Menu Close

Category: Algebra

prove-that-n-2-B-n-n-2-e-3-e-e-1-3-where-B-n-is-the-n-th-bernouli-s-number-

Question Number 195733 by York12 last updated on 09/Aug/23 $${prove}\:{that} \\ $$$$\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\left[\frac{{B}_{\overset{\_} {{n}}} }{\left({n}−\mathrm{2}\right)!}\right]=\frac{{e}\left(\mathrm{3}−{e}\right)}{\left({e}−\mathrm{1}\right)^{\mathrm{3}} } \\ $$$${where}\:{B}_{\overset{\_} {{n}}} \:{is}\:{the}\:{n}−\:{th}\:{bernouli}'{s}\:{number} \\ $$ Answered by…

1-3-f-x-3-f-x-dx-1-3-f-x-3-df-x-1-3-t-3-dt-t-4-4-1-3-f-x-4-4-1-3-

Question Number 195689 by Shomurotovdiyorbek last updated on 07/Aug/23 $$\int_{\mathrm{1}} ^{\mathrm{3}} {f}\left({x}\right)^{\mathrm{3}} {f}'\left({x}\right){dx}=\int_{\mathrm{1}} ^{\mathrm{3}} \left({f}\left({x}\right)^{\mathrm{3}} \right){df}\left({x}\right)=\int_{\mathrm{1}} ^{\mathrm{3}} {t}^{\mathrm{3}} {dt}=\frac{{t}^{\mathrm{4}} }{\mathrm{4}}\int_{\mathrm{1}} ^{\mathrm{3}} =\frac{{f}\left({x}\right)^{\mathrm{4}} }{\mathrm{4}}\int_{\mathrm{1}} \mathrm{3} \\…

f-x-1276-x-1-ln-2-4589-domain-f-x-

Question Number 195647 by mathlove last updated on 06/Aug/23 $${f}\left({x}\right)=\frac{\mathrm{1276}}{\left({x}−\mathrm{1}\right)^{{ln}\frac{\mathrm{2}}{\mathrm{4589}}} } \\ $$$${domain}\:{f}\left({x}\right)=? \\ $$ Answered by Tokugami last updated on 17/Sep/23 $$\frac{\mathrm{1276}}{\left({x}−\mathrm{1}\right)^{\mathrm{ln}\left(\mathrm{2}\right)−\mathrm{ln}\left(\mathrm{4589}\right)} }=\mathrm{1276}\left({x}−\mathrm{1}\right)^{\mathrm{ln}\left(\mathrm{4589}\right)−\mathrm{ln}\left(\mathrm{2}\right)} \\…

an-unsolved-old-question-190875-a-b-c-are-real-roots-of-the-equation-x-3-7x-2-4x-1-0-find-1-a-1-3-1-b-1-3-1-c-1-3-

Question Number 195628 by mr W last updated on 06/Aug/23 $$\underline{{an}\:{unsolved}\:{old}\:{question}\:#\mathrm{190875}} \\ $$$${a},\:{b},\:{c}\:{are}\:{real}\:{roots}\:{of}\:{the}\:{equation} \\ $$$${x}^{\mathrm{3}} −\mathrm{7}{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{1}=\mathrm{0}. \\ $$$${find}\:\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{{a}}}+\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{{b}}}+\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{{c}}}=? \\ $$ Commented by Frix last…

Question-195653

Question Number 195653 by mustafazaheen last updated on 06/Aug/23 Answered by MM42 last updated on 06/Aug/23 $${e}−\mathrm{1}>\mathrm{1}\:\:\: \\ $$$$\Rightarrow{lim}_{{x}\rightarrow\:\mathrm{0}^{+} } \:\left({e}−\mathrm{1}\right)^{\frac{\sqrt{\mathrm{3}}}{{x}}} =\infty \\ $$$$\Rightarrow{lim}_{{x}\rightarrow\:\mathrm{0}^{−} }…

0-lt-x-lt-1-1-1-x-1-2x-1-x-2-4x-3-1-x-4-8x-7-1-x-8-16x-15-1-x-16-evaluate-the-previous-summation-

Question Number 195651 by York12 last updated on 06/Aug/23 $$\mathrm{0}<{x}<\mathrm{1} \\ $$$$\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{1}} }+\frac{\mathrm{2}{x}}{\mathrm{1}+{x}^{\mathrm{2}} }+\frac{\mathrm{4}{x}^{\mathrm{3}} }{\mathrm{1}+{x}^{\mathrm{4}} }+\frac{\mathrm{8}{x}^{\mathrm{7}} }{\mathrm{1}+{x}^{\mathrm{8}} }+\frac{\mathrm{16}{x}^{\mathrm{15}} }{\mathrm{1}+{x}^{\mathrm{16}} }+….+\infty \\ $$$${evaluate}\:{the}\:{previous}\:{summation} \\ $$ Answered…