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Dear Maple experts,

 

I would like to visualize the equation -3*x+2*y+3*z=0  and (with other color) 2*y+3*z =0. I used the following commands:

with(Student[LinearAlgebra]):
infolevel[Student[LinearAlgebra]]:=1:
PlanePlot(-3*x+ 2*y + 3*z = 0, [x,y,z], normaloptions=[shape=harpoon], showbasis);

But I do not know how to show at the same time the second equation (2*y+3*z=0 ).

 

How should I proceed? Any hint?

Thanks for your attention,

 

Jean-Jaques

 

A linear system with three variables has augmented matrix that is row-equivalent to the 
following matrix: 

k + 3 2 k − 4 = 3 
0 2 −9 = 5 
0 0 k^2 + k − 2 = k − 1 
Determine the values of k for which the system has: 
(a) exactly one solution, 
(b) infinitely many solutions, 
(c) no solutions.

I have a matrix with repeated columns

with(LinearAlgebra):

Q := Matrix([1,1,1,0,0,0,-2,-1],[0,0,0,1,1,1,0,-3]);

I'd like to write a loop that defines u[1],...,u[n] to be the unique columns of Q. Is there a way to do this?

hi all.

I have wrore the following program for optimization with bernstein and block pulse hybrid functions.

the program have some errors which i can't understand.

Bernestien1.mws

restart:

alias(C=binomial):
with(LinearAlgebra):
macro(LA= LinearAlgebra):


HybrFunc:=proc(N, M,  tj)               # N=Number of subintervals,  M=Number of functions in subintervals
 
local B, n, m;

global b;

for n from 1 to N do
for m from 0 to M-1 do

B := (i,m,t) -> C(m,i)*(1-t)^(m-i)*t^i:

b[n,m]:=unapply(piecewise(t>=(n-1)*tj/N and t<n*tj/N, B(m,2,N*t-(n-1)*tj), 0), t):
 od:od:


Array(1..N, 0..M-1, (n,m)->b[n,m](t)):

#convert(%,vector);
end proc:

HybrFunc(3, 3, 1);




                                       # End Of Definition
 
g2(t):=t;            #*exp(t-1):                      # Any other function can be replaced here
    

g1(t):=add(add(c[n,m]*b[n,m](t), m=0..2), n=1..3);
Optimization[Minimize](sqrt(int((g2(t)-g1(t))^2, t=0.. 1)));
assign(op(%[2]));
plot([g2(t),g1(t)], t=0..1, 0..5, color=[blue,red],thickness=[1,3],discont, scaling=constrained);

Array(1 .. 3, 0 .. 2, {(1, 0) = piecewise(0 <= t and t < 1/3, (1-3*t)^2, 0), (1, 1) = piecewise(0 <= t and t < 1/3, (6*(1-3*t))*t, 0), (1, 2) = piecewise(0 <= t and t < 1/3, 9*t^2, 0), (2, 0) = piecewise(1/3 <= t and t < 2/3, (2-3*t)^2, 0), (2, 1) = piecewise(1/3 <= t and t < 2/3, (2*(2-3*t))*(3*t-1), 0), (2, 2) = piecewise(1/3 <= t and t < 2/3, (3*t-1)^2, 0), (3, 0) = piecewise(2/3 <= t and t < 1, (3-3*t)^2, 0), (3, 1) = piecewise(2/3 <= t and t < 1, (2*(3-3*t))*(3*t-2), 0), (3, 2) = piecewise(2/3 <= t and t < 1, (3*t-2)^2, 0)}, datatype = anything, storage = rectangular, order = Fortran_order)

g2(t) := t

"g1(t):=c[1,0] ({[[(1-3 t)^2,0<=t and t<1/3],[0,otherwise]])+c[1,1] ({[[6 (1-3 t) t,0<=t and t<1/3],[0,otherwise]])+c[1,2] ({[[9 t^2,0<=t and t<1/3],[0,otherwise]])+c[2,0] ({[[(2-3 t)^2,1/3<=t and t<2/3],[0,otherwise]])+c[2,1] ({[[2 (2-3 t) (3 t-1),1/3<=t and t<2/3],[0,otherwise]])+c[2,2] ({[[(3 t-1)^2,1/3<=t and t<2/3],[0,otherwise]])+c[3,0] ({[[(3-3 t)^2,2/3<=t and t<1],[0,otherwise]])+c[3,1] ({[[2 (3-3 t) (3 t-2),2/3<=t and t<1],[0,otherwise]])+c[3,2] ({[[(3 t-2)^2,2/3<=t and t<1],[0,otherwise]])"

