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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.

 

 

How  we can decide whether a polynomial f is a linear combination of some polynomials g1,...,gm?

For example if f=x^2+y^2 and g1=y+x^2 , g2=y^2-y then f=g1+g2.

Hello there

I'm quite an amature so please don't judge.  I'm trying to use fsolve to solve a system of non-linear equations but Maple is just "spitting" on me the equations with no intention to solve them:

> delta5 := P*(1+mu5)*((1-2*mu5)*x/(sqrt(x^2+zeq^2)*(sqrt(x^2+zeq^2)*x))+x*zeq/sqrt(x^2+zeq^2)^3)/(2*Pi*E5);
print(`output redirected...`); # input placeholder
> shrinkage := P*(1+mu5)*((1-2*mu5)*x/(sqrt(x^2+Zb^2)*(sqrt(x^2+Zb^2)*x))+x*Zb/sqrt(x^2+Zb^2)^3)/(2*Pi*E5)-P*(1+mu5)*((1-2*mu5)*x/(sqrt(x^2+Za^2)*(sqrt(x^2+Za^2)*x))+x*Za/sqrt(x^2+Za^2)^3)/(2*Pi*E5);
> eq10 := subs(x = 1800, delta5)+subs(x = 1800, Zb = z2, Za = z1, shrinkage)+subs(x = 1800, Zb = z3, Za = z2, shrinkage)+subs(x = 1800, Zb = z4, Za = z3, shrinkage)+subs(x = 1800, Zb = z5, Za = z4, shrinkage) = 36.7*10^(-3);
print(`output redirected...`); # input placeholder
> eq9 := subs(x = 1500, delta5)+subs(x = 1500, Zb = z2, Za = z1, shrinkage)+subs(x = 1500, Zb = z3, Za = z2, shrinkage)+subs(x = 1500, Zb = z4, Za = z3, shrinkage)+subs(x = 1500, Zb = z5, Za = z4, shrinkage) = 47.2*10^(-3);
print(`output redirected...`); # input placeholder
> eq8 := subs(x = 1200, delta5)+subs(x = 1200, Zb = z2, Za = z1, shrinkage)+subs(x = 1200, Zb = z3, Za = z2, shrinkage)+subs(x = 1200, Zb = z4, Za = z3, shrinkage)+subs(x = 1200, Zb = z5, Za = z4, shrinkage) = 63.8*10^(-3);
> eq7 := subs(x = 900, delta5)+subs(x = 900, Zb = z2, Za = z1, shrinkage)+subs(x = 900, Zb = z3, Za = z2, shrinkage)+subs(x = 900, Zb = z4, Za = z3, shrinkage)+subs(x = 900, Zb = z5, Za = z4, shrinkage) = 91.1*10^(-3);
print(`output redirected...`); # input placeholder
> eq6 := subs(x = 600, delta5)+subs(x = 600, Zb = z2, Za = z1, shrinkage)+subs(x = 600, Zb = z3, Za = z2, shrinkage)+subs(x = 600, Zb = z4, Za = z3, shrinkage)+subs(x = 600, Zb = z5, Za = z4, shrinkage) = 137.9*10^(-3);
> eq5 := subs(x = 450, delta5)+subs(x = 450, Zb = z2, Za = z1, shrinkage)+subs(x = 450, Zb = z3, Za = z2, shrinkage)+subs(x = 450, Zb = z4, Za = z3, shrinkage)+subs(x = 450, Zb = z5, Za = z4, shrinkage) = 175.2*10^(-3);
> eq4 := subs(x = 300, delta5)+subs(x = 300, Zb = z2, Za = z1, shrinkage)+subs(x = 300, Zb = z3, Za = z2, shrinkage)+subs(x = 300, Zb = z4, Za = z3, shrinkage)+subs(x = 300, Zb = z5, Za = z4, shrinkage) = 230.9*10^(-3);
print(`output redirected...`); # input placeholder
> sys := {eq10, eq5, eq6, eq7, eq8, eq9};
print(`output redirected...`); # input placeholder
> fsolve(sys, {E1 = 1000 .. 2000, E2 = 0 .. 2000, E3 = 0 .. 2000, E4 = 0 .. 2000, E5 = 0 .. 2000, h4 = 100 .. 400});

and this is what Maple gives after the fsolve

 

