u-n-3-u-n-2-u-n-1-u-n-3-n-IN-find-u-n-

Question Number 148645 by metamorfose last updated on 29/Jul/21

u_{n + 3} = \frac{u_{n + 2} + u_{n + 1} + u_{n}}{3}, \forall n \in I N

f i n d u_{n}

Answered by Olaf_Thorendsen last updated on 29/Jul/21

u_{n + 3} = \frac{u_{n + 2} + u_{n + 1} + u_{3}}{3}

3 u_{n + 3} - u_{n + 2} - u_{n + 1} - u_{n} = 0

We solve 3 r^{3} - r^{2} - r - 1 = 0

3 (r - 1) (r^{2} + \frac{2}{3} r + \frac{1}{3}) = 0

3 (r - 1) (r - r_{1}) (r - r_{2}) = 0

r_{1, 2} = - \frac{1 \pm i \sqrt{2}}{3}

u_{n} = λ . 1^{n} + μ r_{1}^{n} + γ r_{2}^{n}

u_{n} = λ + μ r_{1}^{n} + γ r_{2}^{n}

We find λ, μ, γ with u_{0}, u_{1}, u_{2} .

In particular for μ = γ = 0, u_{n} = λ

All constant sequences are solutions .

Commented by Olaf_Thorendsen last updated on 29/Jul/21

r_{1}^{n} = {(- \frac{1 + i \sqrt{2}}{3})}^{n}

r_{1}^{n} = \frac{{(- 1)}^{n}}{3^{n}} 3^{n} e^{i n \arctan \sqrt{2}}

r_{0}^{n} = {(- 1)}^{n} e^{i n \arctan \sqrt{2}}

idem with r_{2}^{n}

So we can write the expression with

{(- 1)}^{n} \cos (n \arctan \sqrt{2})

but we cannot simplify a lot .

Commented by metamorfose last updated on 29/Jul/21

t h a n k u s i r .

Commented by metamorfose last updated on 29/Jul/21

i t h o u g h t t h e e x p r e s s i o n w o u l d b e w i t h c o s