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-lt-1-

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

W e a r e i n C .

G i v e n Z_{0} = 1; Z_{n + 1} = \frac{1}{2} Z_{n} + \frac{1}{2} i

n \in N .

S h o w t h a t \forall n \in N^{*}, ∣ Z_{n} ∣< 1 .

Answered by Olaf last updated on 22/Oct/20

Let U_{n} = Z_{n} - i, n \in N

\Rightarrow U_{0} = 1 - i and

U_{n + 1} = Z_{n + 1} - i = \frac{1}{2} Z_{n} - \frac{1}{2} i = \frac{1}{2} U_{n}

\Rightarrow U_{n} = U_{0} {(\frac{1}{2})}^{n} = \frac{1 - i}{2^{n}}

and Z_{n} = \frac{1 - i}{2^{n}} + i = \frac{1}{2^{n}} (1 + (2^{n} - 1) i)

∣ Z_{n} ∣ = \frac{\sqrt{1 + {(2^{n} - 1)}^{2}}}{2^{n}} < \frac{\sqrt{2^{2 n}}}{2^{n}} = 1, n \in N^{*}

and ∣ Z_{0} ∣ = \sqrt{2}

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

by recurrence n = 1 \Rightarrow z_{1} = \frac{1}{2} z_{0} + \frac{i}{2} = \frac{1}{2} + \frac{i}{2} \Rightarrow∣ z_{1} ∣= \frac{1}{2} ∣ 1 + i ∣

= \frac{\sqrt{2}}{2} < 1 relation true for n = 1 let suppise ∣ z_{n} ∣< 1

we hsve ∣ z_{n + 1} ∣= \frac{1}{2} ∣ z_{n} + i ∣⩽ \frac{1}{2} ∣ z_{n} ∣ + \frac{1}{2} < \frac{1}{2} + \frac{1}{2} = 1 \Rightarrow∣ z_{n + 1} ∣< 1

relation is true at term n + 1

Commented by mathocean1 last updated on 25/Oct/20

Thank you sirs .