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March 31, 2020 15:22
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UNTreF - Análisis Numérico
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| { | |
| "cells": [ | |
| { | |
| "cell_type": "markdown", | |
| "metadata": {}, | |
| "source": [ | |
| "# Análisis Numérico\n", | |
| "\n", | |
| "**Nota final**: 50% (EF) + 30% (EP) + 20% (NC)\n", | |
| "\n", | |
| "- **NC**: nota de concepto. Viene de los TP's que hay que entregar durante el cuatrimestre (en fecha, calidad, prolijidad)\n", | |
| "- **EP**: Examen parcial (quizás domiciliario)\n", | |
| "- **EF**: Examen final: Entrega de un TP final.\n", | |
| "\n", | |
| "Para aprobar la materia.\n", | |
| "\n", | |
| "- Todos los TP's aprobadas.\n", | |
| "- EP aprobado (nota $\\geq$ 4).\n", | |
| "- EF aprobado (nota $\\geq$ 4).\n", | |
| "- Nota final $\\geq$ 4." | |
| ] | |
| }, | |
| { | |
| "cell_type": "markdown", | |
| "metadata": {}, | |
| "source": [ | |
| "$\n", | |
| "\\require{cancel}\n", | |
| "\\newcommand{\\annoterel}[3][]{\\underset{\\substack{#1\\uparrow\\\\#2}}{#3}}\n", | |
| "$\n", | |
| "# Clase 1\n", | |
| "\n", | |
| "## Errores de redondeo\n", | |
| "\n", | |
| "La mayoría de las computadoras trabajan en **sistema binario** (base 2) mientras que los humanos trabajos em **sistema decimal** (base 10).\n", | |
| "\n", | |
| "Vamos a repasar estos dos sistemas y a ver como pasar de uno a otro:\n", | |
| "\n", | |
| "Ejemplo:\n", | |
| "\n", | |
| "$(427,325)_{10} = 4 \\cdot 10^2 + 2 \\cdot 10^1 + 7 \\cdot 10^0 + 3 \\cdot 10^{-1} + 2 \\cdot 10^{-2} + 5 \\cdot 10^{-2}$\n", | |
| "\n", | |
| "Análogamente:\n", | |
| "\n", | |
| "$\n", | |
| "\\begin{align}\n", | |
| "(1010.11101)_2 & = 1 \\cdot 2^3 + 0 \\cdot 2^2 + 1 \\cdot 2^1 + 0 \\cdot 2^0 + 1 \\cdot 2^{-1} + 1 \\cdot 2^{-2} + 1 \\cdot20^{-3} + 0 \\cdot 2^{-4} + 1 \\cdot 2^{-5} \\\\\n", | |
| "& = 8 + 0 + 2 + 0 + \\dfrac{1}{2} + \\dfrac{1}{4} + \\dfrac{1}{8} + 0 + \\dfrac{1}{32} \\\\\n", | |
| "& = 10.90625 = (10.90625)_{10}\n", | |
| "\\end{align}\n", | |
| "$\n", | |
| "\n", | |
| "¿Cómo paso de base 10 a base 2?\n", | |
| "\n", | |
| "Dos pasos:\n", | |
| "\n", | |
| "1. Parte entera (10 en este caso)\n", | |
| "2. Parte decimal (0.90625 en este caso)\n", | |
| "\n", | |
| "**Parte entera**: Algoritmo (división entera)\n", | |
| "\n", | |
| "$\n", | |
| "\\begin{align}\n", | |
| "10/2 & = 5 \\to resto = 0 \\\\\n", | |
| "5/2 & = 2 \\to resto = 1 \\\\ \n", | |
| "2/1 & = 1 \\to resto = 0 \\\\\n", | |
| "1/2 & = 0 \\to resto = 1 \\uparrow (1010)_2\n", | |
| "\\end{align}\n", | |
| "$\n", | |
| "\n", | |
| "$\n", | |
| "\\begin{align}\n", | |
| "10 & = 8 + 2 \\\\\n", | |
| "& = 2^3 + 2^1 \\\\ \n", | |
| "& = (1010)_2\n", | |
| "\\end{align}\n", | |
| "$\n", | |
| "\n", | |
