@Markiyan Hirnyk 

I answer with great pleasure when the person who asked the question does not understand the solution. And I did it each time. Now I do not know what to think.

Is it not clear that

-RootOf(_Z^2*s-s+1)*s/(s-1)

equals

+-  sqrt(s*(s-1))/(s-1)

?

Must I elaborate? Must I prove that Maple is correct? Are you serious?

@Markiyan Hirnyk 

Why mine and not yours?

But more important, why must we waste our time for such a simple problem?

Just apply allvalues as you did in your post.
Or even simpler. use the definition of RootOf.

@Markiyan Hirnyk

@Markiyan Hirnyk 

It was a simple typo. The principle was obviously correct.

[a = s/r, b = -r/s, c = r, r^2-s^2+s=0 ];
eliminate(%,r);

solve(%[2], {a,b,c});

which is the same as the original.

@Markiyan Hirnyk 

You are as usual ready to object without reason.

The correct call is

solve({a*c-s, b*s+c, -c*s^2+c^2+s}, {a,b,c});

which gives the expected result.

@Nikol 

F:= unapply( rsolve({f(n) = .5*f(n-1)+.5*f(n+1)}, f(n)),  n);
solve( {F(a)=0, F(0)=1}, {f(0),f(1)} );
simplify(eval(F(n),%));

@Preben Alsholm 

Ok, ler's call it semi-bug or quasi-bug :-)

But unfortunately

rsolve(f(n+1)=f(n)+c, f(n));

Congratuations, you have found a bug!

The correct answer should be of course f(n) = 1 - n/6.

It seems that rsolve fails for arithmetic progressions!

E.g. also:

rsolve(f(n+1)=f(n)+c, f(n));

Strange! The bug seems to be new.

@Axel Vogt 

0.002744071672397914633171931780975747758...

And symbolically:

@Markiyan Hirnyk 

OK, can you obtain more than 5 corect digits this way?

Edit

exact = 4.0148672762003073023634...*10^9

@Markiyan Hirnyk 

It works indeed but with very poor accuracy: only 3 correct digits in this case.

For this, f must be a formal powerseries (not a series in the Maple sense, which is a truncation, containing O(...)).

So,

f:=convert(exp(k*t)*cos(w*t),FPS,t,n);

g:=convert(f/exp(k*t),FPS,t);

After several conversions, it it possible to simplify g to
But probably for more complicated functions this will not work.

@Preben Alsholm 

This behavior of the roots is known.

If the coefficients are independently and identically distributed with a mean of zero,
the complex roots are on or close to the unit circle.

See:
https://en.wikipedia.org/wiki/Properties_of_polynomial_roots

@baharm31 

Yes, you have found one continuous solution. implicitplot suggests that other solutions exists. They can be obtained by combining distinct branches of the RootOf. You might want to plot all the branches (after giving values to parameters) and inspect the intersection points; a rather tedious task because the RootOfs will depend on x.

Note that your equation can be reduced to a polynomial one, so that the Lagrange expansion applies, see
https://en.wikipedia.org/wiki/Lagrange_inversion_theorem

but this gives only local solutions.