{"id":789,"date":"2021-09-25T05:00:25","date_gmt":"2021-09-25T03:00:25","guid":{"rendered":"https:\/\/www.glc.us.es\/~jalonso\/calculemus\/?p=789"},"modified":"2021-09-24T19:06:08","modified_gmt":"2021-09-24T17:06:08","slug":"suma-de-potencias-de-dos","status":"publish","type":"post","link":"https:\/\/www.glc.us.es\/~jalonso\/calculemus\/suma-de-potencias-de-dos\/","title":{"rendered":"Suma de potencias de dos"},"content":{"rendered":"<p>Demostrar que<\/p>\n<pre lang=\"text\">\n   1 + 2 + 2\u00b2 + 2\u00b3 + ... + 2\u207d\u207f\u207b\u00b9\u207e = 2\u207f - 1\n<\/pre>\n<p>Para ello, completar la siguiente teor\u00eda de Lean:<\/p>\n<pre lang=\"lean\">\nimport algebra.big_operators\nimport tactic\nopen finset nat\n\nopen_locale big_operators\n\nvariable (n : \u2115)\n\nexample :\n  \u2211 k in range n, 2^k = 2^n - 1 :=\nsorry\n<\/pre>\n<p>[expand title=\u00bbSoluciones con Lean\u00bb]<\/p>\n<pre lang=\"lean\">\r\nimport algebra.big_operators\r\nimport tactic\r\nopen finset nat\r\n\r\nopen_locale big_operators\r\nset_option pp.structure_projections false\r\n\r\nvariable (n : \u2115)\r\n\r\nexample :\r\n  \u2211 k in range n, 2^k = 2^n - 1 :=\r\nbegin\r\n  induction n with n HI,\r\n  { simp, },\r\n  { calc \u2211 k in range (succ n), 2^k\r\n         = \u2211 k in range n, 2^k + 2^n\r\n             : sum_range_succ (\u03bb x, 2 ^ x) n\r\n     ... = (2^n - 1) + 2^n\r\n             : congr_arg2 (+) HI rfl\r\n     ... = (2^n + 2^n) - 1\r\n             : by omega\r\n     ... = 2^n * 2 - 1\r\n             : by {congr; simp}\r\n     ... = 2^(succ n) - 1\r\n             : by {congr' 1; ring_nf}, },\r\nend\r\n<\/pre>\n<p>Se puede interactuar con la prueba anterior en <a href=\"https:\/\/leanprover-community.github.io\/lean-web-editor\/#url=https:\/\/raw.githubusercontent.com\/jaalonso\/Calculemus\/main\/src\/Suma_de_potencias_de_dos.lean\" rel=\"noopener noreferrer\" target=\"_blank\">esta sesi\u00f3n con Lean<\/a>.<\/p>\n<p>En los comentarios se pueden escribir otras soluciones, escribiendo el c\u00f3digo entre una l\u00ednea con &#60;pre lang=&quot;lean&quot;&#62; y otra con &#60;\/pre&#62;<br \/>\n[\/expand]<\/p>\n<p>[expand title=\u00bbSoluciones con Isabelle\/HOL\u00bb]<\/p>\n<pre lang=\"isar\">\r\ntheory Suma_de_potencias_de_dos\r\nimports Main\r\nbegin\r\n\r\n(* 1\u00aa demostraci\u00f3n *)\r\nlemma \"(\u2211k\u2264n. (2::nat)^k) = 2^(n+1) - 1\"\r\nproof (induct n)\r\n  show \"(\u2211k\u22640. (2::nat)^k) = 2^(0+1) - 1\"\r\n    by simp\r\nnext\r\n  fix n\r\n  assume HI : \"(\u2211k\u2264n. (2::nat)^k) = 2^(n+1) - 1\"\r\n  have \"(\u2211k\u2264Suc n. (2::nat)^k) =\r\n        (\u2211k\u2264n. (2::nat)^k) + 2^Suc n\"\r\n    by simp\r\n  also have \"\u2026 = (2^(n+1) - 1) + 2^Suc n\"\r\n    using HI by simp\r\n  also have \"\u2026 = 2^(Suc n + 1) - 1\"\r\n    by simp\r\n  finally show \"(\u2211k\u2264Suc n. (2::nat)^k) = 2^(Suc n + 1) - 1\" .