Error, (in Optimization:-NLPSolve) complex value encountered

Error, invalid left hand side in assignment

(1)



Download Bernestien1.mws

 I'll be so grateful if any one can help me.

Mahmood   Dadkhah

Ph.D Candidate

Applied Mathematics Department

 http://en.wikibooks.org/wiki/Linear_Algebra/

Representing_Linear_Maps_with_Matrices

 

how to calculate the first step

(2,0) -> (1,1,1) and (1,4) -> (1,2,0)

how to use maple command to get (1,1,1) and (1,2,0)

how to use maple command to calculate rep(h)

 

to get (0,-1/2,1) and (1,-1,0)

http://en.wikibooks.org/wiki/Linear_Algebra

/Representing_Linear_Maps_with_Matrices

I'm having some trouble maybe someone can point out my error please. I'm using the Maple 18 worksheet to try some basic linear equations. The trouble is in the last step.

 

1.) I start with 2 ordered pairs (2, 14) and (14,18)

Then I put in my formula to discover the slope. I confirm it looks correct in the Variables window.

m := (y2-y1)/(x2-x1);

 

2.)  Next I input the values for my ordered pairs. I also confirm thru the Variables window.

x1 := 2;

y1 := 14;

x2 := 3;

y2 := 18;

 

3.) Now I can type m and expect to get an answer to what my slope is.

m;

4.) Now I want Slope/Intercept form of y=mx+b. When I put in the formula y-y1=m(x-x1) i get a strange result

 

When I execute this formula, the result is y-14=4. (or thru context menu I tell it to solve for y, then I get y=18)

y-y1=m(x-x1) 

When I manually input the values, the output is y-14=4x-8 (or thru context menu I tell it to solve for y, then I get y=4x+6)

y-14 = 4*(x-2)

 

 

 

Why is my equation (y-y1=m(x-x1)) not executing properly?

Hi,

I have a linear system to solve.

 

mm:=proc(a,x,h,i)
local A,Z1,Z2,Z,F,result;  # to declare the local variable
A:=array(1..2,1..2,[[1,1],[a,a+h]]);
Z1:=evalf(int(1/(abs(y-x)+.000000001),y=a..a+h));
Z2:=evalf(int(y/(abs(y-x)+.000000001),y=a..a+h));
Z:=array([Z1,Z2]);
F:=evalm(inverse(A)&*Z);
result:=F[i]
end:

My questions: 

1) My exact Z1 is Z1:=evalf(int(1/(abs(y-x)),y=a..a+h)); but I ask if can I put

Z1:=evalf(int(1/(abs(y-x)+.000000001),y=a..a+h));

the same for Z2.

2) Can I writte in a simple form the vector Z.  Because, later, il have a second system contains Z1,Z2, Z3, Z4,Z5.  The difference between Z1 and Z2 is the variable "y" added in the integral of Z2.

 

Many thinks.

 

Hi,

I have these lines in my code, the function alpha:=(m,n,1)->...... is defined in the last line of the code below, but when I want to calculate alpah(1,1,1), there is no numeric value. Why????  Many thinks

 

 restart:
with(LinearAlgebra):
with(plots):
with(PDEtools):
with(IntegrationTools):
interface(rtablesize=20):
d:=1: N:=2: a:=1: h:=a/N:
Kernel(x,y):=ln(abs(x-y)):

 eq2:=int(Kernel(x,y)*phi(y),y=-a..a)=sum(int(Kernel(x,y)*phi(y),y=n*h..(n+d)*h),n=-N..N-d):

