Question

Solve the recurrence relation of an=3an-3an-2-an-3 n>=3 a0=1 a1=-2 a2=-1

Answer 1

$a_n-3a_n+3a_{n-2}+a_{n-3}=0$

$-2a_n+3a_{n-2}+a_{n-3}=0$

Characteristics equalation

$-2x^3+3x+1=0$

$Q=(a^2-3b)/9=0.5$

$R=(2a^3-9ab+27c)/54=-0.25$

$S=Q^3-R^2=0.06$ >0

$\phi=1/3 arccos(R/\sqrt{Q^3})$

$x_1=-2\sqrt{Q} cos(\phi)-a/3=-1$

$x_2=-2\sqrt{Q} cos(\phi+2/3\pi)-a/3=1.37$

$x_3=-2\sqrt{Q} cos(\phi-2/3\pi)-a/3=-0.37$

$a_n=a(-1)^n+(1.37)^nb+(-0.37)^nc$

$a_0=a+b+c=1$

$a_1=-a+1.37b-0.37c=-2$

$a_2=a+1.8769b+0.1369c=-1$

$a_0+a_1=2.37b+0.63c=-1$

$a_1+a_2=3.2469b-0.2331c=-3$

$b=(-1-0.63c)/2.37$

$3.2469(-1-0.63c)-0.2331c=-3$

$-3.2469-2.045547c-0.2431c=-3$

$-2.288647c=0.2469$

$c=-0.0108$

$b=-0.419$

$a=1-b-c=1+0.0108+0.419=1.43$

$a_n=1.43(-1)^n-0.419(1.37)^n-0.0108(-0.37)^n$