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These are replies submitted by acer

Isn't  n^2+n+17  a good fit?

@mapleatha The following example displays the Array in the Standard Java GUI by using the `print/rtable` procedure to return an RTABLE call.

This can be confirmed by stepping through the `print/rtable` call via the debugger, hence the stopat call below.

restart;
interface(typesetting=standard):
stopat(`print/rtable`);
Array([x]);

It's a pity that you cannot, in turn, provide me with an example of what you meant.

@mapleatha Provide a full example in which using RTABLE for printing purposes ensures that terms in a sum won't be printed out-of-original order (eg, following sort, etc). Perhaps there will be a decent alternative.

[edit] You have since provided an example expression, but you omitted to provide the actual code that calls RTABLE. Upload a worksheet that contains all the output and reproduces both the original problem as well as the purported workaround.

I don't understand what you are trying to say, with, "...show me one file where it is used without being explicitly used in the associated program."

@cicoats See the Help page with topic evalf,int,cuba in Maple.

Some of the offered Cuba methods are probabilistic (Monte Carlo), but I also used the deterministic _CubaCuhre.

When successful, these methods can quite often provide a coarse numeric approximation more quickly than do Maple's other (non-Cuba) methods for multidimensional numeric integration.

@Felipe_123 Do you want something like this?

restart;

kernelopts(version);

`Maple 17.02, X86 64 LINUX, Sep 5 2013, Build ID 872941`

with(StringTools):

B:=(parse~@Vector[row]@Explode)~(Generate(3,"012"));

B := [Vector[row](3, {(1) = 0, (2) = 0, (3) = 0}), Vector[row](3, {(1) = 0, (2) = 0, (3) = 1}), Vector[row](3, {(1) = 0, (2) = 0, (3) = 2}), Vector[row](3, {(1) = 0, (2) = 1, (3) = 0}), Vector[row](3, {(1) = 0, (2) = 1, (3) = 1}), Vector[row](3, {(1) = 0, (2) = 1, (3) = 2}), Vector[row](3, {(1) = 0, (2) = 2, (3) = 0}), Vector[row](3, {(1) = 0, (2) = 2, (3) = 1}), Vector[row](3, {(1) = 0, (2) = 2, (3) = 2}), Vector[row](3, {(1) = 1, (2) = 0, (3) = 0}), Vector[row](3, {(1) = 1, (2) = 0, (3) = 1}), Vector[row](3, {(1) = 1, (2) = 0, (3) = 2}), Vector[row](3, {(1) = 1, (2) = 1, (3) = 0}), Vector[row](3, {(1) = 1, (2) = 1, (3) = 1}), Vector[row](3, {(1) = 1, (2) = 1, (3) = 2}), Vector[row](3, {(1) = 1, (2) = 2, (3) = 0}), Vector[row](3, {(1) = 1, (2) = 2, (3) = 1}), Vector[row](3, {(1) = 1, (2) = 2, (3) = 2}), Vector[row](3, {(1) = 2, (2) = 0, (3) = 0}), Vector[row](3, {(1) = 2, (2) = 0, (3) = 1}), Vector[row](3, {(1) = 2, (2) = 0, (3) = 2}), Vector[row](3, {(1) = 2, (2) = 1, (3) = 0}), Vector[row](3, {(1) = 2, (2) = 1, (3) = 1}), Vector[row](3, {(1) = 2, (2) = 1, (3) = 2}), Vector[row](3, {(1) = 2, (2) = 2, (3) = 0}), Vector[row](3, {(1) = 2, (2) = 2, (3) = 1}), Vector[row](3, {(1) = 2, (2) = 2, (3) = 2})]

M:=Matrix([[2,1,0,0],[0,2,1,0],[0,0,2,1]]);

M := Matrix(3, 4, {(1, 1) = 2, (1, 2) = 1, (1, 3) = 0, (1, 4) = 0, (2, 1) = 0, (2, 2) = 2, (2, 3) = 1, (2, 4) = 0, (3, 1) = 0, (3, 2) = 0, (3, 3) = 2, (3, 4) = 1})

map[2](LinearAlgebra:-Modular:-Multiply,5,B,M);

