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January 14

An elementary homotopy equivalence

Continuing with my attempts to grasp the basics of quotient spaces and homotopy, I find myself quickly running into trouble when trying to show that two simple spaces are homotopically equivalent. Consider, for example, the problem of showing that the two spaces "S² with a straight line joining north and south poles" and "wedge sum of S² with a circle" are homotopically equivalent.

I can intuitively see some vague ways of deforming these spaces into each other; for example, one can take the second space and imagine the circle standing inside S² and joined at the north pole, and then quotient half the circle into the side of the sphere; the result should by rights be homeomorphic to the first space. But when it comes to actually writing down explicit maps or homotopies, I'm lost; the text (linked above) is rather light on examples, so it's not even clear what a correct solution would look like. How do topologists approach problems like this? — merge 12:30, 14 January 2010 (UTC)

This is example 0.8 in Hatcher's Algebraic Topology (relevant chapter here). He does it by collapsing a curve in the sphere from N to S to a point, using some general technical lemmas to make this precise. Algebraist 13:17, 14 January 2010 (UTC)

Hmm. Thanks for the reference, but it doesn't seem of much use for this problem; in Bredon's book, it's posed in Chapter 1 just after introducing the most basic definitions and properties of homotopy. There are certainly no CW-complexes in sight, and no propositions about homotopy equivalence of quotient spaces (at least that I can identify as such). — merge 16:23, 14 January 2010 (UTC)

Here's an approach that might help. Imagine the sphere as a (hollow) cube and line as a semicircular "handle" above two points near the centre of one face of the cube. Now apply a homotopy that moves the two points together while preserving the boundary of the cube - it should be reasonably easy to define an explicit formula to do this. Extend to the whole cube using the identity map on the other faces, and you're done (messy details left as an exercise..) (Disclaimer - it's been a long time since I did this sort of thing) AndrewWTaylor (talk) 17:16, 14 January 2010 (UTC)

Just because the approach of using CW complexes and a few technical lemmas isn't the intended approach of your text doesn't mean it can't be the best approach to such problems. Hatcher clearly thinks it is. Algebraist 17:28, 14 January 2010 (UTC)

I have no opinion on what approach is best; just trying to solve the problem with the tools I've been given. — merge 19:31, 14 January 2010 (UTC)

To AndrewWTaylor: Your homotopy takes place outside of both spaces, so with the same reasoning the OP's first space is homotopic to the sphere, which is impossible because one has a non-trivial fundamental group and the other doesn't. Correct me if I'm wrong I'm a novice. Money is tight (talk) 17:34, 14 January 2010 (UTC)

I recall being messily taught this example in an introductory algebraic topology course - the lecturer used a (non-obvious) deformation retraction of both spaces onto a common space...but I can't fully recall what that was. Not that that matters really, since Brenon seems to want it done using basic definitions. Icthyos (talk) 19:19, 14 January 2010 (UTC)

Heh. This section does contain definitions for deformation retracts and some homotopy properties of mapping cylinders, for what it's worth.  :) If this problem is bad, the next two problems on the page must be much worse... — merge 19:31, 14 January 2010 (UTC)

How about this: If X is the one-point union of the sphere and the circle at point x₀ and Y is the other space, let f:Y → X map the points on the closed left side of the Y sphere to x₀ and the rest of the Y sphere to the X sphere minus x₀, and then the line gets mapped to the circle minus x₀ in the way you'd expect. Then let g:X → Y map the X sphere to the Y sphere identically with the point x₀ at the north pole and then map half of the circle to the line connecting the poles and the other half to a path along the left side of the sphere connecting the poles. I'm pretty sure fg and gf are homotopic to 1 but it might be annoying to show. Rckrone (talk) 22:10, 14 January 2010 (UTC)

I've come up with a couple of ideas along these lines, but it always ends up seeming far from obvious how to even show the maps in question are continuous. — merge 23:35, 14 January 2010 (UTC)

Yeah, that's fairly similar to the attempt at a solution I provided - I was told it wasn't clear that the compositions of f and g were homotopic to the identity maps, then we moved on to some hand-waving. Icthyos (talk) 00:21, 15 January 2010 (UTC)

Consecutive integers with same sum of proper divisors

At one point, I was having trouble remembering the definition of Ruth-Aaron pairs, and that got me into some other interesting questions.

• Are there pairs of consecutive integers with the same sum of all their divisors? Yes. (14,15) and (206,207) are the first two such pairs; this is (sequence A002961 in the OEIS).
• What if we look only at proper divisors? (2,3) works and there are no others under 5,000. Note that this sum is always the same parity as the number itself, unless the number is a square or twice a square, so if we want two consecutive ones to be the same, then one of them must be of that type. Before I spend a lot of time programming this to at least start things off, it would be useful to know if there is already a result in the literature.

Ma^t_chups 14:48, 14 January 2010 (UTC)

A few minutes search with PARI/GP says (2,3) is the only case where n² or 2n² with n < 10⁶ is part of such a pair. I don't know whether others have studied the problem. PrimeHunter (talk) 02:11, 16 January 2010 (UTC)

January 15

Is there a name for this property of sets?

Is there a name for the property of sets that its union set is itself? For example, the natural numbers under the standard construction satisfy this property. --129.116.47.169 (talk) 00:25, 15 January 2010 (UTC)

I'd say that such a set is ordered by $\in$ hence well ordered because it is explicitly required by ZF; it's an ordinal. --pma 09:20, 15 January 2010 (UTC)

No, it doesn't have to be an ordinal (nor do all ordinals qualify). For example V_ω has the desired property, but 1 (understood as $\{\emptyset \}$) does not; its union is just $\emptyset$. It would have to be a transitive set, but not all transitive sets work (e.g. 1 again). I don't know of a specific name for the property. --Trovatore (talk) 09:26, 15 January 2010 (UTC)

ach I realized it and was just on the point of correcting it but you are too fast :-) --pma 09:31, 15 January 2010 (UTC) So we may rephrase the assumption on such a set X saying it's a transitive set without "maximal element" wrto the relation $\in$ (into commas because $\in$ needn't to be transitive as a relation on X; X has no maximal element wrto $\in$ in the sense that for any $x\in X$ there is $y\in X$ such that $x\in y$ ) One could also say X is a transitive set non-well-founded upwards. --pma 09:38, 15 January 2010 (UTC)

ω + 1 is transitive and upwards non-well-founded, but it does not have the property. — Emil J. 18:52, 15 January 2010 (UTC)

Haha!! oops --pma 21:42, 15 January 2010 (UTC)

Unless I'm missing something, none of the finite natural numbers under the standard convention have this property. Since n = {0, ..., n-1}, the union set of n is surely {0, ..., n-2}, or n-1. For example: 3 = {0, 1, 2} = {{}, {0}, {0, 1}}, and thus its union set is {0, 1} = 2. -- The Anome (talk) 17:07, 15 January 2010 (UTC)

The set of all natural numbers satisfies it. I think that is what the OP meant. --Tango (talk) 17:14, 15 January 2010 (UTC)

Ah. That would make sense. -- The Anome (talk) 17:24, 15 January 2010 (UTC)

Thanks for the responses, guys.

Tango is right. I should have made that clearer.

pma's description as a transitive set with no maximal element seems to work; from equating the two sets in my original characterization of X, you get that x is an element of X iff x is an element of an element of X. pma's first condition expresses the backward implication, and his second condition expresses the forward implication.

Am I correct in thinking that pma was wrong when he said the set is required to be well-ordered under $\in$? Isn't V_ω a counterexample? --128.62.37.240 (talk) 01:09, 16 January 2010 (UTC)

Yes, that was part of the first sentence (that the set was an ordinal) and wrong; however the relation $\in$ is well-founded (no infinite descending chains exist) according to the axiom of regularity of ZF (assuming you are working there). --pma 08:40, 16 January 2010 (UTC)

Is a set that contains only itself hereditary?

In the process of copyediting hereditary set, I found myself writing the sentence

In non-well-founded set theories where such objects are allowed, a set that contains only itself is also a hereditary set.

