Kitonum

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These are replies submitted by Kitonum

@Carl Love  Yes of course.

The record  12^6^5   is ambiguous. It can be understood as  (12^6)^5  or as  12^(6^5) .

@Carl Love  To prevent extra zeros in output  [8500., [F = HFloat(50.00000000000008), K = HFloat(149.99999999999991)]], we can solve symbolically using the simplex:-maximize  command:

simplex:-maximize(P.V, {seq(C.V <=~ R), seq(V >=~ 0)});
eval(P.V, %);
                        
 {F = 50, K = 150}
                                 8500
 

 

@Jayaprakash J  In Maple 2018.2, the code works correctly. You probably have an earlier version of Maple. Try the following option:

restart;
f:=1+B*x-(1/12)*B*x^3+0.1666666667e-4*B^3*x^3-4.166666667*10^(-8)*B^4*x^4+(1/160)*B*x^5+8.333333333*10^(-11)*B^5*x^5-0.5000000000e-2*B^2*x^2+0.1666666667e-4*B*x^3*C^2-4.166666667*10^(-8)*B*x^4*C^3+8.333333333*10^(-11)*B*x^5*C^4-0.5000000000e-2*B*C*x^2+0.3333333333e-4*B^2*x^3*C-1.250000000*10^(-7)*B^3*x^4*C-1.250000000*10^(-7)*B^2*x^4*C^2+3.333333333*10^(-10)*B^4*x^5*C+5.000000000*10^(-10)*B^3*x^5*C^2+3.333333333*10^(-10)*B^2*x^5*C^3+0.7291666667e-3*B*x^4*C-0.3333333333e-5*B*x^5*C^2+0.6250000000e-3*B^2*x^4-0.2083333333e-5*B^3*x^5-0.5416666667e-5*B^2*x^5*C:
Explore(plot(f, x=-4..4, view=-10..10), parameters=[B=-5...5., C=-5...5.]);

 

@vv  I use Maple 2018.2  64 bit

@vv  I use Maple 2018.2

@Carl Love  Of course you are right. But maybe my simpler solution for OP will be clearer. 

@minhthien2016  It can be clearly seen from the figure in your post that the segment connecting the origin with the center of the pink circle is perpendicular to the plane of this circle. Therefore, the circle in your example cannot be built by the command  \tdplotCsDrawCircle[style]{r}{alpha}{beta}{epsilon}

@ferago42  Maple simply doesn’t implement step-by-step calculation of integrals well enough. If the program you are linking to does this better, then use it. Better yet, if you study the methods of calculating integrals, do it yourself, using a good tutorial that lists the necessary substitions.

@Christopher2222  Unfortunately, Robert Israel's code will not help you, because your problem is not one-dimensional but two-dimensional Cutting stock problem. It is much more difficult to automate. In a separate answer I give a manual solution to your problem.

@vv  I did it consciously, because I think that the very presence of this minus is a typo. Now at least for  a>0  and  C>0  we get the necessary simplification of the initial.

@emendes 

1.  I do not know the reason. The element-wise operator  appeared shortly before Maple 14 (it seems in Maple 13) and perhaps there was some bugs in its implementation.

2.  As for the second option, the function argument must be specified not by a list, but by a sequence. If you want to use iterations for such a function, then the value of this function should also be a sequence:

f:=(x,y,z)-> op([y,y*z-x,-15*x*y-x*z-x]);
f(1,2,3);
f(f(1,2,3));

 

@Carl Love  OP asked "Suppose I am interested in n-large and, instead of locating one focal point, A, I wish to obtain several,i.e. A, B, C, etc. such that the distances from each of these to their respective neighboring points is also a minimum". I replied referring to the wiki that if the initial n points do not lie on one straight line, then this is impossible.

@minhthien2016   I can of course, but I don't have time now for this.

See Google search on  The condition for the existence of a tetrahedron with given lengths of all edges

Your tetrahedron defined by the lengths of the edges actually exists.

Thank you all! I did not expect the solution to be so simple.

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