Question and Answers Forum

All Questions      Topic List

Relation and Functions Questions

Previous in All Question      Next in All Question      

Previous in Relation and Functions      Next in Relation and Functions      

Question Number 136031 by mathmax by abdo last updated on 18/Mar/21

study the sequence U_n =(√((1+u_(n−1) )/2)) with u_0 =(1/2) and determine lim_(n→+∞) U_n

$$\mathrm{study}\:\mathrm{the}\:\mathrm{sequence}\:\:\mathrm{U}_{\mathrm{n}} =\sqrt{\frac{\mathrm{1}+\mathrm{u}_{\mathrm{n}−\mathrm{1}} }{\mathrm{2}}} \\ $$$$\mathrm{with}\:\mathrm{u}_{\mathrm{0}} =\frac{\mathrm{1}}{\mathrm{2}}\:\mathrm{and}\:\mathrm{determine}\:\mathrm{lim}_{\mathrm{n}\rightarrow+\infty} \mathrm{U}_{\mathrm{n}} \\ $$

Answered by mindispower last updated on 18/Mar/21

u_n ^2 =((1+u_(n−1) )/2) u_n <1 ...easy to see u_0 =0.5≤1 suppose u_n <1 ,∀n u_(n+1) =(√((1+u_n )/2))<(√((1+1)/2))=1⇔u_(n+1) ≤1 by reccursion u_n ≤1 u_n ^2 −u_(n−1) <u_n −u_(n−1) u_n ^2 −u_(n−1) =((1−u_(n−1) )/2_ )>0,⇒u_n −u_(n−1) >0 u_n increase withe bounded 0<u_n <1 ⇒u_n conveege lim_(n→∞) u_n =l⇒l=(√((1+l)/2))⇔2l^2 −l−1 l∈{1,−(1/2)} l=1 ,u_n ∈[(1/2),1[

$${u}_{{n}} ^{\mathrm{2}} =\frac{\mathrm{1}+{u}_{{n}−\mathrm{1}} }{\mathrm{2}} \\ $$$${u}_{{n}} <\mathrm{1}\:...{easy}\:{to}\:{see} \\ $$$${u}_{\mathrm{0}} =\mathrm{0}.\mathrm{5}\leqslant\mathrm{1} \\ $$$${suppose}\:{u}_{{n}} <\mathrm{1}\:,\forall{n} \\ $$$${u}_{{n}+\mathrm{1}} =\sqrt{\frac{\mathrm{1}+{u}_{{n}} }{\mathrm{2}}}<\sqrt{\frac{\mathrm{1}+\mathrm{1}}{\mathrm{2}}}=\mathrm{1}\Leftrightarrow{u}_{{n}+\mathrm{1}} \leqslant\mathrm{1} \\ $$$${by}\:{reccursion}\:{u}_{{n}} \leqslant\mathrm{1} \\ $$$${u}_{{n}} ^{\mathrm{2}} −{u}_{{n}−\mathrm{1}} <{u}_{{n}} −{u}_{{n}−\mathrm{1}} \\ $$$${u}_{{n}} ^{\mathrm{2}} −{u}_{{n}−\mathrm{1}} =\frac{\mathrm{1}−{u}_{{n}−\mathrm{1}} }{\mathrm{2}_{} }>\mathrm{0},\Rightarrow{u}_{{n}} −{u}_{{n}−\mathrm{1}} >\mathrm{0} \\ $$$${u}_{{n}} {increase}\:{withe}\:{bounded}\:\mathrm{0}<{u}_{{n}} <\mathrm{1} \\ $$$$\Rightarrow{u}_{{n}} \:\:{conveege} \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}{u}_{{n}} ={l}\Rightarrow{l}=\sqrt{\frac{\mathrm{1}+{l}}{\mathrm{2}}}\Leftrightarrow\mathrm{2}{l}^{\mathrm{2}} −{l}−\mathrm{1} \\ $$$${l}\in\left\{\mathrm{1},−\frac{\mathrm{1}}{\mathrm{2}}\right\} \\ $$$${l}=\mathrm{1}\:,{u}_{{n}} \in\left[\frac{\mathrm{1}}{\mathrm{2}},\mathrm{1}\left[\right.\right. \\ $$

Commented by mathmax by abdo last updated on 18/Mar/21

thank you sir mind

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir}\:\mathrm{mind} \\ $$

Commented by mindispower last updated on 20/Mar/21

withe pleasur

$${withe}\:{pleasur} \\ $$

Answered by mathmax by abdo last updated on 18/Mar/21

we have u_0 =(1/2)=cos((π/3)) ⇒u_1 =(√((1+cos((π/3)))/2))=cos((π/6)) let suppose u_n =cos((π/(3.2^n ))) ⇒u_(n+1) =(√((1+cos((π/(3.2^n ))))/2)) =cos((π/(3.2^(n+1) ))) ⇒∀n≥1 u_n =cos((π/(3.2^n ))) ⇒lim_(n→+∞) u_n =cos(0)=1

$$\mathrm{we}\:\mathrm{have}\:\mathrm{u}_{\mathrm{0}} =\frac{\mathrm{1}}{\mathrm{2}}=\mathrm{cos}\left(\frac{\pi}{\mathrm{3}}\right)\:\Rightarrow\mathrm{u}_{\mathrm{1}} =\sqrt{\frac{\mathrm{1}+\mathrm{cos}\left(\frac{\pi}{\mathrm{3}}\right)}{\mathrm{2}}}=\mathrm{cos}\left(\frac{\pi}{\mathrm{6}}\right)\:\mathrm{let} \\ $$$$\mathrm{suppose}\:\mathrm{u}_{\mathrm{n}} =\mathrm{cos}\left(\frac{\pi}{\mathrm{3}.\mathrm{2}^{\mathrm{n}} }\right)\:\Rightarrow\mathrm{u}_{\mathrm{n}+\mathrm{1}} =\sqrt{\frac{\mathrm{1}+\mathrm{cos}\left(\frac{\pi}{\mathrm{3}.\mathrm{2}^{\mathrm{n}} }\right)}{\mathrm{2}}} \\ $$$$=\mathrm{cos}\left(\frac{\pi}{\mathrm{3}.\mathrm{2}^{\mathrm{n}+\mathrm{1}} }\right)\:\Rightarrow\forall\mathrm{n}\geqslant\mathrm{1}\:\:\mathrm{u}_{\mathrm{n}} =\mathrm{cos}\left(\frac{\pi}{\mathrm{3}.\mathrm{2}^{\mathrm{n}} }\right)\:\Rightarrow\mathrm{lim}_{\mathrm{n}\rightarrow+\infty} \mathrm{u}_{\mathrm{n}} =\mathrm{cos}\left(\mathrm{0}\right)=\mathrm{1} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com