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Question Number 208418 by alcohol last updated on 15/Jun/24 | ||
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$$\left.{u}_{{n}+\mathrm{1}} \:=\:{u}_{{n}} −{u}_{{n}} ^{\mathrm{3}} \:;\:{u}_{\mathrm{0}} \:\in\:\right]\mathrm{0},\:\mathrm{1}\left[\right. \\ $$$$\left..\:{show}\:{that}\:{u}_{{n}} \:\in\:\right]\mathrm{0},\:\mathrm{1}\left[\right. \\ $$$$.\:{show}\:{that}\:{u}_{{n}} \:{converges}\:{to}\:\mathrm{0} \\ $$$${v}_{{n}} \:=\:\frac{\mathrm{1}}{{u}_{{n}+\mathrm{1}} }\:−\:\frac{\mathrm{1}}{{u}_{{n}} ^{\mathrm{2}} } \\ $$$$.\:{express}\:{v}_{{n}} \:{interms}\:{of}\:{u}_{{n}} \\ $$$$.\:{show}\:{that}\:{v}_{{n}} \:{converges}\:{to}\:\mathrm{2} \\ $$$${f}\left({x}\right)\:=\:\frac{\mathrm{2}−{x}}{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} } \\ $$$$.\:{show}\:{that}\:{f}\:{is}\:{increasing}\:{and}\:{deduce}\:{that}\: \\ $$$${v}_{{n}} \:{is}\:{decreasing} \\ $$$$.\:{show}\:{that}\:{v}_{{n}} \:\geqslant\:\mathrm{2} \\ $$ | ||
Answered by Berbere last updated on 15/Jun/24 | ||
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$${tchek}\:{expression}\:{of}\:{V}_{{n}} =...? \\ $$ | ||