","image/vnd.openai.fileservice2.png","{\"url\": \"file-service://file-5TzkrzNkUyD3wcLDXeyjHQ\", \"image_width\": 1099, \"image_height\": 964, \"bytes\": 230279}","image/vnd.openai.fileservice.png","file-service://file-5TzkrzNkUyD3wcLDXeyjHQ",{"_250":60,"_251":280,"_53":281},{"_253":254,"_255":256},{"_258":282},"idle","timeout_triggered",[285],"85e779b3-c00f-49d3-9642-981b4ef9405d",{"_40":285,"_45":287,"_70":211,"_41":300},{"_40":285,"_47":288,"_32":290,"_53":291,"_60":61,"_62":14,"_63":91,"_51":297,"_68":69},{"_49":146,"_51":289},{},1748345605.209603,{"_55":161,"_161":292,"_174":296},[293],{"_164":171,"_53":294,"_167":295,"_169":12},"Возможно, картинку не удастся увидеть в сети, но, возможно, я смогу вывести её. Однако, если доступ к изображению закрыт, можно попробовать вывести регион, чтобы ориентироваться по приблизительному местоположению. Но если изображения нет, я могу попробовать получить информацию, прочитав текстовый фрагмент. В любом случае, открою изображение и попробую его исследовать.",[],"5e487ce0-c7b2-49ce-ade0-725b0fabf804",{"_177":178,"_179":298,"_181":299,"_140":-5,"_183":184,"_185":184,"_186":211,"_135":136,"_138":139,"_67":24},[],[],[301],"d33660eb-0f26-4e1b-9cd1-011a049d0ca3",{"_40":301,"_45":303,"_70":285,"_41":314},{"_40":301,"_47":304,"_32":306,"_53":307,"_60":61,"_62":14,"_63":91,"_51":309,"_68":209},{"_49":146,"_51":305},{},1748345596.533099,{"_55":195,"_196":197,"_56":308},"plt.figure(figsize=(6,4))\nplt.imshow(img.crop((0,200,800,750)))\nplt.axis('off')",{"_200":310,"_206":12,"_179":312,"_181":313,"_177":178,"_140":-5,"_183":184,"_185":184,"_186":285,"_135":136,"_138":139,"_67":24},{"_27":202,"_203":311},[205],[],[],[315],"d9164c44-cf67-4318-a80d-a5759d653ac5",{"_40":315,"_45":317,"_70":301,"_41":354},{"_40":315,"_47":318,"_32":320,"_34":321,"_53":322,"_60":61,"_63":91,"_51":324,"_68":69},{"_49":215,"_125":209,"_51":319},{},1748345605.0992885,1748345608.280453,{"_55":220,"_56":323},"\n<>(-0.5, 799.5, 549.5, -0.5)",{"_206":12,"_223":325,"_177":178,"_140":-5,"_183":184,"_185":184,"_186":301,"_135":136,"_138":139,"_67":24},{"_60":225,"_226":326,"_228":320,"_34":321,"_195":308,"_229":321,"_230":327,"_232":-5,"_233":-5,"_234":328,"_247":334,"_283":-5},"aa27f9c7-53d7-4811-9e75-4fc37c145a8e","(-0.5, 799.5, 549.5, -0.5)",[329],{"_140":237,"_238":330,"_240":241,"_242":-5,"_243":331,"_112":332,"_114":333},1748345608.1926336,"file-service://file-4z93p3JVSZSp5fVuWPzLsF?shared_conversation_id=683800b8-6584-8004-96d7-8978af7c3453",936,656,[335,338,340,344,351],{"_250":60,"_251":336,"_53":337},{"_253":254,"_255":256},{"_258":259},{"_250":261,"_251":339},{"_253":254,"_255":256},{"_250":264,"_251":341,"_53":342},{"_253":254,"_255":256},{"_28":343},{"_268":327},{"_250":270,"_251":345,"_53":346},{"_253":254,"_255":256},{"_28":347},{"_268":348,"_275":349,"_277":350},"

","{\"url\": \"file-service://file-4z93p3JVSZSp5fVuWPzLsF\", \"image_width\": 936, \"image_height\": 656, \"bytes\": 76273}","file-service://file-4z93p3JVSZSp5fVuWPzLsF",{"_250":60,"_251":352,"_53":353},{"_253":254,"_255":256},{"_258":282},[355],"2bf5136d-1e51-4b73-8409-684c88f3b578",{"_40":355,"_45":357,"_70":315,"_41":371},{"_40":355,"_47":358,"_32":360,"_53":361,"_60":61,"_62":14,"_63":91,"_51":368,"_68":69},{"_49":146,"_51":359},{},1748345615.453517,{"_55":161,"_161":362,"_174":367},[363],{"_164":364,"_53":365,"_167":366,"_169":12},"Работа с изображением","Я думаю, что, возможно, изображение не будет доступно для меня, если оно в интернете, но я могу его отобразить. Однако, чтобы добиться точности, можно попытаться рассчитать регион. Вместо этого можно приблизительно определить его, прочитав текстовый фрагмент. Открою изображение, увеличу его, а затем, возможно, подрежу, чтобы сфокусироваться на нужной части. Так я смогу получить точный фрагмент для