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Question Number 36626 by tanmay.chaudhury50@gmail.com last updated on 03/Jun/18 Commented by tanmay.chaudhury50@gmail.com last updated on 03/Jun/18 $${the}\:{vincci}\:{code}\:{is}\:{about}\:{golden}\:{ratio}..{see}\:{vdo} \\ $$$${in}\:{youtube}\:{to}\:{see}\:{the}\:{magic}.. \\ $$ Terms of Service Privacy…
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Question Number 102142 by Ar Brandon last updated on 06/Jul/20 $$\mathrm{A}\:\mathrm{four}-\mathrm{sidex}\:\mathrm{dice}\:\mathrm{with}\:\mathrm{numbered}\:\mathrm{1},\:\mathrm{2},\:\mathrm{3},\:\mathrm{and}\:\mathrm{4}\:\mathrm{is}\:\mathrm{thrown} \\ $$$$\mathrm{and}\:\mathrm{the}\:\mathrm{number}\:\mathrm{at}\:\mathrm{the}\:\mathrm{base}\:\mathrm{is}\:\mathrm{read}. \\ $$$$\mathrm{The}\:\mathrm{dice}\:\mathrm{is}\:\mathrm{biased}\:\mathrm{such}\:\mathrm{that}\:\mathrm{the}\:\mathrm{probabilies}\:\mathrm{P}_{\mathrm{1}} ,\:\mathrm{P}_{\mathrm{2}} ,\:\mathrm{P}_{\mathrm{3}} , \\ $$$$\mathrm{and}\:\mathrm{P}_{\mathrm{4}} \:\mathrm{to}\:\mathrm{obtain}\:\mathrm{1},\:\mathrm{2},\:\mathrm{3},\:\mathrm{and}\:\mathrm{4}\:\mathrm{respectively}\:\mathrm{are}\:\mathrm{in}\:\mathrm{an} \\ $$$$\mathrm{arithmetic}\:\mathrm{progression}. \\ $$$$\mathrm{1}\backslash\:\mathrm{Given}\:\mathrm{P}_{\mathrm{4}}…
Question Number 36601 by tanmay.chaudhury50@gmail.com last updated on 03/Jun/18 $${i}\:{want}\:{to}\:{share}\:{proof}\:{that}\:{universe}\:{and} \\ $$$${mathematics}\:{co}\:{related}…{golden}\:{ratio}…{is}\:{the} \\ $$$${proof}…{fibanocci}\:{numbers}\:{present}\:{in}\:{resl}\:{life} \\ $$$${if}\:{you}\:{want}\:{i}\:{can}\:{share}… \\ $$ Commented by ajfour last updated on 03/Jun/18…
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Question Number 102089 by Ar Brandon last updated on 06/Jul/20 $$ \\ $$$$\mathrm{The}\:\mathrm{Gamma}\:\mathrm{function}\:\Gamma\left(\alpha\right)\:\mathrm{is}\:\mathrm{defined}\:\mathrm{as}\:\mathrm{follows}; \\ $$$$\Gamma\left(\alpha\right)=\int_{\mathrm{0}} ^{\infty} \mathrm{y}^{\alpha−\mathrm{1}} \mathrm{e}^{−\mathrm{y}} \mathrm{dy}\:,\:\alpha>\mathrm{0} \\ $$$$\mathrm{a}\backslash\:\mathrm{Show}\:\mathrm{that}\:\Gamma\left(\alpha+\mathrm{1}\right)=\alpha\Gamma\left(\alpha\right). \\ $$$$\mathrm{b}\backslash\mathrm{Conclude}\:\mathrm{that}\:\Gamma\left(\mathrm{n}\right)=\left(\mathrm{n}−\mathrm{1}\right)!\:,\:\mathrm{n}=\mathrm{1},\:\mathrm{2},\:\mathrm{3},\:… \\ $$$$\mathrm{c}\backslash\mathrm{Determine}\:\Gamma\left(\mathrm{55}\right).…