Question Number 181323 by KONE last updated on 24/Nov/22 | ||
$$\begin{cases}{{U}_{\mathrm{0}} =\mathrm{1}\:{et}\:{U}_{\mathrm{1}} =\mathrm{2}}\\{{U}_{{n}+\mathrm{2}} =\sqrt{{U}_{{n}} {U}_{{n}+\mathrm{1}} }}\end{cases} \\ $$$${determiner}\:{le}\:{terme}\:{generale}\:{et}\:{sa}\:{nature} \\ $$$${besoin}\:{d}'{aide}\:{avp} \\ $$ | ||
Commented by KONE last updated on 24/Nov/22 | ||
$$\bullet{svp} \\ $$ | ||
Answered by Frix last updated on 24/Nov/22 | ||
$${u}_{\mathrm{0}} =\mathrm{2}^{\mathrm{0}} \\ $$$${u}_{\mathrm{1}} =\mathrm{2}^{\mathrm{1}} \\ $$$${u}_{\mathrm{2}} =\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$${u}_{\mathrm{3}} =\mathrm{2}^{\frac{\mathrm{3}}{\mathrm{4}}} \\ $$$${u}_{\mathrm{4}} =\mathrm{2}^{\frac{\mathrm{5}}{\mathrm{8}}} \\ $$$$... \\ $$$$\mathrm{let}\:{u}_{{n}} =\mathrm{2}^{\frac{{v}_{{n}} }{\mathrm{2}^{{n}−\mathrm{1}} }} \\ $$$${v}_{\mathrm{0}} =\mathrm{0} \\ $$$${v}_{\mathrm{1}} =\mathrm{1} \\ $$$${v}_{\mathrm{2}} =\mathrm{1} \\ $$$${v}_{\mathrm{3}} =\mathrm{3} \\ $$$${v}_{\mathrm{4}} =\mathrm{5} \\ $$$${v}_{{n}} =\mathrm{2}{v}_{{n}−\mathrm{2}} +{v}_{{n}−\mathrm{1}} \\ $$$${v}_{{n}} =\frac{\mathrm{2}^{{n}} −\left(−\mathrm{1}\right)^{{n}} }{\mathrm{3}} \\ $$$$\Rightarrow \\ $$$${u}_{{n}} =\mathrm{2}^{\frac{\mathrm{2}^{{n}} −\left(−\mathrm{1}\right)^{{n}} }{\mathrm{3}×\mathrm{2}^{{n}−\mathrm{1}} }} =\mathrm{2}^{\frac{\mathrm{2}}{\mathrm{3}}\left(\mathrm{1}−\frac{\mathrm{1}}{\left(−\mathrm{2}\right)^{{n}} }\right)} \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:{u}_{{n}} \:=\mathrm{2}^{\frac{\mathrm{2}}{\mathrm{3}}} \\ $$ | ||
Commented by KONE last updated on 25/Nov/22 | ||
$${merci} \\ $$ | ||
Answered by mr W last updated on 24/Nov/22 | ||
$${U}_{{n}+\mathrm{2}} =\sqrt{{U}_{{n}+\mathrm{1}} {U}_{{n}} } \\ $$$$\mathrm{ln}\:{U}_{{n}+\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{ln}\:{U}_{{n}+\mathrm{1}} +\mathrm{ln}\:{U}_{{n}} \right) \\ $$$${let}\:{a}_{{n}} =\mathrm{ln}\:{U}_{{n}} \\ $$$$\mathrm{2}{a}_{{n}+\mathrm{2}} −{a}_{{n}+\mathrm{1}} −{a}_{{n}} =\mathrm{0} \\ $$$${let}\:{a}_{{n}} ={Cp}^{{n}} \\ $$$$\mathrm{2}{Cp}^{{n}+\mathrm{2}} −{Cp}^{{n}+\mathrm{1}} −{Cp}^{{n}} =\mathrm{0} \\ $$$${Cp}^{{n}} \left(\mathrm{2}{p}^{\mathrm{2}} −{p}−\mathrm{1}\right)=\mathrm{0} \\ $$$$\mathrm{2}{p}^{\mathrm{2}} −{p}−\mathrm{1}=\mathrm{0} \\ $$$$\left(\mathrm{2}{p}+\mathrm{1}\right)\left({p}−\mathrm{1}\right)=\mathrm{0} \\ $$$$\Rightarrow{p}=−\frac{\mathrm{1}}{\mathrm{2}}\:{or}\:\mathrm{1} \\ $$$$\Rightarrow{a}_{{n}} ={C}×\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)^{{n}} +{D}×\left(\mathrm{1}\right)^{{n}} =\frac{{C}}{\left(−\mathrm{2}\right)^{{n}} }+{D} \\ $$$${a}_{\mathrm{0}} =\mathrm{ln}\:{U}_{\mathrm{0}} =\mathrm{ln}\:\mathrm{1}=\mathrm{0}=\frac{{C}}{\left(−\mathrm{2}\right)^{\mathrm{0}} }+{D} \\ $$$$\Rightarrow{C}+{D}=\mathrm{0}\:\:\:...\left({i}\right) \\ $$$${a}_{\mathrm{1}} =\mathrm{ln}\:{U}_{\mathrm{1}} =\mathrm{ln}\:\mathrm{2}=\frac{{C}}{\left(−\mathrm{2}\right)^{\mathrm{1}} }+{D}=−\frac{{C}}{\mathrm{2}}+{D} \\ $$$$\Rightarrow−{C}+\mathrm{2}{D}=\mathrm{2ln}\:\mathrm{2}\:\:\:...\left({ii}\right) \\ $$$$\Rightarrow{D}=\frac{\mathrm{2ln}\:\mathrm{2}}{\mathrm{3}}\:\Rightarrow{C}=−\frac{\mathrm{2ln}\:\mathrm{2}}{\mathrm{3}} \\ $$$$\Rightarrow{a}_{{n}} =\frac{\mathrm{2ln}\:\mathrm{2}}{\mathrm{3}}\left(\mathrm{1}−\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{2}^{{n}} }\right)=\mathrm{ln}\:\mathrm{2}^{\frac{\mathrm{2}}{\mathrm{3}}\left(\mathrm{1}−\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{2}^{{n}} }\right)} \\ $$$$\Rightarrow{U}_{{n}} ={e}^{{a}_{{n}} } =\mathrm{2}^{\frac{\mathrm{2}}{\mathrm{3}}\left[\mathrm{1}−\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{2}^{{n}} }\right]} \:\checkmark \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:{U}_{{n}} =\mathrm{2}^{\frac{\mathrm{2}}{\mathrm{3}}} =\sqrt[{\mathrm{3}}]{\mathrm{4}} \\ $$ | ||
Commented by mnjuly1970 last updated on 24/Nov/22 | ||
$$\:\:{sir}\:{W}\:\:\:\left(\:{excuse}\:{me}\:\right) \\ $$$$\:\:\:\:{e}^{\:{a}_{{n}} } \:=\:\sqrt[{\mathrm{3}}]{\mathrm{4}}\:.\:{e}^{\:\mathrm{1}−\frac{\left(−\mathrm{1}\right)^{\:{n}} }{\mathrm{2}^{\:{n}} }} \:??? \\ $$ | ||
Commented by mnjuly1970 last updated on 24/Nov/22 | ||
$$\:\:{thanks}\:{alot}\:{sir}.. \\ $$$$\:\:\:{my}\:\:{mistake} \\ $$$$\:\:\:\:{in}\:{fact}\::\:\:{a}_{\:{n}} \:=\:\mathrm{ln}\left(\mathrm{2}\right)^{\:\frac{\mathrm{2}}{\mathrm{3}}\:\left(\mathrm{1}−\frac{\left(−\mathrm{1}\right)^{\:{n}} }{\mathrm{2}^{\:{n}} }\right)} \\ $$ | ||
Commented by mr W last updated on 24/Nov/22 | ||
$${no}!\: \\ $$$${e}^{{a}_{{n}} } =\left(\sqrt[{\mathrm{3}}]{\mathrm{4}}\right)^{\mathrm{1}−\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{2}^{{n}} }} \\ $$ | ||
Commented by mnjuly1970 last updated on 24/Nov/22 | ||
Commented by KONE last updated on 25/Nov/22 | ||
$${merci} \\ $$ | ||