Wikipedia:Reference desk/Archives/Mathematics/2007 November 8

= November 8 =

trial functions in sturm-liouville eigenvalue problems
how should one pick trial functions when trying to find the upper bound for the lowest eigenvalue? —Preceding unsigned comment added by 141.213.66.214 (talk) 04:50, 8 November 2007 (UTC)


 * I don't know what you mean - eigenvalues in S-L are determined by boundary conditions - I think what you're asking has no real general solution, but rather is specific to the problem at hand. At least, giving us an example will clarify the question a bit more. And is your trial function for p(x), q(x), y(x), or w(x)? Meanwhile, see the article Sturm-Liouville theory - maybe that will help. SamuelRiv 05:51, 9 November 2007 (UTC)

Real numbers vs. Complex Number
Here was a question proposed to me. What are the properties of the complex plane that are different from the properties of the real numbers (or even the plane RxR)? Which properties does the complex plane has that are not inherited by the real numbers even though the reals are a subset of the complex numbers?

Here are some that I already can think of:

1.The complex are not ordered. The reals are ordered.

2.Two complex numbers can be multiplied. Two points in the real plane cannot be multiplied.

3.Meaning that the complex numbers form a field. The real plane is not a field.

4.The trigonometric functions sine and cosine are unbounded in the complex plane. They are always bounded between -1 and 1 in the reals.

5.The exponential function is periodic in the complex plane. The exponential is unbounded in the real numbers.

6.Any polynomial with all complex coefficients has complex roots always. A polynomial with real coefficients may not have any real roots.

7.It is possible for a continuous function to have a derivative which is discontinuous at a point and continuous everywhere else (like the sharp edge of the absolute value function at zero). In the complex, this is not possible.

Is there anything else that I am missing? I am basically trying to concisely compare the complex and the reals so anything would help. This has turned into a debate somewhat among fellow students. And by the way, we are all math graduate students. A Real Kaiser 05:21, 8 November 2007 (UTC)


 * Are you sure about number 7? Doesn't the absolute value function itself, in the complex plane, provide a counterexample?  —Ilmari Karonen (talk) 06:45, 8 November 2007 (UTC)


 * The absolute value function is not differentiable anywhere in the complex plane. --Lambiam 07:00, 8 November 2007 (UTC)


 * When you say things like "two points in the real plane cannot be multiplied" or "the real plane is not a field", the situation is rather that there is no commonly agreed understanding of the requisite operations, not that it is impossible to define such. You could say, in fact, that the complex plane is the real plane plus a slew of operations. Further, while exp is periodic in the imaginary direction, it is also unbounded in the complex plane. To make item 6 fully correct, you need to restrict it to polynomials of degree 1 or higher; 1 is also a polynomial. --Lambiam 07:10, 8 November 2007 (UTC)
 * 8. It is possible for a real function to be differentiable in an open set (even infinitely many times) without being analytic. For a complex function, just knowing it is differentiable once implies that it is analytic. -- Meni Rosenfeld (talk) 08:34, 8 November 2007 (UTC)

There's a good #9 to add (I think)

$$x^2+Y^2=r^2$$

If Y=yi then

$$x^2-y^2=r^2$$

It also works starting with the Hyperbola.

So every complex value x+Yi satifing a particular Hyperbola or Circle, has a point on the Real Plane on the corresponding Circle or Hyperbola respectively. A math-wiki 08:49, 8 November 2007 (UTC)


 * This works not only for circles and hyperbolas, but for any curve: if x+iy is such that F(x,y) = 0, then (x,y) is such that F(x,y) = 0, and conversely. That is a complicated way of stating that there is a bijection between the complex numbers x+iy and the points (x,y) in the real plane. --Lambiam 09:47, 8 November 2007 (UTC)


 * Note that the distinction between C and RxR is one of structure. RxR is a two-dimensional vector space which does not come pre-equipped with a definition of (non-scalar) multiplication - no meaning is attached to (1,2)x(2,1). But you can extend RxR in various ways by attaching a definition of multiplication. If you attach one particular definition of multiplication, you get a structure which is isomorphic to the field of complex numbers C - with this definition, (1,2)x(2,1)=(0,5). However, you can attach other definitions of multiplication instead to get other algebraic structures - for example, the split-complex numbers, in which (1,2)x(2,1)=(4,5), or the algebra of dual numbers, in which (1,2)x(2,1)=(2,5). Gandalf61 10:47, 8 November 2007 (UTC)