анализа.",[],"e6bfe975-db78-4986-85c8-cf8cffaddd80",{"_177":178,"_179":369,"_181":370,"_140":-5,"_183":184,"_185":184,"_186":315,"_135":136,"_138":139,"_67":24},[],[],[372],"e29cd2cb-4e29-486c-87fe-936ff7424552",{"_40":372,"_45":374,"_70":355,"_41":385},{"_40":372,"_47":375,"_32":377,"_53":378,"_60":61,"_62":14,"_63":91,"_51":380,"_68":209},{"_49":146,"_51":376},{},1748345608.389279,{"_55":195,"_196":197,"_56":379},"plt.figure(figsize=(6,4))\nplt.imshow(img.crop((0,430,800,1130)))\nplt.axis('off')",{"_200":381,"_206":12,"_179":383,"_181":384,"_177":178,"_140":-5,"_183":184,"_185":184,"_186":355,"_135":136,"_138":139,"_67":24},{"_27":202,"_203":382},[205],[],[],[386],"cae2b43e-9fbd-4fbb-9ff7-5723c4849800",{"_40":386,"_45":388,"_70":372,"_41":423},{"_40":386,"_47":389,"_32":391,"_34":392,"_53":393,"_60":61,"_63":91,"_51":395,"_68":69},{"_49":215,"_125":209,"_51":390},{},1748345615.3082397,1748345616.468553,{"_55":220,"_56":394},"\n<>(-0.5, 799.5, 699.5, -0.5)",{"_206":12,"_223":396,"_177":178,"_140":-5,"_183":184,"_185":184,"_186":372,"_135":136,"_138":139,"_67":24},{"_60":225,"_226":397,"_228":391,"_34":392,"_195":379,"_229":392,"_230":398,"_232":-5,"_233":-5,"_234":399,"_247":404,"_283":-5},"98e8e84f-d13a-4a4b-ac97-af20953c5f5d","(-0.5, 799.5, 699.5, -0.5)",[400],{"_140":237,"_238":401,"_240":241,"_242":-5,"_243":402,"_112":403,"_114":333},1748345616.3825374,"file-service://file-PQBVupLQTM1pXGgP7Pphbg?shared_conversation_id=683800b8-6584-8004-96d7-8978af7c3453",744,[405,408,410,414,420],{"_250":60,"_251":406,"_53":407},{"_253":254,"_255":256},{"_258":259},{"_250":261,"_251":409},{"_253":254,"_255":256},{"_250":264,"_251":411,"_53":412},{"_253":254,"_255":256},{"_28":413},{"_268":398},{"_250":270,"_251":415,"_53":416},{"_253":254,"_255":256},{"_28":417},{"_268":348,"_275":418,"_277":419},"{\"url\": \"file-service://file-PQBVupLQTM1pXGgP7Pphbg\", \"image_width\": 744, \"image_height\": 656, \"bytes\": 109417}","file-service://file-PQBVupLQTM1pXGgP7Pphbg",{"_250":60,"_251":421,"_53":422},{"_253":254,"_255":256},{"_258":282},[424],"324f7c93-cf3d-4d15-9692-b0c4d7f56317",{"_40":424,"_45":426,"_70":386,"_41":439},{"_40":424,"_47":427,"_32":429,"_53":430,"_60":61,"_62":14,"_63":91,"_51":436,"_68":69},{"_49":146,"_51":428},{},1748345624.160036,{"_55":161,"_161":431,"_174":435},[432],{"_164":364,"_53":433,"_167":434,"_169":12},"Похоже, я не смогу полностью просматривать изображение, если оно ограничено доступом. Однако я могу предположить его местоположение, основываясь на фрагментах текста. Открою изображение, увеличу его, а затем подрежу, чтобы сосредоточиться на нужной части. Думаю, стоит продолжить обрезку, чтобы получить более точные данные, необходимые для дальнейшего анализа.",[],"ee004053-6100-48f5-8cf7-0ac138764590",{"_177":178,"_179":437,"_181":438,"_140":-5,"_183":184,"_185":184,"_186":386,"_135":136,"_138":139,"_67":24},[],[],[440],"f7b34b98-eeb7-4267-b7a6-736f993645d7",{"_40":440,"_45":442,"_70":424,"_41":453},{"_40":440,"_47":443,"_32":445,"_53":446,"_60":61,"_62":14,"_63":91,"_51":448,"_68":209},{"_49":146,"_51":444},{},1748345616.571369,{"_55":195,"_196":197,"_56":447},"plt.figure(figsize=(6,4))\nplt.imshow(img.crop((0,820,800,1420)))\nplt.axis('off')",{"_200":449,"_206":12,"_179":451,"_181":452,"_177":178,"_140":-5,"_183":184,"_185":184,"_186":424,"_135":136,"_138":139,"_67":24},{"_27":202,"_203":450},[205],[],[],[454],"74c3ad4b-5517-40c8-858c-723375519ecf",{"_40":454,"_45":456,"_70":440,"_41":491},{"_40":454,"_47":457,"_32":459,"_34":460,"_53":461,"_60":61,"_63":91,"_51":463,"_68":69},{"_49":215,"_125":209,"_51":458},{},1748345624.0509114,1748345625.0171194,{"_55":220,"_56":462},"\n<>(-0.5, 799.5, 599.5, -0.5)",{"_206":12,"_223":464,"_177":178,"_140":-5,"_183":184,"_185":184,"_186":440,"_135":136,"_138":139,"_67":24},{"_60":225,"_226":465,"_228":459,"_34":460,"_195":447,"_229":460,"_230":466,"_232":-5,"_233":-5,"_234":467,"_247":472,"_283":-5},"288f28bf-bbfa-48a0-8714-1462b413e8bf","(-0.5, 799.5, 599.5, -0.5)",[468],{"_140":237,"_238":469,"_240":241,"_242":-5,"_243":470,"_112":471,"_114":333},1748345624.9223921,"file-service://file-Sc45awP31KxKiy64yYY1Ax?shared_conversation_id=683800b8-6584-8004-96d7-8978af7c3453",861,[473,476,478,482,488],{"_250":60,"_251":474,"_53":475},{"_253":254,"_255":256},{"_258":259},{"_250":261,"_251":477},{"_253":254,"_255":256},{"_250":264,"_251":479,"_53":480},{"_253":254,"_255":256},{"_28":481},{"_268":466},{"_250":270,"_251":483,"_53":484},{"_253":254,"_255":256},{"_28":485},{"_268":348,"_275":486,"_277":487},"{\"url\": \"file-service://file-Sc45awP31KxKiy64yYY1Ax\", \"image_width\": 