Taken from invariantology
217053
If you were to think about it biologically, one answer is the genetic code. All animals- monkeys, insects, bacteria, cows, snakes, sharks, cats, dogs, and humans have very similiar genetic codes.
60578
check those equations out at wikipedia
402985
Since the question is relative to whether it is good or not:\nYes, it is.\nWould you like us to work it out as well? I appreciate that it must come right out of a quesiton from, perhaps, your homework? Otherwise, why the last sentence? "One hint is that it is related to chapter- 'CIRCLES'" Can you attach that entire chapter for us as well?\n\nThanks!
43222
that's right. cause infinit is indefinit.
155241
Please visit http://mathforum.org/library/drmath/view/52240.html
438003
this is a maths board-not a forum for \npolitical correctness
307904
http://www.inthesetimes.com/site/main/article/2487/
464788
Let u=ln t, e^u=t, du=dt/t, dt=t du\nThen:\n∫ln² t dt = ∫t u² du = ∫u² e^u du\nIntegrating by parts:\nu² e^u - ∫2u e^u du\nu² e^u - 2u e^u + ∫2e^u du\nu² e^u - 2u e^u + 2e^u + C\nt ln² t - 2t ln t + 2t +C\nAnd we are done.
79099
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.
234987
As usual the Wikipedia has a whole lot of information on this. There's no point in my repeating all that here - it's not as though I understood any of it. \n\nBe warned, the article has a lot of weird heiroglyphics in it ;-), like this:\nhttp://upload.wikimedia.org/math/f/0/7/f071fb2817d450f44013bcb6373749a1.png
179079
Looks like A> Reining
267992
see http://mathworld.wolfram.com/NaturalLogarithm.html
248859
diagrams and all, see the link
71820
http://www-groups.dcs.st-and.ac.uk/~history/Mathematicians/Dubreil.html
294512
Try looking it up in your math book if you need more clarity
601150
proof positive\nyou can find it here\nhttp://www.scifi.com/proofpositive/
395556
Try Google or Yahoo.  You'll find worked examples there.  I think it needs diagrams to understand; they and the accompanying explanations are too complex to attempt to reproduce here.