Question Number 213548 by universe last updated on 08/Nov/24
$$\mathrm{0}<{c}<\mathrm{1}\:\mathrm{such}\:\mathrm{that}\:\mathrm{the}\:\mathrm{recursive}\:\mathrm{sequence} \\ $$$$\left\{{a}_{{n}} \right\}\:\mathrm{defined}\:\mathrm{by}\:\mathrm{setting}\: \\ $$$$\:\mathrm{a}_{\mathrm{1}\:} =\:\frac{\mathrm{c}}{\mathrm{2}}\:\:,\:{a}_{\mathrm{n}+\mathrm{1}} \:=\:\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{c}+\mathrm{a}_{\mathrm{n}} ^{\mathrm{2}} \right)\:\:\mathrm{for}\:\mathrm{n}\in\:\mathbb{N} \\ $$$$\mathrm{monotonic}\:\mathrm{and}\:\mathrm{convergent} \\ $$
Answered by Berbere last updated on 10/Nov/24
![Let f(z)=(1/2)(c+z^2 ) z∈R_+ ;f′(z)=z≥0 claim ∀n∈N a_n ∈[0,1] a_n =fof.....f(a_1 ) n timez f[0,1]=[f(0),f(1)]=[(c/2),((1+c)/2)] ⊆[0,1];0≤c≤1 ⇒fo.....f([0,1])⊆[0,1]⇒∀n∈N a_n ∈[0,1] a_2 =(c/2)[1+(c/4)]≥(c/2)⇒a_2 ≥a_1 since f increse ⇒∀n∈N a_(n+1) ≥a_n ⇒a_n increase bounded cv to fix pint of f l=lim_(n→∞) a_n is Solution of f(x)=x⇔x^2 −2x+c=0 x=1+(√(1−c))>1;1−(√(1−c)) l=1−(√(1−c))](https://www.tinkutara.com/question/Q213602.png)
$${Let}\:{f}\left({z}\right)=\frac{\mathrm{1}}{\mathrm{2}}\left({c}+{z}^{\mathrm{2}} \right)\:{z}\in\mathbb{R}_{+} ;{f}'\left({z}\right)={z}\geqslant\mathrm{0} \\ $$$${claim}\:\forall{n}\in\mathbb{N}\:\:{a}_{{n}} \in\left[\mathrm{0},\mathrm{1}\right] \\ $$$${a}_{{n}} ={fof}…..{f}\left({a}_{\mathrm{1}} \right)\:\:{n}\:{timez} \\ $$$${f}\left[\mathrm{0},\mathrm{1}\right]=\left[{f}\left(\mathrm{0}\right),{f}\left(\mathrm{1}\right)\right]=\left[\frac{{c}}{\mathrm{2}},\frac{\mathrm{1}+{c}}{\mathrm{2}}\right]\:\subseteq\left[\mathrm{0},\mathrm{1}\right];\mathrm{0}\leqslant{c}\leqslant\mathrm{1} \\ $$$$\Rightarrow{fo}…..{f}\left(\left[\mathrm{0},\mathrm{1}\right]\right)\subseteq\left[\mathrm{0},\mathrm{1}\right]\Rightarrow\forall{n}\in\mathbb{N}\:{a}_{{n}} \in\left[\mathrm{0},\mathrm{1}\right] \\ $$$${a}_{\mathrm{2}} =\frac{{c}}{\mathrm{2}}\left[\mathrm{1}+\frac{{c}}{\mathrm{4}}\right]\geqslant\frac{{c}}{\mathrm{2}}\Rightarrow{a}_{\mathrm{2}} \geqslant{a}_{\mathrm{1}} \:\:{since}\:{f}\:{increse} \\ $$$$\Rightarrow\forall{n}\in\mathbb{N}\:{a}_{{n}+\mathrm{1}} \geqslant{a}_{{n}} \Rightarrow{a}_{{n}} \:{increase}\:{bounded}\:{cv}\:{to}\:{fix}\:{pint}\:{of}\:{f} \\ $$$${l}=\underset{{n}\rightarrow\infty} {\mathrm{lim}}{a}_{{n}} \:{is}\:{Solution}\:{of}\:{f}\left({x}\right)={x}\Leftrightarrow{x}^{\mathrm{2}} −\mathrm{2}{x}+{c}=\mathrm{0} \\ $$$${x}=\mathrm{1}+\sqrt{\mathrm{1}−{c}}>\mathrm{1};\mathrm{1}−\sqrt{\mathrm{1}−{c}} \\ $$$${l}=\mathrm{1}−\sqrt{\mathrm{1}−{c}} \\ $$$$ \\ $$$$ \\ $$
Commented by universe last updated on 10/Nov/24
$$\mathrm{nice}\:\mathrm{solution}\:\mathrm{sir}\: \\ $$$$\mathrm{thank}\:\mathrm{you}\:\mathrm{so}\:\mathrm{much} \\ $$