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These are replies submitted by Kitonum

@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.


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]);


@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.


X := [3, 6, 4, 13]:   Y := [8, 6, 5, 7]:  n:=nops(X):
local D;  
L:=[seq(seq([[[X[i],Y[i]],[X[j],Y[j]]],(i,j)=sqrt((X[i]-X[j])^2+(Y[i]-Y[j])^2)], j=i+1..n), i=1..n-1)];
T:=Matrix(4, {map(t->t[2],L)[],map(t->(lhs(t[2])[2],lhs(t[2])[1])=rhs(t[2]),L)[]}):


@Carl Love  Thank you for your interest. Unfortunately, getting a noticeable gain in time at the cost of obtaining only one solution is impossible. In the branch and bound method  each variant may branch out into several variants, or it may disappear altogether, if it cannot be continued. And this we will not know until we reach the last step. How the method works is clearly seen in the example below (the second example from the worksheet). I brought all the steps to the display:



@DimaBrt  I did not understand your comment. For me, everything works in Maple 2018.2.

@kambiz1199  You can use the  trace  option to show the intermediate positions of a ladder. See my updated answer above.

@Rouben Rostamian   I also do not know a short way. Here is a rather cumbersome way:

test := -2+exp(theta)+exp(-theta);
convert(%, tanh);


@minhthien2016  Of course, this line is only needed to make these assignments  a:=...  and  b:=...  clear.

@acer  and  Carl. Thanks for replies! The trouble removed. 

I tried to test this procedure. I just copied the text of the procedure into a new worksheet. But during initialization, the error occurs:  Error, unexpected neutral operator

Can anyone confirm this situation? What could be the reason? I'm using Maple 2018.2

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