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Question Number 119159 by mathocean1 last updated on 22/Oct/20

We are in C. Given Z_(0 ) =1 ; Z_(n+1 ) =(1/2)Z_(n ) +(1/2)i n ∈ N. Show that ∀ n ∈ N^(∗ ) , ∣Z_n ∣<1.

$${We}\:{are}\:{in}\:\mathbb{C}. \\ $$ $${Given}\:{Z}_{\mathrm{0}\:\:} =\mathrm{1}\:;\:\:\:\:\:{Z}_{{n}+\mathrm{1}\:} =\frac{\mathrm{1}}{\mathrm{2}}{Z}_{{n}\:} +\frac{\mathrm{1}}{\mathrm{2}}{i} \\ $$ $${n}\:\in\:\mathbb{N}. \\ $$ $${Show}\:{that}\:\forall\:{n}\:\in\:\mathbb{N}^{\ast\:} ,\:\mid{Z}_{{n}} \mid<\mathrm{1}. \\ $$

Answered by Olaf last updated on 22/Oct/20

Let U_n = Z_n −i, n∈N ⇒ U_0 = 1−i and U_(n+1) = Z_(n+1) −i = (1/2)Z_n −(1/2)i = (1/2)U_n ⇒ U_n = U_0 ((1/2))^n = ((1−i)/2^n ) and Z_n = ((1−i)/2^n )+i = (1/2^n )(1+(2^n −1)i) ∣Z_n ∣ = ((√(1+(2^n −1)^2 ))/2^n ) < ((√2^(2n) )/2^n ) = 1, n∈N^∗ and ∣Z_0 ∣ = (√2)

$$\mathrm{Let}\:{U}_{{n}} \:=\:{Z}_{{n}} −{i},\:{n}\in\mathbb{N} \\ $$ $$\Rightarrow\:{U}_{\mathrm{0}} \:=\:\mathrm{1}−{i}\:\:\mathrm{and} \\ $$ $${U}_{{n}+\mathrm{1}} \:=\:{Z}_{{n}+\mathrm{1}} −{i}\:=\:\frac{\mathrm{1}}{\mathrm{2}}{Z}_{{n}} −\frac{\mathrm{1}}{\mathrm{2}}{i}\:=\:\frac{\mathrm{1}}{\mathrm{2}}{U}_{{n}} \\ $$ $$\Rightarrow\:{U}_{{n}} \:=\:{U}_{\mathrm{0}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{{n}} \:=\:\frac{\mathrm{1}−{i}}{\mathrm{2}^{{n}} } \\ $$ $$\mathrm{and}\:{Z}_{{n}} =\:\frac{\mathrm{1}−{i}}{\mathrm{2}^{{n}} }+{i}\:=\:\frac{\mathrm{1}}{\mathrm{2}^{{n}} }\left(\mathrm{1}+\left(\mathrm{2}^{{n}} −\mathrm{1}\right){i}\right) \\ $$ $$\mid{Z}_{{n}} \mid\:=\:\frac{\sqrt{\mathrm{1}+\left(\mathrm{2}^{{n}} −\mathrm{1}\right)^{\mathrm{2}} }}{\mathrm{2}^{{n}} }\:<\:\frac{\sqrt{\mathrm{2}^{\mathrm{2}{n}} }}{\mathrm{2}^{{n}} }\:=\:\mathrm{1},\:{n}\in\mathbb{N}^{\ast} \\ $$ $$\mathrm{and}\:\mid{Z}_{\mathrm{0}} \mid\:=\:\sqrt{\mathrm{2}} \\ $$

Answered by mathmax by abdo last updated on 22/Oct/20

by recurrence n=1 ⇒z_1 =(1/2)z_0 +(i/2) =(1/2)+(i/2) ⇒∣z_1 ∣=(1/2)∣1+i∣ =((√2)/2)<1 relation true for n=1 let suppise ∣z_n ∣<1 we hsve ∣z_(n+1) ∣=(1/2)∣z_n +i∣≤(1/2)∣z_n ∣+(1/2)<(1/2)+(1/2)=1 ⇒∣z_(n+1) ∣<1 relation is true at term n+1

$$\mathrm{by}\:\mathrm{recurrence}\:\mathrm{n}=\mathrm{1}\:\Rightarrow\mathrm{z}_{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}}\mathrm{z}_{\mathrm{0}} +\frac{\mathrm{i}}{\mathrm{2}}\:=\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{i}}{\mathrm{2}}\:\Rightarrow\mid\mathrm{z}_{\mathrm{1}} \mid=\frac{\mathrm{1}}{\mathrm{2}}\mid\mathrm{1}+\mathrm{i}\mid \\ $$ $$=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}<\mathrm{1}\:\:\:\mathrm{relation}\:\mathrm{true}\:\mathrm{for}\:\mathrm{n}=\mathrm{1}\:\mathrm{let}\:\mathrm{suppise}\:\mid\mathrm{z}_{\mathrm{n}} \mid<\mathrm{1} \\ $$ $$\mathrm{we}\:\mathrm{hsve}\:\mid\mathrm{z}_{\mathrm{n}+\mathrm{1}} \mid=\frac{\mathrm{1}}{\mathrm{2}}\mid\mathrm{z}_{\mathrm{n}} +\mathrm{i}\mid\leqslant\frac{\mathrm{1}}{\mathrm{2}}\mid\mathrm{z}_{\mathrm{n}} \mid+\frac{\mathrm{1}}{\mathrm{2}}<\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}=\mathrm{1}\:\Rightarrow\mid\mathrm{z}_{\mathrm{n}+\mathrm{1}} \mid<\mathrm{1} \\ $$ $$\mathrm{relation}\:\mathrm{is}\:\mathrm{true}\:\mathrm{at}\:\mathrm{term}\:\mathrm{n}+\mathrm{1} \\ $$

Commented bymathocean1 last updated on 25/Oct/20

Thank you sirs.

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{sirs}. \\ $$ $$ \\ $$

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