A-1-2-B-1-1-A-n-1-A-n-8B-n-B-n-1-A-n-7B-n-find-A-n-B-n-

Question Number 75616 by mr W last updated on 13/Dec/19

A_{1} = 2

B_{1} = 1

A_{n + 1} = A_{n} - 8 B_{n}

B_{n + 1} = A_{n} + 7 B_{n}

f i n d A_{n} = ?, B_{n} = ?

Commented by mind is power last updated on 13/Dec/19

(\begin{matrix} A_{n + 1} \\ B_{n + 1} \end{matrix}) = (\begin{matrix} 1 - 8 \\ 1 7 \end{matrix}) (\begin{matrix} A_{n} \\ B_{n} \end{matrix})

M = (\begin{matrix} 1 - 8 \\ 1 7 \end{matrix})

(\begin{matrix} 1 - x - 8 \\ 1 7 - x \end{matrix}) = 0 \Rightarrow x^{2} - 8 x + 15 = 0

x = 5, x = 3

D = (\begin{matrix} 3 0 \\ 0 5 \end{matrix})

x = 3 \Rightarrow - 2 x - 8 y = 0 \Rightarrow x = - 4 y

U = (- 4, 1)

x = 5 \Rightarrow x + 2 y = 0 \Rightarrow x = - 2 y \Rightarrow V = (- 2, 1)

P = (\begin{matrix} - 4 - 2 \\ 1 1 \end{matrix})

p^{-} = - \frac{1}{2} (\begin{matrix} 1 2 \\ - 1 - 4 \end{matrix})

M = P D P^{-}

M^{n} = {PD}^{n} P^{-}

(\begin{matrix} A_{n} \\ B_{n} \end{matrix}) = {PD}^{n - 1} . P^{-} . (\begin{matrix} A_{1} \\ B_{1} \end{matrix})

{PD}^{n} P^{-} = P . (\begin{matrix} 3^{n} 0 \\ 0 5^{n} \end{matrix}) . - \frac{1}{2} (\begin{matrix} 1 2 \\ - 1 - 4 \end{matrix})

= (\begin{matrix} - 4 - 2 \\ 1 1 \end{matrix}) (\begin{matrix} - \frac{3^{n}}{2} - 3^{n} \\ \frac{5^{n}}{2} 2 . 5^{n} \end{matrix}) = (\begin{matrix} 2 . 3^{n} - 5^{n} 4 . 3^{n} - 4 . 5^{n} \\ - \frac{3^{n}}{2} + \frac{5^{n}}{2} - 3^{n} + 2 . 5^{n} \end{matrix})

(\begin{matrix} A_{n + 1} \\ B_{n + 1} \end{matrix}) = (\begin{matrix} 2 . 3^{n} - 5^{n} 4 . 3^{n} - 4 . 5^{n} \\ - \frac{3^{n}}{2} + \frac{5^{n}}{2} - 3^{n} + 2 . 5^{n} \end{matrix}) (\begin{matrix} 2 \\ 1 \end{matrix})

A_{n + 1} = 8 . 3^{n} - 6 . 5^{n}

B_{n + 1} = - 2 . 3^{n} + 3 . 5^{n}

A_{n} = 8 . 3^{n - 1} - 6 . 5^{n - 1}

B_{n} = - 2 . 3^{n - 1} + 3 . 5^{n - 1}

Commented by mr W last updated on 14/Dec/19

t h a n k s a l o t s i r!

i n e e d s o m e t i m e t o u n d e r s t a n d

y o u r s o l u t i o n a n d w h y y o u d i d s o .

Commented by peter frank last updated on 14/Dec/19

h e l p m e Q n 75602

Commented by mind is power last updated on 14/Dec/19

Hello Sir Mr W, its solve system of series usign lineair Algebre

not sur in french Tap in google

∴ System de suite numerique et diagonilisations Des matrice ∵

you get in french but you can understand and my bee you will

find after that in english sorry my english is not good

Commented by mr W last updated on 14/Dec/19

M^{ERCI} B^{EAUCOUP}!

Commented by mind is power last updated on 14/Dec/19

De Rien :)

Commented by mathmax by abdo last updated on 15/Dec/19

w e h a v e a_{n + 1} = a_{n} - 8 b_{n}

b_{n + 1} = a_{n} + 7 b_{n} \Rightarrow (\begin{matrix} a_{n + 1} \\ b_{n + 1} \end{matrix}) = (\begin{matrix} 1 - 8 \\ 1 7 \end{matrix}) (\begin{matrix} a_{n} \\ b_{n} \end{matrix})

\Rightarrow (\begin{matrix} a_{n + 1} \\ b_{n + 1} \end{matrix}) = A^{n} (\begin{matrix} a_{1} \\ b_{1} \end{matrix}) l e t c a l c u l a t e A^{n}

p_{c} (x) = d e t (A - x I) = | \begin{matrix} 1 - x - 8 \\ 1 7 - x \end{matrix} | = (1 - x) (7 - x) + 8

