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aml_chapter2_exercises.ipynb

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"Problem 2.5\n",
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"<br>\n",
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"<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",
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"<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",
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"<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",
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"<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",
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"<br> $\\Leftrightarrow {N\\choose D+1} \\leq \\frac{N^{D+1}}{(D+1)!} \\leq N^{D+1}(1-\\frac{1}{N})$ \n",
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"<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"

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