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Hello Dear!

I want to solve the system of linear equation but facing some problem please see the attachmen. I am waiting your positive response 

1_(1).mw

Hi,

 

I am trying to solve a simple system of the form AX=0, where A is a N*N matrix, X is an N*1 vector (and the right-hand side of the equation is an N*1 vector of zeros, I apologize for the inexact notation). The difficulty comes from the fact that the values of A are parameterized by 2*N parameters (that I will write as the 2*N vector P), and I would like to get a solution in the form X=f(P).

 

One solution is to try to use LinearAlgebra[LinearSolve], but it only returns the trivial solution X=0, which I am not interested in.

Another solution is to compute analytically the Moore-Penrose pseudoinverse Ag of A, as the general solution is of the form

(I - Ag A)f ;

where f is a vector of free parameters. However, even for a small matrix size (N=4), Maple is still computing after 3 hours on my (fairly powerful) machine, and it is taking more and more memory over time. As the results are polynomial/rational equations in the parameters P, I was actually expecting Maple to be more powerful than other softwares, but for this particular problem, Matlab's symbolic toolbox (muPAD) gives quick solutions until N=6. I need, in the end, to solve additional polynomial/rational equations that are derived from the solutions X=f(P), where Matlab fails. This is why I would really like to be able to solve the above-mentioned problem AX=0 with Maple in order to try to solve the subsequent step of the problem (polynomial system) with Maple.

 

Any suggestions on how to do this would be highly appreciated! Thank you very much for your time and help.

 

Laureline

Hello..  I want to know if there is anny command to show the matrix of linear system.  I recently entred a 64 equations and i solved it by command solve,  but i want to show the matrix of system..  So plz. Help 

I have an equation eq := diff(y(x), x$3)+3*diff(y(x), x$2)+12*y(x);

dsolve(eq, y(x)); gave me a general solution.

I tried to get a particular solution using dsolve({eq, y(0) = a, y'(0)=0, y"(0) = 0}, y(x));

But I got Error, (in dsolve) not a system with respect to the unknowns [y(x)].

Thank you for any help.

Heather

Does any one know if you can extract the linear graph from a Maple Sim model?  And by linear graph I mean the alternative to a a bond graph, not a type of plot.

I solve a linear system of equations which is rank deficient. Naturally, when Maple solves it symbolically, it chooses some of its variables to use them as a basis to express the solution. 

In a specific problem I'm solving, the basis chosen by Maple is -very- smart, showing a good exploitation of the problem structure. 

I'm curious as to what kind of factorization is used by default, or if there's a lot of by hand "black magic" involved, what are its general characteristics. 

 

Best regards

Claudio

After manually working out answer for problem 4-4 in Mathews & Walker's Mathematical Methods of Physics , I tried to check my solution with maple2015. Briefly the problem involves inputs periodic with period T, being transformed into outputs, through a kernal G.  The net result is that all input frequencies omega periodic in T are multiplied by (omega_0/omega)^2, except for constant frequency which transforms to zero.  The problem asks to evaluate the kernal G.

Maple2015 correctly evaluated the integral for a constant input, a cosine input, and a sine input, but gave undefined when I tried an exponential(i*x) input which is just a linear combination of the two previous inputs.  I found this interesting because the integral is finite, well defined, and only has an absolute function (in the kernal), which may cause Maple problems, as it correctly evaluated integral when I split it into two regions.  Interestingly if instead of working with a period of T, I used 2*pi, and redfined my G function accordingly, Maple evaluated the exp input integral without any problems.  So the problem appears to be with the T variable, but I correctly used assumptions of T>0, and 0<t<T, so I am not sure why it would work correctly when I use T=2*pi, but failed when using a general period T.  Any help would be welcome.

 

 

restart

assume(T > 0)

assume(0 < t and t < T)

about(T)

Originally T, renamed T~:

  Involved in the following expressions with properties
    T-t assumed RealRange(Open(0),infinity)
  is assumed to be: real
  also used in the following assumed objects
  [T-t] assumed RealRange(Open(0),infinity)

 

about(t)

Originally t, renamed t~:

  Involved in the following expressions with properties
    T-t assumed RealRange(Open(0),infinity)
  is assumed to be: RealRange(Open(0),infinity)
  also used in the following assumed objects
  [T-t] assumed RealRange(Open(0),infinity)

 

assume(n::integer, n > 0)

about(n)

Originally n, renamed n~:

  is assumed to be: AndProp(integer,RealRange(1,infinity))

 

