{"id":2991,"date":"2017-02-22T06:00:28","date_gmt":"2017-02-22T04:00:28","guid":{"rendered":"http:\/\/www.glc.us.es\/~jalonso\/exercitium\/?p=2991"},"modified":"2022-03-26T11:32:06","modified_gmt":"2022-03-26T09:32:06","slug":"calculo-de-pi-mediante-la-fraccion-continua-de-lange-2017","status":"publish","type":"post","link":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/calculo-de-pi-mediante-la-fraccion-continua-de-lange-2017\/","title":{"rendered":"C\u00e1lculo de pi mediante la fracci\u00f3n continua de Lange"},"content":{"rendered":"<p>En 1999, L.J. Lange public\u00f3 el art\u00edculo <a href=\"http:\/\/www.maa.org\/sites\/default\/files\/pdf\/pubs\/amm_supplements\/Monthly_Reference_11.pdf\">An elegant new continued fraction for \u03c0<\/a>.<\/p>\n<p>En el primer teorema del art\u00edculo se demuestra la siguiente expresi\u00f3n de \u03c0 mediante una fracci\u00f3n continua<br \/>\n<a href=\"https:\/\/i0.wp.com\/www.glc.us.es\/~jalonso\/exercitium\/wp-content\/uploads\/2017\/02\/Calculo_de_pi_mediante_la_fraccion_continua_de_Lange.png\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/i0.wp.com\/www.glc.us.es\/~jalonso\/exercitium\/wp-content\/uploads\/2017\/02\/Calculo_de_pi_mediante_la_fraccion_continua_de_Lange.png?resize=516%2C276\" alt=\"Calculo_de_pi_mediante_la_fraccion_continua_de_Lange\" width=\"516\" height=\"276\" class=\"aligncenter size-full wp-image-2992\" srcset=\"https:\/\/i0.wp.com\/www.glc.us.es\/~jalonso\/exercitium\/wp-content\/uploads\/2017\/02\/Calculo_de_pi_mediante_la_fraccion_continua_de_Lange.png?w=516&amp;ssl=1 516w, https:\/\/i0.wp.com\/www.glc.us.es\/~jalonso\/exercitium\/wp-content\/uploads\/2017\/02\/Calculo_de_pi_mediante_la_fraccion_continua_de_Lange.png?resize=300%2C160&amp;ssl=1 300w, https:\/\/i0.wp.com\/www.glc.us.es\/~jalonso\/exercitium\/wp-content\/uploads\/2017\/02\/Calculo_de_pi_mediante_la_fraccion_continua_de_Lange.png?resize=100%2C53&amp;ssl=1 100w, https:\/\/i0.wp.com\/www.glc.us.es\/~jalonso\/exercitium\/wp-content\/uploads\/2017\/02\/Calculo_de_pi_mediante_la_fraccion_continua_de_Lange.png?resize=150%2C80&amp;ssl=1 150w\" sizes=\"(max-width: 516px) 100vw, 516px\" data-recalc-dims=\"1\" \/><\/a><\/p>\n<p>La primeras aproximaciones son<\/p>\n<pre lang=\"text\">\n   a(1) = 3+1                = 4.0\n   a(2) = 3+(1\/(6+9))        = 3.066666666666667\n   a(3) = 3+(1\/(6+9\/(6+25))) = 3.158974358974359\n<\/pre>\n<p>Definir las funciones<\/p>\n<pre lang=\"text\">\n   aproximacionPi :: Int -> Double\n   grafica        :: [Int] -> IO ()\n<\/pre>\n<p>tales que<\/p>\n<ul>\n<li>(aproximacionPi n) es la n-\u00e9sima aproximaci\u00f3n de pi con la fracci\u00f3n continua de Lange. Por ejemplo,<\/li>\n<\/ul>\n<pre lang=\"text\">\n     aproximacionPi 1     ==  4.0\n     aproximacionPi 2     ==  3.066666666666667\n     aproximacionPi 3     ==  3.158974358974359\n     aproximacionPi 10    ==  3.141287132741557\n     aproximacionPi 100   ==  3.141592398533554\n     aproximacionPi 1000  ==  3.1415926533392926\n<\/pre>\n<ul>\n<li>(grafica xs) dibuja la gr\u00e1fica de las k-\u00e9simas aproximaciones de pi donde k toma los valores de la lista xs. Por ejemplo, (grafica [1..10]) dibuja<br \/>\n<a