It then occurred to me not only that this may or may not be true, but that it might not even be a meaningful statement. Consider the set E = {E}. By the definition of hereditary sets, if E is hereditary, then {E} is hereditary, which merely restates the initial premise. If E isn't hereditary, then E isn't hereditary, again restating the inital premise. I can't see how to get a better handle on this problem. Can anyone help? -- The Anome (talk) 14:59, 15 January 2010 (UTC)

The usual way to unambiguously phrase such definitions in non-well-founded set theories is to define that A is a hereditary xxx iff every object in the transitive closure of {A} is a xxx (note that this is equivalent to the inductive definition if the universe is well-founded). Your E is thus indeed a hereditary set. — Emil J. 15:09, 15 January 2010 (UTC)

Thanks! Could you, or anyone else with good set theory knowledge, please update the hereditary set article to reflect this? I'm afraid I'm outside my area of competence here. -- The Anome (talk) 15:12, 15 January 2010 (UTC)

OK, I've expanded it a bit. — Emil J. 16:05, 15 January 2010 (UTC)

Thanks! -- The Anome (talk) 16:55, 15 January 2010 (UTC)

Continuity of power functions

How to prove the continuity of all functions f: ℝ→ℝ, f(x)=|x|^a, a>0, in x=0? --84.61.165.65 (talk) 15:30, 15 January 2010 (UTC)

Whenever you want to prove a function is continuous at a single point the best method is almost always to just apply the definition. Write down the definition, plug in this function and this point and see what happens. --Tango (talk) 17:10, 15 January 2010 (UTC)

Using monotonicity of both the power function and its inverse may help. Michael Hardy (talk) 16:01, 16 January 2010 (UTC)

Or just take a logarithm, if you have it in your toolbox. --pma 23:10, 16 January 2010 (UTC)

Note that the question is about $x=0$. I doubt the right-continuity of log at 0 is in the OP's toolbox. -- Meni Rosenfeld (talk) 15:32, 17 January 2010 (UTC)

Who knows! It's just about $\scriptstyle \inf _{x>0}\log x=-\infty ,$ not so esoteric. pma 17:23, 17 January 2010 (UTC)

The thing is that applying the term "continuous" here requires moving from interpreting $-\infty$ as a piece of notation to including it as an actual element of the topological space under consideration, which is relatively advanced. -- Meni Rosenfeld (talk) 17:48, 17 January 2010 (UTC)

Well, this is going to be a theoretical debate about the content of the OP's head, since (s)he already evaporated. But saying that a sequence of positive numbers converges to zero if and only if the sequence of their logarithms converges to negative infinity (as a piece of notation, why not), is an elementary fact about limits, even more elementary than the continuity of |x|^a in 0 for a>0, and as far as I know they usually teach it in basic calculus courses. But note that somehow I shared your doubt, since I put an "if" clause. So, if OP doesn't have that in the toolbox, it's a good opportunity to put it in. Actually, we should have started asking the OP what's his definition of t^a, and what's his definition of continuity, in order to give a pertinent answer (Salomonic answer by Tango is good in any case, as usual). I added the log option for completeness, especially in case his definition of t^a was e^a log(t), which is reasonable and commonly adopted. pma 22:44, 17 January 2010 (UTC)

Infinite ordinals

Would the infinite ordinal epsilon zero be expressed using hyperoperations as ω [4] ω or ω [ω] ω? 99.237.180.215 (talk) 17:48, 15 January 2010 (UTC)

The former, if I understand the notation correctly. — Emil J. 18:57, 15 January 2010 (UTC)

What would ω [ω] ω be? 99.237.180.215 (talk) 21:40, 15 January 2010 (UTC)

According to epsilon zero, $\phi _{\omega }(0)=\omega \uparrow ^{\omega }\omega$. I'm not 100% sure about the notation, so I'm not sure that is exactly what you are asking for, but it is close. See Veblen function for an explanation of the phi notation. --Tango (talk) 22:54, 15 January 2010 (UTC)

Thanks. 76.67.74.33 (talk) 05:18, 16 January 2010 (UTC)

Properties of certain symmetric matricies?

Suppose I have a plain-old function $f:\mathbb {N} \to \mathbb {R}$ from the integers to the reals, and a plain-old matrix $M$ with matrix elts $M_{ij}=f(ij)$. Aside from the fact that M is symmetric, are there any other "well-known" properties or theorems that apply to this case? Anything in particular about the eigenvectors or eigenvalues of M? (No, the function f is not a multiplicative function, which would have made M into a outer product of two vectors.) linas (talk) 23:25, 15 January 2010 (UTC)

Is the domain of f the integers or the natural numbers? You've said integers and used the symbol for natural numbers... --Tango (talk) 00:23, 16 January 2010 (UTC)

Presumably the naturals, unless M is a very odd matrix indeed. Algebraist 00:45, 16 January 2010 (UTC)

You mean you don't index your matrix rows starting from -1? --Tango (talk) 00:56, 16 January 2010 (UTC)

I'll take any theorems on any oddly numbered indexes you've got :-) In my case, f is tending to zero, not quite exponentially fast, and is oscillatory.99.153.64.179 (talk) 04:10, 16 January 2010 (UTC)

Something is not quite clear to me. Can you precise more in detail what information you have on the order of the square matrix and on the function f ? If you are talking of a square matrix A_n of a given order n, and if the only information on f is what you wrote, then we are just talking of any symmetric matrix where some entries coincide (for instance if n=6, it's a symmetric matrix where the coefficients may be whatever, with the only constraints a_2,2=a_1,4 and a_2,3=a_1,6). Of course if n is large the information become more relevant, but the fact that f vanishes at infinity never enters, unless what you want is an asymptotic information on the matrices A_n as n → ∞. Or are you thinking to an infinite matrix, that is a e.g. a linear operator on l₂? --pma 08:49, 16 January 2010 (UTC)

Not relevant to your problem since f vanishes at infinity and you're probably aware of this anyway, but if f is a polynomial then the determinant of the matrix will vanish if it has order 2 or more greater than the order of the polynomial. Dmcq (talk) 09:30, 16 January 2010 (UTC)

In case you are interested in the operator on $l_{2}$, it is relevant to know whether your $f$ is in $l_{2}$ (that is $\textstyle \sum _{n=1}^{\infty }f(n)^{2}<\infty$) which for your situation is equivalent to $M(i,j):=f(ij)$ being a Carleman kernel (hence the corresponding operator on $l_{2}$, let's call it $T$, is closed). --pma 10:45, 16 January 2010 (UTC) I'm not sure if $f\in l_{2}$ also implies that the operator $T$ is bounded; it shouldn't be difficult to decide it (but first you have to decide if you really want to know it). If you have something better on $f$, namely $\textstyle \sum _{n=1}^{\infty }f(n)^{2}n\log \log n<\infty$, then $T$ is a Hilbert-Schmidt operator (thus compact, and its spectrum is ahttp://bits.wikimedia.org/skins-1.5/common/images/button_sig.pngn $l_{2}$ sequence of eigenvalues). In conclusion: personally I do not know of results where the above form of a symmetric kernel plays a role (as compared e.g. with $M_{i,j}\,:=f(i-j)$, that produces discrete convolution operators), but it is well possible that it has been introduced and studied. I can imagine that these matrices may appear quite naturally in some situation, e.g. from number theory or harmonic analysis (in fact I suppose they came out from some research problem of yours). And, of course, the special form of your matrices may give rise to some semplification when appling known results; however, this is too general to provide an answer here (I mean: if we were as powerful as Google, or if we had 1000 years and nothing else to do, we could produce as answer 1,000,000 results vaguely related with your matrices, possibly none satisfing your needs). So we really need you to be more specific both on the assumptions and on the conclusions of the results you are looking for.--pma 17:18, 16 January 2010 (UTC)

Heh.The jig is up. Yes, in my case, M is an operator. I've a class of other similar operators (they are all transfer operators, although this is the only symmetric one I have). I can define it on l₂ and also L₂; I've reason to believe it's spectrum in L₂ is continuous; I'm more interested in the l₂ basis where it has a discrete spectrum. What I actually have is an operator G whose asymptotic expansion has M as its first term (as I mentioned above, the f is oscillatory and decreasing; I'm trying to nail down the details now). (My operator G is bounded, I presume my asymptotic term must be also.). A google search search for 'Carleman operator' indicates that it's a condition on the integral kernel, viz, that when then integral kernel K(x,y) has $\int |K(x,y)|^{2}dy<\infty$ then the operator is a Carleman operator. In my case, I don't believe that my kernel is square integrable in this way (the kernel for my G has dirac delta functions in it, I haven't thought about M). So I the Carleman condition is a side-track.

But this is all wandering off-topic -- I figured that the definition I gave above is 'generic' enough that perhaps something is know about the structure of these kinds of operators. Yes, my particular f is square-summable -- it has $f(n)\sim e^{-c{\sqrt {n}})}$. Also, fwiw, I suspect that this may be relevant but may sound far-fetched to you: note that if $z=a+ib$ then $z^{2}=a^{2}-b^{2}+2iab$ and I suspect that reason that my operator $M_{ab}=f(ab)$ has this product ab of the indexes is because there's a hidden z^2 elsewhere in the problem (my operators come from analytic number theory so these kinds of crazy weirdnesses abound). (I can tell you all about my operator G, if you really want to know, its the GKW operator, but that's beside the point ...)

