Download examples.mw

Technically f is a number and writing diff(f, something) is identically null.

Don't you mean "what is the variation delta(f) of f given a variation delta(q) of function q" ?

Look to this, maybe that will help you (or not)

variation.mw

The key code is 

# N = numelems(team):
# n = number of teammates per set
# T = list of sets lists

T := [seq(team[1+(i-1)*n..i*n], i=1..N/n)]:

Download teams_mmcdara.mw

x in RealRange(1, Open(4)):
convert(%, relation)
                    1 <= x < 4

If you realy want to get a single inequality when the interval contains +/- infinity you can use procedure rel below: 

Download Enhanced_convert.mw

restart
f := X^2 + Y^2 - 1:
g := X - Y:
intersections := proc(P, Q, T)
  local ind, R, W, w:

  # check
  ind := indets({P, Q}):
  if not member(T, ind) then
    error cat(sprintf("%s%", T), " is not an indeterminate of {P, Q}")
  elif numelems(ind) = 1 then
    error cat("Only one indeterminate ", sprintf("%s%", op(ind)), " found, 2 needed")
  elif numelems(ind) > 2 then
    error "More than 2 indeterminates found: P and/or Q do not define plane curves"
  end if:

  R := resultant(P, Q, T);
  W := op(ind minus {T});
  
  w := solve(R, W);

  return W = {w}
end proc:
intersections(f, g, X)
                        /  1  (1/2)  1  (1/2)\ 
                  Y =  { - - 2     , - 2      }
                        \  2         2       / 

intersections(f, g, A)
Error, (in intersections) A is not an indeterminate of {P, Q}
intersections(f+A, g, X)
Error, (in intersections) More than 2 indeterminates found: P and/or Q do not define plane curves
intersections(X, X^2, X)
Error, (in intersections) Only one indeterminate X found, 2 needed

Note that
 

solve({f, g}):
allvalues(%);
 /    1  (1/2)      1  (1/2)\            /      1  (1/2)        1  (1/2)\ 
{ X = - 2     , Y = - 2      }    ,     { X = - - 2     , Y = - - 2      }
 \    2             2       /            \      2               2       / 

numelems([%])
                 2

Note: I don't know if you got this code from Mistral AI as you often does, but it's worth saying that a lot of Artificial Intelligences give stupid results specially when used blindly or not trained properly

@GFY @acer

@acer's answer is great, but here is another wayto answer your problem which seems to me more natural (IMO).

The idea is still to use events but in conjunction with option discrete_variables. (see the event help page for more details).
Basically one augment the differential system by two functions A(t) and B(t) (the "discrete variables", not defined from differential equations), each of them being aimed to represent one of the two envelopes.
Specific events are used to build A(t) and B(t).

odeplot can be used directly in the usual way to plot those functions, for instance:

odeplot(solu12, [t, A(t)], t=...)

saopinjifen1230_mmcdara.mw

It is worth mentioning that the computation time for dsolve (without range option) is 0.1 sec and that it takes only 2.5 sec to odeplot the result (40 times faster than @acer's solution).

Note that using the range option and the refine option in odeplot (which can be useful for a better representation of x(t) only) can dramatically increase these times.


Successive-zooms.pdf

As you see below (first figure)  A(t) and B(t), excepted for the early times, are stepwise functions.
This is different from what @acer gets (second figure)

Note that it would quite simple to transform "my" stepwise envelopes into piecewise linear ones

Let P a multidimensional parameter.
You have 

phis := {phin = Fn(G(xi), P) , n=1..7 }

and you want to solve the system 

{Fn(G(xi), P) = 0 , n=1..7 }

wrt to P.

To get P you use the command

unknowns := indets(rhs~(phis), name)

which indeed returns all the names P contains, but also the name (xi) of the independent variable G depends upon.
So the error you get:
Error, (in solve) cannot solve expressions with diff(G(xi), xi) for xi

To get rid of it write

unknowns := indets(rhs~(phis), name) minus {xi};   # this is the "true" P

Now you can compute COEFFS, which is a sequence of 16 solutions:

COEFFS := solve(rhs~(phis), unknowns)

whattype(COEFFS);
                            exprseq

 numelems([COEFFS]);
                               16

Some of them depend on G(xi), which might not (?) be suitable.
If you want to get rid of them execute

GfreeCOEFFS := remove(has, [COEFFS], xi):
numelems(%);
                               9

Remark:

phis contains 7 relations while the parameter vector P has dimension 16. Which mean that 9 parameters are likely to get arbitrary values.

Maybe a better way would be to solve the system  { F_n(G(xi), Q) = 0 , n=1..7 } with Q being a subvector parameter of P with size  7.
Ideally you should chose the 7 parameters which are the most relevant to the use you have in mind.

and ifI understand corrrectly your problem, here is a solution (which is likely not optimal in terms of computation time nor memory use) .
It is fast for N=8 but I didn't checked it for much larger values of N.

Round_Robin_mmcdara.mw

Maple Worksheet - Error

Failed to load the worksheet /maplenet/convert/round_robin_mmcdara.mw .



 

key: use this rewritting rule

isolate(m+1/(diff(G(xi), xi))=Phi, diff(G(xi), xi));
                       d              1   
                      ---- G(xi) = -------
                       dxi         Phi - m

and collect wrt Phi :

Download factoring_mmcdara.mw

Nevertheless here is a real example based on the Grid package and a specific strategy (read carefully the text).
 

Download Example.mw

The piece of code

	
W__points := plottools:-getdata(P)[1, -1]:
t_grid := convert(W__points[..,1], list):
x_grid := [seq(-2..2, 4/N)]:
> 	
T, X := map(mul, [selectremove(has, [op(expand(solnum))], t)])[]:
ST := unapply(eval(T, W(t)=w), w)~(W__points[.., 2]):
SX := evalf(unapply(X, x)~(x_grid)):
STX := Matrix(N$2, (it, ix) -> ST[it]*SX[ix])

I sent you days ago was written to handle T3 expressions of the form F(x)*G(W(t)), which is no longer the case here.

Here is an update for your current T3 expression (in your code you gave w two different meanings: one was a parameter, the other a surrogate name for W(t), which is quite confusing ; even if I no longer use this surrogate in the code below, I thought it would be better to avoid confusion in any future  expressions like eval(..., W(t)=w) and then arbitrarily renamed w as psi).

 

Download S5_mmcdara.mw

The simplest way, IMO, to get this table  is to use the option output=Array in dsolve/numeric:

when    := [seq(0..50, 1)]:
sol     := dsolve(`union`(sys, ic), numeric, output=Array(when)):
sol[1];  # to verify the order of the outputs
results := round~(sol[2][1]):

printf("%-10s %-15s %-15s %-15s\n", "Day", "Infected", "Recovered", "Susceptible");
printf("---------------------------------------------\n");

fmt := "%-10d %-15d %-15d %-15d\n":
map(i -> printf(fmt, entries(results[i+1], nolist)), when):

y := substring(x, 1..1)

To get the value of y:

x := asdf:
a := 3:
y := substring(x, 1..1);
                               a
eval(y)
                               3

This doesn't answer your question, just to say that geom3d provides an elegant way (IMO) to determine if a point is on a line.

restart
with(geom3d):
line(L, [3+2*alpha, 1+6*alpha, 4-5*alpha], alpha):

point(A, 9, 19, -11):
IsOnObject(A, L)
                              true
point(B, 9, 1, -11):
IsOnObject(B, L)
                             false

Plus an animation to illustrate the demonstration
Similar_Triangles_Animation.mw