{"id":34513,"date":"2025-11-26T04:51:41","date_gmt":"2025-11-26T03:51:41","guid":{"rendered":"https:\/\/xmau.com\/wp\/notiziole\/?p=34513"},"modified":"2025-11-26T09:24:51","modified_gmt":"2025-11-26T08:24:51","slug":"armonici-quasi-interi","status":"publish","type":"post","link":"https:\/\/xmau.com\/wp\/notiziole\/2025\/11\/26\/armonici-quasi-interi\/","title":{"rendered":"Armonici quasi interi"},"content":{"rendered":"<div class='__iawmlf-post-loop-links' style='display:none;' data-iawmlf-post-links='[{&quot;id&quot;:222,&quot;href&quot;:&quot;https:\\\/\\\/www.johndcook.com\\\/blog\\\/2025\\\/11\\\/19\\\/closest-consecutive-reciprocal-sum-to-an-integer&quot;,&quot;archived_href&quot;:&quot;http:\\\/\\\/web-wp.archive.org\\\/web\\\/20260118171533\\\/https:\\\/\\\/www.johndcook.com\\\/blog\\\/2025\\\/11\\\/19\\\/closest-consecutive-reciprocal-sum-to-an-integer\\\/&quot;,&quot;redirect_href&quot;:&quot;&quot;,&quot;checks&quot;:[{&quot;date&quot;:&quot;2026-02-11 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22:58:27&quot;,&quot;http_code&quot;:200},&quot;process&quot;:&quot;done&quot;},{&quot;id&quot;:225,&quot;href&quot;:&quot;https:\\\/\\\/www.johndcook.com\\\/blog\\\/2025\\\/11\\\/20\\\/solving-h_n-100&quot;,&quot;archived_href&quot;:&quot;http:\\\/\\\/web-wp.archive.org\\\/web\\\/20260118171538\\\/https:\\\/\\\/www.johndcook.com\\\/blog\\\/2025\\\/11\\\/20\\\/solving-h_n-100\\\/&quot;,&quot;redirect_href&quot;:&quot;&quot;,&quot;checks&quot;:[{&quot;date&quot;:&quot;2026-02-11 17:23:46&quot;,&quot;http_code&quot;:200},{&quot;date&quot;:&quot;2026-02-17 09:11:26&quot;,&quot;http_code&quot;:200},{&quot;date&quot;:&quot;2026-02-20 17:14:54&quot;,&quot;http_code&quot;:200},{&quot;date&quot;:&quot;2026-02-24 23:41:01&quot;,&quot;http_code&quot;:200},{&quot;date&quot;:&quot;2026-03-02 08:38:51&quot;,&quot;http_code&quot;:200},{&quot;date&quot;:&quot;2026-03-13 18:10:03&quot;,&quot;http_code&quot;:200},{&quot;date&quot;:&quot;2026-03-20 01:32:14&quot;,&quot;http_code&quot;:200},{&quot;date&quot;:&quot;2026-03-24 17:19:26&quot;,&quot;http_code&quot;:200},{&quot;date&quot;:&quot;2026-04-01 20:57:06&quot;,&quot;http_code&quot;:200},{&quot;date&quot;:&quot;2026-04-07 08:18:09&quot;,&quot;http_code&quot;:200},{&quot;date&quot;:&quot;2026-04-12 22:13:59&quot;,&quot;http_code&quot;:200},{&quot;date&quot;:&quot;2026-04-20 22:58:28&quot;,&quot;http_code&quot;:200}],&quot;broken&quot;:false,&quot;last_checked&quot;:{&quot;date&quot;:&quot;2026-04-20 22:58:28&quot;,&quot;http_code&quot;:200},&quot;process&quot;:&quot;done&quot;}]'><\/div>\n<p>Nel 1918 J\u00f3zsef K\u00fcrsch\u00e1k dimostr\u00f2 che la somma di reciproci di due o pi\u00f9 numeri consecutivi non pu\u00f2 mai essere un intero. Come corollario, l&#8217;unico valore intero toccato calcolando la serie armonica $H_n = 1 + 1\/2 + 1\/3 + 1\/4 + &#8230;$ \u00e8 1. Ci si pu\u00f2 per\u00f2 oziosamente chiedere quanto vicino si pu\u00f2 arrivare a un intero. John Cook in una successione di post ha mostrato che se ci limitiamo ai numeri da 1 a 100000 <a href=\"https:\/\/www.johndcook.com\/blog\/2025\/11\/19\/closest-consecutive-reciprocal-sum-to-an-integer\/\">abbiamo che<\/a> $$ \\sum_{k=27134}^{73756} \\frac{1}{k} \\approx 