G := proc (x) options operator, arrow; (1/2)*omega0^2*T^2*((1/6)*Pi^2-(1/2)*Pi*abs(2*Pi*x/T)+Pi^2*x^2/T^2)/Pi^2 end proc

proc (x) options operator, arrow; (1/2)*omega0^2*T^2*((1/6)*Pi^2-(1/2)*Pi*abs(2*Pi*x/T)+Pi^2*x^2/T^2)/Pi^2 end proc

(1)

(int(G(t-tp), tp = 0 .. T))/T

0

(2)

(int(G(t-tp)*sin(2*Pi*n*tp/T), tp = 0 .. T))/T

(1/2)*T^2*omega0^2*cos(t*Pi*n/T)*sin(t*Pi*n/T)/(Pi^2*n^2)

(3)

(int(G(t-tp)*cos(2*Pi*n*tp/T), tp = 0 .. T))/T

(1/4)*T^2*omega0^2*(2*cos(t*Pi*n/T)^2-1)/(Pi^2*n^2)

(4)

(int(G(t-tp)*exp((I*2)*Pi*n*tp/T), tp = 0 .. T))/T

undefined/T

(5)

(int(G(t-tp)*(cos(2*Pi*n*tp/T)+I*sin(2*Pi*n*tp/T)), tp = 0 .. T))/T

undefined/T

(6)

simplify((int(G(t-tp)*exp((I*2)*Pi*n*tp/T), tp = 0 .. t))/T+(int(G(t-tp)*exp((I*2)*Pi*n*tp/T), tp = t .. T))/T)

(1/4)*omega0^2*exp((2*I)*t*Pi*n/T)*T^2/(Pi^2*n^2)

(7)

assume(0 < t and t < 2*Pi)

G2 := proc (x) options operator, arrow; 2*omega0^2*((1/6)*Pi^2-(1/2)*Pi*abs(x)+(1/4)*x^2) end proc

proc (x) options operator, arrow; 2*omega0^2*((1/6)*Pi^2-(1/2)*Pi*abs(x)+(1/4)*x^2) end proc

(8)

(int(G2(t-tp)*exp(I*n*tp), tp = 0 .. 2*Pi))/(2*Pi)

omega0^2*exp(I*n*t)/n^2

(9)

 

Download MathewsWalkerProblem4-4.mwMathewsWalkerProblem4-4.mw

 

 

I'm trying to solve the differential equation.

Eq := diff(y(x), x, x) = -(x^2+1)*y(x)+K;

dsolve({Eq, y(-1) = 0, y(1) = 0}, y(x));

But this not work very well.

Best Regards,

Hello all

I am trying to write  a tutorial about systems of linear equations, and I want to demonstrate the idea that when you have a system of 3 euqtions with 3 unknowns, the solution is the intersection point between these planes. Plotting 3 planes in Maple 2015 is fairly easy (you plot one and just drag the others in), but I don't know how to plot the intersection point. Can you help please ?

 

My equations are:

x-2y+z=0

2y-8z=8

-4x+5y+9z=-9

The intersection point is (29,16,3)

 

Thank you !

I have a vector of dimension n with each component being an equation of a linear system.

Can maple convert this Vector to a Matrix-Vector form with the matrix being constant coefficients?

Dear Friends,

I am having trouble in defining a linear differential operator. This is how Maple defines a linear differential operator

"A differential operator L in C(x)[Dx] is an expression a_0*Dx^0+ ... +a_n*Dx^n where a_0, ... , a_n are elements of C(x). So it is a polynomial in Dx with rational functions as coefficients."

Here's the link http://www.maplesoft.com/support/help/maple/view.aspx?path=diffop

where Dx^n implies n-th derivative with respect to x.

Now I want to know how can I apply this operator to a given function. Here is a failed example:

>L :=Dx^2+Dx;
Dx^2 + Dx
>simplify(L(x));
Dx(x)^2 + Dx(x)

I was expecting to get 1 if the given opertor is applied to x. I would really appreciate if someone can help me with this. 

Many thanks.

 

Hello,

I have a problem with MAPLE. I would like to solve a system of 18 inequalities with 4 variables. The variables shall be rational numbers. I should also mention that I am not sure if the system has a solution. Here is my MAPLE code: 