| "**Parte decimal**: Algotirmo, multiplicar por 2 y ver si es major a 1 o no.\n", | |
| "\n", | |
| "$\n", | |
| "\\begin{align}\n", | |
| "0.90625 \\cdot 2 & = 1.8125 & \\geq 1 & \\to 1 \\downarrow (0.11101)_2 \\\\\n", | |
| "0.8125 \\cdot 2 & = 1.625 & \\geq 1 & \\to 1 \\\\\n", | |
| "0.625 \\cdot 2 & = 1.25 & \\geq 1 & \\to 1 \\\\\n", | |
| "0.25 \\cdot 2 & = 0.5 & \\lt 1 & \\to 0 \\\\\n", | |
| "0.5 \\cdot 2 & = 1.0 & \\geq 1 & \\to 1\n", | |
| "\\end{align}\n", | |
| "$\n", | |
| "\n", | |
| "Luego,\n", | |
| "\n", | |
| "$\n", | |
| "\\begin{align}\n", | |
| "0.90625 & = 2^{-1} \\cdot (1 + 0.8125) \\\\\n", | |
| " & = 2^{-1} \\cdot (1 + 2^{-1} \\cdot (1 + 0.625)) \\\\\n", | |
| " & = 2^{-1} \\cdot (1 + 2^{-1} \\cdot (1 + 2^{-1} \\cdot (1 + 0.25))) \\\\\n", | |
| " & = 2^{-1} \\cdot (1 + 2^{-1} \\cdot (1 + 2^{-1} \\cdot (1 + 2^{-1} \\cdot (0 + 0.5)))) \\\\\n", | |
| " & = 2^{-1} \\cdot (1 + 2^{-1} \\cdot (1 + 2^{-1} \\cdot (1 + 2^{-1} \\cdot (0 + 2^{-1} \\cdot (1))))) \\\\\n", | |
| " & = 2^{-1} \\cdot (1 + 2^{-1} \\cdot (1 + 2^{-1} \\cdot (1 + 0 \\cdot 2^{-1} + 1 \\cdot 2^{-2}))) \\\\\n", | |
| " & = 2^{-1} \\cdot (1 + 2^{-1} \\cdot (1 + 1 \\cdot 2^{-1} + 0 \\cdot 2^{-2} + 1 \\cdot 2^{-3})) \\\\\n", | |
| " & = 1 \\cdot 2^{-1} + 1 \\cdot 2^{-2} + 1 \\cdot 2^{-3} + 0 \\cdot 2^{-4} + 1 \\cdot 2^{-5} \\\\\n", | |
| " & = (0.11101)_2\n", | |
| "\\end{align}\n", | |
| "$\n", | |
| "\n", | |
| "### Notación científica normalizada\n", | |
| "\n", | |
| "**Base 10**: Hay que multiplicar por una potencia de 10 (o sea $10^{(\\_)}$) para que todos los números distintos de cero queden a la derecha de la coma y además el primer número después de la coma sea distinto de cero.\n", | |
| "\n", | |
| "Ejemplo: $732.5051 = 0.7325051 \\cdot 10^3$\n", | |
| "\n", | |
| "Claramente, el número tiene que ser distinto de cero ($0$) (para poder escribirlo de esta manera).\n", | |
| "\n", | |
| "Del lado derecho, el número está escrito en notación científica normalizada.\n", | |
| "\n", | |
| "Otro ejemplo: $-0.005612 = -0.5612 \\cdot 10^{-2}$\n", | |
| "\n", | |
| "**En general**: Si $x \\in \\mathbb{R}$, $x \\neq 0$ siempre puedo escribir a $x$ de la forma:\n", | |
| "\n", | |
| "$x = \\pm r \\cdot 10^m$ con $\\dfrac{1}{10} \\leq r \\lt 1$ y $m$ un número **entero** ($m \\in \\mathbb{Z}$)\n", | |
| "\n", | |
| "**Base 2**: con el mismo razonamiento (escribiendo el número en base 2 y multiplicando por $2^{(\\_)}$ una potencia de 2)\n", | |
| "\n", | |
| "Si $x \\in \\mathbb{R}$ y $x \\neq 0$ siempre puedo escribir:\n", | |
| "\n", | |
| "$x = \\pm q \\cdot 2^m$, $\\dfrac{1}{2} \\leq q \\lt 1$, $m \\in \\mathbb{Z}$, $q$ en base 2\n", | |
| "\n", | |
| "**Referencias**:\n", | |
| "\n", | |
| "- $q$: mantisa\n", | |
| "- $m$: exponente\n", | |
| "- $x$: número flotante normalizado\n", | |