861, \"image_height\": 656, \"bytes\": 119959}","file-service://file-Sc45awP31KxKiy64yYY1Ax",{"_250":60,"_251":489,"_53":490},{"_253":254,"_255":256},{"_258":282},[492],"a2511100-ed35-4feb-81d9-a08e92992a4c",{"_40":492,"_45":494,"_70":454,"_41":508},{"_40":492,"_47":495,"_32":497,"_53":498,"_60":61,"_62":14,"_63":91,"_51":505,"_68":69},{"_49":146,"_51":496},{},1748345632.763128,{"_55":161,"_161":499,"_174":504},[500],{"_164":501,"_53":502,"_167":503,"_169":12},"Корректировка изображения","Я не могу полностью просматривать изображение, если оно недоступно, но могу попытаться понять примерно его местоположение. Открою изображение, увеличу и подрежу, чтобы сосредоточиться на нужной области. Но, похоже, обрезка всё еще мешает видеть полный текст, особенно правую часть. Тогда, возможно, стоит увеличить горизонтальную ширину окна для лучшего отображения.",[],"183b9471-580c-40bb-8964-60f83e1354fd",{"_177":178,"_179":506,"_181":507,"_140":-5,"_183":184,"_185":184,"_186":454,"_135":136,"_138":139,"_67":24},[],[],[509],"2e82a9bd-81e0-4c6d-a4f5-4243b5c6d894",{"_40":509,"_45":511,"_70":492,"_41":522},{"_40":509,"_47":512,"_32":514,"_53":515,"_60":61,"_62":14,"_63":91,"_51":517,"_68":209},{"_49":146,"_51":513},{},1748345625.153804,{"_55":195,"_196":197,"_56":516},"plt.figure(figsize=(8,4))\nplt.imshow(img.crop((0,430,1500,1130)))\nplt.axis('off')",{"_200":518,"_206":12,"_179":520,"_181":521,"_177":178,"_140":-5,"_183":184,"_185":184,"_186":492,"_135":136,"_138":139,"_67":24},{"_27":202,"_203":519},[205],[],[],[523],"81ca779d-603a-4a75-9481-8b187a117d1b",{"_40":523,"_45":525,"_70":509,"_41":562},{"_40":523,"_47":526,"_32":528,"_34":529,"_53":530,"_60":61,"_63":91,"_51":532,"_68":69},{"_49":215,"_125":209,"_51":527},{},1748345632.6691613,1748345633.7659712,{"_55":220,"_56":531},"\n<>(-0.5, 1499.5, 699.5, -0.5)",{"_206":12,"_223":533,"_177":178,"_140":-5,"_183":184,"_185":184,"_186":509,"_135":136,"_138":139,"_67":24},{"_60":225,"_226":534,"_228":528,"_34":529,"_195":516,"_229":529,"_230":535,"_232":-5,"_233":-5,"_234":536,"_247":542,"_283":-5},"59fda499-6b36-427c-a099-74c3e6a4fc43","(-0.5, 1499.5, 699.5, -0.5)",[537],{"_140":237,"_238":538,"_240":241,"_242":-5,"_243":539,"_112":540,"_114":541},1748345633.682861,"file-service://file-LTfvsUdBEBt3tCB9A7sAgB?shared_conversation_id=683800b8-6584-8004-96d7-8978af7c3453",1280,618,[543,546,548,552,559],{"_250":60,"_251":544,"_53":545},{"_253":254,"_255":256},{"_258":259},{"_250":261,"_251":547},{"_253":254,"_255":256},{"_250":264,"_251":549,"_53":550},{"_253":254,"_255":256},{"_28":551},{"_268":535},{"_250":270,"_251":553,"_53":554},{"_253":254,"_255":256},{"_28":555},{"_268":556,"_275":557,"_277":558},"

","{\"url\": \"file-service://file-LTfvsUdBEBt3tCB9A7sAgB\", \"image_width\": 1280, \"image_height\": 618, \"bytes\": 159049}","file-service://file-LTfvsUdBEBt3tCB9A7sAgB",{"_250":60,"_251":560,"_53":561},{"_253":254,"_255":256},{"_258":282},[563],"6469b74e-8f97-4b71-8d7c-ce7285ff22d9",{"_40":563,"_45":565,"_70":523,"_41":579},{"_40":563,"_47":566,"_32":568,"_53":569,"_60":61,"_62":14,"_63":91,"_51":576,"_68":69},{"_49":146,"_51":567},{},1748345640.967252,{"_55":161,"_161":570,"_174":575},[571],{"_164":572,"_53":573,"_167":574,"_169":12},"Корректировка изображения для деталей","Я пытаюсь понять, что можно сделать с изображением, чтобы можно было увидеть больше текста. Обрезание всё ещё мешает видеть полную информацию, особенно в правой части. Для улучшения стоит расширить горизонтальную область. Мы видим больше деталей, и теперь нужно увеличить зум, чтобы чётко увидеть, где расположено определение κ_n. 