= 7 - x - 7 x + x^{2} + 8 = x^{2} - 8 x + 15

Δ^{^{'}} = 4^{2} - 15 = 1 \Rightarrow x_{1} = 4 + 1 = 5 a n d x_{2} = 4 - 1 = 3

w e h a v e x^{n} = q p_{c} (x) + u_{n} x + v_{n} \Rightarrow A^{n} = u_{n} A + v_{n} I

w e h a v e 3^{n} = 3 u_{n} + v_{n} a n d 5^{n} = 5 u_{n} + v_{n} \Rightarrow

2 u_{n} = 5^{n} - 3^{n} \Rightarrow u_{n} = \frac{5^{n} - 3^{n}}{2} a l s o v_{n} = 3^{n} - 3 u_{n} = 3^{n} - 3 \times \frac{5^{n} - 3^{n}}{2}

= \frac{2 . 3^{n} - 3 . 5^{n} + 3 . 3^{n}}{2} = \frac{5 . 3^{n} - 3 . 5^{n}}{2} \Rightarrow

A^{n} = \frac{5^{n} - 3^{n}}{2} (\begin{matrix} 1 - 8 \\ 1 7 \end{matrix}) + \frac{5 . 3^{n} - 3 . 5^{n}}{2} (\begin{matrix} 1 0 \\ 0 1 \end{matrix})

= (\begin{matrix} \frac{5^{n} - 3^{n}}{2} - 4 . 5^{n} + 4 . 3^{n} \\ \frac{5^{n} - 3^{n}}{2} \frac{7}{2} (5^{n} - 3^{n}) \end{matrix}) + (\begin{matrix} \frac{5 . 3^{n} - 3 . 5^{n}}{2} 0 \\ 0 \frac{5 . 3^{n} - 3 . 5^{n}}{2} \end{matrix})

= (\begin{matrix} - 5^{n} + 2 . 3^{n} 4 . 3^{n} - 4 . 5^{n} \\ \frac{5^{n} - 3^{n}}{2} 2 . 5^{n} - 3^{n} \end{matrix}) \Rightarrow

(\begin{matrix} a_{n + 1} \\ b_{n + 1} \end{matrix}) = A^{n} (\begin{matrix} 2 \\ 1 \end{matrix}) a n d A^{n} i s k n o w n \dots

Answered by mr W last updated on 14/Dec/19

a t t e m p t a n o t h e r w a y :

l e t C_{n + 1} = A_{n + 1} + {k B}_{n + 1} w i t h k = c o n s t a n t

\Rightarrow C_{n} = A_{n} + {k B}_{n}

i f w e c a n g e t C_{n + 1} = {h C}_{n} w i t h h = c o n s t a n t

t h e n w e c a n s o l v e b o t h A_{n} a n d B_{n} .

C_{n + 1} = A_{n + 1} + {k B}_{n + 1}

C_{n + 1} = (A_{n} - 8 B_{n}) + k (A_{n} + 7 B_{n})

C_{n + 1} = (1 + k) A_{n} + (7 k - 8) B_{n}

C_{n + 1} = (1 + k) [A_{n} + \frac{7 k - 8}{1 + k} B_{n}]

\equiv C_{n + 1} = h [A_{n} + {k B}_{n}] = {h C}_{n}

i . e .

\frac{7 k - 8}{1 + k} = k

\Rightarrow k^{2} - 6 k + 8 = 0

\Rightarrow (k - 2) (k - 4) = 0

\Rightarrow k = 2 o r k = 4

h = 1 + k = 3 o r 5

w i t h k = 2 a n d h = 3 :

C_{1} = A_{1} + {k B}_{1} = 2 + 2 \times 1 = 4

C_{n + 1} = {h C}_{n} = h^{n} C_{1} = 4 \times 3^{n}

i . e . A_{n + 1} + 2 B_{n + 1} = 4 \times 3^{n} \dots (i)

w i t h k = 4 a n d h = 5 :

C_{1} = A_{1} + {k B}_{1} = 2 + 4 \times 1 = 6

C_{n + 1} = {h C}_{n} = h^{n} C_{1} = 6 \times 5^{n}

i . e . A_{n + 1} + 4 B_{n + 1} = 6 \times 5^{n} \dots (i i)

(i i) - (i) :

2 B_{n + 1} = 6 \times 5^{n} - 4 \times 3^{n}

\Rightarrow B_{n + 1} = 3 \times 5^{n} - 2 \times 3^{n}

(i) \times 2 - (i i) :

A_{n + 1} = 2 \times 4 \times 3^{n} - 6 \times 5^{n}

\Rightarrow A_{n + 1} = 8 \times 3^{n} - 6 \times 5^{n}

\Rightarrow A_{n} = 8 \times 3^{n - 1} - 6 \times 5^{n - 1}

\Rightarrow B_{n} = 3 \times 5^{n - 1} - 2 \times 3^{n - 1}

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