how to find [[1,3,2,4],[1,2]] and [2,3][[1,3,2,4],[1,2]] and [2,4],[[1,3,2,4],[1,2]]?

and why Orbits(G) put permutation group into power, how to display the result of this power group?

i use this or quartic polynomials' resolvent

with(GroupTheory):
G := SpecialUnitaryGroup(3, 1);
Orbits(G);
Orbit(1,G);
Orbit(3,G);

with(GroupTheory):
G := Group({[[1, 2]], [[3, 4]]});
Orbit(G);
G := Group({[[1, 2], [3, 4]]});
Orbit(G);

GroupTheory:-PermutationGroup(

{module () local cycles, p, d, work; option object; end module,

module () local cycles, p, d, work; option object; end module},

degree = 4)
Error, invalid input: GroupTheory:-Orbit expects its 1st argument, point, to be of type posint, but received module () local labels, minSupp, maxSupp, suppSize, AtkinsonsAlgorithm, IsSimpleGroupOrder, doDerivedSeries, doLowerCentralSeries, Intersection2, RightCosetRepresentatives, LeftCosetRepresentatives, PRA, `Giant?`, `Even?`, doStab1, doStab, CycleIndexMonomial; export generator_list, n, supergroup, Sylows, pCores, ModulePrint, ModuleDeconstruct, Generators, Orbit, Orbits, IsTransitive, Transitivity, IsPrimitive, GroupOrder, Elements, IsAbelian, IsElementary, IsSimple, ConjugacyClass, ConjugacyClasses, CayleyTable, Centre, DerivedSubgrou...
GroupTheory:-PermutationGroup(

{module () local cycles, p, d, work; option object; end module},

degree = 4)
Error, invalid input: GroupTheory:-Orbit expects its 1st argument, point, to be of type posint, but received module () local labels, minSupp, maxSupp, suppSize, AtkinsonsAlgorithm, IsSimpleGroupOrder, doDerivedSeries, doLowerCentralSeries, Intersection2, RightCosetRepresentatives, LeftCosetRepresentatives, PRA, `Giant?`, `Even?`, doStab1, doStab, CycleIndexMonomial; export generator_list, n, supergroup, Sylows, pCores, ModulePrint, ModuleDeconstruct, Generators, Orbit, Orbits, IsTransitive, Transitivity, IsPrimitive, GroupOrder, Elements, IsAbelian, IsElementary, IsSimple, ConjugacyClass, ConjugacyClasses, CayleyTable, Centre, DerivedSubgrou...

how to calculate Ferrari resolvent of x^4-c1*x^3+c2*x^2-c3*x+c4

I was using Maple18 for the Ideal Membership Problem. While checking it I got the following error

Error, (in F4:-GroebnerBasis) argument `[0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,-48,-48,48,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0]/A1[119295][119295]` is incorrect or out of order

Please tell me, how can I resolve this error ?.

Thank You.

i use normaliser's example's code in maple help file

generators is [50] originally, then i calculated again , it become [51], [52], [53] , i do not know whether virus change my library

https://drive.google.com/file/d/0Bxs_ao6uuBDUb1VzaWQwQlBYLWs/view?usp=sharing

then i use another computer to calculate, the result is [50]

then i further calculate subgroup got error below

with(GroupTheory):
with(group):
G := AlternatingGroup(5);
IsFinite(G);
GroupOrder(G);
spg := SylowSubgroup(5, G);
IsAbelian(spg);
Elements(spg);
lprint(%);
H := Subgroup(Elements(G), spg);
N := Normaliser(G, spg);
#N := Normaliser(spg, G);
Elements(N);
lprint(%);
Elements(G);
H2 := Subgroup({[[5,2],[3,4]]}, G);
H2 := Subgroup(Elements(G), G);
elements2 := convert(Elements(G), 'list');
generators := map(ListTools:-Search, [Perm([[1,2,3]])], elements2);
H2 := Subgroup(generators, G);

H2 := Subgroup(Perm([generators]), G);
Error, invalid input: GroupTheory:-Subgroup expects its 1st argument, generators, to be of type {list, set, identical(undefined)}, but received module () local cycles, p, d, work; option object; end module
H2 := Subgroup(generators, G);
Error, (in Perm:-normalform) invalid input: map expects 2 or more arguments, but received 1

SubgroupMembership(H2, G);

