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Question Number 164478 by mathocean1 last updated on 17/Jan/22
Given  { ((u_0 =α ∈ C)),((u_(n+1) =((u_n +∣u_n ∣)/2))) :} ; n∈ N  where (u_n ) _(n∈N)  is a complex sequence.  Determinate the sequence (Im(u_n )) _(n∈N)   and calculate its limit.  NB: Im(u_n ) is the complex part of u_(n.)
$${Given}\:\begin{cases}{{u}_{\mathrm{0}} =\alpha\:\in\:\mathbb{C}}\\{{u}_{{n}+\mathrm{1}} =\frac{{u}_{{n}} +\mid{u}_{{n}} \mid}{\mathrm{2}}}\end{cases}\:;\:{n}\in\:\mathbb{N} \\ $$$${where}\:\left({u}_{{n}} \right)\:_{{n}\in\mathbb{N}} \:{is}\:{a}\:{complex}\:{sequence}. \\ $$$${Determinate}\:{the}\:{sequence}\:\left({Im}\left({u}_{{n}} \right)\right)\:_{{n}\in\mathbb{N}} \\ $$$${and}\:{calculate}\:{its}\:{limit}. \\ $$$${NB}:\:{Im}\left({u}_{{n}} \right)\:{is}\:{the}\:{complex}\:{part}\:{of}\:{u}_{{n}.} \\ $$$$ \\ $$
Answered by Rasheed.Sindhi last updated on 18/Jan/22
u_n =a_n +ib_n   u_0 =α=a_0 +ib_0   a_(n+1) +ib_(n+1) =((a_n +ib_n +∣a_n +ib_n ∣)/2)  a_(n+1) +ib_(n+1) =((a_n +(√(a_n ^2 +b_n ^2 )))/2)+((ib_n )/2)  im(u_(n+1) ):   b_(n+1) =(b_n /2)  ⇒^(n→n−1)     b_n =(b_(n−1) /2)  b_1 =(b_0 /2)  b_2 =(b_1 /2)=((b_0 /2)/2)=(b_0 /2^2 )  b_3 =(b_2 /2)=(b_0 /2^3 )  ...  b_n =(b_0 /2^n )=((im(α))/2^n )  lim_(n→∞) b_n  = lim_(n→∞) ((im(α))/2^n )=0                                 [∵ im(α) is constant]
$${u}_{{n}} ={a}_{{n}} +{ib}_{{n}} \\ $$$${u}_{\mathrm{0}} =\alpha={a}_{\mathrm{0}} +{ib}_{\mathrm{0}} \\ $$$${a}_{{n}+\mathrm{1}} +{ib}_{{n}+\mathrm{1}} =\frac{{a}_{{n}} +{ib}_{{n}} +\mid{a}_{{n}} +{ib}_{{n}} \mid}{\mathrm{2}} \\ $$$${a}_{{n}+\mathrm{1}} +{ib}_{{n}+\mathrm{1}} =\frac{{a}_{{n}} +\sqrt{{a}_{{n}} ^{\mathrm{2}} +{b}_{{n}} ^{\mathrm{2}} }}{\mathrm{2}}+\frac{{ib}_{{n}} }{\mathrm{2}} \\ $$$${im}\left({u}_{{n}+\mathrm{1}} \right):\:\:\:{b}_{{n}+\mathrm{1}} =\frac{{b}_{{n}} }{\mathrm{2}}\:\:\overset{{n}\rightarrow{n}−\mathrm{1}} {\Rightarrow}\:\:\:\:{b}_{{n}} =\frac{{b}_{{n}−\mathrm{1}} }{\mathrm{2}} \\ $$$${b}_{\mathrm{1}} =\frac{{b}_{\mathrm{0}} }{\mathrm{2}} \\ $$$${b}_{\mathrm{2}} =\frac{{b}_{\mathrm{1}} }{\mathrm{2}}=\frac{\frac{{b}_{\mathrm{0}} }{\mathrm{2}}}{\mathrm{2}}=\frac{{b}_{\mathrm{0}} }{\mathrm{2}^{\mathrm{2}} } \\ $$$${b}_{\mathrm{3}} =\frac{{b}_{\mathrm{2}} }{\mathrm{2}}=\frac{{b}_{\mathrm{0}} }{\mathrm{2}^{\mathrm{3}} } \\ $$$$… \\ $$$${b}_{{n}} =\frac{{b}_{\mathrm{0}} }{\mathrm{2}^{{n}} }=\frac{{im}\left(\alpha\right)}{\mathrm{2}^{{n}} } \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}{b}_{{n}} \:=\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{{im}\left(\alpha\right)}{\mathrm{2}^{{n}} }=\mathrm{0}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left[\because\:{im}\left(\alpha\right)\:{is}\:{constant}\right] \\ $$
Commented by mathocean1 last updated on 19/Jan/22
thanks
$${thanks} \\ $$

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