Wikipedia:Reference desk/Archives/Mathematics/2010 July 31

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July 31

Axiom of choice

Suppose we have an equivalence relation on a set A and we are constructing a partition $\{A_{i}\}_{i\in I}$ of A in the natural way: take x in A and consider the set of all elements related to it. Call that set A_x, and consider A-A_x. Pick y in A-A_x and repeat the procedure. This way keep constructing the equivalence classes and thus get the partition. My question is why, if this procedure isnt finite (for example if the equivalence relation is equality on an infinite set) do we consider it valid. My teacher said its because of the axiom of choice, I want to know how exactly the axiom of choice is being used here, what is the choice function etc. Regards-Shahab (talk) 03:19, 31 July 2010 (UTC)

No you don't need to "repeat" to construct the equivalence classes. For each x, define A_x like you said. Then it can be easily proved that for any x,y in A, if A_x and A_y intersect, then they are equal. So you have a family of pairwise disjoint subsets of A indexed by elements of A. However, when you actually want to turn equivalences class into, say a quotient group, you need to consider all choice functions on the classes and then prove it's a congruence. Also your question looks a bit like transfinite induction (search for that article on wiki), and although that doesn't require the AC, it requires the set to be well ordered and we need AC to prove well ordering theorem. Money is tight (talk) 05:03, 31 July 2010 (UTC)

I have a small doubt in what you said: So you have a family of pairwise disjoint subsets of A indexed by elements of A. If the index set is A, shouldnt each element of A correspond with a unique class, which is clearly not the case. I feel the index set should be smaller. (In reality, I dont know, its just a hunch)-Shahab (talk) 07:07, 31 July 2010 (UTC)

If you don't want the indices to be repeated (which is pretty reasonable), then you have to select one element from each piece of the partition as a representative. This is exactly the axiom of choice. J Elliot (talk) 09:50, 31 July 2010 (UTC)

This is only a problem if for some reason you want the sets of the partition to be indexed by elements of them. If you just want to form the partition as a set of sets (without indexing them), or if you're happy to index them with something else, then this procedure requires no choice. Algebraist 10:20, 31 July 2010 (UTC)

Thats exactly my question JElliot. You are saying a choice function exists. What I want to know, in the application of the axiom of choice, we have non empty sets A_i and an index set I beforehand. We are searching for A_i here, so they are not given beforehand. Also what is the index site I? -11:14, 31 July 2010 (UTC)

As has already been said, forming the equivalence classes is choice-free: if B is a subset of A, then B is an equivalence class iff B is nonempty, any two elements of B are related, and anything in A related to an element of B is already in B. Thus we can form the partition (as a nonindexed set of sets) using just the axioms of powerset and separation. If we want to turn this set into an indexed family for some reason, then we can perform the standard trick of indexing it with itself. Algebraist 11:24, 31 July 2010 (UTC)

So algebraist I think your saying define a function from A to P(A) sending each x to its eq class and take the image of this function? Guess that would be a way Money is tight (talk) 07:29, 1 August 2010 (UTC)

What I have understood from Algebraist is that the original problem was choice free, and for indexing consider the index set to be the partition, (the set of sets) and the axiom of choice guarantees that there is a function which associates each set to an element in it, i.e. f(A_i) is in A_i. Thus we can pick an element from each of the equivalence classes.-Shahab (talk) 07:49, 1 August 2010 (UTC)

problem with inequality

Resolved

This is from a research paper I am reading. Specifically the problem is big, but I am stuck at a small inequality. Let $a_{1},\cdots a_{m},x_{1},\cdots x_{m}$ be natural numbers (all non-zero) and let $t=\min\{a_{1},\cdots a_{m}\}$, $b=a_{1}+\cdots a_{m}-t$. I know that at least one $x_{i}\geq (b+t)^{2}$ and need to show that $\sum a_{i}x_{i}\geq b+t(b+t)^{2}$. I am struggling to show that. Can anyone please help. Thanks--Shahab (talk) 07:03, 31 July 2010 (UTC)

$\sum a_{j}x_{j}\geq \sum _{j\neq i}a_{j}+a_{i}x_{i}=b+t-a_{i}+a_{i}x_{i}\geq b+t-a_{i}+a_{i}(b+t)^{2}=b+(a_{i}-t)((b+t)^{2}-1)+t(b+t)^{2}\geq b+t(b+t)^{2}.$ Rckrone (talk) 07:24, 31 July 2010 (UTC)

Thanks. Is there some proper way to figure these inequalities out or you just used your experience?-Shahab (talk) 07:36, 31 July 2010 (UTC)

I hope I do not sound rude by telling you this, but I think that the inequality here is quite easy to figure out. (It is something with which you become comfortable with experience, I suppose.) The idea is, step-by-step:

1. First observe that you are given some information about x_i for some i, and hence it should be reasonable to isolate this particular term from the sum $\sum a_{j}x_{j}$ as Rckrone has done in the first step.

