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Question Number 431 by 123456 last updated on 25/Jan/15

given x_(n+1) =(1/(3−x_n )) x_0 =2 proof that a.0<x_n ≤2,n∈N b.x_n is decreasing c.lim_(n→∞) x_n

$$\mathrm{given} \\ $$ $${x}_{{n}+\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{3}−{x}_{{n}} } \\ $$ $${x}_{\mathrm{0}} =\mathrm{2} \\ $$ $$\mathrm{proof}\:\mathrm{that} \\ $$ $$\mathrm{a}.\mathrm{0}<{x}_{{n}} \leqslant\mathrm{2},{n}\in\mathbb{N} \\ $$ $$\mathrm{b}.{x}_{{n}} \:\mathrm{is}\:\mathrm{decreasing} \\ $$ $$\boldsymbol{\mathrm{c}}.\underset{\boldsymbol{\mathrm{n}}\rightarrow\infty} {\mathrm{lim}}{x}_{{n}} \\ $$

Answered by prakash jain last updated on 03/Jan/15

If x_(n+1) >2 (1/(3−x_n ))>2 ⇒ 3> x_n >(5/2) If x_(n+1) <0 (1/(3−x_n ))<0 ⇒x_n >3 Hence If 0<x_n ≤2 ⇒ 0<x_(n+1) <2 Since x_0 =2, Result a is proved by induction. If x_(n+1) <x_n ⇒−x_(n+1) >−x_n ⇒3−x_(n+1) >3−x_n ⇒(1/(3−x_(n+1) ))<(1/(3−x_n ))⇒x_(n+2) <x_(n+1) x_(n+1) <x_n ⇒x_(n+2) <x_(n+1) x_1 =1,x_0 =2,x_1 <x_0 ⇒x_2 <x_1 ⇒x_3 <x_2 Result b proved by induction. In the limiting case x_(n+1) =x_n x_(n+1) =x_n =(1/(3−x_n )) x_n ^2 −3x_n +1=0 x_n =((3±(√(9−4)))/2) x_n =((3−(√5))/2) lim_(n→∞) x_n =((3−(√5))/2)

$$\mathrm{If}\:{x}_{{n}+\mathrm{1}} >\mathrm{2} \\ $$ $$\frac{\mathrm{1}}{\mathrm{3}−{x}_{{n}} }>\mathrm{2}\:\Rightarrow\:\:\:\mathrm{3}>\:{x}_{{n}} >\frac{\mathrm{5}}{\mathrm{2}} \\ $$ $$\mathrm{If}\:{x}_{{n}+\mathrm{1}} <\mathrm{0} \\ $$ $$\frac{\mathrm{1}}{\mathrm{3}−{x}_{{n}} }<\mathrm{0}\:\Rightarrow{x}_{{n}} >\mathrm{3} \\ $$ $$\mathrm{Hence}\:\mathrm{If}\:\mathrm{0}<{x}_{{n}} \leqslant\mathrm{2}\:\Rightarrow\:\mathrm{0}<{x}_{{n}+\mathrm{1}} <\mathrm{2} \\ $$ $$\mathrm{Since}\:{x}_{\mathrm{0}} =\mathrm{2},\:\mathrm{Result}\:\mathrm{a}\:\mathrm{is}\:\mathrm{proved}\:\mathrm{by}\:\mathrm{induction}. \\ $$ $$\mathrm{If}\:{x}_{{n}+\mathrm{1}} <{x}_{{n}} \:\Rightarrow−{x}_{{n}+\mathrm{1}} >−{x}_{{n}} \\ $$ $$\Rightarrow\mathrm{3}−{x}_{{n}+\mathrm{1}} >\mathrm{3}−{x}_{{n}} \Rightarrow\frac{\mathrm{1}}{\mathrm{3}−{x}_{{n}+\mathrm{1}} }<\frac{\mathrm{1}}{\mathrm{3}−{x}_{{n}} }\Rightarrow{x}_{{n}+\mathrm{2}} <{x}_{{n}+\mathrm{1}} \\ $$ $${x}_{{n}+\mathrm{1}} <{x}_{{n}} \Rightarrow{x}_{{n}+\mathrm{2}} <{x}_{{n}+\mathrm{1}} \\ $$ $${x}_{\mathrm{1}} =\mathrm{1},{x}_{\mathrm{0}} =\mathrm{2},{x}_{\mathrm{1}} <{x}_{\mathrm{0}} \Rightarrow{x}_{\mathrm{2}} <{x}_{\mathrm{1}} \:\Rightarrow{x}_{\mathrm{3}} <{x}_{\mathrm{2}} \\ $$ $$\mathrm{Result}\:\mathrm{b}\:\mathrm{proved}\:\mathrm{by}\:\mathrm{induction}. \\ $$ $$\mathrm{In}\:\mathrm{the}\:\mathrm{limiting}\:\mathrm{case}\:{x}_{{n}+\mathrm{1}} ={x}_{{n}} \\ $$ $${x}_{{n}+\mathrm{1}} ={x}_{{n}} =\frac{\mathrm{1}}{\mathrm{3}−{x}_{{n}} } \\ $$ $${x}_{{n}} ^{\mathrm{2}} −\mathrm{3}{x}_{{n}} +\mathrm{1}=\mathrm{0} \\ $$ $${x}_{{n}} =\frac{\mathrm{3}\pm\sqrt{\mathrm{9}−\mathrm{4}}}{\mathrm{2}} \\ $$ $${x}_{{n}} =\frac{\mathrm{3}−\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$ $$\underset{{n}\rightarrow\infty} {\mathrm{lim}}{x}_{{n}} =\frac{\mathrm{3}−\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$

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