let-A-1-1-3-1-1-calculate-A-n-2-find-e-A-and-e-A-3-find-cosA-and-sinA-

Question Number 75735 by mathmax by abdo last updated on 16/Dec/19

l e t A = (\begin{matrix} 1 1 \\ 3 - 1 \end{matrix})

1) c a l c u l a t e A^{n}

2) f i n d e^{A} a n d e^{- A}

3) f i n d c o s A a n d s i n A

Commented by abdomathmax last updated on 22/Dec/19

c o s A = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n} A^{2 n}}{(2 n)!}

s i n A = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{(2 n + 1)} A^{2 n + 1}

Commented by mathmax by abdo last updated on 22/Dec/19

P c (x) = d e t (A - x I) = | \begin{matrix} 1 - x 1 \\ 3 - 1 - x \end{matrix} | = - (1 + x) (1 - x) - 3

= x^{2} - 1 - 3 = x^{2} - 4

t h e p r o p e r v a l u e s a r e λ_{1} = 2 a n d λ_{2} = - 2

w e h s v e x^{n} = P_{c} (x) Q (x) + u_{n} x + v_{n} \Rightarrow 2^{n} = 2 u_{n} + v_{n}

a n d {(- 2)}^{n} = - 2 u_{n} + v_{n} \Rightarrow

{\begin{cases} 2 u_{n} + v_{n} = 2^{n} \\ - 2 u_{n} + v_{n} = {(- 2)}^{n} \Rightarrow 4 u_{n} = 2^{n} - {(- 2)}^{n} \Rightarrow u_{n} = \frac{2^{n} - {(- 2)}^{n}}{4} \end{cases}

v_{n} = 2^{n} - 2 u_{n} = 2^{n} - \frac{2^{n} - {(- 2)}^{n}}{2} = \frac{2 . 2^{n} - 2^{n} + {(- 2)}^{n}}{2} = \frac{2^{n} + {(- 2)}^{n}}{2}

a l s o w e h a v e A^{n} = u_{n} A + v_{n} I

= \frac{2^{n} - {(- 2)}^{n}}{4} (\begin{matrix} 1 1 \\ 3 - 1 \end{matrix}) + \frac{2^{n} + {(- 2)}^{n}}{2} (\begin{matrix} 1 0 \\ 0 1 \end{matrix})

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

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

Commented by mathmax by abdo last updated on 22/Dec/19

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

Commented by mathmax by abdo last updated on 22/Dec/19

e^{A} = \sum_{n = 0}^{\infty} \frac{A^{n}}{n!} = \sum_{n = 0}^{\infty} \frac{1}{n!} (\begin{matrix} \frac{3 . 2^{n} + {(- 2)}^{n}}{4} \frac{2^{n} - {(- 2)}^{n}}{4} \\ \frac{3}{4} (2^{n} - {(- 2)}^{n}) \frac{2^{n} + 3 {(- 2)}^{n}}{4} \end{matrix})

= (\begin{matrix} \frac{3}{4} \sum_{n = 0}^{\infty} \frac{2^{n}}{n!} + \frac{1}{4} \sum_{n = 0}^{\infty} \frac{{(- 2)}^{n}}{n!} \frac{1}{4} \sum_{n = 0}^{\infty} \frac{2^{n}}{n!} - \frac{1}{4} \sum_{n = 0}^{\infty} \frac{{(- 2)}^{n}}{n!} \\ \frac{3}{4} \sum_{n = 0}^{\infty} \frac{2^{n}}{n!} - \frac{3}{4} \sum_{n = 0}^{\infty} \frac{{(- 2)}^{n}}{n!} \frac{1}{4} \sum_{n = 0}^{\infty} \frac{2^{n}}{n!} + \frac{3}{4} \sum_{n = 0}^{\infty} \frac{{(- 2)}^{n}}{n!} \end{matrix})

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

w e u s e t h e s a m e w a y t o f i n d e^{- A} = \sum_{n = 0}^{\infty} \frac{{(- A)}^{n}}{n!}