As to having a million years to solve math problems ... well, perhaps in 30 years, we'll be able to attach our brain directly, neuronally, to a calculator having a few exaflops of cpu power...so we'll all be walking Ramanujan's ... exactly what might happen then, who knows. I suppose only rich mathematicians will be able to afford this :-( linas (talk) 18:05, 16 January 2010 (UTC)

Sorry if I wasn't clear: talking of the Carleman condition I referred to $\mathbb {N}$ as a measure space (with the counting measure), so that the kernel is a function $K:\mathbb {N} \times \mathbb {N} \to \mathbb {C}$, which here has the form $K(i,j):=f(ij);$ the Carleman condition you mentioned in this case writes $\sum _{i}|K(i,j)|^{2}<\infty$ for all $j\in \mathbb {N} ,$, and this of course is true if $f\in l^{2}$. But that was just a hint, even because I didn't know much about your $f$. The other condition I wrote just ensures (due to the asymptotics on the divisor function d(n), of course) the much stronger $\sum _{i,j}|K(i,j)|^{2}=\sum _{n=1}^{\infty }d(n)f(n)^{2}<\infty ,$ that is, K is Hilbert-Schmidt operator. More generally, the operator should be in some p-trace class provided some summability on f holds; I guess $f\in O\left(\exp(-c{\sqrt {n}})\right)$ as you say should give you that with no pain, even for p=1 (so the operator is nuclear, which implies that the spectrum is even in $l^{1}$). --pma 19:36, 16 January 2010 (UTC)

Yes, I was hoping for something more specific .. e.g. something analogous to the Hankel matrix or Toeplitz matrix (which becomes a Toeplitz operator) or perhaps some relation to some variant of a companion matrix or Vandermonde matrix or something along those lines. (roughly speaking, the GKW is the shift operator on the space of continued fractions so it should have all sorts of interesting multiplicative structure). linas (talk) 20:39, 16 January 2010 (UTC)

I see the analogy... actually the kernel $f(i-j)$ corresponds to a convolution with $f$, and there is a very nice algebraic structure around, that gives a lot of facilities. The kernel f(ij) may still enjoy some nice algebraic feature, but I can't see it. What I see is more on the side of bounds, e.g., that if $f$ is as you say $\scriptstyle O\left(\exp(-c{\sqrt {n}})\right),$ the operator is nuclear, even if you adopt suitable weighted $l_{2}$ spaces, and this gives you asymptotic estimates for the eigenvalues (I guess they go to 0 like $f$ or so).  :-/ pma 22:03, 16 January 2010 (UTC)

January 17

Archimidean Spiral Segment

What is the area of a segment of an Archimidean spiral r=a+bt between t₀ and t₁, where t₀ and t₁ are separated by no more than 180 degrees?

In fact, what I'm really looking for is the maximum distance between the chord and the piece of the spiral, but I'm guessing that the area is easier to calculate and will do as an approximation for my error calculation.

82.255.173.168 (talk) 17:15, 17 January 2010 (UTC)

${1 \over 2}\int _{t_{0}}^{t_{1}}r^{2}\,d\theta$ will give you the area of a sector; subtract from that the area of a triangle and you get the segment. It shouldn't be difficult to write down the equations for the distance of an arbitrary point (parametrized by t) on the arc from the chord, and then differentiating and equating to 0 to obtain the equation for the maximizing argument. Maybe it even has a nice-looking solution, which you can plug in the the formula for the distance to obtain the maximum distance. -- Meni Rosenfeld (talk) 17:58, 17 January 2010 (UTC)

And what actually is that integral? I notice the article on Archimidean spirals has an expression for the arc length, but not the segment area, which makes me think that the integral is non-trivial to work out (it's certainly beyond me). 82.255.173.168 (talk) 19:55, 17 January 2010 (UTC)

Here the above is ${1 \over 2}\int _{t_{0}}^{t_{1}}(a+b\theta )^{2}\,d\theta$ can't you compute it?--84.220.119.148 (talk) 02:31, 18 January 2010 (UTC)

Unless it's ${{a^{2}t+abt^{2}+b^{2}t^{3}+c} \over 2}$, then no, I can't (I'm a software guy). Is that what it is? 82.255.173.168 (talk) 07:05, 18 January 2010 (UTC)

Almost. The integral is ${1 \over 2}\int _{t_{0}}^{t_{1}}(a+bt)^{2}\,dt={\frac {(a+bt_{1})^{3}-(a+bt_{0})^{3}}{6b}}$, and the area of the triangle is ${\frac {(a+bt_{0})(a+bt_{1})\sin(t_{1}-t_{0})}{2}}$. -- Meni Rosenfeld (talk) 08:19, 18 January 2010 (UTC)

Thanks, that's just what I needed. 66.240.20.113 (talk) 11:26, 18 January 2010 (UTC)

Ooops, it might not be just what I need. As a true empiricist, I tried a few extreme values to find out what happes. In the case of a=1, b=0, The segment evaluates to ${(1+0)^{3}-(1+0)^{3}} \over {6\times 0})$. Would l'Hopital's rule have given me a more realistic value, or is there a blunder? 66.240.20.113 (talk) 13:06, 18 January 2010 (UTC)

Indeed, the formula was written for $b\neq 0$, but this is just a notational issue. You certainly can use l'Hopital's rule to obtain the limit when $b\to 0$, which gives the expected result ${\frac {a^{2}(t_{1}-t_{0})}{2}}$ for the integral. Or you can write the answer as ${1 \over 2}\int _{t_{0}}^{t_{1}}(a+bt)^{2}\,dt={\frac {a^{2}(t_{1}-t_{0})}{2}}+{\frac {ab(t_{1}^{2}-t_{0}^{2})}{2}}+{\frac {b^{2}(t_{1}^{3}-t_{0}^{3})}{6}}$ which works for any $b\geq 0$. -- Meni Rosenfeld (talk) 14:09, 18 January 2010 (UTC)

Identity Matrix Multiplication

I am a little confused on Identity Matrix Multiplication

${\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}}{\begin{bmatrix}3&-2&5\\0&2&-3\\-1&4&0\end{bmatrix}}={\begin{bmatrix}3&-2&5\\0&2&-3\\-1&4&0\end{bmatrix}}$

Whereas, Matrix Multiplaction requires us to go this way:

• matrix 1 row 1 * matrix 2 column 1
• matrix 1 row 1 * matrix 2 column 2
• matrix 1 row 1 * matrix 2 column 3

....

Thus:

• 1(3) + 0(0) + 0(-1) = 3
• 1(-2) + 0(2) + 0(4) = -2
• 1(5) + 0(-3) + 0(0) = 5

Thus:

${\begin{bmatrix}3&-2&5\\.&.&.\\.&.&.\end{bmatrix}}$

However, Identity Matrix Multiplication seems to be:

• matrix 1 row 1 * matrix 2 row 1 = solution matrix row 1?

• and when the element (in matrix 2) is multiplied by 0 it remains the same; i.e, assume it is multiplied by 1?

• and when the element is 0 (matrix 2 row 3 column 3) and it is multiplied by 1 it remains 0?

--33rogers (talk) 20:13, 17 January 2010 (UTC)

No, matrix 1 row 1 * matrix 2 col 1 is answer matrix element in row 1 and column 1. (Multiply a row by a column, by multiplying element by element and adding up the products.) Matrix 1 row 1 * matrix 2 col 2 is answer matrix element in row 1 and column 2. So the answer matrix has the number of rows of the first matrix and the number of columns of the second matrix. And your calculations for the other rows are:

Second row:

• 0(3) + 1(0) + 0(-1) = 0
• 0(-2) + 1(2) + 0(4) = 2
• 0(5) + 1(-3) + 0(0) = 3

Third row:

• 0(3) + 0(0) + 1(-1) = -1
• 0(-2) + 0(2) + 1(4) = 4
• 0(5) + 0(-3) + 1(0) = 0

Get it now? I hope this helped! Remember the paratroopers - they fly in horizontally from the left and parachute down vertically on the right! RupertMillard (Talk) 20:40, 17 January 2010 (UTC)

That is when it normal matrix multiplication. My question is regarding Identity Matrix Multiplication. --33rogers (talk) 20:43, 17 January 2010 (UTC)

Not sure if this is what is causing the confusion, but: 1(-2) + 0(2) + 0(4) = -2 (not +2, as you have above). Abecedare (talk) 20:57, 17 January 2010 (UTC)

Thanks for that (now fixed).--33rogers (talk) 21:21, 17 January 2010 (UTC)

There is no separate such thing as 'Identity Matrix Multiplication'. There is an identity matrix for each size matrix and if you do matrix multiplication using an identity matrix then the result is the same as the other operand. Dmcq (talk) 21:02, 17 January 2010 (UTC)

Thanks.

