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 "cells": [
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# Solving Nonlinear Equations"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "To solve a nonlinear equation, we used the package ``NLsolve``."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 1,
   "metadata": {},
   "outputs": [],
   "source": [
    "using NLsolve"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Our problem is to solve for $r$ in\n",
    "\n",
    "$$\n",
    "   1 - e^{-25 r} = 12.5 \\left( 1 - e^{-r} \\right).\n",
    "$$"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "The above equation is equivalent to\n",
    "\n",
    "$$\n",
    "   1 - e^{-25 r} - 12.5 \\left( 1 - e^{-r} \\right) = 0.\n",
    "$$"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Writing the above equation as $f(x) = 0$, the function below defines $f$ as the input to ``nlsolve`` of the package ``NLsolve``."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 2,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "f!"
      ]
     },
     "execution_count": 2,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "\"\"\"\n",
    "    function f!(F, x)\n",
    "\n",
    "defines F[1] as the input to nlsolve.\n",
    "\"\"\"\n",
    "function f!(F, x)\n",
    "   r = x[1]\n",
    "   F[1] = 1 - exp(-25*r) - 12.5*(1 - exp(-r)) \n",
    "end"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Let us pass $r = 0.05$ as the initial value for the solver."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 3,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "Results of Nonlinear Solver Algorithm\n",
       " * Algorithm: Trust-region with dogleg and autoscaling\n",
       " * Starting Point: [0.05]\n",
       " * Zero: [0.0673745373743581]\n",
       " * Inf-norm of residuals: 0.000000\n",
       " * Iterations: 4\n",
       " * Convergence: true\n",
       "   * |x - x'| < 0.0e+00: false\n",
       "   * |f(x)| < 1.0e-08: true\n",
       " * Function Calls (f): 5\n",
       " * Jacobian Calls (df/dx): 5"
      ]
     },
     "execution_count": 3,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "sol = nlsolve(f!, [0.05])"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "We see that the solution is 0.0673745373743581."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Let us verify."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 4,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "f (generic function with 1 method)"
      ]
     },
     "execution_count": 4,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "f(r) = 1 - exp(-25*r) - 12.5*(1 - exp(-r))"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Let us see what the output type of ``nlsolve`` really is."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 5,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "NLsolve.SolverResults{Float64, Float64, Vector{Float64}, Vector{Float64}}\n",
      "  method: String \"Trust-region with dogleg and autoscaling\"\n",
      "  initial_x: Array{Float64}((1,)) [0.05]\n",
      "  zero: Array{Float64}((1,)) [0.0673745373743581]\n",
      "  residual_norm: Float64 9.36140054363932e-13\n",
      "  iterations: Int64 4\n",
      "  x_converged: Bool false\n",
      "  xtol: Float64 0.0\n",
      "  f_converged: Bool true\n",
      "  ftol: Float64 1.0e-8\n",
      "  trace: NLsolve.SolverTrace\n",
      "    states: Array{NLsolve.SolverState}((0,))\n",
      "  f_calls: Int64 5\n",
      "  g_calls: Int64 5\n"
     ]
    }
   ],
   "source": [
    "dump(sol)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 6,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "1-element Vector{Float64}:\n",
       " 0.0673745373743581"
      ]
     },
     "execution_count": 6,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "sol.zero"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 7,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "-9.36140054363932e-13"
      ]
     },
     "execution_count": 7,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "f(sol.zero[1])"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "The residual is very small, so we accept the solution."
   ]
  }
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