find-the-sequence-u-n-wich-verify-u-n-u-n-1-cosn-

Question Number 134630 by mathmax by abdo last updated on 06/Mar/21

find the sequence u_{n} wich verify u_{n} + u_{n + 1} = cosn

Answered by mathmax by abdo last updated on 07/Mar/21

u_{n} + u_{n + 1} = \cos (n) \Rightarrow \sum_{k = 0}^{n - 1} {(- 1)}^{k} (u_{k} + u_{k + 1}) = \sum_{k = 0}^{n - 1} {(- 1)}^{k} \cos (k) \Rightarrow

u_{0} + u_{1} - (u_{1} + u_{2}) + \dots . + {(- 1)}^{n - 2} (u_{n - 2} + u_{n - 1}) + {(- 1)}^{n - 1} (u_{n - 1} + u_{n})

= \sum_{k = 0}^{n - 1} {(- 1)}^{k} cosk \Rightarrow u_{0} + {(- 1)}^{n - 1} u_{n} = \sum_{k = 0}^{n - 1} {(- 1)}^{k} cosk \Rightarrow

{(- 1)}^{n - 1} u_{n} = \sum_{k = 0}^{n - 1} {(- 1)}^{k} [cosk - u_{0} \Rightarrow

u_{n} = {(- 1)}^{n - 1} \sum_{k = 0}^{n - 1} {(- 1)}^{k} \cos (k) - {(- 1)}^{n - 1} u_{o} \Rightarrow

u_{n} = {(- 1)}^{n} u_{0} - {(- 1)}^{n} \sum_{k = 0}^{n - 1} {(- 1)}^{k} cosk and

\sum_{k = 0}^{n - 1} {(- 1)}^{k} cosk = Re (\sum_{k = 0}^{n - 1} {(- 1)}^{k} e^{ik}) we have

\sum_{k = 0}^{n - 1} {(- e^{i})}^{k} = \frac{1 - {(- e^{i})}^{n}}{1 - (- e^{i})} = \frac{1 - {(- 1)}^{n} e^{ni}}{1 + e^{i}}

= \frac{1 - {(- 1)}^{n} {cosn + isinn}}{1 + \cos 1 + isin 1}

= \frac{(1 - {(- 1)}^{n} (cosn + isinn)) (1 + \cos 1 - isin 1)}{{(1 + \cos (1))}^{2} + \sin^{2} (1)}

rest to extract Re (of this quantity) \dots .