Search results
Results from the WOW.Com Content Network
The following are important identities in vector algebra.Identities that only involve the magnitude of a vector ‖ ‖ and the dot product (scalar product) of two vectors A·B, apply to vectors in any dimension, while identities that use the cross product (vector product) A×B only apply in three dimensions, since the cross product is only defined there.
In Cartesian coordinates, the divergence of a continuously differentiable vector field = + + is the scalar-valued function: = = (, , ) (, , ) = + +.. As the name implies, the divergence is a (local) measure of the degree to which vectors in the field diverge.
Logarithms and exponentials with the same base cancel each other. This is true because logarithms and exponentials are inverse operations—much like the same way multiplication and division are inverse operations, and addition and subtraction are inverse operations.
If the cross product of two vectors is the zero vector (that is, a × b = 0), then either one or both of the inputs is the zero vector, (a = 0 or b = 0) or else they are parallel or antiparallel (a ∥ b) so that the sine of the angle between them is zero (θ = 0° or θ = 180° and sin θ = 0). The self cross product of a vector is the zero ...
Here, vec(X) denotes the vectorization of the matrix X, formed by stacking the columns of X into a single column vector. It now follows from the properties of the Kronecker product that the equation AXB = C has a unique solution, if and only if A and B are invertible ( Horn & Johnson 1991 , Lemma 4.3.1).
The derivative of ln(x) is 1/x; this implies that ln(x) is the unique antiderivative of 1/x that has the value 0 for x = 1. It is this very simple formula that motivated to qualify as "natural" the natural logarithm; this is also one of the main reasons of the importance of the constant e.
In the study of ordinary differential equations and their associated boundary value problems in mathematics, Lagrange's identity, named after Joseph Louis Lagrange, gives the boundary terms arising from integration by parts of a self-adjoint linear differential operator.
We prove commutativity (a + b = b + a) by applying induction on the natural number b. First we prove the base cases b = 0 and b = S(0) = 1 (i.e. we prove that 0 and 1 commute with everything). The base case b = 0 follows immediately from the identity element property (0 is an additive identity), which has been proved above: a + 0 = a = 0 + a.