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Question Number 133856 by mathocean1 last updated on 24/Feb/21

{ ((U_0 =1)),((U_1 =2)),(( U_(n+2) =(3/2)U_(n+1) −(1/2)U_n )) :} Determinate the smallest integer n_0 such that ∀ n≥n_0 we have ∣U_n −3∣≤10^(−4)

$$ \\ $$$$\begin{cases}{{U}_{\mathrm{0}} =\mathrm{1}}\\{{U}_{\mathrm{1}} =\mathrm{2}}\\{\:{U}_{{n}+\mathrm{2}} =\frac{\mathrm{3}}{\mathrm{2}}{U}_{{n}+\mathrm{1}} −\frac{\mathrm{1}}{\mathrm{2}}{U}_{{n}} }\end{cases} \\ $$$${Determinate}\:{the}\:{smallest}\:{integer} \\ $$$${n}_{\mathrm{0}} \:{such}\:{that}\:\forall\:{n}\geqslant{n}_{\mathrm{0}} \:{we}\:{have}\:\mid{U}_{{n}} −\mathrm{3}\mid\leqslant\mathrm{10}^{−\mathrm{4}} \\ $$

Answered by mr W last updated on 25/Feb/21

x^2 −(3/2)x+(1/2)=0 (x−1)(x−(1/2))=0 x=1, (1/2) ⇒U_n =A+(B/2^n ) U_0 =A+B=1 U_1 =A+(B/2)=2 ⇒B=−2, A=3 ⇒U_n =3−(1/2^(n−1) ) lim_(n→∞) U_n =3 ∣U_n −3∣=(1/2^(n−1) )≤10^(−4) 2^(n−1) ≥10^4 n≥1+log_2 10^4 =1+(4/(log 2))≈14.29 ⇒n_0 =15

$${x}^{\mathrm{2}} −\frac{\mathrm{3}}{\mathrm{2}}{x}+\frac{\mathrm{1}}{\mathrm{2}}=\mathrm{0} \\ $$$$\left({x}−\mathrm{1}\right)\left({x}−\frac{\mathrm{1}}{\mathrm{2}}\right)=\mathrm{0} \\ $$$${x}=\mathrm{1},\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow{U}_{{n}} ={A}+\frac{{B}}{\mathrm{2}^{{n}} } \\ $$$${U}_{\mathrm{0}} ={A}+{B}=\mathrm{1} \\ $$$${U}_{\mathrm{1}} ={A}+\frac{{B}}{\mathrm{2}}=\mathrm{2} \\ $$$$\Rightarrow{B}=−\mathrm{2},\:{A}=\mathrm{3} \\ $$$$\Rightarrow{U}_{{n}} =\mathrm{3}−\frac{\mathrm{1}}{\mathrm{2}^{{n}−\mathrm{1}} } \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}{U}_{{n}} =\mathrm{3} \\ $$$$\mid{U}_{{n}} −\mathrm{3}\mid=\frac{\mathrm{1}}{\mathrm{2}^{{n}−\mathrm{1}} }\leqslant\mathrm{10}^{−\mathrm{4}} \\ $$$$\mathrm{2}^{{n}−\mathrm{1}} \geqslant\mathrm{10}^{\mathrm{4}} \\ $$$${n}\geqslant\mathrm{1}+\mathrm{log}_{\mathrm{2}} \:\mathrm{10}^{\mathrm{4}} =\mathrm{1}+\frac{\mathrm{4}}{\mathrm{log}\:\mathrm{2}}\approx\mathrm{14}.\mathrm{29} \\ $$$$\Rightarrow{n}_{\mathrm{0}} =\mathrm{15} \\ $$

Commented by mathocean1 last updated on 27/Feb/21

wow it′s great sir... Can you explain me how you find the first and fourth lines please...

$${wow}\:{it}'{s}\:{great}\:{sir}... \\ $$$${Can}\:{you}\:{explain}\:{me}\:{how}\:{you}\:{find}\:{the} \\ $$$${first}\:{and}\:{fourth}\:{lines}\:{please}... \\ $$

Commented by mr W last updated on 27/Feb/21

we can use “characteristic root method” to solve this kind of problems with “recurrence relation”. you may get some explanation here: (just read the text after example 2.4.5.

$${we}\:{can}\:{use}\:``{characteristic}\:{root}\:{method}'' \\ $$$${to}\:{solve}\:{this}\:{kind}\:{of}\:{problems}\:{with} \\ $$$$``{recurrence}\:{relation}''.\:{you}\:{may}\:{get} \\ $$$${some}\:{explanation}\:{here}: \\ $$$$\left({just}\:{read}\:{the}\:{text}\:{after}\:{example}\:\mathrm{2}.\mathrm{4}.\mathrm{5}.\right. \\ $$

Commented by mr W last updated on 27/Feb/21

Commented by mr W last updated on 27/Feb/21

https://math.libretexts.org/Courses/Saint_Mary's_College_Notre_Dame_IN/SMC%3A_MATH_339_-_Discrete_Mathematics_(Rohatgi)/Text/2%3A_Sequences/2.4%3A_Solving_Recurrence_Relations

Commented by mathocean1 last updated on 27/Feb/21

Super! Thank you very much sir ...

$${Super}! \\ $$$$ \\ $$$${Thank}\:{you}\:\:{very}\:\:{much}\:{sir}\:... \\ $$

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