Homology Theory
I have an assignment due on this tomorrow but the Homology section of our course was badly rushed and I didn't absorb anything in it (the lecturer is probably going to be quite lenient in marking this assignment since it was due well after the end of semester on material we barely covered). I'm trying to do this:
 * Let Δp denote the standard p-simplex in Rp+1 with vertices
 * $$\textstyle { e_0 = (1,0,0,\ldots,0), e_2 = (0,1,0,\ldots,0) , \ldots , e_p = (0,0,0,\ldots,1) } $$
 * If $$\textstyle v_0,v_1,...,v_p \in \mathbb{R}^N $$, then let $$\textstyle < v_0, v_1, ..., v_p > $$ denote the unique affine map $$\textstyle f : \Delta_p \to \mathbb{R}^N $$ taking ei to vi for i = 0, ..., p: thus by definition of an affine map, $$\textstyle f\left( \sum_{i=0}^{p} t_ie_i \right) = \sum_{i=0}^{p} t_if(e_i) = \sum_{i=0}^{p} t_iv_i $$.
 * (a) Compute the boundaries $$\textstyle \partial , \partial  , \partial $$, where vi are points in R3 and graphically illustrate, indicating all face orientations.
 * (b) Compute $$\textstyle \partial\partial  $$ and $$\textstyle \partial\partial  $$ and verify that both are zero. Illustrate these results graphically, showing all face orientations.

I haven't got the faintest idea what the question is about and the various Wikipedia articles seem to contradict each other (sometimes a singular simplex is a set of points, sometimes a map). I can't even work out what the $$\textstyle < v_0, v_1, ..., v_p > $$ things are, nor what the boundaries of them will be or even look like (maps? subsets of RN?). About the only thing I know is what a standard p-simplex is. Please help! Maelin (Talk | Contribs) 11:58, 8 November 2007 (UTC)

Addendum: I'm not asking for a solution, I just want someone to explain what the questions are asking about so I can work the solutions out for myself. The problem is not that I don't know how to do this, it's that I don't know or understand what I'm being asked to do. I know and respect the Refdesk homework policy. Maelin (Talk | Contribs) 12:07, 8 November 2007 (UTC)
 * A singular p-simplex is a continuous map from the standard p-simplex Δp into some space under study. Here we are considering the space $$\mathbb{R}^N $$, and the singular simplice we are considering are very well-behaved, being affine maps Δp→$$\mathbb{R}^N $$. Such a map is determined by its values at the vertices of the standard p-simplex, so we denote it by these values: thus $$\textstyle < v_0, v_1, ..., v_p > $$ is a singular simplex in $$\mathbb{R}^N $$, in other words a continous map from the standard p-simplex to $$\mathbb{R}^N $$. If the p+1 points vi are independent, then this map is an affine embedding of Δp into $$\mathbb{R}^N $$: if not, then the map is not injective, with the p-simplex being 'crushed down' onto an affine subspace of $$\mathbb{R}^N $$ of dimension less than p. Since this is somewhat harder to draw, I think you're probably expected to assume the vi are independent for the purposes of drawing pictures. Thus to draw $$\textstyle $$, say, you just need to draw a tetrahedron (=3-simplex) in $$\mathbb{R}^3 $$ and label its vertices v0 up to v3. Algebraist 14:51, 8 November 2007 (UTC)
 * For the rest, you need the definition of the boundary map in singular homology. This maps a singular p-simplex to a formal sum of singular p-1 simplices with appropriate signs. We think (and draw, as you've been asked to do) of a minus sign as being a change of orientation, so $$-\textstyle < v_0, v_1 > =< v_0, v_1 >$$ for example. Thus for the first case, the boundary of the line segment $$\textstyle < v_0, v_1 >$$ is v1 - v0, i.e. the endpoints of the segment, one of them with a minus sign (not easy to draw, perhaps). The boundary of the triangle $$\textstyle $$ is $$\textstyle  -  + $$, i.e. the three sides of the triangle, with orientations that combine to give a cycle round the triangle (either clockwise or anticlockwise, depending on how you draw it). Similarly, the boundary of a 3-simplex (=tetrahedron) is the 4 2-simplices (= triangles) in its boundary, with compatible orientations.
 * In summary of this over-lengthy response, a p-simplex in R3 is officially a map of the standard p-simplex into R3, and its boundary is a formal sum (with signs) of maps from the p-1 simplex to R3 but (at least when the map is nice and affine and injective), we can think of it (and draw it) as an embedded copy of the p-simplex, and its boundary as the faces/sides/end-points of this embedded copy, with appropriate signs.
 * I hope some of this is actually comprehensible and useful. Please ask further questions if I'm making no sense. It would be much easier if we had pictures of simplices and their boundaries, but I can't find any. Algebraist 15:03, 8 November 2007 (UTC)
 * For current and future reference, I just remembered Allen Hatcher's excellent Alg. Top. textbook is available freely online: . Chapter 2 contains a lot of homology theory, including some pictures. Algebraist 15:17, 8 November 2007 (UTC)


