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Question Number 167181 by mathocean1 last updated on 08/Mar/22

G i v e n x, c \in R .

{(u_{n})}_{n \in N} : {\begin{cases} u_{0} = 0 \\ u_{n + 1} = x s i n (u_{n}) + c \end{cases}

1 . S h o w t h a t ∣ u_{n + 1} - u_{n} ∣⩽∣ c ∣∣ x^{n} ∣ .

2 . S h o w t h a t : (∣ x ∣< 1 e t m ⩾ n \Rightarrow∣ u_{m} - u_{n} ∣⩽ \frac{∣ c ∣∣ x^{n} ∣}{1 - ∣ x ∣}

3 . D e d u c t t h a t u_{n} i s c o n v e r g e n t a n d

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

Answered by mindispower last updated on 09/Mar/22

n = 0

∣ U_{1} - U_{0} ∣=∣ c ∣ T r u e

s u p p o s e \forall n ∣ U_{n + 1} - U_{n} ∣<∣ c ∣∣ x^{n} ∣

w e s h o w T h a t ∣ u_{n + 2} - u_{n + 1} ∣<∣ c ∣∣ x ∣^{n + 1}

\forall n \in N^{*} ∣ U_{n + 2} - U_{n + 1} ∣=∣ x s i n (u_{n + 1}) - x s i n (u_{n}) ∣

=∣ x ∣ . ∣ s i n (U_{n + 1}) - s i n (U_{n}) ∣ \dots . (E)

\forall (a, b) \in R^{2} W e h a v e

∣ s i n (a) - s i n (b) ∣<∣ a - b ∣

(E) ⩽∣ x ∣ . ∣ U_{n + 1} - U_{n} ∣⩽∣ x ∣ . ∣ c ∣ . ∣ x ∣^{n} =∣ c ∣∣ x^{n + 1} t r u e

(2) u s e m = n + k

a n d ∣ U_{m} - U_{n} ∣=∣ \underset{k = n}{\sum^{m - 1}} U_{k + 1} - U_{k} ∣< Σ ∣ U_{k + 1} - U_{k} ∣ a n d Q u a t i o m 1