Kitonum

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@ActiveUser 

1. You can use the  op  command for this: 

Sol:=solve(a*x^2 + b*x + c=0, x, parametric=true, real);
op(Sol);
op(9,Sol);
op(9..10,Sol);


2. I do not know the commands from this package and have never used it.

@Carl Love I do not understand what your doubts are based on. As for the work of the value procedure, I would call it not an error, but simply an unsuccessful design. The essence of the problem is that this procedure does not calculate any derivatives (but only the values of the functions themselves from the system), and returns  0  for any derivatives. Of course, it would be better if in the case of derivatives, it returns the input unevaluated or returns an error message. I believe that spacestep and timestep options are not connected in any way with this problem.

Compare the angular coefficients of the tangents at  y = 0  with the numerical results (this to some extent confirms the correctness of the results):

restart;
a := 0.7: L := 8: HAA := [0, 2, 5, 10]: h:=0.001:
PDE1 := diff(u(y, t), t) = diff(u(y, t), y, y)+diff(diff(u(y, t), y, y), t)-b*u(y, t)+T(y, t):
PDE2 := diff(T(y, t), t) = (1+(1+(a-1)*T(y, t))^3)*(diff(T(y, t), y, y))+(a-1)*(1+(a-1)*T(y, t))^2*(diff(T(y, t), y))^2+T(y, t)*(diff(T(y, t), y, y))+(diff(T(y, t), y))^2:
ICandBC := {T(L, t) = 0, T(y, 0) = 0, u(0, t) = t, u(L, t) = 0, u(y, 0) = 0, (D[1](T))(0, t) = -1}:
for i from 1 to nops(HAA) do
b := HAA[i];
pds[i] := pdsolve({PDE1,PDE2}, ICandBC, numeric);
f:=(y0,t0)->rhs((pds[i]:-value(u(y,t),t=t0)(y0))[3]);
A[i]:=(f(h,1)-f(0,1))/h;
end do:
seq(A[i], i=1..4);
plots:-display(< `<|>`(seq(pds[i]:-plot(u(y,t), y=0..3, t=1, scaling=constrained),i=1..4))>);

 

 

If you need a procedure, then do so (I just rewrote vv's code as a procedure):

Squares:=(M,N)->plots:-display(seq(seq(plottools:-polygon([[m,n],[m+1/2,n+1/2],[m+1,n],[m+1/2,n-1/2]], color=`if`(m::odd,red,blue)), m=0..M),n=0..N), axes=none, scaling=constrained):


Examples of use:
Squares(2,2);
Squares(5,5);
Squares(3,2);

@vv 

sys := [x^2 + y^2 - x*y - 1 = 0, y^2 + z^2 - y*z - a^2 = 0, z^2 + x^2 - x*z - b^2 = 0]:
ab:={a=10,b=104/10}:
solve(eval(sys,ab), explicit):
evalf(%);

           {x = 6.676479229-2.966007548*I, y = 5.939434090+4.226624806*I, z = 12.50085790+.1379423560*I}, {x = -6.676479229+2.966007548*I, y = -5.939434090-4.226624806*I, z = -12.50085790-.1379423560*I}, {x = 6.676479229+2.966007548*I, y = 5.939434090-4.226624806*I, z = 12.50085790-.1379423560*I}, {x = -6.676479229-2.966007548*I, y = -5.939434090+4.226624806*I, z = -12.50085790+.1379423560*I}, {x = .1989718373, y = 1.084528494, z = 10.49807685}, {x = -.1989718373, y = -1.084528494, z = -10.49807685}, {x = 1.141952419, y = .4227976407, z = -9.781889234}, {x = -1.141952419, y = -.4227976407, z = 9.781889234}


Or even simpler:

sys := [x^2 + y^2 - x*y - 1 = 0, y^2 + z^2 - y*z - a^2 = 0, z^2 + x^2 - x*z - b^2 = 0]:
ab:={a=10,b=10.4}:
solve(eval(sys,ab));

Output is the same.

@abdgafartunde  Instead of column matrices, it is better to use vectors.

See the toy example:

N:= 5:
A:=<1,2; 3,4>;
b:=<1, 2>;  C:=<1, 1>;

for i from 0 to N do
      x[i+1] := A%.((i^2+1)*b) + C;
od; 

 

@abdgafartunde  I meant that in the same way, you can assign a name to each output, and then refer to these names when necessary.

@Ahmed111  See update to my answer above. The commands  diff  and  Physics:-diff  calculate any derivatives, both ordinary and partial. See help on these commands.

@Carl Love  See update to my answer. Compare my result with yours. 

@Scot Gould  Thank you for this!  I did not know that Maple solves systems written in vector form.

@amirhadiz  The  Interpolation  package appeared only in the latest versions of Maple. You have an older version, but you can use the code from my answer.

@acer  Thanks for this info. I did not know that such an evident command appeared recently only. Of course, for older versions, we can write
L1:=map(t->[t,ListTools:-Occurrences(t,L)], convert(L,set));
instead of  
L1:=ListTools:-Collect(L);

 but of course, Statistics:-Tally is more convenient. My point was simply to show that there are different ways to solve the same problem.

@Ali2020  It works in Maple 2019 only.

@minhthien2016 

Here is the equation with 6 another solutions:

abs(a*x+b)+abs(c*x+d)-3*x^2+n*x+p=0

@minhthien2016  
 

f:=x-> abs(a*x+b)+abs(c*x+d)-x^2+n*x+p;
solve([f(1) = 0, f(2) = 0, f(3) = 0, f(4) = 0, f(5) = 0, f(6) = 0], [a, b, c, d, n, p]);

     [[a = -2, b = 5, c = -2, d = 9, n = 7, p = -16], [a = -2, b = 5, c = 2, d = -9, n = 7, p = -16], [a = 2, b = -5, c = -2, d = 9, n = 7, p = -16], [a = 2, b = -5, c = 2, d = -9, n = 7, p = -16], [a = -2, b = 9, c = -2, d = 5, n = 7, p = -16], [a = 2, b = -9, c = -2, d = 5, n = 7, p = -16], [a = -2, b = 9, c = 2, d = -5, n = 7, p = -16], [a = 2, b = -9, c = 2, d = -5, n = 7, p = -16]]

@mehran rajabi  See help on  Student:-NumericalAnalysis:-Quadrature  command, spesifically the  method  option.

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