Resolved

--33rogers (talk) 21:21, 17 January 2010 (UTC)

Weird summation notation

Hi all,

I have an article and there's a part I don't quite understand. Here it is, quoted:

---

$v\displaystyle \sum b_{i_{1}}b_{i_{2}}\dots b_{i_{h-1}}b_{i_{1}+\dots +i_{h-1}}$

where the sum is over $i_{1},\dots ,i_{h-1}$ and the index of the last term is to be interpreted modulo v.

---

I don't quite understand what this means: The summation part or the part saying "where the sum is over $i_{1},\dots ,i_{h-1}$". The prior parts of the article make no reference to a double subscripted variable - just $b_{i}$ and $b_{ij}$. Could anyone shed some light on this?

This is the paper of interest: Page 222.

Thanks in advance. x42bn6 Talk Mess 21:27, 17 January 2010 (UTC)

I can't read the paper, but the notation is standard and unambiguous, provided it is clear where each of the $h-1$ indices $i_{1},\dots ,i_{h-1}$ varies; e.g. from 1 to v (or whatever). So you have $v^{h-1}$ of these (h-1)-ples $(i_{1},\dots ,i_{h-1})$, namely $(1,\dots ,1)$ up to $(v,\dots ,v)$. Suppose for instance $v=h=3,$ then there are 9 pairs and the sum is

$b_{1}b_{1}\,b_{2}+b_{1}b_{2}\,b_{3}+b_{1}b_{3}\,b_{1}+b_{2}b_{1}\,b_{3}+b_{2}b_{2}\,b_{1}+b_{2}b_{3}\,b_{2}+b_{3}b_{1}\,b_{1}+b_{3}b_{2}\,b_{2}+b_{3}b_{3}\,b_{3}$--pma 23:08, 17 January 2010 (UTC)

Certain repeating digits

Prob'ly just an amusing bit of recreational math, but tell me if there's something deeper here. Just two examples:

$2^{25}=33554432.\,$

And if one finds the probability of at least 10 successes in 12 independent trials with probability 0.6 of success on each trial one gets (the denominator has the 11th power of 5 rather than the 12th power because of a cancellation with a 5 in the numerator):

${\frac {3^{10}\cdot 69}{5^{11}}}=0.08344332288.$

So why do these odd repetitions happen? Are these both instances of a common phenomenon? (Of course, dividing by 5 is the same as multiplying by 2 and moving a decimal point.) Michael Hardy (talk) 22:12, 17 January 2010 (UTC)

Without knowing the sample space in which you're looking it's impossible to say much. E.g. in all the above you've powers of 2, 3, and 5 and a × 69. Looking at those up to powers of 25 (though your bound could be higher), with the other number going up to say 100, and its over a million possibilities, and that's not counting the number of ways you can arrange them. So if you look at all those sure you're going to see a few odd patterns of numbers, rather like if you look at the first million digits of π. Not sure I'd read anything else into it. --JohnBlackburne^words_deeds 00:03, 18 January 2010 (UTC)

A "sample space", as I usually understand that term, is a probability space. No "sample space" is involved in my first observation; it's just a fact of arithmetic, involving no probability. For the second observation, I was completely explicit in identifying the sample space. If you prefer, I could say we're talking about a binomial distribution, and anyone familiar with probability will know that when they read what I wrote. Michael Hardy (talk) 03:37, 18 January 2010 (UTC)

As I read his comment, John does not refer to the probabilities in the problem you described. Instead, he assumes (or models the situation so) that you have drawn these two formulas from some probability distribution over formulas, and surmises that the set of possible formulas (the sample space) is large enough that the oddity could be adequately explained by mere chance (i.e. given your rate of seeing "random" formulas, you'd expect to encounter a few odd ones once in a while). The probability distribution he means might be something like $\operatorname {Pr} ({\mathcal {F}}=cb^{n})\propto 1$, with $(c,b,n)\in \{1,\ldots ,100\}\times \{2,6\}\times \{1,\ldots ,25\}$, and ${\mathcal {F}}$ denotes a formula.

I'd write your latter example as 8344332288 = 6¹¹×23 to keep it as an integer. This way in both of your examples there is a relatively high power of a small number multiplied by a small number. Some more numbers of this type would be e.g. 1678822119 = 3¹⁷×13, 1451188224 = 6¹¹×4 and 9886633715 = 7¹¹×5. No insight to offer though, sorry. -- Coffee2theorems (talk) 05:07, 18 January 2010 (UTC)

Well, my example of 8344332288 = 6¹¹×23 has four of those pairs and only two digits not participating. So I'm winning. (But don't read this comment.) Michael Hardy (talk) 05:33, 18 January 2010 (UTC)

Heh. I was just a little curious what the other examples look like in case there's an obvious pattern, so I sicced the computer at it. If this were a competition, I guess using a computer would be grounds for disqualification :) -- Coffee2theorems (talk) 14:07, 18 January 2010 (UTC)

Well, the repeated digits are solutions to N = a×11 + b×11×100 + c×11×10⁴ + ... for integers 0<a,b,c<10. I have no particular insight, other than the general experience that patterns are endemic when fiddling with integers in this way. Instead of base-10, how about other bases? Instead of 11, how about other values? Is there a base B and multiplier M (instead of 11) for which this phenomenon does not occur? linas (talk) 01:10, 20 January 2010 (UTC)

January 18

Asians

Why are Asians so good at math? --70.129.187.105 (talk) 01:03, 18 January 2010 (UTC)

Who says they are? --Tango (talk) 01:14, 18 January 2010 (UTC)

The vast majority of students at national math programs like MathCamp and AwesomeMath are Asian. Also, China almost always wins the IMO. --75.33.218.22 (talk) 02:45, 18 January 2010 (UTC)

Asians have a record of doing tremendously well in international mathematics Olympiads, but such says little about how an individual will approach "real mathematics". Olympiads are designed to evoke mathematical interest in high school students, and thus typically they test "problem solving", but with simple mathematical content (thus, the problems are readily understood by an average high school student, but not necessarily easily solved). However, it is quite possible to do very well in Olympiads, but struggle while doing higher level mathematics, and it is also quite possible to be an excellent mathematician, but struggle to solve Olympiads problems (in fact, few contestants in the IMO actually pursue a mathematical career). I would also note that much of the Olympiad subsumes "finite mathematics"; for instance, basic combinatorics, number theory and Euclidean geometry.

With regards to your question, note mathematicians such as Georg Cantor, Felix Hausdorff, Leonhard Euler, Carl Freidrich Gauss, Joseph-Louis Lagrange, Stefan Banach, Niels Henrik Abel, Évariste Galois, Carl Gustav Jacobi and so forth (obviously I have missed many great mathematicians in this list), none of whom were Asian. This does not say that Asians are not "good at mathematics"; in fact, quite the contrary is true. Rather, one does not need to be Asian to be good at mathematics, and a great number of mathematicians of the past were not Asian. If it is not obvious in my post, doing well in mathematics tests and exams says little about mathematical ability; perhaps your question was based on this (incorrect) assumption. And I do not think that classification of "how good someone is at mathematics" is useful. --PST 03:19, 18 January 2010 (UTC)

Please define "Asians". Thanks, hydnjo (talk) 04:16, 18 January 2010 (UTC)

If you were asking this to the OP, he/she might as well direct you to the article Asian. If you were asking this to me, I was merely using "Asian" in the context of the OP's question, the purpose of which I am unable to comprehend. --PST 05:57, 18 January 2010 (UTC)

With regards to why the Chinese and people from other Asian countries seem to be so good at certain types of math, like olympiads, it's probably cultural. Chinese culture and Chinese institutions emphasize certain skills and values that many Western countries don't as much, and vice versa. That influences how people develop and what kinda of goals they pursue. Even people who have emigrated are still highly influenced by the values encouraged by their families. Rckrone (talk) 04:24, 18 January 2010 (UTC)

You might like to look at Race and intelligence which deals with a number of differences in tests including mathematics plus a number of explanations. An interesting bit I saw there was about Ashkenazi Jews scoring high on mathematics but low on visuospatial, it seems surprising those aren't correlated. Dmcq (talk) 13:12, 18 January 2010 (UTC)

Olympiads are not the same as real math. Olympiads typically involves speed, and real math is "sit there and think". Money is tight (talk) 18:56, 18 January 2010 (UTC)

