{"id":4061,"date":"2017-12-10T11:15:47","date_gmt":"2017-12-10T16:15:47","guid":{"rendered":"http:\/\/blog.espol.edu.ec\/matg1013\/?p=4061"},"modified":"2026-04-05T20:52:35","modified_gmt":"2026-04-06T01:52:35","slug":"s3eva2010tii_t4-edo-con-taylor","status":"publish","type":"post","link":"https:\/\/blog.espol.edu.ec\/algoritmos101\/mn-s3eva10\/s3eva2010tii_t4-edo-con-taylor\/","title":{"rendered":"s3Eva2010TII_T4 EDO con Taylor"},"content":{"rendered":"\n<p><em><strong>Ejercicio<\/strong><\/em>: <a href=\"https:\/\/blog.espol.edu.ec\/algoritmos101\/mn-3eva10\/3eva2010tii_t4-edo-con-taylor\/\" data-type=\"post\" data-id=\"786\">3Eva2010TII_T4 EDO con Taylor<\/a><\/p>\n\n\n<span class=\"wp-katex-eq katex-display\" data-display=\"true\"> \\frac{\\delta y}{\\delta x} = \\frac{y^3}{1-2xy^2} <\/span>\n\n\n<span class=\"wp-katex-eq katex-display\" data-display=\"true\"> y(0) = 1, 0 \\leq x \\leq 1 <\/span>\n\n\n\n<p>escrita en forma simplificada<\/p>\n\n\n<span class=\"wp-katex-eq katex-display\" data-display=\"true\"> y' = \\frac{y^3}{1-2xy^2} <\/span>\n\n\n\n<p>tomando como referencia Taylor de 2 t\u00e9rminos m\u00e1s el t\u00e9rmino de error O(h<sup>2<\/sup>)<\/p>\n\n\n<span class=\"wp-katex-eq katex-display\" data-display=\"true\"> y_{i+1} = y_i +\\frac{h}{1!}y'_i + \\frac{h^2}{2!}y''_i <\/span>\n\n\n\n<p>Se usa hasta el segundo t\u00e9rmino para el algoritmo.<\/p>\n\n\n<span class=\"wp-katex-eq katex-display\" data-display=\"true\"> y_{i+1} = y_i +\\frac{h}{1!} <\/span>\n\n\n\n<p>Con lo que se puede realizar el algoritmo<\/p>\n\n\n\n<pre class=\"wp-block-code\"><code>estimado&#091;xi,yi]\n&#091;&#091;0.  1. ]\n &#091;0.2 1.2 ]\n &#091;0.4 2.01509434]\n &#091;0.6 1.28727044]\n &#091;0.8 0.85567954]\n &#091;1.  0.12504631]]<\/code><\/pre>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"640\" height=\"480\" src=\"http:\/\/blog.espol.edu.ec\/algoritmos101\/files\/2017\/12\/s3Eva2010TII_T4_EDO_Taylor_1term.png\" alt=\"s3Eva2010TII_T4 EDO Taylor 2 t\u00e9rmino\" class=\"wp-image-18582\" \/><\/figure>\n\n\n\n<h2 class=\"wp-block-heading\">Algoritmo en Python<\/h2>\n\n\n<div class=\"wp-block-syntaxhighlighter-code \"><pre class=\"brush: python; title: ; notranslate\" title=\"\">\n# 3Eva_IIT2010_T4 EDO con Taylor\n# estima la soluci\u00f3n para muestras espaciadas h en eje x\n# valores iniciales x0,y0\n# entrega arreglo &#x5B;&#x5B;x,y]]\nimport numpy as np\n\ndef edo_taylor2t(d1y,x0,y0,h,muestras):\n    tamano = muestras + 1\n    estimado = np.zeros(shape=(tamano,2),dtype=float)\n    # incluye el punto &#x5B;x0,y0]\n    estimado&#x5B;0] = &#x5B;x0,y0]\n    x = x0\n    y = y0\n    for i in range(1,tamano,1):\n        y = y + h*d1y(x,y)\n        x = x+h\n        estimado&#x5B;i] = &#x5B;x,y]\n    return(estimado)\n\n# PROGRAMA PRUEBA\n# 3Eva_IIT2010_T4 EDO con Taylor\n\n# INGRESO.\n# d1y = y' = f, d2y = y'' = f'\nd1y = lambda x,y: (y**3)\/(1-2*x*(y**2))\nx0 = 0\ny0 = 1\nh = 0.02\na = 0\nb = 1\nmuestras = int((b-a)\/h)\n\n# PROCEDIMIENTO\npuntos = edo_taylor2t(d1y,x0,y0,h,muestras)\nxi = puntos&#x5B;:,0]\nyi = puntos&#x5B;:,1]\n\n# SALIDA\nprint('estimado&#x5B;xi,yi]')\nprint(puntos)\n\n# Gr\u00e1fica\nimport matplotlib.pyplot as plt\nplt.plot(xi&#x5B;0],yi&#x5B;0],'o', color='r', label ='&#x5B;x0,y0]')\nplt.plot(xi&#x5B;1:],yi&#x5B;1:],'o', color='g', label ='y estimada')\nplt.title('EDO: Soluci\u00f3n con Taylor 2 t\u00e9rminos')\nplt.xlabel('x')\nplt.ylabel('y')\nplt.legend()\nplt.grid()\nplt.show()\n<\/pre><\/div>\n\n\n<p><strong>Nota<\/strong>: Revisar los resultados no lineales con los valores de h=0.02<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Ejercicio: 3Eva2010TII_T4 EDO con Taylor escrita en forma simplificada tomando como referencia Taylor de 2 t\u00e9rminos m\u00e1s el t\u00e9rmino de error O(h2) Se usa hasta el segundo t\u00e9rmino para el algoritmo. Con lo que se puede realizar el algoritmo Algoritmo en Python Nota: Revisar los resultados no lineales con los valores de h=0.02<\/p>\n","protected":false},"author":8043,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"wp-custom-template-entrada-mn-ejemplo","format":"standard","meta":{"footnotes":""},"categories":[50],"tags":[58,54],"class_list":["post-4061","post","type-post","status-publish","format-standard","hentry","category-mn-s3eva10","tag-ejemplos-python","tag-mnumericos"],"_links":{"self":[{"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/4061","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/users\/8043"}],"replies":[{"embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/comments?post=4061"}],"version-history":[{"count":3,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/4061\/revisions"}],"predecessor-version":[{"id":23909,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/4061\/revisions\/23909"}],"wp:attachment":[{"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/media?parent=4061"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/categories?post=4061"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/tags?post=4061"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}