for example, how to convert below word relation into graph

a^2 = 1

a*b = b*a

my code can only do for one variable, 

how to make divisible checking for multivariable cases with the ordering such as plex

IsDivisible(LP(h, t), LP(g[i], t), x)

it is not only x when multivariable

f:=LP(y^2*x,plex(x, y))[2];
g:=LP(y*x-y,plex(x, y))[2];
Remainder(f, g, gcd(f,g));
degree(Remainder(f, g, x),x);
degree(g, x);

remainder has error expect its 3rd argument x, to be of type or but received y*x

how to do if have ordering

do it need to check whether both f and g have variable x using indets and then apply remainder?

do it need to check each variable starting from the first variable in the ordering? 

how about if f has variable x but g do not have variable x, or f do not have variable x and g have variable x

if so, i try to replace below code in the bottom code, it has error

Error, (in FindDivisble) cannot determine if this expression is true or false: 0 < Search(x, {x, y})

FindDivisble := proc(g, h, t)
with(ListTools):
result := 0;
for i from 1 to nops(g) do
mainvariable := 0;
for j from 1 to nops(t) do
mainvariable := op(j, t);
if mainvariable <> 0 then
if Search(mainvariable, indets(h)) > 0 and Search(mainvariable, indets(g[i])) > 0 then
if IsDivisible(LP(h,t), LP(g[i],t), mainvariable) = 0 then
return i;
else
result := 0;
end if:
end if:
end if:
od:
od:
return result;
end proc:

with(Groebner):
LP := proc(f, t)
return LeadingTerm(f, t)/LeadingCoefficient(f, t);
end proc:
IsDivisible := proc(f, g, x)
with(Algebraic):
if Remainder(f, g, x) = 0 or degree(Remainder(f, g, x),x) < degree(g, x) then
return 0;
else
return 1;
end if:
end proc:
FindDivisble := proc(g, h, t)
result := 0;
for i from 1 to nops(g) do
if IsDivisible(LP(h, t), LP(g[i], t), x) = 0 then
return i;
else
result := 0;
end if:
od:
return result;
end proc:
MD := proc(f, g, t)
r := 0;
u := Matrix(nops(g), 1);
for j from 1 to nops(g) do
u[j] := 0;
od:
h := f;
while h <> 0 do
i := FindDivisble(g, h, t);
if i > 0 then
u[i] := u[i] + LeadingTerm(h, t)/LeadingTerm(f[i], t);
h := h - LeadingTerm(h, t)/LeadingTerm(f[i], t)*f[i];
else
r := r + LeadingTerm(h, t);
h := h - LeadingTerm(h, t);
end if:
od:
end proc:
f:=y^2*x;
f1 := y*x-y;
f2 := y^2-x;
MD(f,[f1,f2],plex(x, y));

From some scientific experiments, the following set of data is available
(1.0, 2.33),(2.0, 0.0626),(3.0, −2.16),(4.0, −2.45),(5.0, −0.357),(6.0, 2.21),(7.0, 2.75),(8.0, 0.636),(9.0, −2.45).

We need to use least squares method to fit the curve to the data
y = a + b cos(x) + c sin(x) + d cos(2x) + e sin(2x)

Show your solution procedure in Maple and get the best fit for the coefficients. Finally
plot the data and the y curve together on the same graph to visualize the fitting. For the
data, use blue circles while for the y curve, use red solid line style. Use the leastsquares
command we learned in LinearAlgebra package.

Find the eigenvalues and eigenvectors for

A =
2 0.37 0
0 1.1 −4.29
1.6 0 2.2

Can you estimate the largest eigenvalues (in absolute value) using Power method for this
problem? Try to explain why if you can

If I have a list, how can I write a program to see is that list is graphical? So far I have

graphicalseq := proc (L::list)

local i::integer, N;

N := convert(L, `+`);

if type(N, odd) then print("Sequence is not graphical")

else if numelems(L)-1 < L[1] then print("Sequence is not graphical")
end if;

end if;

end proc;

I know I still have to keep going to determine whether the sequence is graphical, but I'm not sure how.

 I was thinking of trying to somehow use Havel-Hakimi's theorem, but again I'm not sure how. Any hints would be appreciated.  I can't use the is Graph Sequence function

Hello,

A flate metal plate lies on an xy plane such that the temperature T at (x,y) is given by T=10(x^2+y^2)^2 , where T is in degrees and x and y are in centimeters.

Find the instantaneous rate of change of T with respect to distance at (1,2) in the direction of

a) the x-axis

b) the y-axis

Hi,

I have to run my model with lots of parameter sets and I want to export the result every time as a csv file.

Is therea possibilty to export the result automatically?

Thanks in advance for any help!

How do I permanently change the current directory on a MAC computer?

