{
 "cells": [
  {
   "cell_type": "markdown",
   "id": "7c8c115e-dc7b-41b3-907a-700b3503fc61",
   "metadata": {},
   "source": [
    "## MT09 - TP3 - Automne 2025\n",
    "### Factorisation de Choleski & méthodes de Newton"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "9c071e68-a309-4a2c-96ad-ecc661d6891a",
   "metadata": {},
   "source": [
    "### 1. Programmation de la factorisation de Choleski sous forme vectorielle\n",
    "Pour une matrice $A$ symétrique définie positive, programmez la factorisation de Choleski \n",
    "\n",
    "\\begin{align*}\n",
    "& b_{11} = \\sqrt{a_{11}},\\quad b_{i1}=\\frac{a_{i1}}{b_{11}}, \\ \\text{ pour } i>1, \\\\[1.5ex]\n",
    "& b_{jj} = \\sqrt{a_{jj}-\\sum_{k=1}^{j-1} b_{jk}^2},\\ \\text{ pour } j\\geq 2,\\quad\n",
    "b_{ij}  =\\frac{1}{b_{jj}} \\left( a_{ij}-\\sum_{k=1}^{j-1} b_{ik} b_{jk}\\right),\\ 1<j<i.\n",
    "\\end{align*}\n",
    "\n",
    "dans une fonction\n",
    "\n",
    "```\n",
    "def choleski(A):\n",
    "    ...\n",
    "    return B\n",
    "```\n",
    "\n",
    "en remplaçant les sommations par des produits scalaires dans le code (attention aux indices démarrant à 0 en python). \n",
    "\n",
    "Question subsidiaire : peut-on s'arranger pour se limiter à une seule boucle ? (indication : en travaillant par bloc ; s'il vous reste du temps, à la fin du TP par exemple)."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 5,
   "id": "cce744c6-bbce-4d53-a077-55b7d6a10aac",
   "metadata": {
    "tags": []
   },
   "outputs": [],
   "source": [
    "import numpy as np\n",
    "\n",
    "def choleski(A):\n",
    "    n,m = A.shape\n",
    "    B = np.zeros((n,n))\n",
    "    B[0,0] = np.sqrt(A[0,0])\n",
    "    for i in range(1,n):\n",
    "        B[i,0] = A[i,0] / B[0,0]\n",
    "    for j in range(1,n):\n",
    "        sum = 0.0\n",
    "        for k in range(0,j):\n",
    "            sum += B[j,k]**2\n",
    "        B[j,j]= np.sqrt(A[j,j] - sum)\n",
    "        for i in range(j+1,n):\n",
    "            sum = 0.0\n",
    "            for k in range(0,j):\n",
    "                sum += B[i,k]*B[j,k]\n",
    "            B[i,j] = (A[i,j] - sum) / B[j,j]\n",
    "    return B"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 9,
   "id": "32d1c3a6-1a96-40f1-910d-4e30529a2ea4",
   "metadata": {
    "tags": []
   },
   "outputs": [],
   "source": [
    "# Semi-vectorisé:\n",
    "# Sommation sur k vectorisee\n",
    "def chol1(A):\n",
    "    m,n = A.shape\n",
    "    B = np.zeros((n,n))\n",
    "    B[0,0] = np.sqrt(A[0,0])\n",
    "    B[1:, 0] = A[1:, 0] / B[0,0]\n",
    "    #\n",
    "    for j in range(1,n):\n",
    "        B[j,j] = np.sqrt( A[j,j] - B[j, 0:j] @ B[j, 0:j] )\n",
    "        for i in range(j+1, n):\n",
    "            B[i,j] = (A[i,j] - B[i, 0:j].T @ B[j, 0:j]) / B[j,j]\n",
    "    #\n",
    "    return B\n"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 10,
   "id": "b1aaa822-589c-4eb9-ba59-90a2b864fdd3",
   "metadata": {
    "tags": []
   },
   "outputs": [],
   "source": [
    "# Version totalement vectorisée\n",
    "def chol2(A):\n",
    "    m,n = A.shape; B = np.zeros((n,n))\n",
    "    B[0,0] = np.sqrt(A[0,0])\n",
    "    B[1:, 0] = A[1:, 0] / B[0,0]\n",
    "    for j in range(1,n):\n",
    "        B[j,j] = np.sqrt( A[j,j] - B[j, 0:j] @ B[j, 0:j] )\n",
