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I know that the sum of the reciprocals of factorials to infinity and beyond is e by using a Taylor Series for e^x, (1+1/1+1/2+1/6+1/24+...=e), but is there a specific formula for just finding them up to a given n? 1+1/1+1/2+1/6+1/24+...+1/100! Just curious.

RainbowRising wrote: I have to write a 2000 word essay on a 'high physics context' subject. I chose QM tunneling in stars because I could include a lot of physics from other areas in order to estimate the temperature of the core of a star, and then show that QM tunneling is necessary. So, anything that we can discuss on that front that could spurn some ideas would be great!

I was just wondering if someone amongst us are working on linear algebra/matrix theory and graph theory....I have four (make that three as one already dropped the course) special problem students working on the eigenvalues of special classes of graphs (as obtained using certain operations on labeled graphs) as defined in the book by Kelp and Knauer... The main question I have is this...what happens if after generating a finite number of examples/graphs involving vertices and edges of Paths and/or Cyclesand still no pattern can be found ---would this be a case that bounds the unsolvable? Specially that most of the software being used by my students can only compute up to 25x25 vertices limiting their samples to a around 8-10 vertices (per class of Path or Cycle)... The main reason is that they have barely three months to find results and failure to do so means they will have to extend for at least a semester (or at most two semesters) to finish the said course (MATH 190-Special Problem).. Advance thanks for those who will respond...

Am posting this topic for the assistance/help fellow Gaussians can give so as we can ahve a reference in the development of curriculum, research topics we're working on, or materials to browse in relation to the lessons we teach or study...

Want more Team matches, or more Vote games? Maybe a private tourney just for this group? Let me know by making suggestions here If enough people show support for something I will make it happen! Thanks and feel free to suggest anything you like

I might make a series of these just for fun. Future winners will very likely receive fun trophies. Whoever answers most of these quotes in 7 days or less will be the winner of the week. This week will not be too hard. First come first serve! Part 1: Guess which famous mathematician is being mentioned in the context of these quotes! Part 2: What is the context? "[T]hey {answer 1, answer 2} believe they have discovered a second Newton, a Hindu clerk in Madras on 20 pounds a year." -{answer 3} "When {answer 4} says he has proved something, I think it is very likely; when {answer 5} says it, it is a fifty-fifty bet; when {answer 6} says it, it is certain." -Carl Jacobi "Sir, (a+b^n)/n =x, hence God exists; reply." -{answer 7}" "If I could prove by logic that you would die in five minutes, I should be sorry if you {answer 8} were going to die, but my sorrow would be very much mitigated by pleasure in the proof." -{answer 8} "Vicimus GEGAN" -{answer 9} "Allez en avant, et la foi vous viendra." -{answer 10}

Does anyone here have an Erdos number? If so, what is yours?

Sup dudes, I understand how something could be injective (one-to-one), and I understand how something could be surjective (onto). But how can something be both? In other words, how can something be bijective? I'm having trouble grasping this simple idea because for some reason, I see injective and bijective as being almost the same thing. I know I'm missing something! Does it mean that a function has both injective and surjective parts? I'm currently piecing together the Schroeder-Bernstein Theorem in my head. So if you also know a fun resource for that, do tell! Thanks Gaussians!

Does anyone know a good website that generates latex math expressions so I could put [x^2+6]/[5+y] and get an image? Please don't say MS Paint.

OK, this is really shameful, but I cannot for the life of me figure out something that ought to be rather simple. Take a circle, X^2+Y^2=r^2, centered at the origin, and a point, (p,q), outside the circle. There are two tangent lines to that circle through that point. Find them. I know that the Pythagorean theorem says that (p^2+q^2)=r^2+m^2, where m is the distance of (p,q) to either tangent point. And certainly given any point on the perimeter of the circle, it is easy to determine if the tangent line through that point passes through (p,q). But going the other way ... what am I missing? Thank you!

