{"id":8334,"date":"2023-11-14T06:00:08","date_gmt":"2023-11-14T04:00:08","guid":{"rendered":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/?p=8334"},"modified":"2024-05-17T18:44:07","modified_gmt":"2024-05-17T16:44:07","slug":"14-nov-23","status":"publish","type":"post","link":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/14-nov-23\/","title":{"rendered":"L\u00edmites de sucesiones"},"content":{"rendered":"

Definir la funci\u00f3n<\/p>\n

\n   limite :: (Double -> Double) -> Double -> Double\n<\/pre>\n
tal que limite f a<\/code> es el valor de f<\/code> en el primer t\u00e9rmino x<\/code> tal que, para todo y entre x+1<\/code> y x+100<\/code>, el valor absoluto de la diferencia entre f(y)<\/code> y f(x)<\/code> es menor que a<\/code>. Por ejemplo,<\/p>\n
\n   limite (\\n -> (2*n+1)\/(n+5)) 0.001  ==  1.9900110987791344\n   limite (\\n -> (1+1\/n)**n) 0.001     ==  2.714072874546881\n<\/pre>\n
<\/p>\n
Soluciones<\/b><\/p>\n
A continuaci\u00f3n se muestran las soluciones en Haskell<\/a> y las soluciones en Python<\/a>.<\/p>\n
<\/a>
\nSoluciones en Haskell<\/b><\/p>\n
\nmodule Limites_de_sucesiones where\n\nimport Test.Hspec (Spec, hspec, it, shouldBe)\n\nlimite :: (Double -> Double) -> Double -> Double\nlimite f a =\n  head [f x | x <- [1..],\n              maximum [abs (f y - f x) | y <- [x+1..x+100]] < a]\n\n-- Verificaci\u00f3n\n-- ============\n\nverifica :: IO ()\nverifica = hspec spec\n\nspec :: Spec\nspec = do\n  it \"e1\" $\n    limite (\\n -> (2*n+1)\/(n+5)) 0.001  `shouldBe`  1.9900110987791344\n  it \"e2\" $\n    limite (\\n -> (1+1\/n)**n) 0.001     `shouldBe`  2.714072874546881\n\n-- La verificaci\u00f3n es\n--    \u03bb> verifica\n--\n--    e1\n--    e2\n--\n--    Finished in 0.1927 seconds\n--    2 examples, 0 failures\n<\/pre>\n
<\/a>
\nSoluciones en Python<\/b><\/p>\n
\nfrom itertools import count\nfrom typing import Callable\n\n# 1\u00aa soluci\u00f3n\n# ===========\n\ndef limite(f: Callable[[float], float], a: float) -> float:\n    x = 1\n    while True:\n        maximum_diff = max(abs(f(y) - f(x)) for y in range(x+1, x+101))\n        if maximum_diff < a:\n            return f(x)\n        x += 1\n\n# 2\u00aa soluci\u00f3n\n# ===========\n\ndef limite2(f: Callable[[float], float], a: float) -> float:\n    x = 1\n    while True:\n        y = f(x)\n        if max(abs(y - f(x + i)) for i in range(1, 101)) < a:\n            break\n        x += 1\n    return y\n\n# 3\u00aa soluci\u00f3n\n# ===========\n\ndef limite3(f: Callable[[float], float], a: float) -> float:\n    for x in count(1):\n        if max(abs(f(y) - f(x)) for y in range(x + 1, x + 101)) < a:\n            r = f(x)\n            break\n    return r\n\n# Verificaci\u00f3n\n# ============\n\ndef test_limite() -> None:\n    assert limite(lambda n :  (2*n+1)\/(n+5), 0.001) ==  1.9900110987791344\n    assert limite(lambda n : (1+1\/n)**n, 0.001)     ==  2.714072874546881\n    assert limite2(lambda n :  (2*n+1)\/(n+5), 0.001) ==  1.9900110987791344\n    assert limite2(lambda n : (1+1\/n)**n, 0.001)     ==  2.714072874546881\n    assert limite3(lambda n :  (2*n+1)\/(n+5), 0.001) ==  1.9900110987791344\n    assert limite3(lambda n : (1+1\/n)**n, 0.001)     ==  2.714072874546881\n    print(\"Verificado\")\n\n# La comprobaci\u00f3n es\n#    >>> test_limite()\n#    Verificado\n<\/pre>\n","protected":false},"excerpt":{"rendered":"
Definir la funci\u00f3n limite :: (Double -> Double) -> Double -> Double tal que limite f a es el valor de f en el primer t\u00e9rmino x tal que, para todo y entre x+1 y x+100, el valor absoluto de la diferencia entre f(y) y f(x) es menor que a. Por ejemplo, limite (\\n ->…<\/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":"default","_kad_post_title":"default","_kad_post_layout":"default","_kad_post_sidebar_id":"","_kad_post_content_style":"default","_kad_post_vertical_padding":"default","_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":[581],"tags":[],"jetpack_featured_media_url":"","jetpack_sharing_enabled":true,"_links":{"self":[{"href":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/wp-json\/wp\/v2\/posts\/8334"}],"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=8334"}],"version-history":[{"count":2,"href":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/wp-json\/wp\/v2\/posts\/8334\/revisions"}],"predecessor-version":[{"id":8573,"href":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/wp-json\/wp\/v2\/posts\/8334\/revisions\/8573"}],"wp:attachment":[{"href":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/wp-json\/wp\/v2\/media?parent=8334"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/wp-json\/wp\/v2\/categories?post=8334"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/wp-json\/wp\/v2\/tags?post=8334"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}