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    "%matplotlib inline\n",
    "import numpy as np;\n",
    "from matplotlib import pyplot as plt"
   ]
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    "$\\LaTeX \\text{ commands here}\n",
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    "\\newcommand{\\inner}[1]{\\langle #1 \\rangle}\n",
    "\\newcommand{\\span}{\\mathrm{span}}\n",
    "\\newcommand{\\proj}{\\mathrm{proj}}\n",
    "$"
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    "<hr style=\"border: 5px solid black\">\n",
    "\n",
    "**Georgia Tech, CS 4540**\n",
    "\n",
    "# L4: Convexity & Vector Calculus\n",
    "\n",
    "Jacob Abernethy\n",
    "\n",
    "*Thursday, August 29, 2019*"
   ]
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   "source": [
    "## Separating Hyperplanes\n",
    "\n",
    "The set of separating hyperplanes. Suppose that $C$ and $D$ are disjoint subsets of $\\R^n$.\n",
    "Consider the set of $(a, b) \\in \\R^{n+1}$ for which\n",
    "$$a^\\top x \\leq b \\text{ for all } x  \\in C$$\n",
    "and\n",
    "$$a^\\top x \\geq b \\text{ for all } x \\in D.$$\n",
    "**Show** that this set is a convex cone (which is the singleton $\\{0\\}$ if there is no hyperplane that separates $C$ and $D$)"
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   "source": [
    "### Separating hyperplane theorem\n",
    "If convex sets $C$ and $D$ are disjoint, then there is vector $w$ and scalar $b$ such that $\\forall x\\in C, w^\\top x \\leq b$ and $\\forall z \\in D, \\; w^\\top z \\geq b$.\n",
    "### Supporting hyperplane theorem\n",
    "Given a non-empty convex set $C$, and some $x_0 \\in \\text{bndry}(C)$, then there is vector $w$ and scalar $b$ such that $w^\\top x_0 = b$ and $\\forall x \\in C, \\; w^\\top x \\leq b$\n",
    "\n",
    "**Prove**: the first theorem implies the second (*Hint*: think about the interior of $C$)"
   ]
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   "source": [
    "**Caratheodory's Theorem**: Let $X \\subset \\R^d$. Then each point in $\\text{conv}(X)$ is a convex combination of at most $d + 1$ points of $X$."
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   "source": [
    "**Proof (start)**: Suppose there is some $y \\in \\text{conv}(X)$ that cannot be written as a convex combination of fewer than $m \\geq d+2$ points in $X$. Then\n",
    "$$ y = \\sum_{i=1}^m \\lambda_i x_i \\text{ with } \\sum_{i=1}^m \\lambda_i = 1 \\text{ and } \\lambda_i > 0 \\forall i.$$\n",
    "Notice that $m \\geq d+2$ points $x_1, \\ldots, x_m$ *must* be affinely dependent (since no more than $d+1$ points in $\\R^d$ can be affinely independent). That means that there are coefficients $\\mu_1, \\ldots, \\mu_m$ so that\n",
    "$$\\sum_{i=1}^m \\mu_i x_i = 0 \\text{ where } \\sum_{i=1}^m \\mu_i = 0, \\text{ and } \\exists \\mu_i > 0.$$\n",
    "Let's combine these two statements, and observe that for any $\\alpha \\in \\R$ we have\n",
    "$$y = y + \\alpha \\cdot 0 = \\sum_{i=1}^m \\lambda_i x_i + \\alpha \\sum_{i=1}^m \\mu_i x_i = \\sum_{i=1}^m (\\lambda_i + \\alpha \\mu_i) x_i$$\n",
    "\n",
    "$\\ldots$\n",
    "\n",
    "$\\ldots$ which is a contradiction!"
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    "#### Answer:\n",
    "\n",
    "We know that there is one $\\mu_j$ so that $\\mu_j > 0$. Set $\\alpha = - \\min_{i : \\mu_i > 0} \\frac{\\lambda_j}{\\mu_j}$, and let $i^*$ be the minimizing $i$ in the previous expression. Also let $\\lambda_i' := \\lambda_i + \\alpha \\mu_i$. Our particular choice of $\\alpha$ ensures two things:\n",
    "1. $\\lambda_{i^*}' = 0$, because $\\lambda_{i^*}' = \\lambda_{i^*} + (-\\frac{\\lambda_{i^*}}{\\mu_{i^*}}) \\mu_{i^*}$\n",
    "2. For any $j$, $\\lambda_j' \\geq 0$. That's because either\n",
    "    + $\\mu_j = 0$, in which case $\\lambda_j' = \\lambda_j$\n",
    "    + OR $\\lambda_j' = \\lambda_j + (-\\frac{\\lambda_{i^*}}{\\mu_{i^*}})\\mu_j \\geq \\lambda_j + (-\\frac{\\lambda_{j}}{\\mu_{j}})\\mu_j = 0$\n",
    "\n",
    "What we have just done is show that we can write $y$ in terms of another convex combination, i.e. $y = \\sum_{i=1}^m \\lambda_i' x_i$. But notice that, for this convex combination, one of the $\\lambda_i'$ is 0! That means we were able to write $y$ as a convex combination of fewer than $m$ points, a contradiction."