Approximate the integral
eq3:=phi->int(Kernel(x,y)*phi(y),y=n*h..(n+d)*h)=add(beta[i,n]*phi((n+i-1)*h),i=1..d+1):
eq4:=int(Kernel(x,y)*phi(y),y=n*h..(n+d)*h)=add(alpha[m,n,i]*phi((n+i-1)*h),i=1..d+1):

Compute the coefficients  beta[i]

Vct_basis:=[seq(x^i,i=0..d+1)]:
fct:=[seq(unapply(Vct_basis[i],x),i=1..d+2)]:


sys:=[seq(eq3(fct[i]),i=1..d+1)]:
x:=m*h:
w := [seq(beta[i,n],i=1..d+1)]:
M,b := GenerateMatrix(sys,w):

M1:=-M: V:=-b:

Vect_beta:=(M1)^(-1).V:# Vect_beta is a vector.

beta[1]:=Vect_beta[1]:  
alpha1:=(n,m)->beta[1];
(n, m) -> beta[1]
alpaha(1,1);
                                alpaha(1, 1)  ??????????????????? No numeric result


assuming a equations likes this:

tan(x)+y*sin(x)=1;

tan(x)-y*sin(x)=0;

I want to solve this equations using LinearSolve function(do not use solve()). I  fisrtly change tan(x)=u,y*sin(x)=v and then solve u,v using LinearSolve. finally, I obtain x=atan(u), y=v/sin(x).

the question is how can I perform the above operates neatly? I know I can achive this by using a lot of subs().but is there any tools in maple do this neatly just like IntegrationTools[Change]?

I’m trying to figure out how to find a basis for a subspace, V, of Rdefined by V = {(x, y, z)l(2x-3y+6z=0)}

 

I’m using the student linear algebra module for maple 17

 

I’ve tried defining the subspace and asking for the basis of V but I always get an error code.

 

I’ve tried consulting the maple website and looking through their help menu, but can’t find anything that answers how to find a basis... At least a basis from the subspace defined in my problem.

I know how to find a basis for the subspace by hand but not with maple.

Any help will be greatly appreciated. 

Hi all 

I have the following segment of maple program which belongs to time delay systems dynamic. here C=X-X0-G.Z-X.Dtau.P+X.Dtau.Z-U.P, is a matrix(vector) which comes from reordering the system terms and my goal is to minimizing J:=X.E.Transpose(X)+U.E.Transpose(U), subject to constraint C=0, but i don't know how to do so.

I will be so grateful if anyone can guide me

best wishes

Mahmood   Dadkhah

Ph.D Candidate

Applied Mathematics Department


restart:
with(Optimization):
with(LinearAlgebra):
macro(LA= LinearAlgebra):
L:=1:  r:=2:  tau:= 1:
interface(rtablesize= 2*r+1):

Z:= Matrix(
     2*r+1, 2*r+1,
     [tau,
      seq(evalf((L/(2*(iz-1)*Pi))*sin(2*(iz-1)*Pi*tau/L)), iz= 2..r+1),
      seq(evalf((L/(2*(iz-1-r)*Pi))*(1-cos(2*(iz-1-r)*Pi*tau/L))), iz= r+2..2*r+1)
      ],
     scan= columns,
     datatype= float[8]
);
                        
Dtau00:= < 1 >:
Dtau01:= Vector[row](r):
Dtau02:= Vector[row](r):
Dtau10:= Vector(r):
Dtau20:= Vector(r):

Dtau1:= LA:-DiagonalMatrix([seq(evalf(cos(2*i*Pi*tau/L)), i= 1..r)]):
Dtau2:= LA:-DiagonalMatrix([seq(evalf(sin(2*i*Pi*tau/L)), i= 1..r)]):
Dtau3:= -Dtau2:
Dtau4:= copy(Dtau1):

Dtau:= < < Dtau00 | Dtau01 | Dtau02 >,
         < Dtau10 | Dtau1  | Dtau2  >,
         < Dtau20 | Dtau3  | Dtau4  > >;
 
