Question Number 200022 by jlewis last updated on 12/Nov/23 $$\mathrm{solve}\:\mathrm{the}\:\mathrm{associated}\:\mathrm{legendre}\:\mathrm{equation} \\ $$$$\lambda={l}\:\left({l}+\mathrm{1}\right)\eta^{\mathrm{2}} \:;{l}=\mathrm{0},\mathrm{1},\mathrm{2}…\:\:\:{and}\:{m}^{\mathrm{2}} \leqslant\:{l}\left({l}+\mathrm{1}\right)\: \\ $$$${which}\:{requires}\:−{l}\leqslant{m}\leqslant{l}\:\mathrm{using}\:\mathrm{power}\:\mathrm{series} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 200019 by hardmath last updated on 12/Nov/23 Answered by mr W last updated on 12/Nov/23 Commented by hardmath last updated on 12/Nov/23 $$\mathrm{very}\:\mathrm{nice}\:\mathrm{dear}\:\mathrm{pfofessor}……
Question Number 200012 by hardmath last updated on 12/Nov/23 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 200013 by hardmath last updated on 12/Nov/23 Commented by mr W last updated on 12/Nov/23 $${geometry}\:{is}\:{not}\:{like}\:{a}\:{novel}.\:{geometry} \\ $$$${is}\:{best}\:{with}\:{diagram},\:{not}\:{with}\:{only} \\ $$$${text}. \\ $$ Commented…
Question Number 200010 by Lekhraj last updated on 12/Nov/23 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 200011 by Lekhraj last updated on 12/Nov/23 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 200004 by Mingma last updated on 12/Nov/23 Answered by aleks041103 last updated on 12/Nov/23 $$−\mathrm{1}<{x}<\mathrm{1} \\ $$$${f}\left({x}\right)=\frac{{x}+\mathrm{2}^{{k}} }{\mathrm{2}^{{k}+\mathrm{1}} }=\frac{{x}}{\mathrm{2}^{{k}+\mathrm{1}} }+\frac{\mathrm{1}}{\mathrm{2}}\Rightarrow\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}^{{k}+\mathrm{1}} }<{f}\left({x}\right)<\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}^{{k}+\mathrm{1}} } \\…
Question Number 200005 by Mingma last updated on 12/Nov/23 Answered by AST last updated on 12/Nov/23 $$\left(\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} }{\mathrm{3}}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \geqslant\left(\frac{{a}+{b}+{c}}{\mathrm{3}}\right)\Rightarrow{a}+{b}+{c}\leqslant\mathrm{3} \\ $$$$\Sigma\frac{\mathrm{1}}{\mathrm{2}{a}+{b}}\geqslant\frac{\mathrm{9}}{\mathrm{3}\left({a}+{b}+{c}\right)}\geqslant\frac{\mathrm{9}}{\mathrm{3}×\mathrm{3}}=\mathrm{1} \\ $$$${Equality}\:{holds}\:{when}\:{a}={b}={c}=\mathrm{1}…
Question Number 200007 by Hummayoun last updated on 12/Nov/23 $${f}\left({x}\right)=\sqrt{{x}} \\ $$$${g}\left({x}\right)=\mathrm{2}{x}+\mathrm{1} \\ $$$${fog}\left({x}\right)={f}\left(\mathrm{2}{x}+\mathrm{1}\right) \\ $$$${fog}\left({x}\right)=\sqrt{\mathrm{2}{x}+\mathrm{1}} \\ $$$${gof}\left({x}\right)={g}\left(\sqrt{\left.{x}\right)}\right. \\ $$$${gof}\left({x}\right)=\mathrm{2}\sqrt{{x}} \\ $$ Terms of Service…
Question Number 199932 by hardmath last updated on 11/Nov/23 $$\mathrm{1}. \\ $$$$\mathrm{If}\:\:\:\mathrm{3}\:\centerdot\:\overline {\mathrm{ab}}\:+\:\overline {\mathrm{bc}}\:=\:\mathrm{115} \\ $$$$\mathrm{Find}:\:\:\:\mathrm{max}\left(\mathrm{a}+\mathrm{b}+\mathrm{c}\right)=? \\ $$$$ \\ $$$$\mathrm{2}. \\ $$$$\mathrm{a},\mathrm{b},\mathrm{c}\in\mathbb{N} \\ $$$$\mathrm{If}\:\:\:\frac{\mathrm{a}}{\mathrm{2}}\:\:+\:\:\frac{\mathrm{b}}{\mathrm{3}}\:\:=\:\:\frac{\mathrm{c}}{\mathrm{4}} \\…