and it contains every number you're ever likely to talk about. Algebraist 02:13, 28 December 2009 (UTC)

I'm not quite sure why you want to drag in provability. What exactly do you mean by saying that the theory proves that the formula describes "the number in question"? You can't talk about the number in question until you have a definition for it.

So presumably what you really mean is that the theory proves that the formula defines a unique real, and then consider the set of all reals that (Platonistically) satisfy some such formula. But then what does the proof add? Why not just consider the set of all reals that Platonistically are the unique real satisfying some formula, whether or not you can prove that there's a unique real satisfying the formula? --Trovatore (talk) 02:59, 28 December 2009 (UTC)

By the way have you come across the Vitali set? Dmcq (talk) 13:29, 29 December 2009 (UTC)

[edit] Largest countable this that and the other

This isn't exactly responsive to what the original poster asked, but it seems similar enough to mention. There are results from effective descriptive set theory that certain lightface pointclasses have largest countable members. For example there is a largest countable $\Pi^0_1$ set of reals. A set of reals is $\Pi^0_1$ if it's the complement of a $\Sigma^0_1$ set; a set is $\Sigma^0_1$ if it's the union of a collection of (open) intervals with rational endpoints, where some computer program, given infinite time, can list all the pairs of rational endpoints involved.

(So a $\Pi^0_1$ set is always closed, but being $\Pi^0_1$ is a stricter notion than being closed. The sense in which it is closed has to be somehow "effective" or "computable". For example there are uncountably many closed sets, but there are only countably many $\Pi^0_1$ sets, because there are only countably many computer programs.)

~~There's a classic paper by Donald A. Martin on this, in Proceedings of the Cabal Seminar, called Largest countable this that and the other.~~

I don't know how to characterize the largest countable $\Pi^0_1$ set, but the largest countable $\Sigma^0_2$ set is simply the set of all reals that are in Goedel's constructible universe. Obviously this requires something beyond ZFC; you have to be able to show that that set is countable. But I think the very modest assumption of zero-sharp suffices. I don't know whether you can prove in ZFC alone that there exists a largest countable set in any of these pointclasses. --Trovatore (talk) 23:14, 27 December 2009 (UTC)

Sorry, I had a total brain freeze. Actually most of what I write above is correct, except that I should be talking about the largest countable $\Pi^1_1$ and $\Sigma^1_2$ sets (note the superscript 1 instead of 0). The definition I gave for $\Pi^0_1$ is correct, but not relevant, because I should have been talking about $\Pi^1_1$ sets, which are much more complicated (as sets) and somewhat harder to explain in elementary terms. The largest countable $\Sigma^1_2$ set (not $\Sigma^0_2$ ) is the set of all reals in Goedel's constructible universe. --Trovatore (talk) 01:34, 28 December 2009 (UTC)

PST's response made me think of a (I hope) interesting question. Let X be the collection of all countable subsets of [0,1], partially ordered by set inclusion. Since there's no maximal element, by Zorn's lemma there exists a chain with no up bound. Can someone exhibit such a chain? I'm thinking about embedding all countable ordinals into [0,1] in some way... Money is tight (talk) 13:05, 29 December 2009 (UTC)

"In some way" — well put. Of course there is an injection from the set of all countable ordinals into [0,1]. If f is such an injection, then letting A_α be {f(β) | β < α}, then the A_α are your desired chain. Good example.

I don't know exactly what you're requiring when you say "exhibit" such a chain, though. It's consistent with ZF, without the axiom of choice, that no such f exists. In fact it follows from the axiom of determinacy, AD, that there is no such f. AD of course is inconsistent with the axiom of choice, or to put it another way, false.

In this context people tend to claim that there is no "explicit" such f. You see this claim all over the WP math articles when discussing objects whose existence can't be proved without AC. It's a problematic claim, as it tends to depend on what you mean by "explicit", which is a term that has no precise accepted definition. It may also depend on such things as whether V=HOD, which (unlike V=L) is not currently known to be ruled out by any accepted large cardinal hypotheses. --Trovatore (talk) 21:19, 29 December 2009 (UTC)

[edit] Math and calculations

I want to do math but I'm terrible with calculations. Is it possible to do math without calculating/solving complicated equations with formulas? If so, how? My friend's a math person and he says that when he does math he thinks and rarely does calculate. He says he researches algebra, but I thought algebra was about calculating??? He also says that math people don't do arithmetic with numbers. My world has collapsed! Help! Or maybe my friend's wrong. Can't calculators solve math? Why does he research it? What sorts of math are there (my friend says there's lots)? Please tell me how I can do math without my calculator! —Preceding unsigned comment added by 122.109.239.199 (talk) 09:54, 27 December 2009 (UTC)

See Mathematics. Some areas of mathematics (such as abstract algebra, as opposed to high-school algebra) don't involve arithmetic calculations at all. Some do, and application of these fields can benefit tremendously from the use of a calculator. Being comfortable around numbers and equations is an important skill for the mathematically literate. -- Meni Rosenfeld (talk) 11:49, 27 December 2009 (UTC)

So you should first learn how to do computations, and then you should learn how not to do computations. --pm a (talk) 12:28, 27 December 2009 (UTC)

Good advice. --PS T 12:44, 27 December 2009 (UTC)