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52827}","file-service://file-CpwuqL91oXZGdwQnrd6DLx",{"_250":60,"_251":1045,"_53":1046},{"_253":254,"_255":256},{"_258":282},[1048],"df4cc6ed-545e-4d66-8246-dd76eebda848",{"_40":1048,"_45":1050,"_70":1010,"_41":1064},{"_40":1048,"_47":1051,"_32":1053,"_53":1054,"_60":61,"_62":14,"_63":91,"_51":1061,"_68":69},{"_49":146,"_51":1052},{},1748345718.894233,{"_55":161,"_161":1055,"_174":1060},[1056],{"_164":1057,"_53":1058,"_167":1059,"_169":12},"Анализ уравнений","Теперь мы можем полностью рассмотреть уравнение. Для \$ \\kappa_n \$ индекс в показателе степени, похоже, описан как минимальное значение из \$ \\xi_1, \\xi_4, \\xi_9, \\dots, \\xi_{\\lfloor\\sqrt{n}\\rfloor^2} \$. Нужно ещё уточнить, что это реально говорят по индексам, так как \$ 1, 4, 9, \\dots \$ указывают на квадраты чисел с корнем из \$ n \$. Я увеличу изображение, чтобы точно разобраться в номерах в подстрочных индексах!",[],"972586c1-8b63-4fcb-be2a-9b8655ccc837",{"_177":178,"_179":1062,"_181":1063,"_140":-5,"_183":184,"_185":184,"_186":1010,"_135":136,"_138":139,"_67":24},[],[],[1065],"758db5e4-7a83-4466-b24f-6ffeb27e76c4",{"_40":1065,"_45":1067,"_70":1048,"_41":1078},{"_40":1065,"_47":1068,"_32":1070,"_53":1071,"_60":61,"_62":14,"_63":91,"_51":1073,"_68":209},{"_49":146,"_51":1069},{},1748345708.130659,{"_55":195,"_196":197,"_56":1072},"plt.figure(figsize=(6,3))\nplt.imshow(img.crop((915,960,1110,1030)))\nplt.axis('off')",{"_200":1074,"_206":12,"_179":1076,"_181":1077,"_177":178,"_140":-5,"_183":184,"_185":184,"_186":1048,"_135":136,"_138":139,"_67":24},{"_27":202,"_203":1075},[205],[],[],[1079],"e19dc82d-3e9e-4d0b-9a81-42ba4aabe1b7",{"_40":1079,"_45":1081,"_70":1065,"_41":1116},{"_40":1079,"_47":1082,"_32":1084,"_34":1085,"_53":1086,"_60":61,"_63":91,"_51":1088,"_68":69},{"_49":215,"_125":209,"_51":1083},{},1748345718.5578098,1748345719.2733834,{"_55":220,"_56":1087},"\n<>(-0.5, 194.5, 69.5, -0.5)",{"_206":12,"_223":1089,"_177":178,"_140":-5,"_183":184,"_185":184,"_186":1065,"_135":136,"_138":139,"_67":24},{"_60":225,"_226":1090,"_228":1084,"_34":1085,"_195":1072,"_229":1085,"_230":1091,"_232":-5,"_233":-5,"_234":1092,"_247":1097,"_283":-5},"ec76b353-b6a3-4c44-a2c8-d4010ff1edd9","(-0.5, 194.5, 69.5, -0.5)",[1093],{"_140":237,"_238":1094,"_240":241,"_242":-5,"_243":1095,"_112":732,"_114":1096},1748345719.199231,"file-service://file-R4jBDRjQVpSHujqmiSF3ZK?shared_conversation_id=683800b8-6584-8004-96d7-8978af7c3453",373,[1098,1101,1103,1107,1113],{"_250":60,"_251":1099,"_53":1100},{"_253":254,"_255":256},{"_258":259},{"_250":261,"_251":1102},{"_253":254,"_255":256},{"_250":264,"_251":1104,"_53":1105},{"_253":254,"_255":256},{"_28":1106},{"_268":1091},{"_250":270,"_251":1108,"_53":1109},{"_253":254,"_255":256},{"_28":1110},{"_268":748,"_275":1111,"_277":1112},"{\"url\": \"file-service://file-R4jBDRjQVpSHujqmiSF3ZK\", \"image_width\": 969, \"image_height\": 373, \"bytes\": 7325}","file-service://file-R4jBDRjQVpSHujqmiSF3ZK",{"_250":60,"_251":1114,"_53":1115},{"_253":254,"_255":256},{"_258":282},[1117],"02a63129-9e05-46b3-85be-eb3a7629d576",{"_40":1117,"_45":1119,"_70":1079,"_41":1137},{"_40":1117,"_47":1120,"_32":1122,"_53":1123,"_60":61,"_62":14,"_63":91,"_51":1134,"_68":69},{"_49":146,"_51":1121},{},1748345727.075994,{"_55":161,"_161":1124,"_174":1133},[1125,1129],{"_164":1126,"_53":1127,"_167":1128,"_169":12},"Анализ экспоненты в уравнении","Кажется, что экспонента в уравнении для \$ \\kappa_n \$ выражена через минимальное значение из \$ \\xi_1, \\xi_4, \\xi_9, \\dots, \\xi_{\\lfloor\\sqrt{n}\\rfloor^2} \$, где индексы — это квадраты чисел. Это подходит, так как они могут быть равны \$ i^2 \$ для \$ i = 1,2,3,\\dots \$. Я подробно проверю это, увеличив изображение и уточнив детали.",[],{"_164":1130,"_53":1131,"_167":1132,"_169":12},"Реконструкция проблемы","Экспонента использует квадраты индексов до \$ \\lfloor \\sqrt{n} \\rfloor^2 \$, а произведение — кубы индексов до \$ \\lfloor n^{1/3} \\rfloor^3 \$. Выводя итоговое формулирование, получаем:\n\n\\[ \\kappa_n = \\left( \\xi_1 \\xi_8 \\xi_{27} \\dots \\xi_{\\lfloor n^{1/3} \\rfloor^3} \\right)^{\\min \\{\\xi_1, \\xi_4, \\xi_9, \\dots, \\xi_{\\lfloor \\sqrt{n} \\rfloor^2}\\}}. \\]\n\nЗадача сводится