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These are replies submitted by Kitonum

@erik10  When solving nonlinear systems at specific intervals, the  DirectSearch  package can be useful.  It can be downloaded freely from the Maple Application Center. 

I think that the position of the rectangle surrounding these legends should be made one of the formal parameters of the procedure . You will specify it each time you use the procedure, depending on the situation (use  plottools:-rectangle  command). The rest can be automated in the procedure body.

@nm  You seem to have misunderstood the meaning of my answer. My code does exactly the same as  tomleslie's one or almost the same as  vv's code. I edited my answer a bit.

@priya_priti  Now the syntax is correct, but Maple simply cannot symbolically solve your system. By the way, your system has the obvious null solution: u(x)=0, v(x)=0, w(x)=0 .

@manju  In the new file, you did not do what I wrote about above.

@Keshav4  If a> = 0 then obviously your function  is increasing and it takes its smallest value at  x = 0 . For negative values of the parameter  a , the example can be solved only numerically.

@Prakash J 

plot(abs@h, 1..2, labels=["t","h(t)"]);


@Scot Gould  The disadvantage is that the syntax is longer, and you yourself have already written about the advantages. I usually don't use this option because it is quite possible to do without it.

@vv  and @Carl Love  Thank you for answers.

You can arrange this calculation without assignments and conversions:

Sum(GAMMA(2*b-1+2*n)*GAMMA(2*n-s)*(b-1/2+2*n)/(GAMMA(2*b+2*n+s)*GAMMA(2*n+1)), n = 0 .. infinity); 
subs(Sum = sum, %) = eval(%, Sum = sum);



@nm  Omega &x v(t)  means the cross product of the vectors  Omega  and  v(t)  in LinearAlgebra package.

Upload your worksheet (or a link to it) here for localizing the problem (using the bold green arrow in the MaplePrimes editor). As a last resort, just copy and paste here the code in which this error occurs.

@minhthien2016  Geometrically, all the examples vv provided differ little from each other. All of these tetrahedrons are similar to your original example  (0, 0, 0), (3, 0, 0), (0, 5, 0), (0, 0, 14) . They are obtained from your original example by shifting by some vector, stretching and rotating in space. The easiest way to see this is by plotting. The code below plots your example and the first example from vv's code:

plots:-display(< A | B >, scaling=constrained);

Below is an example of a tetrahedron that is not in my previous answer. Its inscribed sphere has  unit radius and its center is [4, 2, 1] :

B:=plottools:-sphere([4, 2, 1], 1):
plots:-display(A,B, scaling=constrained, transparency=0.3, axes=normal, view=[0..5.5,0..5.5,0..4.3]);

I don't know of any other way to get substantially different examples than "brute force".

@mehran rajabi 

asympt(1/(x^2-1), x, 10);

# Or
Sum(1/x^(2*n), n=1..infinity);

# Or automatically
convert(1/(x^2-1), FormalPowerSeries, x=infinity);



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