    "        B[j+1:n,j] = ( A[j+1:n,j] - B[j+1:n, 0:j] @ B[j,0:j]) / B[j,j]\n",
    "    return B"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "410ae58f-39ff-4930-be8a-fa85dda9478d",
   "metadata": {},
   "source": [
    "Appliquez la factorisation de Choleski à la matrice\n",
    "\n",
    "$$\n",
    "A = \\begin{pmatrix}\n",
    "2 & -1 & 0 \\\\\n",
    "-1 & 2 & -1 \\\\\n",
    "0 & -1 & 2\n",
    "\\end{pmatrix}.\n",
    "$$\n",
    "\n",
    "et vérifiez que $B B^T = A$."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 14,
   "id": "4e443018-de10-4232-b56c-cfba74b3c1fa",
   "metadata": {
    "tags": []
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "B = \n",
      " [[ 1.41421356  0.          0.        ]\n",
      " [-0.70710678  1.22474487  0.        ]\n",
      " [ 0.         -0.81649658  1.15470054]]\n",
      "BB^T - A = \n",
      " [[0. 0. 0.]\n",
      " [0. 0. 0.]\n",
      " [0. 0. 0.]]\n"
     ]
    }
   ],
   "source": [
    "import numpy as np\n",
    "A = np.array([[2, -1, 0],[-1, 2, -1],[0, -1, 2]])\n",
    "B = choleski(A)\n",
    "# Verif\n",
    "print('B = \\n', B)\n",
    "C = B @ B.T\n",
    "ERR = np.abs(C-A)\n",
    "ERR[np.abs(ERR)<1e-15] = 0\n",
    "print('BB^T - A = \\n', ERR)"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "ec5644d2-541b-4fcc-b5eb-8b33df6cbfec",
   "metadata": {},
   "source": [
    "### Comparaison des performances"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 16,
   "id": "4dec625e-e952-497a-ba71-65d48b97af7e",
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "[[ 2. -1.  0.  0.  0.]\n",
      " [-1.  2. -1.  0.  0.]\n",
      " [ 0. -1.  2. -1.  0.]\n",
      " [ 0.  0. -1.  2. -1.]\n",
      " [ 0.  0.  0. -1.  2.]]\n"
     ]
    }
   ],
   "source": [
    "N = 400\n",
    "A = 2*np.diag(np.ones(N)) - np.diag(np.ones(N-1),-1) - np.diag(np.ones(N-1),+1)\n",
    "print(A[:5,:5])"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 17,
   "id": "e7bbc7dd-90b4-4a35-9926-7d2f0cfd34f3",
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "6.96 s ± 63.6 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)\n"
     ]
    }
   ],
   "source": [
    "%timeit B = choleski(A)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 18,
   "id": "b551ac14-05ac-48d9-86e3-1ba00cf4d7a1",
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "515 ms ± 42.7 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)\n"
     ]
    }
   ],
   "source": [
    "%timeit B = chol1(A)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 19,
   "id": "1d6743a4-3c7c-449f-9fe4-90d2c103f15e",
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "60.6 ms ± 3.04 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)\n"
     ]
    }
   ],
   "source": [
    "%timeit B = chol2(A)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 64,
   "id": "07e08a35-c57c-43ed-af04-3cec803251e3",
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "[[ 1.41421356  0.          0.         ...  0.          0.\n",
      "   0.        ]\n",
      " [-0.70710678  1.22474487  0.         ...  0.          0.\n",
      "   0.        ]\n",
      " [ 0.         -0.81649658  1.15470054 ...  0.          0.\n",
      "   0.        ]\n",
      " ...\n",
      " [ 0.          0.          0.         ...  1.00125549  0.\n",