P00:= < L/2 >:
P01:= Vector[row](r):
P02:= Vector[row](r, j-> evalf(-L/j/Pi), datatype= float[8]):
P10:= Vector(r):
P20:= Vector(r, i-> evalf(L/2/i/Pi)):
P1:= Matrix(r,r):
P2:= LA:-DiagonalMatrix(P20):
P3:= LA:-DiagonalMatrix(-P20):
P4:= Matrix(r,r):

P:= < < P00 | P01 | P02 >,
      < P10 | P1  | P2  >,
      < P20 | P3  | P4  > >;

interface(rtablesize=2*r+1):    # optionally
J:=Vector([L, L/2 $ 2*r]):      # Matrix([[...]]) would also work here

E:=DiagonalMatrix(J);

X:=  Vector[row](2*r+1,symbol=a);
U:=Vector[row](2*r+1,symbol=b);

X0:= Vector[row](2*r+1,[1]);
G:=Vector[row](2*r+1,[1]);
C:=simplify(X-X0-G.Z-X.Dtau.P+X.Dtau.Z-U.P);

Z := Matrix(5, 5, {(1, 1) = 1., (1, 2) = 0., (1, 3) = 0., (1, 4) = 0., (1, 5) = 0., (2, 1) = 0., (2, 2) = 0., (2, 3) = 0., (2, 4) = 0., (2, 5) = 0., (3, 1) = 0., (3, 2) = 0., (3, 3) = 0., (3, 4) = 0., (3, 5) = 0., (4, 1) = 0., (4, 2) = 0., (4, 3) = 0., (4, 4) = 0., (4, 5) = 0., (5, 1) = 0., (5, 2) = 0., (5, 3) = 0., (5, 4) = 0., (5, 5) = 0.})

Dtau := Matrix(5, 5, {(1, 1) = 1, (1, 2) = 0, (1, 3) = 0, (1, 4) = 0, (1, 5) = 0, (2, 1) = 0, (2, 2) = 1., (2, 3) = 0, (2, 4) = 0., (2, 5) = 0, (3, 1) = 0, (3, 2) = 0, (3, 3) = 1., (3, 4) = 0, (3, 5) = 0., (4, 1) = 0, (4, 2) = -0., (4, 3) = -0., (4, 4) = 1., (4, 5) = 0, (5, 1) = 0, (5, 2) = -0., (5, 3) = -0., (5, 4) = 0, (5, 5) = 1.})

P := Matrix(5, 5, {(1, 1) = 1/2, (1, 2) = 0, (1, 3) = 0, (1, 4) = -.318309886100000, (1, 5) = -.159154943000000, (2, 1) = 0, (2, 2) = 0, (2, 3) = 0, (2, 4) = .1591549430, (2, 5) = 0, (3, 1) = 0, (3, 2) = 0, (3, 3) = 0, (3, 4) = 0, (3, 5) = 0.7957747152e-1, (4, 1) = .1591549430, (4, 2) = -.159154943000000, (4, 3) = 0, (4, 4) = 0, (4, 5) = 0, (5, 1) = 0.7957747152e-1, (5, 2) = 0, (5, 3) = -0.795774715200000e-1, (5, 4) = 0, (5, 5) = 0})

E := Matrix(5, 5, {(1, 1) = 1, (1, 2) = 0, (1, 3) = 0, (1, 4) = 0, (1, 5) = 0, (2, 1) = 0, (2, 2) = 1/2, (2, 3) = 0, (2, 4) = 0, (2, 5) = 0, (3, 1) = 0, (3, 2) = 0, (3, 3) = 1/2, (3, 4) = 0, (3, 5) = 0, (4, 1) = 0, (4, 2) = 0, (4, 3) = 0, (4, 4) = 1/2, (4, 5) = 0, (5, 1) = 0, (5, 2) = 0, (5, 3) = 0, (5, 4) = 0, (5, 5) = 1/2})

X := Vector[row](5, {(1) = a[1], (2) = a[2], (3) = a[3], (4) = a[4], (5) = a[5]})

