Why not simply:

Heaviside(0):=0: Heaviside(0.0):=0.0:

?

@Axel Vogt 

I have also obtained a symbolic result using some conversions such as
Expand(%); Change(%,xp=t^(1/3));
subsindets(%,'specfunc(anything,int)', Split, [infinity]);

It's a pitty that one has to do such acrobatics instead of a simple convert(%,GAMMA).

use e.g.

'f()' $i=1..5;

@acer 

@H-R 

A shorter formatted proof:

by definition, for any branch.

@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