к нахождению предела распределения \$ \\kappa_n \$ при \$ n \\to \\infty \$.",[],"e49b9ae0-8c3a-42d4-9997-ae966a212cc6",{"_177":178,"_179":1135,"_181":1136,"_140":-5,"_183":184,"_185":184,"_186":1079,"_135":136,"_138":139,"_67":24},[],[],[1138],"fd0b41ec-2eab-4503-8f77-a86138d74e7b",{"_40":1138,"_45":1140,"_70":1117,"_41":1151},{"_40":1138,"_47":1141,"_32":1143,"_53":1144,"_60":61,"_62":14,"_63":91,"_51":1147,"_68":69},{"_49":146,"_51":1142},{},1748345805.554871,{"_55":1145,"_53":1146},"reasoning_recap","Думал на протяжении 4m 10s",{"_177":1148,"_1149":1150,"_140":-5,"_183":184,"_185":184,"_186":1117,"_135":136,"_138":139,"_67":24},"reasoning_ended","finished_duration_sec",250,[1152],"6eb991e2-5057-4a2f-8d71-148761dd16fc",{"_40":1152,"_45":1154,"_70":1138,"_41":1166},{"_40":1152,"_47":1155,"_32":1157,"_53":1158,"_60":61,"_62":12,"_63":91,"_51":1161,"_68":69,"_1164":1165},{"_49":146,"_51":1156},{},1748345719.379586,{"_55":56,"_57":1159},[1160],"```latex\n\\section*{В. Минимальный минимум}\n\nПусть $\\xi_{1},\\xi_{2},\\dots$ — независимые положительные случайные величины, \nчья плотность пропорциональна\n\\[\np(x)=\\min\\{\\sqrt{x},1\\}\\,\\cdot\\,\\min\\!\\Bigl\\{\\tfrac{1}{x^{2}},\\,1\\Bigr\\},\\qquad x>0 .\n\\]\n\nНайдите предел по распределению при $n\\to\\infty$ случайных величин \n\\[\n\\kappa_{n}\n=\\Bigl(\\xi_{1}\\,\\xi_{8}\\,\\xi_{27}\\,\\dotsm\\,\\xi_{\\lfloor n^{1/3}\\rfloor^{3}}\\Bigr)^{\\displaystyle\n \\min\\{\\xi_{1},\\xi_{4},\\xi_{9},\\dots,\\xi_{\\lfloor\\sqrt{n}\\rfloor^{2}}\\}},\n\\]\nто есть функцию распределения такой случайной величины $\\kappa$, что \n\\[\n\\kappa_{n}\\xrightarrow{\\,d\\,}\\kappa .\n\\]\n\nВ качестве ответа требуется указать значение функции распределения $\\kappa$ \nв точке $x_{0}=\\pi$ либо число $2112$, если последовательность \n$\\kappa_{n}$ не сходится по распределению. \nВведённое число должно отличаться от правильного ответа не более чем на~$10^{-9}$.\n```",{"_179":1162,"_181":1163,"_140":-5,"_183":184,"_185":184,"_186":1138,"_135":136,"_138":139,"_67":24},[],[],"channel","final",[1167],"aa71f322-b91b-4da4-9609-752c7ccd3127",{"_40":1167,"_45":1169,"_70":1152,"_41":1178},{"_40":1167,"_47":1170,"_32":1172,"_53":1173,"_60":61,"_63":91,"_51":1176,"_68":69},{"_49":86,"_51":1171},{},1748347588.074,{"_55":56,"_57":1174},[1175],"Напиши решение задачи.",{"_135":1177,"_137":-5,"_138":139,"_140":-5,"_67":24},"946562eef92efd15-SEA",[1179],"bdca21ab-be76-4e0a-879a-c658e13a8efe",{"_40":1179,"_45":1181,"_70":1167,"_41":1207},{"_40":1179,"_47":1182,"_32":1184,"_53":1185,"_60":61,"_62":14,"_63":91,"_51":1204,"_68":69},{"_49":146,"_51":1183},{},1748347610.66006,{"_55":161,"_161":1186,"_174":1203},[1187,1191,1195,1199],{"_164":1188,"_53":1189,"_167":1190,"_169":12},"Решение задачи","Задано множество независимых положительных случайных величин ξ_i с плотностью, пропорциональной p(x) = min(√x, 1) * min(1/x^2, 1) для x > 0. Мы определяем κ_n как сочетание произведений и минимальных значений этих переменных. Задача заключается в нахождении предела распределения κ_n, когда n стремится к бесконечности, а также вычислении функции распределения на x0 = π. Для этого подбираем последовательности индексов и анализируем их поведение на больших значениях n.",[],{"_164":1192,"_53":1193,"_167":1194,"_169":12},"Рассматриваем решение","Формула для κ_n включает произведение и минимумы из случайных величин, где есть пересечения индексов, которые являются как кубами, так и квадратами. Это влияет на финальные вычисления, но обработка этих пересечений manageable. Нужно проанализировать распределение минимальных величин для большого n и поведение функции плотности около 0. По задаче, при n → ∞, возможно, будет наблюдаться тенденция к 0 для показателя. Результат может привести к тому, что весь продукт со степенью сходится к 1.",[],{"_164":1196,"_53":1197,"_167":1198,"_169":12},"Анализ распределения κ_n","Мы рассматриваем распределение переменной κ_n, основанной на произведении и минимуме случайных величин. Нужно вычислить поведение продукта и экспоненты, включая возможные пределы для