Question Number 150550 by bekzodjumayev last updated on 13/Aug/21 Commented by bekzodjumayev last updated on 13/Aug/21 $$\boldsymbol{{H}}{elp}?? \\ $$ Commented by liberty last updated on…
Question Number 150548 by mathdanisur last updated on 13/Aug/21 $$\mathrm{For}\:\:\boldsymbol{\mathrm{k}}<\mathbb{N}\:\:\mathrm{fixed}\:\:\mathrm{and}\:\:\boldsymbol{\alpha}>\mathrm{0}\:\:\mathrm{then}: \\ $$$$\underset{\boldsymbol{\mathrm{n}}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{1}}{\:\sqrt{\mathrm{n}^{\boldsymbol{\alpha}} }}\:\centerdot\:\left(\frac{\underset{\boldsymbol{\mathrm{i}}=\mathrm{1}} {\overset{\boldsymbol{\mathrm{k}}} {\prod}}\left(\mathrm{n}+\mathrm{k}+\mathrm{i}\right)}{\underset{\boldsymbol{\mathrm{i}}=\mathrm{1}} {\overset{\boldsymbol{\mathrm{k}}} {\prod}}\left(\mathrm{n}+\mathrm{i}\right)}\right)^{\boldsymbol{\mathrm{n}}^{\boldsymbol{\alpha}} } \\ $$ Terms of Service Privacy…
Question Number 150539 by mathdanisur last updated on 13/Aug/21 $$\mathrm{Prove}\:\mathrm{or}\:\mathrm{disprove}\:\mathrm{the}\:\mathrm{foolowing}: \\ $$$$\underset{\boldsymbol{\mathrm{n}}=\mathrm{1}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{\frac{\boldsymbol{\mathrm{n}}^{\mathrm{2}} +\boldsymbol{\mathrm{n}}+\mathrm{2}}{\mathrm{2}}} \:\mathrm{e}^{−\boldsymbol{\pi\mathrm{n}}^{\mathrm{2}} \boldsymbol{\mathrm{x}}} \:=\:\underset{\boldsymbol{\mathrm{n}}=\mathrm{1}} {\overset{\infty} {\sum}}\mathrm{e}^{−\boldsymbol{\pi\mathrm{n}}^{\mathrm{2}} \boldsymbol{\mathrm{x}}} \\ $$ Answered by…
Question Number 150531 by mathdanisur last updated on 13/Aug/21 $$\mathrm{Find}\:\:\mathrm{x};\mathrm{y}\:\:;\:\:\mathrm{x}\in\mathrm{Q}\:\:\mathrm{and}\:\:\mathrm{y}\in\mathrm{Z}\:\:\mathrm{such}\:\mathrm{that}: \\ $$$$\mathrm{2020}\left(\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{y}^{\mathrm{2}} \right)\:+\:\mathrm{2019}\left(\mathrm{x}\:+\:\mathrm{y}\right)\:=\:\mathrm{2021xy} \\ $$ Commented by Rasheed.Sindhi last updated on 14/Aug/21 $$\underset{\smile} {\overset{\frown}…
Question Number 19455 by Tinkutara last updated on 11/Aug/17 $$\mathrm{Prove}\:\mathrm{that}\:\mathrm{if}\:{z}\:=\:\mathrm{cos}\:\mathrm{6}°\:+\:{i}\:\mathrm{sin}\:\mathrm{6}°,\:\mathrm{then} \\ $$$$\frac{\mathrm{1}}{{z}^{\mathrm{2}} \:+\:\mathrm{1}}\:−\:\frac{{iz}}{{z}^{\mathrm{4}} \:−\:\mathrm{1}}\:+\:\frac{{iz}^{\mathrm{3}} }{{z}^{\mathrm{8}} \:−\:\mathrm{1}}\:+\:\frac{{iz}^{\mathrm{7}} }{{z}^{\mathrm{16}} \:−\:\mathrm{1}}\:=\:\mathrm{0}. \\ $$ Answered by ajfour last updated…
