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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 only complaint I have regards your inability to get over it. Seriously it was just a throw away comment made to jibe at Dmcqs outstanding unhelpfulness. I really couldn't care for any of this overblown fallout.

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 it 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)

You guys have been awesome, thanks so much! I don't know how I ever managed without you :-) 82.6.96.22 (talk) 08:33, 22 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)

How do you find that using only the metric tensor? From what I managed to find, there isn't generally a closed-form solution. Is it simple to find a taylor series solution? — DanielLC 19:59, 22 January 2010 (UTC)

Even with simple figures you find integrals of square roots coming in, you might get a trig function but you're quite liable to get something much worse. what are you expecting really? Try expressing even the great circles of a sphere easily and that's as simple as they come. Dmcq (talk) 20:26, 22 January 2010 (UTC)

I'm just asking if you could find the taylor series. It seems like it would be simple enough. I doubt there's any difficulty in just finding the geodesic curvature. I can't imagine finding the change in geodesic curvature would be much harder. Keep doing that and you get the taylor series. I've only been able to find out how to get the geodesic curvature using an actual surface. — DanielLC 02:05, 23 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)

Plus I guess he means Fractional coloring rather than the straight chromatic number but I've not come across fractional coloring before. Dmcq (talk) 18:02, 21 January 2010 (UTC)

Fractional chromatic number redirects to Fractional coloring, which does define the term. --Tango (talk) 18:09, 21 January 2010 (UTC)

4 is an upper bound on both numbers: File:Map of USA with state names.svg. That's hardly surprising, given that the graph is planar. — Emil J. 18:26, 21 January 2010 (UTC)

Why is the chromatic number of the United States not smaller than 4? --84.61.165.65 (talk) 18:43, 21 January 2010 (UTC)

West Virginia and the adjoining states are already not three-colourable. Algebraist 18:49, 21 January 2010 (UTC)

Well the fractional version is ≤7/2. I don't really want to try 3-fold coloring by hand though. :) Edit: Since the graph H with 1 point surrounded by 5 has χ_f(H) = 7/2, and the US map has H as a subgraph, the fractional chromatic number of the U.S. map is exactly 7/2. Rckrone (talk) 20:01, 21 January 2010 (UTC)

I want a proof of the fact that the fractional chromatic number of the U.S. map is not greater than 7/2. --84.61.165.65 (talk) 20:27, 21 January 2010 (UTC)

Try 2-fold coloring it. Rckrone (talk) 20:37, 21 January 2010 (UTC)

Expected value problem

I've been looking at some self-generated 2-dimensional probability problems, and have got stuck. I'm considering a kind of random walk, specifically the case of a particle initially being at a particular distance $d$ (≥1) from the origin, then moving by unit distance in a uniformly-distributed direction. If it is then distance $r$ from the origin, what is E[$r$]? For $d=1$ the answer comes fairly readily to be 4/π via the pdf of $r$, but for general $d$ I got totally lost, and for the particular case of $d=2$ the integration was beyond me. Numerical simulation showed that E[$r$] always exceeds $d$, with the ratio decreasing towards 1 as $d$ increases, but didn't help in getting a general result - is there one?→86.155.184.123 (talk) 19:11, 21 January 2010 (UTC)

According to Wolfram Alpha, the answer is $2{\frac {d+1}{\pi }}E\left({\frac {4d}{(d+1)^{2}}}\right)$, where E is the complete elliptic integral of the second kind, whatever that may be. Algebraist 19:33, 21 January 2010 (UTC)

Wolfram uses a different notation from our article; the result is $2{\frac {d+1}{\pi }}E\left({\frac {2{\sqrt {d}}}{d+1}}\right)$ in our notation.

It may also be interesting to know that the answer is $d+{\frac {1}{4d}}+{\frac {1}{64d^{3}}}+O\left({\frac {1}{d^{5}}}\right)$ for large d and $1+{\frac {d^{2}}{4}}+{\frac {d^{4}}{64}}+O(d^{6})$ for small d. -- Meni Rosenfeld (talk) 19:54, 21 January 2010 (UTC)

can you explain turbo codes OR LDPC?

