{"id":4529,"date":"2017-12-09T08:05:53","date_gmt":"2017-12-09T13:05:53","guid":{"rendered":"http:\/\/blog.espol.edu.ec\/matg1013\/?p=4529"},"modified":"2026-04-05T20:54:31","modified_gmt":"2026-04-06T01:54:31","slug":"s3eva2009ti_t2-edo-taylor-senox","status":"publish","type":"post","link":"https:\/\/blog.espol.edu.ec\/algoritmos101\/mn-s3eva10\/s3eva2009ti_t2-edo-taylor-senox\/","title":{"rendered":"s3Eva2009TI_T2 EDO Taylor Seno(x)"},"content":{"rendered":"\n

Ejercicio<\/strong>: 3Eva2009TI_T2 EDO Taylor Seno(x)<\/a><\/p>\n\n\n\n

La ecuaci\u00f3n del problema es:<\/p>\n\n\n xy'+ 2y = \\sin (x) <\/span>\n\n\n \\frac{\\pi}{2} \\leq x \\leq \\frac{3\\pi}{2} <\/span>\n\n\n y\\Big(\\frac{\\pi}{2} \\Big) = 1<\/span>\n\n\n\n

Para el algoritmo se escribe separando y':<\/p>\n\n\n y' = \\frac{\\sin (x)}{x} - 2\\frac{y}{x} <\/span>\n\n\n\n

tomando como referencia Taylor de 3 t\u00e9rminos m\u00e1s el t\u00e9rmino de error O(h3<\/sup>)<\/p>\n\n\n y_{i+1} = y_i +\\frac{h}{1!}y'_i + \\frac{h^2}{2!}y''_i + \\frac{h^3}{3!}y'''_i<\/span>\n\n\n\n

Se usa hasta el tercer t\u00e9rmino para el algoritmo.<\/p>\n\n\n y_{i+1} = y_i +\\frac{h}{1!}y'_i + \\frac{h^2}{2!}y''_i<\/span>\n\n\n\n

Se determina que se requiere la segunda derivada para completar la aproximaci\u00f3n. A partir de la ecuaci\u00f3n del problema se aplica en cada t\u00e9rmino:<\/p>\n\n\n \\Big(\\frac{u}{v}\\Big)' = \\frac{u'v-uv' }{v^2}<\/span>\n\n\n y'' = \\frac{\\cos (x) x - sin(x)}{x^2} - 2\\frac{y'x-y}{x^2} <\/span>\n\n\n y'' = \\frac{\\cos (x) x - sin(x)}{x^2} - 2\\frac{\\Big(\\frac{\\sin (x)}{x} - 2\\frac{y}{x} \\Big)x-y}{x^2} <\/span>\n\n\n\n

Con lo que se realizan las iteraciones para llenar la tabla<\/p>\n\n\n\n

iteraci\u00f3n 1<\/strong><\/em><\/p>\n\n\n\n

x0<\/sub> = \u03c0\/2= 1.57079633<\/p>\n\n\n\n

y0<\/sub>= 1<\/p>\n\n\n y' = \\frac{\\sin (\u03c0\/2)}{\u03c0\/2} - 2\\frac{1}{\u03c0\/2} = -0.40793719<\/span>\n\n\n y'' = \\frac{\\cos (\u03c0\/2) \u03c0\/2 - sin(\u03c0\/2)}{(\u03c0\/2)^2}+<\/span>\n\n\n - 2\\frac{(-0.40793719)(\u03c0\/2)-1}{(\u03c0\/2)^2}= 0.48531359<\/span>\n\n\n y_{1} = 1 +\\frac{\\pi\/10}{1!}(-0.40793719) + \\frac{(\\pi\/10)^2}{2!}(0.48531359) = 0.86<\/span>\n\n\n\n

x1<\/sub> = x0<\/sub>+h = 1.57079633 + \u03c0\/10 =1.88495559<\/p>\n\n\n\n

iteraci\u00f3n 2<\/strong><\/em><\/p>\n\n\n\n

x1<\/sub> = 1.88495559<\/p>\n\n\n\n

y1<\/sub> = 0.86<\/p>\n\n\n y' = \\frac{\\sin (1.88495559)}{1.88495559} - 2\\frac{0.86}{1.88495559} = -0.31947719 <\/span>\n\n\n y'' = \\frac{\\cos (1.88495559)(1.88495559) - sin(1.88495559)}{(1.88495559)^2}<\/span>\n\n\n - 2\\frac{(-0.31947719) (1.88495559)-y}{(1.88495559)^2} = 0.16854341 <\/span>\n\n\n y_{2} = 0.86 +\\frac{\\pi \/10}{1!}(-0.31947719) + \\frac{(\\pi \/10)^2}{2!}(0.16854341) = 0.75579202<\/span>\n\n\n\n