"Your world has collapsed"? That should not be the case! Almost everyone is told that they are wrong about something, at some point in their lives; such is an important experience. With regards to computations, I have a couple of remarks. Firstly, I think that the mathematical brain, by default, has ability to do computations; that is, if you are able to do the "mathematics without computations", you should have the ability to do the "mathematics with computations". Basic computations really do not require extreme intelligence to carry out, save silly human errors. On the other hand, much of mathematics is not really concerned with computations; rather, it is concerned with deeper problems which may require computations (at some stage) to solve. In abstract algebra, for instance, one uses substructures to encapsulate computations abstractly (ideals in ring theory, subgroups in group theory, etc...). In another branch of mathematics, topology, much of the goal is to attain a strong intuition of closeness; although seemingly simple, this is very deep. Differential topology sometimes employs differential geometry for this purpose. Hope this answers your question(s). However, if this is an attempt to mock mathematics, please do not; mathematics has developed tremendously over the past few hundred years, and not surprisingly, it is difficult for many to comprehend the extent of this development (I should add that it was even difficult for me to comprehend during my first exposure to formal mathematics). --PS T 12:43, 27 December 2009 (UTC)

Why do you want to do maths if you don't have some feel for what it is about? Being very good at calculating isn't that important though it is quite useful. I've known what maths is since I was a child stringing up a frame when I realized I didn't get a circle and I wondered how to arrange the pegs to get a circle. I can't see any basis for your desire to do maths. Dmcq (talk) 13:12, 27 December 2009 (UTC)

Skill with numbers is likely to develop with practice. The more you work with arithmetic and elementary algebra, the more proficient you will become. A solid basis in those areas is needed to progress to more advanced mathematics. Many of the branches of mathematics which do not work with computations are are too complex to be meaningful without the foundations. —Anonymous Dissident^Talk 13:29, 27 December 2009 (UTC)

I want to become a mathematician but I hate equations and formulas, and my friend says that's not necessary for mathematics. You tell me something, and probably that was what my friend said but I wasn't paying attention. Thanks! But why do people research mathematics? People research physics because it's important to know about the universe. People research medicine because it helps our survival. What's the use of mathematics without calculating??? I want to do math because my math teacher said to me when I was in high school that it's purpose is to shape the world, and that mathematicians are society's core, because of their mind boggling human-calculator ability. And if math's not about numbers, what is it about? My high school teacher said that math is about understanding different properties about numbers, such as prime and composite and he has a math degree. Many thanks for the responses. Please tell me why people think math though! It hurts my brain and isn't fun. But I need it to be part of society's finest. —Preceding unsigned comment added by 122.109.239.199 (talk) 13:48, 27 December 2009 (UTC)

You want to become a mathematician but you hate some aspects of it; it hurts your brain and isn't fun. Have you considered a career in masochism?→→86.160.104.185 (talk) 16:15, 27 December 2009 (UTC)

Please don't be disrespectful to the OP; he or she doesn't understand what mathematics is about and would like to better understand it, I'm sure you can provide some insight that would be useful to him or her. Eric. 131.215.159.171 (talk) 21:59, 27 December 2009 (UTC)

By now it should be considered very unlikely that the OP is asking these questions in good faith. 86's response is appropriate. -- Meni Rosenfeld (talk) 10:32, 28 December 2009 (UTC)

Considering the OP's more recent questions, you are probably right. However there is still no use to mocking him or her. Eric. 131.215.159.171 (talk) 01:40, 29 December 2009 (UTC)

It's possible you will find Lockhart's Lament interesting; it's an article about the teaching of high school mathematics in the US. Although it does not spend much time explaining what mathematics is, it does address some misconceptions of mathematics and explains what mathematics is not. Eric. 131.215.159.171 (talk) 22:05, 27 December 2009 (UTC)

The best way to judge whether you can become a mathematician is probably to simply find out if more time spent with it still leaves you feeling dislike toward it. In the meantime, you will become at least good enough at math to make a living doing something with math that doesn't involve actually being a mathematician. This just seems like a reasonable piece of advice/opinion for the OP. One thing about mathematics that is appealing and repellent at the same time is that it involves abstract objects directly and entirely, and its only connection to the 'real world' is through the filter of modelling. Doing mathematics per se involves ignoring the real world for extended stretches of time, but there is a dividend personally and in the large for doing so that makes it worthwhile (virtually always, despite some mathematics seeming ten steps removed from reality).Julzes (talk) 05:12, 28 December 2009 (UTC)

[edit] Limit of a sequence

Let {x_n} be a sequence with $\lim_{n\rarr\infty}\frac{x_n}n=g$ with g>0. How to prove the fact that $\lim_{n\rarr\infty}\sqrt[n]{x_n}=1$ ? --84.62.197.235 (talk) 11:24, 27 December 2009 (UTC)

First do it in the case

x n : = n

(to this end you may write the inequality of arithmetic and geometric means with the n numbers $a_1=\dots =a_{n-2}:=1, a_{n-1}=a_n:=\sqrt{n})$ . For the more general case observe that $\frac{g}{2}n\leq x_n\leq 2g\,n$ for large n; then use the former case and a sandwich argument. --pm a (talk) 11:42, 27 December 2009 (UTC)

