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Question Number 167183 by alcohol last updated on 09/Mar/22
Commented by ArielVyny last updated on 09/Mar/22
i let question (f) for your attention
$${i}\:{let}\:{question}\:\left({f}\right)\:{for}\:{your}\:{attention} \\ $$
Answered by ArielVyny last updated on 09/Mar/22
a)showing that (U_n ) and (V_n ) are strictly positive sequences. reasonnig by reccurence on we have U_0 =1 and V_0 =2 also are strictly positive we suppose that (U_n ) ans (V_n ) are strictly positive then we have U_n >0 and V_n >0 now we will sbowing that U_(n+1) and V_(n+1) are strictly posirive V_(n+1) =((U_n +V_n )/2) according that U_n and V_n are stritly positive and 2>0 (evidence) we have V_(n+1) >0 and U_(n+1) =((U_n V_n )/V_(n+1) ) we know that V_(n+1) >0 and U_n V_n is strictly positive because it′s a product of two positives values
$$\left.{a}\right){showing}\:{that}\:\left({U}_{{n}} \right)\:{and}\:\left({V}_{{n}} \right)\:{are}\:{strictly} \\ $$$${positive}\:{sequences}. \\ $$$${reasonnig}\:{by}\:{reccurence}\:{on}\: \\ $$$${we}\:{have}\:{U}_{\mathrm{0}} =\mathrm{1}\:{and}\:{V}_{\mathrm{0}} =\mathrm{2}\:\:{also}\:{are}\:{strictly} \\ $$$${positive} \\ $$$${we}\:{suppose}\:{that}\:\left({U}_{{n}} \right)\:{ans}\:\left({V}_{{n}} \right)\:{are}\:{strictly}\: \\ $$$${positive}\:{then}\:{we}\:{have}\:{U}_{{n}} >\mathrm{0}\:\:{and}\:{V}_{{n}} >\mathrm{0} \\ $$$${now}\:{we}\:{will}\:{sbowing}\:{that}\:{U}_{{n}+\mathrm{1}} \:{and}\:{V}_{{n}+\mathrm{1}} \\ $$$${are}\:{strictly}\:{posirive} \\ $$$${V}_{{n}+\mathrm{1}} =\frac{{U}_{{n}} +{V}_{{n}} }{\mathrm{2}}\:\:{according}\:{that}\:{U}_{{n}} \:{and}\:{V}_{{n}} \:{are} \\ $$$${stritly}\:{positive}\:{and}\:\mathrm{2}>\mathrm{0}\:\left({evidence}\right)\:{we} \\ $$$${have}\:{V}_{{n}+\mathrm{1}} >\mathrm{0} \\ $$$${and}\:{U}_{{n}+\mathrm{1}} =\frac{{U}_{{n}} {V}_{{n}} }{{V}_{{n}+\mathrm{1}} }\:\:{we}\:{know}\:{that}\:{V}_{{n}+\mathrm{1}} >\mathrm{0} \\ $$$${and}\:{U}_{{n}} {V}_{{n}} \:{is}\:{strictly}\:{positive}\:{because}\:{it}'{s}\:{a} \\ $$$${product}\:{of}\:{two}\:{positives}\:{values} \\ $$$$ \\ $$
Answered by ArielVyny last updated on 09/Mar/22
(b) shosing that 0≤W_(n+1) ≤(1/2)W_n w_n =v_n −u_n →w_(n+1) =v_(n+1) −u_(n+1) →w_(n+1) =((u_n +v_n )/2)−((2u_n v_n )/(u_n +v_n )) =(((u_n +v_n )^2 )/(2(u_n +v_n )))−((4u_n v_n )/(2(u_n +v_n ))) =((u_n ^2 +v_n ^2 −2u_n v_n )/(2(u_n +v_n )))=(((u_n −v_n )^2 )/(2(u_n +v_n )))≥0 (question a) 0≤w_(n+1) →w_(n+1) −(w_n /2)=((u_n +v_n )/2)−((2u_n v_n )/(u_n +v_n ))−(v_n /2)+(u_n /2) =u_n −((2u_n v_n )/(u_n +v_n ))=u_n (1−((2v_n )/(u_n +v_n ))) w_(n+1) ≥0→w_n ≥0 and v_n ≥u_n ((2v_n )/(u_n +v_n ))=(v_n /v_(n+1) )≥(1/2)→2v_n ≥v_(n+1) →2v_n −(u_n /2)−(v_n /2)≥0 →(3/2)v_n −(u_n /2)≥0 (true) because v_n >u_n then we show that 1−((2v_n )/(u_n +v_n ))≤0 then w_(n+1) −(w_n /2)≤0 conclusion 0≤w_(n+1) ≤(w_n /2)