| "\n", | |
| "**Observaciones**: en una computadora binaria los números se representan en base 2, con restricciones en $q$ y en $m$.\n", | |
| "\n", | |
| "Ejemplo de computadora: 32 bits (digitos binarios).\n", | |
| "\n", | |
| "Sea $x$ un _número flotante normalizado_. $x \\neq 0$. Los 32 bits son alocados de la siguiente manera para representar a $x$\n", | |
| "\n", | |
| "- 1 bit: signo de $x$\n", | |
| "- 1 bit: signo del exponente ($m$)\n", | |
| "- 7 bits: exponente ($|m|$ número natural)\n", | |
| "- 23 bits: mantisa ($|q| \\in \\mathbb{R}$)\n", | |
| "\n", | |
| "**Observaciones**:\n", | |
| "\n", | |
| "- Sabemos que el primer digito de $q$ es un $1$, entonces los 23 bits son guardados para los digitos 2do, 3ro, ..., 24avo de $q$. O sea, es como tener 24 bits para la mantisa.\n", | |
| "- Si $|m|$ ocupa 7 bits y $|q|$ 24 bits, entonces es un _número de máquina_ y puede ser representado por la computadora de manera exacta.\n", | |
| "- Si tenemos un `INPUT` que **no** es un número de máquina, entonces vamos a tener un error al representar este número por un número de máquina\n", | |
| "- $\n", | |
| "\\begin{align}\n", | |
| "|m| \\leq (1111111)_2 & = 2^0 + 2^1 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 \\\\\n", | |
| "& = 2^7 - 1 = 127\n", | |
| "\\end{align}\n", | |
| "$ \n", | |
| "\n", | |
| " pues:\n", | |
| "\n", | |
| " $\n", | |
| "\\begin{align}\n", | |
| "1 + 2^0 + 2^1 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 & = \\\\\n", | |
| "(2 \\cdot 2^0) + 2^1 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 & = \\\\\n", | |
| "2^1 + 2^1 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 & = \\\\\n", | |
| "(2 \\cdot 2^1) + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 & = \\\\\n", | |
| "2^2 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 & = \\\\\n", | |
| "(2 \\cdot 2^2) + 2^3 + 2^4 + 2^5 + 2^6 & = \\\\\n", | |
| "2^3 + 2^3 + 2^4 + 2^5 + 2^6 & = \\\\\n", | |
| "(2 \\cdot 2^3) + 2^4 + 2^5 + 2^6 & = \\\\\n", | |
| "2^4 + 2^4 + 2^5 + 2^6 & = \\\\\n", | |
| "(2 \\cdot 2^4) + 2^5 + 2^6 & = \\\\\n", | |
| "2^5 + 2^5 + 2^6 & = \\\\\n", | |
| "(2 \\cdot 2^5) + 2^6 & = \\\\\n", | |
| "2^6 + 2^6 & = \\\\\n", | |
| "(2 \\cdot 2^6) & = 2^7\n", | |
| "\\end{align}\n", | |
| "$ \n", | |
| "\n", | |
| "- Esto puede ser suficiente para algunos calculos científicos $\\implies$ doble precisión.\n", | |
| "\n", | |
| "### Distribución de Números Flotantes de Máquina (en la recta de números reales)\n", | |
| "\n", | |
| "Misma cantidad de números entre dos potencias de 2.\n", | |
| "\n", | |
| "Ejemplo: entre $2^{-1}$ y $2^0$ hay igual cantidad de números que entre $2^0$ y $2^1$.\n", | |
| "\n", | |
| "Luego, hay una distribución **NO** uniforme (en $\\mathbb{R}$, entre dos potencias de 2 sí), con más densidad cerca del $0$.\n", | |