href=\"https:\/\/i0.wp.com\/www.glc.us.es\/~jalonso\/exercitium\/wp-content\/uploads\/2017\/02\/Calculo_de_pi_mediante_la_fraccion_continua_de_Lange_2.png\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/i0.wp.com\/www.glc.us.es\/~jalonso\/exercitium\/wp-content\/uploads\/2017\/02\/Calculo_de_pi_mediante_la_fraccion_continua_de_Lange_2.png?resize=624%2C474\" alt=\"Calculo_de_pi_mediante_la_fraccion_continua_de_Lange_2\" width=\"624\" height=\"474\" class=\"aligncenter size-full wp-image-2993\" srcset=\"https:\/\/i0.wp.com\/www.glc.us.es\/~jalonso\/exercitium\/wp-content\/uploads\/2017\/02\/Calculo_de_pi_mediante_la_fraccion_continua_de_Lange_2.png?w=624&amp;ssl=1 624w, https:\/\/i0.wp.com\/www.glc.us.es\/~jalonso\/exercitium\/wp-content\/uploads\/2017\/02\/Calculo_de_pi_mediante_la_fraccion_continua_de_Lange_2.png?resize=300%2C227&amp;ssl=1 300w, https:\/\/i0.wp.com\/www.glc.us.es\/~jalonso\/exercitium\/wp-content\/uploads\/2017\/02\/Calculo_de_pi_mediante_la_fraccion_continua_de_Lange_2.png?resize=100%2C75&amp;ssl=1 100w, https:\/\/i0.wp.com\/www.glc.us.es\/~jalonso\/exercitium\/wp-content\/uploads\/2017\/02\/Calculo_de_pi_mediante_la_fraccion_continua_de_Lange_2.png?resize=150%2C113&amp;ssl=1 150w\" sizes=\"(max-width: 624px) 100vw, 624px\" data-recalc-dims=\"1\" \/><\/a><br \/>\n(grafica [10..100]) dibuja<br \/>\n<a href=\"https:\/\/i0.wp.com\/www.glc.us.es\/~jalonso\/exercitium\/wp-content\/uploads\/2017\/02\/Calculo_de_pi_mediante_la_fraccion_continua_de_Lange_3.png\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/i0.wp.com\/www.glc.us.es\/~jalonso\/exercitium\/wp-content\/uploads\/2017\/02\/Calculo_de_pi_mediante_la_fraccion_continua_de_Lange_3.png?resize=632%2C474\" alt=\"Calculo_de_pi_mediante_la_fraccion_continua_de_Lange_3\" width=\"632\" height=\"474\" class=\"aligncenter size-full wp-image-2994\" srcset=\"https:\/\/i0.wp.com\/www.glc.us.es\/~jalonso\/exercitium\/wp-content\/uploads\/2017\/02\/Calculo_de_pi_mediante_la_fraccion_continua_de_Lange_3.png?w=632&amp;ssl=1 632w, https:\/\/i0.wp.com\/www.glc.us.es\/~jalonso\/exercitium\/wp-content\/uploads\/2017\/02\/Calculo_de_pi_mediante_la_fraccion_continua_de_Lange_3.png?resize=300%2C225&amp;ssl=1 300w, https:\/\/i0.wp.com\/www.glc.us.es\/~jalonso\/exercitium\/wp-content\/uploads\/2017\/02\/Calculo_de_pi_mediante_la_fraccion_continua_de_Lange_3.png?resize=100%2C75&amp;ssl=1 100w, https:\/\/i0.wp.com\/www.glc.us.es\/~jalonso\/exercitium\/wp-content\/uploads\/2017\/02\/Calculo_de_pi_mediante_la_fraccion_continua_de_Lange_3.png?resize=150%2C112&amp;ssl=1 150w\" sizes=\"(max-width: 632px) 100vw, 632px\" data-recalc-dims=\"1\" \/><\/a><br \/>\ny (grafica [100..200]) dibuja<br \/>\n<a href=\"https:\/\/i0.wp.com\/www.glc.us.es\/~jalonso\/exercitium\/wp-content\/uploads\/2017\/02\/Calculo_de_pi_mediante_la_fraccion_continua_de_Lange_4.png\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/i0.wp.com\/www.glc.us.es\/~jalonso\/exercitium\/wp-content\/uploads\/2017\/02\/Calculo_de_pi_mediante_la_fraccion_continua_de_Lange_4.png?resize=633%2C472\" alt=\"Calculo_de_pi_mediante_la_fraccion_continua_de_Lange_4\" width=\"633\" height=\"472\" class=\"aligncenter size-full