Sentence examples for matrix of equation from inspiring English sources

Thus, the augmented matrix of equation (3.7) becomes [ W G G ]. (3.8).

This strategy is based on the analysis of the distillation column phase equilibrium sensitivity and leads to an unique matrix of equation of state dominant binary parameters for the whole fractionation train.

Accordingly, we use STM to modify the eigenvalue of Jacobian matrix of Equation (5) under the condition and get the controlled MCA Equation (8) without changing the value and location of unstable fixed points.

where γ ′ ∈Γ is an arbitrary channel into which the incident electron wave scatters, and c l (r n,z γ,E γ ) denotes the the eigenvector of the lead dynamical matrix of Equation 6 for the inequivalent site n at z γ and E γ.

If ( x 0, y 0 ) ∈ S 2, then 〚 x N 2, y N 2 〛 ∈ S 3 for some N 2. Remark 3.9 The Jacobian matrix of equation (3.10) at ϑ equals D T ϑ as defined in equation (3.2), and then Theorem 3.5 remains true if r 1 a 21 − r 2 a 11 > 0, r 2, and θ in Theorem 3.5 are replaced with r 1 a 22 − r 2 a 12 > 0, − r 2, and ϑ, respectively.

If the linear system (2.1) admits an exponential dichotomy, the almost periodic system x'(t)=A t)x t)+g(t) has a unique almost periodic solution (x t)), and x t)= int_{-infty }^{t} X t PX^{-1}(s)f(s),ds- int_{t}^{infty } X t PX^{-1} ^{-1}(s)f(s),ds, where (X(t)) is the fundamental solution matrix of equation (2.1).

The first set of numbers plugged into the matrix of equations is always an educated guess, said Dr. Curtis C. Covey, a physicist at the Lawrence Livermore National Laboratory who compares the performance of various models.

We used a quasi-Newtonian approximation obtained by computing the Hessian matrix of Equations 28-29, zeroing the cross-derivatives, and solving for w.

Since (L:= gamma frac{z(p)}{w(p)} - frac{z(1)}{w(1)} + frac{z(0)}{w(1)} - gamma frac{z(0)}{w(p)}) is the determinant of the coefficient matrix of equations (14) and (15), when (Lneq 0), equations (14) and (15) have a unique nonzero solution.

where x s s is the equilibrium solution of (5a, 5b); i.e., (7) C 0 − D 0 x ss − D x ss 3 Γ + x ss = 0. Condition (6) is derived by computing the Jacobian matrix of Equations (5a) and (5b) at the equilibrium point and determining when the eigenvalues have a nonzero imaginary part.

As a result, the matrix version of equation (5) is: y = ρWy + βX + ε (6).