PLEASE HELP. Does anyone know the Example 1 info on page 595 Discrete Maths and its Applications by K. Rosen?
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http://highered.mcgraw-hill.com/webintegration/WebIntegrationServlet?call=PurchaseInfoWeblet\n\nYou can buy the book on-line
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There is a primer here:\nhttp://www.phon.ucl.ac.uk/courses/spsci/matlab/primer40.pdf\n\nand an online course here:\nhttp://www.phon.ucl.ac.uk/courses/spsci/matlab/\n\nI may just be reviewing them myself!
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Lots of places, actually.  Algebraic number theory, Galois theory, linear algebra, discrete geometry, and other fields have shown tremendous usefulness in producing hard-to-break cryptography and digital signature schemes.\n\nAlgebraic geometry, abstract algebra, some real analysis, discrete geometry, linear algebra, Galois theory, and other fields are integral in the design of error correcting codes used currently in CDs, DVDs, satellite communications, your telephone and broadband lines, and many other places where reliable communication is needed.\n\nComputer science (I hope the applications of this are obvious!) essentially grew out of mathematical logic, with lots of influence from universal algebra and category theory.\n\nArtificial intelligence research also borrows heavily from several mathematical fields, most especially mathematical logic.  Though the true "thinking computer" is still science fiction, some of the results of this research is already used in word processors (including the dreaded "Clippy"!!!! but more useful places too), voice recognition software, automatic optimization algorithms for electronic circuit designs,  and other software, and don't forget the Roomba!\n\nAbstract analysis has solved many problems in fluid dynamics, with applications to aerodynamic design, electronic circuit design, Internet traffic management, and even road system design!  (Cars on the road behave a lot like fluid in pipes).\n\nLie algebras, tensor algebras, manifold theory, algebraic topology, systems theory, and several other abstract mathematical topics come into play with the General Theory of Relativity (GPS satellites take into account the General Theory of Relativity for their calculations--or else they would give badly inaccurate results!  Not only their motion, but the Earth's gravity has an effect on measuring time, which is essential to the process of geolocation) as well as quantum theory (which gave us the transistor, the laser, superconductors, and other current technology....and far more future technology) and very importantly, at least from an intellectual viewpoint, much progress on "how the universe works" fundamentally (we don't know for sure yet!).
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This geometric trick is very involving:\nYou can use it as a starting point for geometric construction of square roots, equivalence of areas and isometric transforms\nhttp://answers.yahoo.com/question/?qid=1006030501625\n\nThis other about divisibility and prime factoring of integers has deep mathematical concepts beyond it: starting from it you can speak about prime factoring, bynary count, combinatory calculus and mathematical induction\nhttp://answers.yahoo.com/question/?qid=1006022300298&r=w\n\nI'm going to use them in my school classes
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Go to this page: http://britton.disted.camosun.bc.ca/jbfunpatt.htm\n\nGo down to topic #2 "Clock or Modular arithmetic"\n\nCheck out the links "Modular art" and "Cayley Quilt Maker"\n\nThere are also many other links to interesting recreational math projects, as well as explanations of the mathematics behind modular arithmetic art.
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In late 2002 and 2003 Grigori Perelman of the Steklov Institute of Mathematics, Saint Petersburg posted 3 papers on the arXiv:[7]\n\n    * The entropy formula for the Ricci flow and its geometric applications, 2002\n    * Ricci flow with surgery on three-manifolds, 2003\n    * Finite extinction time for the solutions to the Ricci flow on certain three-manifolds, 2003\n\nIn these papers he sketched a proof of the Poincaré conjecture and a more general conjecture, Thurston's geometrization conjecture, completing the Ricci flow program outlined earlier by Richard Hamilton.\n\nFrom May to July 2006, several groups presented papers that filled in the details of Perelman's proof of the Poincaré conjecture, as follows:\n\n    * Bruce Kleiner and John Lott, Notes on Perelman's Papers (May 2006) filled in the details of Perelman's proof of the geometrization conjecture.\n    * Zhu Xiping and Cao Huaidong, A Complete Proof of the Poincaré and Geometrization Conjectures - Application of the Hamilton-Perelman theory of the Ricci flow (Asian Journal of Mathematics Volume 10, Number 2, p. 165-492 (June 2006)).\n    * John Morgan and Gang Tian Ricci Flow and the Poincaré Conjecture (July 2006) gave a detailed proof of just the Poincaré Conjecture (which is somewhat easier than the full geometrization conjecture).\n\nOn August 22, 2006, the ICM awarded Perelman the Fields Medal for his work on the conjecture, but Perelman refused the medal.[8][9][10] John Morgan spoke at the ICM on the Poincaré conjecture on August 24, 2006, declaring that "in 2003, Perelman solved the Poincaré Conjecture."[11]\n\nThe August 2006 issue of The New Yorker contains an article, titled "Manifold Destiny", on some of the issues surrounding Perelman's work.