| "\n", | |
| "### Representación de números enteros\n", | |
| "\n", | |
| "En una máquina de 32 bits, solo guardo 1 bit para el signo.\n", | |
| "\n", | |
| "$\n", | |
| "\\annoterel{\\text{1 bit}\\\\\\text{signo}}{\\pm}\n", | |
| "\\annoterel[\\Big]{\\text{31 lugares}}{(11\\dots11)_2}\n", | |
| "$, de $-(2^{31} - 1)$ a $(2^{31} - 1)$\n", | |
| "\n", | |
| "### Error en la representación de números flotantes\n", | |
| "\n", | |
| "Supongamos $x \\gt 0$ (si $x \\lt 0$, se hace de forma análoga), y que tenemos 32 bits.\n", | |
| "\n", | |
| "$x = q \\cdot 2^m$; $\\dfrac{1}{2} \\leq q \\lt 1$ y $\\annoterel{\\text{x es un número}\\\\\\text{de maquina}}{|m| \\lt 127}$\n", | |
| "\n", | |
| "¿Cuál es el número de máquina más cercano?\n", | |
| "\n", | |
| "$x = (0.a_1a_2\\dots a_{24}a_{25}a_{26}\\dots)_2 \\cdot 2^m$\n", | |
| "\n", | |
| "```\n", | |
| "<----------|----|----|----------> R\n", | |
| " x' x x''\n", | |
| "```\n", | |
| "\n", | |
| "Llamo $x'$ al número de la izquierda más cercano y $x''$ al de la derecha más cercano.\n", | |
| "\n", | |
| "Es claro que:\n", | |
| "\n", | |
| "$x' = (0.a_1a_2\\dots a_{24})_2 \\cdot 2^m$\n", | |
| "\n", | |
| "$x' = ((0.a_1a_2\\dots a_{24})_2 + 2^{-24})\\cdot 2^m$\n", | |
| "\n", | |
| "**Observaciones**:\n", | |
| "\n", | |
| "$x'' - x' = 2^{-24} \\cdot 2^m$\n", | |
| "\n", | |
| "Miro $x - x'$ y $x'' - x$ y me quedo con el $x'$/$x''$ que me dé más chica la cuenta.\n", | |
| "\n", | |
| "**Observaciones**:\n", | |
| "\n", | |
| "Si el más cercanoes $x'$ entonces:\n", | |
| "\n", | |
| "$|x - x'| \\leq \\dfrac{1}{2} \\cdot |x'' - x'| = \\dfrac{1}{2} \\cdot 2^{-24} \\cdot 2^m = 2^{-25}$\n", | |
| "\n", | |
| "**Nombres**:\n", | |
| "\n", | |
| "- $|x - x'|$ es el `ERROR`\n", | |
| "- $\\dfrac{|x - x'|}{|x|}$ es el `ERROR RELATIVO`\n", | |
| "\n", | |
| "Además:\n", | |
| "\n", | |
| "$\\dfrac{|x - x'|}{|x|} \\leq \\annoterel{q \\geq \\frac{1}{2}}{\\dfrac{2^{n - 25}}{q \\cdot 2^m}} \\leq \\dfrac{2^{-25}}{2^{-1}} = 2^{-24}$\n", | |
| "\n", | |
| "- Si el más cercado es $x''$ hago lo mismo. ¿Sí $x$ está fuera del rango de la computadora?\n", | |
| "- Si $m \\gt 0$ y $|m| > 127$ $\\implies$ `OVERFLOW` (en muchas computadoras `NaN`).\n", | |
| "- Si $m \\lt 0$ y $|m| > 127$ $\\implies$ `UNDERFLOW` (en general devuelve $0$ y sigue...).\n", | |
| "\n", | |
| "**Resumen**:\n", | |
| "\n", | |
| "Si $x \\neq 0$ número flotane en el rango de la máquina. Llamamos y notamos:\n", | |
| "\n", | |
| "$x* = fl(x)$; al número de máquina más cercano\n", | |
| "\n", | |
| "**Observaciones**:\n", | |
| "\n", | |
| "$x* = x \\cdot (1 + \\delta)$; con $\\delta = \\dfrac{x* - x}{x}$\n", | |
| "\n", | |
| "y vimos que $|\\delta| \\leq 2^{-24}$; (en 32 bits).\n", | |
| "\n", | |
| "**Nombres**:\n", | |
| "\n", | |