I faced a very large eigenproblem during my research. The square matrix under consideration is of size more than 2^30 times 2^30. I have tried to deal with this problem by the QR algorithm with double implicit shift (more precisely, the Francis double step QR algorithm). I'm a very beginner of programming, but I tried as follows:

--------------------------------------------------------------------------------------------------

A := Matrix([[7, 3, 4, -11, -9, -2], [-6, 4, -5, 7, 1, 12], [-1, -9, 2, 2, 9, 1], [-8, 0, -1, 5, 0, 8], [-4, 3, -5, 7, 2, 10], [6, 1, 4, -11, -7, -1]]):
H := HessenbergForm(A):
p:=6:  
for p while p>2 do: 
q:=p-1: 
s:=H(q,q)+H(p,p):  
t:=H(q,q)*H(p,p)-H(q,p)*H(p,q): 
x:=(H(1,1))^(2)+H(1,2)*H(2,1)-s*H(1,1)+t: 
y:=H(2,1)*(H(1,1)+H(2,2)-s): 
z:=H(2,1)*H(3,2): 
for k from 0 to p-3 do:  
V:=Vector([x,y,z]):   
P:=Transpose(HouseholderMatrix(1/(Norm(V+exp(argument(V(1))*I)*Norm(V,2)*Vector(3,shape=unit[1]),2))*(V+exp(argument(V(1))*I)*Norm(V,2)*Vector(3,shape=unit[1])))):   
r:=max(1,k):
H[k+1..k+3,r..6]:=MatrixMatrixMultiply(Transpose(P),SubMatrix(H,[k+1..k+3],[r..6])):  
r:=min(k+4,6):
H[1..r,k+1..k+3]:=MatrixMatrixMultiply(SubMatrix(H,[1..r],[k+1..k+3]),P):   
x:=H(k+2,k+1):
y:=H(k+3,k+1):   
if k<3 then z:=H(k+4,k+1):   
end if: 
od: 
P:=GivensRotationMatrix(Vector([x,y]),1,2): 
H[q..p,p-2..6]:=MatrixMatrixMultiply(Transpose(P),SubMatrix(H,[q..p],[p-2..6])): 
H[1..p,p-1,p]:=MatrixMatrixMultiply(SubMatrix(H,[1..p],[p-1,p]),P): 
if abs(H(p,q))<10^(-20)*(abs(H(q,q))+abs(H(p,p))) then    H(p,q):=0: p:=p-1:q=p-1:  
elif abs(H(p-1,q-1))<10^(-20)*(abs(H(q-1,q-1))+abs(H(q,q))) then    H(p-1,q-1):=0: p:=p-2:q:=p-1:  
end if:  od:
--------------------------------------------------------------------------------------------------

It seemed that replacing 0 in a Hessenberg matrix by a non-zero element is not allowed. How can I remedy this?

Plus, can anyone tell me the problem of the above thing(it's not really a programming...;( ), please?

I would also appreciate it if someone let me know a better idea for a huge eigenproblem.

Thanks in advance.

I have some preferences for viewing. Specially, I prefer atomic variables to be showed with a different color. So I check mark the related option for this in the view menu. But every time I close and reopen the Maple, it just restore my setting for atomic variables. Suprisingly, this does not happen for other viewing options in the view menu! What should I do? Why is it like this?

Hi,

I was wondering how to go about plotting a Fourier Tranform in Maple.

My assignment is to plot a simple harmonic equation as a Fourier transform, depicting amplitude against fequency.

I've been given: x'' + w^2 x = 0

And want to obtain both the f(x) = a0 sin(wt) + b0cos(wt) form, and a graph of the the amplitude (c^2 =a0^2 + b0^2) against frequency.

I know how to do this on paper but not in Maple, so any help with line commands and layout would be very much appreciated.

Thanks

Hi,

I have an issue calculating an electronics circuit with Maple, using units. I have a current source that I know, and I want to determin the voltage in a capacitor by solving an ODE (except that the current source is defined piecewise). And to make sure I have all the units and scales right, I use the standard unit package. All my variables have their units defined.

Except that Maple doesn't want to solve the equation. It seems to me that it assumes that the function I am trying to solve is unitless, and therefore refuses to solve. 

