Let $a \in (-1,1)$ and $x\in {\mathbb R}$ and consider a following sequence of rational functions: \begin{eqnarray} R^{(0)}[x]&:=& \frac{x (a x-1)}{a x-x^2+x-1}\\ R^{(1)}[x]&:=& -\frac{(x-1) x^3 \left(\left(a^2+a-1\right) x^2-2 a x+1\right)}{(a x-1) \left(a x-x^2+x-1\right)^3}\\ R^{(2)}[x]&:=&(x-1) x^4 \frac{\left( -1+ (5 a-3) x+ \left(-10 a^2+5 a+10\right) x^2+ \left(10 a^3+a^2-30 a-1\right) x^3+ \left(-5 a^4-5 a^3+30 a^2+9 a-4\right) x^4+ \left(a^5+2 a^4-12 a^3-9 a^2+2 a+2\right) x^5+ \left(2 a^4+2 a^3-a^2+a-1\right)x^6 \right) }{(a x-1)^3 \left(a x-x^2+x-1\right)^5}\\ \vdots \end{eqnarray} In general we have: \begin{eqnarray} R^{(n)}[x]:=\sum\limits_{p=1}^n \frac{x^{p+1} (1-x)}{(1-a x)^p (1-(1+a) x+x^2)}\cdot (x \frac{d}{d x})^p R^{(n-p)}[x] \end{eqnarray} for $n=1,2,\cdots$. Those rational functions are Z-transforms of certain wave-forms in the time domain. In other words we have: \begin{equation} r_k^{(m)} := \left. \frac{1}{k!} \frac{d^k}{d x^k} R^{(m)}[x] \right|_{x=0} \end{equation} for $m=0,1,\cdots$ and $k\in {\mathbb N}$. Here $m$ is the order of the waveform.
Now, by decomposing the rational functions into partial fractions we have derived the following results: \begin{eqnarray} &&r^{(0)}_k = \frac{\sqrt{\Delta} \left(a^2+3 a+1\right) \sin ((k+1) \phi )}{a+3}+\left(a^2+a-1\right) \cos ((k+1) \phi )\\ \hline\\ &&r^{(1)}_k= \frac{a^k}{1-a}+\\ &&\mbox{$-\frac{\left(a^2-2\right) \cos ((k+1) \phi )}{a-1}-\frac{\left(-a^4-4 a^3+a^2+10 a-8\right) \sin ((k+1) \phi )}{(a-1) (a+3) \sqrt{\Delta}}$}+\\ &&\mbox{$\binom{k+1}{1} \left(\frac{\left(-2 a^4-11 a^3-13 a^2+9 a+10\right) \sin ((k+2) \phi )}{(a+3) \sqrt{\Delta}}+\frac{\left(2 a^4+7 a^3-a^2-11 a+2\right) \cos ((k+2) \phi )}{(a-1) (a+3)}\right)$}+\\ &&\mbox{$-\binom{k+2}{2} \left(\frac{\left(a^4+5 a^3+6 a^2-a-2\right) \cos ((k+3) \phi )}{a+3}-\frac{a \left(a^3+3 a^2-3\right) \sin ((k+3) \phi )}{\sqrt{\Delta}}\right)$}\\ \hline\\ &&r^{(2)}_k= \frac{(3 a-1) a^{k+1}}{(a-1)^2}+\frac{\left(3 a^2-3 a+1\right) (k+1) a^{k-1}}{(a-1)^2}+\frac{(k+1) (k+2) a^{k-1}}{2 (a-1)}+\\ &&\mbox{$-\frac{(3 a-1) \cos ((k+1) \phi )}{(a-1)^2}-\frac{\left(3 a^5+29 a^4+92 a^3+80 a^2-89 a-113\right) \sin ((k+1) \phi )}{(a-1)^2 (a+3)^3 \sqrt{-a^2-2 a+3}}+$}\\ &&\mbox{$\binom{k+1}{1} \left(-\frac{\sqrt{\Delta} \left(2 a^6+12 a^5+9 a^4-44 a^3-26 a^2+69 a-29\right) \sin ((k+2) \phi )}{(a-1)^3 (a+3)^3}-\frac{\left(2 a^6+16 a^5+33 a^4-22 a^3-94 a^2+31 a+97\right) \cos ((k+2) \phi )}{(a-1)^2 (a+3)^3}\right)+$}\\ &&\mbox{$-\binom{k+2}{2} \left(-\frac{\sqrt{\Delta} \left(7 