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Ex:  Solve the recurrance relation:  $h_n + h_{n-2} = 0$

$h_0 = 3, h_1 = 5$

$q^n + q^{n-2} = 0$

$q^2 + 1 = 0$

$q = \pm i$

$h_n = c_1 i^n + c_2 (-i)^n$

$c_1 + c_2 = 3$ implies $c_2 = 3 - c_1$ 

$c_1i - c_2i = 5$

$-c_1 + c_2 = 5i$

$-c_1 + 3 - c_1 = 5i$

$-2c_1 + 3 = 5i$

$c_1 = {3 - 5i \over 2}$

$c_2 = 3 - ({3 - 5i \over 2}) = {3 + 5i\over 2}$

$h_n = ({3 - 5i \over 2}) i^n + ({3 + 5i \over 2}) (-i)^n$


$h_{2j} =  ({3 - 5i \over 2}) i^{2j} + ({3 + 5i \over 2}) (-i)^{2j}$

$h_{2j+1} = ({3 - 5i \over 2}) i^{2j+1} + ({3 + 5i \over 2}) (-i)^{2j+1}$

$= ({3i + 5 \over 2}) (-1)^j + ({-3i + 5 \over 2}) (-1)^{j}$

Ex:  Solve the recurrance relation,
$h_n - 2h_{n-1} + h_{n-3} - h_{n-4} = 0$ \hfil \break
$h_0 = 3, h_1 = 3, h_2 = 7, h_3 = 15$,


$(x - 1)^3(x + 1)$

$=(x^3 - 3x^2 + 3x - 1)$

$=(x^4 - 2x^3 + 2x - 1)$

q = 1 , 1, 1, -1


$h_n = c_1 + c_2n + c_3n^2 + c_4(-1)^n$


$h_0 = c_1 + c_4$

$h_1 = c_1 + c_2 + c_3 - c_4$

$h_2 = c_1 + 2c_2 + 4c_3 + c_4$

$h_3 = c_1 + 3c_2 + 9c_3 - c_4$


$\matrix{
1 & 0 & 0 &  1 & a \cr
1 & 1 & 1 & -1 & b \cr
1 & 2 & 4 &  1 & c \cr
1 & 3 & 9 & -1 & d }$

$\matrix{
1 & 0 & 0 &  1 & a \cr
0 & 1 & 1 & -2 & b-a \cr
0 & 2 & 4 &  0 & c-a \cr
0 & 3 & 9 & -2 & d-a }$


$\matrix{
1 & 0 & 0 &  1 & a \cr
0 & 1 & 1 & -2 & b-a \cr
0 & 0 & 2 &  4 & c-a - 2(b-a) \cr
0 & 0 & 6 & 4 & d-a - 3(b-a) }$


$\matrix{
1 & 0 & 0 &  1 & a \cr
0 & 1 & 1 & -2 & b-a \cr
0 & 0 & 2 &  4 & c-a - 2(b-a) \cr
0 & 0 & 0 & -8 & d-a - 3(b-a) - 3[c-a - 2(b-a)]}$

$\matrix{
1 & 0 & 0 &  1 & a \cr
0 & 1 & 1 & -2 &  \cr
0 & 0 & 2 &  4 & c-a  \cr
0 & 0 & 0 & -8 & d-a  - 3[c-a]}$


$\matrix{
1 & 0 & 0 &  1 & 3 \cr
0 & 1 & 1 & -2 & 0 \cr
0 & 0 & 2 &  4 & 4  \cr
0 & 0 & 0 & -8 & d- 3  - 12}$

$h_0 = 3, h_1 = 3, h_2 = 7, h_3 = 15$,

$\matrix{
1 & 0 & 0 &  1 & 3 \cr
0 & 1 & 1 & -2 & 0 \cr
0 & 0 & 1 &  2 & 2  \cr
0 & 0 & 0 & 1 & 0}$

$\matrix{
1 & 0 & 0 &  0 & 3 \cr
0 & 1 & 1 &  0 & 0 \cr
0 & 0 & 1 &  0 & 2  \cr
0 & 0 & 0 &  1 & 0}$

$\matrix{
1 & 0 & 0 &  0 & 3 \cr
0 & 1 & 0 &  0 & -2 \cr
0 & 0 & 1 &  0 & 2  \cr
0 & 0 & 0 &  1 & 0}$


$=    ~~~~~~~~   (x^3 - 3x^2 + 3x - 1)$

$=(x^4 - 3x^3 + 3x^2 - x)$


\end

16:  
$x^n - 3x^{n-2} + 2x^{n-3}$

$x^3 - 3x + 2 = 0

1 - 3 + 2 = 0


(x - 1)(x^2 + x - 2) = (x - 1)(x - 1)(x + 2)




\end