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capacitive electric load: low power factor in range of 0.1 to 0.3; high ignition voltage 1–10 kV; huge amount of energy stored in electric field – requirement of energy recovery if DBD is not driven continuously; voltages and currents during discharge event have major influence on discharge behaviour (filamented, homogeneous).
The structure is simply reduced by a and disorder dependent term because all disorder of the first-kind does is smear out the scattering planes, effectively reducing the form factor. In three dimensions the effect is the same, the structure is again reduced by a multiplicative factor, and this factor is often called the Debye–Waller factor .
These equations describe boundary-value problems, in which the solution-function's values are specified on boundary of a domain; the problem is to compute a solution also on its interior. Relaxation methods are used to solve the linear equations resulting from a discretization of the differential equation, for example by finite differences.
Some authors allow any real , [1] [2] whereas others require that not be 0 or 1. [ 3 ] [ 4 ] The equation was first discussed in a work of 1695 by Jacob Bernoulli , after whom it is named. The earliest solution, however, was offered by Gottfried Leibniz , who published his result in the same year and whose method is the one still used today.
Here, m=2 and there are 10 subsets of 2 indices, however, not all of them are bases: the set {3,5} is not a basis since columns 3 and 5 are linearly dependent. The set B={2,4} is a basis, since the matrix = is non-singular.
Once the fundamental solution is found, it is straightforward to find a solution of the original equation, through convolution of the fundamental solution and the desired right hand side. Fundamental solutions also play an important role in the numerical solution of partial differential equations by the boundary element method.
This problem has a graph-theoretic solution in which a graph with four vertices labeled B, G, R, W (for blue, green, red, and white) can be used to represent each cube; there is an edge between two vertices if the two colors are on the opposite sides of the cube, and a loop at a vertex if the opposite sides have the same color. Each individual ...
For the fourth time through the loop we get y = 1, z = x + 2, R = (x + 1)(x + 2) 4, with updates i = 5, w = 1 and c = x 6 + 1. Since w = 1, we exit the while loop. Since c ≠ 1, it must be a perfect cube. The cube root of c, obtained by replacing x 3 by x is x 2 + 1, and calling the