Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 181323 by KONE last updated on 24/Nov/22

 { ((U_0 =1 et U_1 =2)),((U_(n+2) =(√(U_n U_(n+1) )))) :}  determiner le terme generale et sa nature  besoin d′aide avp

$$\begin{cases}{{U}_{\mathrm{0}} =\mathrm{1}\:{et}\:{U}_{\mathrm{1}} =\mathrm{2}}\\{{U}_{{n}+\mathrm{2}} =\sqrt{{U}_{{n}} {U}_{{n}+\mathrm{1}} }}\end{cases} \\ $$$${determiner}\:{le}\:{terme}\:{generale}\:{et}\:{sa}\:{nature} \\ $$$${besoin}\:{d}'{aide}\:{avp} \\ $$

Commented by KONE last updated on 24/Nov/22

•svp

$$\bullet{svp} \\ $$

Answered by Frix last updated on 24/Nov/22

u_0 =2^0   u_1 =2^1   u_2 =2^(1/2)   u_3 =2^(3/4)   u_4 =2^(5/8)   ...  let u_n =2^(v_n /2^(n−1) )   v_0 =0  v_1 =1  v_2 =1  v_3 =3  v_4 =5  v_n =2v_(n−2) +v_(n−1)   v_n =((2^n −(−1)^n )/3)  ⇒  u_n =2^((2^n −(−1)^n )/(3×2^(n−1) )) =2^((2/3)(1−(1/((−2)^n ))))   lim_(n→∞)  u_n  =2^(2/3)

$${u}_{\mathrm{0}} =\mathrm{2}^{\mathrm{0}} \\ $$$${u}_{\mathrm{1}} =\mathrm{2}^{\mathrm{1}} \\ $$$${u}_{\mathrm{2}} =\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$${u}_{\mathrm{3}} =\mathrm{2}^{\frac{\mathrm{3}}{\mathrm{4}}} \\ $$$${u}_{\mathrm{4}} =\mathrm{2}^{\frac{\mathrm{5}}{\mathrm{8}}} \\ $$$$... \\ $$$$\mathrm{let}\:{u}_{{n}} =\mathrm{2}^{\frac{{v}_{{n}} }{\mathrm{2}^{{n}−\mathrm{1}} }} \\ $$$${v}_{\mathrm{0}} =\mathrm{0} \\ $$$${v}_{\mathrm{1}} =\mathrm{1} \\ $$$${v}_{\mathrm{2}} =\mathrm{1} \\ $$$${v}_{\mathrm{3}} =\mathrm{3} \\ $$$${v}_{\mathrm{4}} =\mathrm{5} \\ $$$${v}_{{n}} =\mathrm{2}{v}_{{n}−\mathrm{2}} +{v}_{{n}−\mathrm{1}} \\ $$$${v}_{{n}} =\frac{\mathrm{2}^{{n}} −\left(−\mathrm{1}\right)^{{n}} }{\mathrm{3}} \\ $$$$\Rightarrow \\ $$$${u}_{{n}} =\mathrm{2}^{\frac{\mathrm{2}^{{n}} −\left(−\mathrm{1}\right)^{{n}} }{\mathrm{3}×\mathrm{2}^{{n}−\mathrm{1}} }} =\mathrm{2}^{\frac{\mathrm{2}}{\mathrm{3}}\left(\mathrm{1}−\frac{\mathrm{1}}{\left(−\mathrm{2}\right)^{{n}} }\right)} \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:{u}_{{n}} \:=\mathrm{2}^{\frac{\mathrm{2}}{\mathrm{3}}} \\ $$

Commented by KONE last updated on 25/Nov/22

merci

$${merci} \\ $$

Answered by mr W last updated on 24/Nov/22

U_(n+2) =(√(U_(n+1) U_n ))  ln U_(n+2) =(1/2)(ln U_(n+1) +ln U_n )  let a_n =ln U_n   2a_(n+2) −a_(n+1) −a_n =0  let a_n =Cp^n   2Cp^(n+2) −Cp^(n+1) −Cp^n =0  Cp^n (2p^2 −p−1)=0  2p^2 −p−1=0  (2p+1)(p−1)=0  ⇒p=−(1/2) or 1  ⇒a_n =C×(−(1/2))^n +D×(1)^n =(C/((−2)^n ))+D  a_0 =ln U_0 =ln 1=0=(C/((−2)^0 ))+D  ⇒C+D=0   ...(i)  a_1 =ln U_1 =ln 2=(C/((−2)^1 ))+D=−(C/2)+D  ⇒−C+2D=2ln 2   ...(ii)  ⇒D=((2ln 2)/3) ⇒C=−((2ln 2)/3)  ⇒a_n =((2ln 2)/3)(1−(((−1)^n )/2^n ))=ln 2^((2/3)(1−(((−1)^n )/2^n )))   ⇒U_n =e^a_n  =2^((2/3)[1−(((−1)^n )/2^n )])  ✓  lim_(n→∞)  U_n =2^(2/3) =(4)^(1/3)