In fact, I would say that Olympiads have little to do with real mathematics, even in terms of content. Many problems in the IMO really subsume finite mathematics, if anything, but even that "finite mathematics" is a poor excuse for real "finite mathematics". Sure, Olympiads require a particular kind of intelligence, but it cannot be concluded that that intelligence is mathematical intelligence. Typically, Olympiads test one's ability to apply known techniques to solve a problem, whereas real mathematics focuses on inventing new techniques to develop intuition. --PST 02:30, 19 January 2010 (UTC)

Indians are clearly better at math than Western people:

http://www-history.mcs.st-and.ac.uk/Projects/Pearce/Chapters/Ch9_3.html

http://en.wikipedia.org/wiki/Srinivasa_Ramanujan

This is probably due to genetics. Indians have lived in tolerant civilizations continuously for more than 3000 years. In Europe people were still living in caves 2000 years ago. In the Middle Ages, people in Europe were persecuted, there was little room for free thought. Count Iblis (talk) 23:31, 18 January 2010 (UTC)

I strongly disagree; there is no concrete criteria using which we may compare the mathematical abilities of two different races. Could you please state what exactly you mean by "better at math"? --PST 02:30, 19 January 2010 (UTC)

I think (hope) this is trolling. Rckrone (talk) 04:35, 19 January 2010 (UTC)

Well, there are more people from China than people from any other country so, assuming fairly equal access to olympiads, you should expect more Chinese winners. Other differences might be explained by differences in culture, education, nutrition etc. See Race and intelligence for more. Zain Ebrahim (talk) 10:31, 20 January 2010 (UTC)

It may be a matter of prestige of math and specific educational system. E.g. 20 years ago USSR score on IMOs was better Russia, i suppose. My school was the best in math in my city (it provided IMO medalists literally every year). A half or even more of children there were of mixed Jewish-Russin descent. It cannt be explained in terms of genetics: a half or even more of children who tried but failed to pass through exams to our school were also of the same Ashkenazi descent. Well, may be the genes made them _interested_ in math education. I bet it were their parents but not the genes.--95.84.241.53 (talk) 19:09, 20 January 2010 (UTC)

An IMO winner is not necessarily "good at math"; there is a wealth of qualities that one must possess, not only ability to "solve problems", to become a mathematician. As I noted, IMO can only comprise subject matter relating to grade 12 mathematics at the most; it does not subsume many branches of mathematics studied at university, for instance. In terms of intelligence, one needs to be well acquianted with the techniques used to solve IMO problems, and often this requires intensive practice just like any other skill. But knowing techniques and applying them to a given situation is not what "real mathematics" is about. --PST 02:32, 21 January 2010 (UTC)

OP asked about math-education and competitions and IMOs specifically. There was a joke: the three major competitors at IMOs were Russian team (jews taught by jews), American team (chinese taught by jews) and China team (Chinese taught by chinese). I think, one have to be smart enough (yes, i cannt define 'smart') to win and one should love math to participate. Both qualities help one with math. At least there shoud be correlation. May be (maybe), such competitions aren't good as educational tools, the problems lack beauty and sence IMHO. Probably they require essentially different kind of creativity than 'pure math' does. Back to OP's question - I just see how moscow jewish children tend to concentrate around sciences and math-education whether they are good at math or bad. There exist cultural reasons. Possible underlying genetics is obscured by them as it usually happens. BTW, abilities necessary to get it through exams into some prestigeous university are never inentical to those one must possess to be good at anything--95.84.241.53 (talk) 08:14, 21 January 2010 (UTC)

I guess there exist some papers relevant to subject. Stats, sociology, etc. If nobody can point to one, OP may repost the question in humanities section. Another my guess: american chinese may seem 'better at math' than chinese chinese. Let's consider a child exellent in problems solving. Whether the child will be interested in advancing and whether (s)he will be informed of appropriate educational possibilities and whether (s)he will decide to participate in the aforementioned national math-programs depends on the parents, the teachers, the classmates etc. A talented child belonging to chinese diaspora may have better chances to be noticed and to notice him/herself and to realize value and importance of his/her interest at math than 'common' chinese or american child. A 'common' child have chances to be ignored or to chose conventional ways of getting education and work or just to never consider it cool to be a nerd. BTW excuse my English:)

differential

for shm equation $P''+k^{2}P=0$, when $k=0$, $P=c\phi +d$ is a solution. However if you are solving some sort of problem in polar coordinates, and have the restriction $P(\phi )=P(\phi +2m\pi )$ for an integer m, is the solution $P=2\pi \times floor({\frac {c\phi +d}{2\pi }})$ a solution? —Preceding unsigned comment added by 129.67.39.49 (talk) 01:03, 18 January 2010 (UTC)

They are not solutions. The solution has sine waves. See Simple harmonic motion for solutions. Dmcq (talk) 08:06, 18 January 2010 (UTC)

In your differential equation $P''+k^{2}P=0$, what exactly are you differentiating with respect to ? If this is simple harmonic motion then $P''$ normally denotes ${\frac {d^{2}P}{dt^{2}}}$, but you seem to be using it to denote ${\frac {d^{2}P}{d\phi ^{2}}}$. Gandalf61 (talk) 10:47, 18 January 2010 (UTC)

To respnd to both of you in one post;

Dmcq, I was considering only k=0, or $P''=0$, the sine wave is not a solution.

Gandalf, yes I am actually looking at laplaces equation in spherical polars, which when seperated gives $P''+k^{2}P=0$ where is defined by $V(r,\theta ,\phi )=R(r)T(\theta )P(\phi )$where k is the seperation constant. Since when solving equations involves summing over all k, I was considering the case where k is zero. (It was my understanding that $P''$ denoted a non specific differnetial, with context usually used to give the variable, and that it was ${\ddot {P}}$ denoted specifically $P_{tt}$). Anyway, any answers would be lovely! —Preceding unsigned comment added by 129.67.39.49 (talk) 16:15, 18 January 2010 (UTC)

(edit conflict) Actually $P=c\phi +d$ is the correct solution, not a sine wave: the OP specified that $k=0$ so $P''=0$. However, that's not what SHM or simple harmonic motion refers to: SHM means $P''+k^{2}P=0$ with non-zero $k$ and gives sine-waves as Dmcq says.

As for your actual question: differentiating shows us that it solves the differential equation except at multiples of $\pi$ where the sudden jump in $P$ means that the derivative is undefined: there's no sensible answer to the question 'what's the derivative at this point?'. It also fails your requirement that $P(\phi )=P(\phi +2k\pi )$ - for example, if $c=1$ and $d=2\pi$ then $P(0)=2\pi$ but $P(2\pi )=4\pi$.

The most general solution is in fact just $P(\phi )=d$, since any non-zero $c$ will give discontinuity problems as above.

Also, Gandalf: I disagree that $P'$ usually indicates the time-derivative - I'd say it would mean differential with respect to whatever variable $P$ is a function of, or the non-time variable if it includes two dependencies. Still, it's best to always say what variable you're differentiating with respect to to avoid confusion. Olaf Davis (talk) 16:32, 18 January 2010 (UTC)

(after edit conflict)Well, the reference to shm was somewhat confusing. Anyway, the only solutions to $P''=0$ that also satisfy $P(\phi )=P(\phi +2m\pi )$ are the constant functions $P(\phi )=d$. The proposed solution $P=2\pi \times floor({\frac {c\phi +d}{2\pi }})$ is piecewise constant, but it has discontinuities at $\phi ={\frac {2m\pi -d}{c}}$, where it is not differentiable, and it does not in any case satisfy $P(\phi )=P(\phi +2m\pi )$ for all values of φ. Gandalf61 (talk) 16:43, 18 January 2010 (UTC)

Yes I didn't read the k=0 part. That's a strange way to state the problem. And yes the general solution is P=$P=c\phi +d$. You can consider k=0 as meaning an infinite wavelength, perhaps they were trying to make it periodic? For that you'd have m infinite in $P(\phi )=P(\phi +2m\pi )$ which doesn't lead you anywhere fast. Dmcq (talk) 17:33, 18 January 2010 (UTC)

Sorry my apologies all, but I have just realised I have made a rather bad typo. I would actually like to consider the solution $P=c\phi +d-2\pi \times floor({\frac {c\phi +d}{2\pi }})$. This I believe does fit the boundary $P(\phi )=P(\phi +2m\pi )$. And fulfills the equation P′′=0. What is the behavious at the discontinuity, does it negate it as a possible solution? —Preceding unsigned comment added by 129.67.39.49 (talk) 19:08, 18 January 2010 (UTC)

That doesn't solve the differential equation. It has discontinuities. Why do you keep thinking it is a solution? It doesn't have a deriviative at the discontinuity so it doesn't have a second deriviative and the differential equation doesn't hold. The second deriviative is not equal to 0 at the discontinuities, if it was there wouldn't be a discontinuity. Dmcq (talk) 19:57, 18 January 2010 (UTC)

Maybe the OP is confusing having a periodic domain with having a periodic codomain. For instance if this was a function P:S¹ → S¹, then for c an integer P(φ) = cφ + d = 2π⌊(cφ + d)/2π⌋ would be continuous and a solution to P′′=0. But there's not any reason to think the codomain is anything other than R unless that's a detail the OP neglected to mention. Rckrone (talk) 03:46, 19 January 2010 (UTC) changed some apostrophes into primes to fix the formatting in mine and another post

I don't see that. It would still be discontinuous at φ = 0 on S<sup[>1 unless it is just the constant c. Dmcq (talk) 11:06, 19 January 2010 (UTC)

Oh you're right sorry. I was thinking of the fractional part instead of the floor. I think the OP made that mistake too though. What I meant then was P(φ) = cφ + d = cφ + d - 2π⌊(cφ + d)/2π⌋ in S¹.

