deletions | additions       diff --git a/section_Density_of_states_In__.tex b/section_Density_of_states_In__.tex  index e314485..ac71205 100644  --- a/section_Density_of_states_In__.tex  +++ b/section_Density_of_states_In__.tex 
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|\sin k(N+1)-\sin kN e^{iq}}|^2 = \sin^2 k(N+1)+\sin k N^2 -2\cos q \sin kN\sin K(N+1)   \end{equation}  with 2\cos q =2\cos k-(V_g/t), then   \begin{multline}  \sin^2 k(N+1)+\sin k N^2 -2\cos q \sin kN\sin K(N+1)   &=& \sin^2 k(N+1)+\sin k N^2 -\cos k \sin kN\sin K(N+1) +\frac{V_g}{t} \sin kN\sin K(N+1)   \end{multline}  Now we simplify the following term  \begin{equation}  \sin^2 k(N+1)+\sin k N^2 -2\cosq \sin kN\sin K(N+1) = \sin^2 k(N+1)+\sin k N^2 -2\cos  k \sin kN\sin K(N+1) +\frac{V_g}{t} =\sin k(N+1) \left[  \sin kN\sin K(N+1) k(N+1)-2 \cos k \sin kN \right]+\sin k N^2  \end{equation}  Now then  we simplify split  the following 2\cos k in two contributions (we do not consider the last  term $\sin k N^2$ for these simplifications)  \begin{equation}  \sin^2 k(N+1)+\sin \sin k(N+1) \left[ \sin k(N+1)- \cos k \sin kN \right] =\sin k(N+1) \left[ \sin kN \cos  k N^2 -2\cos +\cos kN \sin k- \cos  k \sin kN\sin K(N+1) kN \right]  \end{equation}  Then we have  \begin{multline}  \sin k(N+1) \left[ \sin kN \cos k +\cos kN \sin k- \cos k \sin kN \right] -\sin k(N+1) \sin kN \cos k   \\  =\sin k(N+1) (\cos kN \sin k- \cos k \sin kN )  \end{multline}  However we need to add the term $\sin k N^2$ to simplify the whole denominator, then  \begin{multline}  \sin^2 k(N+1)+\sin k N^2 -\cos k \sin kN\sin K(N+1) =\sin k(N+1) (\cos kN \sin k- \cos k \sin kN ) + \sin k N^2 =  \\  \sin k(N+1) \cos kN \sin k + \sin k N (\sin kN-\cos k\sin k(N+1))  \end{mutline}  Using the fact that $\sin k(N+1)=\sin kN \cos k +\cos kN \sin k$, then the whole expression is simplified to  \begin{multline}  \sin k(N+1) \cos kN \sin k + \sin k N (\sin kN-\cos k\sin k(N+1)) = \sin^ k  \end{mutline}  Therefore the final result reads  \begin{equation}  |c_1|^2 = D^2\frac{\sin^2 q \sin ^2 k}{\sin^k -\frac{V_g}{t}\sin kN \sin k(N+1)}  \end{equation}  Finally the density of states coincides with the one in \cite{Park_2013}. Using the fact that we are using $k$ and $q$ wavevectors in units of $a$, the lattice spacing, we have  \begin{equation}  \rho(E)=\frac{\sin q a\sin^ka}{(\pi a t)(\left\sin^2(ka)+\frac{eV_g}{t}\sin{k(N+1)a}\right)\sin kNa}  \end{equation} 
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