@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.

You use here the term "solution" for what in mathematics is called "continuous selections" for a set-valued function
x |--> {y : f(x,y)=0}; this also appear in implicit function theorems.

solve is not designed for such tasks. It finds (in principle) the set {y : f(x,y)=0} and it is on your side to "assemble" the selections. Note that this could be a difficult task and the number of selections could be infinite.

On the other side, implicitplot simply plots the set of points {(x,y) : f(x,y)=0} and it is your eye which detects the selections. This plot could then be used to effectively assemble some solutions (if they actually exist, because the plot could be misleading).

Edit. Example:

How many continuous solutions y = y(x) do you detect for (y - sin(x))*(y - cos(x)) = 0, 0 <= x <= 4*Pi ?
(I am sure that you will miss some of them.)

I just suggested that if you want eigenvectors than it is better to use Eigenvectors instead of solving the characteristic polynomial. This is the case with your 1.mw.

Your coefficiens are in the range 10^(-25) .. 10^36 (approx). The degree of the polynomial is 50.
This will produce huge roundoff errors.
As acer said, Digits must be increased, but only if you are sure that the initial coefficients are very accurate [their precission should be also increased].
Otherwise the result will be useless.
Usually in such cases the function must be pre-processed and normalized; and an error estimate is recommended.