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These are replies submitted by vv

@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);


@Stretto I don't see how you could make ArgK to work. As it is, after correcting the Is problem (see acer's comment), it would work only when computing the values in an array, but not randomly (in any language!). I'd suggest to take the dsolve approach.

It is interesting to reformulate eq2  and see what happens.




eq2:=int(diff(y(x),x)*x,x)+int(y(x),x) +1 = 0;

int((diff(y(x), x))*x, x)+int(y(x), x)+1 = 0


u(x)+v(x)+1 = 0,
diff(u(x),x) = diff(y(x),x)*x,
diff(v(x),x) = y(x)


{u(x) = _C2*ln(x)+_C1, v(x) = -_C2*ln(x)-_C1-1, y(x) = -_C2/x}


odetest(sol, sys2);



eval(eq2, sol); # due to integration constants!

1 = 0



@Carl Love  A and B having polynomial entries, A.B = Id ==> A invertible, that's all.
BTW, your potential objection is also valit for the provided answer, if A = <p,0;0,1/p>, where det(A) gives 1. But if we disect such a simple problem, everything becomes complicated (more or less artificially).

Actually, you solved the problem! For a)  it's enough to compute  simplify(A . B) , obtaining the unit matrix, without any assumption on p. (Of course it's very useful to know what a singular matrix is.)

@fhn2022 A smooth discontinuous function? Analytic? In mathematics such things do not exist
[f analytic ==> f smooth ==> f continuous].

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