Hi,

Let, fixed an integer i and 1<=j<=2^{i}-1, for each x and y in [0,1] let the following mapping

Then, with the above procedure we can obtained, for a fixed i, all the mappings for j=1,...,2^{i}-1

However, How can I to evalute the "components" of the above procedure? For instance, I can not to compute CreaF(2)[1](0.35,0.465) (i.e., the first function in the "vector" CreaF(2), in x=0.35, y=0.465). 

Thanks very much for your time.

Hi, I am new to maple, but I think that my question should be simple.

I have a matrix where each element is an expression. I want to compute the matrix for different constant and to save it without crushing the previous matrix. 

If the file that I joined, I have a first part where the constant are defined. In the second part the expression of the matrix is defined. Finally, I compute each matrix with different constant. Each results is called C_p0, C_s0, C_g0. When I called them back, only the last matrix computed remains.

I would like to be able to save each matrix to performed operation on them later.

Thank you. 

Forum_Question1.mw

equidistant_curve_MP.mw  Equidistant curves to the curves on the surface. (Without any sense, but real.)

I have the following expression (generated by some other procedure):

This does not have a taylor expansion in pV[6] in the general case because the square roots can become negative:

taylor(xpr,pV[6]);
Error, does not have a taylor expansion, try series()

But I can get an expansion by restrictig the range of pV[6]:

taylor(xpr,pV[6]) assuming -0.01<pV[6],pV[6]<0.01;

So far things are perfectly fine. But when I try mtaylor:

mtaylor(xpr,pV[6]) assuming -0.01<pV[6],pV[6]<0.01;
Error, (in assuming) when calling 'mtaylor'. Received: 'does not have a taylor expansion, try series()'

So the assumption seems to be ignored. I can work around this by expanding in pV[6] first, using taylor, and then expanding the result from that using mtaylor (I really also want the expansions in the other pV components; 6 in total although in this example some do not show up). I'll have to convince myself that this work-around gives the correct result but I think it does. However, I don't particularly like it.

I consider this a bug and am tempted to submit an SCR. But before I do that; is there anything obvious I am missing here?

Thanks,

M.D.

PS: This was done using Maple 15. I'll check newer versions later.

mtaylor_assuming.mw

     Example of the equidistant surface at a distance of 0.25 to the surface
x3-0.1 * (sin (4 * x1) + sin (3 * x2 + x3) + sin (2 * x2)) = 0
Constructed on the basis of universal parameterization of surfaces.

equidistant_surface.mw 

When I was editing the head of the question (? instead of .), its body disappeared. Please, insert it again.

Regard,

Markiyan Hirnyk

I want to ask., I put delta as my constant in maple program and I want the answer are in delta as well., but the thing is., when running., it let delta=0, delta=-1, and delta=delta.,
the condition is we cannot let delta=1 or delta=0 because it is just same for s5 and s7.,.(delta is refer to the s8). How can I get answer as delta? with the condition? here I attach my maple programme..

> derivation := proc (A, n)
local i, j, k, t, s5, s7, s8, m, D,
sols5, sols7, sols8, eqns5, eqns7, eqns8,
BChange5, BChange7, BChange8; eqns5 := {}; eqns7 := {}; eqns8 := {};
D := matrix(n, n);
BChange5 := matrix(n, n); BChange7 := matrix(n, n); BChange8 := matrix(n, n);
for i to n do for j to n do for m to n do
s5 := sum(0*A[i, j, k]*D[m, k], k = 1 .. n)-(sum(A[k, j, m]*D[k, i]+A[i, k, m]*D[k, j], k = 1 .. n));
s7 := sum(0*A[i, j, k]*D[m, k], k = 1 .. n)-(sum(A[k, j, m]*D[k, i]+0*A[i, k, m]*D[k, j], k = 1 .. n));
s8 := sum(0*A[i, j, k]*D[m, k], k = 1 .. n)-(sum(A[k, j, m]*D[k, i]+delta*A[i, k, m]*D[k, j], k = 1 .. n));
eqns5 := `union`(eqns5, {s5}); eqns7 := `union`(eqns7, {s7}); eqns8 := `union`(eqns8, {s8})
end do end do end do;
sols5 := [solve(eqns5)]; sols7 := [solve(eqns7)]; sols8 := [solve(eqns8)];
t := nops(sols5); t := nops(sols7); t := nops(sols8);
for i to t do for j to n do for k to n do
BChange5[k, j] := subs(sols5[i], D[k, j]);
BChange7[k, j] := subs(sols7[i], D[k, j]);
BChange8[k, j] := subs(sols8[i], D[k, j])
end do end do;
print("eqns&Assign;", eqns5); print("sols:=", sols5); print("BChange5:=", BChange5);
print("eqns&Assign;", eqns7); print("sols:=", sols7); print("BChange8:=", BChange7);
print("eqns&Assign;", eqns8); print("sols:=", sols8); print("BChange8:=", BChange8)
end do end proc;

