f:=series( exp(k*t)*cos(w*t),t=0);
series(f/exp(k*t),t=0);

For the 2nd sheet you should use Eigenvectors which is designed exactly for this purpose.

A:=eval(q,omega=0);
C:=eval(A-q,omega^2=1);
Eigenvectors(A, C);

(the eigenvalues correspond now for lambda = omega^2)

If p is not available, then use:

f:=3*(r/sqrt(a)) + (r/sqrt(a))^2:
v:= r/sqrt(a):
diff(f,r)/diff(v,r);

You must change Y with Y(r1,r2,r3) and it will work.
Also, the inverse transforms are not necessary (being very simple), so it's enough:

PDEtools[dchange]({r1=v/2+w/4, r2=u/2+w/4, r3=u/2+v/2},  del_1 + del_2);

The system is very special. If we try solve, we get:

sys := {-6-4*y-x-(1+y)*x+sqrt((4*(1+y))*(2+x)*(4+2*y+x)+(-(1+y)*x+2+x)^2), (2*(4+2*y+x))*(1+y)-(1+y)*x+2+x+sqrt((4*(1+y))*(2+x)*(4+2*y+x)+(-(1+y)*x+2+x)^2)-(2+y)*(-(1+y)*x+2+x+sqrt((4*(1+y))*(2+x)*(4+2*y+x)+(-(1+y)*x+2+x)^2))};

solve(sys);

This means that the system reduces to an identity in a certain region of the plane (x,y); a warning would have been very useful!.
Unfortunately, in such cases:
- Maple in unable to find this region
- fsolve does not work.

So, we will have to maniputate the system, or help Maple.

simplify(sys) assuming x*y+2*x+4*y+6>=0;
                              {0}
simplify(sys) assuming x*y+2*x+4*y+6<0;

solve(sys) assuming x*y+2*x+4*y+6>=0;

solve(sys) assuming x*y+2*x+4*y+6<0;

Conclusion: the system reduces to identities in the region x*y+2*x+4*y+6>=0;
(it is the "interior" of a hyperbola):

In the "exterior" of this hyperbola, the solution is on a curve (also a hyperbola). You may want to plot it.

For the function x : [0,1] --> R you probably want the Fourier expansion on each of the N intervals [(n-1)/N, n/N], 1<=n<=N.

In this case, X(m,n) must be the Fourier coefficient associated to the restriction of x() to this interval.

v:=unapply(H*sin(w*t),t):
solve(v(t)=0,t,allsolutions);

#assuming w>0, the solutions in [a,b] are
sol:=%$_Z1=ceil(a*w/Pi)..floor(b*w/Pi):

eval(sol, [w=10,a=0.5,b=2]);

evalf(%);

v[i]:=unapply(H*sin(w*t+phi[i]), t);

(in this way, i will be evaluated)

As you know, the integral (antiderivative) is defined up to an additive constant; this constant coud be even complex.

In your case:

f:=int(u/(1-u),u);

g:=-u-ln(1-u):

combine(g-f) assuming u>1;

combine(g-f) assuming u<1;

simplify(%);

As you see, f and g differ by a complex constant and this constant depends on the domain.

Notice that for u>1 it is more convenient Maple's answer.

 Edit.

In calculus, the integral appears usually as -u-ln|u-1|
It is better suited for real u, but has the disavantage of being wrong if u is complex.

My previous assertion (the hint) is not correct.
One can only prove that if two of the three pairs of polynomials have a common root then the third pair also has a common root. But the three polynomials do not necessarily have a common root.
In my (incorrect) solution, Maple splitted the problem in 15 sub-cases which had to be verified, which I did, but being too lazy, only for half of them. Unfortunately, one of the other cases provided a counterexample for a,b,c.
The exact expressions for a,b,c are very complicated (many RootOfs), so I shall give only the approximations.

p:=a*x^11+b*x^4+c; q:=b*x^11+c*x^4+a; r:=c*x^11+a*x^4+b:
Digits:=20:
abc:=[a = -0.23463440749094831025+0.20142703184123511503*I,
           b = -0.45433251158473825754+0.92621981907035507390*I,
           c =  1]:

commonz:=proc(p,q,eps:=10^(-Digits+4))
local zp,zq,a,b,res:=NULL;
zp:=[fsolve(p,complex)]; zq:=[fsolve(q,complex)];
for a in zp do for b in zq do
if abs(a-b)<eps then res:=res,a fi;
od;od;
[res]
end:  # find the common approx roots of two polynomials

pp:=eval(p,abc): qq:=eval(q,abc): rr:=eval(r,abc):
zpq:=commonz(pp,qq); eval(pp,x=zpq[1]),eval(qq,x=zpq[1]);
      [0.97952994125249449394 + 0.20129852008866007914 I]
                     
zpr:=commonz(pp,rr); eval(pp,x=zpr[1]),eval(rr,x=zpr[1]);
       [0.34730525284482028554 - 0.93775213214708045843 I] 
     

zqr:=commonz(qq,rr); eval(qq,x=zqr[1]),eval(rr,x=zqr[1]);
      [0.52896401032696245736 + 0.84864425749475095047 I]
    
                 
So, any two polynomials have a unique common root but they are distinct!

It can be shown a little more, namely:
If two  (of the three) pairs of the polynomials p,q,r have a common root then all the three polynomials have a common root.
[Or, in an other formulation: if there exist complex x,y  such that p(x)=q(x)=0=p(y)=r(y)  then there exists z such that p(z)=q(z)=r(z)=0].

Note that a Maple solution takes almost the same amount of time as a "by hand" one.

I'll give only a hint.
There are two cases.
  a) If {a,b,c} is (a multiple of) the set {z: z^3=1} (i.e. roots of the unity of order 3), then the three polynomials are equivalent (they are constant multiples of a same polynomial) and there is nothing to prove.
  b)  Otherwise, show that a common root of the three polynomials is a root of the unity of order 3.

Just use simplify to get rid of the negative terms.

But a good idea would be to tell us what mathematical result do you want to use in order to obtain the convergence.

You have a sum of a series, not an equation.

The sum cannot be computed exactly, but you can approximate it with any precision: just use

evalf[20](sum((factorial(n)/factorial(2*n))^n, n = 1 .. infinity));

to get 20 decimals.

Note that your series converges very fast: the first 4 terms are enough for 20 correct digits!

I'd say that the solution is straightforward.

a*(x-m)+b*(y-n)=0 is a general equation of a line thru [m,n]; it can be normalized by a^2+b^2=1
and one may suppose f(0,0)>=0 (because f=0 and -f=0 produce the same line).

The distance from (0,0) to the line is exactly |f(0,0)|/sqrt(a^2+b^2) = f(0,0). So, the tangency condition is just f(0,0)=r. That's all.

Happy New Year!

Why not the exact solution?

sol1:=dsolve({ode,ics}):
plot(rhs(sol1),x=0..1);