Forum

We have been challenged to a match (3 days/move) by a newly formed group called Chess Bay. The group has a very high average rating (over 2100) so it would be helpful to know how many of the stronger players in this group would be interested before accepting the challenge. The proposed match is for anyone over 1700. Any interest?

We haven't started any new matches in a while, so I accepted another challenge from our old friends, The Power of Chess. Here's hoping a few good players are in the mood for a game! http://www.chess.com/groups/team_match.html?id=44054

Of course one can easily fill a 12x12 grid of squares with "dominoes", e.g. pieces that measure 2x1 squares. But can you still do it if you remove two 1x1 corner squares, say the bottom left and bottom right corner? And what happens if you remove the bottom left and top right corner squares? Can you still fill it with dominoes? Same Q can be posed for triominoes, e.g. 3x1 pieces, on a 12x12 board. When can you fill the board with triominoes, if three 1x1 squares have randomly been removed? What about 4x1 pieces? Or nx1 pieces on an kn*kn board?

arccos x + arcsin x = ? Find the answer . First that give the solution and in end the right answer , i will give a nice chess.com trophy .

Russian mathematician rejects $1 million prize AP – This undated file photo released by the International Mathematician Congress shows Grigory Perelman. … By MALCOLM RITTER, AP Science Writer Malcolm Ritter, Ap Science Writer – Thu Jul 1, 12:13 pm ET NEW YORK – He said nyet to $1 million. Grigory Perelman, a reclusive Russian mathematics genius who made headlines earlier this year for not immediately embracing a lucrative math prize, has decided to decline the cash. Perelman's decision was announced Thursday by the Clay Mathematics Institute in Cambridge, Mass., which had awarded Perelman its Millennium Prize. The award honors his solving of the Poincare (pwan-kah-RAY) conjecture, which deals with shapes that exist in four or more dimensions. Jim Carlson, institute president, said Perelman's decision was not a complete surprise, since he had declined some previous math prizes. Carlson said Perelman had told him by telephone last week of his decision and gave no reason. But the Interfax news agency quoted Perelman as saying he believed the prize was unfair. Perelman told Interfax he considered his contribution to solving the Poincare conjecture no greater than that of Columbia University mathematician Richard Hamilton. "To put it short, the main reason is my disagreement with the organized mathematical community," Perelman, 43, told Interfax. "I don't like their decisions, I consider them unjust." Attempts by The Associated Press to reach Perelman, a resident of St. Petersburg, were unsuccessful. Carlson said institute officials will meet this fall to decide what to do with the prize money. "We have some ideas in mind," he said. "We want to consider that carefully and make the best use possible of the money for the benefit of mathematics."

There is a theorem that if a natural number n is a sum of two squares of rational numbers, then n is also a sum of two squares of integers. A proof could involve Fermat's theorem on sums of two squares, but the proof of that theorem in itself is quite lengthy, so that the whole proof will be quite lengthy if you don't just take Fermat's theorem for granted.Does anyone know if a short(er) proof exists that if n is the sum of two rational squares, it's also the sum of two integral squares? Thanks.

You are stuck in a dream of golf, and the only way to get out is by getting a hole in one. Of course, you're not a golf pro, so you don't know how to hit the ball over walls or spin it. When you hit it, there is little friction, so it keeps going until you move to it to pick it up. You start at point A and the hole is at B. Due to low friction, any amount of power is sufficient so long as B is in its path. What angle should you hit it at?

Hi, I am preparing for the Gauss math competition at my school, and I am have problems with one problem. Here it is: The sum of all the digits of the integers from 98to 101 is 9+8+9+9+1+0+0+1+0+1=38 What is the sum of all the digits of the integers from 1 to 2008? Does anyone know of a quick and easy way to caculate this? Thanks for all of your help.

Week 352 Lenny Conundrum: There is a circle O (yellow), radius 1000. There are 10 congruent medium sized circles (brown) on the exterior of O, all tangent to O. They are each tangent to two others as well. There are 10 more small circles (blue), all congruent, tangent to O, as well as 2 medium sized circles. What is the total area covered by all of the small circles? In other words, the yellow circle is of radius 1000, what is the area shaded in blue?