(Edit Conflict) Let

g > 0

and let

x n = g n

for all $n\in \mathbb{N}$ . Do you accept that $\lim_{n\rarr\infty}\sqrt[n]{x_n}=1$ (Proof: $\lim_{n\rarr\infty}\sqrt[n]{x_n}=\lim_{n\rarr\infty}\sqrt[n]{g}\sqrt[n]{n}=\lim_{n\rarr\infty}\sqrt[n]{n}\cdot\lim_{n\rarr\infty}\sqrt[n]{g}=1\cdot 1=1$ )? Now apply this intuition to prove your claim. --PS T 11:44, 27 December 2009 (UTC)

One of the problems with your approach is that $x_n\not = gn$ (Igny (talk) 15:33, 27 December 2009 (UTC))

It was not intended to be a formal proof; rather it was intended to guide the intuition (and my x_n was not intended to be the same sequence as the OP's x_n; I defined a specific sequence). There is really no purpose in proving the value of a limit unless you have some idea as to how to obtain the value. The sequence x_n that I defined, and its limit, motivates the OP's question, and provides some intuition. But of course, this is guided with the assumption that the OP did not have any idea as to how to solve the problem initially (so pma's response, in that case, might have been what the OP was looking for). --PS T 01:16, 28 December 2009 (UTC)

Let {x_n} be a sequence with $\lim_{n\rarr\infty}x_n=g$ with g>0. How to prove the fact that $\lim_{n\rarr\infty}\sqrt[n]{x_n}=1$ ? --84.62.197.235 (talk) 14:22, 27 December 2009 (UTC)

Since x_n converges to g>0, you have ε<x_n<1/ε for some number ε>0 and for all n large enough, hence also ε^1/n<x_n^1/n<1/(ε^1/n). As you can see, the sandwich principle allows you to reduce your problem to the case of x_n=c>0 a positive constant. Is that clear? Then, can you do it for the case x_n=c>0? Take a logarithm for instance. --pm a (talk) 13:52, 28 December 2009 (UTC)

[edit] Computationally expensive problems

Could you give examples of things that are expensive to solve but easy to check. I'd like to know the best solve/check ratio even if it's not well defined, but just any examples would be useful. --93.106.33.14 (talk) 14:32, 27 December 2009 (UTC)

Anyone with a piece of paper and pencil can calculate a product of two integers to verify the solution to a factorization problem (given time and persistence that can be done even for numbers with thousands of digits). The factorization problem for big integers however requires a lot of computational power, and modern supercomputers start chocking at factorization of 100 digit numbers. (Igny (talk) 15:57, 27 December 2009 (UTC))

All NP-complete problems (probably) have this property. Algebraist 16:55, 27 December 2009 (UTC)

Here's a stab at a well-defined ratio: RSA numbers says that the RSA-200 factoring challenge was solved using "the equivalent of 75 years work for a single 2.2 GHz Opteron". The solution was found 14 years after the problem was presented. Verifying its correctness takes less than 3 milliseconds on my computer. So that's a ratio of at least 75*365*24*60*60*1000/3 = ~788 billion in the amount of processor time used. 98.226.122.10 (talk) 03:00, 28 December 2009 (UTC)

[edit] The all-ones vector, and how to notate it

What's the most common way of writing the all-ones vector, that is, the vector, when projected onto each standard basis vector of a given vector space, has length one? The zero vector is frequently written $\vec{0}$ , so I'm partial to writing the all-ones vector as $\vec{1}$ , but I don't know how popular this is, and I don't know if a reader might confuse it with the identity matrix. I'm writing for a graph theory audience, if that helps pick a notation. --Bkkbrad (talk) 20:36, 27 December 2009 (UTC)

(1, 1, 1, ..., 1) ? I just had need of one in sixth dimension and actually wrote (1, 1, 1, 1, 1, 1). They don't come up very often, depending on you having made a choice of basis whereas most vector theory is designed to be independent of basis. I've not done any graph theory for a long time though. checking Adjacency matrix#Properties there's one like my first example, so maybe graph theory is not that different.--JohnBlackburne (talk) 20:53, 27 December 2009 (UTC)

I've mostly seen the notation $\underline{1}$ , $\vec{1}$ or $\mathbf{1}$ (depending on your style for vectors). I believe I've seen $\boldsymbol{\iota}$ (bolded iota) used once or twice as well. Another alternative notation may simply be $\mathbf{e}$ (depending on your preferred vector style of course) - the rationale being that $\mathbf{e_1}=(1, 0, \dots, 0)$ and $\mathbf{e_2}=(0, 1, 0, \dots, 0)$ and so on - although I've seldom encountered it and would avoid it as $\mathbf{e}$ is not of unit length by this definition. I do not think any of these will be confused with the identity matrix, which is usually

I

. All this is mostly from a statistical background. x42bn6 Talk Mess 02:46, 29 December 2009 (UTC)

[edit] December 28

[edit] Lebesgue measure versus Borel measure

Hi. I was searching the web recently for an example of a set of real numbers that is Lebesgue measurable but not Borel measurable. Such sets must exist, because I know that the cardinality of the family of Lebesgue measurable sets is greater than the cardinality of Borel measurable sets ( $2^\mathfrak{c} > \mathfrak{c}$ ). However, this argument does not furnish an example.

