latex differences

Author: githubz0r

aml_chapter2_exercises.ipynb

@@ -122,8 +122,8 @@
     "<br>\n",
     "<br> We show that $\\sum^D_{i=0} {N\\choose i} \\leq N^D +1 \\Leftrightarrow \\sum^D_{i=1} {N\\choose i} \\leq N^D $ by induction.\n",
     "<br> It's easy to see that it's true for $D=1$ and $N=1$ etc., so assume it works for $D$. Then we have \n",
-    "<br> $\\sum^{D+1}_{i=1} {N\\choose i} = \\sum^{D}_{i=1} {N\\choose i} + {N\\choose D+1} \\leq N^D + {N\\choose D+1}$ i.e. we want $N^D + {N\\choose D+1} \\leq N^{D+1} \\Leftrightarrow {N\\choose D+1} \\leq N^{D+1} - N^D$ \n",
-    "<br> $\\Leftrightarrow {N\\choose D+1} \\leq \\frac{N^{D+1}}{(D+1)!} \\leq N^{D+1}(1-\\frac{1}{N})$ \n",
+    "<br> $\\sum^{D+1}_{i=1} {N\\choose i} = \\sum^{D}_{i=1} {N\\choose i} + {N\\choose {D+1}} \\leq N^D + {N\\choose {D+1}}$ i.e. we want $N^D + {N\\choose {D+1}} \\leq N^{D+1} \\Leftrightarrow {N\\choose {D+1}} \\leq N^{D+1} - N^D$ \n",
+    "<br> $\\Leftrightarrow {N\\choose {D+1}} \\leq \\frac{N^{D+1}}{(D+1)!} \\leq N^{D+1}(1-\\frac{1}{N})$ \n",
     "<br> Assume that $N\\geq2$ and $D\\geq1$ and then observe that $\\frac{1}{(D+1)!} \\leq \\frac{1}{2} \\leq (1-\\frac{1}{N}) < 1$ thus the inequality holds and the last step of the induction holds.\n"
    ]
   },