Another question that has been bugging me for a while: Consider a vector space of all piecewise continuous functions (more specifically, those that satisfy the Dirichlet conditions) defined on an interval of the reals, for example f:(-pi, pi) => R. This space has (I think!) a number of dimensions equal to the cardinality of the continuum, alef[1]. However, consider the Fourier series approximation: g(x) = a[0] + sum(n = 1, 2, ...)(a[n]*cos(nx) + b[n]*sin(nx)) There are a countable number of parameters for g, namely: a[0], a[1], ..., and b[1], b[2], ...; in other words, the space of all such g's has aleph[0] dimensions. How is it possible that for every such f, there is a corresponding choice of the a's and b's? Treated a slightly different way: There are alef[2] such f's, and there are only alef[1]^alef[0] = alef[1]. I asked one of my prof's, and he pointed out that although the set of rationals is smaller than the set of reals, any real number can be approximated arbitrarily close by the rationals. Using that analogy, he argued that any f(x) can be approximated arbitrarily close by a g(x). However, there is a ton of hand-waving here, and I'm not sure whether I'm convinced. Another resolution to this issue would be if there are only countably many dimensions for the vector space of all f's satisfying the Dirichlet conditions. However, I think that, at least in the case where the domain for f is the whole number line, there are uncountably many dimensions; please correct me if that is my error.

Do any of you ever visit other math-related forums or discussion boards? I'd be interested to hear what they are, since I don't really know any, and this group's forum is not really that big and active.

I know there must exist a bijection between X = (0,1) and Y = X U {1}, because they have the same cardinality, but I just can't find one. Can anyone show me an example?

Hello i am hosting a couple of tournaments. please sign up. thanks. http://www.chess.com/tournament/random-chosen-position-i-a00-shy-attack http://www.chess.com/tournament/raadical-ruy if you have any trouble signing up for either tournament message me.

I don't quite remember the integrals needed to find the surface area of a curve, say z=sqrt(1-x^2-y^2), the top half of the unit sphere. I do remember the triple integrals needed to find the volume, but what was needed to find the surface areas? Also, I don't remember the surface area for the equation p=1, in spherical coordinates, so knowing that would also help, thanks! Likewise for cylindrical coordinates.

Problem: Given three numbers, x, y, z we know that they sum up to a certain number, say 12. Is there a way to prove that to minimize x^2+y^2+z^2, you have to set x=y=z=4? It's simple doing it with 2 numbers, that's just substitution, but with 3 or more variables it seems trickier. Any ideas? As a generalization, given n variables, a(1), a(2), a(3), ..., a(n), that sum to S, show that to minimize [a(1)]^2+[a(2)]^2+[a(3)]^2+...+[a(n)]^2, all a(1), a(2), a(3)..., a(n) must equal S/n. Thanxx!

I would like to invite you to a new, no holds barred tournament. You can bring out whatever you can find to help you in this tournament, be it chess engines, grandmaster friends, entire chess schools... anything! Fritz Friendly! - II

Hi everyone, I am hosting a rapid chess tournament. If you want to join here is the link below. http://www.chess.com/tournament/super-rapid-chess-openno-vacation

evaluate (all the limits are approaching 0 couldnt get the zeros to show for some reason) I know this is proly really easy but can some one help me out I am tutoring this girl in calc and have forgotten how to do this!! and she is paying me $25 an hour. I am meeting her in like 8 hours if some one can tell me the trick I would appreciate it :D

How can I find some info about the L-Series of a given elliptical equation? A simple example would be helpful...thanks in advance!

Hello everybody, I am currently trying to use MatCont 2.5.1 to perform a bifrucation analysis of a 5-Dim system. I got a bit of the way using matcont 2.4, but ran into some problems, thus I downloaded 2.5.1 instead hoping that this would remedy said problems. However, with the newer version I can't even solve my system numerically!! Have any of you smart people worked with MatCont and if so, do you have any advice? The error I am getting is something like: Error using vertcat ...