$${U}_{{n}+\mathrm{2}} =\sqrt{{U}_{{n}+\mathrm{1}} {U}_{{n}} } \\ $$$$\mathrm{ln}\:{U}_{{n}+\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{ln}\:{U}_{{n}+\mathrm{1}} +\mathrm{ln}\:{U}_{{n}} \right) \\ $$$${let}\:{a}_{{n}} =\mathrm{ln}\:{U}_{{n}} \\ $$$$\mathrm{2}{a}_{{n}+\mathrm{2}} −{a}_{{n}+\mathrm{1}} −{a}_{{n}} =\mathrm{0} \\ $$$${let}\:{a}_{{n}} ={Cp}^{{n}} \\ $$$$\mathrm{2}{Cp}^{{n}+\mathrm{2}} −{Cp}^{{n}+\mathrm{1}} −{Cp}^{{n}} =\mathrm{0} \\ $$$${Cp}^{{n}} \left(\mathrm{2}{p}^{\mathrm{2}} −{p}−\mathrm{1}\right)=\mathrm{0} \\ $$$$\mathrm{2}{p}^{\mathrm{2}} −{p}−\mathrm{1}=\mathrm{0} \\ $$$$\left(\mathrm{2}{p}+\mathrm{1}\right)\left({p}−\mathrm{1}\right)=\mathrm{0} \\ $$$$\Rightarrow{p}=−\frac{\mathrm{1}}{\mathrm{2}}\:{or}\:\mathrm{1} \\ $$$$\Rightarrow{a}_{{n}} ={C}×\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)^{{n}} +{D}×\left(\mathrm{1}\right)^{{n}} =\frac{{C}}{\left(−\mathrm{2}\right)^{{n}} }+{D} \\ $$$${a}_{\mathrm{0}} =\mathrm{ln}\:{U}_{\mathrm{0}} =\mathrm{ln}\:\mathrm{1}=\mathrm{0}=\frac{{C}}{\left(−\mathrm{2}\right)^{\mathrm{0}} }+{D} \\ $$$$\Rightarrow{C}+{D}=\mathrm{0}\:\:\:...\left({i}\right) \\ $$$${a}_{\mathrm{1}} =\mathrm{ln}\:{U}_{\mathrm{1}} =\mathrm{ln}\:\mathrm{2}=\frac{{C}}{\left(−\mathrm{2}\right)^{\mathrm{1}} }+{D}=−\frac{{C}}{\mathrm{2}}+{D} \\ $$$$\Rightarrow−{C}+\mathrm{2}{D}=\mathrm{2ln}\:\mathrm{2}\:\:\:...\left({ii}\right) \\ $$$$\Rightarrow{D}=\frac{\mathrm{2ln}\:\mathrm{2}}{\mathrm{3}}\:\Rightarrow{C}=−\frac{\mathrm{2ln}\:\mathrm{2}}{\mathrm{3}} \\ $$$$\Rightarrow{a}_{{n}} =\frac{\mathrm{2ln}\:\mathrm{2}}{\mathrm{3}}\left(\mathrm{1}−\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{2}^{{n}} }\right)=\mathrm{ln}\:\mathrm{2}^{\frac{\mathrm{2}}{\mathrm{3}}\left(\mathrm{1}−\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{2}^{{n}} }\right)} \\ $$$$\Rightarrow{U}_{{n}} ={e}^{{a}_{{n}} } =\mathrm{2}^{\frac{\mathrm{2}}{\mathrm{3}}\left[\mathrm{1}−\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{2}^{{n}} }\right]} \:\checkmark \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:{U}_{{n}} =\mathrm{2}^{\frac{\mathrm{2}}{\mathrm{3}}} =\sqrt[{\mathrm{3}}]{\mathrm{4}} \\ $$

Commented by mnjuly1970 last updated on 24/Nov/22

  sir W   ( excuse me )      e^( a_n )  = (4)^(1/3)  . e^( 1−(((−1)^( n) )/2^( n) ))  ???

$$\:\:{sir}\:{W}\:\:\:\left(\:{excuse}\:{me}\:\right) \\ $$$$\:\:\:\:{e}^{\:{a}_{{n}} } \:=\:\sqrt[{\mathrm{3}}]{\mathrm{4}}\:.\:{e}^{\:\mathrm{1}−\frac{\left(−\mathrm{1}\right)^{\:{n}} }{\mathrm{2}^{\:{n}} }} \:??? \\ $$

Commented by mnjuly1970 last updated on 24/Nov/22

  thanks alot sir..     my  mistake      in fact :  a_( n)  = ln(2)^( (2/3) (1−(((−1)^( n) )/2^( n) )))

$$\:\:{thanks}\:{alot}\:{sir}.. \\ $$$$\:\:\:{my}\:\:{mistake} \\ $$$$\:\:\:\:{in}\:{fact}\::\:\:{a}_{\:{n}} \:=\:\mathrm{ln}\left(\mathrm{2}\right)^{\:\frac{\mathrm{2}}{\mathrm{3}}\:\left(\mathrm{1}−\frac{\left(−\mathrm{1}\right)^{\:{n}} }{\mathrm{2}^{\:{n}} }\right)} \\ $$

Commented by mr W last updated on 24/Nov/22

no!   e^a_n  =((4)^(1/3) )^(1−(((−1)^n )/2^n ))

$${no}!\: \\ $$$${e}^{{a}_{{n}} } =\left(\sqrt[{\mathrm{3}}]{\mathrm{4}}\right)^{\mathrm{1}−\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{2}^{{n}} }} \\ $$

Commented by mnjuly1970 last updated on 24/Nov/22

Commented by KONE last updated on 25/Nov/22

merci

$${merci} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com