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    "%matplotlib inline\n",
    "import numpy as np;\n",
    "from matplotlib import pyplot as plt"
   ]
  },
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    "$\\LaTeX \\text{ commands here}\n",
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    "$"
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    "<hr style=\"border: 5px solid black\">\n",
    "\n",
    "**Georgia Tech, CS 4540**\n",
    "\n",
    "# L5: Convex Functions & Multivariate Differentiation\n",
    "\n",
    "Jacob Abernethy\n",
    "\n",
    "*Tuesday, September 03, 2019*\n",
    "\n",
    "**Quiz password**: longweekend"
   ]
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   "source": [
    "## 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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   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "#### Answer\n",
    "\n",
    "We can see that\n",
    "+ $D_h[y] = \\nabla h(y) = [-1/y_1, \\ldots, -1/y_m]$\n",
    "+ $D_g[x] = \\nabla g(x) = A$, a constant matrix\n",
    "\n",
    "So this means\n",
    "\\begin{eqnarray*}\n",
    "D_f[x] & = & D_h[g(x)] \\cdot D_g[x] = [-1/(A_1 x - b_1), \\ldots, -1/(A_m x - b_m)] \\cdot A \\\\\n",
    "& = & \\sum_{i=1}^m \\frac{-1}{A_i x - b_i}A_i\n",
    "\\end{eqnarray*}\n",
    "\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "slideshow": {
     "slide_type": "slide"
    }
   },
   "source": [
    "#### Problem:\n",
    "\n",
    "Let $f(x) = A(2x + \\mathbf{1})$ where $A$ is a matrix, and $\\mathbf{1}$ is the all ones vector.\n",
    "\n",
    "Let's write $f(x) = h \\circ g (x)$ where $g(x) = 2x + \\mathbf{1}$ and $h(x) = A x$. What is the gradient of $f$?"
   ]
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   "metadata": {
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   "source": [
    "#### Answer\n",
    "\n",
    "$D_g[x] = \\nabla g(x) = 2I$\n",
    "\n",
    "$D_h[x] = \\nabla h(x) = A$\n",
    "\n",
    "$D_f[x] = D_h[g(x)] \\cdot D_g[x] = 2AI = 2A$"
   ]
  },
  {
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   "metadata": {
    "slideshow": {
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   },
   "source": [
    "### Some Calc. Problems\n",
    "\n",
    "Let $w$ be an $n$-dim vector. Calculate the first and second derivatives of the following functions:\n",
    "+ $f(w) = \\frac 1 2 w^\\top w$\n",
    "+ $f(w) = w^\\top v$ where $v$ is some fixed vector"
   ]
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   "metadata": {
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   "source": [
    "#### Answer\n",
    "\n",
    "* $f(w) = \\frac 1 2 w^\\top w$:\n",
    "    + $\\nabla f(w) = w$\n",
    "    + $\\nabla^2 f(w) = I$, the identity matrix\n",
    "* $f(w) = w^\\top v$\n",
    "    + $\\nabla f(w) = v$\n",
    "    + $\\nabla^2 f(w) = 0$, the zero matrix"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "slideshow": {
     "slide_type": "slide"
    }
   },
   "source": [
    "### More Calc. Problems\n",
    "\n",
    "Let $w$ be an $n$-dim vector. Calculate the derivatives of the following functions:\n",
    "+ $f(w) = \\|w\\|_2$\n",
    "+ $f(w) = \\| w \\|_1$"
   ]
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   "metadata": {
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   "source": [
    "#### Answer\n",
    "+ $f(w) = \\|w\\|_2 \\implies \\nabla f(w) = \\frac{w}{\\|w\\|_2}$\n",
    "+ $f(w) = \\| w \\|_1 \\implies \\nabla f(w) = [\\text{sign}(w_i)]_{i=1\\ldots n}$"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "slideshow": {
     "slide_type": "slide"
    }
   },
   "source": [
    "### More Calc. Problems\n",
    "\n",
    "Let $w$ be an $n$-dim vector. Calculate the derivatives of the following functions:\n",
    "+ $f(w) = \\frac 1 2 \\|w - v\\|_2^2$ where $v$ is some fixed vector\n",
    "+ $f(w) = \\frac 1 2 w^\\top M w$ (calculate the hessian for this one as well!)\n",
    "+ $f(w) = \\exp(w^\\top M w)$ where $M$ is some symmetric $n \\times n $ matrix"
   ]
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   "metadata": {
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   "source": [
    "#### Answer\n",
    "\n",
    "+ $f(w) = \\frac 1 2 \\|w - v\\|_2^2 \\implies \\nabla f(w) = (w - v)^\\top w$\n",