      "   0.        ]\n",
      " [ 0.          0.          0.         ... -0.99874608  1.00125235\n",
      "   0.        ]\n",
      " [ 0.          0.          0.         ...  0.         -0.99874922\n",
      "   1.00124922]]\n"
     ]
    }
   ],
   "source": [
    "B = choleski(A)\n",
    "print(B)"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "e971acbf-213f-49d9-a934-8dc77b717e4d",
   "metadata": {
    "tags": []
   },
   "source": [
    "### 2. Méthode de Newton-Raphson (méthode de Newton, cas vectoriel)\n",
    "\n",
    "La méthode de Newton-Raphson permet de résoudre numériquement un système d'équations algébriques non linéaires écrit sous forme condensée vectorielle\n",
    "\n",
    "$$\n",
    "\\mathbf{F}(\\mathbf{x}) = \\mathbf{0}.\n",
    "$$\n",
    "\n",
    "La méthode itérative s'écrit\n",
    "\n",
    "$$\n",
    "\\mathbf{x}^{(k+1)} = \\mathbf{x}^{(k)} - [J(\\mathbf{x}^{(k)})]^{-1}\\, \\mathbf{F}(\\mathbf{x}^{(k)})\n",
    "$$\n",
    "\n",
    "en partant de $\\mathbf{x}^{(0)}$ donné (initialisation). La matrice\n",
    "$J(\\mathbf{x}^{(k)})$ est la matrice jacobienne de $\\mathbf{F}$ au point \n",
    "$\\mathbf{x}^{(k)}$.\n",
    "\n",
    "En pratique, on procède en deux étapes: on résout d'abord le système linéaire\n",
    "\n",
    "$$\n",
    "J(\\mathbf{x}^{(k)}) \\, \\mathbf{w}^{(k)} = -\\mathbf{F}(\\mathbf{x}^{(k)}),\n",
    "$$\n",
    "\n",
    "puis on fait l'affectation\n",
    "\n",
    "$$\n",
    "\\mathbf{x}^{(k+1)} = \\mathbf{x}^{(k)} + \\mathbf{w}^{(k)}.\n",
    "$$\n",
    "\n",
    "Ecrire une fonction python\n",
    "```\n",
    "xout, kout, Fout = newtonRaphson(x0, foncjac, kmax, tol)\n",
    "```\n",
    "qui résout l'équation $F(x)=0$ avec la méthode de Newton-Raphson avec ```x0``` l'initial guess, la fonction `python` générique\n",
    "```\n",
    "F, J = foncjac(x)\n",
    "``` \n",
    "étant supposée retourner la fonction $F$ ainsi que sa matrice jacobienne $J(\\mathbf{x})=DF(\\mathbf{x})=(\\partial F_i/\\partial x_j)_{ij}$ au point trouvé, $k_{max}$ est le nombre max d'itérations de Newton,  $tol$ est la tolérance de convergence : on fera le test\n",
    "\n",
    "$$\n",
    "\\frac{\\|\\mathbf{F}(\\mathbf{x}^{(k)})\\|}{\\|\\mathbf{F}(\\mathbf{x}^{(0)})\\|} > tol\n",
    "$$\n",
    "\n",
    "(si oui et si $k<k_{max}$ on continue, sinon on s'arrête).\n",
    "\n",
    "La fonction ```newtonRaphson()``` doit retourner $x_{out}$ la solution trouvée, $k_{out}$ le numéro de dernière itération et $\\mathbf{F}_{out}=\\mathbf{F}(\\mathbf{x}_{out})$ le 'résidu vecteur'.\n",
    "\n",
    "NB : on remarquera que ```python``` autorise le passage de fonctions (ici ```foncjac()```) en arguments de fonctions python. Il s'agit bien de l'objet 'fonction', et non de l'évaluation de la fonction en un point."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 13,
   "id": "ac627cd6-9fe7-4e63-a5b7-ffae92f03876",
   "metadata": {
    "tags": []
   },
   "outputs": [],
   "source": [
    "import numpy as np\n",
    "import numpy.linalg as la\n",
    "\n",
    "def newtonRaphson(x0, foncjac, kmax, tol):\n",
    "    k=0\n",
    "    xk = x0\n",
    "    F, J = foncjac(x0)\n",
    "    res0 = la.norm(F)\n",
    "    res  = 1e15\n",
    "    xk   = x0\n",
