Unanswered Questions

This page lists MaplePrimes questions that have not yet received an answer

and how to prove this make x and y are conjugate by an element of  N?

with(GroupTheory):
with(group):
G := AlternatingGroup(3);

IsFinite(G);

GroupOrder(G);

spg := SylowSubgroup(3, G);

IsAbelian(spg);

Elements(spg);
lprint(%);

H := Subgroup(G, spg);
got error, invalid input here,

GroupTheory:-SylowSubgroup(3, 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, DerivedSubgroup,

IsPerfect, DerivedSeries, LowerCentralSeries, NilpotencyClass\

, IsNilpotent, doUpperCentralSeries, UpperCentralSeries,

SylowSubgroup, IsSubgroup, IsNormal, Core, NormalClosure,

Normaliser, Conjugator, AreConjugate, Centraliser,

Intersection, `intersect`, LeftCoset, RightCoset,

RightCosets, LeftCosets, Factor, RandomElement, IsAlternating\

, IsSymmetric, PCore, FittingSubgroup, FrattiniSubgroup,

MatrixRepresentation, Stabiliser, CycleIndexPolynomial,

properties; option object; end module)
Error, invalid input: GroupTheory:-Subgroup expects its 1st argument, generators, to be of type {list, set, identical(undefined)}, 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, ...


N := Normaliser(G, spg);

H2 := Subgroup(G, G);

how to find the subgroup G which is finite group here?


elist := Elements(H2);
AreConjugate(elist[1], elist[2], N);

originally x and y are not conjugate,

how to prove this make x and y are conjugate by an element of  N if spg is abelian

where x and y are elements of H2 which is subgroup of G, which is finite group

 

i guess find subgroup with following command, however, normaliser N can not

make elements of x and y conjugate

H2 := Subgroup(Elements(G), G);
elist := Elements(H2);
AreConjugate(elist[2], elist[3], N); #N*elist[1]*N^(-1) = elist[2]

but it is false,

i use G := AlternatingGroup(5); it is true,

does it mean that this theorem is not for all cases?

Good Morning,

This is the script that I have used to to an integration, as the command ''int'' didn't work because the coputation was really heavy.


p_old := 0;

g := 0; #index

E_tot_net_Point := 0;

for g to 2 do

 h := g+1;

 loopstart := SP_time[g];

 loopend := SP_time[h];

  for tt from loopstart by 0.01 to loopend do

   t := tt;

   P_tot_net_Fkt := t -> C_motore[g]*omega_motore[g]

if t≠SP_time[g] and t ≠ SP_time[h] then

 P_tot_net_Point := P_tot_net_Fkt(t);

 P_tot_net_Point_next := P_tot_net_Fkt(t+0.001);

 plt_P_tot_net_plt[t] := pointplot([t, P_tot_net_Point]);

 E_tot_net_Point := E_tot_net_Point+(P_tot_net_Point*0.01);

 plt_E_tot_net_plt[t] := pointplot([t, E_tot_net_Point]);

 p_old := P_tot_net_Point;

end if;      

  end do:

end do:

plots:-display(entries(plt_E_tot_net_plt, 'nolist'), title = "Energy", color = "Red", symbol = point, symbolsize = 20, gridlines = true, size = [1000, 500], labels = ['[s]', '[Watt*s]'], font = [axes, 12])


I have plotted the function as a series of points over the orizontal axis (t).

 

Now I would like to calculate the last value of the curve (which is itself a point) as a function of the orizontal axes t. How can I do it?

Something like E_tot_net_Point = f(t)

 

Thank you for the help

                                                                                                                                         

I want to build a custom component by the material’s stress-strain curve ,for the component signal input is the x(t)(displacement-time curve), then I can simulate the material’s acceleration-time curve,the stress-strain equation , parameter and the custom component I built has been given in the picture,but the x(t) is same as the s[rel](t),so how to solve this problem? Thanks first.

 

 

 

 

 

we have positive number from 1 to 1000. how many time we write number 3?

How to increase the size of toolbar icons in Maple 2015 32-bit Classic on Windows?