> AS1 := array(sparse, 1 .. 2, 1 .. 2, 1 .. 2, [(1, 1, 2) = 1]);
> derivation(AS1, 2);

> AS2 := array(sparse, 1 .. 2, 1 .. 2, 1 .. 2, [(1, 1, 1) = 1, (1, 2, 2) = 1]);
> derivation(AS2, 2);

> AS3 := array(sparse, 1 .. 2, 1 .. 2, 1 .. 2, [(1, 1, 1) = 1, (2, 1, 2) = 1]);
> derivation(AS3, 2);

> AS4 := array(sparse, 1 .. 2, 1 .. 2, 1 .. 2, [(1, 1, 1) = 1, (2, 2, 2) = 1]);
> derivation(AS4, 2);

> AS5 := array(sparse, 1 .. 2, 1 .. 2, 1 .. 2, [(1, 1, 1) = 1, (1, 2, 2) = 1, (2, 1, 2) = 1]);
> derivation(AS5, 2);

> AS1 := array(sparse, 1 .. 3, 1 .. 3, 1 .. 3, [(1, 3, 2) = 1, (3, 1, 2) = 1]);
> derivation(AS1, 3);

> AS2 := array(sparse, 1 .. 3, 1 .. 3, 1 .. 3, [(1, 3, 2) = 1, (3, 1, 2) = alpha]);
> derivation(AS2, 3);

> AS3 := array(sparse, 1 .. 3, 1 .. 3, 1 .. 3, [(1, 1, 2) = 1, (1, 2, 3) = 1, (2, 1, 3) = 1]);
> derivation(AS3, 3);

> AS4 := array(sparse, 1 .. 3, 1 .. 3, 1 .. 3, [(1, 3, 2) = 1, (2, 3, 2) = 1, (3, 3, 3) = 1]);
> derivation(AS4, 3);

> AS5 := array(sparse, 1 .. 3, 1 .. 3, 1 .. 3, [(2, 3, 2) = 1, (3, 1, 1) = 1, (3, 3, 3) = 1]);
> derivation(AS5, 3);

> AS6 := array(sparse, 1 .. 3, 1 .. 3, 1 .. 3, [(3, 1, 2) = 1, (3, 2, 2) = 1, (3, 3, 3) = 1]);
> derivation(AS6, 3);

> AS7 := array(sparse, 1 .. 3, 1 .. 3, 1 .. 3, [(1, 2, 1) = 1, (2, 2, 2) = 1, (3, 1, 1) = 1, (3, 3, 3) = 1]);
> derivation(AS7, 3);

> AS8 := array(sparse, 1 .. 3, 1 .. 3, 1 .. 3, [(1, 3, 1) = 1, (2, 3, 2) = 1, (3, 1, 1) = 1, (3, 3, 3) = 1]);
> derivation(AS8, 3);

> AS9 := array(sparse, 1 .. 3, 1 .. 3, 1 .. 3, [(2, 3, 2) = 1, (3, 1, 1) = 1, (3, 2, 2) = 1, (3, 3, 3) = 1]);
> derivation(AS9, 3);

> AS10 := array(sparse, 1 .. 3, 1 .. 3, 1 .. 3, [(1, 3, 1) = 1, (2, 3, 2) = 1, (3, 1, 1) = 1, (3, 2, 2) = 1, (3, 3, 3) = 1]);
> derivation(AS10, 3);

> AS11 := array(sparse, 1 .. 3, 1 .. 3, 1 .. 3, [(1, 3, 2) = 1, (2, 3, 2) = 1, (3, 1, 2) = 1, (3, 2, 2) = 1, (3, 3, 3) = 1]);
> derivation(AS11, 3);

> AS12 := array(sparse, 1 .. 3, 1 .. 3, 1 .. 3, [(1, 1, 2) = 1, (1, 3, 1) = 1, (2, 3, 2) = 1, (3, 1, 1) = 1, (3, 2, 2) = 1, (3, 3, 3) = 1]);
> derivation(AS12, 3);