    "    while (res/res0 > tol) and (k <= kmax):\n",
    "        Fk, Jk = foncjac(xk)\n",
    "        res = la.norm(Fk)\n",
    "        xk -= la.solve(Jk, Fk)\n",
    "        k += 1\n",
    "    xout = xk\n",
    "    kout = k\n",
    "    Fout = Fk\n",
    "    return xout, kout, Fout"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "25e5ddbf-e125-4f15-8173-8fc19c821ab3",
   "metadata": {},
   "source": [
    "Vérification : écrire la fonction ```foncjac()``` et tester la fonction ```newtonRaphson()``` pour la fonction à 2 variables\n",
    "$x=(x_1,x_2)$ :\n",
    "\n",
    "$$\n",
    "f(x) = \\begin{pmatrix} x_1^2 - 2 \\\\ x_1^4 + x_2^2 - 5 \\end{pmatrix} \n",
    "$$\n",
    "\n",
    "en prenant $x^0=(4,2)$, $k_{max}=16$, $tol=10^{-15}$."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 15,
   "id": "b46104d6-a672-430f-9ffa-50d8da16315a",
   "metadata": {
    "tags": []
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "xout =  [1.41421356 1.        ] \n",
      "kout =  12 \n",
      "Résidu =  [-4.4408921e-16  0.0000000e+00]\n",
      "Erreur = 0.0 4.440892098500626e-16\n"
     ]
    }
   ],
   "source": [
    "# Exemple\n",
    "def foncjac(x):\n",
    "    F = np.array([x[0]**2 - 2.0, x[0]**4 + x[1]**2 - 5.0], dtype=np.float64)\n",
    "    J = np.array([ [2*x[0], 0], [4*x[0]**3, 2*x[1]] ])\n",
    "    return F, J\n",
    "x0 = np.array([4, 2], dtype=np.float64)\n",
    "kmax = 16\n",
    "tol = 1e-15\n",
    "\n",
    "xout, kout, Fout = newtonRaphson(x0, foncjac, kmax, tol)\n",
    "print('xout = ', xout, \"\\nkout = \", kout, \"\\nRésidu = \", Fout)\n",
    "# Verif\n",
    "print('Erreur =', np.sqrt(2) - xout[0], 1.0 - xout[1])"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "3eff893a-0cd2-45df-808b-d68e578eae56",
   "metadata": {},
   "source": [
    "### 3. Problème inverse, identification paramétrique, solutions de systèmes d'équations algébriques, \n",
    "\n",
    "Un modèle simplifié de croissance cellulaire avec saturation donne l'évolution de la densité de cellules $\\rho(t)$ au cours du temps $t$ :\n",
    "$$\n",
    "\\rho(t) = \\rho_M\\, \\frac{e^{\\alpha t}}{\\frac{\\rho_M}{\\rho_0}-1 + e^{\\alpha t}},\n",
    "\\qquad t>0,\n",
    "$$\n",
    "\n",
    "où $\\rho_0$ est la densité initiale, $\\rho_M$ est la densité maximale $(\\rho_M\\gg \\rho_0)$, et $\\alpha$ est le taux de croissance cellulaire dans la phase exponentielle. \n",
    "\n",
    "Tracer la fonction $t\\mapsto \\rho(t)$ sur l'intervalle $[0,10]$ pour les paramètres\n",
    "$(\\rho_0,\\,\\rho_M,\\,\\alpha)=(3, 1000, 1.2)$."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 26,
   "id": "6e475dde-fca2-4864-8c1a-29e685a2842b",
   "metadata": {},
   "outputs": [
    {
     "data": {
      "image/png": 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",
      "text/plain": [
       "<Figure size 640x480 with 1 Axes>"
      ]
     },
     "metadata": {},
     "output_type": "display_data"
    }
   ],
   "source": [
    "def rho(t):\n",
    "    rho0 = 3\n",
    "    rhoM = 1000\n",
    "    alpha = 1.2\n",
    "    y = rhoM * np.exp(alpha*t)/(rhoM/rho0 - 1.0 + np.exp(alpha*t))\n",
    "    return y\n",
    "\n",
    "t = np.linspace(0, 10, 1000)\n",
    "import matplotlib.pyplot as plt\n",