1$$ con un errore dell&#8217;ordine di $10^{-11}$, e questa \u00e8 la migliore approssimazione possibile a 1. <\/p>\n<p>Limitandoci alle porzioni di serie armonica, si arriva rapidamente a un problema: i numeri in virgola mobile non sono abbastanza precisi per fare tutte le addizioni. Fortunatamente abbiamo a nostra disposizione un&#8217;approssimazione molto buona: $H_n \\approx \\log n + \\gamma + \\frac{1}{2n} &#8211; \\frac{1}{12n^2}$, dove $\\gamma$ \u00e8 la <a href=\"https:\/\/it.wikipedia.org\/wiki\/Costante_di_Eulero-Mascheroni\">costante di Eulero-Mascheroni<\/a> pari a circa 0,57721. (Curiosit\u00e0: non \u00e8 noto se sia o no un numero irrazionale, ma tutti credono di s\u00ec, anche perch\u00e9 in caso contrario il suo denominatore dovrebbe avere almeno $10^{242080}$ cifre&#8230;). Notate come l&#8217;approssimazione usata da Cook sia molto pi\u00f9 accurata di quella usuale $H_n \\approx \\log n + \\gamma$, per andare pi\u00f9 sul sicuro. Cook si \u00e8 divertito a <a href=\"https:\/\/www.johndcook.com\/blog\/2025\/11\/19\/closest-harmonic-number-to-an-integer\/\">scrivere un programmino Python<\/a> per trovare il termine della serie armonica pi\u00f9 vicino a un numero (non necessariamente intero) dato, scoprendo per esempio che $H_12366 \\approx 9,99996214846655$.<\/p>\n<p>Ma anche questo programma, pur essendo ben fatto, ha dei problemi di arrotondamento se il numero cercato \u00e8 molto grande. Per esempio dice che se vogliamo arrivare ad approssimare 100 dobbiamo sommare 15092688622113830917200248731913020965388288 termini, e l&#8217;errore relativo \u00e8 dell&#8217;ordine di $3 \\times 10^{-15}$. Ma usando Mathematica e l&#8217;approssimazione $n \\approx \\rm{exp}(m \u2212 \\gamma)$, dove $m$ \u00e8 il numero che vogliamo approssimare e $n$ il numero di termini richiesti, Cook <a href=\"https:\/\/www.johndcook.com\/blog\/2025\/11\/20\/solving-h_n-100\/\">mostra<\/a> che la vera quantit\u00e0 di termini che dobbiamo sommare \u00e8 15092688622113788323693563264538101449859497; insomma i due numeri divergono dalla quattordicesima cifra, il che ha senso visto che siamo ai limiti della precisione dei numeri a 64 bit. Per curiosit\u00e0, per arrivare a 1000 occorrono un bel po&#8217; di termini, cio\u00e8 <\/p>\n<p>110611511026604935641074705584421138393028001852577373936470952377218354575172401275457597579044729873152469512963401398362087144972181770571895264066114088968182356842977823764462179821981744448731785408629116321919957856034605877855212667092287520105386027668843119590555646814038787297694678647529533718769401069269427475868793531944696435696745559289326610132208504257721469829210704462876574915362273129090049477919400226313586033<\/p>\n<p>(un numero di 435 cifre). Che possiamo dedurre da tutto questo? Due cose. La prima \u00e8 che la serie armonica cresce <b>molto<\/b> lentamente; la seconda \u00e8 che bisogna sempre sapere qual \u00e8 il modo migliore per fare un conto, tenendo conto delle limitazioni dei computer&#8230;<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Come si pu\u00f2 trovare l&#8217;approssimazione di un intero sommando la serie armonica?