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Question-129892

Question Number 129892 by mohammad17 last updated on 20/Jan/21 Answered by Ar Brandon last updated on 20/Jan/21 $$\mathrm{GP}\:\mathrm{with}\:\mathrm{U}\left(\mathrm{1}\right)=\frac{\mathrm{11x}}{\mathrm{12}}\:\mathrm{and}\:\mathrm{r}=\frac{\mathrm{1}}{\mathrm{12}} \\ $$$$\mathrm{S}_{\infty} =\frac{\frac{\mathrm{11}}{\mathrm{12}}\mathrm{x}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{12}}}=\mathrm{x} \\ $$ Terms of…

Question-129903

Question Number 129903 by 0731619177 last updated on 20/Jan/21 Commented by Dwaipayan Shikari last updated on 20/Jan/21 $$\int{e}^{{x}^{\mathrm{2}} } {dx}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}!\left(\mathrm{2}{n}+\mathrm{1}\right)}{x}^{\mathrm{2}{n}+\mathrm{1}} =\frac{{x}}{\mathrm{1}}+\frac{{x}^{\mathrm{3}} }{\mathrm{3}}+\frac{{x}^{\mathrm{5}} }{\mathrm{10}}+\frac{{x}^{\mathrm{7}}…

Question-129890

Question Number 129890 by 0731619177 last updated on 20/Jan/21 Answered by Olaf last updated on 20/Jan/21 $$\underset{{x}\rightarrow\mathrm{3}} {\mathrm{lim}}\frac{\Gamma\left({x}+\mathrm{1}\right)−\mathrm{6}}{{xx}−\mathrm{33}} \\ $$$$=\:\underset{{x}\rightarrow\mathrm{3}} {\mathrm{lim}}\frac{\Gamma\left({x}+\mathrm{1}\right)−\mathrm{6}}{\mathrm{11}\left({x}−\mathrm{3}\right)} \\ $$$$=\:\underset{{x}\rightarrow\mathrm{3}} {\mathrm{lim}}\frac{\Gamma'\left({x}+\mathrm{1}\right)}{\mathrm{11}} \\…

Is-the-vector-1-2-1-an-eigen-vector-of-3-6-7-3-3-7-5-6-5-if-sp-find-the-corresponding-eigen-value-

Question Number 129887 by zarawan last updated on 20/Jan/21 $${Is}\:{the}\:{vector}\:\begin{bmatrix}{\mathrm{1}}\\{−\mathrm{2}}\\{\mathrm{1}}\end{bmatrix}{an}\:{eigen}\:{vector}\:{of}\:\begin{bmatrix}{\mathrm{3}}&{\mathrm{6}}&{\mathrm{7}}\\{\mathrm{3}}&{\mathrm{3}}&{\mathrm{7}}\\{\mathrm{5}}&{\mathrm{6}}&{\mathrm{5}\:}\end{bmatrix}?\:{if}\:{sp}\:{find}\:{the}\:{corresponding}\:{eigen}\:{value}? \\ $$ Answered by Olaf last updated on 20/Jan/21 $$\mathrm{AX}\:=\:\begin{bmatrix}{\mathrm{3}}&{\mathrm{6}}&{\mathrm{7}}\\{\mathrm{3}}&{\mathrm{3}}&{\mathrm{7}}\\{\mathrm{5}}&{\mathrm{6}}&{\mathrm{5}}\end{bmatrix}\begin{pmatrix}{\mathrm{1}}\\{−\mathrm{2}}\\{\mathrm{1}}\end{pmatrix}\:=\:\begin{pmatrix}{−\mathrm{2}}\\{\mathrm{4}}\\{−\mathrm{2}}\end{pmatrix} \\ $$$$=\:−\mathrm{2}\begin{pmatrix}{\mathrm{1}}\\{−\mathrm{2}}\\{\mathrm{1}}\end{pmatrix}\:=\:−\mathrm{2X} \\ $$$$\exists\lambda\backslash\:\mathrm{AX}\:=\:\lambda\mathrm{X} \\…

Question-129883

Question Number 129883 by Algoritm last updated on 20/Jan/21 Answered by Olaf last updated on 20/Jan/21 $$\:{y}''+\mathrm{2}{y}'+{y}\:=\:\mathrm{2}{x}+\mathrm{1}−\mathrm{2}{e}^{−{x}} \:\left(\mathrm{1}\right) \\ $$$$\mathrm{Let}\:{y}\:=\:{u}\left({x}\right){e}^{−{x}} +{v}\left({x}\right) \\ $$$${y}'\:=\:\left({u}'−{u}\right){e}^{−{x}} +{v}' \\…