Question Number 208445 by alcohol last updated on 16/Jun/24
![u_(n+1) = u_n −u_n ^3 , u_0 ∈]0,1[ v_n = (1/u_(n+1) ^2 )−(1/u_n ^2 ) = f(u_n ^2 ) ; f(x) = ((2−x)/((1−x)^2 )) v_n converges to 2, v_n is decreasing . show that v_n ≥ 2 x_n =(1/(n+1))Σ_(m=0) ^m (v_m ) . show that x_0 ≥x_n ≥v_n . show that x_n is decreasing and lim_(n→∞) x_n = l ≥2 . show that 2x_(n+1) −x_n ≤v_(n+1) and deduce l . express x_(n+1) −x_n interms of u_n . deduce lim_(n→∞) nu_n ^2](https://www.tinkutara.com/question/Q208445.png)
$$\left.{u}_{{n}+\mathrm{1}} \:=\:{u}_{{n}} −{u}_{{n}} ^{\mathrm{3}} ,\:{u}_{\mathrm{0}} \in\right]\mathrm{0},\mathrm{1}\left[\right. \\ $$$${v}_{{n}} \:=\:\frac{\mathrm{1}}{{u}_{{n}+\mathrm{1}} ^{\mathrm{2}} }−\frac{\mathrm{1}}{{u}_{{n}} ^{\mathrm{2}} }\:=\:{f}\left({u}_{{n}} ^{\mathrm{2}} \right)\:;\:{f}\left({x}\right)\:=\:\frac{\mathrm{2}−{x}}{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} } \\ $$$${v}_{{n}} \:{converges}\:{to}\:\mathrm{2},\:{v}_{{n}} \:{is}\:{decreasing} \\ $$$$.\:{show}\:{that}\:{v}_{{n}} \:\geqslant\:\mathrm{2} \\ $$$${x}_{{n}} =\frac{\mathrm{1}}{{n}+\mathrm{1}}\underset{{m}=\mathrm{0}} {\overset{{m}} {\sum}}\left({v}_{{m}} \right) \\ $$$$.\:{show}\:{that}\:{x}_{\mathrm{0}} \geqslant{x}_{{n}} \geqslant{v}_{{n}} \\ $$$$.\:{show}\:{that}\:{x}_{{n}} \:{is}\:{decreasing}\:{and}\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}{x}_{{n}} =\:{l}\:\geqslant\mathrm{2} \\ $$$$.\:{show}\:{that}\:\mathrm{2}{x}_{{n}+\mathrm{1}} −{x}_{{n}} \leqslant{v}_{{n}+\mathrm{1}} \:{and}\:{deduce}\:{l} \\ $$$$.\:{express}\:{x}_{{n}+\mathrm{1}} −{x}_{{n}} \:{interms}\:{of}\:{u}_{{n}} \\ $$$$.\:{deduce}\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}{nu}_{{n}} ^{\mathrm{2}} \\ $$
Answered by Berbere last updated on 16/Jun/24
![u_(n+1) =u_n −u_n ^3 =g(u_n ) g(x)=x−x^3 ⇒g′=1−3x^2 g(0)=g(1)⇒g(x)∈[0,g((1/( (√3))))] =[0(2/(3(√3)))]⊂[0,1] u_n =g^n (u_0 )⊂[0,1] ⇒∀n∈N U_n ∈[0,1] u_(n+1) −u_n =−u_n ^3 <0⇒u_n decrease Bounded Cv lim_(n→∞) u_n =l ; solution of g(x)=x⇒l=0 v_n =(1/u_(n+1) ^2 )−(1/u_n ^2 )=(1/(u_n ^2 (1−u_n ^2 )^2 ))−(1/u_n ^2 ) u_n ^2 =w_n ⇒v_n =(1/w_n )[((1−(1−w_n )^2 )/((1−w_n )^2 ))]=((2w_n −w_n ^2 )/(w_n (1−w_n )^2 ))=((2−w_n )/((1−w_n )^2 )) =f(w_n )=f(u_n ^2 ) lim_(n→∞) V_n =lim_(n→∞) f(u_n ^2 )=f(lim_(n→∞) u_n ^2 )=f(0)=2 by continuity f in 2 f(x)=((2−x)/((1−x)^2 ))=((−(1−x)^2 +2(1−x)(2−x))/((1−x)^4 ))= (((1−x)(−(1−x)+2(2−x))/((1−x)^4 ))=(((1−x)(3−x))/((1−x)^4 ))>0 f(u_n ^2 )∈]2,∞[⇒v_n >2 u_(n+1) <u_n ⇒f(u_(n+1) ^2 )<f(u_n ^2 )⇒v_(n+1) <v_n ;v_n decrease x_n =(1/(n+1))Σ_(k=0) ^n v_m ;∀k∈[0,n+1] v_n ≤v_k ≤v_0 ⇒^(v_n decrease) v_n ≤x_n ≤v_0 x_(n+1) −x_n =(1/(n+2))Σ_(k=0) ^(n+1) v_k −(1/(n+1))Σ_(k=0) ^n v_k =(((n+1)Σv_k −(n+2)Σv_k )/((n+2)(n+1))) =(((n+1)v_(n+1) −Σ_(k=0) ^n v_k )/((n+1)(n+2)))=((Σ(v_(n+1) −v_k ))/((n+1)(n+2)))<0 ∀k∈[0,n] v_k >v_(n+1) v_n decrease x_n decrease bounded Cv l ;x_n ≥v_n ⇒l≥2 2x_(n+1) −x_n =(2/(n+2))Σ_(k=0) ^(n+1) v_k −(1/(n+1))Σ_(k=0) ^n v_k =((2(n+1)Σ^(n+1) v_k −(n+2)Σ_(k=0) ^n v_k )/((n+2)(n+1)))=((2v_(n+1) +2nv_(n+1) +nΣ_(k=0) ^n v_k )/) ≥(((2+2n)v_(n+1) +n(n+1)v_(n+1) )/((n+1)(n+2)))≥v_(n+1) not 2x_(n+1) −x_n <v_(n+1) l=2 by cesaro x_n =(1/(n+1))(Σ_(k=0) ^n (1/u_(k+1) ^2 )−(1/u_k ^2 ))=(1/(n+1))((1/u_(n+1) ^2 )−(1/u_0 ^2 )) ⇒lim_(n→∞) x_n =2=lim_(n→∞) (1/((n+1)u_(n+1) ^2 )) ⇒lim_(n→∞) (n+1)u_(n+1) ^2 =(1/2) w_n =(n+1)u_(n+1) ^2 cv⇒lim_(n→∞) W_(n−1) =lim_(n→∞) W_n =(1/2) nU_n ^2 →(1/2)](https://www.tinkutara.com/question/Q208449.png)
$${u}_{{n}+\mathrm{1}} ={u}_{{n}} −{u}_{{n}} ^{\mathrm{3}} ={g}\left({u}_{{n}} \right) \\ $$$${g}\left({x}\right)={x}−{x}^{\mathrm{3}} \Rightarrow{g}'=\mathrm{1}−\mathrm{3}{x}^{\mathrm{2}} \\ $$$${g}\left(\mathrm{0}\right)={g}\left(\mathrm{1}\right)\Rightarrow{g}\left({x}\right)\in\left[\mathrm{0},{g}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)\right]\:=\left[\mathrm{0}\frac{\mathrm{2}}{\mathrm{3}\sqrt{\mathrm{3}}}\right]\subset\left[\mathrm{0},\mathrm{1}\right] \\ $$$${u}_{{n}} ={g}^{{n}} \left({u}_{\mathrm{0}} \right)\subset\left[\mathrm{0},\mathrm{1}\right]\:\Rightarrow\forall{n}\in\mathbb{N}\:{U}_{{n}} \in\left[\mathrm{0},\mathrm{1}\right] \\ $$$${u}_{{n}+\mathrm{1}} −{u}_{{n}} =−{u}_{{n}} ^{\mathrm{3}} <\mathrm{0}\Rightarrow{u}_{{n}} \:{decrease}\:{Bounded}\:{Cv} \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}{u}_{{n}} ={l}\:\:\:;\:{solution}\:{of}\:{g}\left({x}\right)={x}\Rightarrow{l}=\mathrm{0} \\ $$$${v}_{{n}} =\frac{\mathrm{1}}{{u}_{{n}+\mathrm{1}} ^{\mathrm{2}} }−\frac{\mathrm{1}}{{u}_{{n}} ^{\mathrm{2}} }=\frac{\mathrm{1}}{{u}_{{n}} ^{\mathrm{2}} \left(\mathrm{1}−{u}_{{n}} ^{\mathrm{2}} \right)^{\mathrm{2}} }−\frac{\mathrm{1}}{{u}_{{n}} ^{\mathrm{2}} } \\ $$$${u}_{{n}} ^{\mathrm{2}} ={w}_{{n}} \Rightarrow{v}_{{n}} =\frac{\mathrm{1}}{{w}_{{n}} }\left[\frac{\mathrm{1}−\left(\mathrm{1}−{w}_{{n}} \right)^{\mathrm{2}} }{\left(\mathrm{1}−{w}_{{n}} \right)^{\mathrm{2}} }\right]=\frac{\mathrm{2}{w}_{{n}} −{w}_{{n}} ^{\mathrm{2}} }{{w}_{{n}} \left(\mathrm{1}−{w}_{{n}} \right)^{\mathrm{2}} }=\frac{\mathrm{2}−{w}_{{n}} }{\left(\mathrm{1}−{w}_{{n}} \right)^{\mathrm{2}} } \\ $$$$={f}\left({w}_{{n}} \right)={f}\left({u}_{{n}} ^{\mathrm{2}} \right) \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}{V}_{{n}} =\underset{{n}\rightarrow\infty} {\mathrm{lim}}{f}\left({u}_{{n}} ^{\mathrm{2}} \right)={f}\left(\underset{{n}\rightarrow\infty} {\mathrm{lim}}{u}_{{n}} ^{\mathrm{2}} \right)={f}\left(\mathrm{0}\right)=\mathrm{2} \\ $$$${by}\:{continuity}\:{f}\:{in}\:\mathrm{2} \\ $$$${f}\left({x}\right)=\frac{\mathrm{2}−{x}}{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} }=\frac{−\left(\mathrm{1}−{x}\right)^{\mathrm{2}} +\mathrm{2}\left(\mathrm{1}−{x}\right)\left(\mathrm{2}−{x}\right)}{\left(\mathrm{1}−{x}\right)^{\mathrm{4}} }= \\ $$$$\frac{\left(\mathrm{1}−{x}\right)\left(−\left(\mathrm{1}−{x}\right)+\mathrm{2}\left(\mathrm{2}−{x}\right)\right.}{\left(\mathrm{1}−{x}\right)^{\mathrm{4}} }=\frac{\left(\mathrm{1}−{x}\right)\left(\mathrm{3}−{x}\right)}{\left(\mathrm{1}−{x}\right)^{\mathrm{4}} }>\mathrm{0} \\ $$$$\left.{f}\left({u}_{{n}} ^{\mathrm{2}} \right)\in\right]\mathrm{2},\infty\left[\Rightarrow{v}_{{n}} >\mathrm{2}\right. \\ $$$${u}_{{n}+\mathrm{1}} <{u}_{{n}} \Rightarrow{f}\left({u}_{{n}+\mathrm{1}} ^{\mathrm{2}} \right)<{f}\left({u}_{{n}} ^{\mathrm{2}} \right)\Rightarrow{v}_{{n}+\mathrm{1}} <{v}_{{n}} \:;{v}_{{n}} \:{decrease} \\ $$$${x}_{{n}} =\frac{\mathrm{1}}{{n}+\mathrm{1}}\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}{v}_{{m}} ;\forall{k}\in\left[\mathrm{0},{n}+\mathrm{1}\right]\: \\ $$$$\:{v}_{{n}} \leqslant{v}_{{k}} \leqslant{v}_{\mathrm{0}} \overset{{v}_{{n}} \:{decrease}} {\Rightarrow}{v}_{{n}} \leqslant{x}_{{n}} \leqslant{v}_{\mathrm{0}} \\ $$$${x}_{{n}+\mathrm{1}} −{x}_{{n}} =\frac{\mathrm{1}}{{n}+\mathrm{2}}\underset{{k}=\mathrm{0}} {\overset{{n}+\mathrm{1}} {\sum}}{v}_{{k}} −\frac{\mathrm{1}}{{n}+\mathrm{1}}\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}{v}_{{k}} =\frac{\left({n}+\mathrm{1}\right)\Sigma{v}_{{k}} −\left({n}+\mathrm{2}\right)\Sigma{v}_{{k}} }{\left({n}+\mathrm{2}\right)\left({n}+\mathrm{1}\right)} \\ $$$$=\frac{\left({n}+\mathrm{1}\right){v}_{{n}+\mathrm{1}} −\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}{v}_{{k}} }{\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)}=\frac{\Sigma\left({v}_{{n}+\mathrm{1}} −{v}_{{k}} \right)}{\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)}<\mathrm{0} \\ $$$$\forall{k}\in\left[\mathrm{0},{n}\right]\:{v}_{{k}} >{v}_{{n}+\mathrm{1}} \:{v}_{{n}} \:{decrease}\: \\ $$$${x}_{{n}} \:{decrease}\:\:{bounded}\:{Cv}\:\:{l}\:;{x}_{{n}} \geqslant{v}_{{n}} \Rightarrow{l}\geqslant\mathrm{2} \\ $$$$\mathrm{2}{x}_{{n}+\mathrm{1}} −{x}_{{n}} =\frac{\mathrm{2}}{{n}+\mathrm{2}}\underset{{k}=\mathrm{0}} {\overset{{n}+\mathrm{1}} {\sum}}{v}_{{k}} −\frac{\mathrm{1}}{{n}+\mathrm{1}}\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}{v}_{{k}} \\ $$$$ \\ $$$$=\frac{\mathrm{2}\left({n}+\mathrm{1}\right)\overset{{n}+\mathrm{1}} {\sum}{v}_{{k}} −\left({n}+\mathrm{2}\right)\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}{v}_{{k}} }{\left({n}+\mathrm{2}\right)\left({n}+\mathrm{1}\right)}=\frac{\mathrm{2}{v}_{{n}+\mathrm{1}} +\mathrm{2}{nv}_{{n}+\mathrm{1}} +{n}\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}{v}_{{k}} }{} \\ $$$$\geqslant\frac{\left(\mathrm{2}+\mathrm{2}{n}\right){v}_{{n}+\mathrm{1}} +{n}\left({n}+\mathrm{1}\right){v}_{{n}+\mathrm{1}} }{\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)}\geqslant{v}_{{n}+\mathrm{1}} \\ $$$${not}\:\mathrm{2}{x}_{{n}+\mathrm{1}} −{x}_{{n}} <{v}_{{n}+\mathrm{1}} \\ $$$${l}=\mathrm{2}\:{by}\:{cesaro} \\ $$$${x}_{{n}} =\frac{\mathrm{1}}{{n}+\mathrm{1}}\left(\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{{u}_{{k}+\mathrm{1}} ^{\mathrm{2}} }−\frac{\mathrm{1}}{{u}_{{k}} ^{\mathrm{2}} }\right)=\frac{\mathrm{1}}{{n}+\mathrm{1}}\left(\frac{\mathrm{1}}{{u}_{{n}+\mathrm{1}} ^{\mathrm{2}} }−\frac{\mathrm{1}}{{u}_{\mathrm{0}} ^{\mathrm{2}} }\right) \\ $$$$\Rightarrow\underset{{n}\rightarrow\infty} {\mathrm{lim}}{x}_{{n}} =\mathrm{2}=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right){u}_{{n}+\mathrm{1}} ^{\mathrm{2}} } \\ $$$$\Rightarrow\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left({n}+\mathrm{1}\right){u}_{{n}+\mathrm{1}} ^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${w}_{{n}} =\left({n}+\mathrm{1}\right){u}_{{n}+\mathrm{1}} ^{\mathrm{2}} \:{cv}\Rightarrow\underset{{n}\rightarrow\infty} {\mathrm{lim}}{W}_{{n}−\mathrm{1}} =\underset{{n}\rightarrow\infty} {\mathrm{lim}}{W}_{{n}} =\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${nU}_{{n}} ^{\mathrm{2}} \rightarrow\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$