From 4430d5767b9dc2282f0a0c307b6921089621076a Mon Sep 17 00:00:00 2001
From: Lars kleyn Winkel <l.kleynwinkel@student.tudelft.nl>
Date: Thu, 5 Mar 2020 15:32:59 +0000
Subject: [PATCH] Update 8_many_atoms_solutions.md

---
 src/8_many_atoms_solutions.md | 2 +-
 1 file changed, 1 insertion(+), 1 deletion(-)

diff --git a/src/8_many_atoms_solutions.md b/src/8_many_atoms_solutions.md
index b20a3a3b..7abc6677 100644
--- a/src/8_many_atoms_solutions.md
+++ b/src/8_many_atoms_solutions.md
@@ -25,7 +25,7 @@ Optical branch corresponds with (+) in the equation given in the lecture notes.
 
 ### Subquestion 3
 
-Density of states is given as $g(\omega) = \frac{dN}{d\omega} = \frac{dN}{dk}\frac{dk}{d\omega}$. We know $\frac{dN}{dk} = 2\frac{L}{2\pi} = \frac{L}{\pi}$ since we have 1D and positive and negative k-values. $\frac{dk}{d\omega}$ can be computed using the group velocity: $$\frac{dk}{d\omega} = \bigg(\frac{d\omega}{dk} \bigg)^{-1} = (v_g)^{-1}$$
+Density of states is given as $g(\omega) = \frac{dN}{d\omega} = \frac{dN}{dk}\frac{dk}{d\omega}$. We know $\frac{dN}{dk} = \frac{L}{2\pi}$ since we have 1D. $\frac{dk}{d\omega}$ can be computed using the group velocity: $$\frac{dk}{d\omega} = \bigg(\frac{d\omega}{dk} \bigg)^{-1} = (v_g)^{-1}$$
 
 ## Exercise 2: the Peierls transition
 
-- 
GitLab