базы и показателя. Рассмотрим распределение переменной E_n, которая определяет минимумы, и вычислим ее асимптотику. Также важен процесс нормализации плотности. После детального анализа математических выражений мы получаем приближенное распределение для κ_n, которое стремится к экспоненциальному виду с определенной нормализацией.",[],{"_164":1200,"_53":1201,"_167":1202,"_169":12},"Вычисление ожидания","Необходимо вычислить E[ln ξ] = (3/5)( ∫_0^1 √x ln x dx + ∫_1^∞ (ln x)/ x^2 dx ). Первый интеграл, вероятно, отрицателен, второй — положителен, но маленький. Я проведу численное вычисление этих интегралов, используя приближённое интегрирование. Примерно оценив результаты интегралов, смогу вычислить значение μ численно.",[],"9c64f605-1451-4d6d-9096-b2c5d2f5d24c",{"_177":178,"_179":1205,"_181":1206,"_140":-5,"_183":184,"_185":184,"_186":1167,"_135":1177,"_138":139,"_67":24},[],[],[1208],"10b651f5-6e5c-4c7e-b650-8c3ddc698271",{"_40":1208,"_45":1210,"_70":1179,"_41":1221},{"_40":1208,"_47":1211,"_32":1213,"_53":1214,"_60":61,"_62":14,"_63":91,"_51":1216,"_68":209},{"_49":146,"_51":1212},{},1748347589.48868,{"_55":195,"_196":197,"_56":1215},"import mpmath as mp, math, sys",{"_200":1217,"_206":12,"_179":1219,"_181":1220,"_177":178,"_140":-5,"_183":184,"_185":184,"_186":1179,"_135":1177,"_138":139,"_67":24},{"_27":202,"_203":1218},[205],[],[],[1222],"5869c143-dae2-4e27-a94e-7e2388e4a701",{"_40":1222,"_45":1224,"_70":1208,"_41":1235},{"_40":1222,"_47":1225,"_32":1227,"_53":1228,"_60":61,"_62":14,"_63":91,"_51":1230,"_68":209},{"_49":146,"_51":1226},{},1748347709.943084,{"_55":195,"_196":197,"_56":1229},"f1 = mp.quad(lambda x: mp.sqrt(x)*mp.log(x), [0,1])\nf2 = mp.quad(lambda x: (mp.log(x))/ (x**2), [1, mp.inf])\nmu = (3/5)*(f1+f2)\nmu",{"_200":1231,"_206":12,"_179":1233,"_181":1234,"_177":178,"_140":-5,"_183":184,"_185":184,"_186":1208,"_135":1177,"_138":139,"_67":24},{"_27":202,"_203":1232},[205],[],[],[1236],"0ab35992-b085-4b57-b2b3-9d8f1b0c367c",{"_40":1236,"_45":1238,"_70":1222,"_41":1262},{"_40":1236,"_47":1239,"_32":1241,"_34":1242,"_53":1243,"_60":61,"_63":91,"_51":1245,"_68":69},{"_49":215,"_125":209,"_51":1240},{},1748347723.0644367,1748347724.6387808,{"_55":220,"_56":1244},"mpf('0.33333333333333332099752194860935')",{"_206":12,"_223":1246,"_177":178,"_140":-5,"_183":184,"_185":184,"_186":1222,"_135":1177,"_138":139,"_67":24},{"_60":225,"_226":1247,"_228":1241,"_34":1242,"_195":1229,"_229":1242,"_230":1244,"_232":-5,"_233":-5,"_234":1248,"_247":1249,"_283":-5},"69e69375-0472-462d-bcde-c5002a89466c",[],[1250,1253,1255,1259],{"_250":60,"_251":1251,"_53":1252},{"_253":254,"_255":256},{"_258":259},{"_250":261,"_251":1254},{"_253":254,"_255":256},{"_250":264,"_251":1256,"_53":1257},{"_253":254,"_255":256},{"_28":1258},{"_268":1244},{"_250":60,"_251":1260,"_53":1261},{"_253":254,"_255":256},{"_258":282},[1263],"35166372-5afe-4df6-8fb3-967c39bfbbfd",{"_40":1263,"_45":1265,"_70":1236,"_41":1287},{"_40":1263,"_47":1266,"_32":1268,"_53":1269,"_60":61,"_62":14,"_63":91,"_51":1284,"_68":69},{"_49":146,"_51":1267},{},1748347742.616287,{"_55":161,"_161":1270,"_174":1283},[1271,1275,1279],{"_164":1272,"_53":1273,"_167":1274,"_169":12},"Вычисление математического ожидания","Необходимо вычислить математическое ожидание: \$E[\\ln \\xi] = \\frac{3}{5} \\left( \\int_0^1 \\sqrt{x} \\ln(x) \\,dx + \\int_1^\\infty \\frac{\\ln(x)}{x^2} \\,dx \\right)\$. Первый интеграл отрицателен, второй, вероятно, положителен, но мал. Я начну с численного приближения, а затем попробую провести символическое вычисление. Для первого интеграла использую подстановку \$x = t^2\$, что даст ответ \$-4/9\$.",[],{"_164":1276,"_53":1277,"_167":1278,"_169":12},"Вычисление математического ожидаемого значения","Я вычислил математическое ожидаемое значение для \$ \\xi \$. Получается, что \$ \\mu = \\frac{1}{3} \$, что является точным значением. Это говорит о том, что типичное значение \$ \\xi \$ больше 1 (так как \$ e^{1/3} > 1 \$). В результате, распределение для \$ \\kappa_n \$ имеет экспоненциальную форму, где \$ \\kappa \\geq 1 \$ почти