How would you solve for the closed form solution of a(n) given the general form of the third order linear homogenous recurrence relation with real constant coefficients.
$a_n=Pa_{n-1}+Qa_{n−2}+Ra_{n−3}$
with the initial terms of a_1, a_2, and a_3
given that the roots of the characteristic equations have
1)two repeated roots and a real root
2)three repeated roots
(can you give answers for both cases please)
When I search the web I get these results
1)$a_n = nAx_1^n + Bx_1^n + Cx_2^n$,for the case when there are two repeated roots
and
2)$a_n = n^2Ax^n + nBx^n + Cx^n$, for the case when there are three repeated roots
Can anyone help derive the closed form of each case in order to get such results?
Please help
I'm new to the system so i didn't quite know how to get the symbols right (sorry) if you're uncertain about anything please ask
Let the generating function $$g(x)=\sum_{n=0}^\infty a_nx^n$$ with the recurrence $a_n-pa_{n-1}-qa_{n-2}-ra_{n-3}=0$
Now consider $$(1-px-qx^2-rx^3)g(x)=a_0+(a_1-pa_0)x+(a_2-pa_1-qa_2)x^2=A(x)$$ Note that all the other terms go to zero because of the recurrence. We then have $$g(x)=\frac{A(x)}{(1-px-qx^2-rx^3)}$$
$A(x)$ is quadratic (and we have an explicit expression for it). If the denominator factors as $(1-sx)^2(1-tx)$ we have the partial fraction decomposition$$g(x)=\frac B{(1-sx)^2}+\frac C{1-sx}+ \frac D{1-tx}$$ [$B,C,D$ are constant]
We expand using the binomial theorem and equate coefficients - the coefficients involving $n$ come from the quadratic factor in the denominator.
If the denominator factors as $(1-ex)^3$ we have the partial fraction decomposition$$g(x)=\frac B{(1-sx)^3}+\frac C{(1-sx)^2}+ \frac D{1-sx}$$ The cubic factor gives the $n^2$ factor which appears as a coefficient in the expression for $a_n$.
Note: this was adjusted in the light of Brian's comment, which highlighted a careless basic error in the original text.