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**Example text**

5). Since both of these potentials have long range, they should be grouped in such a way that at large distance the negative and positive terms cancel each other. Usually, this is done by dividing ve (r) + vH (r) into components due to the charges from inside and from outside of the cell at R. The intra-cell part of the electrostatic potential then becomes I vR (rR ) = − 2 ZR +2 rR ΩR nR (rR ) dr . 58). 3 The Poisson Equation 31 where nRL0 (rR ) is the L0 = (0, 0) partial component of the charge density near site R.

Therefore, subtracting D˙ a (z)/Da (z) removes the nonphysical pole z0 of g a (z)K˙ a (z). In the second step, we have to restore the real poles of D˙ a (z)/Da (z) due to the zeros of the logarithmic derivative function. We denote by D a real energy where Da (z) vanishes. , and thus D˙ a (z)/Da (z) ≈ 1/(z − D ). 2 Electron Density 25 no poles due to the zeros of the logarithmic derivative function. 46) gives no contribution to N ( F ) if D˙ a (z) is an analytic function of z inside the complex energy contour.

Energies above the Fermi level are needed when searching for the Fermi level for the next iteration. 4 Ry. , and their alloys at ambient conditions. 8 Ry. Using an average w2 = 10 Bohr2 , the energy interval of interest turns out to be ±8 around ω = 0. For these energies, we expand the slope matrix in Taylor series around ω0 , viz. S a (ω) = S a (ω0 ) + 1 dS a (ω) 1 d2 S a (ω) (ω − ω0 )2 + .... 21) (ω − ω0 ) + 1! dω 2! dω 2 where, for simplicity, we have dropped the RL subscripts. 18). In practice, the expansion center ω0 is chosen somewhere close to 0.