наверняка. Я также проверил, что \$ E_n S_n \$ сходится к \$ \\mu \\cdot Y \$, что подтверждает правильность вычислений.",[],{"_164":1280,"_53":1281,"_167":1282,"_169":12},"Вычисление функции распределения","Я заметил, что нижняя граница для \$ \\kappa \$ равна 1. Поэтому функция распределения \$ F_\\kappa(t) \$ равна 0 для \$ t < 1 \$. Для \$ t \\geq 1 \$ она имеет вид \$ F_\\kappa(t) = 1 - \\exp\\left(-\\frac{2}{5} \\left(\\frac{\\ln t}{\\mu}\\right)^{3/2}\\right) \$. С учётом \$ \\mu = 1/3 \$, я вычислил значение \$ F_\\kappa(\\pi) \$. Это дает \$ 1 - \\exp\\left(-\\frac{2}{5} (3 \\ln \\pi)^{3/2}\\right) \$.",[],"b8e97d88-83ca-4f96-9d34-e89b13decfa7",{"_177":178,"_179":1285,"_181":1286,"_140":-5,"_183":184,"_185":184,"_186":1236,"_135":1177,"_138":139,"_67":24},[],[],[1288],"f7ae93cc-3cd1-49ff-9ed2-2d4ba6b7fc4e",{"_40":1288,"_45":1290,"_70":1263,"_41":1301},{"_40":1288,"_47":1291,"_32":1293,"_53":1294,"_60":61,"_62":14,"_63":91,"_51":1296,"_68":209},{"_49":146,"_51":1292},{},1748347724.973412,{"_55":195,"_196":197,"_56":1295},"import math,mpmath as mp\nval = 1 - math.exp( - (2/5) * (3*math.log(math.pi))**1.5 )\nval",{"_200":1297,"_206":12,"_179":1299,"_181":1300,"_177":178,"_140":-5,"_183":184,"_185":184,"_186":1263,"_135":1177,"_138":139,"_67":24},{"_27":202,"_203":1298},[205],[],[],[1302],"fad7f337-0dd2-4a5f-b433-81ddd69c7249",{"_40":1302,"_45":1304,"_70":1288,"_41":1328},{"_40":1302,"_47":1305,"_32":1307,"_34":1308,"_53":1309,"_60":61,"_63":91,"_51":1311,"_68":69},{"_49":215,"_125":209,"_51":1306},{},1748347781.7740755,1748347782.1919582,{"_55":220,"_56":1310},"0.9215768875297303",{"_206":12,"_223":1312,"_177":178,"_140":-5,"_183":184,"_185":184,"_186":1288,"_135":1177,"_138":139,"_67":24},{"_60":225,"_226":1313,"_228":1307,"_34":1308,"_195":1295,"_229":1308,"_230":1310,"_232":-5,"_233":-5,"_234":1314,"_247":1315,"_283":-5},"17f567e2-a209-4d11-ba2c-f8670a25564c",[],[1316,1319,1321,1325],{"_250":60,"_251":1317,"_53":1318},{"_253":254,"_255":256},{"_258":259},{"_250":261,"_251":1320},{"_253":254,"_255":256},{"_250":264,"_251":1322,"_53":1323},{"_253":254,"_255":256},{"_28":1324},{"_268":1310},{"_250":60,"_251":1326,"_53":1327},{"_253":254,"_255":256},{"_258":282},[1329],"6af9f7e5-a154-4343-b92f-00ab3958533f",{"_40":1329,"_45":1331,"_70":1302,"_41":1345},{"_40":1329,"_47":1332,"_32":1334,"_53":1335,"_60":61,"_62":14,"_63":91,"_51":1342,"_68":69},{"_49":146,"_51":1333},{},1748347790.133127,{"_55":161,"_161":1336,"_174":1341},[1337],{"_164":1338,"_53":1339,"_167":1340,"_169":12},"Вычисление CDF для \$ \\kappa \$","Я понимаю, что нижняя граница для экспоненты равна 1, поэтому \$ \\kappa \\geq 1 \$. Для \$ t < 1 \$ функция распределения \$ F_\\kappa(t) = 0 \$, а для \$ t \\geq 1 \$ она принимает вид \$ F_\\kappa(t) = 1 - \\exp\\left( -\\frac{2}{5} (3 \\ln t)^{3/2} \\right) \$. Для \$ t = \\pi \$ проверяю численно: значение примерно \$ 0.9215768875 \$. Нужно вычислить с высокой точностью.",[],"cac545f8-4d65-41c3-9c90-3c6b6665e2ac",{"_177":178,"_179":1343,"_181":1344,"_140":-5,"_183":184,"_185":184,"_186":1302,"_135":1177,"_138":139,"_67":24},[],[],[1346],"5c9217a8-4a91-46cb-a44d-a7311f801630",{"_40":1346,"_45":1348,"_70":1329,"_41":1359},{"_40":1346,"_47":1349,"_32":1351,"_53":1352,"_60":61,"_62":14,"_63":91,"_51":1354,"_68":209},{"_49":146,"_51":1350},{},1748347782.315621,{"_55":195,"_196":197,"_56":1353},"mp.mp.dps = 50\nval = mp.mpf(1) - mp.e**( - mp.mpf(2)/5 * (3* mp.log(mp.pi))**(mp.mpf(3)/2) )\nval",{"_200":1355,"_206":12,"_179":1357,"_181":1358,"_177":178,"_140":-5,"_183":184,"_185":184,"_186":1329,"_135":1177,"_138":139,"_67":24},{"_27":202,"_203":1356},[205],[],[],[1360],"7d6f7a81-b1f1-4e56-af68-dbbc3a710bb1",{"_40":1360,"_45":1362,"_70":1346,"_41":1386},{"_40":1360,"_47":1363,"_32":1365,"_34":1366,"_53":1367,"_60":61,"_63":91,"_51":1369,"_68":69},{"_49":215,"_125":209,"_51":1364},{},1748347790.9821937,1748347791.3721685,{"_55":220,"_56":1368},"mpf('0.92157688752973032364527260905648961756559339337152599')",{"_206":12,"_223":1370,"_177":178,"_140":-5,"_183":184,"_185":184,"_186":1