final check DOS

notebook/density_of_states.ipynb

@@ -144,10 +144,11 @@
     "<p style=\"text-align: justify;font-size:15px\">\n",
     "    Finally, the third method we describe here is the\n",
     "    linear tetrahedron interpolation (LTI). In this method, the volume in \n",
-    "    reciprocal space is splitted into small tetrahedra. The energy at each corner \n",
-    "    is computed similarly to the previous case.\n",
-    "    Typically, a regular grid is chosen also in this case, and each small volume - that typycally has a shape of a cube or, more generally, a parallelepiped - is split into tetrahedra: the image below demonstrates how to split \n",
-    "    a cubic reciprocal space volume into six tetrahedra.\n",
+    "    reciprocal space is split into small tetrahedra. The energy at each corner \n",
+    "    is computed similarly to the previous case. Typically, a regular grid is chosen \n",
+    "    also in this case, and each small volume - that typically has a shape of a cube or, \n",
+    "    more generally, a parallelepiped - is split into tetrahedra: the image below \n",
+    "    demonstrates how to split a cubic reciprocal space volume into six tetrahedra.\n",
     "</p>\n",
     "\n",
     "<div class=\"container\" style=\"text-align: center; width: 500px;\">\n",
@@ -159,11 +160,12 @@
     "    \n",
     "<p style=\"text-align: justify;font-size:15px\">    \n",
     "    Then, the method assumes that, within a tetrahedron, the energy behaves linearly; therefore a \n",
-    "    linear interpolation is employed to obtain the value of the energy in any point inside the tetrahedron,\n",
-    "    knowing the values of the energy at its fours corners.\n",
-    "    Thanks to this, it is possible to calculate much more accurately the portion of volume of each tetrahedron that\n",
-    "    is above or below a given energy, making the resulting DOS much more accurate than a simple histogram\n",
-    "    obtained from the value of the energy at its four corners.\n",
+    "    linear interpolation is employed to obtain the value of the energy in any point inside the\n",
+    "    tetrahedron, knowing the values of the energy at its fours corners.\n",
+    "    Thanks to this, it is possible to calculate much more accurately the portion of the \n",
+    "    volume of each tetrahedron that is above or below a given energy, making the resulting \n",
+    "    DOS much more accurate than a simple histogram obtained from the value of the energy at \n",
+    "    its four corners.\n",
     "</p>\n",
     "\n",
     "</details>"
@@ -177,7 +179,7 @@
     "\n",
     "<ol style=\"text-align: justify;font-size:15px\">\n",
     "    <li> \n",
-    "    Investigate the number of the kpoints' influence on the results of the calculated DOS\n",
+    "    Investigate the number of the kpoints' influence on the results of the calculated DOS.\n",
     "    <details style=\"color: blue\">\n",
     "    <summary>Hints</summary>\n",
     "    In the right panel, the blue line is the analytical solution for the DOS. By choosing \n",
@@ -430,7 +432,7 @@
     "    kpts = np.dot(monkhorst_pack(shape), G).reshape(shape + (3,))\n",
     "    eigs = _compute_dos(kpts, G, 0)\n",
     "    \n",
-    "    gx = np.linspace(0, 5, 500)\n",
+    "    gx = np.linspace(-0.03, 5, 500)\n",
     "    gy = 0.0*gx\n",
     "    \n",
     "    for eig in eigs.ravel():\n",
@@ -462,8 +464,8 @@
     "    analy_y = 1.0/(2.0*np.pi**2)*2.0**0.5*analy_x**0.5*(alat_bohr / 2.0)**3.0;\n",
     "    lanaly, = ax.plot(analy_y, analy_x, 'b', label='Analytical solution')\n",
     "    \n",
-    "    ax.set_ylim([0, 0.5])\n",
-    "    ax.set_xlim([0, analy_y.max() + 0.1])\n",
+    "    ax.set_ylim([-0.03, 0.5])\n",
+    "    ax.set_xlim([0, analy_y.max() + 3.1])\n",
     "    ax.legend(loc=1, bbox_to_anchor=(1.3, 1.0))\n",
     "    ax.yaxis.tick_right()\n",
     "    ax.yaxis.set_label_position(\"right\")\n",