vv

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These are replies submitted by vv

@whtitefall 

Use:
l:=1/2;
evalf(DD(1)); 

That's why I have suggested to define R := (x,h) -> ...  etc

 

@one pound 

It is useful for asymptotics; my objection was for using "=" instead of "~".

@Adam Ledger 

f(oo) = oo implies f(k)>0 for k large enough, so your question is answered.
A bound can be computed using x -1 < floor(x) <= x; ==> S subset {1,...,527}. Then use Maple.

 

In (I) you use divisibility in Q* which is unusual; anyway, the third implication is false.

For (II)  a counterexample is expected.

@tomleslie 

The 'elegant' approach appears in

Blumenthal L.M. - Theory and Applications of Distance Geometry. Chelsea, 1970

theorem 43.1 (due to Schoenberg).

 

Please note that I also have a mistake in the "Clarification" reply. The fM procedure there is wrong. It always returns `false`.
(I forgot to use the transform from IsTetra).
The correct one is:

 

fMok:=proc(a,b,c,x,y,z) #AB=c,AC=b,BC=a,DA=x,DB=y,DC=z
<2*x^2, -c^2+x^2+y^2, -b^2+x^2+z^2; -c^2+x^2+y^2, 2*y^2, -a^2+y^2+z^2; -b^2+x^2+z^2, -a^2+y^2+z^2, 2*z^2>;
LinearAlgebra:-IsDefinite(%, query = 'positive_definite')
end proc:

 

 

So, your positive_definite version is basically correct. I said that it's wrong because I missed the different notations.
Namely, if ABCD is the tetrahedron then you used AB=cbar, AC=bbar, BC=abar, DA=a, DB=b, DC=c. Mine were reversed.

So, apologies  for this!

Also, IsTetra in my answer is OK (because the notations are clear) but istetra should be:

 

istetraok:=proc(a,b,c,x,y,z) #AB=c,AC=b,BC=a,DA=x,DB=y,DC=z
is(`and`(       #((
a>0,b>0,c>0,x>0,y>0,z>0,
2*max(a,b,c)<a+b+c,
2*max(a,y,z)<a+y+z,
2*max(x,b,z)<x+b+z,
2*max(x,y,c)<x+y+c,
LinearAlgebra:-Determinant(
<0,x^2,y^2,z^2,1; x^2,0,c^2,b^2,1; y^2,c^2,0,a^2,1; z^2,b^2,a^2,0,1; 1,1,1,1,0>) >0
))
end:

 

 

Download distance-geometry.mw

@kuwait1 

It is easy to see that the integrand always has a pole in [0,Pi] for n::posint.

@tomleslie 

That was the purpose of the example. There is no tetrahedron!
(A triangle cannot have the sides 1,1,3).
You should turn both disagree into agree :-)

 

@Carl Love 

I know, but S[n] etc have huge values and for larger n the situation is worse.

Maybe the approximation is valid only near t=0. It's a kind of Picard iteration.

The "homotopy" S_n approch leads to something like:

restart;
N:=5e7;
beta:=0.00001;
mu:=0.02;
sigma:=45.6;
gamma1:=73;

S[0]:=t -> 12500000;
E[0]:=t -> 50000;
J[0]:=t -> 30000;
R[0]:=t -> 37420000;

for n from 0 to 2 do
S[n+1]:= unapply( S[n](t)-int(D(S[n])(w)- mu*N+beta*S[n](w)*J[n](w)+mu*S[n](w), w=0..t), t):
E[n+1]:= unapply( E[n](t)-int(D(E[n])(w) - beta*S[n](w)*J[n](w)+mu*E[n](w)+sigma*E[n](w), w=0..t), t);
J[n+1]:= unapply( J[n](t)-int(D(J[n])(w)- sigma*E[n](w)+mu*J[n](w)+gamma1*J[n](w), w=0..t), t);
R[n+1]:= unapply( R[n](t)-int(D(R[n])(w)-gamma1*J[n](w)+mu*R[n](w), w=0..t), t);
od;

plot(S[3], 0..70);

but does not seem to approximate the solution.

@tomleslie 

1. If you don't add the triangle inequalities, your seq test does not give a correct answer.
You can check this for  test:=[1,1,3,5,1,3];

2. The positive_definite approach is correct only for the Cayley-Menger matrix; it does not work for your  McCrea.
So the correct procedure is

fM:=proc(a,b,c,d,e,f) # a>0 etc
<0,a^2,b^2,c^2,1; a^2,0,f^2,e^2,1; b^2,f^2,0,d^2,1; c^2,e^2,d^2,0,1; 1,1,1,1,0>;
LinearAlgebra:-IsDefinite(%, query = 'positive_definite')
end;

So, it's not a matter of excessive bother.

Best regards,
V.A.

 

S_0(t)  etc  are needed. Probably you mean  S_0(t) = 12500000 etc.

Why don't you write your equations in Maple to be copy-pasted? Do you prefer to be typed by the forum members?

@digerdiga 

Maple 2016: For infinite sums, Maple is now more careful regarding potentially divergent parametric sums.

@digerdiga 

Mathematically the equality is not true without restrictions. The default option is now formal=false.

@nm 

But e.g. dsolve cannot solve the simple:

dsolve(diff(y(x),x)+Int((y(x)),x=0..x)=x,y(x));

and also

dsolve(diff(y(x),x)+Int((y(x)),x=0..1)=x,y(x));

@acer 

For an ODE, y(x) under Int should be (probably) also rejected.

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