2π⌊(cφ + d)/2π⌋ is continuous in S¹ also, it's just uniformly zero. Rckrone (talk) 15:56, 19 January 2010 (UTC)

Alright dmcq, since you seem to have fun only by lampooning my attempts, you do my fucking homework for me. Is there a solution to P''=0, P'≠0 when restrained to P(φ)=P(φ+2kπ)?

The above post is vulgar, in any context whatsoever, and should be removed (the edit was made by the OP although it is unsigned). You are strongly discouraged to post comments of this nature; if you feel that you have been unfairly treated by Dmcq, note it politely. I do not see Dmcq using vulgar language so neither should you. --PST 02:47, 20 January 2010 (UTC)

Comments generally should not be removed. Eric. 131.215.159.171 (talk) 06:08, 20 January 2010 (UTC)

The perils of trying to help someone with their homework. Dmcq (talk) 10:48, 20 January 2010 (UTC)

This issue need go no further, but I would not accuse Dmcq of being unfair, at best tactless and at worst arrogant. Whether these were justified is hard to tell, as tonal context is lost. However, I would not call my response a personal attack, as Dmcq did when threatening me with a ban, but I would call it rude; as rude as I percieved his, if more vulgar. Obviously the sarcasm was lost on him as well. —Preceding unsigned comment added by 129.67.39.49 (talk) 14:38, 20 January 2010 (UTC)

I do not see where you got the idea I was making fun of you. I believe I answered your query politely. I pointed out to you on your talk page that attacking people personally can lead to you being banned as per No personal attacks. If you wish to use the reference desk or contribute to wikipedia you need to try and follow the sites policies. If you wish to take it further then do so, the appropriate place is WP:WQA in the first instance or WP:ANI if you feel strongly but I believe you would be making a bad mistake. If you wanted to complain here you should have pointed out the place where you thought I was uncivil and you should not have used words like 'fucking'. Dmcq (talk) 15:19, 20 January 2010 (UTC)

Bien. —Preceding unsigned comment added by 129.67.39.49 (talk) 22:25, 20 January 2010 (UTC)

Please sign your post by typing four tildas (~~~~) after your comment, so that we can readily identify you. With regards to your post, I have nothing to say more than common sense should forbid you from posting such comments. It is also forbidden per WP:CIVIL, so if you continue to use such language, you would not receive support from other users at WP:AN/I, should you file a complaint. --PST 02:24, 21 January 2010 (UTC)

The second derivative being zero means it must be linear. The only linear functions which are periodic are constants, which you rule out, so no. I'm sure you could have worked that out yourself if you had tried. If not, you shouldn't be taking whatever course this is from - you need to go back and learn the prerequisites. --Tango (talk) 04:16, 20 January 2010 (UTC)

What does this symbol/suffix mean?

I have the equation for working out the uptake of CO2 in concrete, where the maximum depth is D.

D = K * the square root of t.

To work out K, I need to multiply k1 * k2 * k3. k1 is a particular figure with the suffix (year)-1/2 (where the "-1/2" is always in superscript. (k2 & k3 are simply other figures).

So an example equation is given as

K = 6 * 0.7 * 1.1 * (years)-1/2 = 4.6mm (years)-1/2

What does the (years)-1/2 mean? I know the (years) is the time the concrete exists for (or a derivative thereof) but the "-1/2" is a mystery - I assume it's the square root or 1/the square root but am unsure.

Thanks. 157.203.42.175 (talk) —Preceding undated comment added 10:34, 18 January 2010 (UTC).

Do you mean years^-1/2. In that case it means years to the power of -0.5. So it means divide my the square root of years. Taemyr (talk) 11:26, 18 January 2010 (UTC)

Yes, y^-1/2 = 1/(y^1/2) = $1/{\sqrt[{2}]{y^{1}}}$ = $1/{\sqrt {y}}$. In general, y^-n = 1/yⁿ and y^a/b = ${\sqrt[{b}]{y^{a}}}$. PrimeHunter (talk) 12:16, 18 January 2010 (UTC)

The would only make sense if the other constants have units that make the product come out in units of distance per square root time. For example cm/year^1/2. The D value must come out in the end to be in units of distance. See Dimensional analysis.--RDBury (talk) 14:43, 18 January 2010 (UTC)

Thanks all, that's really useful :) 157.203.42.175 (talk) 16:32, 18 January 2010 (UTC)

January 19

Policy recommendation for comparing citation totals or h-index for any mathematician

On Talk:Steve Shnider, a discussion on testing his notability has included calculating his h-index. Using the Web of Knowledge has been challenged and MathSciNet/Mathematics Reviews suggested. Is there a policy for how we compare citation totals or h-index for mathematicians and where the data should come from?

PS. If anyone happens to be familiar with Shnider's work, a comment on notability would be helpful on the talk page.—Ash (talk) 09:44, 19 January 2010 (UTC)

/dev/random is an excelent source of such data. Freely available, too. — Emil J. 11:43, 19 January 2010 (UTC)

Talk:Steve Shnider#RfC Using citation totals in articles on academics has been raised; perhaps your critical viewpoint (I assume your comment relates to the idea of using an h-index) would be helpful in reaching a consensus...—Ash (talk) 11:47, 19 January 2010 (UTC)

Variation on equals sign

I came across a variation on the well-known defining rule $Z=Z^{2}+c$, which rather than the simple equals sign had one with short extra bits - the upper line bent backwards at its RH end to head northwest, as it were, while the lower line bent backwards at its LH end to head southeast. I can't find an image to display, hence the description. Does this symbol have a standard meaning, say an assignment statement applied repeatedly?→86.132.164.11 (talk) 11:08, 19 January 2010 (UTC)

This ≅ ? It's Unicode, so ≅ and ≅ should display it too.--JohnBlackburne^words_deeds 11:28, 19 January 2010 (UTC)

(e/c) Do you mean $\rightleftharpoons$? I don't think it has a standard meaning in mathematics. I've seen somebody use it for definitions, and even for equivalence, but neither seems to be widespread usage. A Google search suggests that it means something in chemistry, but that's probably not what you want. $Z\rightleftharpoons Z^{2}+c$ looks like the transformation rule whose iteration defines the Mandelbrot set, so it may mean just that: a transformation rule (though I would rather write it either as $Z\mapsto Z^{2}+c$ or $Z_{n+1}=Z_{n}^{2}+c$ or something like that). You should check the context, it may define the symbol somewhere. — Emil J. 11:36, 19 January 2010 (UTC)

It was exactly this $Z\rightleftharpoons Z^{2}+c$, thanks. It appeared in a TV programme so the meaning was just up to the viewer. I assumed that it had a particular significance, but it seems that the producer just appropriated it to mean iteration of the transformation rule.→86.132.164.11 (talk) 12:24, 19 January 2010 (UTC)

Films and TV shows routinely use nonsensical formulas whenever they want to impress the audience with the characters' awesome mathiness. The people who created the formulas for the show like to use lots of symbols which look exotic, even when they have no meaning. So somebody probably said "hey- let's use that crazy hooked equal sign- it'll look cool". Staecker (talk) 13:54, 19 January 2010 (UTC)

And now that I read it more carefully it sounds like maybe this was a legitimate documentary-style math program? About fractals? In that case forget what I said. The hooks are probably implying iteration, like you said. But like the others said, this isn't a standard meaning. Staecker (talk) 13:56, 19 January 2010 (UTC)

Even if it was indeed an informational program, I suspect that they still favored formulae by virtue of their perceived mathiness. -- Meni Rosenfeld (talk) 15:48, 19 January 2010 (UTC)