| "$2^{-24}$ se llama **error de redondeo** de la unidad o épsilon de máquina.\n", | |
| "\n", | |
| "Ejemplo: calcular $x*$ en 32 bits para $x = \\dfrac{1}{10}$.\n", | |
| "\n", | |
| "$\n", | |
| "\\begin{align}\n", | |
| "\\dfrac{1}{10} & = (0.0001100110011001100110011001100110011001100110011001101)_2 \\\\\n", | |
| "& = (0.1100110011001100110011001100110011001100110011001101)_2 \\cdot 2^{-3}\n", | |
| "\\end{align}\n", | |
| "$\n", | |
| "\n", | |
| "$\n", | |
| "\\begin{align}\n", | |
| "x' & = (0.\\annoterel{\\text{24 lugares}}{110011001100110011001100})_2 \\cdot 2^{-3}\n", | |
| "\\end{align}\n", | |
| "$\n", | |
| "\n", | |
| "$\n", | |
| "\\begin{align}\n", | |
| "x'' & = (0.110011001100110011001101)_2 \\cdot 2^{-3}\n", | |
| "\\end{align}\n", | |
| "$\n", | |
| "\n", | |
| "$\n", | |
| "\\begin{align}\n", | |
| "x - x' & = (0.\\annoterel{\\text{24 lugares}}{000000000000000000000000}~1100110011001100110011001101)_2 \\cdot 2^{-3} \\\\\n", | |
| "& = (0.1100110011001100110011001101)_2 \\cdot 2^{-24} \\cdot 2^{-3} \\\\\n", | |
| "& = \\boxed{\\dfrac{1}{10} \\cdot 2^{-24}}\n", | |
| "\\end{align}\n", | |
| "$\n", | |
| "\n", | |
| "$\n", | |
| "\\begin{align}\n", | |
| "x'' - x & = (x'' - x') - (x - x') \\\\\n", | |
| "& = 2^{-24} \\cdot 2^{-3} - \\dfrac{1}{10} \\cdot 2^{-24} \\\\\n", | |
| "& = (\\dfrac{1}{8} - \\dfrac{1}{10}) \\cdot 2^{-24} \\\\\n", | |
| "& = \\boxed{\\dfrac{1}{40} \\cdot 2^{-24}}\n", | |
| "\\end{align}\n", | |
| "$ \n", | |
| "\n", | |
| "Luego, $x* = x''$\n", | |
| "\n", | |
| "### Operaciones\n", | |
| "\n", | |
| "Vamos a asumir que la computadora opera con exactitud, luego normaliza y luego redondea.\n", | |
| "\n", | |
| "- Si $x$, $y$ son números de máquina:\n", | |
| "\n", | |
| " $fl(x \\odot y) = [x \\odot y] \\cdot (1 + \\delta)$ con $|\\delta| \\leq \\epsilon$\n", | |
| " \n", | |
| " $\\odot = +, -, \\times, \\div$. $\\epsilon$ (el épsilon de la máquina).\n", | |
| " \n", | |
| "- Si $x$, $y$ **no** son números de máquina\n", | |
| "\n", | |
| " $fl(fl(x) \\odot fl(y)) = [(x \\cdot (1 + \\delta_1)) \\odot (y \\cdot (1 + \\delta_2))] \\cdot (1 + \\delta_3)$ con $|\\delta_1| \\cdot |\\delta_2| \\cdot |\\delta_3| = \\epsilon$\n", | |
| "\n", | |
| "Ejemplo: $x$, $y$, $z$ números de máquina.\n", | |
| "\n", | |
| "$$\n", | |
| "\\begin{align}\n", | |
| "fl(x \\cdot (y + z)) & = x \\cdot fl(y + z) \\cdot (1 + \\delta_1) && |\\delta_1| \\leq \\epsilon \\\\\n", | |
| "& = x \\cdot (y + z) \\cdot (1 + \\delta_2) \\cdot (1 + \\delta_1) && |\\delta_2| \\leq \\epsilon \\\\\n", | |
| "& = x \\cdot (y + z) \\cdot (1 + \\delta_1 + \\delta_2 + \\delta_2\\delta_1) && \\delta_2\\delta_1 \\lt\\lt \\delta_2,\\delta_1 \\text{ lo podemos despreciar} \\\\\n", | |