    "plt.plot(t, rho(t), '.-'); plt.grid()"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "c1d12d01-4a45-4259-a237-303f12325903",
   "metadata": {},
   "source": [
    "Les quantités $\\rho_M$ et $\\alpha$ sont d'intérêt biologique, mais ne sont pas des observables (ne peuvent pas être mesurées directement).\n",
    "Expérimentalement, on peut compter le nombre de cellules sur un élément de surface, on peut donc mesurer la densité à un temps de prolifération donné.\n",
    "\n",
    "À partir de trois mesures de densité de cellules à trois instants différents, on peut\n",
    "espérer déterminer les trois paramètres $\\rho_0$, $\\rho_M$ et $\\alpha$ du modèle.\n",
    "\n",
    "Supposons que les trois mesures sont les suivantes :\n",
    "\n",
    "| Temps $t^k$        | Densité $\\rho(t^k)$ | \n",
    "| :---------------- | :------: | \n",
    "| 2.0               |   32.104 | \n",
    "| 4.0               |  267.726 | \n",
    "| 8.0               |  977.987 | \n",
    "\n",
    "À partir des mesures, déterminer les paramètres $\\rho_0$, $\\rho_M$ et $\\alpha$.\n",
    "Pour cela, dans un premier temps, écrivez le système d'équations  $\\mathbf{F}(\\mathbf{x})=0$ à résoudre.\n",
    "\n",
    "En utilisant `scipy.optimize.fsolve()` (voir l'aide associée), résoudre le problème. On pourra utiliser l'initialisation :\n",
    "\n",
    "```\n",
    "X0 = np.array([1.0, 2000.0, 1.0]) # X0=(rho0, rhoM, alpha) # initial guess \n",
    "```\n",
    "\n",
    "Affichez les valeurs de  $\\rho_0$, $\\rho_M$ et $\\alpha$ ainsi obtenues.\n",
    "Arrondir les valeurs à deux décimales en utilisant `np.round(...,2)`.\n"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 30,
   "id": "a1825b00-e204-4dce-88ff-4efbc8e1f6d7",
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "{'nfev': 17, 'fjac': array([[1., 0., 0.],\n",
      "       [0., 1., 0.],\n",
      "       [0., 0., 1.]]), 'r': array([-0.,  0.,  0., -0.,  0.,  0.]), 'qtf': array([ 1.71812045e-04,  1.02358534e-02, -3.59713507e-05]), 'fvec': array([ 1.71812045e-04,  1.02358534e-02, -3.59713507e-05])}\n"
     ]
    }
   ],
   "source": [
    " def F(X):\n",
    "    rho0, rhoM, alpha = X\n",
    "    t1 = 2.0\n",
    "    t2 = 4.0\n",
    "    t3 = 8.0\n",
    "    rho1 = 32.104\n",
    "    rho2 = 267.726\n",
    "    rho3 = 977.987\n",
    "    f1 = rhoM * np.exp(alpha*t1)/(rhoM/rho0 - 1.0 + np.exp(alpha*t1)) - rho1\n",
    "    f2 = rhoM * np.exp(alpha*t2)/(rhoM/rho0 - 1.0 + np.exp(alpha*t2)) - rho2\n",
    "    f3 = rhoM * np.exp(alpha*t3)/(rhoM/rho0 - 1.0 + np.exp(alpha*t3)) - rho3\n",
    "    return np.array([f1,f2,f3])\n",
    "\n",
    "from scipy.optimize import fsolve\n",
    "X0 = np.array([1.0, 2000.0, 1.0])\n",
    "Xsol, infodict, ier, mesg = fsolve(F, X0, full_output=True)\n",
    "print(infodict)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 31,
   "id": "d587d448-8b7c-4bf3-b020-5e243ab73c45",
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "rho0 =  1.0\n",
      "rhoM =  2000.0\n",
      "alpha =  1.0\n"
     ]
    }
   ],
   "source": [
    "rho0, rhoM, alpha = Xsol\n",
    "print('rho0 = ', np.round(rho0,2) )\n",
    "print('rhoM = ', np.round(rhoM,2) )\n",
    "print('alpha = ', np.round(alpha,2) )"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 32,
   "id": "b491b1d9-dd6b-4562-8455-c4faedf3eb75",