How do you put labels on individual columns in ColumnGraph?

a1:= f(x) :
> T1 :=simplify((taylor(a1,x=alpha,N+3))):
> E1:=subs([seq(((D@@i)(f))(alpha) = 0,i=1..m-1),f(alpha)=0,x=e[n]+alpha],T1):
> g1 :=(convert(simplify(series((E1,e[n]=0,N))),polynom));

 

 



thanks. I played around, and had problems implementing your ideas for one of the systems I'm interested in.I don't see a difference between this and what you had advised me on, but it gets an error.

any idea why?
or how to fix it?

thing1 := diff(B[1](t), t) = piecewise(t <= 500, 0.3e-2-(63/10000)*B[1](t)-(3/500)*B[2](t), -(3/10000)*B[1](t)):
thing2 := diff(B[1](t), t) = piecewise(t <= 500, 0.1e-1-(1/50)*B[1](t)-(13/625)*B[2](t), -(1/1250)*B[2](t)):
sol := dsolve({thing1, thing2, B[1](0) = 0, B[2](0) = 0}, {B[1](t), B[2](t)}, numeric, output = listprocedure); plots:-odeplot(sol, [B[1](t), B[2](t)], t = 450 .. 550);

Error, (in dsolve/numeric/DAE/explicit) unable to obtain the standard form of the DAE system due to the presence of leading dependent variables/derivatives in the piecewise: piecewise(t <= 500, 1/100-(1/50)*B[1](t)-(13/625)*B[2](t), -(1/1250)*B[2](t))-piecewise(t <= 500, 3/1000-(63/10000)*B[1](t)-(3/500)*B[2](t), -(3/10000)*B[1](t))
Error, (in plots/odeplot) curve is not fully specified in terms of the ODE solution, found additional unknowns {B[1](t), B[2](t)}


how to do differentiation of an ideal in maple?

availables variables : a,b,c

case 1 : all are independent variebles, a,b,c

case 2 : only one independent variable, a

case 3: only one dependent variable a

 

i find this, but i do not know respect to which variable when differentiate an ideal which has 3 variables and 3 equations

http://www.maplesoft.com/support/help/maple/view.aspx?path=DifferentialAlgebra%2fTools%2fDifferentiate

I try to repeat lines (25)-(28) at

 

http://www.maplesoft.com/support/help/maple/view.aspx?path=Physics%2fTrace#commentform

 

I use Maple 14. However, instead of (28) I get the following result:

 

It means that Maple 14 does not perceive p_\mu, k_\nu and m as scalar quantities. I would like to ask how to define these variables correctly.

 

Thank you in advance!

hi.please help me for solve this equations

thanks...

Tur.mw

Non dimensionalisation is a vary common task, and I was suprised that I couldn't find a maple tool to automate it . Has anyone developed their own package for it?

I want to automatically do it to the system equations for some Dynamical systems to make some of the other processing I do with them easier.

I was hoping to start with somehting in the form of 

Diff(x[1],t)=f[1](p[1]....p[n],x[1]...x[m])

...

Diff(x[m],t)=f[m](p[1]....p[n],x[1]...x[m])

where each f[i] is some kind of quotient of multivariate polynomials in the variables and parameters:
and end up with something like

Diff(y[1],s)=f[1](q[1]....q[p],y[1]...y[m])

...

Diff(y[m],s)=f[m](q[1]....q[p],y[1]...y[m])

where p<n

Hi everyone...!

Can somebody tell me how to express this equation in Maple? 

xij <= zkl ; ∀ i ∈ I: S(i)=k, ∀ j ∈ B: R(j)=l; 

Currently I'm dealing with containerization problem and have 4 indexes in the constraints (namely: i for item, j for container, k for shipment, l for route, S for Set of Shipment, and R for Set of Route) while x and z are binary variables. What I want to express is: (for example), item 1,2,3 are in shipment 1, item 4,5 are in shipment 2, etc etc. SO, if i = 1,2,3 then the value of k will be 1. If i = 4,5 then the value of k will be 2, etc. Same thing goes to j and l, (for example) if j = 1,2 then the value of l will be 1, etc etc. Further depcition is more or less like this:

S(i) = k

S(1) = 1

S(2) = 1

S(3) = 1

S(4) = 2

S(5) = 2

 

Thank you very much for the help.

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.

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