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These are replies submitted by vv

It would be nice to tell us how the equation was obtained.

Here is a (simpler) parametric one (Klimek G.&M. - Discovering curves and surfaces with Maple, Springer 1997.)

      x=-Pi .. Pi, y=-Pi .. Pi, color=[sin(x+1)+cos(y), sin(y), 0.3],
      light=[85, 50,  0.6, 0.8, 0.2], orientation=[47, 55],
      style=patchnogrid, axes=none);


Your function of Pe also depends on several variables and some arbitrary functions!
How can you imagine that the limit could be computed?

@mmcdara For me, it's not about assume or assuming. 
For the same assumptions, assumptions := a>0, b>0, a*b < 1:
gives the same results.
Also, in Maple 2022, assumptions := a>0, b>0, b < 1/a: 
==> all answers are true (But it seems that in 2023, one answer is FAIL !!).

So, everything seems to be about weakness!

@Carl Love  Unfortunately the assume facility is very weak here.

# _EnvTry := true;  # useless
assume(a>0, b>0, a*b<1);
is( (1 + sqrt(1-a^2*b^2))/(a*b) >0 ); # FAIL   ?
is( (1 + sqrt(1-a*b))/(a*b) >0 );     # FAIL   ?
is( (1 + sqrt(1-a^2*b^2)) >0 );       # false ??
is( (1 - sqrt(1-a^2*b^2)) >0 );       # false ??
is( (1 - sqrt(1-a*b)) >0 );           # true
is( (1 + sqrt(1-a*b)) >0 );           # true


@nm The general solution of the ode  y'' + y = 0  is y = c1*sin(x) + c2*cos(x)  but also
 y = c1*(2*sin-7*cos(x)) - c2*(sin(x)-3*cos(x)).

@dharr a) A square complex matrix C has not a square root iff its Jordan form has a Jordan block having dimension > 1 and eigenvalue 0.

b) For C as in the file (40x40), M := evalf[n](MatrixPower(C, 1/2)), eps = Norm(M^2-C)  then:
n=14-->eps=0.003;  n=16-->eps=2e-4;   n=18-->eps=1e-6;  n=25--> eps=2e-13;  n=30-->eps= 2e-15.

Use odetest to check the solution. Yes, it's a bug.

ode:=[diff(x(t), t, t) + 2*diff(x(t), t)/sqrt(L*C) + x(t)/(L*C)];
s5 := dsolve(ode, [x(t)]);
odetest(s5, ode);
s6 := dsolve(ode, [x(t)], method = laplace);
odetest(s6, ode);   # bug


@Christian Wolinski The problem is that {a>0, b>0, c>0, d>0} is far for being equivalent to {e1>0, e2>0, e3>0, e4>0}.

G is the centroid of the (inscribed) quadrilateral P1P2P3P4.
E is the Euler center of  the quadrilateral P1P2P3P4  defined as the point in the intersection of the four Euler circles of the triangles P1P2P3, ..., P4P1P2. (E exists!).

Note that OG = (OP1+OP2+OP3+OP4)/4, OE = (OP1+OP2+OP3+OP4)/2,  OPk  being vectors.

@Alex0099  1. The order of summation does not matter, the summand being symmetric.

2. The proof could be very tricky. Probably here the Residue Theorem should be useful.  

@Thomas Dean  In Windows, after a restart everything is OK. Sorry, I don't use Linux.

@emendes  procedure is a type, so, its use is covered.

AFAIK the Draghilev method is to find a natural parametric representation of a curve given as an intersection of n surfaces in (n+1) unknowns. That is what Rouben did, but using direct computations.

(I was never able to find the original article of Draghilev, but there are many places -- even in this site -- where the method is used).   

@lcz Replace:

first:= fsolve((f(t)-x0)^2+(g(t)-y0)^2=0.02^2, t=0..1);
last := fsolve((f(t)-x3)^2+(g(t)-y3)^2=0.02^2, t=0..1);


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