V__out := 3*Unit('kV');

C__out := 2*Unit('nF');
R__blead := 520*Unit('`k&Omega;`');

I__fly := proc (t) options operator, arrow; Unit('A')*piecewise(t < 3.25*Unit('us'), (1+(-1)*t/(3.25*Unit('us')))*.2, 0) end proc;

dsolve({I__fly(t*Unit('s'))-V__C(t*Unit('s'))/R__blead = C__out*(diff(V__C(t*Unit('s')), t)), V__C(0*Unit('s')) = V__out}, V__C(t*Unit('s')));
Error, (in Units:-Standard:-+) the units `A` and `S` have incompatible dimensions

Is there a way to make Maple assume the unit of what it's trying to solve ? I need it to understands that V__C is in Unit('V') ...

Thanks

It took me a while to figure out what was causing this unusual sort result;

I loaded the MTM package because I want to convert numbers to integers for indexing. This sort result seem like it could be a bug.

Download MTMsort.mw

Variable exists but is not shown when using save/read with extension .m

My example maple code is

 

a:=b^2:

save a, "test1.m"

restart:

read "test1.m"

a;

after the read command, I can access the variable, but it is not shown under "Variables".

This lead to some confusion when debugging the worksheet. Can I change this somehow?

Using the input type file format is not a solution, since then reading takes forever for complicated expressions.

Further, in the read command documentation it says "This functionality is not intended for end users" for saving the file as .m. What does that mean?

Hi everyone,

Consider this maple 18 doc: Euler18.mw

The code is regular code for Julia sets of the exponential.

To see how the Julia set behaves at infinity, I apply the transform mu(z)=1/z.

The plot3d command correctly plots the Julia set at an appropriate neighborhood of infinity, but:

1) Axes are not transformed

2) Saving as .eps produces an empty plot, sans the axes (plot is saved correctly, when not applying mu(z))

Is there any trick to force the axes to also show correctly with the transformed ranges?

Seems that this misalignment is bothering the .eps renderer, which probably plots the graph in twilight zone, given the false ranges of the untransformed axes.

Any ideas on how to force the saveas .eps to work in this case?

Many thanks,

Yiannis

I would like to plot the following singular double integral, but I cannot due to singularities...

where x>0, t=0.2 and m=0.2.

I defined f(y) function as f:=y->exp(-(y-4.68)^2/0.4):

I attached my file:
1st_try.mw

Thank you !

i have this problem -> f'^2 -ff''=f'''-k1(2f'f'''-ff''''-f''^2)+Ha^2(E1-f') with boundary conditions f(0)=0, f'(0)=1, f'(∞)=0.

since it is a fourth order equation, but only three bcs, it does not produce unique solution. so the solution of the equation may be seek in form of f=f0(eta)+k1f1(eta).

thus the equation will become 

f0'^2-f0f0''=f0'''+Ha^2(E1-f0')

and

f1'''-Ha^2f1'-2f0'f1'+f0f1''+f1f0''=2f0'f0'''-f0f0''''-f0''^2.

boundary conditions are 

f0(0)=0,f0'(0)=1,f0'(∞)=0

f1(0)=0,f1'(0)=0,f1'(∞)=0.

i had been clueless in solving this problem. please somebody help me with this problem.

Given the sequence defined by the recursive relation a[n+1] = r*a[n](1-a[n])
You need to use the procedure iterate.
Throughout this problem you should choose initial values in the interval 0<a0<1.
(a) Let r=3/2. Calculate a moderate number of terms in the sequence (between 10 and 20). Does the sequence appear to be converging? If so to what value? Does the limit depend upon your choice of initial value? Plot the terms you have calculated
(b) Let r=2.8. Calculate a moderate number of terms in the sequence (between 10 and 20). Does the sequence appear to be converging? If so to what value? Does the limit
depend upon your choice of initial value? Plot the terms you have calculated How does this sequence differ from that in part (a).
(c) Let r=3.2. Calculate a moderate number of terms in the sequence (between 10 and 20). Show that the sequence does not appear to converging. Plot the terms you have calculated and describe how the sequence behaves in this case.
(d) Consider intermediate values between 2.8 and 3.2 to determine more precisely where the transition in behaviour takes place. Provide a few plots (no more than 4) showing the values you have investigated.
(e) Consider the values of r in the range 3.43<r<3.46. Determine as accurately as you can the value of r for which the period of oscillation doubles.
(f) As r increase further period doubling occurs. Try to find the when the sequence appears to oscillate between 8 values.
(g) Let r =3.65 and calculate a considerable number of terms (at least a few hundred) and plot your values.
(h) For r=3.65 choose a0=0.3 and then a0=0.301. Find and plot some terms in the sequence for each initial value. Determine how long the terms in the two sequences remain close together and when they begin to depart significantly from each other.