So, I found a website claiming that the following set is Lebesgue measurable but not Borel measurable: The set of real numbers with continued fraction expansions of the form $\left[ a_0; a_1,a_2,\ldots\right]$ such that there is some sequence of integers $i_0 < i_1 < \cdots$ with $a_{i_{n-1}}$ dividing $a_{i_n}$ for all $n\geq 1$ .

Evidently, this set can be seen to be Lebesgue measurable because it is an analytic set, which I don't really understand. I'm not sure how to see that it's not a Borel set, although there is some talk about it here, which I again don't really understand. On to my question: Can someone either help me better understand the example I've presented, or else demonstrate another example where we can see why it's Lebesgue-but-not-Borel measurable? Thanks in advance for any illumination anyone can provide. -GTBacchus^(talk) 00:50, 28 December 2009 (UTC)

So if all you want to know is that there is some Lebesgue measurable set that's not Borel, here's a cheapo way of doing it: Look at the almost-bijection between the Cantor set (thought of, let's say, as the set of infinite sequences of zeroes and ones) and the interval [0,1] given by putting a binary point before the infinite sequence and reading it off in binary notation. I say almost bijection because there are countably many points in [0,1] that have two distinct binary representations and therefore are mapped to by two elements of the Cantor set.

Now do the Vitali set construction inside [0,1], and pull it back to the Cantor set. The resulting set cannot be Borel, because, if it were, the Vitali set would also be Borel, and therefore Lebesgue measurable (there are a couple of details here that I'm confident you can check for yourself).

But if you now think of the Cantor set as the usual middle-thirds construction inside the reals, well, that's a set of measure zero, so the pullback of the Vitali set is a subset of a set of measure zero, and is therefore Lebesgue measurable (specifically, measure zero).

This is almost cheating. I'd encourage you to pursue the more interesting stuff about an analytic set that's not Borel. Unfortunately this requires much more theory. It's theory well worth learning, though. --Trovatore (talk) 02:13, 28 December 2009 (UTC)

That's very good; thank you. I can understand that construction, and relate it to my study partner, who was asking me about this. As for the analytic set business... yeah. It looks like a lot of point-set topology. I know that stuff's good for the soul, but I'm a number theorist! One of these days. :) -GTBacchus^(talk) 02:29, 28 December 2009 (UTC)

(e/c) Since the Cantor space 2^ω is homeomorphic to a closed measure-zero subset of the real line (namely, the middle-thirds Cantor set K), you can use any nonmeasurable set in Cantor space to achieve the same goal. This is because a homeomorphism between two spaces gives an isomorphism of the Borel algebras of the spaces. So any standard example of a nonmeasurable set inside Cantor space – an uncountable set with no perfect subset, a non-determined set, etc. – will correspond to a non-Borel, Lebesgue measurable subset of K. This avoids the "almost-injective" argument.

The point of the "continued fraction" thing in the original post is to use this same trick: the map that sends a real number to its continued fraction expansion is a homeomorphism from the irrationals to ω^ω.

The original question seems to be how to show that the set in question is analytic, and how to show that it is not Borel. To show that a subset of Cantor space or Baire space is analytic, you just check that the set is definable by a logical formula of the correct complexity, namely $\Sigma^1_1$ in the analytical hierarchy. This is more complicated for the real line, which is one reason descriptive set theorists avoid working with the real line directly. But the desired result here can be obtained pretty easily using the isomorphism fact I just mentioned and the standard fact that the continuous image of an analytic set is still analytic.

To show that the set described in the original post is not Borel is harder, and I don't know what the original proof could have been. The way I would prove it is by showing that some standard $\Sigma^1_1$ -universal set reduces to the set in question. That would just be a standard textbook exercise, but as Trovatore says you would have to invest some time in the textbook to do it. The two places this sort of thing comes up are in descriptive set theory and in generalized recursion theory. Kechris' book Classical descriptive set theory has detailed information about universal $\Sigma^1_1$ sets. — Carl (CBM · talk) 02:57, 28 December 2009 (UTC)

I've seen this "divisibility" argument many times, and it always seemed like it should be trivial to check that it's just encoding either the problem of whether a linear order is a wellordering, or whether a tree is wellfounded. When I actually started thinking about it, I realized it's not quite as easy as maybe I expected.

But it's still not hard. I think this works: Assume given a transitive relation R on the naturals, that we want to check whether it's wellfounded. Code R as the following element of Baire space:

f (n) = p n Π k < n, n R k f (k)

, where p_n is the nth prime. Then for k<n, f(k) divides f(n) just in case nRk. So the continued fraction given by f is in the set if and only if R is illfounded. --Trovatore (talk) 08:23, 28 December 2009 (UTC)

Yes, that will work. The difficulty I was referring to is in showing

(*): the set of (codes of) non-well-founded trees is $\Sigma^1_1$ complete

in the first place, which you would need to do to really believe that you had solved the original problem. In jargon, (*) is just the normal form theorem for $\Sigma^1_1$ formulas plus the normal form theorem for $\Pi^0_1$ formulas. None of this is truly hard, but it takes a little time to go through it all. I think it could be done very well to a general mathematical audience in two 50-minute seminars. — Carl (CBM · talk) 13:36, 28 December 2009 (UTC)

