{"id":196,"date":"2020-01-30T13:45:37","date_gmt":"2020-01-30T11:45:37","guid":{"rendered":"https:\/\/www.glc.us.es\/~jalonso\/calculemus\/?p=196"},"modified":"2021-08-21T12:57:37","modified_gmt":"2021-08-21T10:57:37","slug":"teorema-de-nicomaco","status":"publish","type":"post","link":"https:\/\/www.glc.us.es\/~jalonso\/calculemus\/teorema-de-nicomaco\/","title":{"rendered":"Teorema de Nic\u00f3maco"},"content":{"rendered":"<p>Demostrar el <a href=\"http:\/\/bit.ly\/2OaJI7q\">teorema de Nic\u00f3maco<\/a> que afirma que la suma de los cubos de los n primeros n\u00fameros naturales es igual que el cuadrado de la suma de los n primeros n\u00fameros naturales; es decir, para todo n\u00famero natural n se tiene que<\/p>\n<pre lang=\"text\">\n1\u00b3 + 2\u00b3 + ... + n\u00b3 = (1 + 2 + ... + n)\u00b2\n<\/pre>\n<h4>Soluciones con Isabelle\/HOL<\/h4>\n<pre lang=\"isar\">\ntheory Teorema_de_Nicomaco\nimports Main\nbegin\n\n(* (suma n) es la suma de los primeros n\u00fameros naturales. *)\nfun suma :: \"nat \u21d2 nat\" where\n  \"suma 0 = 0\"\n| \"suma (Suc n) = suma n + Suc n\"\n\n(* (sumaCubos n) es la suma de los cubos primeros n\u00fameros naturales. *)\nfun sumaCubos :: \"nat \u21d2 nat\" where\n  \"sumaCubos 0 = 0\"\n| \"sumaCubos (Suc n) = sumaCubos n + (Suc n)^3\"\n\n(* F\u00f3rmula para la suma *)\nlemma \"2 * suma n = n^2 + n\"\nproof (induct n)\n  show \"2 * suma 0 = 0^2 + 0\" by simp\nnext\n  fix n\n  assume \"2 * suma n = n^2 + n\"\n  then have \"2 * suma (Suc n) = n^2 + n + 2 + 2 * n\"\n    by simp\n  also have \"\u2026 = (Suc n)^2 + Suc n\"\n    by (simp add: power2_eq_square)\n  finally show \"2 * suma (Suc n) = (Suc n)^2 + Suc n\"\n    by this\nqed\n\n(* Demostraci\u00f3n autom\u00e1tica de la propiedad anterior *)\nlemma formula_suma:\n  \"2 * suma n = n^2 + n\"\n  by (induct n) (auto simp add: power2_eq_square)\n\nlemma \"4 * sumaCubos n = (n^2 + n)^2\"\nproof (induct n)\n  show \"4 * sumaCubos 0 = (0^2 + 0)^2\"\n    by simp\nnext\n  fix n\n  assume \"4 * sumaCubos n = (n^2 + n)^2\"\n  then have \"4 * sumaCubos (Suc n) = (n^2 + n)^2 + 4 * (Suc n)^3\"\n    by simp\n  then show \"4 * sumaCubos (Suc n) = ((Suc n)^2 + Suc n)^2\"\n  by (simp add: algebra_simps\n                power2_eq_square\n                power3_eq_cube )\nqed\n\n(* Demostraci\u00f3n autom\u00e1tica de la propiedad anterior *)\nlemma formula_sumaCubos:\n  \"4 * sumaCubos n = (n^2 + n)^2\"\n  by (induct n) (auto simp add: algebra_simps\n                                power2_eq_square\n                                power3_eq_cube)\n\n(* Lema auxiliar *)\nlemma aux: \"4 * (m::nat) = (2 * n)^2 \u27f9 m = n^2\"\n  by (simp add: power2_eq_square)\n\n(* Teorema de Nic\u00f3maco *)\ntheorem teorema_de_Nicomaco:\n  \"sumaCubos n = (suma n)^2\"\n  by (simp only: formula_suma formula_sumaCubos aux)\n\nend\n<\/pre>\n<h4>Otras soluciones<\/h4>\n<ul>\n<li>Se pueden escribir otras soluciones en los comentarios.\n<li>El c\u00f3digo se debe escribir entre una l\u00ednea con &#60;pre lang=&quot;isar&quot;&#62; y otra con &#60;\/pre&#62;\n<\/ul>\n","protected":false},"excerpt":{"rendered":"<p>Demostrar el teorema de Nic\u00f3maco que afirma que la suma de los cubos de los n primeros n\u00fameros naturales es igual que el cuadrado de la suma de los n primeros n\u00fameros naturales; es decir, para todo n\u00famero natural n se tiene que 1\u00b3 + 2\u00b3 + &#8230; + n\u00b3 = (1 + 2 + &#8230; + n)\u00b2 Soluciones con Isabelle\/HOL theory Teorema_de_Nicomaco imports Main begin (* (suma n) es la suma de los primeros n\u00fameros naturales. *) fun suma :: \u00abnat \u21d2 nat\u00bb where \u00absuma 0 = 0\u00bb | \u00absuma (Suc n) = suma n + Suc n\u00bb (* (sumaCubos n) es la suma de los cubos primeros n\u00fameros naturales. *) fun sumaCubos :: \u00abnat \u21d2 nat\u00bb where \u00absumaCubos 0 = 0\u00bb | \u00absumaCubos&#8230;<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"jetpack_post_was_ever_published":false,"_kad_post_transparent":"","_kad_post_title":"","_kad_post_layout":"","_kad_post_sidebar_id":"","_kad_post_content_style":"","_kad_post_vertical_padding":"","_kad_post_feature":"","_kad_post_feature_position":"","_kad_post_header":false,"_kad_post_footer":false,"_jetpack_newsletter_access":"","_jetpack_dont_email_post_to_subs":false,"_jetpack_newsletter_tier_id":0,"_jetpack_memberships_contains_paywalled_content":false,"_jetpack_memberships_contains_paid_content":false,"footnotes":""},"categories":[105,25],"tags":[],"jetpack_featured_media_url":"","jetpack_sharing_enabled":true,"_links":{"self":[{"href":"https:\/\/www.glc.us.es\/~jalonso\/calculemus\/wp-json\/wp\/v2\/posts\/196"}],"collection":[{"href":"https:\/\/www.glc.us.es\/~jalonso\/calculemus\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.glc.us.es\/~jalonso\/calculemus\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.glc.us.es\/~jalonso\/calculemus\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.glc.us.es\/~jalonso\/calculemus\/wp-json\/wp\/v2\/comments?post=196"}],"version-history":[{"count":5,"href":"https:\/\/www.glc.us.es\/~jalonso\/calculemus\/wp-json\/wp\/v2\/posts\/196\/revisions"}],"predecessor-version":[{"id":2530,"href":"https:\/\/www.glc.us.es\/~jalonso\/calculemus\/wp-json\/wp\/v2\/posts\/196\/revisions\/2530"}],"wp:attachment":[{"href":"https:\/\/www.glc.us.es\/~jalonso\/calculemus\/wp-json\/wp\/v2\/media?parent=196"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.glc.us.es\/~jalonso\/calculemus\/wp-json\/wp\/v2\/categories?post=196"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.glc.us.es\/~jalonso\/calculemus\/wp-json\/wp\/v2\/tags?post=196"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}