I don't really care which one, which proves that this isn't homework (and it's not) but could someone explain EITHER turbo codes or LDPC in a way I can actually understand? I'm having so much trouble following either articles... thanks! 84.153.235.239 (talk) 19:31, 21 January 2010 (UTC)

p.s. I don't mean the effect, the net outcome, but the math: how it's actually done. Thanks again.

Error detection and correction is a huge subject and it's hard to know where you want to start. If you haven't heard of Shannon's theorem then I'd suggest getting a book on the subject.--RDBury (talk) 00:12, 22 January 2010 (UTC)

Try this: you've got a set of M possible messages to be encoded into an n-dimensional codeword space. If you map each message to a single codepoint in n-dimensional codeword space, and transmit that message into your channel, you can perform error detection by finding the nearest message codepoint to the received codeword point. If the encoded message points are far enough from one another, you can decode to the correct original message even if the received codeword is quite far away from the transmitted codeword.

The problem then becomes a very simple one: how do you pack M points into n-dimensional space while maximizing the distances between those points, yet still retaining a simple, fast algorithm for finding the nearest codepoint given a codeword? All the different error-correcting codes are ways of addressing this single problem for a variety of different constraints, and metrics for what is considered to be "near".

Shannon's noisy-channel coding theorem establishes a best-case bound on what is possible, by considering a very general scenario; LDPC and Turbo codes are ingenious attempts at solving the problem above that can get very close to the Shannon limit, providing n is sufficiently large for any given M.

Understanding Shannon's coding theorem is the key to understanding. There is a really good treatment of the underlying reasoning here, which is not unduly mathematical, and, if I recall correctly, rather similar to the treatment used in Shannon's original paper. -- The Anome (talk) 08:24, 22 January 2010 (UTC)

Sorry, the above is ALMOST easy enough for me to understand, but you could you simplify exactly what "n-dimensional" codespace is? How am I supposed to visualize that? Thanks!! 82.113.121.203 (talk) 03:23, 23 January 2010 (UTC)

n-dimensional codespace is the set of n-tuples. For n=1, 2, or 3 it is visualized as line, plane or space. For n>3 it is not actually visualized, but the concepts of point and distance are used. Bo Jacoby (talk) 08:15, 23 January 2010 (UTC).

January 22

Weaker versions of Fermat's Last Theorem

Andrew Wiles proved that Fermat's Last Theorem is true, i.e. that for the equation $a^{n}+b^{n}=c^{n}$, there are no solutions where a, b, c and n are all natural numbers and n is greater than 2. What if we weaken the preconditions? If we allow at least one of a, b, and c to be any positive real, then it is trivially true that $a^{n}+b^{n}=c^{n}$ has solutions, because it reduces to checking whether $(a^{n}+b^{n})^{1/n}$ is a positive real, and such a positive real always exists. But what if we say that a, b, and c have to be natural numbers, but n can be any positive real? What can we say about what solutions exist for what values of n? JIP | Talk 07:04, 22 January 2010 (UTC)

For all natural numbers (a, b, c) such that c > a and c > b and c < a + b there exists a real number n such that $a^{n}+b^{n}=c^{n}$. I've discovered a lovely little proof of this, but this edit window is too small to contain it. Dragons flight (talk) 08:04, 22 January 2010 (UTC)

Let me try: ⊿. --Stephan Schulz (talk) 08:25, 22 January 2010 (UTC)

:-) Nice actually you don't need c<a+b (for real numbers 0<a≤b<c or 0<c<a≤b there exists a unique real n; if 0<a≤c≤b no such real n exists). --pma 09:49, 22 January 2010 (UTC)

6³+7³>8³ but 6⁴+7⁴<8⁴, so there is a solution with a=6, b=7, c=8, and n somewhere between 3 and 4.--RDBury (talk) 13:35, 22 January 2010 (UTC)

I don't have a proof but just thinking it through I'm sure there's an infinity of a solution for n: take e.g. the above and double all the numbers. The solution will be the same but you can also modify a, b and c to get a slightly different equation with a different solution. Rinse and repeat with ever bigger numbers. There should be $\aleph _{0}$ equations with solutions and so $\aleph _{0}$ values for n. But there are $\aleph _{1}$ real numbers in any interval, so an n picked at random will not be a solution. I don't know if you can say anything about the values for n other than this. --JohnBlackburne^words_deeds 13:49, 22 January 2010 (UTC)