| "& \\approx x \\cdot (y + z) \\cdot (1 + \\delta_1 + \\delta_2) && \\delta_1 + \\delta_2 = \\delta_3 \\text{ y } |\\delta_3|\\leq 2 \\cdot \\epsilon\n", | |
| "\\end{align}\n", | |
| "$$\n", | |
| "\n", | |
| "**Observaciones**: en una máquina de 32 bits no siempre se puede hacer con exactitud $x \\odot y$. Pero se pueden usar más bits que los que se usan para los números de máquina.\n", | |
| "\n", | |
| "**Guard-bits**: temporalmente y con más presición\n", | |
| "\n", | |
| "### Algoritmo para estimar él número de máquina\n", | |
| "\n", | |
| "**Observaciones**: si $fl(1 + \\epsilon) = 1 \\implies \\dfrac{\\epsilon}{2} \\leq \\epsilon_M$ (y vamos a tomar $\\epsilon_M \\approx \\dfrac{\\epsilon}{2}$)\n", | |
| "\n", | |
| "**Desarrollo**:\n", | |
| "\n", | |
| "$$\n", | |
| "\\begin{align}\n", | |
| "fl(1 + \\epsilon) = (1 + \\epsilon) \\cdot (1 + \\delta) & = 1 && \\text{con } |\\delta| \\leq \\epsilon_M \\\\\n", | |
| "1 + \\delta + \\epsilon + \\epsilon\\delta & = 1 \\\\\n", | |
| "\\delta \\cdot (1 + \\epsilon) & = -\\epsilon \\\\\n", | |
| "\\delta & = \\dfrac{-\\epsilon}{(1 + \\epsilon)}\n", | |
| "\\end{align}\n", | |
| "$$\n", | |
| "\n", | |
| "luego: $|\\delta| = \\dfrac{-\\epsilon}{(1 + \\epsilon)} \\annoterel{\\text{vamos a}\\\\\\text{tomar }\\epsilon \\leq 1}{\\geq} \\dfrac{\\epsilon}{2}$\n", | |
| "\n", | |
| "$\\implies \\dfrac{\\epsilon}{2} \\leq |\\delta| \\leq \\epsilon_M$\n", | |
| "\n", | |
| "$\\implies \\boxed{\\dfrac{\\epsilon}{2} \\leq \\epsilon_M}$\n", | |
| "\n", | |
| "Tomo $\\epsilon = 2^{-1}, 2^{-2}, \\dots$ hasta que se verifique $fl(1 + \\epsilon) = 1$" | |
| ] | |
| }, | |
| { | |
| "cell_type": "code", | |
| "execution_count": 1, | |
| "metadata": {}, | |
| "outputs": [ | |
| { | |
| "name": "stdout", | |
| "output_type": "stream", | |
| "text": [ | |
| "e_M: 5.551115123125783e-17\n" | |
| ] | |
| } | |
| ], | |
| "source": [ | |
| "e = 2**(-1)\n", | |
| "\n", | |
| "while True:\n", | |
| " if (1 + e) == 1:\n", | |
| " print('e_M:', e / 2)\n", | |
| " break\n", | |
| " else:\n", | |
| " e = e / 2" | |
| ] | |
| }, | |
| { | |
| "cell_type": "markdown", | |
| "metadata": {}, | |
| "source": [ | |
| "### Errores evitables\n", | |
| "\n", | |
| "El error de redondero es un error inevitable, pero hay otros errores que sí son evitables.\n", | |
| "\n", | |
| "Ejemplo: computadora decimal con 5 digitos de mantisa.\n", | |
| "\n", | |
| "$\n", | |
| "\\begin{align}\n", | |
| "x & = 0.3721478693 && \\to fl(x) = 0.37215 \\\\\n", | |
| "y & = 0.3720230572 && \\to fl(y) = 0.37202 \\\\\n", | |
| "\\implies x - y & = 0.000124812 && \\to fl(x) - fl(y) = 0.00013\n", | |
| "\\end{align}\n", | |
| "$\n", | |
| "\n", | |
| "$\\left|\\dfrac{(x - y) - (fl(x) - fl(y))}{x - y}\\right| = \\dfrac{0.000005188}{0.000124812} = 0.041\\dots$\n", | |
| "\n", | |
| "$\\implies$ error relativo $\\approx 4\\%$\n", | |