Oh, well, I'm used to seeing $\Sigma^1_1$ defined as projections of trees, which makes your (*) pretty trivial. If you have some other characterization in mind, sure, there's a bit of work in showing all the characterizations are equivalent. --Trovatore (talk) 19:11, 28 December 2009 (UTC)

[edit] Universal enveloping algebra

Is every (associative) K-algebra a universal enveloping algebra of some Lie algebra (over K)? -- Taku (talk) 05:23, 28 December 2009 (UTC)

No. Universal enveloping algebras are infinite-dimensional (see Poincaré–Birkhoff–Witt theorem). 86.15.141.42 (talk) 16:09, 28 December 2009 (UTC)

Right. Assume associative algebras are infinite-dimensional too. Or how about a quotient of a universal enveloping algebra? (By the way, I tend to think the answer is yes (if the question is formulated correctly)) -- Taku (talk) 22:07, 28 December 2009 (UTC)

You are probably going to have to work hard to make the answer yes. Universal enveloping algebras have PBW-type bases, this is very special. The question about quotients is easy because the UEA of a free Lie algebra is a free algebra. 86.15.141.42 (talk) 01:59, 29 December 2009 (UTC)

Good point. So, the question really boils down to which associative algebra admits a PWB-type basis; that's certainly not the case in general. (I guess a quotient one is trivial.) Anyway, thanks. -- Taku (talk) 06:52, 29 December 2009 (UTC)

[edit] How to solve math poblems without calculating

Many thanks for all responses! My high school teacher said though that there isn't math after 12th grade; I read the Lockhart's lament suggested, but my math teacher has a math degree and he's always right. He also says that he knows all math like all math teachers. He won many math competitions and came first place, and he says math is about problem solving. He also says that all problems have been solved, and that's why it matters to show working in math. Who's the world best mathematician? Who's won the most math competitions? I understand that math is not about calculating but how else do you solve math problems like x + 2 = 5? Thanks but please don't insult me though. —Preceding unsigned comment added by 122.109.239.199 (talk) 09:06, 28 December 2009 (UTC)

If there's no math after 12th grade, how did your teacher get a math degree? -- Meni Rosenfeld (talk) 10:33, 28 December 2009 (UTC)

By doing high school math, and solving complex math problems? I don't know. That's sort of what I'm asking. Why's there math after 12th grade and what math's there? My high school teacher says there isn't and he's always right about math (he's a math nerd as my friends at school used to say). I'm trying to understand what sort of math is done if calculating isn't there. My math teacher is an expert on calculus, and isn't calculus the highest math? He said that calculus was taught at uni. So does he have a degree in calculus or something? Please explain? Many of youy say my high school teacher's wrong, but how's that? He has a math degree. Many thanks. —Preceding unsigned comment added by 122.109.239.199 (talk) 11:18, 28 December 2009 (UTC)

Look, we've already answered your questions. We told you that mathematics does involve calculations, but there's much more to it than only that.

Your teacher may have a math degree, but the people who have responded to your query have about a dozen math degrees between them. If you trust him so much, and since he appears eager to discuss these matters with you, and you also have an algebraist friend, then what are you asking us for?

Personally I don't believe you are asking these questions seriously. But if I'm wrong and you are, please read our replies carefully and don't ask us to repeat ourselves. -- Meni Rosenfeld (talk) 12:25, 28 December 2009 (UTC)

Yeah, so what math's there higher than 12th grade if it isn't calculating. Please give an example of such math. Give a mind-boggling math problem to me please. I'm dead serious about these questions. I just wnat to understand the purpose of math, and what really math is after 12th grade math. My high school teacher says that teachers know most about math since that's the main profesional math job. Applied math's there also. That's why I'm confused why my math teacher's wrong. —Preceding unsigned comment added by 122.109.239.199 (talk) 13:05, 28 December 2009 (UTC)

Ask your teacher to define "math." I've heard it said the exact opposite way at the beginning of calculus. Some calc teachers say that everything before calc is not math at all. Without having a discrete definition of what math means, you are just having a stupid semantic argument that gets nowhere. Further, nobody here appears to care what your teacher thinks. So, attempting to form an argument between your teacher and everyone else is a trollish waste of time. As for math beyond 12th grade - there is a lot of it. For example, given two vectors of unequal size, how to you find the optimal alignment of the shorter one against the longer one? That is nothing but a bunch of calculations. Have you done vectors or vector alignment? It is very important if you get into health informatics (which is a blend of math and computer science and health). -- kainaw ™ 13:18, 28 December 2009 (UTC)

[ec] I think it's safe to say Mathematical logic is higher than 12th grade and isn't calculating. For mind-boggling math problems, you needn't look further than the Millennium Prize Problems, which involve varying amounts of calculations.

