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How can I derive the inverse z-transform of:

$$\sqrt{\frac{1-a^2}{1-\frac{a}{z}}}$$

If Maple is not the way, how to derive manually?

With Maple code I encounter some problems

seriestoratpoly(series(simplify(LaguerreL(5, a, x)),x,6),['egf']);

restart;
alpha := 0;
s := x;
tau := alpha + 1 - x;
source := s*Diff(y(x),x$2)+tau*Diff(y(x),x)+n*y(x)=0;
sol := dsolve(source,y(x),method=laplace);
zsol := ztrans(sol,x,z);
convert(zsol, ratpoly);

sol := y(x) = _C1*invlaplace((_s1-1)^n*_s1^(-1-n), _s1, x)
zsol := ztrans(y(x), x, z) = _C1*invlaplace((_s1-1)^n*_s1^(-1-n)*z/(z-1), _s1, x)
Error, (in convert/ratpoly) expecting a Laurent series or a Chebyshev series

what are _C1 and _s1?


> with(gfun); ratp := seriestoratpoly(dsolve(source, y(x), series), ['egf']);
Error, (in gfun:-seriestoratpoly) not a series, y(x) = _C1*(series(1-n*x+((1/4)*(n-1)*n)*x^2-((1/36)*(-2+n)*(n-1)*n)*x^3+((1/576)*(-3+n)*(-2+n)*(n-1)*n)*x^4-((1/14400)*(n-4)*(-3+n)*(-2+n)*(n-1)*n)*x^5+O(x^6),x,6))+_C2*(ln(x)*(series(1-n*x+((1/4)*(n-1)*n)*x^2-((1/36)*(-2+n)*(n-1)*n)*x^3+((1/576)*(-3+n)*(-2+n)*(n-1)*n)*x^4-((1/14400)*(n-4)*(-3+n)*(-2+n)*(n-1)*n)*x^5+O(x^6),x,6))+(series((1+2*n)*x+(-(1/2)*n+1/4-(3/4)*(n-1)*n)*x^2+((1/36)*(n-1)*n+(1/36)*(-2+n)*n+(1/36)*(-2+n)*(n-1)+(11/108)*(-2+n)*(n-1)*n)*x^3+(-(1/576)*(-2+n)*(n-1)*n-(1/576)*(-3+n)*(n-1)*n-(1/576)*(-3+n)*(-2+n)*n-(1/576)*(-3+n)*(-2+n)*(n-1)-(25/3456)*(-3+n)*(-2+n)*(n-1)*n)*x^4+...

do not know _C1 and _C2
if assume _C1 and _C2 = 1

with(gfun);
ratp := seriestoratpoly(subs(_C1=1,subs(_C2=1,dsolve(source, y(x), series))), ['egf']);
output FAILED
share|improve this question
    
Is n to be taken as a positive integer? –  acer Jul 22 '12 at 3:35
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