(OP) The programme was called The Secret Life of Chaos on UK BBC and though fairly undemanding was pitched at a respectable level. The narrator was Jim Al-Khalili, Professor of Physics at the University of Surrey. As said above, the notation was obviously used to mean iteration, but nobody here seems to recognise it as a standard usage.→86.132.164.11 (talk) 16:24, 19 January 2010 (UTC)

Yes, it seems quite common to use this sign in TV programs to indicate iteration. I saw that program, but it wasn't the first time I saw that symbol used in a TV program. I agree with EmilJ, one standard notation to use in place would be $z_{n+1}=z_{n}^{2}+c$ --Xedi^Talk 10:51, 20 January 2010 (UTC)

P.G.F. question

I am stuck on part of a question about probability-generating functions. In part (i), you had to derive the P.G.F. of $X\sim {\mbox{Poisson}}\left(\lambda \right)$, and I got $G_{X}(t)=e^{\lambda \left(t-1\right)}$, which is right, I believe. Then you need to derive the distribution for $X+Y$, where $Y\sim {\mbox{Poisson}}\left(\mu \right)$, via $G_{X+Y}\left(t\right)$; I got $X+Y\sim {\mbox{Poisson}}\left(\lambda +\mu \right)$, so that $P\left(X+Y=n\right)={\frac {e^{-\left(\lambda +\mu \right)}\cdot \left(\lambda +\mu \right)^{n}}{n!}}$. Now, part (iv) with which I'm having problems:

Use the distributions of $X$, $Y$ and $X+Y$ to find the conditional probability that $X=x$ given that $X+Y=n$, where $n$ is a non-negative integer. Deduce that the conditional distribution of $X$ given that $X+Y=n$ is binomial with parameters $n$ and ${\frac {\lambda }{\lambda +\mu }}$.

I've tried working forwards, then working backwards, but my answers don't quite meet in the middle!

Working forwards

$P\left(X=x|X+Y=n\right)$

$={\frac {P\left[\left(X=x\right)\cap \left(X+Y=n\right)\right]}{P\left(X+Y=n\right)}}$

Next, the only interpretation I can find of $=P\left[\left(X=x\right)\cap \left(X+Y=n\right)\right]$ is to say that if $X=x$ and $X+Y=n$ then $Y=n-x$. Thus I get:

${\frac {P\left(Y=n-x\right)}{P\left(X+Y=n\right)}}$

$={\frac {\frac {e^{-\mu }\cdot \mu ^{n-x}}{\left(n-x\right)!}}{\frac {e^{-\left(\lambda +\mu \right)}\cdot \left(\lambda +\mu \right)^{n}}{n!}}}$

$={\frac {e^{-\mu }\cdot \mu ^{n-x}}{\left(n-x\right)!}}\cdot {\frac {n!}{e^{-\left(\lambda +\mu \right)}\cdot \left(\lambda +\mu \right)^{n}}}$ $={\frac {e^{-\mu }\mu ^{n-x}n!}{\left(n-x\right)!e^{-\left(\lambda +\mu \right)}\left(\lambda +\mu \right)^{n}}}$

${\color {blue}={\frac {e^{\lambda }\mu ^{n-x}n!}{\left(n-x\right)!\left(\lambda +\mu \right)^{n}}}}$, and that's as far as I'm able to simplify it. Next,

Working backwards

If $X\sim B\left(n,{\frac {\lambda }{\lambda +\mu }}\right)$ then $P\left(X=x\right)={_{n}C_{x}}\left({\frac {1}{1}}-{\frac {\lambda }{\lambda +\mu }}\right)^{n-x}\left({\frac {\lambda }{\lambda +\mu }}\right)^{x}$

$={\frac {n!}{x!\left(n-x\right)!}}\left({\frac {\lambda +\mu -\lambda }{\lambda +\mu }}\right)^{n-x}\left({\frac {\lambda }{\lambda +\mu }}\right)^{x}$

$={\frac {n!}{x!\left(n-x\right)!}}\left({\frac {\mu }{\lambda +\mu }}\right)^{n-x}\left({\frac {\lambda }{\lambda +\mu }}\right)^{x}$

$={\frac {n!}{x!\left(n-x\right)!}}\cdot {\frac {\mu ^{n-x}}{\left(\lambda +\mu \right)^{n-x}}}\cdot {\frac {\lambda ^{x}}{\left(\lambda +\mu \right)^{x}}}$

$={\frac {n!\mu ^{n-x}\lambda ^{x}}{x!\left(n-x\right)!\left(\lambda +\mu \right)^{n-x}\left(\lambda +\mu \right)^{x}}}$ ${\color {blue}={\frac {n!\mu ^{n-x}\lambda ^{x}}{x!\left(n-x\right)!\left(\lambda +\mu \right)^{n}}}}$, and here end my powers of simplification.

As you can see, some of the terms are the same in the two expressions I manage to arrive at, but I am out by a few factors. Please advise me on where I have gone wrong. Thanks in advance. It Is Me Here ^{t / c} 17:27, 19 January 2010 (UTC)

In your working forward step, you make a very crucial mistake. In the numerator of your conditional probability, note that $P((X=x)\cap (X+Y=n))=P(X=x)P(Y=n-x)$ by the independence of $X$ and $Y$. Everything should work out fine starting from there. Nm420 (talk) 18:51, 19 January 2010 (UTC)

Thanks very much! It Is Me Here ^{t / c} 21:39, 19 January 2010 (UTC)

See also Cumulant#Cumulants of some discrete probability distributions. Bo Jacoby (talk) 22:53, 19 January 2010 (UTC).

3 rotations about the vertices of a triangle: identity map

Hi all,

I've just started a geometry course this term and I'm trying out a few questions but I'm struggling with the following:

Let R(P,$\theta$) denote the clockwise rotation of $\mathbb {R} ^{2}$ through an angle $\theta$ about a point P. If A, B, C are the vertices, labeled clockwise, of a triangle in $\mathbb {R} ^{2}$, prove that R(A,$\theta$)R(B,$\phi$)R(C,$\psi$) is the identity if and only if $\theta =2\alpha$, $\phi =2\beta$ and $\psi =2\gamma$, where $\alpha ,\,\beta$ and $\gamma$ are the angles at the vertices A, B and C of the triangle.

Now I think I can show the 'if', just by showing that the rotations reflect A in BC then back again, and similarly for B and C, so the rotation fixes 3 points so must be the identity (I hope!) - however, I'm completely stuck on the 'iff'. I can't see any nice way to approach this at all, and I'd rather not churn through mounds of vectors if anyone can suggest an elegant proof I might utilise? I can't seem to get my head around it!

Many thanks, 82.6.96.22 (talk) 18:02, 19 January 2010 (UTC)

The circle round B radius AB and around C radius CA intersect at only two points. Which incidentally shows there's another set of angles satisfying the conditions - set them all to 0. Dmcq (talk) 18:16, 19 January 2010 (UTC)

January 20

Eigenvalues and eigenvectors

I am working on some project which in part involves derivatives of eigenvalues and eigenvectors and associated problems. The following requires additional insight from someone else (or from some classic textbook, I don't know). Consider a matrix $A(t)$, linear with respect to complex argument $t\in \mathbb {C}$, such that $A(0)$ has an eigenvalue of multiplicity two. I could identify three completely different scenarios illustrated by examples below

$A=\left({\begin{array}{cc}1+t&0\\0&1-t\end{array}}\right)$

where $A(0)$ is in fact diagonalizable.