<\/p>\n","protected":false},"author":2,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_lmt_disableupdate":"no","_lmt_disable":"no","jetpack_post_was_ever_published":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,"activitypub_content_warning":"","activitypub_content_visibility":"","activitypub_max_image_attachments":3,"activitypub_interaction_policy_quote":"anyone","activitypub_status":"federated","footnotes":"","jetpack_publicize_message":"","jetpack_publicize_feature_enabled":true,"jetpack_social_post_already_shared":true,"jetpack_social_options":{"image_generator_settings":{"template":"highway","default_image_id":0,"font":"","enabled":false},"version":2}},"categories":[1005,214],"tags":[],"class_list":["post-34513","post","type-post","status-publish","format-standard","hentry","category-matelight-2025","category-matematica_light"],"modified_by":".mau.","jetpack_publicize_connections":[],"jetpack_featured_media_url":"","jetpack_shortlink":"https:\/\/wp.me\/p6hcSh-8YF","jetpack-related-posts":[{"id":29041,"url":"https:\/\/xmau.com\/wp\/notiziole\/2024\/06\/05\/la-serie-di-kempner\/","url_meta":{"origin":34513,"position":0},"title":"La serie di Kempner","author":".mau.","date":"2024-06-05","format":false,"excerpt":"La serie armonica diverge cos\u00ec lentamente che non le si pu\u00f2 togliere troppa roba","rel":"","context":"In &quot;mate-light-2024&quot;","block_context":{"text":"mate-light-2024","link":"https:\/\/xmau.com\/wp\/notiziole\/category\/matematica_light\/matelight-2024\/"},"img":{"alt_text":"la serie armonica","src":"https:\/\/i0.wp.com\/xmau.com\/wp\/notiziole\/wp-content\/uploads\/sites\/6\/2024\/06\/HarmonicNumbers.svg_-300x240.png?resize=350%2C200&ssl=1","width":350,"height":200},"classes":[]},{"id":18993,"url":"https:\/\/xmau.com\/wp\/notiziole\/2019\/08\/18\/quizzino-della-domenica-somme-armoniche\/","url_meta":{"origin":34513,"position":1},"title":"Quizzino della domenica: somme armoniche","author":".mau.","date":"2019-08-18","format":false,"excerpt":"La funzione armonica H(n) \u00e8 definita come la somma delle frazioni 1 + 1\/2 + 1\/3 + ... + 1\/n. Dimostrate che a parte H(1) la funzione non potr\u00e0 mai avere un valore intero. (un aiutino lo trovate sul mio sito, alla pagina http:\/\/xmau.com\/quizzini\/p398.html; la risposta verr\u00e0 postata l\u00ec il\u2026","rel":"","context":"In &quot;giochi&quot;","block_context":{"text":"giochi","link":"https:\/\/xmau.com\/wp\/notiziole\/category\/giochi\/"},"img":{"alt_text":"","src":"https:\/\/i0.wp.com\/xmau.com\/wp\/notiziole\/wp-content\/uploads\/sites\/6\/2019\/07\/q398a.png?resize=350%2C200","width":350,"height":200},"classes":[]},{"id":7749,"url":"https:\/\/xmau.com\/wp\/notiziole\/2010\/03\/11\/fibonacci_e_lin\/","url_meta":{"origin":34513,"position":2},"title":"Fibonacci e l&#8217;induzione","author":".mau.","date":"2010-03-11","format":false,"excerpt":"un teorema sui numeri di Fibonacci dimostrabile con l'induzione","rel":"","context":"In &quot;matematica_light&quot;","block_context":{"text":"matematica_light","link":"https:\/\/xmau.com\/wp\/notiziole\/category\/matematica_light\/"},"img":{"alt_text":"","src":"","width":0,"height":0},"classes":[]},{"id":34584,"url":"https:\/\/xmau.com\/wp\/notiziole\/2025\/12\/03\/come-dimostrare-che-e-e-irrazionale\/","url_meta":{"origin":34513,"position":3},"title":"Come