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Give-f-x-x-5-5x-4-4x-3-3x-2-2x-1-amp-2-1-3-5-3-3-1-3-2-1-3-5-3-3-1-3-Find-f-

Question Number 144327 by SOMEDAVONG last updated on 24/Jun/21 $$\mathrm{Give}\:\mathrm{f}\left(\mathrm{x}\right)=\mathrm{x}^{\mathrm{5}} −\mathrm{5x}^{\mathrm{4}} +\mathrm{4x}^{\mathrm{3}} −\mathrm{3x}^{\mathrm{2}} +\mathrm{2x}−\mathrm{1},\& \\ $$$$\alpha=\:\sqrt[{\mathrm{3}}]{\mathrm{2}}\left(\sqrt[{\mathrm{3}}]{\mathrm{5}+\mathrm{3}\sqrt{\mathrm{3}}}−\:\frac{\sqrt[{\mathrm{3}}]{\mathrm{2}}}{\:\sqrt[{\mathrm{3}}]{\mathrm{5}+\mathrm{3}\sqrt{\mathrm{3}}}}\right).\mathrm{Find}\:\mathrm{f}\left(\alpha\right)? \\ $$ Answered by Rasheed.Sindhi last updated on 25/Jun/21…

Question-13237

Question Number 13237 by Abbas-Nahi last updated on 17/May/17 Commented by RasheedSindhi last updated on 17/May/17 $$\frac{\mathrm{36}}{\left(\mathrm{n}+\mathrm{3}\right)!}−\frac{\mathrm{1}}{\left(\mathrm{n}+\mathrm{1}\right)!}−\frac{\mathrm{1}}{\mathrm{n}!}=\mathrm{0} \\ $$$$\frac{\mathrm{1}}{\mathrm{n}!}\left(\frac{\mathrm{36}}{\left(\mathrm{n}+\mathrm{3}\right)\left(\mathrm{n}+\mathrm{2}\right)\left(\mathrm{n}+\mathrm{1}\right)}−\frac{\mathrm{1}}{\left(\mathrm{n}+\mathrm{1}\right)}−\mathrm{1}\right)=\mathrm{0} \\ $$$$\frac{\mathrm{1}}{\mathrm{n}!}=\mathrm{0}\:\mid\:\left(\frac{\mathrm{36}}{\left(\mathrm{n}+\mathrm{3}\right)\left(\mathrm{n}+\mathrm{2}\right)\left(\mathrm{n}+\mathrm{1}\right)}−\frac{\mathrm{1}}{\left(\mathrm{n}+\mathrm{1}\right)}−\mathrm{1}\right)=\mathrm{0} \\ $$$$\frac{\mathrm{1}}{\mathrm{n}!}=\mathrm{0}\Rightarrow\mathrm{n}\rightarrow\infty \\ $$…

Question-144291

Question Number 144291 by SOMEDAVONG last updated on 24/Jun/21 Answered by som(math1967) last updated on 24/Jun/21 $${I}=\int_{\mathrm{0}} ^{\mathrm{2021}} \frac{\left(\mathrm{2021}−{x}\right)^{\mathrm{2021}} }{\left(\mathrm{2021}−{x}\right)^{\mathrm{2021}} +\left(\mathrm{2021}−\mathrm{2021}+{x}\right)^{\mathrm{2021}} }{dx} \\ $$$$\mathrm{2}{I}=\int_{\mathrm{0}} ^{\mathrm{2021}}…

S-n-n-1-n-1-2-k-tanh-1-2-k-

Question Number 144272 by SOMEDAVONG last updated on 24/Jun/21 $$\mathrm{S}_{\mathrm{n}} =\underset{\mathrm{n}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{k}} }\mathrm{tanh}\left(\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{k}} }\right)=? \\ $$ Answered by Olaf_Thorendsen last updated on 24/Jun/21 $$\mathrm{S}_{{n}}…