The amount of mathematical knowledge currently in existence is too great for even a brilliant individual to learn in a lifetime. So no, neither your teacher nor any other math teacher knows most of what there is to know in math. -- Meni Rosenfeld (talk) 13:23, 28 December 2009 (UTC)

Whoever claims there is no math beyond highschool level is like a kid that believes that the universe ends just out his little garden. There are thousands of streets instead, towns, mountains, seas and oceans, woods and deserts as large as a state, and rivers and lakes.... and this is just what we know; then there are planets stars and galaxies... The only thing that gives me a comparable feeling of immensity is the stupidity of your teacher's claim! :-) --pm a (talk) 15:16, 28 December 2009 (UTC)

Have you seen Pleasantville? --Trovatore (talk) 21:52, 28 December 2009 (UTC)

I haven't; after reading the plot summary it sounds a nice movie --pm a (talk) 15:26, 29 December 2009 (UTC)

Is it not obvious that our esteemed original poster is amusing himself by seeing how much we try to enlighten a fool? Michael Hardy (talk) 01:55, 29 December 2009 (UTC)

Yes, an obvious timewaster. As the masochism poster, I feel that those who've replied since are exhibiting that very thing.→→86.152.78.244 (talk) 11:27, 29 December 2009 (UTC)

You people shouldn't insult the OP so bad. When I was in high school I thought just like him "I know all the math, I'm the best in the world". Things started to change after I finished year 12 and now I feel like my knowledge is a needle in the ocean (and in reality it's less than a bacterium). It takes time for people to realize and appreciate the real stuff. And for a mind boggling problem, how about theory building? It's all so easy when you read it but trying to develop the tools for the first time is, to me, only a genius can do. Money is tight (talk) 12:39, 29 December 2009 (UTC)

Actually I do not think that the OP's antics surprise anyone, since it is well known that a majority hold the belief that mathematics is "arithmetical computations" (educated people and high school teachers inclusive). What is somewhat awkward is that the OP expresses his ignorance openly. Most people are so confident that they are "right about mathematics", that they believe it is not necessary to explicitly note its purpose. Rather thay feel they may implicitly assume its "arithmetical nature" in their assertions. That said, I agree with you, and especially with your last statement (Galois being an excellent example of this). --PS T 13:21, 29 December 2009 (UTC)

You are missing the point. Of course most people are confused to some extent about mathematics. It's the specifics of what the OP said and how that triggered our troll-detectors. Things like

Asking us what his algebraist friend does, instead of asking him.
Telling us he wants to be a mathematician although he knows little about it and hates what little he knows.
Explaining that his teacher is always right. Virtually nobody believes that some other individual is always right.
Reporting that his math teacher said there is no math after 12th grade and that math teachers know all math, when of course anyone with a math degree would not say that.
Not updating his questions based on our answers - e.g. "[Without calculating, how] do you solve math problems like x + 2 = 5" after we explained that there are calculations in math, and asking for examples of higher math after we've given them.
And of course other subtleties in his style.

-- Meni Rosenfeld (talk) 15:27, 29 December 2009 (UTC)

Why feed the trolls (per Don't feed the trolls)? If a majority feels that the OP's posts were not made in full seriousness, the posts should simply be ignored. In fact, a better alternative would be to remove the posts altogether. Although I certainly agree with your judgement of the OP's questions, insulting him is counterproductive; whatever the circumstance (anyway, I think he is enjoying our insults, as Michael points out). --PS T 02:34, 30 December 2009 (UTC)

Well I guess he is somewhat trolling. I was just trying to "assume good faith" but Meni made his point. Money is tight (talk) 03:30, 30 December 2009 (UTC)

Assuming good faith is something I completely agree with - so much so that I have the userbox on my page. But the point of WP:AGF is to assume good faith when lacking evidence to the contrary. In this case there is certainly no lack of evidence.

In my view, the other point of AGF is to always consider, if possible, the distant possibility that the posts are genuine, give the OP a chance to explain himself, and not do anything that would be incredibly offensive (which is why I disagree with PST's suggestion of removal). Of course, in particularly disruptive cases this is not an option.

Anyway, posts like those of 86 and Michael Hardy are not meant to insult the OP, only to signal to others that they should not waste their time. -- Meni Rosenfeld (talk) 08:55, 30 December 2009 (UTC)

And avoid the fate that nearly befell Matilda's aunt in Matilda Who told Lies - and was Burned to Death. :) Dmcq (talk) 09:19, 30 December 2009 (UTC)

After all I think is OK being polite with maths trolls, especially at christmas times... Go and look at a troll's pub... all the other trolls treat them like scum you know, I mean real trolls working in categories such as history politics sexuality social sciences women studies...--pm a (talk) 12:07, 30 December 2009 (UTC)

Considering that two interesting queries (Wikipedia:Reference desk/Mathematics#transcendental_functions and Wikipedia:Reference desk/Mathematics#No_math_after_12th_grade.3F) have resulted in the troll's silly questions, I would say that he has done more good than he intended... (of course, it would have been more productive if his posts were ignored sooner, but I think that he has left permanently by now). --PS T 12:32, 30 December 2009 (UTC)

[edit] Sum and Difference of Two Tangents

Hello. Can a sum or difference of two tangents be expressed as a product or quotient? For example,

$\sin x - \sin y = 2\cos\left(\frac{x+y}{2}\right)\sin\left(\frac{x-y}{2}\right).$

Thanks in advance. --Mayfare (talk) 21:03, 28 December 2009 (UTC)