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "\u001b[1;31mSignature:\u001b[0m\n",
       "\u001b[0mfsolve\u001b[0m\u001b[1;33m(\u001b[0m\u001b[1;33m\n",
       "\u001b[0m    \u001b[0mfunc\u001b[0m\u001b[1;33m,\u001b[0m\u001b[1;33m\n",
       "\u001b[0m    \u001b[0mx0\u001b[0m\u001b[1;33m,\u001b[0m\u001b[1;33m\n",
       "\u001b[0m    \u001b[0margs\u001b[0m\u001b[1;33m=\u001b[0m\u001b[1;33m(\u001b[0m\u001b[1;33m)\u001b[0m\u001b[1;33m,\u001b[0m\u001b[1;33m\n",
       "\u001b[0m    \u001b[0mfprime\u001b[0m\u001b[1;33m=\u001b[0m\u001b[1;32mNone\u001b[0m\u001b[1;33m,\u001b[0m\u001b[1;33m\n",
       "\u001b[0m    \u001b[0mfull_output\u001b[0m\u001b[1;33m=\u001b[0m\u001b[1;36m0\u001b[0m\u001b[1;33m,\u001b[0m\u001b[1;33m\n",
       "\u001b[0m    \u001b[0mcol_deriv\u001b[0m\u001b[1;33m=\u001b[0m\u001b[1;36m0\u001b[0m\u001b[1;33m,\u001b[0m\u001b[1;33m\n",
       "\u001b[0m    \u001b[0mxtol\u001b[0m\u001b[1;33m=\u001b[0m\u001b[1;36m1.49012e-08\u001b[0m\u001b[1;33m,\u001b[0m\u001b[1;33m\n",
       "\u001b[0m    \u001b[0mmaxfev\u001b[0m\u001b[1;33m=\u001b[0m\u001b[1;36m0\u001b[0m\u001b[1;33m,\u001b[0m\u001b[1;33m\n",
       "\u001b[0m    \u001b[0mband\u001b[0m\u001b[1;33m=\u001b[0m\u001b[1;32mNone\u001b[0m\u001b[1;33m,\u001b[0m\u001b[1;33m\n",
       "\u001b[0m    \u001b[0mepsfcn\u001b[0m\u001b[1;33m=\u001b[0m\u001b[1;32mNone\u001b[0m\u001b[1;33m,\u001b[0m\u001b[1;33m\n",
       "\u001b[0m    \u001b[0mfactor\u001b[0m\u001b[1;33m=\u001b[0m\u001b[1;36m100\u001b[0m\u001b[1;33m,\u001b[0m\u001b[1;33m\n",
       "\u001b[0m    \u001b[0mdiag\u001b[0m\u001b[1;33m=\u001b[0m\u001b[1;32mNone\u001b[0m\u001b[1;33m,\u001b[0m\u001b[1;33m\n",
       "\u001b[0m\u001b[1;33m)\u001b[0m\u001b[1;33m\u001b[0m\u001b[1;33m\u001b[0m\u001b[0m\n",
       "\u001b[1;31mDocstring:\u001b[0m\n",
       "Find the roots of a function.\n",
       "\n",
       "Return the roots of the (non-linear) equations defined by\n",
       "``func(x) = 0`` given a starting estimate.\n",
       "\n",
       "Parameters\n",
       "----------\n",
       "func : callable ``f(x, *args)``\n",
       "    A function that takes at least one (possibly vector) argument,\n",
       "    and returns a value of the same length.\n",
       "x0 : ndarray\n",
       "    The starting estimate for the roots of ``func(x) = 0``.\n",
       "args : tuple, optional\n",
       "    Any extra arguments to `func`.\n",
       "fprime : callable ``f(x, *args)``, optional\n",
       "    A function to compute the Jacobian of `func` with derivatives\n",
       "    across the rows. By default, the Jacobian will be estimated.\n",
       "full_output : bool, optional\n",
       "    If True, return optional outputs.\n",
       "col_deriv : bool, optional\n",
       "    Specify whether the Jacobian function computes derivatives down\n",
       "    the columns (faster, because there is no transpose operation).\n",
       "xtol : float, optional\n",
       "    The calculation will terminate if the relative error between two\n",
       "    consecutive iterates is at most `xtol`.\n",
       "maxfev : int, optional\n",
       "    The maximum number of calls to the function. If zero, then\n",
       "    ``100*(N+1)`` is the maximum where N is the number of elements\n",
       "    in `x0`.\n",
       "band : tuple, optional\n",