2. In the second step, you want to get rid of the "messy term" $\sum _{j\neq i}a_{j}$, but you immediately notice that this is just $b+t-a_{i}$. (And you want to get the b in there somehow, anyway, since the actual inequality that you are trying to prove has a b in it.)

3. For the third step, recall the reason we isolated x_i from the sum: we wanted to bound x_i from below by $(b+t)^{2}$ so let us do exactly this.

4. The final steps follow from observing what you are trying to prove and attempting to somehow obtain the term on the right of the inequality which you are trying to prove.

Of course, what I did above was nothing more that expalin the intuitions step-by-step with the prior knowledge that the result was true. Research mathematics is often not like that, of course. I think in general it is difficult to say "how one gets an idea" or "what is the inspiration behind certain proofs". Once you have intutions in certain areas it seems all too easy to tell others that it is not really difficult! That being said, getting an intution for figuring out inequalities (or doing virtually any other mathematics) is not something that you cannot do with enough perseverence.

Is the research paper for a PhD, or are you just reading it out of interest or for independent research purposes? If the latter, I think that it is worthwhile to try to broaden your knowledge in related areas since that will definitely improve your ability to quickly grasp things in research papers without having to think too much. (I am not saying that you cannot do that already; just that sometimes it is easy to make the mistake to begin research too early when you would probably benefit more by actually broadening your knowledge in these areas first.) By all means, continue reading research papers, but there are some excellent books that might be useful especially in this area: Hardy, Littlewood and Pólya's Inequalities, Rudin's two excellent books Principles of Mathematical Analysis and Real and Complex Analysis. I especially recommend the first one and the second one (i.e., baby Rudin) if you are primarily interested in inequalities and not so much in real, functional, Fourier etc. analysis. But since inequalities do in a sense lie within analysis, it certainly would be extremely valuable to get Real and Complex Analysis as well. (Fermat's Last Theorem is a classic example of so many tools from different areas being used to prove something that can be readily understood to even laymen.) PST 08:30, 31 July 2010 (UTC)

Thank you (and its not rude at all). Since I have not studies mathematics properly often, easy things seem difficult to me. In this case, the idea that got entrenched in my head was that after the second step, (i.e. $\sum a_{j}x_{j}\geq \sum _{j\neq i}a_{j}+a_{i}x_{i}$ since $a_{i}x_{i}\geq t(b+t)^{2}$ so I need to establish $\sum _{j\neq i}a_{j}\geq b$. I couldnt get anywhere with that and try as I might I couldnt get rid of this idea that this is what I need to establish. So this sort of blocked everything else off! I'll look in the books you recommended, especially Inequalities. Thanks-Shahab (talk) 09:09, 31 July 2010 (UTC)

Simplicial Homology

A torus with the generators of H₁(T,Z)

I've been trying to get an intuition for simplicial homology. I know all of the algebraic definitions, and can calculate explicit examples. But I don't have a feel for what they mean. For example, let T denote the torus, we have

$H_{0}(T,\mathbb {Z} )\cong \mathbb {Z} ,\ H_{1}(T,\mathbb {Z} )\cong \mathbb {Z} ^{2},\ H_{2}(T,\mathbb {Z} )\cong \mathbb {Z} .$

Is it correct that two points are in the same equivalence class of H₀(T,Z) if the points are the boundary of a 1-simplex, i.e. a path starting at one point and ending at the other? Does the fact that H₀(T,Z) ≅ Z mean that the torus is made up from a single path connected component?

