Question Number 210439 by universe last updated on 09/Aug/24
Commented by universe last updated on 09/Aug/24
$${prove}\:{that}\:{the}\:{sequence}\:{is}\:{bounded}\:{and}\: \\ $$$${convergent} \\ $$
Answered by aleks041103 last updated on 09/Aug/24
![f(x) = (x+2)2^(−x) f ′(x) = 2^(−x) −ln(2)(x+2)2^(−x) = =(1−ln(2)(x+2))2^(−x) = =−ln(2)2^(−x) (x+(2−(1/(ln(2))))) ⇒x<(1/(ln(2)))−2 → f(x)↗ x>(1/(ln(2)))−2 → f(x)↘ ⇒f(x)≤f((1/(ln(2)))−2)=(1/(ln(2)))2^(2−(1/(ln(2)))) = =(4/(e ln(2))) <(4/((5/2) (2/3)))=2.4 ⇒x_(n+1) =3−f(x_n )>3−2.4>0 x_1 =1>0 ⇒x_n >0 f(x_n >0)<f(0)=2⇒f(x_n )<2⇒x_(n+1) >1 f(x_n >1)>lim_(x→∞) f(x)=0⇒f(x_n >1)>0 ⇒x_n <3 x_n ∈[1,3) by virtue of the graph of the function we can argue that x_n →2 to prove it we will use the fixed point theorem 2−x_(n+1) =2−(3−((x_n +2)/2^x_n ))=((x_n +2)/2^x_n )−1= =((4−(2−x_n ))/2^(2−(2−x_n )) )−1 let a_n =2−x_n , then a_n ∈(−1,1] ⇒a_(n+1) =((4−a_n )/4)2^a_n −1=2^a_n −1−(1/4)a_n 2^a_n = = a_n ( ((2^a_n −1)/a_n ) − (2^a_n /4) ) g(x)=((2^x −1)/x)−(2^x /4) it is known from the series expansion of 2^x that ((2^x −1)/x) is monotonically increasing. for x∈(−1,1) we have g(x)>((2^(−1) −1)/(−1))−(2^x /4)=(1/2)−(2^x /4)>(1/2)−(2/4)=0 also g(x)<((2^1 −1)/1)−(2^x /4)=1−(2^x /4)<1−(2^(−1) /4)=(7/8) ⇒∣a_(n+1) ∣≤(7/8)∣a_n ∣⇒∣a_n ∣≤∣a_1 ∣((7/8))^(n−1) →0 ⇒a_n →0 ⇒x_n →2](https://www.tinkutara.com/question/Q210453.png)
$${f}\left({x}\right)\:=\:\left({x}+\mathrm{2}\right)\mathrm{2}^{−{x}} \\ $$$${f}\:'\left({x}\right)\:=\:\mathrm{2}^{−{x}} −{ln}\left(\mathrm{2}\right)\left({x}+\mathrm{2}\right)\mathrm{2}^{−{x}} = \\ $$$$=\left(\mathrm{1}−{ln}\left(\mathrm{2}\right)\left({x}+\mathrm{2}\right)\right)\mathrm{2}^{−{x}} = \\ $$$$=−{ln}\left(\mathrm{2}\right)\mathrm{2}^{−{x}} \left({x}+\left(\mathrm{2}−\frac{\mathrm{1}}{{ln}\left(\mathrm{2}\right)}\right)\right) \\ $$$$\Rightarrow{x}<\frac{\mathrm{1}}{{ln}\left(\mathrm{2}\right)}−\mathrm{2}\:\rightarrow\:{f}\left({x}\right)\nearrow \\ $$$$\:\:\:\:\:{x}>\frac{\mathrm{1}}{{ln}\left(\mathrm{2}\right)}−\mathrm{2}\:\rightarrow\:{f}\left({x}\right)\searrow \\ $$$$\Rightarrow{f}\left({x}\right)\leqslant{f}\left(\frac{\mathrm{1}}{{ln}\left(\mathrm{2}\right)}−\mathrm{2}\right)=\frac{\mathrm{1}}{{ln}\left(\mathrm{2}\right)}\mathrm{2}^{\mathrm{2}−\frac{\mathrm{1}}{{ln}\left(\mathrm{2}\right)}} = \\ $$$$=\frac{\mathrm{4}}{{e}\:{ln}\left(\mathrm{2}\right)}\:<\frac{\mathrm{4}}{\frac{\mathrm{5}}{\mathrm{2}}\:\frac{\mathrm{2}}{\mathrm{3}}}=\mathrm{2}.\mathrm{4}\: \\ $$$$\Rightarrow{x}_{{n}+\mathrm{1}} =\mathrm{3}−{f}\left({x}_{{n}} \right)>\mathrm{3}−\mathrm{2}.