Question Number 150515 by mathdanisur last updated on 13/Aug/21 Answered by ajfour last updated on 13/Aug/21 $${x}={t}+{h} \\ $$$$\mathrm{7}\sqrt{\mathrm{4}{t}^{\mathrm{2}} +\mathrm{8}{ht}+\mathrm{4}{h}^{\mathrm{2}} +\mathrm{5}{t}+\mathrm{5}{h}−\mathrm{1}} \\ $$$$−\mathrm{7}\sqrt{\mathrm{4}{t}^{\mathrm{2}} +\mathrm{8}{ht}+\mathrm{4}{h}^{\mathrm{2}} −\mathrm{12}{t}−\mathrm{12}{h}+\mathrm{12}}…
Question Number 150516 by mathdanisur last updated on 13/Aug/21 $$\mathrm{Compare}: \\ $$$$\boldsymbol{\mathrm{x}}\:=\:\mathrm{2}^{\mathrm{3}^{\mathrm{2}^{\mathrm{3}} } } \:\:\:\:\:\mathrm{and}\:\:\:\:\:\boldsymbol{\mathrm{y}}\:=\:\mathrm{3}^{\mathrm{2}^{\mathrm{3}^{\mathrm{2}} } } \\ $$ Commented by mr W last updated…
Question Number 19434 by Tinkutara last updated on 11/Aug/17 $$\mathrm{If}\:\left(\frac{\mathrm{1}\:+\:{i}\sqrt{\mathrm{3}}}{\mathrm{1}\:−\:{i}\sqrt{\mathrm{3}}}\right)^{{n}} \:\mathrm{is}\:\mathrm{an}\:\mathrm{integer},\:\mathrm{then}\:{n}\:\mathrm{is} \\ $$ Answered by ajfour last updated on 11/Aug/17 $$\left(\frac{\mathrm{1}+{i}\sqrt{\mathrm{3}}}{\mathrm{1}−{i}\sqrt{\mathrm{3}}}\right)^{\mathrm{n}} =\left(\frac{\mathrm{e}^{{i}\pi/\mathrm{6}} }{\mathrm{e}^{−{i}\pi/\mathrm{6}} }\right)^{\mathrm{n}} =\left[\mathrm{x}\right]…
Question Number 19435 by Tinkutara last updated on 11/Aug/17 $$\mathrm{If}\:\alpha\:=\:\mathrm{cos}\:\frac{\mathrm{2}\pi}{\mathrm{5}}\:+\:{i}\:\mathrm{sin}\:\frac{\mathrm{2}\pi}{\mathrm{5}}\:,\:\mathrm{then}\:\mathrm{find}\:\mathrm{the} \\ $$$$\mathrm{value}\:\mathrm{of}\:\alpha\:+\:\alpha^{\mathrm{2}} \:+\:\alpha^{\mathrm{3}} \:+\:\alpha^{\mathrm{4}} . \\ $$ Answered by ajfour last updated on 11/Aug/17 $$\underset{\mathrm{r}=\mathrm{1}}…
Question Number 19433 by Tinkutara last updated on 11/Aug/17 $$\frac{\sqrt{\mathrm{5}\:+\:\mathrm{12}{i}}\:+\:\sqrt{\mathrm{5}\:−\:\mathrm{12}{i}}}{\:\sqrt{\mathrm{5}\:+\:\mathrm{12}{i}}\:−\:\sqrt{\mathrm{5}\:−\:\mathrm{12}{i}}}\:= \\ $$ Answered by ajfour last updated on 11/Aug/17 $$=\frac{\left(\sqrt{\mathrm{5}+\mathrm{12i}}+\sqrt{\mathrm{5}−\mathrm{12i}}\right)^{\mathrm{2}} }{\mathrm{24i}} \\ $$$$=\frac{\mathrm{10}+\mathrm{2}\sqrt{\mathrm{25}+\mathrm{144}}}{\mathrm{24i}}=\frac{\mathrm{3}}{\mathrm{2i}}=−\frac{\mathrm{3}}{\mathrm{2}}\mathrm{i}\:. \\ $$…