x2<\/sub> = x1<\/sub>+ h = 1.88495559 + \u03c0\/10 = 2.19911486<\/p>\n\n\n\n

iteraci\u00f3n 3<\/strong><\/em><\/p>\n\n\n\n

x2<\/sub> = 2.19911486<\/p>\n\n\n\n

y2<\/sub> = 0.75579202<\/p>\n\n\n y' = \\frac{\\sin (2.19911486)}{2.19911486} - 2\\frac{y}{2.19911486} = -0.29431724 <\/span>\n\n\n y'' = \\frac{\\cos (2.19911486)(2.19911486) - sin(2.19911486)}{(2.19911486)^2} +<\/span>\n\n\n - 2\\frac{(-0.29431724)(2.19911486)-0.75579202}{(2.19911486)^2} = 0.0294177<\/span>\n\n\n y_{3} = 0.75579202 +\\frac{\\pi\/10}{1!}(-0.29431724) + \\frac{(\\pi \/10)^2}{2!}(0.0294177)<\/span>\n\n\n\n

x3<\/sub> = x3<\/sub>+h = 2.19911486 + \u03c0\/10 = 2.19911486<\/p>\n\n\n\n

Con lo que se puede realizar el algoritmo, obteniendo la siguiente tabla:<\/p>\n\n\n\n

estimado\n[xi,          yi,         d1y,        d2y       ]\n[[ 1.57079633 1. -0.40793719 0.48531359] \n [ 1.88495559 0.86 -0.31947719 0.16854341] \n [ 2.19911486 0.75579202 -0.29431724 0.0294177 ] \n [ 2.51327412 0.66374258 -0.29583247 -0.02247944] \n [ 2.82743339 0.5727318 -0.30473968 -0.02730493] \n [ 3.14159265 0.47868397 -0.31027009 -0.00585871] \n [ 3.45575192 0.38159973 -0.30649476 0.02930236] \n [ 3.76991118 0.28383639 -0.29064275 0.06957348] \n [ 4.08407045 0.18899424 -0.26221809 0.10859762] \n [ 4.39822972 0.10111944 -0.22243506 0.14160656] \n [ 4.71238898 0.02410027 0. 0. ]]<\/code><\/pre>\n\n\n\n
\"3Eva_IT2009_T2<\/a><\/figure>\n\n\n\n
Algoritmo en Python<\/h2>\n\n\n
\n# 3Eva_IT2009_T2 EDO Taylor Seno(x)\n# EDO. M\u00e9todo de Taylor 3 t\u00e9rminos \n# estima solucion para muestras separadas h en eje x\n# valores iniciales x0,y0\n# entrega arreglo [[x,y]]\nimport numpy as np\n\ndef edo_taylor3t(d1y,d2y,x0,y0,h,muestras):\n    tamano = muestras + 1\n    estimado = np.zeros(shape=(tamano,4),dtype=float)\n    # incluye el punto [x0,y0]\n    estimado[0] = [x0,y0,0,0]\n    x = x0\n    y = y0\n    for i in range(1,tamano,1):\n        y = y + h*d1y(x,y) + ((h**2)\/2)*d2y(x,y)\n        x = x+h\n        estimado[i-1,2:]= [d1y(x,y),d2y(x,y)]\n        estimado[i,0:2] = [x,y]\n    return(estimado)\n\n# PROGRAMA PRUEBA\n# 3Eva_IT2009_T2 EDO Taylor Seno(x).\n\n# INGRESO.\n# d1y = y', d2y = y''\nd1y = lambda x,y: np.sin(x)\/x - 2*y\/x\nd2y = lambda x,y: (x*np.cos(x)-np.sin(x))\/(x**2)-2*(x*(np.sin(x)\/x - 2*y\/x)-y)\/(x**2)\nx0 = np.pi\/2\ny0 = 1\nh = np.pi\/10\nmuestras = 10\n\n# PROCEDIMIENTO\npuntos = edo_taylor3t(d1y,d2y,x0,y0,h,muestras)\nxi = puntos[:,0]\nyi = puntos[:,1]\n\n# SALIDA\nprint('estimado[xi, yi, d1yi, d2yi]')\nprint(puntos)\n\n# Gr\u00e1fica\nimport matplotlib.pyplot as plt\nplt.plot(xi[0],yi[0],'o', color='r', label ='[x0,y0]')\nplt.plot(xi[1:],yi[1:],'o', color='g', label ='y estimada')\nplt.title('EDO: Soluci\u00f3n con Taylor 3 t\u00e9rminos')\nplt.xlabel('x')\nplt.ylabel('y')\nplt.legend()\nplt.grid()\nplt.show()\n<\/pre><\/div>","protected":false},"excerpt":{"rendered":"
Ejercicio: 3Eva2009TI_T2 EDO Taylor Seno(x) La ecuaci\u00f3n del problema es: Para el algoritmo se escribe separando y': tomando como referencia Taylor de 3 t\u00e9rminos m\u00e1s el t\u00e9rmino de error O(h3) Se usa hasta el tercer t\u00e9rmino para el algoritmo. Se determina que se requiere la segunda derivada para completar la aproximaci\u00f3n. A partir de la […]<\/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-4529","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\/4529","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=4529"}],"version-history":[{"count":4,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/4529\/revisions"}],"predecessor-version":[{"id":23914,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/4529\/revisions\/23914"}],"wp:attachment":[{"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/media?parent=4529"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/categories?post=4529"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/tags?post=4529"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}