Annonymouse

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These are replies submitted by Annonymouse

@Annonymouse a work around could be to use the solutions you got and plot each of them in 3d space where z is tswitch.

I tried this but couldn't get it to work

@tomleslie I would really prefer a 3d plot; a 3d plot can be coloured in various ways to illustrate the map tswitch-> solution curve.

Significantly I have a differential polynomial that is solved on the diaganal of the surface (t=tswitch) for the solution and its derivatives. If i can get code for the 3d plot i can also make 3d plots of the derivatives - and then see graphs of all the variables that are being fed into the differential polynomial.

@Annonymouse i needed to take absolute values

@Carl Love For some reason when i do that my bar chart ends up missing bars

the numbers

             [                          -14  
             [y(t) = 1.73194791841524 10   ,

                                             -11  
               y[0](t) = -1.98951966012828 10   ,

               y[1](t) = -0.00304885510956351,

                                            -8  
               y[2](t) = 3.07160927189243 10  ,

                                             -12  
               y[3](t) = -4.88302273709004 10   ,

                                            -14  
               y[4](t) = 9.20694116760555 10   ,

                                             -15]
               y[5](t) = -1.85145481241699 10   ]

end up looking like

could you look at my worksheet?
(just press enter on every line)
Visualising_numerous_derivatives_of_the_L2_model.mw


In the end I just used a program that i had previusly written to iteratively take the lie-derivatives of the variable i am interested in, and then wrote some ugly plotting commands to get the graphs i wanted.

Visualising_numerous_derivatives_of_the_L2_model.mw

This at least solves half of my problem; but is probably not useful for anyone who finds this page through google

 

@vv 

"I don't understand why not simply plot the solutions of the system."

i thought that was what i was doing - could you explain?

@Kitonum Thats often true, but not always true, and generally unhelpful. Consider the function:

f(x,y)=x^2 +y^2;

the level set for  f(x,y)=0 is a  point
whereas the other level sets are all closed curves circles- by contrast

f(x,y)=x*y;

has all of its level sets as curves or sets of curves; none of which  are closed.

These two cases illustrate thaty we can understand the shape of functions by understanding their level sets.

If someone were to ask about finding the level sets of a scalar function from R^2 it would therefore be both inaccurate and unhelpful to say that they were all 1 dimensional surfaces.


Perhaps a better way of visualising it would be as a set of triangles in 2d space; with the x coordinates of each triangle corner being the dependant variables, and the corresponding y co ordinates being the independant variables.

@Annonymouse is there a way of making the explore polts bigger?

@acer Thats really helpful

@vv I'm just rewriting what I think you've said, so you can confirm or correct it (notably I'm not used to this usage of the word primative).

The  procedure  Basis  in  the  Groebner  package  computes  the  reduced Groebner  basis,  but after  this  computation  it  displays the polynomials without dividing through by the leading term of each.

So, for
B := Basis(J,T) ;  
the "true"  reduced Groebner basis is
B/~LeadingCoefficient(B, T);

 


Maple have confirmed it creates a reduced GB, and are updating the help page

I've just Emailed Maplesoft about this - when I have a response I will confirm here whether Basis produces reduced GBs

based on @tomleslie s answer to a related question, I've found a work around:

allvalues(eval(problem expression)); gives the roots that RootOf represents, and then the problem can be compared for each case

@tomleslie Well spotted, I've updated the worksheet. Thanks for taking a look

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