Question Number 221981 by MrGaster last updated on 14/Jun/25 $$\mathrm{Prove}:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{n}^{\mathrm{3}} }{{e}^{\mathrm{2}\pi{n}} −\mathrm{1}}=\frac{\Gamma\left(\frac{\mathrm{1}}{\mathrm{4}}\right)^{\mathrm{8}} }{\mathrm{5120}\pi^{\mathrm{6}} }−\frac{\mathrm{1}}{\mathrm{240}}=\frac{\mathrm{1}}{\mathrm{80}}\left(\frac{\varpi}{\pi}\right)^{\mathrm{4}} −\frac{\mathrm{1}}{\mathrm{240}} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 221939 by Ismoiljon_008 last updated on 13/Jun/25 $$ \\ $$$$\:\:\:{sin}^{\mathrm{2}} \mathrm{1}°\:+\:{sin}^{\mathrm{2}} \mathrm{3}°\:+\:{sin}^{\mathrm{2}} \mathrm{5}°\:+\:…\:+\:{sin}^{\mathrm{2}} \mathrm{269}°\:=\:? \\ $$$$\:\:\:\mathcal{H}{elp}\:{me},\:\:{please} \\ $$$$ \\ $$ Commented by Rasheed.Sindhi…
Question Number 221919 by fantastic last updated on 12/Jun/25 $${Acable}\:{can}\:{stand}\:{a}\:{maximum}\:{weight}\:{of}\:\mathrm{25kg}\: \\ $$$${If}\:{the}\:{length}\:{of}\:{wire}\:{is}\:{halved}\:{what}\:{is}\:{the} \\ $$$${maximum}\:{weight}\:{thsn}\:{can}\:{be}\:{hung}\:{from}\:{it}?? \\ $$ Commented by mr W last updated on 12/Jun/25 $$\sim\:\mathrm{50}\:{kg}…
Question Number 221843 by wewji12 last updated on 11/Jun/25 $$\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} \:\:{x}^{\mathrm{2}} \mathrm{csc}^{\mathrm{2}} \left({x}\right)\mathrm{d}{x} \\ $$$$\int\:\:−\frac{\mathrm{4}{z}^{\mathrm{2}} }{\left({e}^{\boldsymbol{{i}}{z}} −{e}^{−\boldsymbol{{i}}{z}} \right)^{\mathrm{2}} }\:\mathrm{d}{z}=\int\:\:\frac{{z}^{\mathrm{2}} {e}^{\mathrm{2}\boldsymbol{{i}}{z}} }{\left({e}^{\mathrm{2}\boldsymbol{{i}}{z}} −\mathrm{1}\right)^{\mathrm{2}} }\:\mathrm{d}{z} \\…
Question Number 221786 by wewji12 last updated on 10/Jun/25 $$\:\mathrm{Express}\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \:{z}^{{z}^{{z}^{\iddots} } } \:\mathrm{d}{z}\:\:\mathrm{closed}\:\mathrm{form}… \\ $$$${z}\upuparrows^{\infty} ={z}^{{z}^{{z}^{\iddots} } } \\ $$ Terms of Service…
Question Number 221721 by ahmed2025 last updated on 09/Jun/25 Answered by wewji12 last updated on 09/Jun/25 $$\frac{\mathrm{d}{y}}{\mathrm{dln}\left({t}\right)}=\frac{\frac{\mathrm{d}{y}}{\mathrm{d}{t}}}{\frac{\mathrm{d}\left\{\mathrm{ln}\left({t}\right)\right\}}{\mathrm{d}{t}}}\:\left(\mathrm{Warning}\:\:\frac{\mathrm{d}{y}}{\mathrm{d}{x}}\:\mathrm{is}\:\mathrm{NOT}\:\mathrm{Fraction}!!!\right) \\ $$$$\frac{\frac{\mathrm{d}\:\:}{\mathrm{d}{t}}\left\{{t}^{\mathrm{2}} \mathrm{ln}\left({t}\right)\right\}}{\frac{\mathrm{d}\:\:}{\mathrm{d}{t}}\left\{\mathrm{ln}\left({t}\right)\right\}}=\frac{\mathrm{2}{t}\mathrm{ln}\left({t}\right)+{t}}{\frac{\mathrm{1}}{{t}}}={t}^{\mathrm{2}} +\mathrm{2}{t}^{\mathrm{2}} \mathrm{ln}\left({t}\right)={t}^{\mathrm{2}} \left(\mathrm{1}+\mathrm{2ln}\left({t}\right)\right) \\ $$$$\frac{\mathrm{d}{y}\left({t}\right)}{\mathrm{d}\left\{\mathrm{ln}\left({t}\right)\right\}}={t}^{\mathrm{2}}…
Question Number 221697 by fantastic last updated on 09/Jun/25 $${Is}\:\sqrt{{i}}\:{an}\:{imaginary}\:{number}\:\left({i}=\sqrt{−\mathrm{1}}\right)\:{answer}\:{with}\:{logic} \\ $$ Answered by Ghisom last updated on 09/Jun/25 $$\mathrm{i}=\mathrm{e}^{\mathrm{i}\frac{\pi}{\mathrm{2}}} \\ $$$${z}={r}\mathrm{e}^{\mathrm{i}\theta} \:\Rightarrow\:\sqrt{{z}}=\sqrt{{r}}\mathrm{e}^{\mathrm{i}\frac{\theta}{\mathrm{2}}} \\ $$$$\Rightarrow\:\sqrt{\mathrm{i}}=\mathrm{e}^{\mathrm{i}\frac{\pi}{\mathrm{4}}}…
Question Number 221620 by fantastic last updated on 08/Jun/25 $${Solve}\:{for}\:{x} \\ $$$$\sqrt[{\mathrm{3}}]{\mathrm{7}{x}}=\sqrt{{x}}\left[{x}\neq\mathrm{0}\right] \\ $$ Answered by fantastic last updated on 08/Jun/25 $${or}\:\left(\mathrm{7}{x}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} ={x}^{\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$${or}\:\sqrt[{\mathrm{3}}]{\mathrm{7}}.{x}^{\frac{\mathrm{1}}{\mathrm{3}}}…
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Question Number 221498 by wewji12 last updated on 07/Jun/25 $$\mathrm{Solve}\:\mathrm{differantial}\:\mathrm{Equation} \\ $$$$\frac{{d}\:\:}{{dt}}\left[\frac{{dy}\left({t}\right)}{{dt}}\right]+{ty}\left({t}\right)=\mathrm{0} \\ $$ Answered by MrGaster last updated on 07/Jun/25 $${y}\left({t}\right)={C}_{\mathrm{1}} \underset{{m}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{m}}…