| "\n", | |
| "**Regla**: evitar (en lo posible) restar dos números parecidos.\n", | |
| "\n", | |
| "Ejemplo: $y = \\underbrace{\\sqrt{x^2 + 1}}_{A} - \\underbrace{1}_{B}$\n", | |
| "\n", | |
| "Si calculo $A$, después $B$, y después resto genera grandes errores si $x \\sim 0$\n", | |
| "\n", | |
| "Otra forma:\n", | |
| "\n", | |
| "$$\n", | |
| "\\left(\\sqrt{x^2 + 1} - 1\\right) \\cdot \\dfrac{\\left(\\sqrt{x^2 + 1} + 1\\right)}{\\left(\\sqrt{x^2 + 1} + 1\\right)} = \\dfrac{x^2 + \\cancel 1 - \\cancel 1}{\\sqrt{x^2 + 1} + 1} = \\dfrac{\\overbrace{x^2}^{A}}{\\underbrace{\\sqrt{x^2 + 1} + 1}_{B}}\n", | |
| "$$\n", | |
| "\n", | |
| "Calculo $A$, después $B$, después divido:\n", | |
| "\n", | |
| "### Procesos estables e inestables\n", | |
| "\n", | |
| "Ejemplo: calcular la siguiente sucesión recursiva.\n", | |
| "\n", | |
| "$$\n", | |
| "\\begin{cases}\n", | |
| "x_0 = 1 \\\\\n", | |
| "x_1 = \\dfrac{1}{3} \\\\\n", | |
| "x_{n+1} = \\dfrac{13}{3} \\cdot x_n - \\dfrac{4}{3} \\cdot x_{n - 1}\n", | |
| "\\end{cases}\n", | |
| "$$\n", | |
| "\n", | |
| "Se puede ver que $x_n = \\left(\\dfrac{1}{3}\\right)^n$\n", | |
| "\n", | |
| "Calcular la sucesión de manera recursiva es de por sí inestable (arrastro el error de los pasos anteriores) y además amplía el error por $\\frac{13}{3}$!!\n", | |
| "\n", | |
| "Mejor, ver que vale $x_n = \\left(\\dfrac{1}{3}\\right)^n$\n", | |
| "\n", | |
| "Se demuestra de manera inductiva:\n", | |
| "\n", | |
| "- Vale para $n = 0$ y $n = 1$\n", | |
| "- $n \\implies n + 1$, si vale $\\forall n \\lt n + 1$ entonces también vale para $n + 1$.\n", | |
| "\n", | |
| "$$\n", | |
| "\\begin{align}\n", | |
| "x_{n + 1} & = \\dfrac{13}{3} \\cdot \\left(\\dfrac{1}{3}\\right)^n - \\dfrac{4}{3} \\cdot \\left(\\dfrac{1}{3}\\right)^{n -1} \\\\\n", | |
| "& = \\left(\\dfrac{1}{3}\\right)^{n - 1} \\cdot \\left[\\dfrac{13}{3} \\cdot \\dfrac{1}{3} - \\dfrac{4}{3}\\right] \\\\\n", | |
| "& = \\left(\\dfrac{1}{3}\\right)^{n - 1} \\cdot \\dfrac{1}{3} \\cdot \\left[\\dfrac{13}{3} - \\dfrac{4}{3}\\right] \\\\\n", | |
| "& = \\left(\\dfrac{1}{3}\\right)^n \\cdot \\left(\\dfrac{13 - 12}{3}\\right) \\\\\n", | |
| "& = \\left(\\dfrac{1}{3}\\right)^n \\cdot \\left(\\dfrac{1}{3}\\right) \\\\\n", | |
| "& = \\boxed{\\left(\\dfrac{1}{3}\\right)^{n + 1}}\n", | |
| "\\end{align}\n", | |
| "$$\n", | |
| "\n", | |
| "¿Cómo sé que $x_n = \\left(\\dfrac{1}{3}\\right)^n$?\n", | |
| "\n", | |
| "Me fijo si existen $A, B, \\lambda, \\mu \\in \\mathbb{R}$ que satisfagan:\n", | |
| "\n", | |
| "$$x_n = A \\cdot \\lambda^n + B \\cdot \\mu^n \\tag{*}\\label{*}$$\n", | |
| "\n", | |
| "Y de la ecuación de recurrencia despejo $\\lambda$, $\\mu$, (luego de los datos $x_0$, $x_1$, voy a sacar $A$ y $B$).\n", | |
| "\n", | |
| "Si supongo que vale $\\eqref{*}$, entonces, en la fórmula de recurrencia:\n", | |
