@Fancypants Initially, all Lij were known. Now you say that some are not known. It's not the same thing!
In your picture, P0, P2, P3 seem to be collinear. If true, the number of needed Lij known could be decreased by 1. In principle this number must be >= 10 for a unique solution, but not any 10 will guarantee the uniqueness. Anyway, in this case a symbolic solution is improbable. If your problem is numeric you have the possibility to use DirectSearch to solve the system.

@michalkvasnicka 

If you try to solve f = 0, where

f := x^2 + 2*x + floor(x^2 + x)-25;

Mathematica answers:  This system cannot be solved with the methods available to Solve.

In Maple do (using the presented method):

seq(solve({eval(f,[floor(x^2 + x)=n]), n<=x^2 + x, x^2 + x<n+1}), n=-1..20);

      {x = -1 + sqrt(15)}

@Carl Love

Yes, thank you. Actually I have an enhanced version. It was written immediatley after the module version, but I forgot to replace it. I did it now.

@vs140580  I don't understand what you are trying to obtain. After relabelling the graph looks as in your picyure. I use GraphTheory for computations, not for drawings.
 

@Kitonum It must be noted that identify usually must be used with higher precision and works only for very simple results.
E.g. for eq := x^2+floor(x)-3*Pi  it fails.
 

@Carl Love   Agree but without brackets.

@DarkMath 

Maple does not document when Dirac(f(x))  is defined; Maple accepts f(x)=x^2-1  but rejects f(x)=x^4-1. It works formally, that's why the user must be careful. The usual conditions (f smooth and f' <> 0 at the roots of f) could be difficult to verify.

@DarkMath

The situation is much more delicate and fundamental. For example f(x) = exp(-1/x^2)*sin(1/x)  (with f(0)=0) is very smooth but int(Dirac(f(x)),  x=-1 .. 1) is problematic. Should it be infinity?
Note that in this context int  is not a "regular" integral.

What exactly represent your computations? What is the relation with GF(2^m)? 
I see that f_m mod 2 is irreducible in Z_2[t], so the field  Z_2[t]/(f_t mod 2) is isomorphic to GF(2^m), but why do you take mod 2^N?

@Carl Love 

Note also that with OP's notation, any expression can be written mathematically as a*b in infinitely many ways.

@janhardo 

What you have is a calculus problem, not a Maple one, because Maple computes your series
and you want a solution by hand.
So, you need to read a good Calculus textbook (+some exercises) to be able to manage such tasks  [***].
I'll show you a solution, but it will be useless without [***].

restart;
S:= Sum((-1)^k/((k + 2)*k!), k = 0 .. infinity);
#define
f := Sum(x^(k+1)/((k + 2)!), k = 0 .. infinity);
# it should be obvious that  x*f = exp(x) -1 - x, so
ff := (exp(x) -1 - x)/x; # = f
f1 := diff(f,x); # ==> S = tha value of f1 at x=-1
'S' = eval(diff(ff,x), x=-1);

@janhardo 

It's a standard iterated integral. Use:

int(1/(1-x), [x=0..y, y=0..z]) assuming -1<z,z<1;

[strangely, abs(z)<1 does not work now].
You can also compute int(int(1/(1-x),x),x) + c1*x+c2; and adjust the constants such that f(0)=f'(0)=0.
 

@janhardo 

@janhardo 

It is not "included"; it can be "deduced from".

Actually, the standard metod is the following:

You have the series S:=Sum((-1)^(n+1)/(n*(n+1)),n = 1 .. infinity):

Now you associate to it the power series

f := x -> Sum(x^(n+1)/(n*(n+1)),n = 1 .. infinity):

So, S = f(-1).

Now observe that f''(x) = Sum(x^(n-1), n = 1 .. infinity):

But this is a geometrical series, so f''(x) = 1/(1-x).

It remains to integrate twice (and obtain f in terms of ln) and then make x=-1.

@janhardo 

This makes sense if you are doing the computations by hand (but then you don't need Maple), of if the Maple version is not able to compute the series directly (which is not your case -- Maple 2020).

BTW, the series can be computed by hand easier starting from sum((-1)^(k-1)/k, k=1..infinity) = ln(2)
or from limit(sum(1/k, k=1..n) - ln(n), n=infinity) = gamma