 * Ouch. The bad news is you are completely disoriented; the good news is you are not far into the forest.
 * Simplex, for geometry, means a point, line segment, triangular area, solid tetrahedron, and so on; that is, a simplex is a generalized triangle with n independent points as vertices in a Euclidean space of dimension at least n+1. Simplex, for topology, means any continuous image of a geometric simplex. Lucky you, in the problem at hand the differences between simplex as geometry, topology, or map can basically be ignored, with one caveat: do assume that the image vertices are also independent, so that the triangle does not collapse to a line segment or a point.
 * Boundary is at the heart of algebraic topology, and something you must get deep in your bones. The boundary of a simplex is easy, but do work on absorbing the meaning, not just the manipulation. We must first inject a little order; specificially, we list the vertices of the simplex in some fixed order. The boundary of an n-simplex will be a "sum" of (n−1) simplexes, and each simplex of the boundary will inherit a vertex order. Consider a 2-simplex (a triangle), &lang;v0,v1,v2&rang;. Intuitively, its boundary is its edges, and each edge is a 1-simplex. We go through a little ritual to list them all and to ensure that each has the proper vertex order. What we want is &lang;v0,v1&rang;−&lang;v0,v2&rang;+&lang;v1,v2&rang;. In this case we have merely walked around the perimeter, listing pairs as we go, and using a minus if the order is wrong. For a tetrahedron, it's harder to make that intuitive, which is why we want our ritual.
 * Chain is what we call this triangle boundary, this "sum". To extend boundary to chains, as we must to take the boundary of a boundary, we need two steps. The first step is to distribute the boundary operation over the simplices (or simplexes, as you like) of the sum. For the 2-simplex (the triangle), this gives a sum of 0-simplexes (points). If we do everything right, the pluses and minuses in the boundary of a boundary will cancel out (we define boundary so that will happen!), and we are left with "0".
 * All that is left is to learn the ritual for creating the boundary of a simplex. It has two parts. First, we take turns omitting one vertex at a time. Each omission from n independent vertices leaves n−1 independent vertices. For example, one face of a tetrahedron &lang;v0,v1,v2,v3&rang; omits v2, producing the triangle &lang;v0,v1,v3&rang;. For an n-simplex we can omit each of the n+1 vertices, producing a sum of n+1 terms, each an (n−1)-simplex. The second step is to get the signs right. The pattern is really simple, and I strongly urge you to do this yourself. (But if you get stuck ask again, or see page 105 of Hatcher.) Compare to the triangle example, and be sure the boundary of a boundary will vanish for the tetrahedron.
 * I have every confidence you can manage this exercise. It's amazing how much mathematics can flow from such trivial, essentially combinatoric, calculations. Good luck. --KSmrqT 19:33, 8 November 2007 (UTC)

Okay, I think I've got the hang of the boundary map (thanks KSmrq and Algebraist for the above responses, very helpful!). The double boundaries all came out as zero, as expected, but I couldn't really see any deep meaning in the boundary operation. It might be because I can't see anything meaningful in the idea of chains. The formal sum thing for chains is a little confusing and I think that might be part of the problem. It seems like a chain just a sum with no actual meaningful addition - equivalent to a set of ordered pairs of integers and simplices in the space, and they seem to be pretty devoid of any real meaning. So then we've got Cp(X), the set of chains (these formal sums), and it forms a group with that quasi-addition operation. It has a subgroup Zp(X), the set of cycles - chains with zero boundary, and that has a subgroup Bp(X), the set of chains that are themselves boundaries of p+1 simplices. Then we form Hp(X) as the quotient group of cycles over boundaries. I can't really get my head around Hp(X), though. I can't entirely picture the cosets, nor see what it means for two chains to be equivalent modulo the boundaries. Could anyone give me an idea of what this stuff is about? Maelin (Talk | Contribs) 03:33, 10 November 2007 (UTC)


 * I never really got homology for topological spaces, but homotopy intuitively made a lot of sense to me. Formals sums of spheres?? Crazy.  However, looking at spheres inside your space, especially the ones that didn't contain their middles, that's crazy cool.  On the other hand, calculating homotopy was a bit of a spectral sequence nightmare, and calculating homology was just finitely generated abelian groups and row reducing integer matrices.  At any rate, the way I view homology is simply that it is a convenient approximation to counting holes in spaces, since really counting holes in spaces via homotopy is hard.  Alternatively, homology is what you get when you want homotopy to form a ring.


