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These are replies submitted by sand15


Thanks Tom, this is indeed a good idea.


@Carl Love 

I just read your reply, thanks for the clarification.

@Carl Love 

Thanks a lot

PS: No hard feelings, glad to read you again

Hi again
(sand15 is my professional login and mmcdara my private one)

I've done this very simple test:

F := [ seq(randpoly([x,y], homogeneous), k=1..3) ];
G := Basis(F, plex(x,y)):
LeadingCoefficient~(G, plex(x,y));

The GBs are not reduced (except in exceptional cases)

I've tried to use other algorithms, for instance
G := Basis(F, plex(x,y), method=maplef4): 
(it seems that one can write maplef4 or "maplef4" ,  see the head of  showstat(Basis) where `method` is defined as type string ot symbol).
For all the methods I tested we obtain the same unreduced GB.

But, very surprisingly, you can write  
G := Basis(F, plex(x,y), method=""):
and get no error message ??? 

@Carl Love 


Sorry for the last reply: the screen capture doesn't appear even if it was correctly correctly displayed in the preview window.
Il will try again when I am home


And I repeat too that your solution is {x(t)=sin(t), y(t)=cos(t)}
See also Preben's answer


I hope you have understood why your solution x(t) is just a sine wave (if not look to my answer to Kitonum).
The solution of SYS2 is just the one of SYS={diff(x(t), t)=y(t), diff(y(t), t)=-x(t), x(0)=0, y(0)=1}:

But, if you are still unconvinced, just do this

SYS := {diff(x(t), t)=y(t), diff(y(t), t)=-x(t), x(0)=0, y(0)=1}:
GlobalPlot := NULL:
for n from 1 to floor(Tmax/8) do
   sol := dsolve(SYS, numeric, range=8*(n-1)..8*n):
   GlobalPlot := GlobalPlot, odeplot(sol, [t, x(t)], 8*(n-1)..8*n):
   X:=subs(sol(8*n), x(t)):
   Y:=subs(sol(8*n), y(t)):
   SYS :=  {diff(x(t), t)=y(t), diff(y(t), t)=-x(t), x(8*n)=X, y(8*n)=Y}:
end do:

Still not convinced?
Then do this

gate := 1 - ( Heaviside(t - (8*n - epsilon)) - Heaviside(t - (8*n + epsilon)) ):
SYS := {diff(x(t), t) = y(t)*gate, diff(y(t), t) = -x(t)*gate, x(0)=0, y(0)=1}:
FormalSolution := solve(limit~(SYS, epsilon=0), {x(t}, y{t]);

FormalSolution is the solution of
{diff(x(t), t)=piecewise(t<>8*n, y(t), 0), diff(y(t), t)=piecewise(t<>8*n, -x(t), 0), x(0)=0, y(0)=1}
whatever the value of n.
Note FormalSolution is {x(t)=sin(t), y(t)=cos(t)}, meaning the discontinuity at t=8*n has absolutly no influence.

If you have several such rhs discontinuities, FormalSolution still represents the solution: the argument is that one discontinuity at t=8*n having no influence then a numerable number of them do not have neither



Consider the following problem:

gate := 1 - ( Heaviside(t - (8*n - epsilon)) - Heaviside(t - (8*n + epsilon)) ):
SYS := {diff(x(t), t) = y(t)*gate, diff(y(t), t) = -x(t)*gate, x(0)=0, y(0)=1}:

Would you say this system has no solution?
Easy to check that Maple returns one, at least one n and epsilon has recieved numerical values; for instance
dsolve(subs({n=1, epsilon=0.1}, SYS), {x(t), y(t)})

Now it is also obvious that the rhs of SYS tend to the rhs of SYS2 when epsilon --> 0.
And thus, at least for epsilon <> 0, the limit of the solution of SYS tend to the one of SYS2.