$$\left({b}\right)\:{shosing}\:{that}\:\mathrm{0}\leqslant{W}_{{n}+\mathrm{1}} \leqslant\frac{\mathrm{1}}{\mathrm{2}}{W}_{{n}} \\ $$$${w}_{{n}} ={v}_{{n}} −{u}_{{n}} \rightarrow{w}_{{n}+\mathrm{1}} ={v}_{{n}+\mathrm{1}} −{u}_{{n}+\mathrm{1}} \\ $$$$\rightarrow{w}_{{n}+\mathrm{1}} =\frac{{u}_{{n}} +{v}_{{n}} }{\mathrm{2}}−\frac{\mathrm{2}{u}_{{n}} {v}_{{n}} }{{u}_{{n}} +{v}_{{n}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\left({u}_{{n}} +{v}_{{n}} \right)^{\mathrm{2}} }{\mathrm{2}\left({u}_{{n}} +{v}_{{n}} \right)}−\frac{\mathrm{4}{u}_{{n}} {v}_{{n}} }{\mathrm{2}\left({u}_{{n}} +{v}_{{n}} \right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{{u}_{{n}} ^{\mathrm{2}} +{v}_{{n}} ^{\mathrm{2}} −\mathrm{2}{u}_{{n}} {v}_{{n}} }{\mathrm{2}\left({u}_{{n}} +{v}_{{n}} \right)}=\frac{\left({u}_{{n}} −{v}_{{n}} \right)^{\mathrm{2}} }{\mathrm{2}\left({u}_{{n}} +{v}_{{n}} \right)}\geqslant\mathrm{0}\:\left({question}\:{a}\right) \\ $$$$\mathrm{0}\leqslant{w}_{{n}+\mathrm{1}} \\ $$$$\rightarrow{w}_{{n}+\mathrm{1}} −\frac{{w}_{{n}} }{\mathrm{2}}=\frac{{u}_{{n}} +{v}_{{n}} }{\mathrm{2}}−\frac{\mathrm{2}{u}_{{n}} {v}_{{n}} }{{u}_{{n}} +{v}_{{n}} }−\frac{{v}_{{n}} }{\mathrm{2}}+\frac{{u}_{{n}} }{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:={u}_{{n}} −\frac{\mathrm{2}{u}_{{n}} {v}_{{n}} }{{u}_{{n}} +{v}_{{n}} }={u}_{{n}} \left(\mathrm{1}−\frac{\mathrm{2}{v}_{{n}} }{{u}_{{n}} +{v}_{{n}} }\right) \\ $$$${w}_{{n}+\mathrm{1}} \geqslant\mathrm{0}\rightarrow{w}_{{n}} \geqslant\mathrm{0}\:{and}\:{v}_{{n}} \geqslant{u}_{{n}} \\ $$$$\frac{\mathrm{2}{v}_{{n}} }{{u}_{{n}} +{v}_{{n}} }=\frac{{v}_{{n}} }{{v}_{{n}+\mathrm{1}} }\geqslant\frac{\mathrm{1}}{\mathrm{2}}\rightarrow\mathrm{2}{v}_{{n}} \geqslant{v}_{{n}+\mathrm{1}} \rightarrow\mathrm{2}{v}_{{n}} −\frac{{u}_{{n}} }{\mathrm{2}}−\frac{{v}_{{n}} }{\mathrm{2}}\geqslant\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\rightarrow\frac{\mathrm{3}}{\mathrm{2}}{v}_{{n}} −\frac{{u}_{{n}} }{\mathrm{2}}\geqslant\mathrm{0}\:\left({true}\right)\:{because}\:{v}_{{n}} >{u}_{{n}} \\ $$$${then}\:{we}\:{show}\:{that}\:\mathrm{1}−\frac{\mathrm{2}{v}_{{n}} }{{u}_{{n}} +{v}_{{n}} }\leqslant\mathrm{0}\:{then} \\ $$$${w}_{{n}+\mathrm{1}} −\frac{{w}_{{n}} }{\mathrm{2}}\leqslant\mathrm{0} \\ $$$${conclusion}\:\mathrm{0}\leqslant{w}_{{n}+\mathrm{1}} \leqslant\frac{{w}_{{n}} }{\mathrm{2}} \\ $$
Answered by ArielVyny last updated on 09/Mar/22