Is it correct that two loops (closed simplexes) are in the same equivalence class of H₁(T,Z) if they are the boundary of a 2-simplex? In the picture, the two red circles lie in different equivalence classes of H₁(T,Z). Is that because there is no 2-simplex with them as its boundary? If a 1-simplex is not closed, does it correspond to [0] ∈ H₁(T,Z)? — Fly by Night (talk) 11:50, 31 July 2010 (UTC)

(edit conflict) The main mistake you seem to be making is in confusing simplices with chains (finite formal sums) of simplices. Elements of the homology groups are represented by chains, not by simplices. Thus a typical element of H₀ is represented by a finite sum of points; since the space is path-connected, a point in one place is equivalent to (the negation of) a point at another place, so we can assume all our points are in the same place, and all that matters is how many there are, which is just an integer, so we get that H₀ is just Z. Your red loops are in different classes of H₁ because there is no chain of 2-simplices with them as its boundary.

You also seem to be talking about singular homology (which applies to topological spaces) rather than simplicial homology (which applies to simplicial complexes), but that's more doing things the right way than making a mistake. Algebraist 12:03, 31 July 2010 (UTC)

(after ec) If a 1-simplex (or chain of 1-simplices) is not closed, then it doesn't correspond to anything in H₁ at all. Homology groups are closed chains (aka cycles) modulo boundaries. Algebraist 12:06, 31 July 2010 (UTC)

Oh, you seem to have some signs the wrong way round also. Remember that, for example, 1-simplices have direction, and if two loops are the boundary of a 2-chain, then their sum is zero in homology; that is, one is equivalent to minus the other. Algebraist 12:09, 31 July 2010 (UTC)

Thanks for the reply. Let's focus on H₀(T,Z) for now. You are saying that this consists of chains of complexes. Let's say I have two different chains. These chains are formal sums of points of the torus with integer coefficients. What do we need for the two chains to be in the same equivalence class of H₀(T,Z)? When you say that a point is equivalent to the negation of another point is that because the boundary operator takes a 1-simplex to an end-point minus the other end-point? Given a 0-chain, say $\scriptstyle c\,=\,\lambda _{1}p_{1}\,+\,\cdots \,+\,\lambda _{n}p_{n},$ is it true that $\scriptstyle c$ corresponds to $\scriptstyle \lambda _{1}\,+\,\cdots \,+\,\lambda _{n}\,\in \,\mathbb {Z} ?$ — Fly by Night (talk) 12:28, 31 July 2010 (UTC)

I got confused with negations myself there, sorry. If u and v are points on the torus, then there's a path from u to v, which is a (singular) 1-simplex with boundary v-u. Thus v-u is 0 in homology, so u and v are equivalent. Since a single point is not the boundary of any 1-chain, this gives us that H₀ is isomorphic to Z, with the isomorphism you have given. Algebraist 12:36, 31 July 2010 (UTC)

Washing line mathematics

A question I've been pondering for some time: A washing line of length l2, between two poles of equal height, distance l1 apart, will describe a catenary, if I understand the article correctly (see diagram). There are a lot of formulae on that page, but I can't find anything that relates to the length of the curve. My query is this - is it possible to determine the amount of dip (d) in terms of l1 and l2? Does it also depend on g and the weight per unit length of the line? The reason I ask is that I was tightening up a line recently, and was surprised how little change it took in the length of the line to remove a large dip. —  Tivedshambo  (t/c) 18:42, 31 July 2010 (UTC)

Short answer: The dip is purely a function of the l1 and l2, so the only way g and the weight might come in is that they might cause the line to stretch, thus changing l2.

Why is this? Recall that the formula for a catenary is $y=a\cosh(x/a)$ for some positive $a$ (positive because the curve is concave up). The dip then is $y(l_{1}/2)-y(0)$, which has only one possible value for a fixed $a$ and $l_{1}$. Employing the formula for arc-length, we can get that $l_{2}=2a\sinh(l_{1}/(2a))$. So the question becomes, for fixed values $l_{1}$, $l_{2}$, are there multiple values of $a$ that satisfy this equation?

Consider the function $x\sinh(c/x)$. If we show that this function is injective (on positive inputs), it will ensure that there can only be one such $a$. An easy way to do that is to show that the derivative always has the same (non-zero) sign. The derivative is $\sinh(c/x)-c/x\cosh(c/x)$. Make the substitution $z=c/x$, and this becomes $\sinh(z)-z\cosh(z)$. But for positive $z$, this is always negative (look at the Taylor series).