$A=\left({\begin{array}{cc}1+t&1\\0&1-t\end{array}}\right)$

where $A(0)$ is not diagonalizable, and even though eigenvalues $1\pm t$ are analytic functions, eigenvectors have a singularity (a pole) at $t=0$

$A=\left({\begin{array}{cc}1+t&1-t\\t&1-t\end{array}}\right)$

where $A(0)$ is not diagonalizable, and eigenvalues $1\pm {\sqrt {t}}$ as well as eigenvectors have a branch point at $t=0$

There may be other scenarios as well (which I do not see at the moment). In a more general setting, $A(t)=X+tY$ (which later will even become multidimensional), how can I determine which scenario I should expect just by looking at matrices X and Y (their SVD or other decompositions are all at my disposal). (Igny (talk) 04:41, 20 January 2010 (UTC))

You can determine the svd by taking the root of the function and integrating wrt a constant wher the value of F(x) does not change and remains constant while the determinant of the diagonalizable form may come in handy--123.237.193.11 (talk) 14:54, 21 January 2010 (UTC)

---

Riemann integral

Resolved

When, if ever, is $\int _{a}^{b}|f(x)|dx=0\Rightarrow f(x)=0\forall x\in [a,b]$ true? Thanks-Shahab (talk) 08:09, 20 January 2010 (UTC)

For starters, this obviously holds if f is continuous. -- Meni Rosenfeld (talk) 08:27, 20 January 2010 (UTC)

Why? Can you explain.-Shahab (talk) 09:21, 20 January 2010 (UTC)

Think of it in terms of area-under-the-curve. That area can only be zero if either the function is identically zero or if the area above the x-axis equals the area below it (or the function is discontinuous). |f(x)| is, by definition, positive, so the area below the x-axis must be zero. That leaves you with only f(x)=0 as a possibility. --Tango (talk) 10:14, 20 January 2010 (UTC)

[ec] If $f(x_{0})\neq 0$, then $|f(x_{0})|=\epsilon >0$. Since f is continuous, so is $|f|$, and there is a neighborhood around $x_{0}$ where $|f(x)|>\epsilon /2$. The integral of $|f|$ is positive over that region and non-negative elsewhere. -- Meni Rosenfeld (talk) 10:20, 20 January 2010 (UTC)

I think that's if and only if. If f isn't continuous, then it can't be identically zero. (Obviously, it holds for discontinuous functions where the LHS isn't true, but that isn't interesting.) --Tango (talk) 08:44, 20 January 2010 (UTC)

The real result is of course that, if $\int _{a}^{b}|f(x)|dx=0$, then f is almost everywhere zero. Algebraist 13:19, 20 January 2010 (UTC)

Thanks all-Shahab (talk) 17:11, 20 January 2010 (UTC)

A few more geometry problems!

Hi all,

Sorry to pop up again, I asked a question yesterday (thankyou very much for the help with that, incidentally) and I'm really having some trouble getting my head around some of this geometry stuff, my lecturer is a brilliant mathematician but not so competent when it comes to explanation and I don't want to hand in no work whatsoever because I've not been able to do any of it, but the only book which is recommended on my course (Curved Spaces by P.M.H Wilson) hasn't been a great deal of help - in fact, most of these problems come from the exercises at the end of the chapters, which have no hints or solutions.

I don't expect you to do all these for me, but if I could just get some general guidance on the following points that'd be an amazing help!

1) I'm looking at finite subgroups G of the isometries of $\mathbb {R} ^{n}$, and I've been told to show G fixes some point of $\mathbb {R} ^{n}$ by considering the barycentre (average) of the orbit of the origin under G. Now if we let G act on some point x, and note that gG=G (fairly simple to prove), then for $c={\frac {\sum _{g\in G}gx}{|G|}}$, we have gc=c$\forall g\in G$, right? Since every element in G can be written as $g^{-1}j$, for some $j\in G$, so we end up with exactly the same sum and c is a fixed point. My question is, unless my argument is wrong, why do we need to consider specifically the origin to find this fixed point? Why does it not work with an arbitrary point x? I also want to classify all finite subgroups of isometries in $\mathbb {R} ^{2}$ into cyclic and dihedral groups. Could anyone give me any suggestions to get started?

2) Here I'm looking at isometries of the unit sphere $S^{2}$ in $\mathbb {R} ^{3}$, and I want to show that any matrix A$\in O(3,\mathbb {R} )$ can be written as the product of at most 3 reflections in planes through the origin, so I can then deduce that an isometry of $S^{2}$ can be written as the product of at most 3 reflections in spherical lines. Then, I need to work out which isometries are obtained as the product of 2, and which are obtained as exactly 3 reflections. Should I look at which isometries are orientation preserving or does this not come in to play with spherical reflections? How else would I go about it, in that case?

3) I want to show that for every spherical line l and point P on $S^{2}$ there is a spherical line l' such that P$\in$ l' and l and l' are perpendicular at their intersection(s). I always find that with problems like this I have to rotate the sphere so that l is in the x-y plane and P has y-coordinate 0 and use some 'WLOG' argument before I can actually make any headway, and then I don't really seem to be proving anything at all. Can anyone get me going on this with any sort of more elegant proof? I want to show also that the distance from P to any point Q on the line l is minimized when Q is one of the 2 intersections of l and l', and whilst this seems intuitively obvious (as with much of geometry) I'm getting stuck actually proving it rigorously. I also want to prove that l' is unique if this minimum distance is < ${\frac {\pi }{2}}$, but I'll probably be able to give that a go myself once I've made a dent in the earlier parts of the question.

I appreciate that I'm asking a lot and of course any answers you could give me will probably be quite time consuming so don't feel like you have to help me with everything in one go (or at all!) - I did spend most of yesterday trying and failing to make much progress with these problems and knowing geometry isn't my forté will probably just make me want to learn it more, so please please do lend a hand!

Many thanks in advance, 82.6.96.22 (talk) 15:16, 20 January 2010 (UTC)

Ad 1: your argument is correct, and it does work with an arbitrary point x, there's nothing special about the origin. — Emil J. 15:23, 20 January 2010 (UTC)

Thanks appreciated for previous problem. I haven't time at the moment for this, in fact I shouldn't be even looking here at the moment, but can I suggest that saying your lecturer sucks might not be the best approach with some people here ;-) Dmcq (talk) 17:09, 20 January 2010 (UTC)

Noted ;) (same person, temporarily different I.P. address!) 131.111.185.75 (talk) 18:50, 20 January 2010 (UTC)

For (2), if you choose an orthonormal basis, the isometries are characterized by where they send those basis elements. The reflection that sends a to b is the one through the axis a-b. You can send the basis elements to their new spots with reflections one at a time and show that the later reflections don't mess with the basis elements that were already set. You're right that reflections switch the orientation, so isometries that preserve orientation will be made of an even number of reflections and the orientation reversing ones will be odd. Then it's just a matter of deciding when an isometry requires 1 or 3 reflections. There might be something more clever, but that's how I'd do it. Rckrone (talk) 20:08, 20 January 2010 (UTC)

For (3), a spherical line is characterized by the axis it goes around. In this way, try rephrasing the problem in terms of finding perpendicular vectors. Rckrone (talk) 20:20, 20 January 2010 (UTC)

For the second part of (3), the distance between two points a, b on the sphere is the same as their angle, so a.b = cos(dist(a,b)), so minimizing the distance is maximizing the dot product. The basis formed by the axis of l, the axis of l' and the direction orthogonal to them is convenient to work in. Rckrone (talk) 21:03, 20 January 2010 (UTC)

Geometry question

Is there a way of finding the radius of a circle whose area is equal to the area of a rectangle of given dimensions? (Kind of like the geometric mean describes the size of a square whose area is equal to the area of a rectangle of given dimensions.) Edgeweyes (talk) 18:50, 20 January 2010 (UTC)

An a by b rectangle has area ab, corresponding to a circle of radius ${\sqrt {\frac {ab}{\pi }}}$, so it's just a (constant multiple of a) geometric mean. Algebraist 18:55, 20 January 2010 (UTC)

I see now. I should have thought about it a little more... πr² = ab is not tough to solve. Thanks! Edgeweyes (talk) 19:28, 20 January 2010 (UTC)

January 21

Riemannian geometry

How do you calculate a straight line in Riemannian geometry using a given coordinate system? For example, if I was using the surface of a sphere in spherical coordinates, what series of steps would I have to go through to find the equation of a great circle?

Will it tend to be feasible to solve symbolically, or would you have to do it numerically?

Alternately, what could I read to teach me to do that?

I know up to linear algebra and differential equations. — DanielLC 06:11, 21 January 2010 (UTC)

The closest analogue of a straight line on a curved manifold is a geodesic. If a string is made to stay on the surface then the geodesic is the one where you stretch the string. On a sphere like the world it is a great circle, planes fly along them. Dmcq (talk) 11:32, 21 January 2010 (UTC)

You need the geodesic curvature to be zero along the whole of the curve. Fly by Night (talk) 13:25, 21 January 2010 (UTC)

Fractional chromatic number

What is the fractional chromatic number of the United States? --84.61.165.65 (talk) 17:21, 21 January 2010 (UTC)

Sounds like the Six degrees of separation, but it might also be generated by a new AI. It does help to just invest that extra minute or two making things clearer if you're going ask people to spend time on something. Dmcq (talk) 17:33, 21 January 2010 (UTC)

What is the chromatic number of the United States? --84.61.165.65 (talk) 17:42, 21 January 2010 (UTC)

The US isn't a graph. What do you mean? You mean the states as vertices with edges between those that border each other? Or the resident people as vertices with edges between those that know each other? Or what? --Tango (talk) 17:44, 21 January 2010 (UTC)

I mean the states as vertices with edges between those that border each other. --84.61.165.65 (talk) 17:58, 21 January 2010 (UTC)