Just from the definition of tan, we have

$\tan x \pm \tan y = \left(\frac{\sin x}{\cos x}\right) \pm \left(\frac{\sin y}{\cos y}\right) = \frac{\sin x\cos y \pm \sin y \cos x}{\cos x \cdot \cos y} = \frac{\sin(x\pm y)}{\cos x \cdot \cos y} .$

Is this the kind of thing you had in mind? ~~ Dr Dec (Talk) ~~ 23:37, 28 December 2009 (UTC)

Yes. Thank you very much Dr Dec. Have a happy new year. --Mayfare (talk) 18:34, 29 December 2009 (UTC)

[edit] December 29

[edit] Hypervolume of an 8-sphere

Does anyone know the formula for the hypervolume of an 8-sphere? This is so I can remove all those question marks from Nine-dimensional space... 4 = 2 + 2 04:23, 29 December 2009 (UTC)

See n-sphere. PrimeHunter (talk) 04:32, 29 December 2009 (UTC)

[edit] transcendental functions

The article Lockhart's Lament, linked above has the sentence "The measurement of triangles will be discussed without mention of the transcendental nature of the trigonometric functions." Can someone please explain to me what the transcendental nature of trigonometric functions has to do with measurement of triangles. Thanks-Shahab (talk) 07:09, 29 December 2009 (UTC)

Could you please provide the context for that remark (the exact number of the page on which it was made)? I read that article earlier this year and do not recall such a statement being made. But it is quite possible that I have forgotten its presence. --PS T 07:36, 29 December 2009 (UTC)

It's on the last page (Pg 25) of the article. The section deals with an "honest catalog" of the curriculum(The trigonometry course). The complete paragraph runs as follows:

Two weeks of content are stretched to semester length by masturbatory definitional runarounds. Truly interesting and beautiful phenomena, such as the way the sides of a triangle depend on its angles, will be given the same emphasis as irrelevant abbreviations and obsolete notational conventions, in order to prevent students from forming any clear idea as to what the subject is about. Students will learn such mnemonic devices as “SohCahToa” and “All Students Take Calculus” in lieu of developing a natural intuitive feeling for orientation and symmetry. The measurement of triangles will be discussed without mention of the transcendental nature of the trigonometric functions, or the consequent linguistic and philosophical problems inherent in making such measurements. Calculator required, so as to further blur these issues.

Regards-Shahab (talk) 07:52, 29 December 2009 (UTC)

Not sure what it's about. Trigonometric functions of any rational number besides 0 are always transcendental when you use radians, perhaps he was just referring to the problems of incommensurables the Greeks had with things like the square root of 2? Dmcq (talk) 08:35, 29 December 2009 (UTC)

I think he means that in many high school problems given, the lengths of the sides of the triangles involved are usually assumed to be simple values such as 1 or $\sqrt{2}$ , and this convolutes one's understanding of more general triangles. --PS T 09:49, 29 December 2009 (UTC)

I think he means that learning how to calculate the length of a certain side given an angle and the length of another side is not interesting maths (it's just computation) whereas learning (or, even better, deriving) that the trigonometric functions are transcendental is interesting maths. --Tango (talk) 14:56, 29 December 2009 (UTC)

[edit] No math after 12th grade?

A question about let me thinking. Is it too awkward to say that there is no math before 12th grade? Until 12th grade you are only performing calculations, you are more a human calculator than a real mathematician. ProteanEd (talk) 12:52, 29 December 2009 (UTC)

The word "mathematics" refers to a wide spectrum of activities. It includes things such as addition and multiplication, learned in grade school, and also includes esoteric facts about more complicated objects that are learned in graduate school. These are not fundamentally different topics, but are simply different expressions of the same subject. You have to crawl before you can walk, and later you learn to drive; all of these are forms of locomotion. — Carl (CBM · talk) 12:56, 29 December 2009 (UTC)

I disagree (in fact, I believe that there is not mathematics even in the 12th grade; at least in the current American education system). Mathematics is, after all, creativity. It is not about attaining tools such as computational ability, but rather about using the tools for a particular purpose. Thus mindless computation, on its own, is not mathematics, unless it is done in a creative manner. Even calculus, in the mechanical sense of solving problems, is not mathematical unless one does it creatively. Thus, while the routine technique of finding the maximum values of polynomials of degree two on a compact interval loses its mathematical value at a certain stage, inventing new techniques in numerical analysis can be safely referred to as mathematics. In another context, one could refer to painting blue dots on a white sheet of paper as "painting", but such would not be considered to be of true artistic value. Combining a wide range of colors with different tone, depth and emotion to create a painting is true art. --PS T 13:10, 29 December 2009 (UTC)

You're free to argue that the English language should be changed. At the present, "mathematics" clearly refers both to the things we learn in kindergarten and the things we learn in graduate school. Similarly, finger-painting is indeed painting. Maybe you're talking about real mathematics, but that sort of discussion will always devolve into polemics. — Carl (CBM · talk) 13:16, 29 December 2009 (UTC)

I guess you are right. --PS T 13:22, 29 December 2009 (UTC)

That definition is entirely subjective. In particular, at which point mathematics becomes creative is not properly ascertainable. If "creative" means "original", many mathematicians never end up doing mathematics. In the same way that the composition of a short letter and the composition of a classic piece of fiction are both termed "writing", everything from basic addition to the discovery of new theorems in esoteric fields is mathematics. So, yes, mathematics is done in high school and before. —Anonymous Dissident^Talk 13:30, 29 December 2009 (UTC)