    "+ $f(w) = \\frac 1 2 w^\\top M w \\implies \\nabla f(w) = \\frac 1 2 w^\\top(M + M^\\top)$\n",
    "+ $f(w) = \\exp(w^\\top M w) \\implies \\nabla f(w) = \\exp(w^\\top M w)Mw$"
   ]
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   "source": [
    "## Positive Semidefinite Matrices\n",
    "\n",
    "**Definition**: A matrix $M \\in \\R^{n\\times n}$ is *positive semidefinite* if $x^\\top M x \\geq 0\\; \\forall x \\in \\R^n$\n",
    "\n",
    "Commonly, we work with matrices $M$ that are *symmetric* (that is, $M = M^T$). In this case, the following are equivalent:\n",
    "\n",
    "1. $M$ is positive semidefinite\n",
    "2. The eigenvalues of $M$ are all $\\geq 0$\n",
    "3. The matrix $M$ can be written as $B^\\top B$ for some $B \\in R^n$\n",
    "\n",
    "**Problem**: Show that (3) $\\implies$ (1) $\\implies$ (2)"
   ]
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   "metadata": {},
   "source": [
    "#### Answer\n",
    "\n",
    "If $M$ can be written as $B^TB$ for $B \\in R^n$, then:\n",
    "\n",
    "$$x^TMx = x^TB^TBx = (Bx)^T Bx = (Bx)^2 \\geq 0$$\n",
    "\n",
    "This shows (1).\n",
    "\n",
    "To show (1) implies (2), consider the eigenvectors of $M$ $v_i$ such that $Mv_i = \\lambda_i v_i$. Since $M$ is positive semidefinite:\n",
    "\n",
    "$$v_i^T M_i v_i = v_i^T \\lambda_i v_i = \\lambda v_i^T v_i = \\lambda \\|v_i\\|_2^2 \\geq 0$$\n",
    "\n",
    "Since $\\|v_i\\|_2^2 \\geq 0$, then it must be the case $\\lambda_i \\geq 0$."
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   "source": [
    "## Convex Functions\n",
    "\n",
    "A function $f : \\R^d \\to \\R$ is convex if, for any $x,y \\in \\text{dom}(f)$ and any $\\theta \\in [0,1]$, we have\n",
    "$$f(\\theta x + (1-\\theta)y) \\leq \\theta f(x) + (1-\\theta) f(y)$$\n",
    "\n",
    "**First-order Condition** When $f$ is differentiable, then $f$ is convex if and only if\n",
    "$$f(y) \\geq f(x) + \\nabla f(x)^\\top(y - x) \\quad \\text{ for any } x,y \\in \\text{dom}(f)$$\n",
    "\n",
    "**Second-order Condition** When $f$ is twice differentiable, then $f$ is convex if and only if\n",
    "$$u^\\top (\\nabla^2f(x)) u \\geq 0 \\quad \\text{ for any } x \\in \\text{dom}(f) \\text{ and any } u \\in \\R^d$$\n",
    "This last condition $\\equiv$ \"$f$ is convex if its hessian is always positive semi-definite\""
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   "source": [
    "### Problem\n",
    "\n",
    "**First-order Condition** When $f$ is differentiable, then $f$ is convex if and only if\n",
    "$$f(y) \\geq f(x) + \\nabla f(x)^\\top(y - x) \\quad \\text{ for any } x,y \\in \\text{dom}(f)$$\n",
    "\n",
    "Use the first order condition to show that, for any convex and differentiable $f$, we have\n",
    "$$(\\nabla f(y) - \\nabla f(x))^\\top(y - x) \\geq 0 \\text{ for any } x,y \\in \\text{dom}(f)$$\n"
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   "source": [
    "#### Answer\n",
    "\n",
    "Let's apply the first order condition twice, once at $x$ and once at $y$:\n",
    "\\begin{eqnarray*}\n",
    "f(y) & \\geq & f(x) + \\nabla f(x)^\\top(y - x) \\\\\n",
    "f(x) & \\geq & f(y) + \\nabla f(y)^\\top(x - y).\n",
    "\\end{eqnarray*}\n",
    "Add these two inequalities together, and subtract $f(x) + f(y)$ from both sides and you are done!"
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   "source": [
    "### Problem\n",
    "\n",
    "**Second-order Condition** When $f$ is twice differentiable, then $f$ is convex if and only if\n",
    "$$u^\\top (\\nabla^2f(x)) u \\geq 0 \\quad \\text{ for any } x \\in \\text{dom}(f) \\text{ and any } u \\in \\R^d$$\n",
    "\n",
    "Find conditions under which the following function is convex $f(x) = \\frac 1 2 x^\\top A x$ for a symmetric matrix $A \\in \\R^{d \\times d}$"
   ]
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     "slide_type": "subslide"
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   "source": [
    "#### Answer\n",
    "\n",
    "As we computed earlier in lecture, for $f(x) = \\frac 1 2 x^\\top A x$, the hessian is $\\nabla^2 f(x) = A$. We know that a twice-differentiable function is convex if its hessian is positive semi-definite. Therefore, $f$ is convex if and only if $A$ is positive semi-definite."
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