dimostrare che e \u00e8 irrazionale","author":".mau.","date":"2025-12-03","format":false,"excerpt":"Una dimostrazione alla portata di uno studente liceale.","rel":"","context":"In &quot;mate-light-2025&quot;","block_context":{"text":"mate-light-2025","link":"https:\/\/xmau.com\/wp\/notiziole\/category\/matematica_light\/matelight-2025\/"},"img":{"alt_text":"","src":"","width":0,"height":0},"classes":[]},{"id":31571,"url":"https:\/\/xmau.com\/wp\/notiziole\/2025\/02\/26\/il-rapporto-superaureo-2\/","url_meta":{"origin":34513,"position":4},"title":"Il rapporto superaureo &#8211; 2","author":".mau.","date":"2025-02-26","format":false,"excerpt":"Continuiamo a vedere le propriet\u00e0 del rapporto superaureo, e le somiglianze e differenze con il rapporto aureo.","rel":"","context":"In &quot;mate-light-2025&quot;","block_context":{"text":"mate-light-2025","link":"https:\/\/xmau.com\/wp\/notiziole\/category\/matematica_light\/matelight-2025\/"},"img":{"alt_text":"spirale superaurea","src":"https:\/\/i0.wp.com\/xmau.com\/wp\/notiziole\/wp-content\/uploads\/sites\/6\/2025\/02\/Supergolden_spiral.svg_.png?resize=350%2C200&ssl=1","width":350,"height":200,"srcset":"https:\/\/i0.wp.com\/xmau.com\/wp\/notiziole\/wp-content\/uploads\/sites\/6\/2025\/02\/Supergolden_spiral.svg_.png?resize=350%2C200&ssl=1 1x, https:\/\/i0.wp.com\/xmau.com\/wp\/notiziole\/wp-content\/uploads\/sites\/6\/2025\/02\/Supergolden_spiral.svg_.png?resize=525%2C300&ssl=1 1.5x"},"classes":[]},{"id":32915,"url":"https:\/\/xmau.com\/wp\/notiziole\/2025\/06\/25\/il-problema-di-langford-ii\/","url_meta":{"origin":34513,"position":5},"title":"Il problema di Langford (II)","author":".mau.","date":"2025-06-25","format":false,"excerpt":"Ancora sul problema di Langford","rel":"","context":"In &quot;mate-light-2025&quot;","block_context":{"text":"mate-light-2025","link":"https:\/\/xmau.com\/wp\/notiziole\/category\/matematica_light\/matelight-2025\/"},"img":{"alt_text":"","src":"","width":0,"height":0},"classes":[]}],"jetpack_likes_enabled":true,"jetpack_sharing_enabled":true,"_links":{"self":[{"href":"https:\/\/xmau.com\/wp\/notiziole\/wp-json\/wp\/v2\/posts\/34513","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/xmau.com\/wp\/notiziole\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/xmau.com\/wp\/notiziole\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/xmau.com\/wp\/notiziole\/wp-json\/wp\/v2\/users\/2"}],"replies":[{"embeddable":true,"href":"https:\/\/xmau.com\/wp\/notiziole\/wp-json\/wp\/v2\/comments?post=34513"}],"version-history":[{"count":7,"href":"https:\/\/xmau.com\/wp\/notiziole\/wp-json\/wp\/v2\/posts\/34513\/revisions"}],"predecessor-version":[{"id":34524,"href":"https:\/\/xmau.com\/wp\/notiziole\/wp-json\/wp\/v2\/posts\/34513\/revisions\/34524"}],"wp:attachment":[{"href":"https:\/\/xmau.com\/wp\/notiziole\/wp-json\/wp\/v2\/media?parent=34513"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/xmau.com\/wp\/notiziole\/wp-json\/wp\/v2\/categories?post=34513"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/xmau.com\/wp\/notiziole\/wp-json\/wp\/v2\/tags?post=34513"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}