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These are replies submitted by vv

@Tokoro For n=5, R is the same R=8/sqrt(15) and the edges of the 5-gon: 2,3,2,3,2.

@Tokoro  I had in mind the existence of any number of points at integer distances on a circle or sphere (not a trivial result). The minimal radius depends of course on the dimension. But you could have formulated the problem in R^d in the first post.

@nm You can simply define then

Dy := t -> D(y)(t);
D2y := t -> (D@@2)(y)(t);

Another approach would be to write a proc, converting (via readlineode.mpl to trueode.mpl, by replacing y'(u) with D(y)(u) etc.

Why don't you add this version to the previous very similar one? Probably a more strict admin will delete this.
You have received answers to the previous question, but you seem to ignore them.
Note that the answer to the circle variant is automatically an answer to the sphere one!


@Carl Love The distance between any pair of points should be integer!

@bstuan t = 3 is not an assignment; it is an equation and the variable t remains free.
The assignment operator is :=.
So, only after t := 7;  the exprassion 10*t+5; will be evaluated to 75.

@tomleslie The situation is even worse because both solutions are wrong; they are valid only for b<=0.

sys:=[a + d = 0, -c + d = 0, a - b = sqrt(a^2 + b^2 + c^2)]:
sol:=solve(%) assuming real:
eval(sys,sol[1]), eval(sys,sol[2]) ;

        [0 = 0, 0 = 0, -b = (b^2)^(1/2)], [0 = 0, 0 = 0, -3*b = 9^(1/2)*(b^2)^(1/2)]

G := 0.04361098108*x^2 + 0.4810001561*x*y + 1.326278064*y^2 - 0.7320831383*x - 2.656083763*y + 1;






A,B,C := coeff(G,x^2), coeff(coeff(G,x),y), coeff(G,y^2);
a,b,c := coeff(g,x^2), coeff(coeff(g,x),y), coeff(g,y^2);

0.4361098108e-1, .4810001561, 1.326278064


2143/49139, 9253/19237, 19613/14788


4*A*C-B^2, 4*a*c-b^2=4.*a*c-b^2

0., 10343424/67228040758449227 = 0.2e-9


g:=subs(c*y^2=c1*y^2, g);





_EnvHorizontalName := 'x': _EnvVerticalName := 'y':



Vertex = [-1.315442026, 1.256440837]


Focus = [-1.271192031, 1.248416783]




Download parab-rat.mw

@Preben Alsholm I mean: cannot simplify to  2*Pi/(p*sin(Pi/p))

Edit. Converting first to GAMMA it works.

@Preben Alsholm A human would reduce the integral to a Beta function using the change of variables x^p = t.
Unfortunately, Maple cannot compute it. 
Actually, including the assumption p>1 (mandatory for convergence), the verbose int finds this:

[cook = 2*Beta(1/p, -1/p + 1)/p, FAILS = (distribution, piecewise, series, o, polynomial, ln, lookup, ratpoly, elliptic, elliptictrig, meijergspecial, improper, asymptotic, ftoc, contour), ftocms = 2*GAMMA((p + 1)/p)*GAMMA((p + 1)/p - 2/p), meijerg = 2*Pi/(p*sin(Pi/p))]

@Carl Love Thank you. I was aware that the proc can be improved but I have chosen the simplicity, and anyway a Dynkin system is used mainly for infinite families and in this case Maple is useless.
My opinion is that a Maple code implementing a more complicated algorithm (not the case here) should be presented in two flavours: a simple one and an optimized (speed and/or memory) version.

@JAMET But why?

@dharr Yes, it's a group but the group operation is symmetric difference, not union.

However, the group structure is not enough to obtain the algebra; we need the ring having  "+" = `symmetric difference` and "." = `intersection`.

You are confusing a Dynkin system with a monotone class.

In your example it is possible because the space is finite (so we actually have an algebra instead of a sigma-algebra).
We just need a procedure which starts with the family C and takes finite unions of finite intersections of the sets in C and their complements,  until it stabilizes. 

Unfortunately such a proc is not very useful because the resulting (sigma)algebra is in general HUGE.

BTW, for your example X := {1, 2, 3}, C := {{1}, {2}},  the algebra generated by C is obviously the powerset P(X) of X (i.e. it is maximal, containing 2^3 sets). 

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