\mathrm{4}>\mathrm{0} \\ $$$${x}_{\mathrm{1}} =\mathrm{1}>\mathrm{0} \\ $$$$\Rightarrow{x}_{{n}} >\mathrm{0} \\ $$$${f}\left({x}_{{n}} >\mathrm{0}\right)<{f}\left(\mathrm{0}\right)=\mathrm{2}\Rightarrow{f}\left({x}_{{n}} \right)<\mathrm{2}\Rightarrow{x}_{{n}+\mathrm{1}} >\mathrm{1} \\ $$$${f}\left({x}_{{n}} >\mathrm{1}\right)>\underset{{x}\rightarrow\infty} {{lim}f}\left({x}\right)=\mathrm{0}\Rightarrow{f}\left({x}_{{n}} >\mathrm{1}\right)>\mathrm{0} \\ $$$$\Rightarrow{x}_{{n}} <\mathrm{3} \\ $$$$ \\ $$$${x}_{{n}} \in\left[\mathrm{1},\mathrm{3}\right) \\ $$$$ \\ $$$${by}\:{virtue}\:{of}\:{the}\:{graph}\:{of}\:{the}\:{function}\:{we}\:{can} \\ $$$${argue}\:{that}\:{x}_{{n}} \rightarrow\mathrm{2} \\ $$$${to}\:{prove}\:{it}\:{we}\:{will}\:{use}\:{the}\:{fixed}\:{point}\:{theorem} \\ $$$$\mathrm{2}−{x}_{{n}+\mathrm{1}} =\mathrm{2}−\left(\mathrm{3}−\frac{{x}_{{n}} +\mathrm{2}}{\mathrm{2}^{{x}_{{n}} } }\right)=\frac{{x}_{{n}} +\mathrm{2}}{\mathrm{2}^{{x}_{{n}} } }−\mathrm{1}= \\ $$$$=\frac{\mathrm{4}−\left(\mathrm{2}−{x}_{{n}} \right)}{\mathrm{2}^{\mathrm{2}−\left(\mathrm{2}−{x}_{{n}} \right)} }−\mathrm{1} \\ $$$${let}\:{a}_{{n}} =\mathrm{2}−{x}_{{n}} ,\:{then}\:{a}_{{n}} \in\left(−\mathrm{1},\mathrm{1}\right] \\ $$$$\Rightarrow{a}_{{n}+\mathrm{1}} =\frac{\mathrm{4}−{a}_{{n}} }{\mathrm{4}}\mathrm{2}^{{a}_{{n}} } −\mathrm{1}=\mathrm{2}^{{a}_{{n}} } −\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4}}{a}_{{n}} \mathrm{2}^{{a}_{{n}} } = \\ $$$$=\:{a}_{{n}} \:\left(\:\frac{\mathrm{2}^{{a}_{{n}} } −\mathrm{1}}{{a}_{{n}} }\:−\:\frac{\mathrm{2}^{{a}_{{n}} } }{\mathrm{4}}\:\right) \\ $$$${g}\left({x}\right)=\frac{\mathrm{2}^{{x}} −\mathrm{1}}{{x}}−\frac{\mathrm{2}^{{x}} }{\mathrm{4}} \\ $$$${it}\:{is}\:{known}\:{from}\:{the}\:{series}\:{expansion}\:{of} \\ $$$$\mathrm{2}^{{x}} \:{that}\:\frac{\mathrm{2}^{{x}} −\mathrm{1}}{{x}}\:{is}\:{monotonically}\:{increasing}. \\ $$$${for}\:{x}\in\left(−\mathrm{1},\mathrm{1}\right)\:{we}\:{have} \\ $$$${g}\left({x}\right)>\frac{\mathrm{2}^{−\mathrm{1}} −\mathrm{1}}{−\mathrm{1}}−\frac{\mathrm{2}^{{x}} }{\mathrm{4}}=\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{2}^{{x}} }{\mathrm{4}}>\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{2}}{\mathrm{4}}=\mathrm{0} \\ $$$${also} \\ $$$${g}\left({x}\right)<\frac{\mathrm{2}^{\mathrm{1}} −\mathrm{1}}{\mathrm{1}}−\frac{\mathrm{2}^{{x}} }{\mathrm{4}}=\mathrm{1}−\frac{\mathrm{2}^{{x}} }{\mathrm{4}}<\mathrm{1}−\frac{\mathrm{2}^{−\mathrm{1}} }{\mathrm{4}}=\frac{\mathrm{7}}{\mathrm{8}} \\ $$$$\Rightarrow\mid{a}_{{n}+\mathrm{1}} \mid\leqslant\frac{\mathrm{7}}{\mathrm{8}}\mid{a}_{{n}} \mid\Rightarrow\mid{a}_{{n}} \mid\leqslant\mid{a}_{\mathrm{1}} \mid\left(\frac{\mathrm{7}}{\mathrm{8}}\right)^{{n}−\mathrm{1}} \rightarrow\mathrm{0} \\ $$$$\Rightarrow{a}_{{n}} \rightarrow\mathrm{0} \\ $$$$\Rightarrow{x}_{{n}} \rightarrow\mathrm{2} \\ $$
Commented by universe last updated on 09/Aug/24
$${thank}\:{you}\:{so}\:{much}\:{sir} \\ $$