i've this recurrence relation to which i need to find the general numeric function,
$a_r$ - $5a_r$$_-$$_1$ + $8a_r$$_-$$_2$ - $4a_r$$_-$$_3$ = $r*2^r$ $...(i)$
i get the homogeneous solution as,
$a_r$$^($$^h$$^)$ = $(A_1*r + A_2) + A_3$. $...(ii)$
and, particular solution as,
$a_r$$^($$^p$$^)$ = $r^3/3$ $*2^r$$^-$$^1$. $(iii)$ $//r^2(P_1*r + P_2)2^r$ $,$ $P_1 = 1/6, P_2 = 0$
Total solution : $a_r^($$^p$$^)$ + $a_r^($$^h$$^)$
i'm not getting the correct solution. Plus i'm taking boundary condition randomly.
Do they have to be specific? Or, i can take any 3 values for $a_0,$ $a_1,$ $a_2.$
And put them in $(i)$, to calculate $a_3.$ And, also put them in $(ii)$ to calculate $A_1, A_2, A_3.$
If we take them randomly, can you solve it for $a_0 = 1,$ $a_1 = 5,$ $a_3 = 11.$
Thanks in advance.
Edit:
I've solved it upto.,
$\frac{1-8z+25z^2-12z^3-12z^4}{(1-z)(2z-1)^4}$ $=$ $\frac{A}{1-z}+\frac{B}{(2z-1)^4}+\frac{C}{(2z-1)^3}+\frac{D}{(2z-1)^2}+\frac{E}{(2z-1)}$
which then,
$\frac{1-8z+25z^2-12z^3-12z^4}{(1-z)(2z-1)^4}$ $=$ $\frac{A}{1-z}+\frac{B}{(1-2z)^4}-\frac{C}{(1-2z)^3}+\frac{D}{(1-2z)^2}-\frac{E}{(1-2z)}$ (can i do this?!)
with boundary conditions $a_0 = 1, a_1=1, a_2=2$(seems easier from those which i asked to use earlier), which then resolved to.,
$a_r = A+\frac{B*(r+1)(r+2)(r+3)*2^r}{3!}-\frac{C*(r+1)(r+2)*2^r}{2!}+D*(r+1)*2^r-E*2^r$ $...(iii)$
with, $A=-6, B=2$ ,$C=$$\frac{-25}{2}$, $D=-18$ $or$ $\frac{81}{2},$ i've calculated by equating with constant and with coefficient of $z^3$ respectively, $E=$$\frac{-21}{2}.$
I put the value of boundary condition $(a_0=1, a_1=1, a_2=2)$ in $(i)$ to calculate $a_3=30,$ but the result is not similar with $eq^n$ $3$.
Which i suspected beforehand since i'm getting 2 results for D.
- i'm suspecting again and again that for a recurrence relation we can not take any arbitrarily random boundary conditions, they should be specific, can you insist on this point.
I'm pretty sure that my solution is correct, Coz i $\left.c_r\right.$ $=$ $(re-)^rchecked,$ $0\le $ r$\le7$ or $,$ $otherwise.$
@cesareo Can you look into it? Thanks.
The homogeneous solution is
$$ a_r^h = c_1+c_2 2^r+c_3 2^r r $$
A particular solution can be obtained by plugin
$$ a_r^p = \left(u_3 r^3+u_2 r^2+u_1 r\right)2^r $$
into the complete recurrence equation giving
$$ 2^r (u_2 + r (-1 + 3 u_3)) = 0 $$
and then with $u_2=0, u_3 = \frac 13$ we have
$$ a_r^p = \left(\frac 13 r^3+u_1 r\right) 2^r $$
and the solution is
$$ a_r = a_r^h + a_r^p = c_1+\left(c_2+(c_3+u_1)r + \frac 13 r^3\right)2^r $$
or
$$ a_r = c_1+\left(c_2+c'_3r + \frac 13 r^3\right)2^r $$
NOTE
The solution for the homogeneous equation can be obtained by substituting $a_r^h = \beta^r$ into the homogeneous equation obtaining
$$ (\beta -2)^2 (\beta -1) \beta ^{r-3} = 0 $$
and then due to the double root we have
$$ a_r^h = c_1 1^r+c_2 2^r +c_3 2^r r $$