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Hello! Hope every is fine. I want to expand the following expression

exp(2*c*t+2*d*n-d)*alpha*c*a[0]*b[1]^2-exp(2*c*t+2*d*n-d)*alpha*c*a[1]*b[0]*b[1]-exp(2*c*t+2*d*n-d)*alpha*a[0]*a[1]*b[1]+exp(2*c*t+2*d*n-d)*alpha*a[1]^2*b[0]+exp(2*c*t+2*d*n)*alpha*a[0]*a[1]*b[1]-exp(2*c*t+2*d*n)*alpha*a[1]^2*b[0]+exp(c*t+d*n)*alpha*c*a[0]*b[0]*b[1]-exp(c*t+d*n)*alpha*c*a[1]*b[0]^2-exp(2*c*t+2*d*n-d)*a[0]*b[1]^2+exp(2*c*t+2*d*n-d)*a[1]*b[0]*b[1]-exp(c*t+d*n-d)*alpha*a[0]^2*b[1]+exp(c*t+d*n-d)*alpha*a[0]*a[1]*b[0]+exp(2*c*t+2*d*n)*a[0]*b[1]^2-exp(2*c*t+2*d*n)*a[1]*b[0]*b[1]+exp(c*t+d*n)*alpha*a[0]^2*b[1]-exp(c*t+d*n)*alpha*a[0]*a[1]*b[0]-exp(c*t+d*n-d)*a[0]*b[0]*b[1]+exp(c*t+d*n-d)*a[1]*b[0]^2+exp(c*t+d*n)*a[0]*b[0]*b[1]-exp(c*t+d*n)*a[1]*b[0]^2

 

like this 

exp(2*c*t+2*d*n)*exp(-d)*alpha*c*a[0]*b[1]^2-exp(2*c*t+2*d*n)*exp(-d)*alpha*c*a[1]*b[0]*b[1]-exp(2*c*t+2*d*n)*exp(-d)*alpha*a[0]*a[1]*b[1]+exp(2*c*t+2*d*n)*exp(-d)*alpha*a[1]^2*b[0]+exp(2*c*t+2*d*n)*alpha*a[0]*a[1]*b[1]-exp(2*c*t+2*d*n)*alpha*a[1]^2*b[0]+exp(c*t+d*n)*alpha*c*a[0]*b[0]*b[1]-exp(c*t+d*n)*alpha*c*a[1]*b[0]^2-exp(2*c*t+2*d*n)*exp(-d)*a[0]*b[1]^2+exp(2*c*t+2*d*n)*exp(-d)*a[1]*b[0]*b[1]-exp(c*t+d*n)*exp(-d)*alpha*a[0]^2*b[1]+exp(c*t+d*n)*exp(-d)*alpha*a[0]*a[1]*b[0]+exp(2*c*t+2*d*n)*a[0]*b[1]^2-exp(2*c*t+2*d*n)*a[1]*b[0]*b[1]+exp(c*t+d*n)*alpha*a[0]^2*b[1]-exp(c*t+d*n)*alpha*a[0]*a[1]*b[0]-exp(c*t+d*n)*exp(-d)*a[0]*b[0]*b[1]+exp(c*t+d*n)*exp(-d)*a[1]*b[0]^2+exp(c*t+d*n)*a[0]*b[0]*b[1]-exp(c*t+d*n)*a[1]*b[0]^2

i.e., expand exp(2*c*t+2*d*n-d) into exp(2*c*t+2*d*n)*exp(-d) 

waiting your kind response 

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

Dear Maple T.A. users

I have just begun using Maple T.A. I have access to a number of questions, some of which involves placing points in a coordinate system. It works for most students, but for a few, including myself, it doesn't work. I am not able to place those points in the coordinate system at hand when leftclicking. What can be the reason for this issue?

Erik

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!

I am interested in dynamic systems that changes system equations at a given point in time. So i often want to plot graphs that shows what would happen in the first 500 seconds, then using the point reached after 500 seconds as the starting point show what happens over the next 500 seconds.

For example my equations might innitially

diff(x,t)=x+p*y

diff(y,t)=x/y

and then after 500 seconds switch to 

diff(x,t)=x-p*y

diff(y,t)=x/y

simply estimating where the system is and feeding that into the other equation isn't an option because these equations have lots of parameters which p is representing in the above, and generally i want too use these graphs to illustrate the behaveious of the systems with the given parameters.

So far i use display and DEplot to make these grpahs.

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

I want to cancel some expressions in numerator and denominator of a quotient. But Maple deos not cancel it!

Please see the attached file.


Simplifying_Radicals.mw

Hello everyone, how can this error be corrected

pde.mw

Thanks.

Hello,

I have a vector with 4 components. Each component is a polynom.

N := Vector[row](4, {(1) = -(1/2)*t^3+t^2-(1/2)*t, (2) = (3/2)*t^3-(5/2)*t^2+1, (3) = -(3/2)*t^3+2*t^2+(1/2)*t, (4) = (1/2)*t^3-(1/2)*t^2});
N[1]+N[2]+N[3]+N[4];
sum(N[i], i = 1..4);

I don't why the function sum doesn't work in this case. If i do the sum like this N[1]+N[2]+N[3]+N[4];, it works but it doesn't work with the use of the sum function.

Thank you for your help.

Hello,

I have a little question which is in the title.

It seems to me at the view of this example :

[x[i], y[i]] $ i=1..4;

seq([x[i], y[i]],i=1..4);

May you give me your feedback to be sure ?

Thank you 

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.

What does one use to duplicate a procedure along with its remember table, so that they are distinct instances of said?

in LinearAlgebra Eigenvectors calculation.

Maple 2015 Error

 

 

So the above output startled me.  I have used the Maple Linear Algebra Eigenvalues, Eigenvectors commands many times with no problem.  Can any one explain to me what is going on.  The program correctly calculates the eigenvalues for the matrix which are all distinct for a real symmetric matrix, and thus should have three distinct non-zero eigenvectors, yet the eigenvectore command returns only zeros for the eigenvectors.  I calculated an eigenvector by hand corresponding to the eigenvalue of 1 and obtained (1, -sqrt(2)/sqrt(3), -1/sqrt(3).

 

So this is either a serious bug or I am going completely insane. 

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