c)show that 0≤w_(n+1) ≤(1/2^n ) by reccurence on n 0≤w_0 ≤1 with w_0 =1 suppose thant 0≤w_n ≤(1/2^(n−1) ) knowing that (1/2^(n−1) )≥(1/2^n ) and w_(n+1) ≤(w_n /2)≤w_n 0≤w_n ≤(1/2^(n−1) )→0≤(w_n /2)≤(1/2^n )→0≤w_(n+1) ≤(1/2^n ) wd deduce that lim w_(n+1) =0 (by gendarm′s theorem) and lim w_(n+1) =lim w_n =0
$$\left.{c}\right){show}\:{that}\:\mathrm{0}\leqslant{w}_{{n}+\mathrm{1}} \leqslant\frac{\mathrm{1}}{\mathrm{2}^{{n}} } \\ $$$${by}\:{reccurence}\:{on}\:{n} \\ $$$$\mathrm{0}\leqslant{w}_{\mathrm{0}} \leqslant\mathrm{1}\:{with}\:{w}_{\mathrm{0}} =\mathrm{1} \\ $$$${suppose}\:{thant}\:\mathrm{0}\leqslant{w}_{{n}} \leqslant\frac{\mathrm{1}}{\mathrm{2}^{{n}−\mathrm{1}} } \\ $$$${knowing}\:{that}\:\frac{\mathrm{1}}{\mathrm{2}^{{n}−\mathrm{1}} }\geqslant\frac{\mathrm{1}}{\mathrm{2}^{{n}} }\:{and}\:\:{w}_{{n}+\mathrm{1}} \leqslant\frac{{w}_{{n}} }{\mathrm{2}}\leqslant{w}_{{n}} \\ $$$$\mathrm{0}\leqslant{w}_{{n}} \leqslant\frac{\mathrm{1}}{\mathrm{2}^{{n}−\mathrm{1}} }\rightarrow\mathrm{0}\leqslant\frac{{w}_{{n}} }{\mathrm{2}}\leqslant\frac{\mathrm{1}}{\mathrm{2}^{{n}} }\rightarrow\mathrm{0}\leqslant{w}_{{n}+\mathrm{1}} \leqslant\frac{\mathrm{1}}{\mathrm{2}^{{n}} } \\ $$$${wd}\:{deduce}\:{that}\:{lim}\:{w}_{{n}+\mathrm{1}} =\mathrm{0}\:\left({by}\:{gendarm}'{s}\:{theorem}\right) \\ $$$${and}\:{lim}\:{w}_{{n}+\mathrm{1}} ={lim}\:{w}_{{n}} =\mathrm{0} \\ $$
Answered by ArielVyny last updated on 09/Mar/22
d)v_0 =2 and v_1 =((1+2)/2)=(3/2) u_1 =(4/3) after coputing u_1 and v_1 we admot that (v_n ) decrease and (u_n ) increase and lim w_n =lim(v_n −u_n )=0 and sequenses (u_n ) an (v_n ) are adjacent
$$\left.{d}\right){v}_{\mathrm{0}} =\mathrm{2}\:{and}\:{v}_{\mathrm{1}} =\frac{\mathrm{1}+\mathrm{2}}{\mathrm{2}}=\frac{\mathrm{3}}{\mathrm{2}} \\ $$$${u}_{\mathrm{1}} =\frac{\mathrm{4}}{\mathrm{3}}\:{after}\:{coputing}\:{u}_{\mathrm{1}} \:{and}\:{v}_{\mathrm{1}} \:{we}\:{admot}\:{that} \\ $$$$\left({v}_{{n}} \right)\:{decrease}\:{and}\:\left({u}_{{n}} \right)\:{increase} \\ $$$${and}\:\:\:{lim}\:{w}_{{n}} ={lim}\left({v}_{{n}} −{u}_{{n}} \right)=\mathrm{0} \\ $$$${and}\:{sequenses}\:\:\left({u}_{{n}} \right)\:{an}\:\left({v}_{{n}} \right)\:{are}\:{adjacent} \\ $$
Answered by ArielVyny last updated on 09/Mar/22
u_(n+1) =((u_n v_n )/v_(n+1) )→u_(n+1) v_(n+1) =u_n v_n for all n we deduce that (u_n v_n ) is constant
$${u}_{{n}+\mathrm{1}} =\frac{{u}_{{n}} {v}_{{n}} }{{v}_{{n}+\mathrm{1}} }\rightarrow{u}_{{n}+\mathrm{1}} {v}_{{n}+\mathrm{1}} ={u}_{{n}} {v}_{{n}} \:{for}\:{all}\:{n}\:{we}\:{deduce} \\ $$$${that}\:\left({u}_{{n}} {v}_{{n}} \right)\:{is}\:{constant} \\ $$

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