| "\n", | |
| "$$\n", | |
| "\\begin{align}\n", | |
| "A \\cdot \\lambda^{n + 1} + B \\cdot \\mu^{n + 1} = \\left(A \\cdot \\lambda^n + B \\cdot \\mu^n\\right) - \\dfrac{4}{3} \\cdot \\left(A \\cdot \\lambda^{n - 1} + B \\cdot \\mu^{n - 1}\\right) \\\\\n", | |
| "A \\cdot \\left(\\lambda^{n + 1} - \\dfrac{13}{3} \\cdot \\lambda^n + \\dfrac{4}{3} \\cdot \\lambda^{n - 1}\\right) + B \\cdot \\left(\\mu^{n + 1} - \\dfrac{13}{3} \\cdot \\mu^n + \\dfrac{4}{3} \\cdot \\mu^{n - 1}\\right) = 0 \\\\\n", | |
| "A \\cdot \\lambda^{n - 1} \\cdot \\left(\\lambda^2 - \\dfrac{13}{3} \\cdot \\lambda + \\dfrac{4}{3}\\right) + B \\cdot \\mu^{n - 1} \\cdot \\left(\\mu^2 - \\dfrac{13}{3} \\cdot \\mu + \\dfrac{4}{3}\\right) = 0\n", | |
| "\\end{align}\n", | |
| "$$\n", | |
| "\n", | |
| "esto tiene que valer $\\forall n \\in \\mathbb{N} \\implies \\lambda, \\mu$ soluciones de:\n", | |
| "\n", | |
| "$$\n", | |
| "x^2 - \\dfrac{13}{3} \\cdot x + \\dfrac{4}{3} = 0 \\\\\n", | |
| "3 \\cdot x^2 - 13 \\cdot x + 4 = 0 \\\\\n", | |
| "x_{1,2} = \\dfrac{13 \\pm \\sqrt{13^2 - 4 \\cdot 3 \\cdot 4}}{2 \\cdot 3} = \\begin{cases}\n", | |
| "x_1 = 4 \\\\\n", | |
| "x_2 = \\dfrac{1}{3}\n", | |
| "\\end{cases} \\\\\n", | |
| "\\implies x_n = A \\cdot 4^n + B \\cdot \\left(\\dfrac{1}{3}\\right)^n\n", | |
| "$$\n", | |
| "\n", | |
| "Usando $x_0$ y $x_1$, que son datos:\n", | |
| "\n", | |
| "$$\n", | |
| "\\begin{align}\n", | |
| "x_0 = A \\cdot 4^0 + B \\cdot \\left(\\dfrac{1}{3}\\right)^0 & = A + B = 1 \\tag{1}\\label{1}\\\\\n", | |
| "x_1 = A \\cdot 4^1 + B \\cdot \\left(\\dfrac{1}{3}\\right)^1 & = A \\cdot 4 + B \\cdot \\dfrac{1}{3} = \\dfrac{1}{3} \\tag{2}\\label{2}\n", | |
| "\\end{align}\n", | |
| "$$\n", | |
| "\n", | |
| "Resolvemos el sistema de ecuaciones:\n", | |
| "\n", | |
| "$$\n", | |
| "\\begin{align}\n", | |
| "A = 1 - B \\implies \\left(1 - B\\right) \\cdot 4 + B \\cdot \\dfrac{1}{3} & = \\dfrac{1}{3} \\\\\n", | |
| "4 - B \\cdot 4 + B \\cdot \\dfrac{1}{3} & = \\dfrac{1}{3} \\\\\n", | |
| "B \\cdot \\left(\\cancel{\\dfrac{1}{3} - 4}\\right) & = \\underbrace{\\cancel{\\dfrac{1}{3} - 4}}_1 \\implies \\boxed{B = 1} \\implies \\boxed{A = 0} \\\\\n", | |
| "\\boxed{x_n = \\left(\\dfrac{1}{3}\\right)^n}\n", | |
| "\\end{align}\n", | |
| "$$" | |
| ] | |
| }, | |
| { | |
| "cell_type": "markdown", | |
| "metadata": {}, | |
| "source": [ | |
| "### Notas de la clase\n", | |
| "\n", | |
| "- Programaremos en Python.\n", | |
| "- Entrega 1 TP hasta el ejercicio 11: El próximo sábado, puede ser después de la clase." | |
| ] | |
| } | |
| ], | |
| "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": "3.7.6" | |
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| "nbformat_minor": 2 | |
| } |
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