Based on the diagram in the following URL: http://wwf.edula.com/content/topics/figure1.jpg\n\nI make the conclusion:\nVARIANT IS THE ORDER OF INVARIANTS.\n\nIs my conclusion correct?
132200
Not quite correct, consider the example of two sets of numbers:\n{1,2,3}, {2,1,3}.\nAccording to your conclusion, these sets are variants with different order of invariants. But are we always  interested in the order of invariants? If yes, order becomes no more an invariant itself. So you should reformulate your statement a bit:\n\nVARIANT IS DEFINED BY THE TYPES OF INVARIANTS IT IS BEING COMPOSED OF, AND THEIR MUTUAL ORDER IF THEIR ORDER IS NOT AN INVARIANT ITSELF. (i.e, when order matters it becomes a variant too, as in your drawings).\n\nHave further questions? Mail me: ingameswetrust@yahoo.com\nAlex.
604774
The explanation is really simple, really!\n\nOkay, take a look at the mystic ball chart right now. Take a look at the numbers 0, 9, 18, 27, 36, 45, 54, 63, 72, and 81. They are all the same symbol, right? And they are all multiples of 9! Now, before doing anything, just click on the mystic ball. You'll notice that the symbol that pops up is the same symbol that the numbers mentioned above show. Click on the blue link to try it all again. You should get another sheet with different symbols. Now, look at the very same numbers that you looked at before. They all share the same symbol, but now it is a different one, right?\n\nThis is the trick. No matter what number you choose from 0 to 99, the final answer will always be a multiple of 9. Try it out yourself! Go through each number and figure out what the final number of each number would be. Figure out what the final symbol would be, too. You will see that no matter what number you choose, the same symbol will keep popping up.\n\nIf you want the equation, it would be 10a + b - (a+b). 'a' represents the first digit of the number you chose, and 'b' represents the second digit of the number you chose. '10a + b' represents the number as a whole (for example, let's choose the number 32. a = 3, b = 2. 10(3) + 2 = 32. Make sense?). the '- (a+b)' part represents the time when you add the digits together and subtract it from the original number to get your final answer. After reducing the equation, you get 9a, which is why the answers all come out to be a multiple of 9.\n\nHopefully this makes sense!
584917
(v - v0) / (t - t0)\n\nv - final velocity\nv0 - initial velocity\n\nt - final time\nt0 - initial time
153427
Im not entirely sure if this is correct but, I know that a relation is a partial ordering if it is reflexive, transitive and anti-symmetric.  \n\nSo to satisfy the reflexive part we need that x>=x, which is trivially satisfied by f(x)>=f(x).  For the transitive part we need that if x>=y and y>=z then x>=z, which will be true if for f(x)>=f(y) and f(y)>=f(z) then f(x)>=f(z).\nFor the anti-symmetric part we need that if x>=y and y>=x then x=y, so when f(x)>=f(y) and f(y)>=f(x) then x=y\n\nSo this would suggest that the necessary and sufficient conditions on f are that: (1) when f(x)>=f(y) and f(y)>=f(z) then f(x)>=f(z).  (2) if f(x)>=f(y) and f(y)>=f(x) then x=y.  So f would appear to be one to one.\n\nI hope this helps
412217
By "Vactorial" I assume you mean "factorial"?  Then you can just as well say (2n)! / (n!)^2 for the right hand side.\n\nI wasted an hour of my life trying induction to prove this but I'm going around in circles.  The ravages of aging...
449214
HI: I tried to answer your email but it said that your email address has not been confirmed and so the email didn't go through.\n\nalpha and beta are scalers for the vector space V.  This problem is actually so easy that people don't believe what they've done.  You really just need to show that Span S behaves like V.  \n\nSpan S must be a vector space itself.  Two elements from the span is \nu=a1*v1 + a2*v2...ak*vk\nand \nw=b1*v1 + b2*v2...bk*vk\n\nshow that alpha*u+beta*w is in Span S where alpha, beta are scalers for your vector space.\n\nalpha*u+beta*w = (alpha*a1+beta*b1)*v1 + ....\nis a linear combination of the vectors in Span S, thus it's in Span S.  \n\nSo, Span S is not empty and is closed, thus its a subspace.
