Minimum and Maximum Variance Analysis
... well determined but any pair of vectors perpendicular to x 3 , i.e., any vectors lying in the equatorial plane of the discus, may serve as x 1 and x 2 . This degeneracy, therefore, does not limit the utility of MVAB for normal-vector and normal-field-component determinations, provided λ3 λ2 ' λ1 . ...
... well determined but any pair of vectors perpendicular to x 3 , i.e., any vectors lying in the equatorial plane of the discus, may serve as x 1 and x 2 . This degeneracy, therefore, does not limit the utility of MVAB for normal-vector and normal-field-component determinations, provided λ3 λ2 ' λ1 . ...
K-HOMOLOGY AND FREDHOLM OPERATORS I: DIRAC
... the index theorem for Dirac operators. In [5] we reduce the general elliptic operator case to the Dirac case. Finally, in [6] we reduce the case of hypoelliptic operators on contact manifolds to the elliptic case. The unifying theme of these papers is that K-homology provides the topological foundat ...
... the index theorem for Dirac operators. In [5] we reduce the general elliptic operator case to the Dirac case. Finally, in [6] we reduce the case of hypoelliptic operators on contact manifolds to the elliptic case. The unifying theme of these papers is that K-homology provides the topological foundat ...
Integral Vector Theorems
... The sense of dS is linked to the direction of travel along C by a right hand screw rule. (b) Both sides of the equation are scalars. (c) The theorem is often a useful way of calculating a line integral along a contour composed of several distinct parts (e.g. a square or other figure). (d) ∇ × F is a ...
... The sense of dS is linked to the direction of travel along C by a right hand screw rule. (b) Both sides of the equation are scalars. (c) The theorem is often a useful way of calculating a line integral along a contour composed of several distinct parts (e.g. a square or other figure). (d) ∇ × F is a ...
Three Dimensional Geometry
... Two lines with direction ratios a 1, b 1, c1 and a 2, b 2, c2 are (i) perpendicular i.e. if q = 90° by (1) a 1a 2 + b 1b 2 + c1c2 = 0 (ii) parallel i.e. if q = 0 by (2) ...
... Two lines with direction ratios a 1, b 1, c1 and a 2, b 2, c2 are (i) perpendicular i.e. if q = 90° by (1) a 1a 2 + b 1b 2 + c1c2 = 0 (ii) parallel i.e. if q = 0 by (2) ...
Chapter 6 Orthogonal representations II: Minimal dimension - D-MATH
... The first non-degeneracy condition we study is general position: we assume that any d of the representing vectors in Rd are linearly independent. A result of Lovász, Saks and Schrijver [6] finds an exact condition for this type of geometric representability. Theorem 1.2 A graph with n nodes has a g ...
... The first non-degeneracy condition we study is general position: we assume that any d of the representing vectors in Rd are linearly independent. A result of Lovász, Saks and Schrijver [6] finds an exact condition for this type of geometric representability. Theorem 1.2 A graph with n nodes has a g ...
Secret-Sharing Schemes Based on Self-dual Codes
... enjoy some design properties for codewords of given weight. We will see that 1−designs play an important role in our study when enumerating access structures by group size. Thirdly, their weight enumerators have strong invariance properties that allow us to use invariant theory to study them. In par ...
... enjoy some design properties for codewords of given weight. We will see that 1−designs play an important role in our study when enumerating access structures by group size. Thirdly, their weight enumerators have strong invariance properties that allow us to use invariant theory to study them. In par ...
Quaternions and isometries
... in terms of matrices, but then we will need to prove that every rotation has an axis. As a rotation of R2 has one fixed point and no axis, the proof cannot be entirely trivial (in fact, it depends on the fact that every real cubic polynomial has a real root). Each rotation of an odd-dimensional space ...
... in terms of matrices, but then we will need to prove that every rotation has an axis. As a rotation of R2 has one fixed point and no axis, the proof cannot be entirely trivial (in fact, it depends on the fact that every real cubic polynomial has a real root). Each rotation of an odd-dimensional space ...
