CDF of Binomial(6, 1/2) is OK  but not CDF of Binomial(6, 1/4) 
The step at location [0, 1[  (or ]0, 1] because french and english customs do not agree on this point) is not present!

X := RandomVariable(Binomial(6, 1/4)):   # or Binomial(6, 0.25) if you prefer
F := CDF(X, s):
plot(F, s=-1..7, discont=true, gridlines=true, axis[1]=[gridlines=7]);


The CDF has an uncomprehensible form invoking the Psi function. 
Could anyone from the development team restore what was done in Maple 2015?

@acer 
I'm so sorry acer for not having replied sooner.
After this inexcusable delay I greatly thank you for all these precious informations.

@vv

Thanks vv. 

By the way: limit returns infinity ... It would have been better to recieve a FAIL or undefined answer

@acer 

Very good!

Do I "want additional aspects like color, or fonts/color/weight on 2D Math"? Surely!

By the way, I had thought using Typesetting because I had already saw a few commands using "mn", mrow", somewhere here.
But I wasn't able to retrieve them from the Typesetting help page: are these functions documented in some place?

This text corrected 20 miutes later ...
I've found useful informations from the example given in the InertForm[Typeset] help page!!!

Thanks for the help

@tomleslie 

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:

restart:
with(Groebner):
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="WhatTheF.ck"):
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

@radaar 

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

@radaar 

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}:
Tmax:=600:
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:
plots:-display(GlobalPlot):


Still not convinced?
Then do this

restart:
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

@Kitonum 

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)

Volterra2.mw

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 ????

@maple2015 

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