334909
The Brown University pattern theory group is working with the belief that the world is complex, and to understand it, or a part of it, requires realistic representations of knowledge about it. We create such representations using a mathematical formalism, pattern theory, that is compositional in that the representations are built from simple primitives, combined into (often) complicated structures according to rules that can be deterministic or random. This is similar to the formation of molecules from atoms connected by various forms of bonds. \n\nPattern theory is transformational in that groups or semigroups of transformations operate on the primitives. These transformations express the invariances of the worlds we are looking at. \n\nPattern theory is variational in that it describes the variability of the phenomena observed in different applications in terms of probability measures that are used with a Bayesian interpretation. This leads to inferences that will be realized by computer algorithms. Our aim is to realize them through codes that can be executed on currently available hardware.
540250
I was able to draw 16 different forms (8 with their symmetrical reversed forms).  There may be more but I don't have the time to work on it.\n\nI took another look at the problem and realized that in addition to the left-right symmetry there is also a top-bottom symmetry that adds at least another 14 answers to the 16 that I came up with earlier.  So there are at least 30.\n\nCould be more, but it is my bed time.
352040
I think it's supposed to be symmetrical.  I think the point of the question is to see if you can list the points on the graph as symmetrical pairs.  The 'if' is there so it doesn't give away that it's supposed to be symmetrical for anyone who graphed it wrong
473879
G/H = {xH: x is in G}.\n\nAction: g.(xH) = (gx)H for every g and x in G.\n\nThis is indeed a left action because: (i) the group operation is associative (so (g''g').(xH) = g''.(g'.(xH)) for every g' and g'' in G and (ii) ex= x (so e.(xH) = xH).\n\nIt is transitive because, to satisfy g.(xH) = yH, it is enough to take g = y(x^-1)).\n\nNote: need H to be normal if the notation G/H is to be legit because it means the quotient group. H need not be normal for G to left-act on the *set* {xH: x is in G}, but this set won't be a group w/o normality of H.
382980
Nicole I think that to understand what is to be done here I suggest to assume that a coordinate system be attached to one of the objects. I-don’t-remember-who says, but << any reference system at rest or moving with linear constant speed has no privilege with others the like>>.\n(x) m2=4kg is reference system #2, where v2=0, and m1=2kg is moving with speed v=8+12=20 ms, impact by m1 on m2 M1=m1*v=40kg*m/s\n(y) m1=2kg is reference system #1, where v1=0, and m2=4kg is moving with speed v=8+12=20 ms, impact by m2 on m1 M2=m2*v=80kg*m/s, so R=40/80=0.5
207693
„input harvmac „overfullrule=0pt {„it Exam} (1) Prove via path \n„def„R{{„bf R}} „def„S{{„bf S}} (a) Work in four-dimensional (flat) Euclidean space $„R^4$ and show that conformal invariance implies...\nhttp://www.math.ias.edu/QFT/fall/exam1.tex \n\n \nhttp://www.math.ias.edu/QFT/spring/stringPS8.tex\n...def„det{{„rm det}} „def„Det{{„rm Det}} „def„tr ... B$ of the Wess-Zumino-Witten model, satisfy the equations for conformal invariance and for\nhttp://www.math.ias.edu/QFT/spring/stringPS8.tex \n\n \nComments: Latex(2e), 34 pages, 4 figures (eps) On some UNIX machines\nSome private shortcuts: „def„be{„begin{equation}} „def„ee{„end{equation}} „def„bea{„begin{eqnarray}} „def„eea{„end{eqnarray}} „def...