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    "## Derivatives, and multivariable functions\n",
    "\n",
    "For a differentiable function $f : \\R \\to \\R$, the derivative is defined as\n",
    "$$f'(x) = \\lim_{\\delta \\to 0} \\frac{ f(x + \\delta) - f(x)}{\\delta}.$$\n",
    "\n",
    "For a function $f : \\R^n \\to \\R$, the partial derivative w.r.t. $x_i$ is\n",
    "$$\\frac{\\partial f(x)}{\\partial x_i} = \\lim_{\\delta \\to 0} \\frac{ f(x + \\delta e_i ) - f(x)}{\\delta}$$\n",
    "where $e_i$ is the $i$th basis vector.\n"
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    "## Perspectives on the gradient\n",
    "\n",
    "\n",
    "The *gradient* of $f$ can be viewed as the vector of partial derivatives\n",
    "$$ \\nabla_x f = \\nabla f(x) =  \\left[\\frac{\\partial f(x)}{\\partial x_1}, \\ldots, \\frac{\\partial f(x)}{\\partial x_n}\\right]$$\n",
    "\n",
    "But another way to view the gradient is that it is a *linear transformation* which has the same domain and co-domain of $f$. That is, $\\nabla_x f$ is a linear map $\\R^n \\to \\R$, where\n",
    "$$ \\nabla_x f [u] =  \\left[\\frac{\\partial f(x)}{\\partial x_1}, \\ldots, \\frac{\\partial f(x)}{\\partial x_n}\\right] \\cdot u.$$\n",
    "\n",
    "But this is indeed the same as the *directional derivative* of $f$ in the direction of $u$. That is,\n",
    "$$ \\nabla_x f[u] = \\lim_{\\delta \\to 0} \\frac{ f(x + \\delta u ) - f(x)}{\\delta}$$"
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   "source": [
    "## The Jacobian\n",
    "\n",
    "If $f$ is a function mapping $\\R^n \\to \\R^m$, then we often call the gradient the *Jacobian*, which is the matrix of partial derivatives:\n",
    "$$D_f := \\begin{pmatrix} \n",
    "  \\frac{\\partial f_1}{\\partial x_1}   & \\cdots & \\frac{\\partial f_1}{\\partial x_n}\\\\ \n",
    "  \\vdots & & \\vdots \\\\\n",
    "  \\frac{\\partial f_m}{\\partial x_1}   & \\cdots & \\frac{\\partial f_m}{\\partial x_n}\n",
    "\\end{pmatrix}\n",
    "$$\n",
    "The *hessian* of a function $f : \\R^n \\to \\R$ is defined as the matrix of partial derivatives:\n",
    "$$ \\nabla^2 f = D^2_f := \\begin{pmatrix} \n",
    "  \\frac{\\partial^2 f}{\\partial x_1 \\partial x_1}   & \\cdots & \\frac{\\partial^2 f}{\\partial x_1 \\partial x_n}\\\\ \n",
    "  \\vdots & & \\vdots \\\\\n",
    "  \\frac{\\partial^2 f}{\\partial x_m \\partial x_1}   & \\cdots & \\frac{\\partial^2 f}{\\partial x_n \\partial x_n}\n",
    "\\end{pmatrix}\n",
    "$$"
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    "## Gradient and Hessian problems\n",
    "\n",
    "Define a function $f(x) = \\log \\sum_{i=1}^n \\exp(x_i)$.\n",
    "\n",
    "#### Problem\n",
    "\n",
    "1. What is $\\nabla_x f$?\n",
    "1. What is $\\nabla_x f[\\mathbf{1}]$ where $\\mathbf{1}$ is the vector of all 1s.\n",
    "1. What is $\\nabla^2 f(x)$?"
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    "$f(x) = \\log \\sum_{i=1}^n \\exp(x_i)$.\n",
    "\n",
    "#### Answer\n",
    "\n",
    "1. $\\nabla_x f = \\frac{1}{\\sum_{i=1}^n \\exp(x_i)} \\cdot [\\exp(x_1) \\exp(x_2) \\dots \\exp(x_n)]$\n",
    "2. $\\nabla_x f[\\mathbf{1}] = \\sum_{i = 1}^n \\frac{\\exp(x_i)}{\\sum_{i=1}^n \\exp(x_i)} = 1$\n",
    "3. $\\nabla^2 f(x) = - \\frac{1}{(\\sum_{i=1}^n \\exp(x_i))^2} A$ where for matrix $A$, $a_{i,i} = \\exp(x_i)$, and $a_{i,j} = 0$ if $i \\neq j$."
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    "## Multivariate Chain Rule\n",
    "\n",
    "For a good description, see [these notes](https://math.berkeley.edu/~nikhil/courses/121a/chain.pdf).\n",
    "\n",
    "Imagine we have $f : \\R^n \\to \\R^m$ and $g : \\R^m \\to \\R^k$. Then we want to compute the gradient (Jacobian) of the function $f \\circ g$. Is there any easy way to do this? The answer is YES, according to the **multivariate chain rule**:\n",
    "$$ D_{f \\circ g}[x] = D_f[g(x)] \\cdot D_g[x]$$\n",
    "This is a powerful statement, although it may not be clear why. It says we can obtain the gradient of the composed functions, by simply composing the gradients, treated as *linear transformations*."
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   "source": [
    "#### Problem \n",
    "Let $f(x) = \\sum_{i=1}^m -\\log (A_i \\cdot x - b_i)$ where $A$ is an $m\\times n$ matrix. Notice that we can write $f(x) = h(g(x)) = h \\circ g (x)$ where $h(y) = -\\sum \\log y_i$ and $g(x) = A x - b$\n",
    "\n",
    "Use the multivariate chain rule to compute $\\nabla_x f$."
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