[Vector[row](4, {(1) = 0, (2) = 0, (3) = 0, (4) = 0}), Vector[row](4, {(1) = 0, (2) = 0, (3) = 2, (4) = 1}), Vector[row](4, {(1) = 0, (2) = 0, (3) = 4, (4) = 2}), Vector[row](4, {(1) = 0, (2) = 2, (3) = 1, (4) = 0}), Vector[row](4, {(1) = 0, (2) = 2, (3) = 3, (4) = 1}), Vector[row](4, {(1) = 0, (2) = 2, (3) = 0, (4) = 2}), Vector[row](4, {(1) = 0, (2) = 4, (3) = 2, (4) = 0}), Vector[row](4, {(1) = 0, (2) = 4, (3) = 4, (4) = 1}), Vector[row](4, {(1) = 0, (2) = 4, (3) = 1, (4) = 2}), Vector[row](4, {(1) = 2, (2) = 1, (3) = 0, (4) = 0}), Vector[row](4, {(1) = 2, (2) = 1, (3) = 2, (4) = 1}), Vector[row](4, {(1) = 2, (2) = 1, (3) = 4, (4) = 2}), Vector[row](4, {(1) = 2, (2) = 3, (3) = 1, (4) = 0}), Vector[row](4, {(1) = 2, (2) = 3, (3) = 3, (4) = 1}), Vector[row](4, {(1) = 2, (2) = 3, (3) = 0, (4) = 2}), Vector[row](4, {(1) = 2, (2) = 0, (3) = 2, (4) = 0}), Vector[row](4, {(1) = 2, (2) = 0, (3) = 4, (4) = 1}), Vector[row](4, {(1) = 2, (2) = 0, (3) = 1, (4) = 2}), Vector[row](4, {(1) = 4, (2) = 2, (3) = 0, (4) = 0}), Vector[row](4, {(1) = 4, (2) = 2, (3) = 2, (4) = 1}), Vector[row](4, {(1) = 4, (2) = 2, (3) = 4, (4) = 2}), Vector[row](4, {(1) = 4, (2) = 4, (3) = 1, (4) = 0}), Vector[row](4, {(1) = 4, (2) = 4, (3) = 3, (4) = 1}), Vector[row](4, {(1) = 4, (2) = 4, (3) = 0, (4) = 2}), Vector[row](4, {(1) = 4, (2) = 1, (3) = 2, (4) = 0}), Vector[row](4, {(1) = 4, (2) = 1, (3) = 4, (4) = 1}), Vector[row](4, {(1) = 4, (2) = 1, (3) = 1, (4) = 2})]

 

Download vecsplode_2.mw

Are you ok with a series solution? I don't think that you're going to find a closed form answer in terms of elementary functions.

@Liiiiz Your text emoji is amusing in context.

@vv Rank computes by comparing singular values, for floats (but not by trying to compute the rref, which would suffer similar difficulties as mentioned).

@nm You are wrong in describing that as a bug. It is just the wrong syntax for the stated purpose. 

@mapleatha That's not what I wrote!

I deliberately did not suggest what you just tried, as that would certainly insert floats for the tickmark values (and not the symbolic rational multiples of Pi/2).

Please try what I actually suggested ,(which inserts the symbolic multiples in at least as far back as Maple 16).

@ogunmiloro I have now tagged your Question as being for Maple 18 (six major releases ago, released in the year 2014), as I did previously for your earlier questions.

Has not the version of your Maple been an issue for code suggested in some previous query by you? Is there some reason why you cannot tag your Questions with your version details when you post them?

@nm You might look at the Explore command, eg.

Explore(plot(sin(a*x),x=-2*Pi..2*Pi),
        a=0.0..1.0,initialvalues=[a=0.4]):

That will embed an interactive application within the current Worksheet or Document.

If you'd prefer a popup application (but with only a unimodal operation) then you could look at the Maplets package.

@MapleEnthusiast I have not forgotten your followup refinement, but I just haven't had free time to consider it properly, sorry. I will try and find some spare moments. These are tricky times.

@Carl Love This question has arisen a few times: How to get the effect of a smooth curve (ie. adequate number of points, preferably using default adaptive plotting) and only a modest number of symbols overlaid.

The option style=pointline ties the number of symbols to the generated data points, unfortunately. Hence these solutions to forcibly overlay two "plots" -- one for the curve with adaptive plotting and one for the symbols at a modest number of points without adaptive plotting.

It might be useful for it to be automatically available, so that people don't have to figure out the overlaid scheme as suggested. Also, the forced overlay results in additional data in the constructed plot structure, and that might make external export of the data more fiddly.

Perhaps this kind of thing would be a better (and more commonly desired) result from the style=pointline option itself? Is the current behaviour really more desirable? If so could it benefit from some option modification so as to specify the number of rendered symbols separately from the specification of the number of curve data points? And how about requesting their spacing evenly by arclength rather than by value of the independent parameter? Comments?

@nm I have sent you messages (though this forum) about splitting and duplicating topics. Your email server is rejecting them as spam. I get the bounced email session transcript.

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