       "    If set to a two-sequence containing the number of sub- and\n",
       "    super-diagonals within the band of the Jacobi matrix, the\n",
       "    Jacobi matrix is considered banded (only for ``fprime=None``).\n",
       "epsfcn : float, optional\n",
       "    A suitable step length for the forward-difference\n",
       "    approximation of the Jacobian (for ``fprime=None``). If\n",
       "    `epsfcn` is less than the machine precision, it is assumed\n",
       "    that the relative errors in the functions are of the order of\n",
       "    the machine precision.\n",
       "factor : float, optional\n",
       "    A parameter determining the initial step bound\n",
       "    (``factor * || diag * x||``). Should be in the interval\n",
       "    ``(0.1, 100)``.\n",
       "diag : sequence, optional\n",
       "    N positive entries that serve as a scale factors for the\n",
       "    variables.\n",
       "\n",
       "Returns\n",
       "-------\n",
       "x : ndarray\n",
       "    The solution (or the result of the last iteration for\n",
       "    an unsuccessful call).\n",
       "infodict : dict\n",
       "    A dictionary of optional outputs with the keys:\n",
       "\n",
       "    ``nfev``\n",
       "        number of function calls\n",
       "    ``njev``\n",
       "        number of Jacobian calls\n",
       "    ``fvec``\n",
       "        function evaluated at the output\n",
       "    ``fjac``\n",
       "        the orthogonal matrix, q, produced by the QR\n",
       "        factorization of the final approximate Jacobian\n",
       "        matrix, stored column wise\n",
       "    ``r``\n",
       "        upper triangular matrix produced by QR factorization\n",
       "        of the same matrix\n",
       "    ``qtf``\n",
       "        the vector ``(transpose(q) * fvec)``\n",
       "\n",
       "ier : int\n",
       "    An integer flag.  Set to 1 if a solution was found, otherwise refer\n",
       "    to `mesg` for more information.\n",
       "mesg : str\n",
       "    If no solution is found, `mesg` details the cause of failure.\n",
       "\n",
       "See Also\n",
       "--------\n",
       "root : Interface to root finding algorithms for multivariate\n",
       "       functions. See the ``method='hybr'`` in particular.\n",
       "\n",
       "Notes\n",
       "-----\n",
       "``fsolve`` is a wrapper around MINPACK's hybrd and hybrj algorithms.\n",
       "\n",
       "Examples\n",
       "--------\n",
       "Find a solution to the system of equations:\n",
       "``x0*cos(x1) = 4,  x1*x0 - x1 = 5``.\n",
       "\n",
       ">>> import numpy as np\n",
       ">>> from scipy.optimize import fsolve\n",
       ">>> def func(x):\n",
       "...     return [x[0] * np.cos(x[1]) - 4,\n",
       "...             x[1] * x[0] - x[1] - 5]\n",
       ">>> root = fsolve(func, [1, 1])\n",
       ">>> root\n",
       "array([6.50409711, 0.90841421])\n",
       ">>> np.isclose(func(root), [0.0, 0.0])  # func(root) should be almost 0.0.\n",
       "array([ True,  True])\n",
       "\u001b[1;31mFile:\u001b[0m      c:\\users\\fdevu\\anaconda3\\lib\\site-packages\\scipy\\optimize\\_minpack_py.py\n",
       "\u001b[1;31mType:\u001b[0m      function"
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