Sentence examples for can solve systems of from inspiring English sources

We present a Chebyshev multidomain method that can solve systems of hyperbolic equations in conservation form on an unrestricted quadrilateral subdivision of a domain.

This is of importance if for instance M and A are large sparse bandmatrices, since then one avoids increasing bandwidths in matrix products and one can solve systems of linear combinations of M and A more efficiently than for higher order polynomial combinations.

The first or the second moments can be obtained by solving systems of difference equations (3) and (12) or (3) and (13) respectively, for example, by using a numerical method.

This problem can be reduced to solving systems of linear equations (Lacroix et al., 2008).

Newton's method for solving systems of nonlinear equations can be highly sensitive to the initial guess of the solution.

We can calculate the offspring distribution (4) by solving systems of linear equations (2,3), as outlined earlier for the case of households of size one.

This is because i) active-set methods are usually the fastest ones for quadratic and linear programmes of small and median size; and ii) expensive computations, like solving systems of linear equations, can be shared in the active-set method.

It is shown in [12] that given coarse spectral envelope of sources, one can solve a system of equations for calculation of de-mixing matrix in regular mixtures.

The threshold γ and the number of samples N that are required to reach given P F and P D can be found by solving system of equations formed by (45) and (46) left{ begin{array}{ll} sqrt{2} sigma_{E,H_{0}} text{erfc}^{-1} 2P_{F}) = left[gamma - E(w_{E} |H_{0})right] sqrt{N} sqrt{2} sigma_{E,H_{0}} text{erfc}^{-1} 2P_{D}) = leF}[gamma - E(w_{E}|H_{1})right] sqrt{N}end{array}.

With the aid of Maple or Mathematica we can solve this system of algebraic equations to obtain the following sets of solutions: Case 1. a 0 = − 4 b K 2 L e 1 − K − L 2 a K, a 2 = − 6 K b L e 2 a, a 4 = − 6 K b L e 0 a, C 1 = 1 4 a K { − K 2 − 2 L K + 192 b 2 K 4 L 2 e 0 e 2 − L 2 + 16 b 2 K 4 L 2 e 1 2 }, a 1 = a 3 = 0, (26).

Thus applying scheme (5.14), we can solve the system of equations (6.13) and (6.14).