LinearMultivariateSystem({0 < (1/20)*b11, 0 < (1/20)*b1818, 0 < (1/20)*b22, 0 < (1/20)*b33, 0 < -653385574770525739/313841848320000+(1001/20)*b33+(3003/5)*b22+4004*b11-(91/5)*b1818, 0 < -476383516463665673/69742632960000+(3003/20)*b33-(1001/10)*b1818+(27027/2)*b11+(3861/2)*b22, 0 < -372810037848242383/52306974720000+(72072/5)*b11+(3003/20)*b33-(858/5)*b1818+2002*b22, 0 < -302968656462848461/125536739328000+(1001/20)*b33-(1001/10)*b1818+5005*b11+(1365/2)*b22, 0 < -94060277895192911/627683696640000+(91/20)*b33+273*b11-(7/10)*b1818+(91/2)*b22, 0 < -3219528868317343/14944849920000+468*b11+(91/20)*b33-(91/5)*b1818+63*b22, 0 < -1167616840098623/627683696640000+(7/10)*b22+(1/20)*b33+(21/4)*b11-(7/10)*b1818, 0 < 6620337745005653/9510359040000+(91/20)*b1818-(91/5)*b33-(6552/5)*b11-(819/4)*b22, 0 < 10321214321183681/627683696640000-(21/4)*b22-(7/10)*b33-28*b11+(1/20)*b1818, 0 < 19939504442621873/627683696640000-(7/10)*b33-(39/4)*b22-(364/5)*b11+(91/20)*b1818, 0 < 21128314477665001/24141680640000-(91/5)*b33-1848*b11+(1001/20)*b1818-(1001/4)*b22, 0 < 30458564958023749/6340239360000-(1001/10)*b33+(3003/20)*b1818-9828*b11-(27027/20)*b22, 0 < 78768022311702133/17933819904000-(1001/10)*b33-8580*b11+(1001/20)*b1818-(5005/4)*b22, 0 < 418747163878248241/52306974720000-(858/5)*b33+(3003/20)*b1818-16016*b11-(9009/4)*b22}, [b11, b22, b33, b1818])

I am sorry for the writing style but I do not know how to write the command in MAPLE-style in this forum:-)

The first 4 inequalities shall ensure that all four variables b11, b22, b33, b1818 are positive. When entering the command i get the following error:

Can anybody help me please?:-)

Best regards,

Lucas

Hi everybody,

i'm trying to do an elliptic regularization but i don't know how to proceed ?

Is someone know how to achieve to do that  with an example ?

thanks a lot !

 

PS: i know only how to do a linear regularisation.

 

 

I have a linear problem with 4 variables (p0, p1,p2, p3) and a list of inequality constraints (shown below).  I would like to plot a polyhedral in 3 dimensions (p1,p2, p3 and omitting p0) showing the region that satisfies the inequalities.  That is, something similar to plots[inequal] but in 3d.  Any pointers would be appreciated.

/* Constraints */
+p0 <= 60;
-p0 +p1 >= 4;
-p0 +p2 >= 5;
-p0 +p3 >= -12;
+p0 -p2 >= -33;
+p1 -p2 >= -36;
+p2 <= 67;
-p2 +p3 >= -35;
+p0 -p3 >= 2;
+p1 -p3 >= 0;
+p2 -p3 >= 11;
+p3 <= 57;
+p0 -p1 >= -7;
+p1 <= 43;
-p1 +p2 >= 0;
-p1 +p3 >= -9;

 

I want to test linearly dependence of a polynomial f on a list of polynomials F by additional condition on parametric coefficients of linear parametric polynomial (linear for variables not parameters). Please note that:

  1. The polynomialand the members of are always homogenous in the variables.
  2. The coefficients of f, the coefficients of the members of F are all always polynomials in the parameters or contant and the members of N and W are all always polynomials in the parameters.

 

For example let

and

(a,b,c,d,e,h are parameters and A1,A2,A3 are variables).

If I use PolyLinearCombo(F,f,{A1,A2,A3}) (see http://www.mapleprimes.com/questions/204469-How-Can-I-Find-The-Coefficients-Of-Linear#comment217621)then its output is false,[].

Now we let to condition sets for parameters as the following:

N:=[ebc+ahd]

W:=[a,c]

The elements of N must be zero means that ebc+ahd=0

and the elements of W are non-zero that is a<>0 and c <>0.

Let a=b=c=d=h=1 and e=-1. This specialization satisfy in the above condition sets N and W. By this specialization we have:

and

Now if I use PolyLinearCombo(F,f,{A1,A2,A3}) then its output is true,[-1,1].

By this additional two condition sets I have to check that whether f is linearly independent of F or not. How can I do this without specialization? In fact I want an algorithm that its input is (null condition N, not-null condition W, list of polynomials F, a polynomial f, the set of variables) and its output is true and coefficients if f is linearly dependent of F w.r.t. null and not-null conditions N and W, else its output is false.

If the name of new procedure is ExtPolyLinearCombo and 

N:=[ebc+ahd]

W:=[a,c]

I want the output of

ExtPolyLinearCombo(N,W,F,f,{A1,A2,A3}) be true,[coefficients]

Thank you very much in advance.

 

 

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