{"id":10646,"date":"2014-03-03T11:42:09","date_gmt":"2014-03-03T16:42:09","guid":{"rendered":"http:\/\/blog.espol.edu.ec\/ccpg1001\/?p=10646"},"modified":"2026-04-05T17:54:11","modified_gmt":"2026-04-05T22:54:11","slug":"s2eva2003tiii_t2-raiz-cuadrada-por-newton-recursiva","status":"publish","type":"post","link":"https:\/\/blog.espol.edu.ec\/algoritmos101\/fp-s2eva10\/s2eva2003tiii_t2-raiz-cuadrada-por-newton-recursiva\/","title":{"rendered":"s2Eva2003TIII_T2 Ra\u00edz cuadrada por Newton, recursiva"},"content":{"rendered":"\n<p><em><strong>Ejercicio<\/strong><\/em>: <a href=\"https:\/\/blog.espol.edu.ec\/algoritmos101\/fp-2eva10\/2eva2003tiii_t2-raiz-cuadrada-por-newton-recursiva\/\" data-type=\"post\" data-id=\"2871\">2Eva2003TIII_T2 Ra\u00edz cuadrada por Newton<\/a><\/p>\n\n\n\n<p>Recuerde que toda funci\u00f3n recursiva requiere un valor inicial para la primera iteraci\u00f3n. Empezando para n=0 con el valor de <strong>x<\/strong>:<\/p>\n\n\n<span class=\"wp-katex-eq katex-display\" data-display=\"true\"> f(1) = \\frac{x}{2} <\/span>\n\n\n\n<p>En el caso que n&gt;1, se usa la expresi\u00f3n recursiva:<\/p>\n\n\n<span class=\"wp-katex-eq katex-display\" data-display=\"true\"> f(n) = 0.5\\Bigg(f(n-1) + \\frac{x}{f(n-1)}\\Bigg)<\/span>\n\n\n\n<p>Los valores tabulados para x=9 y n=10<\/p>\n\n\n\n<pre class=\"wp-block-code\"><code>ingrese x: 9\naproximaci\u00f3n n-esima: 10\ni , f(i)\n0 nan\n1 4.5\n2 3.25\n3 3.0096153846153846\n4 3.000015360039322\n5 3.0000000000393214\n6 3.0\n7 3.0\n8 3.0\n9 3.0\n10 3.0\n&gt;&gt;&gt; \n<\/code><\/pre>\n\n\n\n<p>Observe que a partir de la iteraci\u00f3n 6, ya no muestra diferencia entre resultados consecutivos.<\/p>\n\n\n\n<p>Para x= 9.5 con x=10<\/p>\n\n\n\n<pre class=\"wp-block-code\"><code>ingrese x: 9.5\naproximaci\u00f3n n-esima: 10\ni , f(i)\n0 nan\n1 4.75\n2 3.375\n3 3.0949074074074074\n4 3.082233060472589\n5 3.0822070015946474\n6 3.0822070014844885\n7 3.0822070014844885\n8 3.0822070014844885\n9 3.0822070014844885\n10 3.0822070014844885\n&gt;&gt;&gt; <\/code><\/pre>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\" \/>\n\n\n\n<h2 class=\"wp-block-heading\">Algoritmo en Python<\/h2>\n\n\n\n<p>La funci\u00f3n no debe admitir valores de x negativos o cero, de darse el caso se responde np.NaN que corresponde a no es un n\u00famero o 'Not a Number'.<\/p>\n\n\n<div class=\"wp-block-syntaxhighlighter-code \"><pre class=\"brush: python; title: ; notranslate\" title=\"\">\n# 2Eva_IIIT2003_T2 Ra\u00edz cuadrada por Newton\nimport numpy as np\n\n# literal a. funci\u00f3n recursiva\ndef raizNewton(x,n)\n    if n&lt;=0 or x&lt;=0:\n        # no se admiten negativos o cero\n        f = np.NaN\n    else:\n        if n == 1:\n            f = x\/2  # valor inicial\n        if n&gt;1:\n            f = 0.5 *(raizNewton(x,n-1)+x\/raizNewton(x,n-1))\n    return (f)\n\n# literal b. Programa ---------------\n# INGRESO\nx = float(input('ingrese x: '))\nn = int(input('aproximaci\u00f3n n-esima: '))\n\n# PROCEDIMIENTO\nprint(' i , f(i)')\nfor i in range(0,n+1,1):\n    print(i , raizNewton(x,i))\n<\/pre><\/div>","protected":false},"excerpt":{"rendered":"<p>Ejercicio: 2Eva2003TIII_T2 Ra\u00edz cuadrada por Newton Recuerde que toda funci\u00f3n recursiva requiere un valor inicial para la primera iteraci\u00f3n. Empezando para n=0 con el valor de x: En el caso que n&gt;1, se usa la expresi\u00f3n recursiva: Los valores tabulados para x=9 y n=10 Observe que a partir de la iteraci\u00f3n 6, ya no muestra [&hellip;]<\/p>\n","protected":false},"author":8043,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"wp-custom-template-entrada-fp-ejemplos","format":"standard","meta":{"footnotes":""},"categories":[131],"tags":[58,157],"class_list":["post-10646","post","type-post","status-publish","format-standard","hentry","category-fp-s2eva10","tag-ejemplos-python","tag-fundamentos-programacion"],"_links":{"self":[{"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/10646","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=10646"}],"version-history":[{"count":3,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/10646\/revisions"}],"predecessor-version":[{"id":23663,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/10646\/revisions\/23663"}],"wp:attachment":[{"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/media?parent=10646"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/categories?post=10646"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/tags?post=10646"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}