Kind regards

Per Kirkegaard

Hello,In short: I try to get the eigenvalues of a 16x16 complex matrix with one variable B__z. I know, the vales are very small (~10^-24), so I multiply with 10^24 and collect B__z. Now, when I use Eigenvalues(H) [H is the matrix] the result is just wrong:

test.mw

(same with original values) - compared to Mathematica which solves and plots in about 1 sec:

Where is the problem? What should I do? Any suggestions?

is it possible to assume element of matrix 0 or 1

how to write?

after write this, is it possible to display possible matrices which each element is 0 or 1

with(LinearAlgebra):
GetRing := proc(sol)
ringequation := 0;
mono1 := 0;
for j from 1 to 3 do
mono1 := 1;
for i from 1 to nops(sol[1][j]) do
mono1 := mono1*op(i, sol[1][j]);
od:
ringequation := ringequation + mono1;
od:
return ringequation;
end proc;
M := Matrix([[a1,a2,a3],[a4,a5,a6],[a7,a8,a9]]);
sys := a*b+a = GetRing(MatrixMatrixMultiply(Matrix([[a,b,c]]), M));
solve(sys);

I am running a huge symbolic problem in Maple 18. At the moment, I dont even know if there is enough computational power to suceed the computation.

However, I'd like to save a progress. And maybe to continue computation later.

Is there any way to save progress?

Hi,

I use Maple on my laptop computer (windows 8.1) with a small screen. My problem is that the text in the menus, palettes, plots etc. are so small so I can´t read it. I have set Large toolbar icons under the menu tools-options-interface and I use the zoom buttom on the toolbar but this only magnify the text in a document. I also use the magnifying glass in windows, but I find it very inflexible. Is there any better solution to this problem?

Kind Regards

Leif Jonson

I am not sure I know what is going on. I expected to obtain the controllable canonical form in this example, but I am not. May be I am not using it correctly. 

Given A,B, I wanted to transfer the state space to controllable canonical form

http://www.maplesoft.com/support/help/Maple/view.aspx?path=DynamicSystems%2FSSTransformation

In this form, the A matrix will have 1 on the super diagonal, and the last row will have the coefficients of the charaterestic polynomial in reverse order with a minus sign. The B matrix will have all zeros, except for the last entry.  This is what the example on the above page actually shows. 

But when I tried it on my A,B, I do not get this form for the new B matrix. Here is a MWE

restart;
with(DynamicSystems):
A:=Matrix([[0,0,1,0],[0,0,0,1],[-2,-1,0,0],[1,-1,0,0]]);
B:=Matrix([[0],[0],[1],[0]]);
C:=Matrix([[0,0,0,0]]):
D0:=Matrix([[0]]):
sys:=StateSpace(A,B,C,D0): #just to see the polynomial
CharacteristicPolynomial(sys, s);

SSTransformation(A,B,C,D0,form=ControlCanon,output=['A','B']);

The above should be

I am sure I am doing something wrong, but what?

Maple 18.01, windows 7

Hello,

I have a question about the CycleIndexPolynomial command contained in the Group Theory package. The answers it gives for computing the cycle index of the dihedral group and symmetric group appear to has a missing term in its answer. If my knowledge of how the cycle index is computed, it appears to be missing a term that occurs when cycles of length greater than one is combined with terms with cycles of length one. For example, here is what I get when I use Maple 18 on my Windows 8.1 machine to compute the cycle index of the following dihedral groups:

> with(GroupTheory):

> CycleIndexPolynomial(DihedralGroup(3), [x || (1 .. 3)]);

> CycleIndexPolynomial(DihedralGroup(4), [x || (1 .. 4)]);

> CycleIndexPolynomial(DihedralGroup(5), [x || (1 .. 5)]);

 Now if you compare these answers from the cycle index I found at the link:

 http://mathworld.wolfram.com/DihedralGroup.html

 you will see the answers is given as

  =

 =

 =

The disagreement occurs at the terms that contain representations of cycles of length two x2 multiplied to the one-cycle representation term x1.

The same issue happens with the symmetric group. Here is what I get for the following Maple commands:

> with(GroupTheory):

> CycleIndexPolynomial(SymmetricGroup(3), [x || (1 .. 3)]);

> CycleIndexPolynomial(SymmetricGroup(4), [x || (1 .. 4)]);

> CycleIndexPolynomial(SymmetricGroup(5), [x || (1 .. 5)]);

However, as can be seen at the link

http://mathworld.wolfram.com/SymmetricGroup.html

these answers are not in agreement with these:

  =

 =

 =

Again, the difference seems to be with terms combined with the one cycle term representation x1.

Is there something I am not interpreting correctly? Thank you for your help.

Neil Sigmon