 * The fundamental group is a nice way to distinguish spaces up to certain sorts of stretching and contracting, and it distinguishes compact surfaces nicely, basically because it can count the types of holes formed by circles that cannot be shrunk. Higher dimensional spaces have weirder holes and need higher dimensional circles (spheres), so you look at class of maps of spheres into your space to get the higher homotopy groups.  It turns out by Hurewicz's theorem that doing it this way, the "right way", is not so much different from doing it the easy way, which as you called it, are meaningless formal sums of  maps of spheres (simplices).


 * So homology is measuring how spheres (cycles) and balls (their boundaries) differ inside your space, but in the simplest way possible, as the quotient of free abelian groups. Homotopy does it correctly, and handles homotopy classes of maps of spheres into your space, since contracting a sphere with a homotopy requires it to be the boundary of a ball, but addition of spheres is handled topologically, not algebraically.


 * The first homology and the fundamental group are a clearcut case of this: the first homology is the abelianized version of the fundamental group. Nonabelian groups are too hard, but never fear, we can use homology instead and work with the abelian version.  Sure we lose a lot of structure, but we couldn't calculate the fine grained structure anyways.


 * Another completely different view is topology is crazy and homology is for group theorists because it allows one to write down the difference between the formal laws a group obeys and the actual laws a specific group obeys (the Schur multiplier for instance does this in a few different ways). JackSchmidt 04:50, 10 November 2007 (UTC)


 * I have replied on your talk page; those who want to learn more can look there. --KSmrqT 07:20, 10 November 2007 (UTC)

normalizing a zero set
Percent is often used to normalize values in a set such that if you have 50 item and 20 of them are black then you can normalize to 2/5, 4/10, .4 * 100 or 40 percent. A bucket with 60 red items and 60 white items and 60 black items can be normalized by calculating a total of 180 items, 60/180, (100/(60+60+60)* 60), or 33.33...3 percent red items. With 50 red items and 50 white items and 0 black items we would then have (100/(50+50+0))* 0 or zero percent black items. Using the formula (100/(r+w+b))* b, however, we may have any number of total items except zero. In other words at least one item must not be zero percent. What formula would allow us to normalize to zero percent a set of items with no members? Dichotomous 12:04, 8 November 2007 (UTC)


 * The formal result 0/0 is a solution to the equation 0x=0, but any number x solves that equation, so 0/0 does not define a number. Having 0 blacks out of 0 items makes it correct to say that 100% of the items are black, and equally correct to say that 0% of the items are black, because 100% of 0 is 0, and 0% of 0 is also 0. Bo Jacoby 12:13, 8 November 2007 (UTC).


 * Yes but the problem is here: (100/(0 +0 +0))* 0, where 100/0 is not allowed. Dichotomous 16:57, 9 November 2007 (UTC)
 * Exactly. This is an essential problem. "How much percentage is 0 out of 0?" is not a meaningful question. -- Meni Rosenfeld (talk) 18:55, 9 November 2007 (UTC)
 * And if you switch to an algebraic structure in which it does make sense, that answer is just what Bo Jacoby gave: 0/0. You might meaningfully interpret this as "nothing out of nothing", but it's generally not possible to simplify it further, either arithmetically or "by common sense".  —Ilmari Karonen (talk) 00:21, 11 November 2007 (UTC)

Problem with calculating area
I think I need to go back to school. See if you can help me with this...

I'm trying to fathom area calculation.

Consider the following area calculations:

4m x 4m = 16m² <-- Notice how the RHS is GREATER (16 > 4)

0.1m x 0.1m = 0.01m <-- Notice how the RHS is SMALLER (0.01 < 0.1)

What I now need to know is, what is 100mm x 100mm?

My calculation is 100mm x 100mm = 10000mm² (which is what, 10m². Am I right?) Yet Google refuses to give me the answer in mm, instead it shows me 100mm x 100mm = 0.01m²

This is like saying 100mm x 100mm = 10mm² ????

And the inverse would be: 10mm² = 10mm x 10mm !!!!

