From ced984cebd1d3f613aa764e14c3cb59300a6aabb Mon Sep 17 00:00:00 2001
From: marcellus <msimon_fr@hotmail.com>
Date: Sun, 11 May 2025 20:31:18 +0200
Subject: update: exo PCVM

---
 PVCM/cama.md                                       |   3 +-
 .../fr/ma26 Vecteurs propres -- Exercices.ipynb    | 146 +++++++++++++++------
 2 files changed, 105 insertions(+), 44 deletions(-)

(limited to 'PVCM')

diff --git a/PVCM/cama.md b/PVCM/cama.md
index 21e49cf..ce46dac 100755
--- a/PVCM/cama.md
+++ b/PVCM/cama.md
@@ -288,8 +288,7 @@ Une autre vidéo, avec un peu plus d'exemple est qui est tout aussi bien:
 https://www.youtube.com/watch?v=pkVwUVEHmfI&t=785s
 
 La man page de help est aussi très complète.
-La manière de faire la plus simple reste d'écrire un pseudo code avec des boucles for,
-puis de re-écrire avec la notation Einstein.
+La manière de faire la plus simple reste d'écrire un pseudo code avec des boucles for, puis de re-écrire avec la notation Einstein.
 
 #### Formule essentiels + exemple
 
diff --git a/PVCM/cama/fr/ma26 Vecteurs propres -- Exercices.ipynb b/PVCM/cama/fr/ma26 Vecteurs propres -- Exercices.ipynb
index 88d82df..c62c0b1 100644
--- a/PVCM/cama/fr/ma26 Vecteurs propres -- Exercices.ipynb	
+++ b/PVCM/cama/fr/ma26 Vecteurs propres -- Exercices.ipynb	
@@ -1,25 +1,4 @@
 {
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python 3 (ipykernel)",
-   "language": "python",
-   "name": "python3"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.12"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 4,
  "cells": [
   {
    "cell_type": "markdown",
@@ -65,14 +44,14 @@
    "source": [
     "Comme on a 2 termes à droite du signe égal, écrivons Fibonnacci sous forme d'un système matriciel $2\\times 2$ :\n",
     "\n",
-    "$\n",
+    "$$\n",
     "\\begin{align}\n",
     "x_{n-1} &= x_{n-1} \\\\\n",
     "x_n &= x_{n-2} + x_{n-1} \\\\\n",
     "\\end{align}\n",
-    "$\n",
+    "$$\n",
     "ce qui donne\n",
-    "$\n",
+    "$$\n",
     "\\begin{bmatrix}\n",
     "x_{n-1}\\\\\n",
     "x_n  \\\\\n",
@@ -86,7 +65,7 @@
     "x_{n-2}\\\\\n",
     "x_{n-1}  \\\\\n",
     "\\end{bmatrix}\n",
-    "$\n",
+    "$$\n",
     "\n",
     "Écrire la matrice F ci-dessous et vérifier que le 6e élément de la suite de Fibonacci est 8 en n'utilisant que \n",
     "la matrice F et les valeurs initiales de la suite. La fonction `numpy.linalg.matrix_power` pourra vous être utile."
@@ -94,17 +73,47 @@
   },
   {
    "cell_type": "code",
-   "execution_count": null,
+   "execution_count": 3,
    "metadata": {},
-   "outputs": [],
-   "source": []
+   "outputs": [
+    {
+     "data": {
+      "text/plain": [
+       "array([[0, 1],\n",
+       "       [1, 1]])"
+      ]
+     },
+     "execution_count": 3,
+     "metadata": {},
+     "output_type": "execute_result"
+    }
+   ],
+   "source": [
+    "F = np.array([[0,1],[1,1]])\n",
+    "F"
+   ]
   },
   {
    "cell_type": "code",
-   "execution_count": null,
+   "execution_count": 6,
    "metadata": {},
-   "outputs": [],
-   "source": []
+   "outputs": [
+    {
+     "data": {
+      "text/plain": [
+       "array([[ 5,  8],\n",
+       "       [ 8, 13]])"
+      ]
+     },
+     "execution_count": 6,
+     "metadata": {},
+     "output_type": "execute_result"
+    }
+   ],
+   "source": [
+    "F6 = lin.matrix_power(F,6)\n",
+    "F6"
+   ]
   },
   {
    "cell_type": "markdown",
@@ -125,7 +134,9 @@
   {
    "cell_type": "markdown",
    "metadata": {},
-   "source": []
+   "source": [
+    "$F^n = F \\times F^{n-1} = P\\times D^n \\times P^{-1}$"
+   ]
   },
   {
    "cell_type": "markdown",
@@ -136,17 +147,35 @@
   },
   {
    "cell_type": "code",
-   "execution_count": null,
+   "execution_count": 47,
    "metadata": {},
    "outputs": [],
-   "source": []
+   "source": [
+    "def fibo(n):\n",
+    "    F = np.array([[0,1],[1,1]], dtype=int)\n",
+    "    D, P = lin.eig(F)\n",
+    "    return (P @ lin.matrix_power(np.diag(D),n) @ lin.inv(P))[0]"
+   ]
   },
   {
    "cell_type": "code",
-   "execution_count": null,
+   "execution_count": 48,
    "metadata": {},
-   "outputs": [],
-   "source": []
+   "outputs": [
+    {
+     "data": {
+      "text/plain": [
+       "array([5., 8.])"
+      ]
+     },
+     "execution_count": 48,
+     "metadata": {},
+     "output_type": "execute_result"
+    }
+   ],
+   "source": [
+    "fibo(6)"
+   ]
   },
   {
    "cell_type": "markdown",
@@ -158,10 +187,22 @@
   },
   {
    "cell_type": "code",
-   "execution_count": null,
+   "execution_count": 51,
    "metadata": {},
-   "outputs": [],
-   "source": []
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "84.8 μs ± 7.75 μs per loop (mean ± std. dev. of 7 runs, 10,000 loops each)\n",
+      "109 μs ± 6.71 μs per loop (mean ± std. dev. of 7 runs, 10,000 loops each)\n"
+     ]
+    }
+   ],
+   "source": [
+    "%timeit fibo(5)\n",
+    "%timeit fibo(1000)"
+   ]
   },
   {
    "cell_type": "code",
@@ -301,5 +342,26 @@
    "outputs": [],
    "source": []
   }
- ]
-}
\ No newline at end of file
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python 3",
+   "language": "python",
+   "name": "python3"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 3
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython3",
+   "version": "3.13.2"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 4
+}
-- 
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