Given-u-0-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-co

Question Number 164478 by mathocean1 last updated on 17/Jan/22

G i v e n {\begin{cases} u_{0} = α \in C \\ u_{n + 1} = \frac{u_{n} + ∣ u_{n} ∣}{2} \end{cases}; n \in N

w h e r e (u_{n})_{n \in N} i s a c o m p l e x s e q u e n c e .

D e t e r m i n a t e t h e s e q u e n c e (I m (u_{n}))_{n \in N}

a n d c a l c u l a t e i t s l i m i t .

N B : I m (u_{n}) i s t h e c o m p l e x p a r t o f u_{n .}

Answered by Rasheed.Sindhi last updated on 18/Jan/22

u_{n} = a_{n} + {i b}_{n}

u_{0} = α = a_{0} + {i b}_{0}

a_{n + 1} + {i b}_{n + 1} = \frac{a_{n} + {i b}_{n} + ∣ a_{n} + {i b}_{n} ∣}{2}

a_{n + 1} + {i b}_{n + 1} = \frac{a_{n} + \sqrt{a_{n}^{2} + b_{n}^{2}}}{2} + \frac{{i b}_{n}}{2}

i m (u_{n + 1}) : b_{n + 1} = \frac{b_{n}}{2} \overset{n \to n - 1}{\Rightarrow} b_{n} = \frac{b_{n - 1}}{2}

b_{1} = \frac{b_{0}}{2}

b_{2} = \frac{b_{1}}{2} = \frac{\frac{b_{0}}{2}}{2} = \frac{b_{0}}{2^{2}}

b_{3} = \frac{b_{2}}{2} = \frac{b_{0}}{2^{3}}

\dots

b_{n} = \frac{b_{0}}{2^{n}} = \frac{i m (α)}{2^{n}}

\lim_{n \to \infty} b_{n} = \lim_{n \to \infty} \frac{i m (α)}{2^{n}} = 0

[∵ i m (α) i s c o n s t a n t]

Commented by mathocean1 last updated on 19/Jan/22

t h a n k s