The solution x(t)  is "globally" a sine function, excepted in the range [8*n=epsilon, 8*n+epsilon] where it is flat with a constant value equal to x(8*n-epsilon).
More precisely x(t) is made of:

  1. a sine function sin(t)  from t=0 to t=8*n-epsilon (assuming n > 0),
  2. a plateau x(t)= x(8*n-epsilon) from t=8*n=epsilon to t= 8*n+epsilon,
  3. and finally a translated sine function sin(t+2*epsilon) beyond t=8*n+epsilon,

As epsilon  --> 0  this plateau shrinks at t=8*n and, when epsilon=0, the solution (the one of SYS2) is a continuous sine curve (a cosine cuve for y(t))

The numerical solution can easily be obtained with 
sol := dsolve(SYS, numeric, parameters=[n, epsilon])



Another point of view:

  1. Solve SYS from 0 to T=8*n-epsilon
    let's note sol1 the corresponding solution and XT and YT the values of the solution at timùe T for x(t) and y(t) respectivey
  2. Solve {diff(x(t), t) = y(t)*gate, diff(y(t), t) = -x(t)*gate, x(T)=XT, y(T)=YT} up to t=8*n+epsilon
    Obviously this second solution is x(t)=XT, y(t)=YT
    Let's note TT =8*n+epsilon
  3. Solve {diff(x(t), t) = y(t)*gate, diff(y(t), t) = -x(t)*gate, x(TT)=XT, y(TT)=YT}
  4. Connect these 3 solutions

@Mariusz Iwaniuk 

Is this one correct ?
(I think one should plot func and not ifunc)


It's me again...

I used CurveFitting:-Spline with degrees 1 and 3, and CurveFitting:-PolynomialInterpolation with different values for the option 'form'.
In all the cases i get no error but the method seems to diverge.
More precisely, in my first answer I had plotted only  [sinh(x), ifunc(1)(x)]  and  [sinh(x), ifunc(2)(x)]  and it looked well ( ifunc(2)(x) was closer too sinh(x) than  ifunc(1)(x) was).
I pushed a little bit forward by plotting  [sinh(x), ifunc(4)(x)]  ... and  ifunc(1)(x)  looks like some kind of parabola A*x^2 with A < 0 !!!

Maybe I did inadvertently some mistake when I modified your code ????


With the correct values of X and Y fsolve('DetKFF(z)', z=0.1) should return 0.362284....



As vv said "You are not allowed to expand like this.".

More generaly, a sequence S of terms can be convergent even the sequences build from parts of the terms are not.

Take this far-fetched exemple :
S[n]=n-n  for n >=0 ...  obviously S "converges" to 0
Write S as U+V with
U[n]=n and V[n]=-n 
Then U and V both diverge and S=U+V seems to be undefinite (+infinity)+(-infinity)



My mistakes:

  •  You also have a syntax error in f()  [a ":" after proc()  not accepted in 1d].
    Appologies, I rewrote the content of a pdf file; in the code the ":" are not present after proc()
    (furthermore it generates an error if there is a delcration statement after the "proc" line, for instance local ....)
  • I effectively made a mistake by tipping args(k) instead of instead of arg[k] (furthermore I knew the sorrect syntax
    Strangely it seems to be correctly interpreted by Maple 2015: printing the quantity named "objet" returns the correct name
  • Last but not least: One more time I have forgotten that names constructed with "||" are global !!


Then the answer is: either I use the awfull sequence
    if n=1 then print(Object1)
  elif n=2 then print(Object1)
  end if

or either I construt a list ObjectList:=[Object1, ...ObjectN] and a list ListA := [A] and I do this

 Choice := proc(a::list)
  local n, object:
  n := ListTools:-Search(true, a):
  object := ObjectList[n]
end proc:

Thank's for your answer, even if it is not always pleasant to be remembered we are an airhead !


"  The advantage(?) of doing it this  ...": right, it seems a better (the only one?) way to manage "pseudo axis"

By the way: I did a mistake in my original post.
I wrote

  • the empirical marginal distribution of X (Histogram), put below the line "y = min({y[n], n=1..N} )" and rotated by 180 degrees

In fact I should have written:

  • the empirical marginal distribution of X (Histogram), put below the line "y = min({y[n], n=1..N} )" and SYMMETRIZED around the x axis

A simple task to do with plottools:-transform
reflection := plottools:-transform((x,y) -> [x, -y])

From now on the two histograms are placed correctly by using only plottools:-transform ... you deinitely gave me a good advice.


It's good enough for me.

By the way, where did you find this stylesheet = [vertexcolor = "White",vertexborder=false]) info ?

Thanks again

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