Devlin's Angle here is something for the holiday not sure its original source December 2000 The Mathematics of Christmas by Keith Devlin I guess it was an early sign that I was heading for a career in mathematics that, when I was a young child, the run-up to Christmas always presented me with a numerical puzzle. How could Santa Claus possibly visit all children at midnight on the same night? I never did get a satisfactory answer from my parents, whose stock response was "No one knows; he just does." These days, the adult me can address the question in a mathematically more sophisticated way. Just how big is the task facing Santa on Christmas Eve? Let's assume that Santa only visits those who are children in the eyes of the law, that is, those under the age of 18. There are roughly 2 billion such individuals in the world. However, Santa started his annual activities long before diversity and equal opportunity became issues, and as a result he doesn't handle Muslim, Hindu, Jewish and Buddhist children. That reduces his workload significantly to a mere 15% of the total, namely 378 million. However, the crucial figure is not the number of children but the number of homes Santa has to visit. According to the most recent census data, the average size of a family in the world is 3.5 children per household. Thus, Santa has to visit 108,000,000 individual homes. (Of course, as everyone knows, Santa only visits good children, but we can surely assume that, on an average, at least one child of the 3.5 in each home meets that criterion.) That's quite a challenge. However, by traveling east to west, Santa can take advantage of the different time zones, and that gives him 24 hours. Santa can complete the job if he averages 1250 household visits per second. In other words, for each Christian household with at least one good child, Santa has 1/1250th of a second to park his sleigh, dismount, slide down the chimney, fill the stockings, distribute the remaining presents under the tree, consume the cookies and milk that have been left out for him, climb back up the chimney, get back onto the sleigh, and move on to the next house. To keep the math simple, let's assume that these 108 million stops are evenly distributed around the earth. That means Santa is faced with a mean distance between households of around 0.75 miles, and the total distance Santa must travel is just over 75 million miles. Hence Santa's sleigh must be moving at 650 miles per second -- 3,000 times the speed of sound. A typical reindeer can run at most 15 miles per hour. That's quite a feat Santa performs each year. What happens when we take into account the payload on the sleigh? Assuming that the average weight of presents Santa delivers to each child is 2 pounds, the sleigh is carrying 321,300 tons -- and that's not counting Santa himself, who, judging by all those familiar pictures, is no lightweight. On land, a reindeer can pull no more than 300 pounds. Of course, Santa's reindeer can fly. (True, no known species of reindeer can fly. However, biologists estimate that there are some 300,000 species of living organisms yet to be classified, and while most of these are insects and germs, we cannot rule out flying reindeer.) Now, there is a dearth of reliable data on flying reindeer, but let's assume that a good specimen can pull ten times as much as a normal reindeer. This means that Santa needs 214,200 reindeer. Thus, the total weight of this airborne transportation system is in excess of 350,000 tons, which is roughly four times the weight of the Queen Elizabeth. Now, 350,000 tons traveling at 650 miles per second creates enormous air resistance, and this will heat the reindeer up in the same fashion as a spacecraft re-entering the earth's atmosphere. The two reindeer in the lead pair will each absorb some 14.3 quintillion joules of energy per second. In the absence of a NASA-designed heat shield, this will cause them to burst into flames spontaneously, exposing the pair behind them. The result will be a rapid series of deafening sonic booms, as the entire reindeer team is vaporized within 4.26 thousandths of a second. Meanwhile, Santa himself will be subjected to centrifugal forces 17,500 times greater than gravity. That should do wonders for his waistline. Christmas is indeed a magical time.

What exactly is the difference between a monad and a galaxy of a hyperreal number? How do I determine the monad and galaxy of a given hyperreal number? I know, for example, what the monad and galaxy of 0 (the only real hyperreal) is, but I don't know how it is reached. Help please!

The best known piece covering problem is the 8 queens problem, which already gives away the solution: It is possible to place 8 independent queens (independent = not attacking eachother) on the board, but not more than 8. The same can be done for other pieces, so the questions are: How many independent kings/rooks/bishops/knights can you place on a board? But we can also ask for the opposite: not to place as many independent pieces as possible (inevitably covering the whole board), but to place as few pieces as possible such that they attack (or occupy) each square. So how many kings/queens/rooks/bishops/knights are needed to cover the whole board?

Maybe some of you have already seen these puzzles before, but for those who didn't, here are some fun brain twisters: (1) Suppose you take an 8x8 chessboard, and you remove two diagonally opposite corner squares of the board (say a1, h8). Is it possible to cover the remaining 62 squares with 31 1x2 dominoes? If so, how, and if not, why not? (2) What if we take an nxn board (where n is even), remove two opposite corners and try to do the same? (3) And what if, instead of removing two squares in opposite corners, we remove two random squares of the 8x8 (or nxn) board? When can the remaining squares of such a board be covered with dominoes?

This is similar to the domino problem (I also posted this in the forum at 'I luv 2 do maths', but they don't seem to be very keen at solving it): We have a chess board (64 squares) and 21 triominos (63 squares, the triominos has the "I" shape). It is possible to cover the chess board with all the triominos, meaning only one square will be uncovered. Prove that this square is always one of the red squares: Just to make things clear, these are triominos: http://upload.wikimedia.org/wikipedia/commons/2/22/Trominoes.svg The shape we are using is the left one.

[As pasted from The Emperor's Newest Minds] Having read and heard a little bit about Causal Dynamical Triangulations, I was wondering if anyone here has also even heard of it or knows more about it than I do (which is very likely if so!). Links to movies about this: One homepage (the other link was broken) And Thanks in advance! [Edit: It appears that the links showed up only partially, but they still work]

The World Champion Chess Match of 1921 was held at a resort in the US Virgin Islands and pitted against each other those two masters, Yeti and Zugswang. The winner won seven games and lost six. They took turns playing white, with Yeti having white in the final game which decided the match. The black pieces won five times. Who won the match? -by Martin Hollis

What is the meaning of Lewis Carroll's chess problem. Can it be done, seems to make no sense.

Do any of you know how we could about making the chess.com staff implement a LaTeX feature, so we could write our mathematical discussions properly!?

Who can forget the beautiful Mandelbrot set? It is extraordinary simple yet infinitely complex. Named after Benoit Mandelbrot, the formula behind it is as follows: Z = Z ^ 2 + C C must always remain bounded. The Mandelbrot set was discovered in 1980. While this may seem surprising by how simple the formula is, it required supercomputers to accurately give a picture of the complexity of infinity.