wp-image-2995\" srcset=\"https:\/\/i0.wp.com\/www.glc.us.es\/~jalonso\/exercitium\/wp-content\/uploads\/2017\/02\/Calculo_de_pi_mediante_la_fraccion_continua_de_Lange_4.png?w=633&amp;ssl=1 633w, https:\/\/i0.wp.com\/www.glc.us.es\/~jalonso\/exercitium\/wp-content\/uploads\/2017\/02\/Calculo_de_pi_mediante_la_fraccion_continua_de_Lange_4.png?resize=300%2C223&amp;ssl=1 300w, https:\/\/i0.wp.com\/www.glc.us.es\/~jalonso\/exercitium\/wp-content\/uploads\/2017\/02\/Calculo_de_pi_mediante_la_fraccion_continua_de_Lange_4.png?resize=100%2C74&amp;ssl=1 100w, https:\/\/i0.wp.com\/www.glc.us.es\/~jalonso\/exercitium\/wp-content\/uploads\/2017\/02\/Calculo_de_pi_mediante_la_fraccion_continua_de_Lange_4.png?resize=150%2C111&amp;ssl=1 150w\" sizes=\"(max-width: 633px) 100vw, 633px\" data-recalc-dims=\"1\" \/><\/a><\/li>\n<\/ul>\n<p>Nota: Este ejercicio ha sido propuesto por Antonio Morales.<\/p>\n<h4>Soluciones<\/h4>\n<pre lang=\"haskell\">\nimport Graphics.Gnuplot.Simple\n \n-- fraccionPi es la representaci\u00f3n de la fracci\u00f3n continua de pi como un\n-- par de listas infinitas.\nfraccionPi :: [(Integer, Integer)]\nfraccionPi = zip (3 : [6,6..]) (map (^2) [1,3..])\n\n-- (aproximacionFC n fc) es la n-\u00e9sima aproximaci\u00f3n de la fracci\u00f3n\n-- continua fc (como un par de listas).  \naproximacionFC :: Int -> [(Integer, Integer)] -> Double\naproximacionFC n =\n  foldr (\\(a,b) z -> fromIntegral a + fromIntegral b \/ z) 1 . take n\n\naproximacionPi :: Int -> Double\naproximacionPi n =\n  aproximacionFC n fraccionPi\n \ngrafica :: [Int] -> IO ()\ngrafica xs = \n    plotList [Key Nothing]\n             [(k,aproximacionPi k) | k <- xs]\n<\/pre>\n","protected":false},"excerpt":{"rendered":"<p>En 1999, L.J. Lange public\u00f3 el art\u00edculo An elegant new continued fraction for \u03c0. En el primer teorema del art\u00edculo se demuestra la siguiente expresi\u00f3n de \u03c0 mediante una fracci\u00f3n continua La primeras aproximaciones son a(1) = 3+1 = 4.0 a(2) = 3+(1\/(6+9)) = 3.066666666666667 a(3) = 3+(1\/(6+9\/(6+25))) = 3.158974358974359 Definir las funciones aproximacionPi ::&#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,"footnotes":"","_jetpack_memberships_contains_paid_content":false},"categories":[7],"tags":[94,183,376,10,11,309,47,9],"jetpack_featured_media_url":"","jetpack_sharing_enabled":true,"_links":{"self":[{"href":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/wp-json\/wp\/v2\/posts\/2991"}],"collection":[{"href":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/wp-json\/wp\/v2\/comments?post=2991"}],"version-history":[{"count":4,"href":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/wp-json\/wp\/v2\/posts\/2991\/revisions"}],"predecessor-version":[{"id":3035,"href":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/wp-json\/wp\/v2\/posts\/2991\/revisions\/3035"}],"wp:attachment":[{"href":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/wp-json\/wp\/v2\/media?parent=2991"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/wp-json\/wp\/v2\/categories?post=2991"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/wp-json\/wp\/v2\/tags?post=2991"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}