a^6+54 a^5+113 a^4-34 a^3-249 a^2-29 a+107\right) \sin ((k+3) \phi )}{(a-1)^2 (a+3)^3}-\frac{\left(7 a^6+40 a^5+33 a^4-114 a^3-87 a^2+117 a+11\right) \cos ((k+3) \phi )}{(a-1)^2 (a+3)^2}\right)+$}\\ &&\mbox{$\binom{k+3}{3} \left(-\frac{\sqrt{\Delta} \left(8 a^6+43 a^5+39 a^4-84 a^3-77 a^2+52 a+14\right) \sin ((k+4) \phi )}{(a-1)^2 (a+3)^2}-\frac{\left(8 a^6+59 a^5+125 a^4+10 a^3-175 a^2-62 a+34\right) \cos ((k+4) \phi )}{(a-1) (a+3)^2}\right)+$}\\ &&\mbox{$-\binom{k+4}{4} \left( -\frac{3 \sqrt{\Delta} \left(a^6+7 a^5+15 a^4+6 a^3-11 a^2-6 a+1\right) \sin ((k+5) \phi )}{(a-1) (a+3)^2} -\frac{3 \left(a^6+5 a^5+5 a^4-6 a^3-7 a^2+2 a+1\right) \cos ((k+5)\phi)}{(a-1) (a+3)}\right)\quad$}\\ \end{eqnarray} where $\phi:=\arccos[(1+a)/2]$ and $\Delta:=3-2 a-a^2$.
To convince the reader that there are no typos I include a piece of Mathematica code which verifies the above results. We have:
Clear[RR]; Clear[RR];
RR[x_] := {(
x (-1 + a x))/(-1 + x + a x -
x^2), -(((-1 + x) x^3 (1 - 2 a x - x^2 + a x^2 + a^2 x^2))/((-1 +
a x) (-1 + x + a x - x^2)^3))};
rr[k_] := {-(1 - a - a^2) Cos[(1 + k) phi] + (
Sqrt[3 - 2 a - a^2] (1 + 3 a + a^2) Sin[(1 + k) phi])/(3 + a),
a^k/(1 - a) - ((-2 + a^2) Cos[(1 + k) phi])/(-1 +
a) - ((-8 + 10 a + a^2 - 4 a^3 - a^4) Sin[(1 + k) phi])/((-1 +
a) (3 + a) Sqrt[
3 - 2 a -
a^2]) + (1 +
k) (((2 - 11 a - a^2 + 7 a^3 + 2 a^4) Cos[(2 + k) phi])/((-1 +
a) (3 + a)) + ((10 + 9 a - 13 a^2 - 11 a^3 -
2 a^4) Sin[(2 + k) phi])/((3 + a) Sqrt[3 - 2 a - a^2])) -
1/2 (1 + k) (2 +
k) (((-2 - a + 6 a^2 + 5 a^3 + a^4) Cos[(3 + k) phi])/(
3 + a) - (a (-3 + 3 a^2 + a^3) Sin[(3 + k) phi])/Sqrt[
3 - 2 a - a^2])};
l1 = Simplify[
Table[D[Together[RR[x][[1 + n]]], {x, k}]/k!, {n, 0, 1}, {k, 0,
10}] /. x :> 0];
l2 = Simplify[
FunctionExpand[
Table[rr[k][[1 + n]] /. phi :> ArcCos[(1 + a)/2] , {n, 0, 1}, {k,
0, 10}]]];
MatrixForm[{l1, l2}]
l1 - l2
After running the code above we get the following output:
and the left hand side matches the right hand side as it should be.
Now, if we take higher orders ($m\ge 2$) then both the rational functions and the corresponding wave-forms in the time domain become to big to be written down not to mention to be dealt with.
My question is therefore is there any other way of inverting the Z-transforms in question ? In other words do we always have to resort to partial fraction decomposition if we want to compute the $k$th derivative of a rational function at the origin, or is there some other way of deriving the result?