\r\nqed\r\n\r\n(* 2\u00aa demostraci\u00f3n *)\r\nlemma \"(\u2211k\u2264n. (2::nat)^k) = 2^(n+1) - 1\"\r\nproof (induct n)\r\n  show \"(\u2211k\u22640. (2::nat)^k) = 2^(0+1) - 1\"\r\n    by simp\r\nnext\r\n  fix n\r\n  assume HI : \"(\u2211k\u2264n. (2::nat)^k) = 2^(n+1) - 1\"\r\n  have \"(\u2211k\u2264Suc n. (2::nat)^k) =\r\n        (2^(n+1) - 1) + 2^Suc n\"\r\n    using HI by simp\r\n  then show \"(\u2211k\u2264Suc n. (2::nat)^k) = 2^(Suc n + 1) - 1\"\r\n    by simp\r\nqed\r\n\r\n(* 3\u00aa demostraci\u00f3n *)\r\nlemma \"(\u2211k\u2264n. (2::nat)^k) = 2^(n+1) - 1\"\r\nproof (induct n)\r\n  case 0\r\n  then show ?case by simp\r\nnext\r\n  case (Suc n)\r\n  then show ?case by simp\r\nqed\r\n\r\n(* 4\u00aa demostraci\u00f3n *)\r\nlemma \"(\u2211k\u2264n. (2::nat)^k) = 2^(n+1) - 1\"\r\nby (induct n) simp_all\r\n<\/pre>\n<p>En los comentarios se pueden escribir otras soluciones, escribiendo el c\u00f3digo entre una l\u00ednea con &#60;pre lang=&quot;isar&quot;&#62; y otra con &#60;\/pre&#62;<br \/>\n[\/expand]<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Demostrar que 1 + 2 + 2\u00b2 + 2\u00b3 + &#8230; + 2\u207d\u207f\u207b\u00b9\u207e = 2\u207f &#8211; 1 Para ello, completar la siguiente teor\u00eda de Lean: import algebra.big_operators import tactic open finset nat open_locale big_operators variable (n : \u2115) example : \u2211 k in range n, 2^k = 2^n &#8211; 1 := sorry [expand title=\u00bbSoluciones con Lean\u00bb] import algebra.big_operators import tactic open finset nat open_locale big_operators set_option pp.structure_projections false variable (n : \u2115) example : \u2211 k in range n, 2^k = 2^n &#8211; 1 := begin induction n with n HI, { simp, }, { calc \u2211 k in range (succ n), 2^k = \u2211 k in range n, 2^k + 2^n : sum_range_succ (\u03bb x, 2 ^ x) n &#8230; = (2^n &#8211; 1)&#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":[105,280],"tags":[],"jetpack_featured_media_url":"","jetpack_sharing_enabled":true,"_links":{"self":[{"href":"https:\/\/www.glc.us.es\/~jalonso\/calculemus\/wp-json\/wp\/v2\/posts\/789"}],"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=789"}],"version-history":[{"count":1,"href":"https:\/\/www.glc.us.es\/~jalonso\/calculemus\/wp-json\/wp\/v2\/posts\/789\/revisions"}],"predecessor-version":[{"id":790,"href":"https:\/\/www.glc.us.es\/~jalonso\/calculemus\/wp-json\/wp\/v2\/posts\/789\/revisions\/790"}],"wp:attachment":[{"href":"https:\/\/www.glc.us.es\/~jalonso\/calculemus\/wp-json\/wp\/v2\/media?parent=789"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.glc.us.es\/~jalonso\/calculemus\/wp-json\/wp\/v2\/categories?post=789"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.glc.us.es\/~jalonso\/calculemus\/wp-json\/wp\/v2\/tags?post=789"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}