> AS13 := array(sparse, 1 .. 3, 1 .. 3, 1 .. 3, [(1, 1, 1) = 1, (2, 2, 2) = 1, (3, 3, 3) = 1]);
> derivation(AS13, 3);

> AS14 := array(sparse, 1 .. 3, 1 .. 3, 1 .. 3, [(1, 2, 1) = 1, (2, 1, 1) = 1, (2, 2, 2) = 1, (3, 3, 3) = 1]);
> derivation(AS14, 3);

> AS15 := array(sparse, 1 .. 3, 1 .. 3, 1 .. 3, [(1, 2, 1) = 1, (2, 2, 2) = 1, (3, 3, 3) = 1]);
> derivation(AS15, 3);

> AS16 := array(sparse, 1 .. 3, 1 .. 3, 1 .. 3, [(2, 1, 1) = 1, (2, 2, 2) = 1, (3, 3, 3) = 1]);
> derivation(AS16, 3);

> AS17 := array(sparse, 1 .. 3, 1 .. 3, 1 .. 3, [(1, 1, 2) = 1, (3, 3, 3) = 1]);
> derivation(AS17, 3);
>

Rotational motion mechanism with quasi stops      
02rep.pdf
DIMA.mw

Sarrus linkage
https://en.wikipedia.org/wiki/Sarrus_linkage 

Sarrus_linkage.mw 

I have two polynomials f(x,y,z) and g(x,y,z) and ask MAPLE to find conditions on the coefficients of f and g such that the Jacobian determinant in x and y is purely a polynomial in z. MAPLE finds 4 solutions, one of which is g=0, but does not find the solution f=0. I attach the relevant MAPLE worksheet.

The mechanism of transport of the material of the sewing machine M 1022 class: mathematical animation.   BELORUS.mw 

Hi, currently im using maple 15

the coding did work but it is not the same with the answer
here, i attach the coding with the answer

coding:
derivation := proc (A, n)
local i, j, k, t, s1, s2, m, D, sols, eqns, Andre;
eqns := {};
D := matrix(n, n);
Andre := matrix(n, n);
for i to n-1 do
for j from i+1 to n do
for m to n do
s1 := sum(A[i, j, k]*D[m, k], k = 1 .. n);
s2 := sum(A[k, j, m]*D[k, i]+A[i, k, m]*D[k, j], k = 1 .. n);
eqns := `union`(eqns, {s1 = s2})
end do end do end do;
sols := [solve(eqns)];
t := nops(sols);
for i to t do
for j to n do
for k to n do
Andre[k, j] := subs(sols[i], D[k, j])
end do end do;
print(Andre)
end do end proc

the maple result showing:

> AS1 := array(sparse, 1 .. 2, 1 .. 2, 1 .. 2, [(1, 1, 2) = 1]);

> derivation(AS1, 2);
[D11 0]
[D21 D22]

> AS2 := array(sparse, 1 .. 2, 1 .. 2, 1 .. 2, [(1, 1, 1) = 1, (1, 2, 2) = 1]);
> derivation(AS2, 2);
[0 D12]
[D21 D22]

the maple should showing

> derivation(AS1, 2);
[D11 0]
[D21 2D11]

> AS2 := array(sparse, 1 .. 2, 1 .. 2, 1 .. 2, [(1, 1, 1) = 1, (1, 2, 2) = 1]);
> derivation(AS2, 2);
[0 0]
[D21 D22]

please help., thank you

how to field plot this system?

restart;
with(Physics[Vectors]);
with(DEtools);
with(VectorCalculus);
eq2 := ...;
eq3 := ...;
eq4 := ...;
with(DynamicSystems);
sys := DiffEquation([eq2 = t, eq3 = t], inputvariable = [b(t)], outputvariable = [a(t), c(t)]);
ts := .1;
in_t := t;
sol := Simulate(sys, [in_t]);

with(DEtools):
dfieldplot([...],[a(t),b(t),c(t)],t=−2..2,a=−1..2,b=−1..2,c=−1..2,arrows=SLIM,color=black,dirfield=[10,10]);

  Continuation.
  One way to get rolling without slipping animation in 3d. The trajectory and circle are divided into segments of equal length. In the next segment of the trajectory we construct circle, taking into account the fact that it turned on one segment. Rolling sphere or cylinder can be simulated, if we take plottools templates of the same radius, and replace them on the site of our circle.

ROLLING_WITHOUT_3d.mw

not the same ordering every time of monomials after determinant and map sign positive and op in maple 15

sometimes i need to use Reverse or Rotate List to adjust.

why ordering is different in list of monomials?

is it caused by virus?