Question Number 196828 by ERLY last updated on 01/Sep/23 Answered by Skabetix last updated on 01/Sep/23 $$\left.\mathrm{2}.{a}\right)\:{U}_{{n}+\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{1}}{{U}_{{n}} } \\ $$$${Comme}\:{U}_{{n}} >\mathrm{0}\rightarrow{U}_{{n}} \geqslant\frac{\mathrm{1}}{{U}_{{n}} }\rightarrow{U}_{{n}+\mathrm{1}} \leqslant\frac{\mathrm{1}}{\mathrm{2}}{U}_{{n}}…
Question Number 196793 by MrGHK last updated on 31/Aug/23 Answered by Mathspace last updated on 02/Sep/23 $${B}\left({x},{y}\right)=\frac{\Gamma\left({x}\right)\Gamma\left({y}\right)}{\Gamma\left({x}+{y}\right)} \\ $$$$\Rightarrow\frac{\partial{B}}{\partial{x}}\left({x},{y}\right)=\Gamma\left({y}\right).\frac{\Gamma^{'} \left({x}\right)\Gamma\left({x}+{y}\right)−\Gamma\left({x}\right)\Gamma^{'} \left({x}+{y}\right)}{\Gamma^{\mathrm{2}} \left({x}+{y}\right)} \\ $$$$=\Gamma\left({y}\right)\frac{\Psi\left({x}\right)\Gamma\left({x}\right)\Gamma\left({x}+{y}\right)−\Gamma\left({x}\right)\Psi\left({x}+{y}\right)\Gamma\left({x}+{y}\right)}{\Gamma^{\mathrm{2}} \left({x}+{y}\right)}…
Question Number 196761 by Humble last updated on 31/Aug/23 $$\mathrm{A}\:\mathrm{tightly}\:\mathrm{wound}\:\mathrm{toroidal}\:\mathrm{coil}\:\mathrm{with}\:\mathrm{a}\:\mathrm{square} \\ $$$$\mathrm{cross}\:\mathrm{section}\:\mathrm{and}\:\mathrm{an}\:\mathrm{inner}\:\mathrm{radius}\:\mathrm{of}\:\mathrm{15cm}\:\mathrm{has} \\ $$$$\mathrm{500}\:\mathrm{turns}\:\mathrm{of}\:\mathrm{copper}\:\mathrm{wire}\:\mathrm{and}\:\mathrm{carries}\:\mathrm{an}\:\mathrm{insulated} \\ $$$$\mathrm{filamentary}\:\mathrm{ccurrent}\:\mathrm{of}\:\mathrm{0}.\mathrm{800A}.\:\mathrm{what}\:\mathrm{is}\:\mathrm{the} \\ $$$$\mathrm{strength}\:\mathrm{of}\:\mathrm{the}\:\mathrm{magnetic}\:\mathrm{field}\:\mathrm{inside}\:\mathrm{the} \\ $$$$\mathrm{toroid}\:\mathrm{at}\:\mathrm{the}\:\mathrm{inner}\:\mathrm{radius}? \\ $$ Terms of Service…
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Question Number 196757 by MrGHK last updated on 31/Aug/23 Answered by qaz last updated on 31/Aug/23 $$\sqrt{\pi}\Sigma\frac{\left(\frac{\Gamma\left({n}+\mathrm{1}\right)}{\Gamma\left({n}+\frac{\mathrm{3}}{\mathrm{2}}\right)}\right)'}{{n}+\mathrm{1}}=\Sigma\frac{\mathrm{1}}{{n}+\mathrm{1}}\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{n}} \left(\mathrm{1}−{x}\right)^{−\mathrm{1}/\mathrm{2}} {lnxdx} \\ $$$$=−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{lnxln}\left(\mathrm{1}−{x}\right)}{{x}\sqrt{\mathrm{1}−{x}}}{dx}=−\frac{\partial^{\mathrm{2}}…
Question Number 196758 by sonukgindia last updated on 31/Aug/23 Answered by qaz last updated on 31/Aug/23 $$\mathrm{1}+\left(\mathrm{1}+{x}\right)+\left(\mathrm{1}+{x}+{x}^{\mathrm{2}} \right)+\left(\mathrm{1}+{x}+{x}^{\mathrm{2}} +{x}^{\mathrm{3}} \right)+…\left(\mathrm{1}+{x}+{x}^{\mathrm{2}} +…+{x}^{{n}−\mathrm{1}} \right) \\ $$$$=\frac{\mathrm{1}−{x}}{\mathrm{1}−{x}}+\frac{\mathrm{1}−{x}^{\mathrm{2}} }{\mathrm{1}−{x}}+\frac{\mathrm{1}−{x}^{\mathrm{3}}…
Question Number 196775 by Tawa11 last updated on 31/Aug/23 A stream is flowing at 0.75ms−¹ and a boat heading perpendicular for the stream landed at…
Question Number 196735 by sonukgindia last updated on 30/Aug/23 Answered by qaz last updated on 31/Aug/23 $$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{\mathrm{7}} −{x}^{\mathrm{3}} }{{lnx}}{dx}=\int_{\mathrm{0}} ^{\mathrm{1}} {dx}\int_{\mathrm{3}} ^{\mathrm{7}} {x}^{{t}}…
Question Number 196734 by sonukgindia last updated on 30/Aug/23 Answered by qaz last updated on 31/Aug/23 $${y}''{y}+\left({y}'\right)^{\mathrm{2}} =\left({y}'{y}\right)'=\frac{\mathrm{1}}{\mathrm{2}}\left(\left({y}^{\mathrm{2}} \right)'\right)'=\mathrm{3}{x} \\ $$$$\Rightarrow\left({y}^{\mathrm{2}} \right)'=\mathrm{3}{x}^{\mathrm{2}} +{C}_{\mathrm{1}} \:\:\:\:{y}^{\mathrm{2}} ={x}^{\mathrm{3}}…
Question Number 196725 by sonukgindia last updated on 30/Aug/23 Answered by Frix last updated on 30/Aug/23 $$\left({c}\right) \\ $$$${t}^{\mathrm{8}} +\frac{\mathrm{1}}{{t}^{\mathrm{8}} }={m} \\ $$$${t}^{\mathrm{8}} +\frac{\mathrm{1}}{{t}^{\mathrm{8}} }+\mathrm{2}={m}+\mathrm{2}…