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This is sampling without replacement and without regard to order.\n\nIt's given by the binomial coefficient:\n\nfor n items, choosing k at a time\n\nnCk means n choose k \n\nor \n\n    n!\n-----------  (this is a division)\n(n-k)!k!\nNote that you can get these from Pascal's triangle, nth row, kth column.\n\nSo, 4C2 is 6:\n\nAB\nBC\nCD\nDA\nAC\nBD
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Try searching http://www.knowledgestorm.com/ and http://www.bitpipe.com/\n\nhttp://www.icasit.org/km/resources/kmcases.htm\nhttp://www.a1-termpaper.com/bus-cse.shtml\nhttp://www.sauder.ubc.ca/bcom/docs/course_outlines/comm363.pdf\nhttp://www.nrel.gov/docs/fy00osti/28053.pdf
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http://www.cs.ucdavis.edu/~haungs/paper/node11.html
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http://www-groups.dcs.st-and.ac.uk/~history/Mathematicians/Dubreil.html
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You can get it from following link.Best of luck in the seminar.\nhttp://www.mendelweb.org/Mendel.html
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http://www.convertit.com/Go/ConvertIt/Reference/AMS55.ASP?Res=150\n\nhttp://www.math.sfu.ca/~cbm/aands/subj.htm\n\nhttp://clusterfie.epn.edu.ec/ibernal/html/CURSOS/AbrilAgosto06/Metodos/Libros/Fundamentals%20Of%20Numerical%20Computing%20-%201997%20-Wiley%20%20(By%20Laxxuss).pdf
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We wish to show that:\n[k=1, n]∑kk! = (n+1)! - 1\n\nClearly, this is true for n=1, since 1*1! = 2! - 1 = 1. Suppose it is true for some n. Then for n+1:\n[k=1, n+1]∑kk!\n(n+1)*(n+1)! + [k=1, n]∑kk!\n(n+1)*(n+1)! + (n+1)! - 1\n(n+2)*(n+1)! - 1\n(n+2)! - 1\n\nWhich is what we would get if we substituted n+1 for n in our formula. So by induction, the formula holds for all n. Q.E.D.
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this page has lots of links:\n\nhttp://www.mathtools.net/C_C__/Algorithms_and_Data_structures/index.html\n\nmy personal favourite (although very academic) book:\n\nhttp://www.brpreiss.com/books/opus4/
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wow, that's pretty amazing.  i was convinced that was wrong until i tried drawing the "counterexample" i had in my head.  other than reading through that proof (props to the guy who posted that link, btw) i guess the best way to realize this is true is to show that for any five closed shapes in two dimensions, there will always be at least two shapes that do not touch one another.  because of this, you can color those two shapes the same color, so you will have a maximum of four colors.\n\nthis problem is reducible to a graph with five nodes, where each node represents a shape and each edge represents a shared boundary.  (showing that this is reducible to this is probably the difficult part of the proof, which is why i'm glossing over it).  you cannot connect each node to every other node without there being at least one set of intersecting edges.  if edges intersect, it indicates that the shapes (or countries) represented by the two nodes connected by one of the intersecting edges do not share a border.\n\nhere's my suggested proof (assuming we accept the reduction to this graph problem):\nstart with a graph G with three nodes, {A, B, C}.  There are also edges {AB, BC, AC}.  So, G is a closed graph with three nodes and three edges where each node is connected to each other node in the graph (i.e. G is a triangle).  If our reduction of the coloring problem is correct, then there exists a map reducible to this graph that requires three colors to be colored in the described way.\n\nNext, construct a graph H by adding a node D to graph G.  There are two possible ways to add D: 1) D can be on the interior of closed graph G; or 2) D can be outside the interior of G.\n\nIn case 1, let the new graph with nodes A,B,C,D be called H.  We need to show that we can make H such that D is connected to A, B, and C without any intersecting edges.  This is simple enough; straight edges to A, B, and C from D do not intersect each other or the existing edges of the graph G.  So, H demonstrates that a map can require four colors.  H now has four separate enclosed interior regions: triangles ABD, ACD, BCD, and ABC.\n\nFrom H, construct a graph I by adding another node E.  Repeating the above procedure, E can be placed in four places: a) in the interior of ABD; b) in the interior of ACD; c) in the interior of BCD; or d) on the exterior of ABC.\n\nCASE a: As with constructing H, edges can be drawn from E to A, B, and D.  However, C is on the exterior of ABD, so any edge from E to C must cross either AB, BD, or AD.  So, in case a, there must be an intersection, so four colors is still an option.\nCASES b and c follow similarly.\nCASE d: E is on the exterior of ABC, but D is on the interior of ABC.  Any edge between E and D must intersect AB, BC, or AC.  So, coloring with four colors is possible in this case.  Since it is possible in all cases, it is possible for any map with five regions with a sub-map reducible to a graph like H.\n\nCase 2) Instead of placing D in the interior of G, make a graph H' by placing D on the exterior of triangle ABC.  Any quick drawing will show that D is easily connected to A, B, and C.  So there is a map with four regions that requires four colors.  To make the rest of this easy, note that H' also produces four enclosed regions: ABD, ABC, BCD, and ADC.  I won't go through the trouble to prove this, but clearly H' is isomorphic to H.  That being said, whatever is true of the general graph H is also true of the general graph H'.  Therefore, a graph I' made by adding a node to H' cannot be complete without there being at least one set of intersecting edges.\n\nSince this is true for all cases, any map reducible to a graph with five nodes is colorable with just four colors.  QED\n\nSorry this is so long.  I hope it's helpful.
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Here is a link you may find helpful.\n\nhttp://www.math.buffalo.edu/mad/00.INDEXmad.html
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well.. here is a really wonderful book for this :\n\nDesign and Analysis of algorithms by Anany Levitin.. try getting this book.\n\nAnd : \nhere's a dictionary of algorithms:\n\nhttp://www.nist.gov/dads/\nhere, each and every algorithm's details have been given. And links to other sites, where simulations of algorithms are put up as JAVA APPLETS\n\nfor eg:\nhere's a link to quick sort simultaion using java applet, which i got from NIST . gov :\n\nhttp://www.cosc.canterbury.ac.nz/people/mukundan/dsal/QSort.html\n\nN JOY ; )