By Kupershmidt B.A.

It's well known that jet language is the ordinary technique to communicate with the neighborhood difficulties of differentiable arithmetic. to quote a couple of examples, you can actually check with differential equations ([1], [4], [6], [8], [9], [16]), singularities ([7]), calculus of diversifications and box conception ([5], [10], [11] - [14]).

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**Example text**

Dim ∂H = n − 1 (92) so a hyperplane is a point in R , a line in R2 , a plane in R3 , and so on. Every hyperplane can be described as the intersection of complementary halfspaces; [232, 19] ∂H = H− ∩ H+ = {y | aTy ≤ b , aTy ≥ b} = {y | aTy = b} (93) a halfspace-description. 16 We will later find this expression for y in terms of nullspace of aT (more generally, of matrix AT (122)) to be a useful device for eliminating affine equality constraints, much as we did here. 16 64 CHAPTER 2. CONVEX GEOMETRY a= 1 −1 (a) 1 1 1 −1 1 b= −1 1 −1 −1 −1 {y | bTy = −1} {y | aTy = 1} {y | bTy = 1} {y | aTy = −1} c= (c) 1 1 −1 1 −1 (b) −1 1 1 −1 −1 d= (d) 1 −1 {y | dTy = −1} {y | cTy = 1} {y | cTy = −1} {y | dTy = 1} e= −1 {y | eTy = −1} 1 0 1 (e) {y | eTy = 1} Figure 17: (a)-(d) Hyperplanes in R2 (truncated).

1, nonempty intersections of hyperplanes). 1. 2. 2 45 Vectorized-matrix inner product Euclidean space Rn comes equipped with a linear vector inner-product ∆ y , z = y Tz (26) We prefer those angle brackets to connote a geometric rather than algebraic perspective. Two vectors are orthogonal (perpendicular ) to one another if and only if their inner product vanishes; y, z = 0 y⊥z ⇔ (27) A vector inner-product defines a norm y ∆ 2 = yT y , y 2 =0 ⇔ y=0 (28) When orthogonal vectors each have unit norm, then they are orthonormal.

Every hyperplane can be described as the intersection of complementary halfspaces; [232, 19] ∂H = H− ∩ H+ = {y | aTy ≤ b , aTy ≥ b} = {y | aTy = b} (93) a halfspace-description. 16 We will later find this expression for y in terms of nullspace of aT (more generally, of matrix AT (122)) to be a useful device for eliminating affine equality constraints, much as we did here. 16 64 CHAPTER 2. CONVEX GEOMETRY a= 1 −1 (a) 1 1 1 −1 1 b= −1 1 −1 −1 −1 {y | bTy = −1} {y | aTy = 1} {y | bTy = 1} {y | aTy = −1} c= (c) 1 1 −1 1 −1 (b) −1 1 1 −1 −1 d= (d) 1 −1 {y | dTy = −1} {y | cTy = 1} {y | cTy = −1} {y | dTy = 1} e= −1 {y | eTy = −1} 1 0 1 (e) {y | eTy = 1} Figure 17: (a)-(d) Hyperplanes in R2 (truncated).

### Geometry of jet bundles and the structure of Lagrangian and Hamiltonian formalisms by Kupershmidt B.A.

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