Strictly speaking there are at least $\aleph _{1}$ real numbers in any interval. AndrewWTaylor (talk) 14:15, 22 January 2010 (UTC)

The homogeneity of the equation suggests an alternative proof to the one which (I assume) Dragons flight had in mind. Show that for any real u, v such that 0 < u < 1 and 0 < v < 1 there is a real w > 0 such that $u^{w}+v^{w}=1$. Then take u = a/c and v = b/c. Gandalf61 (talk) 14:31, 22 January 2010 (UTC)

If w = 0 then $u^{w}+v^{w}=2$. The limit as w→+∞ is 0. So by continuity there is a value of w between 0 and +∞ where the expression =1. There are only countably many choices for u and v so the number of possible values for w is countable. In particular Fermat's last theorem says w can't be an integer greater than 3. It probably wouldn't be hard to show the possible ws are dense in some interval though. So even though you can't have w=3 you could get w arbitrarily close to 3.--RDBury (talk) 18:04, 22 January 2010 (UTC)

A variation on these lines: for a given positive integer n it's not hard to find countably many solutions in positive integers x,y,z of the Diophantine equation x^1/n+y^1/n=z^1/n. But can we characterize all solutions? In particular, is it necessary that x,y,z are perfect n-th powers? --84.220.119.131 (talk) 22:58, 22 January 2010 (UTC)

Bounds

Let us assume that a(t) and b(t) are, over ]0,∞[, both real valued and infinitely differentiable functions of the variable t. Furthermore, let us assume that the function c(t) is defined by c(t) = a(t) / b(t). Now, let us assume that the limit of c(t) as t tends towards positive infinity is k, where k is a positive real number. What can we say about a(t) and b(t)? What can we say about anything? Fly by Night (talk) 18:42, 22 January 2010 (UTC)

What sort of things do you want to say? We can say, for example, that b is eventually nonzero (that's the only constraint on b alone, though) and that a and b are eventually of the same sign. Algebraist 18:48, 22 January 2010 (UTC)

I wanted something more substantial than that. If we have two non-zero integers, say a and b, then we can form the number a/b. The integers cross the integers form a quotient space, namely the rational numbers: we say that (a,b) is in the same class as (c,d) if and only if ad = bc. Is their some kind of classification of pair of functions given a limit of their quotient? Fly by Night (talk) 19:05, 22 January 2010 (UTC)

There might be two equivalence relations. Define A(b, k) the set of functions which have limit k when divided by b, define B(a, k) the set of functions which have limit k when multiplied by a. My guess is that for every k, for every triplett of functions b, x, y it is true that: x in A(b, k) and y in A(b, k) => B(x, k) = B(y, k). Then B(A(b, k), k) could be an equivalence class for b. Maybe k is also irrelevant, so we get the class "all functions with similar limit behaviour as b". I don't know if that is actually true, but it sounds interesting and superficially plausible.

Simplify by substituting t=1/x because an infinite t is more confusing than a zero x. Pick an arbitrary positive continuous function C(x) with C(0)=k. Pick an arbitrary positive continuous function B(x). Define A(x)=B(x)C(x). All you can say is that lim_x→0A(x)/B(x)=k. Bo Jacoby (talk) 22:25, 22 January 2010 (UTC).

January 23

Finding a surface on which integrals of multiple functions are 0

Given a set of N linearly independent functions $f_{n}(x,y)$ on some bounded simply-conected 2D surface $S$, under what conditions does there exist another surface $s\subset S$ on which the integrals of all $f_{n}(x,y)$ are 0? $\forall n:\int _{s}f_{n}(x,y)dA=0$

$s$ does not need to be connected.

I wrote a simple program to find $s$ given $f_{n}(x,y)$, and the results indicate that it is sufficient that all $f_{n}(x,y)$ are continuous functions that change sign somewhere on $S$. I know this is not a necessary condition, but I am most interested in knowing if this condition is indeed sufficient. 83.134.167.153 (talk) 08:32, 23 January 2010 (UTC)