Some explanation would be helpful. Thanks Rfwoolf 19:47, 8 November 2007 (UTC)


 * One square meter is not 1000 square millimeters. When you square the units, you have to square the conversion factor, too. So 1 square meter is actually 1,000,000 square millimeters, as 1,000,000 = 1000 squared. Thus, 100mm x 100mm = 10,000mm² = 10,000/1,000,000 m² = .01 m². And also, I think you can say "in square millimeters" to get what you want in google, but I could be wrong. Gscshoyru 19:53, 8 November 2007 (UTC)


 * Ayup. -- LarryMac  | Talk  19:54, 8 November 2007 (UTC)
 * So... do you understand now? Or not? I'm not sure if you do. Gscshoyru 19:57, 8 November 2007 (UTC)
 * Rfwoolf ≠ LarryMac. -- Meni Rosenfeld (talk) 20:06, 8 November 2007 (UTC)
 * Hm. No they are not. The basic question still remains, however. Do you understand? Gscshoyru 20:09, 8 November 2007 (UTC)

Yes I think I've got it. Seems like the use of measurement becomes a distortion of the decimal system. A mm is 1000th of a m. Thus 0.01m² is 0.01 * 1000 * 1000 = 10 000mm². Starts to make sense.

Thanks! Can you by some chance confirm the equivalent units of measurements with inches,

Example: In the metric system we talk of square millimeters, square centimeters, and square meters. In the US system we talk of ???, ???, ???

Thanks again. Rfwoolf 20:17, 8 November 2007 (UTC)


 * Well... ignoring acres and other weird historical measurements of area, we also talk about square inches, square feet, and square miles... and there are 144 square inches in a square foot, as 12 squared is 144. You have to square the conversion factor when converting between squared units, cube it when converting between cubed units, etc. Did I answer your question? I'm not entirely sure what it is. Gscshoyru 20:21, 8 November 2007 (UTC)

Thanks again, I believe you have. I'm now going to look up what's smaller than inches in the system. And, can area in inches be calculated using decimals, or base 12? Back to school I go... ;) Rfwoolf 20:23, 8 November 2007 (UTC) What I've tried to find out, and what you've told me, is something anecdotal, i.e. not some encyclopedic list of units of measurements, but instead what the people in America actually use when referring to, say, material that they'd order. In the metric system I'd say "Can I have 5 square metres of plastic" - and I assume the Americans would say "Can I have 6 square feet of plastic", and the metric people can say "The material costs $5 per square centimeter" the Americans would say "The material costs $6 per square inch". That's basically what I want to know. Of course in the metric system we also have square millimeters. Rfwoolf 20:32, 8 November 2007 (UTC)


 * Hi, it's me who is not you (for all known values of LarryMac and Rfwoolf). We don't really have any standard measure of length smaller than the inch, so most often we talk in fractions - "1/2 inch plywood" for example.  (which is probably only 3/8" thick, but that's a whole other discussion).  And sure, area in inches can be calculated, an item that is 12.2" x 6.5" is 79.3 square inches, or a bit more than 0.55 square feet.  -- LarryMac  | Talk  20:28, 8 November 2007 (UTC)
 * Jack: Who are you?
 * Evil Jack: I am you.
 * Jack: If you are me, then who am I?
 * Evil Jack: You are also you.
 * -- Meni Rosenfeld (talk) 20:40, 8 November 2007 (UTC)


 * I'm only evil sometimes. Usually I'm a really nice lovable guy.  :)  --  JackofOz 23:37, 9 November 2007 (UTC)
 * Worry not, your namesake is of a similarly amicable nature. -- Meni Rosenfeld (talk) 00:48, 10 November 2007 (UTC)
 * And mathematical bases are just representations of numbers -- using any base throughout instead of another in no way changes any calculation or the result. Gscshoyru 20:33, 8 November 2007 (UTC)
 * Thanks LarryMac. I was writing a small application that would have the person input their width and height in millimeters, and have it calculate square centimeters, and apply the price per square centimeter. Now if I enable the US system, they will have to input in inches, calculate in square inches, and apply price per square inch. I was really hoping there would be something smaller than a square inch. But I also know that even if there is, if it's not normally used, it's wiser to change the metric input to centimeters-centimeters-centimers. Thanks Rfwoolf 20:37, 8 November 2007 (UTC)

.9 (Repeating) = EXACTLY 1?
I am basing this on the rule that, to convert a repeating decimal to a fraction, you just take only the group of numbers that repeats over a number of "9's" equaling the number of numbers that repeats, i.e., .732732... (Repeating) = 732/999. And, if so, then what number comes right before .9 (Repeating) in terms of size? —Preceding unsigned comment added by Pitman6787 (talk • contribs) 20:10, 8 November 2007 (UTC)


 * Right before? There is no "right before," for any number in the reals. For every pair of numbers in the reals, there is another between them ( (x1 + x2)/2, for instance) so if you stated that a specific number came "right before" some other number, it would be false, as there is some number that falls between them. And yes, .999... = 1, see 0.999.... Gscshoyru 20:14, 8 November 2007 (UTC)