Question Number 124870 by Algoritm last updated on 06/Dec/20
Answered by mindispower last updated on 06/Dec/20
![ln(x_(n+1) )=(1/4)ln(81)−(1/4)ln(x_n ) W_n =ln(x_n ) W_(n+1) =ln(3)−(w_n /4) find w_n withe n than x_n =e^w_n solve x_n ≤3 the answer will bee n∈[α,+∞[ n_0 =[α]+1](https://www.tinkutara.com/question/Q124874.png)
$${ln}\left({x}_{{n}+\mathrm{1}} \right)=\frac{\mathrm{1}}{\mathrm{4}}{ln}\left(\mathrm{81}\right)−\frac{\mathrm{1}}{\mathrm{4}}{ln}\left({x}_{{n}} \right) \\ $$$${W}_{{n}} ={ln}\left({x}_{{n}} \right) \\ $$$${W}_{{n}+\mathrm{1}} ={ln}\left(\mathrm{3}\right)−\frac{{w}_{{n}} }{\mathrm{4}} \\ $$$${find}\:{w}_{{n}} \:{withe}\:{n}\:{than}\:{x}_{{n}} ={e}^{{w}_{{n}} } \\ $$$${solve}\:{x}_{{n}} \leqslant\mathrm{3}\:{the}\:{answer}\:{will}\:{bee}\:{n}\in\left[\alpha,+\infty\left[\right.\right. \\ $$$${n}_{\mathrm{0}} =\left[\alpha\right]+\mathrm{1} \\ $$
Commented by Algoritm last updated on 06/Dec/20
$$\mathrm{W}_{\mathrm{n}} −\mathrm{lambert}\:? \\ $$
Answered by mr W last updated on 09/Dec/20
/5)) ln (x_n )=[1−(−(1/4))^n ](((4 ln 3)/5)) ⇒x_n =3^((4/5)[1−(−(1/4))^n ]) <3 for n≥2](https://www.tinkutara.com/question/Q125231.png)
$${x}_{{n}+\mathrm{1}} =\left(\frac{\mathrm{81}}{{x}_{{n}} }\right)^{\frac{\mathrm{1}}{\mathrm{4}}} \\ $$$$\mathrm{ln}\:\left({x}_{{n}+\mathrm{1}} \right)=\frac{\mathrm{ln}\:\mathrm{81}}{\mathrm{4}}−\frac{\mathrm{ln}\:\left({x}_{{n}} \right)}{\mathrm{4}} \\ $$$${let}\:{a}_{{n}} =\mathrm{ln}\:\left({x}_{{n}} \right),\:{p}=\frac{\mathrm{ln}\:\mathrm{81}}{\mathrm{4}}=\mathrm{ln}\:\mathrm{3} \\ $$$$\Rightarrow{a}_{{n}+\mathrm{1}} ={p}−\frac{{a}_{{n}} }{\mathrm{4}} \\ $$$${let}\:{a}_{{n}} ={b}_{{n}} +{c} \\ $$$${b}_{{n}+\mathrm{1}} +{c}={p}−\frac{{b}_{{n}} +{c}}{\mathrm{4}} \\ $$$${b}_{{n}+\mathrm{1}} ={p}−\frac{\mathrm{5}{c}}{\mathrm{4}}−\frac{{b}_{{n}} }{\mathrm{4}} \\ $$$${let}\:{p}−\frac{\mathrm{5}{c}}{\mathrm{4}}=\mathrm{0}\:\Rightarrow{c}=\frac{\mathrm{4}{p}}{\mathrm{5}}=\frac{\mathrm{4}\:\mathrm{ln}\:\mathrm{3}}{\mathrm{5}} \\ $$$$\Rightarrow{b}_{{n}+\mathrm{1}} =−\frac{\mathrm{1}}{\mathrm{4}}{b}_{{n}} \:\:\:\Rightarrow{G}.{P}. \\ $$$$\Rightarrow{b}_{{n}} =\left(−\frac{\mathrm{1}}{\mathrm{4}}\right)^{{n}} {b}_{\mathrm{0}} \\ $$$${b}_{\mathrm{0}} ={a}_{\mathrm{0}} −{c}=\mathrm{ln}\:\left({x}_{\mathrm{0}} \right)−{c}=\mathrm{ln}\:\mathrm{1}−\frac{\mathrm{4}\:\mathrm{ln}\:\mathrm{3}}{\mathrm{5}}=−\frac{\mathrm{4}\:\mathrm{ln}\:\mathrm{3}}{\mathrm{5}} \\ $$$${a}_{{n}} =\left(−\frac{\mathrm{1}}{\mathrm{4}}\right)^{{n}} \left(−\frac{\mathrm{4}\:\mathrm{ln}\:\mathrm{3}}{\mathrm{5}}\right)+\frac{\mathrm{4}\:\mathrm{ln}\:\mathrm{3}}{\mathrm{5}} \\ $$$${a}_{{n}} =\left[\mathrm{1}−\left(−\frac{\mathrm{1}}{\mathrm{4}}\right)^{{n}} \right]\left(\frac{\mathrm{4}\:\mathrm{ln}\:\mathrm{3}}{\mathrm{5}}\right) \\ $$$$\mathrm{ln}\:\left({x}_{{n}} \right)=\left[\mathrm{1}−\left(−\frac{\mathrm{1}}{\mathrm{4}}\right)^{{n}} \right]\left(\frac{\mathrm{4}\:\mathrm{ln}\:\mathrm{3}}{\mathrm{5}}\right) \\ $$$$\Rightarrow{x}_{{n}} =\mathrm{3}^{\frac{\mathrm{4}}{\mathrm{5}}\left[\mathrm{1}−\left(−\frac{\mathrm{1}}{\mathrm{4}}\right)^{{n}} \right]} \:<\mathrm{3}\:{for}\:{n}\geqslant\mathrm{2} \\ $$