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Christian Wolinski

MaplePrimes Activity


These are replies submitted by Christian Wolinski

@smithss What you describe is a permutation of colors on lines x=1, x=2, ... Collect all the colors on the line into a list and use combinat[randperm] or combinat[permute], whichever you prefer.

@Rouben Rostamian  You meant: 

return  z((-x+sqrt(3)*y)/2,(-sqrt(3)*x-y)/2);

@Carl Love I used gfun. Nice form, thank you.

Also, more general answer:
[(1/2*I*(1+3*X(0)+2*Y(0))+1/2*X(0)+1/2)/GAMMA(k+1)*(-I)^k+(-1/2*I*(1+3*X(0)+2*Y(0))+1/2*X(0)+1/2)/GAMMA(k+1)*I^k-1/GAMMA(k+1), (1/2*I*(-2-5*X(0)-3*Y(0))+1/2*Y(0)-1/2)/GAMMA(k+1)*(-I)^k+(-1/2*I*(-2-5*X(0)-3*Y(0))+1/2*Y(0)-1/2)/GAMMA(k+1)*I^k+1/GAMMA(k+1)]
 

General solution is:

G:=[proc (n) options operator, arrow; (((-9/2+9/2*(-1)^n)*RootOf(_Z^2+1)+3/2*(-1)^n+3/2)*RootOf(_Z^2+1)^n-1)/GAMMA(n+1) end, proc (n) options operator, arrow; ((15/2-15/2*(-1)^n)*RootOf(_Z^2+1)*RootOf(_Z^2+1)^n+1)/GAMMA(n+1) end];
map(evala@G, [$0..10]);

@Carl Love Which Maple versions support this form?

@C_R Here is a formula slightly better than the other:

plot((2^(1/2)*EllipticF((1-1/(sin(x0)+1))^(1/2)*2^(1/2),1/2*2^(1/2)*(sin(x0)+1)^(1/2))), x0 = 0 .. Pi/2, view = [0 .. Pi/2, 0 .. 10]);
this is:
Int(1/sqrt(sin(x0)-sin(x)),x=0..x0) = 2^(1/2)*EllipticF((1-1/(sin(x0)+1))^(1/2)*2^(1/2),1/2*2^(1/2)*(sin(x0)+1)^(1/2));
 

@Axel Vogt Thank You. That sounds promising. Would this be a textbook reference?

@Carl Love I was working with Groebner Basis and the result suggested symmetric polynomials as basis. To my surprise I found cyclic polynomials among them and also the fact that Maple has no refrences for these two classes of polynomials (is this correct?)

Anyhow. The case is 3 variable case and so I find cyclic polynomials can be reduced to a linear combination of cyclic polynomials of the form x^n*y+y^n*z+z^n*x, coefficients in symmetric polynomials.

@vv I meant algebraic relationships akin to those of symmetric polynomials. I take it matrix operations are unavoidable.

@vv Thank You. Are you saying this function relates that webpage?

But what are you simplifying?

You're meant to present the file on the forum. Use the green arrow to upload the worksheet.

@mmcdara I apologize. In my haste it seems I have not included the requirement on the generators to be mutually disjoint. Minimal base would have least count. You can always subdivide non trivial basis elements, but that would increase the count.

Also I should point out that intersection and union are both n-ary operators, rather than simply binary. Every combination of sets is considered for intersection.

@Carl Love I believe so. It would be a minimal basis of disjoint elements, such that every intersection of elements of S is presentable as a union of elements of said basis.

An example:   F({{1, 2, 3}, {1, 4, 5}, {2, 3, 4}}) = {{}, {1}, {4}, {5}, {2, 3}}; and F({{1, 2, 3, 6}, {1, 4, 5, 6}, {2, 3, 4, 6}}) = {{1}, {4}, {5}, {6}, {2, 3}};

So it seems I forgot to include these criterions for T:
3. every intersection of elements of S is presentable as a union of elements of T uniquely
4. T is minimal

(I think the above means we can skip 1.)

 

Instead of removing my own answer, I accidentally removed someone else's.

You have already spotted the essential difference. The unevaluate is needed when using the $.

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