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Question-213680

Question Number 213680 by issac last updated on 13/Nov/24 Commented by Frix last updated on 13/Nov/24 $$\mathrm{WolframAlpha} \\ $$$$\mathrm{It}'\mathrm{s}\:\mathrm{worthless}\:\mathrm{if}\:\mathrm{you}\:\mathrm{don}'\mathrm{t}\:\mathrm{understand}\:\mathrm{how} \\ $$$$\mathrm{to}\:\mathrm{get}\:\mathrm{there}. \\ $$ Commented by…

Q213662-Not-easy-p-F-q-z-a-b-is-hypergeometric-function-Li-z-is-Dilogarithm-function-

Question Number 213679 by issac last updated on 13/Nov/24 $${Q}\mathrm{213662} \\ $$$$…… \\ $$$$\mathrm{Not}\:\mathrm{easy}……. \\ $$$$\:\:_{{p}} \boldsymbol{\mathrm{F}}_{{q}} \left({z};\cancel{\underbrace{ }}\:\boldsymbol{\mathrm{a}},\boldsymbol{\mathrm{b}}\right)\:\mathrm{is}\:\mathrm{hypergeometric}\:\mathrm{function} \\ $$$$\mathrm{Li}_{\nu} \left({z}\right)\:\mathrm{is}\:\mathrm{Dilogarithm}\:\mathrm{function}. \\ $$ Terms…

Question-213656

Question Number 213656 by efronzo1 last updated on 12/Nov/24 Answered by golsendro last updated on 13/Nov/24 $$\:\:\left(\mathrm{i}\right)\:\mathrm{g}\left(\mathrm{4}−\mathrm{x}\right)=\:−\mathrm{g}\left(\mathrm{x}\right)\: \\ $$$$\:\:\:\:\:\:\:\:\underset{\mathrm{0}} {\overset{\mathrm{4}} {\int}}\:\mathrm{g}\left(\mathrm{x}\right)\mathrm{dx}\:=\:\underset{\mathrm{0}} {\overset{\mathrm{4}} {\int}}\:\mathrm{g}\left(\mathrm{4}−\mathrm{x}\right)\mathrm{dx}\: \\ $$$$\:\:\:\:\:\:\:\:\underset{\mathrm{0}}…

Question-213641

Question Number 213641 by Abdullahrussell last updated on 12/Nov/24 Answered by Ghisom last updated on 03/Dec/24 $$\alpha_{{k}} \in\mathbb{Z} \\ $$$${p}\left({x}\right)=\underset{{k}=\mathrm{3}} {\overset{{n}} {\sum}}\alpha_{{k}} {x}^{{k}} \\ $$$${x}\in\mathbb{Z}\:\Rightarrow\:{p}\left({x}\right)\in\mathbb{Z}…

Question-213642

Question Number 213642 by universe last updated on 12/Nov/24 Answered by Berbere last updated on 12/Nov/24 $${a}_{{n}} =\underset{{N}\rightarrow\infty} {\mathrm{lim}}\underset{{k}={n}} {\overset{{N}} {\sum}}\frac{\mathrm{1}}{{k}^{\mathrm{2}} };\forall{k}\geqslant{n}>\mathrm{1}\:\:{k}\left({k}−\mathrm{1}\right)\:\leqslant{k}^{\mathrm{2}} \leqslant{k}\left({k}+\mathrm{1}\right)\Rightarrow \\ $$$$\underset{{N}\rightarrow\infty}…