find the sequence U_n wich verify U_n +U


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Question Number 67023 by mathmax by abdo last updated on 21/Aug/19

$f i n d t h e s e q u e n c e U_{n} w i c h v e r i f y U_{n} + U_{n + 1} = s i n (n) \forall n f r o m n$
Commented by mathmax by abdo last updated on 23/Aug/19

$w e h a v e u_{n} + u_{n + 1} = s i n (n) \Rightarrow \sum_{k = 0}^{n - 1} (u_{k} + u_{k + 1}) {(- 1)}^{k} = \sum_{k = 0}^{n} {(- 1)}^{k} s i n (k)$ $\Rightarrow u_{0} + u_{1} - u_{1} - u_{2} + u_{2} + u_{3} + . . . . + {(- 1)}^{n - 1} (u_{n - 1} + u_{n})$ $= \sum_{k = 0}^{n} {(- 1)}^{k} s i n (k) \Rightarrow u_{0} + {(- 1)}^{n - 1} u_{n} = \sum_{k = 0}^{n} {(- 1)}^{k} s i n (k) \Rightarrow$ $u_{0} - \sum_{k = 0}^{n} {(- 1)}^{k} s i n (k) = {(- 1)}^{n} u_{n} \Rightarrow u_{n} = {(- 1)}^{n} u_{0} - {(- 1)}^{n} \sum_{k = 0}^{n} {(- 1)}^{k} s i n (k)$ $l e t f i n d w_{n} = \sum_{k = 0}^{n} {(- 1)}^{k} s i n) k)$ $w_{n} = I m (\sum_{k = 0}^{n} {(- 1)}^{k} e^{i k}) = I m (\sum_{k = 0}^{n} {(- e^{i})}^{k})$ $a n d \sum_{k = 0}^{n} {(- e^{i})}^{k} = \frac{1 - {(- e^{i})}^{n + 1}}{1 + e^{i}} = \frac{1 + {(- 1)}^{n} e^{i (n + 1)}}{1 + e^{i}}$ $= \frac{1 + {(- 1)}^{n} e^{i (n + 1)}}{1 + c o s (1) + i s i n (1)} = \frac{(1 + c o s (1) - i s i n (1)) (1 + {(- 1)}^{n} e^{i (n + 1)})}{{(1 + c o s (1))}^{2} + {s i n}^{2} 1}$ $= \frac{(1 + c o s (1) - i s i n (1)) (1 + {(- 1)}^{n} c o s (n + 1) + i {(- 1)}^{n} s i n (n + 1))}{{(1 + c o s (1))}^{2} + {s i n}^{2} 1}$ $= \frac{(1 + c o s (1)) (1 + {(- 1)}^{n} c o s (n + 1)) + i {(- 1)}^{n} s i n (n + 1) (1 + c o s (1)) - i s i n (1) (1 + {(- 1)}^{n} c o s (n + 1)) + . . . .}{{(1 + c o s (1))}^{2} + {s i n}^{2} (1)}$ $\Rightarrow w_{n} = \frac{({- 1)}^{n} s i n (n + 1) (1 + c o s (1)) - s i n (1) (1 + {(- 1)}^{n} c o s) n + 1))}{2 + 2 c o s (1)}$ $\Rightarrow u_{n} = {(- 1)}^{n} u_{0} - {(- 1)}^{n} w_{n}$
	Answered by mind is power last updated on 23/Aug/19

	$l e t v_{n + 1} + v_{n} = c o s (n)$ $a n d Z n = v_{n} + {i u}_{n}$ $z_{n + 1} + z_{n} = c o s (n) + i s i n (n) = e^{i n}$ $==> \frac{z_{n + 1}}{e^{i (n)}} + \frac{z_{n}}{e^{i n}} = 1 = e^{i} \frac{z_{n + 1}}{e^{i (n + 1)}} + \frac{Z_{n}}{e^{i n}} = 1$ $l e t W_{n} = \frac{z_{n}}{e^{i n}} ===> e^{i} W_{n + 1} + W n = 1$ $e i x + 1 = 0 ==> x = - \frac{1}{e^{i x}} = - e^{- i x}$ $W_{n} = c {(- 1)}^{n} e^{- i n x} + t_{n}$ $w i t h e t n p a r t i c u l a r s o l u t i o n o f e^{i} t_{n + 1} + t_{n} = 1$ $l e t t_{n} = a n + b ==> e^{i} (a n + a + b) + a n + b = 1 \forall n \in I N$ $==> a n (e^{i} + 1) = 0 a n d b (e^{i} + 1) = 1 ==> a = 0 a n d b = \frac{1}{1 + e^{i}}$ $==> t_{n} = \frac{1}{1 + e^{i}}$ $W_{n} = {(- 1)}^{n} {c e}^{- i n} + \frac{1}{1 + e^{i}} . c \in I C .$ $==> Z_{n} = e^{i n} W_{n} = c {(- 1)}^{n} + \frac{e^{i n}}{1 + e^{i}} = c {(- 1)}^{n} + \frac{e^{i n} + e^{i (n - 1)}}{2 + 2 c o s (1)}$ $u_{n} = I M (Z_{n}) = k {(- 1)}^{n} + \frac{1}{2 + 2 c o s (1)} [s i n (n) + s i n (n - 1)]$ $s i n (a) + s i n (b) = 2 c o s (\frac{a - b}{2}) s i n (\frac{a + b}{2}) . a n d 2 + 2 c o s (1) = 2 + 2 c o s (2 . \frac{1}{2}) = 4 {c o s}^{2} (\frac{1}{2})$ $w e f i n d u_{n} = k {(- 1)}^{n} + \frac{2 c o s (\frac{1}{2}) s i n (n - \frac{1}{2})}{4 {c o s}^{2} (\frac{1}{2})} = k {(- 1)}^{n} + \frac{s i n (n - \frac{1}{2})}{2 c o s (\frac{1}{2})}$
	Commented by mathmax by abdo last updated on 23/Aug/19

	$t h a n k y o u s i r .$
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