Jump to content

alext87

Senior Members
  • Content Count

    87
  • Joined

  • Last visited

Community Reputation

10 Neutral

About alext87

  • Rank
    Meson
  1. Is this correct? Let G be a group and H a subgroup of G. Suppose that G acts on the (non-empty) set A. Then there exists an element of A which has H as its stabilizer? The converse is easily shown. As if G acts on a set A and then pick any w belonging to A then the the stabilizer of w is a subgroup of G. I.e. it there a one-one correspondence between the subgroups of G and the stabilizers of the elements of A? Any help would be great cheer.s
  2. Can anybody describe to be "in basic" terms what is a Platonic solid? And why are there only five? Also what does it mean to say that the faces, angles and edges are all congruent? Thanks
  3. To clear up the confusion. It is not actually a theorem by it is called the Banach–Tarski paradox. And it is a consequence of accepting the axiom of choice.
  4. I have worked out how to do it now. There was no need to work out the integral (in fact it cannot be done analytically). You need to note that the bivariate normal distribution is centrally symmetric about the origin if X and Z are independent and N(0,1). Hence the angle was proportion to the angle the area made with the origin. Thanks for your help. AT
  5. Does anybody know of a one-one homomorphism from A4 (alternating group) into the symmetric group S6 which is not a consequence of a natural embedding of S4 into S6? Does one even exist?? AT
  6. Okay I can show that they are independent... That they are N(0,1) comes from the marginal distributions. I need to integrate over the first quadrant but the integration is what I am struggling on... One method I tried was to change variables to x and z and then separate as they are independent. However, I can't get the answer. Do you what the limits are of z? Do you know a substitution I can make in the integrand for z?
  7. Let (X,Y) have a bivariate normal pdf with correlation coefficient p and variance 1 for both X and Y. So the joint density is: f(x,y) = (1/((2*PI)SQRT(1-p^2)))*exp(-0.5*(1-p^2)^-1 * (x^2-2pxy+y^2)) Show that X and Z=(Y-pX)/SQRT(1-p^2) are independent N(0,1) variables, and deduce that P(X>0,Y>0) = 1/4 +arcsin(p)/((2*PI)) ?? Does anybody know how to do this!?
  8. For every n-dimensional vector space there does exist a basis consisting of semi-definite matrices. Proof: Take the canonical basis of the vector space. This is where we have matrices {E11,E12,...,E1n,E21,E22,...,E2n,...,...,En1,En2,...,Enn} where Eij is a matrices with all zero elements except the (i,j) entry. All these matrices are semi-definite. As the characteristic polynomial is x^n for all Eij where i is not equal to j and the characteristic poly is x^(n-1) * (x-1) for Eii 1<=i<=n. Now all eigenvalues of a matrix are the roots of the characteristic poly i.e (in this case) just 0 and 1. So the matrices are all semi-definite. Hope this helps. AT
  9. A particle is dropped from the top of a tower on the Earth's equator. As a result of the Earth's rotation, does it land slightly to the east, or slightly to the west of the tower?
  10. Can anybody explain how to and solve the equation: d2005y/dx2005 - y = 0 for y? I have tried ancillary equs but are there complex roots. I know the ans it y=exp(x) but need to be able to show it.
  11. This is like the question does 0/0=0 as 0/n=0; or 0/0=1 as n/n=1; or 0/0=infinite as n/0=infinite. This questions hardly have a definite answer but I like to think that the answer is all three at the same time - its more fun that way!!!!
  12. Yx - aYx-1 = cx + d This is one of the different equations and the particular function is rather complicated! Have a go at solving it please. Thanks!
  13. I am having a bit of problem with finding the particular function for linear difference equations. The method of finding the particular function is similar to the method for finding the particular integral in linear differiational equations. Can anybody help?
  14. Geometric problem traingle.doc
×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.