how can you define "trace" for a tensor (in tensor analysis )of rank more than two?
927290
For a simple matrix 'trace ' is the sum of all main diagonal elements.\n\nFor a tensor, it is a scalar given by the contracted mixed tensor equal to T and it satisfies the conditon\n\nTr[M(x)^(-1)  d/dx^(lambda) M(x)] = d/dx^(lambda) ln det[M(x)]\n\nThe derivative is partial, in this case.\n\nI suggest you write this equation by hand, since its not easy for me to write legibly without equation editor. Sorry for that.
769753
Without knowing the relationship between successive elements, I think the best you can do is this:\n[k=0, n-1]∑(x_(k+1) - x_k)²\n[k=0, n-1]∑(x_(k+1)² - 2x_k*x_(k+1) + x_k²)\n[k=0, n-1]∑x_(k+1)² - 2[k=0, n-1]∑(x_k*x_(k+1)) + [k=0, n-1]∑x_k²\n[k=1, n]∑x_k² - 2[k=0, n-1]∑(x_k*x_(k+1)) + [k=0, n-1]∑x_k²\n\nx_0 + x_n + 2[k=1, n-1]∑x_k² - 2[k=0, n-1]∑(x_k*x_(k+1))
412598
There won't be a book specifically for tensor analysis. Depending on your background, you're most likely to find what you want in either a General Relativity book or a differential geometry book. For a non-axiomatic approach, you probably want the Relativity book. Warner's book is quite good.\n\nOne difficulty with the subject is that mathematicians and physicists do things very differently when it comes to tensors. In physics, a coordinate system is often assumed and the thing that defines a tensor is the transformation properties between coordinate systems. For mathematicians, the crucial thing is multi-linearity on some vector space. Because of this, there is a *lot* of confusion when math and physics people talk about tensors. GOOD LUCK!
406468
D = {x | x < -3 U x > 4 }
146293
I know of one application in an abstruse area of physics: The sinh( ) of a scalar field occurs in gravitational theories where, for whatever reason, one emplys a metric disformally related to the 'physical' metric tensor.\n\nAlso, the shape of a freely hanging level chain is cosh( ) of displacement along the line joining ends
245176
u*(u-2)+v*(u-2)\n(u+v)*(u-2)
735674
Multiple Internal reflections , they r better referred to as the total internal reflection
216212
Yes, but you have to be more clear what you want in general.\n\nSuppose we are working in n-dimensional Euclidean space. Suppose also that you have a hyperplane going through the origin. If you take n-1 independent vectors in that hyperplane and do a determinant construction similar to that for the cross product, you will get a vector normal to your hyperplane. But you will need n-1 *independent* vectors and you will have to do a nxn determinant (with the top row consisting of the unit tangent vectors along the axes).\n\nIf you don't have a full hyperplane, you will be wanting a basis for the normal subspace. Other than the previously mentioned Graham-Schmidt process, I don't know a good way to do that. Hmmm.....
773251
The Wikipedia entry discusses the linear algebraic proof of the existence of the SVD factorization (if that is what you want)\n\nhttp://en.wikipedia.org/wiki/Singular_value_decomposition\n\nA few other easy-to-understand links:\n\nFrom netlib:\nhttp://www.netlib.org/lapack/lug/node53.html\n\nFrom Numerical Recipes:\nhttp://www.nrbook.com/a/bookfpdf/f2-6.pdf
776321
the question is unclear.\nDo you want an explicit formula for Y[n or do you just want to write out the convolution]
77120
It doesn't appear to simply significantly from the original expression.  I was able to get it to the following:\n\n(tan^3-tan)*sec\n\nDo the multipling by csc-1 top-n-bottom and then use the fact that cot^2=csc^2-1.  Bring the cot^2=cos^2/sin^2 to the top to get the above.\nGood luck.
669580
-E-N-G-L-A-N-D- 2 POINTS
441742
Assume there is a pair a_i, b_j with a non zero product.\nWrite fg= (a_i x^i + other terms) (b_j x^j + other terms) =\na_i b_j x^(i+j) is non zero.\n\nContradiction.
202581
The principal diagonal is the most useful diagonal in the study of matrix algebra because it is related to several invariants. \nIf M and N are two matrices of the same linear transformation with respect to different bases, then their trace (the sum of the entries on the principal diagonal) is the same.\nA transformation that can be represented as a scaling of n independent vectors in an n-dimensional space has a diagonal representation in that space under a suitable change of bases (namely to those vectors it scales, called eigenvectors). The product of the entries of the diagonal in that representation is another invariant, called the determinant (it is the signed volume of a parallelepiped made up of the basis vectors after the transformation). The determinant is another important invariant that you will see all over the place in differential geometry.\nThe other diagonals do have names (the superdiagonal is on top of the principal, subdiagonal is on the bottom of the principal, etc.) but they are so rarely important that they have fallen out of fashion. See the source for more details.
839609
Curl, divergence, gradient.
982925
in this order:\n\nprominent\nquaint\nsnare\neradicate\nvengeance\nmakeshift
224978
associative, commutative, identity, inverse