U := Vector[row](5, {(1) = b[1], (2) = b[2], (3) = b[3], (4) = b[4], (5) = b[5]})

X0 := Vector[row](5, {(1) = 1, (2) = 0, (3) = 0, (4) = 0, (5) = 0})

G := Vector[row](5, {(1) = 1, (2) = 0, (3) = 0, (4) = 0, (5) = 0})

C := Vector[row](5, {(1) = 1.500000000*a[1]-2.-.1591549430*a[4]-0.7957747152e-1*a[5]-.5000000000*b[1]-.1591549430*b[4]-0.7957747152e-1*b[5], (2) = a[2]+.1591549430*a[4]+.1591549430*b[4], (3) = a[3]+0.7957747152e-1*a[5]+0.7957747152e-1*b[5], (4) = a[4]+.3183098861*a[1]-.1591549430*a[2]+.3183098861*b[1]-.1591549430*b[2], (5) = a[5]+.1591549430*a[1]-0.7957747152e-1*a[3]+.1591549430*b[1]-0.7957747152e-1*b[3]})

(1)

J:=X.E.Transpose(X)+U.E.Transpose(U);

J := a[1]^2+(1/2)*(a[2]^2)+(1/2)*(a[3]^2)+(1/2)*(a[4]^2)+(1/2)*(a[5]^2)+b[1]^2+(1/2)*(b[2]^2)+(1/2)*(b[3]^2)+(1/2)*(b[4]^2)+(1/2)*(b[5]^2)

(2)

Minimize(J,{C=0});






Error, (in Optimization:-NLPSolve) invalid arguments

 

#XP:=-.015+X[1]+add(X[l+1]*f1(l)+X[r+l+1]*f2(l), l= 1..r):
#plot([XP,T1], t= 0..1);#,legend= "Solution Of x(t) with r=50"):

 

 

 

 

 

 

Download work1.mwswork1.mws

Hi all

In matlab software we have a command namely fmincon which minimizes any linear/nonlinear algebric equations subject to linear/nonlinear constraints.

Now my question is that: what is the same command in maple?or how can we minimize linear/nonlinear function subject to linear/nonlinear constraints in maple?

thanks a lot

Mahmood   Dadkhah

Ph.D Candidate

Applied Mathematics Department

The result of the following lines in Maple is V := [5., 3.,5. 11., 6.] that I think should be V := [5., 3.,5. 0., 6.], is there somthing wrong?

 

v1:= Vector([-28., -63., -17., -55., 17.], datatype= float[8]):

V:= LinearAlgebra:-Modular:-Mod(11, v1, float[8]);

bug.mw



Hi,

I'm currently writing a programme for synchronising automatas, its creates an array and adds words or matrices to the array that aren't already in. I am currently getting an error when i try and run my procedure though and i'm unsure of the problem, any help would be appreciated, here is my code so far.

Thanks!

proc_cerny1:=proc(A::Matrix,B::Matrix,C::Vector[row],N)
local x, S, i, j, T, R, y, found;
x:=2^N;
S:=Array(0..(x-1));
S[0]:=C;
i:=0;
j:=1;
found:=false;

while (i<(x-1) and i<>j) do
T:=S[i].A;
for y from 0 to j do
if LinearAlgebra:-Equal(S[y],T) then
found:=true;
if (found=false) then
S[j]:=T;
j:=j+1;
end if:
end if:
od:

R:=S[i].B;
for y from 0 to j do
if LinearAlgebra:-Equal(S[y],T) then
found:=true;
if (found=false) then
S[j]:=R;
j:=j+1;
i:=i+1;

end if:
end if:
od:
od:
print(S);
end proc:

 

The error i'm getting is when i input this:

proc_cerny1(Matrix([[1,0,0],[1,0,0],[1,0,0]]),Matrix([[0,1,0],[0,0,1],[0,0,1]]),Vector(1..3,1,orientation=row),3);

and the error is:

Error, (in LinearAlgebra:-Equal) invalid input: LinearAlgebra:-Equal expects its 1st argument, X, to be of type {Matrix, Vector} but received 0

 

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