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Category: Algebra

Given-that-I-n-0-1-x-1-x-n-dx-n-Z-show-that-n-2-I-n-nI-n-1-n-1-

Question Number 127090 by physicstutes last updated on 26/Dec/20 $$\mathrm{Given}\:\mathrm{that}\: \\ $$$$\:\:\mathcal{I}_{{n}} \:=\:\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}{x}\left(\mathrm{1}−{x}\right)^{{n}} {dx}\:,\:{n}\:\in\:\mathbb{Z}^{+} \\ $$$$\:\mathrm{show}\:\mathrm{that}\:\:\left({n}+\mathrm{2}\right)\mathcal{I}_{{n}} \:=\:{nI}_{{n}−\mathrm{1}} ,\:{n}\:\geqslant\:\mathrm{1}. \\ $$ Answered by Dwaipayan…

Question-192596

Question Number 192596 by York12 last updated on 22/May/23 Answered by a.lgnaoui last updated on 24/May/23 $$\left(\mathrm{x}_{\mathrm{m}} +\mathrm{iy}_{\mathrm{m}} \right)^{\mathrm{2}\left(\mathrm{n}+\frac{\mathrm{1}}{\mathrm{2}}\right)} =\mathrm{1}\Leftrightarrow\left[\left(\mathrm{x}_{\mathrm{m}} +\mathrm{iy}_{\mathrm{m}} \right)^{\mathrm{n}+\frac{\mathrm{1}}{\mathrm{2}}} −\mathrm{1}\right]×\left[\left(\mathrm{x}_{\mathrm{m}} +\mathrm{iy}_{\mathrm{m}} \right)^{\mathrm{n}+\frac{\mathrm{1}}{\mathrm{n}}}…

Solve-for-n-D-A-1-P-1-i-n-i-1-i-n-1-P-1-i-r-i-1-i-r-1-R-i-1-n-i-1-i-r-i-1-i-r-1-1-n-i-1-i-n-i-1-i-n-1-1-0-

Question Number 61526 by Jarbas last updated on 04/Jun/19 $${Solve}\:{for}\:{n}:\:{D}/{A}×\left\{\mathrm{1}−\frac{{P}×\left(\frac{\left(\mathrm{1}+{i}\right)^{{n}} ×{i}}{\left(\mathrm{1}+{i}\right)^{{n}} −\mathrm{1}}\right)}{\left({P}×\left(\frac{\left(\mathrm{1}+{i}\right)^{{r}} ×{i}}{\left(\mathrm{1}+{i}\right)^{{r}} −\mathrm{1}}\right)\right)−\frac{{R}}{{i}}×\left[\left(\frac{\mathrm{1}}{{n}}+{i}\right)×\left(\frac{\left(\mathrm{1}+{i}\right)^{{r}} ×{i}}{\left(\mathrm{1}+{i}\right)^{{r}} −\mathrm{1}}\right)−\left(\frac{\mathrm{1}}{{n}}+{i}\right)×\left(\frac{\left(\mathrm{1}+{i}\right)^{{n}} ×{i}}{\left(\mathrm{1}+{i}\right)^{{n}} −\mathrm{1}}\right)\right]}\right\}−\mathrm{1}=\mathrm{0} \\ $$$$ \\ $$ Terms of Service…

Question-61495

Question Number 61495 by bhanukumarb2@gmail.com last updated on 03/Jun/19 Commented by bhanukumarb2@gmail.com last updated on 03/Jun/19 $${prove}\:{second}\:{in}\:{which}\:{book}\:{i}\:{can}\:{get}\: \\ $$$${these}\:{type}\:{approximation} \\ $$ Commented by bhanukumarb2@gmail.com last…

a-a-x-a-a-x-2x-this-is-the-solution-Sir-Aifour-and-me-found-trivial-solution-a-x-0-a-x-R-x-2-8-r-r-2-4-2-4a-1-r-2-r-r-2-4-with-r-2-4a-3-3-sin-1-

Question Number 61490 by MJS last updated on 03/Jun/19 $$\sqrt{{a}−\sqrt{{a}+{x}}}+\sqrt{{a}+\sqrt{{a}−{x}}}=\mathrm{2}{x} \\ $$$$\mathrm{this}\:\mathrm{is}\:\mathrm{the}\:\mathrm{solution}\:\mathrm{Sir}\:\mathrm{Aifour}\:\mathrm{and}\:\mathrm{me}\:\mathrm{found} \\ $$$$ \\ $$$$\mathrm{trivial}\:\mathrm{solution}\:{a}={x}=\mathrm{0} \\ $$$$ \\ $$$${a},\:{x}\:\in\mathbb{R} \\ $$$$ \\ $$$${x}=\frac{\sqrt{\mathrm{2}}}{\mathrm{8}}\left({r}+\sqrt{{r}^{\mathrm{2}} +\mathrm{4}}\right)\sqrt{\mathrm{2}\left(\mathrm{4}{a}−\mathrm{1}\right)−{r}^{\mathrm{2}}…