\nhttp://www.ma.utexas.edu/mp_arc/e/96-436.latex.mpmix \n\n \n=4 19| >0 63| >1 65| >2 434| >4 525| 0 1773| 1 630| 1/2 2530| 10 \n\nhttp://www-library.desy.de/lists/spihep.k \n\n \nStardrive.org - - Article/Content Template\nStardrive.org is dedicated to fostering increased public awareness and education in the theoretical and technological aspects of interstellar travel.\nhttp://www.stardrive.org/Sarmail1-3-01.shtml \n\n \n„documentstyle[12pt]{article} %„renewcommand{„baselinestretch}{2.0}\n...oddsidemargin 0 mm „def„ee{„end{equation}} „def ... correlation functions have a power law behaviour, indicates that conformal invariance may be\nhttp://theory.ipm.ac.ir/papers/src/ipm63.tex \n\n \n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%...\n...def „pa {„partial} „def „ti {„widetilde} „def „se {{„prime„prime}} „def ... „subsection{Pre-big bang inflation and conformal frames}\nhttp://www.ba.infn.it/~gasperin/tex/ba437.tex \n\n \nhttp://www.ba.infn.it/~gasperin/tex/cer96267.tex\n...def „pa {„partial} „def „ra {„rightarrow} „def ... discussed in „cite{6a}, to the conditions of $„sg$-model conformal invariance represented by\nhttp://www.ba.infn.it/~gasperin/tex/cer96267.tex \n\n \n%%%%%%%%% LATEX FILE FOR FORMATTING GUIDELINES \n...def„xp{X^{+}} „def„xm{X^{-}} „def„Yp{Y^+} „def„Ym ... must be obeyed (in a conformal gauge this is the same as maintaining conformal\nhttp://theory.tifr.res.in/group/colloq/das.tex \n\n \n„documentstyle[twoside,psfig]{article}\n...def„ds{„displaystyle} „parindent=0pt „def ... out that this model is advantageous for the study of the conformal invariance conjecture for\nhttp://www.math.utah.edu/~ejpecp/EcpVol1/paper8.tex
604677
One problem you are having is that you have decided there should only be two terms for the symmetry here. Also, it is good to write z=f(x,y) so we get all three variables into the picture.\n\nThe symmetry f(-x,-y)=f(x,y) is a symmetry through the line x=y=0. This is a vertical line through the origin.\n\nThe symmetry f(-x,-y)=-f(x,y) is a symmetry through the origin (0,0,0).\n\nThe symmetry f(-x,y)=f(x,y) is a symmetry through the yz-plane.\n\nThe symmetry f(-x,y)=-f(x,y) is a symmetry through the line x=z=0.\n\nTo do a more extensive investigation would require some group theory. Groups are ways of describing the symmetries of a situation and are generally studied in an abstract algebra course.
469735
Yes.\n\nSkew symmetry means that A^T=-A where A^T is the transpose of A. If (a_ij ) are the entries of A, this will say that\na_ji=-a_ij.\nWith i=j,\na_ii=-a_ii,\nso \na_ii=0.
263546
There are two interpretations I can see here. The first is that the homology groups of the 0-dimensional sphere are needed. For this, the answer is trivial: you just get Hn(pt)+Hn(pt) (direct sum). Over a ring R, Hn(pt)=R if n=0 and is 0 if n>0. So the answer is R+R for n=0 and 0 otherwise.\n\nIf the asker actually means SO(k), the special orthogonal group, this is much trickier. It is *not true* that SO(k) is homeomorphic to S^(k-1) (this only happens if k=2; for k=3 you get a projective space). In fact, the dimension of the manifold SO(k) is k(k-1)/2. In this case, I don't know the answer off the top of my head.
514903
Hello,\n\nThe symbols in relationship to the numbers change each time. \nThe symbol you choose is the correct one  for all answers. No matter the number you choose there is one correct symbol for all answers.
386617
Ok, so the relative co-efficient of x is 1, so the unrelative co-hedrent homeomorphic must obviously be 2 to the power of Y-as Y is homeomorphic to X's relative undescribed co-efficient.\nUnderstand my friend. I hope I helped.