346,"_135":1177,"_138":139,"_67":24},{"_60":225,"_226":1371,"_228":1365,"_34":1366,"_195":1353,"_229":1366,"_230":1368,"_232":-5,"_233":-5,"_234":1372,"_247":1373,"_283":-5},"2d5b1d1d-6321-484c-9ecd-13facd2eba38",[],[1374,1377,1379,1383],{"_250":60,"_251":1375,"_53":1376},{"_253":254,"_255":256},{"_258":259},{"_250":261,"_251":1378},{"_253":254,"_255":256},{"_250":264,"_251":1380,"_53":1381},{"_253":254,"_255":256},{"_28":1382},{"_268":1368},{"_250":60,"_251":1384,"_53":1385},{"_253":254,"_255":256},{"_258":282},[1387],"6a09a477-3993-40f2-97e2-3fb05bc02a3c",{"_40":1387,"_45":1389,"_70":1360,"_41":1403},{"_40":1387,"_47":1390,"_32":1392,"_53":1393,"_60":61,"_62":14,"_63":91,"_51":1400,"_68":69},{"_49":146,"_51":1391},{},1748347801.043925,{"_55":161,"_161":1394,"_174":1399},[1395],{"_164":1396,"_53":1397,"_167":1398,"_169":12},"Вычисление значений распределения","Я подвел итог, что \$ \\kappa \\geq 1 \$, и теперь вычисляю CDF для \$ t \\geq 1 \$. Формула выглядит как \$ F_\\kappa(t) = 1 - \\exp\\left(-\\frac{2}{5}(3 \\ln t)^{3/2}\\right) \$. Когда \$ t = \\pi \\approx 3.14159 \$, значение вычисляется как примерно \$ 0.9215768875 \$. Я проверяю точность. Далее нужно вычислить решение с высокой точностью, вычислить предельное распределение и учесть независимость и пересечение.",[],"8af15486-8355-451d-9350-c19fb2c56493",{"_177":178,"_179":1401,"_181":1402,"_140":-5,"_183":184,"_185":184,"_186":1360,"_135":1177,"_138":139,"_67":24},[],[],[1404],"16d2ea65-e53b-425a-81b2-a2613f75d0db",{"_40":1404,"_45":1406,"_70":1387,"_41":1414},{"_40":1404,"_47":1407,"_32":1409,"_53":1410,"_60":61,"_62":14,"_63":91,"_51":1412,"_68":69},{"_49":146,"_51":1408},{},1748347810.690973,{"_55":1145,"_53":1411},"Думал на протяжении 3m 38s",{"_177":1148,"_1149":1413,"_140":-5,"_183":184,"_185":184,"_186":1387,"_135":1177,"_138":139,"_67":24},218,[1415],"70f4353a-b908-4cdb-bdc6-cdfc22d8af59",{"_40":1415,"_45":1417,"_70":1404,"_41":1427},{"_40":1415,"_47":1418,"_32":1420,"_53":1421,"_60":61,"_62":12,"_63":91,"_51":1424,"_68":69,"_1164":1165},{"_49":146,"_51":1419},{},1748347791.486649,{"_55":56,"_57":1422},[1423],"### Решение\n\n#### 1. Нормировка плотности \nДля \$x>0\$\n\\[\np(x)=\\min\\{\\sqrt x,1\\}\\,\\min\\!\\Bigl\\{\\tfrac1{x^{2}},1\\Bigr\\}=\n\\begin{cases}\n\\sqrt x, & 0 x\\}=\\overline F(x)^{N_n}\n\\sim\\exp\\!\\bigl(-N_n\\cdot\\tfrac25x^{3/2}\\bigr).\n\\]\n\nПоложив \n\\[\nm_n = N_n^{-2/3}Y_n,\\qquad \n\\mathbb P\\{Y_n\\le y\\}\\xrightarrow{}1-\\exp\\!\\Bigl(-\\tfrac25\\,y^{3/2}\\Bigr)=:F_Y(y),\n\\]\nполучаем сходимость \$m_n\\stackrel{d}{\\sim} n^{-1/3}Y\$, где \$Y\$ имеет распределение \$F_Y\$.\n\n---\n\n#### 4. Предельное распределение \$\\kappa_n\$\n\n\\[\n\\kappa_n\n=\\Bigl(\\prod_{k=1}^{M_n}\\xi_{k^{3}}\\Bigr)^{m_n}\n= \\exp\\!\\bigl(S_n\\,m_n\\bigr).\n\\]\n\nТак как \n\\[\nS_n\\,m_n\n=\\Bigl(\\tfrac13\\,n^{1/3}+o(n^{1/3})\\Bigr)\\;\n\\Bigl(n^{-1/3}Y+o(n^{-1/3})\\Bigr)\n\\xrightarrow{d}\\frac13\\,Y,\n\\]\nто \n\\[\n\\boxed{\\;\\ln\\kappa \\;=\\;\\frac13\\,Y\\;},\\qquad \n\\kappa=\\exp\\!\\Bigl(\\tfrac13\\,Y\\Bigr).\n\\]\n\n---\n\n#### 5. Функция распределения предела \n\nДля \$t<1\$ вероятность нулевая (так как \$Y\\ge 0\$). \nДля \$t\\ge 1\$:\n\\[\n\\mathbb P\\{\\kappa\\le t\\}\n=\\mathbb P\\!\\Bigl(Y\\le 3\\ln t\\Bigr)\n=1-\\exp\\!\\Bigl(-\\tfrac25\\,(3\\ln t)^{3/2}\\Bigr).\n\\]\n\nИтак \n\\[\nF_\\kappa(t)=\n\\begin{cases}\n0, & t<1,\\\\[4pt]\n1-\\exp\\!\\bigl(-\\dfrac25\\,(3\\ln t)^{3/2}\\bigr), & t\\ge 1.\n\\end{cases}\n\\]\n\n---\n\n#### 6. Значение в точке \$x_0=\\pi\$\n\n\\[\nF_\\kappa(\\pi)=1-\\exp\\!\\Bigl(-\\tfrac25\\,(3\\ln\\pi)^{3/2}\\Bigr)\n\\approx 0.9215768875297303236.\n\\]\n\n---\n\n### Ответ \n\nПредел \$\\kappa_n\\xrightarrow{d}\\kappa\$ существует. \nФункция распределения \$\\kappa\$ имеет вид \n\\[\nF_\\kappa(t)=\n\\begin{cases}\n0, & t<1,\\\\[4pt]\n1-\\exp\\!\\bigl(-\\dfrac25\\,(3\\ln t)^{3/2}\\bigr), & t\\ge 1,\n\\end{cases}\n\\]\nи \n\\[\nF_\\kappa(\\pi)\\;=\\;0.9215768875297303236\\ (\\pm10^{-12}),\n\\]\nчто и следует ввести в 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