As I mean it, doing mathematics is not necessarily creating original maths, nor one needs to discover original results to enjoy it greatly, in the same way that doing music is not necessarily composing. When I was 6 I had a wonderful teacher, which was about 19, so sweet. She made us solve nice addition problems drawing colored balls... when she would come to me and lean over my little desk, how happy I was! I scarcely listened towhat she would tell me, but only hear the sound of her sweet voice and sniff the perfume of her black hairs. Maybe that was no original maths, but how exciting!--pm a (talk) 15:39, 29 December 2009 (UTC)

[edit] Symmetries and perturbations

What do you think about this statement:

Suppose f is a continuous function and h is a (continuous) symmetry for f i.e.

$f \circ h=h \circ f$

Then for any ε there exists δ such that if d(f,f ')<δ then there exists h ' such that h ' is a simmetry for f ' and d(h,h ')<ε.

assuming d is the standard distance given by the uniform norm.

Is it true or false?--Pokipsy76 (talk) 13:57, 29 December 2009 (UTC)

I see it more false than true in general, though I don't have a counterexample. Could you specify the domain? If f₀ is an Anosov diffeomorphism and f₁ is (sufficiently) C¹ close to f₀ then f₁ is Anosov and conjugated to f₀ with a conjugation H which is C⁰ close to the identity, together with its inverse H^-1. So by conjugation with the same H you do have a h₁ C⁰ close to h₀ doing the job. --pm a (talk) 14:36, 29 December 2009 (UTC)

I didn't want to be specific about the domain, I wanted to ask the question in general, however I'm curious in particolar for functions in R and R². I supopose that a first step to find a counterexample could be to find a function which doesn't have any nontrivial simmetry... are there such functions?--Pokipsy76 (talk) 15:26, 29 December 2009 (UTC)

Consider the following counter-example on $\R.$ Let

f 0 : = id

and $h_0\in C^0(\R,\R),$ any function, with $h_0\neq\mathrm{id}.$ Say $\|h_0-\mathrm{id}\|_{\infty}>r>0.$ Then

f 0

and

h 0,

of course, do commute, and I claim that there exists a function

f

as close in the uniform distance to

f 0

as we want (in fact, even in the $C^\infty$ distance if you wish), with the property that any $h\in C^0(\R,\R)$ that commutes with

f

necessarily has a uniform distance $\|h_0-h\|_{\infty}\geq r.$

Indeed, let $a\in\R$ such that $|a-h_0(a)|\ge r.$ There exists a function

f

such that

a

is a globally attractive fixed point of

f

, meaning that

f (a) = a

and $f^{\,n}(x)\to a$ as $n\to\infty$ for all $x\in\R;$ such an

f

may be chosen arbitrarily close to

f 0

(take e.g. $f(x):=x-\epsilon\,\arctan(x-a)$ for a small

ε > 0

). Let $h\in C^0(\R,\R)$ a function that commutes with

f

. Then we have $h(a)=h(f^n(a))=f^n(h(a))\to a$ as $n\to\infty$ so

h (a) = a

and $\|h-h_0\|_\infty\geq |h(a)-h_0(a)|=|a-h_0(a)|\ge r.$ Is that clear? Actually this argument could be adapted in order to show that the original property fails to hold with

f 0 : = id

and $h_0\neq\mathrm{id}$ with any topological manifold as a domain. --pm a (talk) 20:28, 29 December 2009 (UTC)

[edit] December 30

[edit] non polynomial difference

When using master theorem, f(n) and n^log_ba must have a polynomial difference. The example shown is f(n)=n/log n. With a=2 and b=2, n^log_ba=n. So, the claim is that n/log n and just n have a non-polynomial difference. Another example I saw changes a to 4 so n^log_ba=n². The claim is that n/log n and n² have a polynomial difference. I'm left wondering exactly what the "polynomial difference" is. Is it taking (n/log n)-(n) and claiming that is non-polynomial? -- kainaw ™ 02:30, 30 December 2009 (UTC)

"Difference" here is being used multiplicatively -- the quotient of n and

n / log n

, which is

log n

, is not polynomial. Formally, the master theorem requires $f(n) \in O(n^{\log_a(b) - \epsilon})$ for some positive

ε

; or equivalently, requires $\frac {f(n)}{n^{\log_a(b)}} \in O(n^{-\epsilon})$ or $\frac {n^{\log_a(b)}}{f(n)} \in \Omega(n^\epsilon)$ . Eric. 131.215.159.171 (talk) 03:21, 30 December 2009 (UTC)

In any case, what everybody would do as a first step with this

T (n) = a T (n / b) + f (n)

is writing

n = b k

and turning the recurrence into the form

τ(k + 1) = a τ(k) + φ(k);

solutions of the latter have an immediate representation in terms of discrete convolutions, and a whole machinery for growth estimates is available, to bound the solution in terms of

φ(k).

As I see it, in these cases it should be better not to make everything into a theorem (especially with such a name), that makes things more rigid. --pm a (talk) 10:09, 30 December 2009 (UTC)

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