So $a$ is determined once you pick $l_{1}$ and $l_{2}$, and then the dip is determined from that.--203.97.79.114 (talk) 05:52, 1 August 2010 (UTC)

So what your saying is because it follows a sinh curve, it has to be a specific shape. That makes sense, thanks. However, what I'm really after is the actual formula for d in terms of l1 and l2. Any takers? —  Tivedshambo  (t/c) 07:30, 1 August 2010 (UTC)

If I'm not mistaken you have $l_{2}=2a\sinh \left({\frac {l_{1}}{2a}}\right)$ and $d=a\left(\cosh \left({\frac {l_{1}}{2a}}\right)-1\right)$. I don't know if this can be solved algebraically for d in terms of $l_{1},l_{2}$, but you can solve it numerically. Also, you can find arbitrarily good approximations for specified ranges. For example, for $a\approx \infty$, in which case $l_{1}\approx l_{2}$, you have $a\approx {\sqrt {\frac {l_{1}^{3}}{24(l_{2}-l_{1})}}}$ and $d\approx {\sqrt {{\tfrac {3}{8}}l_{1}(l_{2}-l_{1})}}$. So you can see that in this case, d goes like the square root of the slack, so increasing it slightly greatly increases d. -- Meni Rosenfeld (talk) 08:39, 1 August 2010 (UTC)

I don't understand that last sentence. If I double the slack, the dip only increases by about 40%, how is that "greatly increased"? --Tango (talk) 15:52, 1 August 2010 (UTC)

He means the dip relative to the slack, not relative to the previous dip. Dmcq (talk) 16:42, 1 August 2010 (UTC)

Yes, $\left.{\frac {d{\sqrt {x}}}{dx}}\right|_{x=0}=\infty$. Derivative is how rapidly the function changes with the variable. So if for example $l_{1}=2m$ and $l_{2}$ starts at 2m, then an increase of only 1mm in $l_{2}$ will cause an increase of 27mm in the dip. This is what the OP asked about. -- Meni Rosenfeld (talk) 05:09, 2 August 2010 (UTC)

Also, you have $d={\sqrt {a^{2}+l_{2}^{2}/4}}-a$. When $a\approx 0$, $d\approx l_{2}/2-a$. Denoting $r=2\log(l_{2}/l_{1})\;\!$, you have $a\approx {\frac {l_{1}}{r+2\log r+(4\log r)/r}}$. -- Meni Rosenfeld (talk) 11:50, 1 August 2010 (UTC)

Simplify the problem by first considering the special case L1=1. After having solved that, the general case is obtained by multiplying all lengths by L1. Bo Jacoby (talk) 07:36, 2 August 2010 (UTC).

graph of l2 (x axis) against d (y axis)

Thanks for the help. I've created a spreadsheet in Excel, showing l2 against d for various values of a, for unit length l1 - it confirms my observations and the formulae above, in that a small increase in the length of line does indeed create a large dip. See graph. —  Tivedshambo  (t/c) 20:42, 2 August 2010 (UTC)

Weighing scales

I have a person weighing scales with weights. The bar goes to 7 oz. The weights are 8 stone (x2), 4 stone, 2 stone, 1 lb and 1 ounce. Which weights am I missing? Kittybrewster ☎ 19:38, 31 July 2010 (UTC)

I don't know these balance scales but you may not be missing any weights. I'd have thought you should be able to measure a range of weights by moving the weights that are on the two bars across. If you can adjust to weigh up to 2 stone that way then the big ones are to get to the right sort of range. The small weights may be adjustments or test weights though I'd have thought it would have a screw adjustment to ensure it balanced when no-one was on it. Dmcq (talk) 16:35, 1 August 2010 (UTC)

Yes there is a screw. Kittybrewster ☎ 11:09, 3 August 2010 (UTC)

The answer seems to me to be 8lbs, 4lbs, 2lbs and 8oz. 91.104.18.212 (talk) 21:48, 4 August 2010 (UTC)

Summation image

I see the images of summations at the summation article but cannot make my own (tried editing the markup but got an error). Could someone help me make an image of a summation of m^n from n=0 to n=infinity? —Preceding unsigned comment added by 218.186.8.227 (talk) 21:31, 31 July 2010 (UTC)

Here - $\sum _{n=0}^{\infty }{m^{n}}$. Readro (talk) 21:52, 31 July 2010 (UTC)