fsolve({(3937.500000*(.2/(202500+(146.0507832*(E1/E5)^(1/3)+197.1094212*(E2/E5)^(1/3)+295.6641318*(E3/E5)^(1/3)+1.*h4*(E4/E5)^(1/3))^2)+(450*(146.0507832*(E1/E5)^(1/3)+197.1094212*(E2/E5)^(1/3)+295.6641318*(E3/E5)^(1/3)+1.*h4*(E4/E5)^(1/3)))/(202500+(146.0507832*(E1/E5)^(1/3)+197.1094212*(E2/E5)^(1/3)+295.6641318*(E3/E5)^(1/3)+1.*h4*(E4/E5)^(1/3))^2)^(3/2)))/E5-0.3888888889e-2/E5+(3937.500000*(.2/(202500+(650+h4)^2)+(450*(650+h4))/(202500+(650+h4)^2)^(3/2)))/E5 = .1752000000, (3937.500000*(.2/(360000+(146.0507832*(E1/E5)^(1/3)+197.1094212*(E2/E5)^(1/3)+295.6641318*(E3/E5)^(1/3)+1.*h4*(E4/E5)^(1/3))^2)+(600*(146.0507832*(E1/E5)^(1/3)+197.1094212*(E2/E5)^(1/3)+295.6641318*(E3/E5)^(1/3)+1.*h4*(E4/E5)^(1/3)))/(360000+(146.0507832*(E1/E5)^(1/3)+197.1094212*(E2/E5)^(1/3)+295.6641318*(E3/E5)^(1/3)+1.*h4*(E4/E5)^(1/3))^2)^(3/2)))/E5-0.2187500000e-2/E5+(3937.500000*(.2/(360000+(650+h4)^2)+(600*(650+h4))/(360000+(650+h4)^2)^(3/2)))/E5 = .1379000000, (3937.500000*(.2/(810000+(146.0507832*(E1/E5)^(1/3)+197.1094212*(E2/E5)^(1/3)+295.6641318*(E3/E5)^(1/3)+1.*h4*(E4/E5)^(1/3))^2)+(900*(146.0507832*(E1/E5)^(1/3)+197.1094212*(E2/E5)^(1/3)+295.6641318*(E3/E5)^(1/3)+1.*h4*(E4/E5)^(1/3)))/(810000+(146.0507832*(E1/E5)^(1/3)+197.1094212*(E2/E5)^(1/3)+295.6641318*(E3/E5)^(1/3)+1.*h4*(E4/E5)^(1/3))^2)^(3/2)))/E5-0.9722222220e-3/E5+(3937.500000*(.2/(810000+(650+h4)^2)+(900*(650+h4))/(810000+(650+h4)^2)^(3/2)))/E5 = 0.9110000000e-1, (3937.500000*(.2/(1440000+(146.0507832*(E1/E5)^(1/3)+197.1094212*(E2/E5)^(1/3)+295.6641318*(E3/E5)^(1/3)+1.*h4*(E4/E5)^(1/3))^2)+(1200*(146.0507832*(E1/E5)^(1/3)+197.1094212*(E2/E5)^(1/3)+295.6641318*(E3/E5)^(1/3)+1.*h4*(E4/E5)^(1/3)))/(1440000+(146.0507832*(E1/E5)^(1/3)+197.1094212*(E2/E5)^(1/3)+295.6641318*(E3/E5)^(1/3)+1.*h4*(E4/E5)^(1/3))^2)^(3/2)))/E5-0.5468750000e-3/E5+(3937.500000*(.2/(1440000+(650+h4)^2)+(1200*(650+h4))/(1440000+(650+h4)^2)^(3/2)))/E5 = 0.6380000000e-1, (3937.500000*(.2/(2250000+(146.0507832*(E1/E5)^(1/3)+197.1094212*(E2/E5)^(1/3)+295.6641318*(E3/E5)^(1/3)+1.*h4*(E4/E5)^(1/3))^2)+(1500*(146.0507832*(E1/E5)^(1/3)+197.1094212*(E2/E5)^(1/3)+295.6641318*(E3/E5)^(1/3)+1.*h4*(E4/E5)^(1/3)))/(2250000+(146.0507832*(E1/E5)^(1/3)+197.1094212*(E2/E5)^(1/3)+295.6641318*(E3/E5)^(1/3)+1.*h4*(E4/E5)^(1/3))^2)^(3/2)))/E5-0.3500000000e-3/E5+(3937.500000*(.2/(2250000+(650+h4)^2)+(1500*(650+h4))/(2250000+(650+h4)^2)^(3/2)))/E5 = 0.4720000000e-1, (3937.500000*(.2/(3240000+(146.0507832*(E1/E5)^(1/3)+197.1094212*(E2/E5)^(1/3)+295.6641318*(E3/E5)^(1/3)+1.*h4*(E4/E5)^(1/3))^2)+(1800*(146.0507832*(E1/E5)^(1/3)+197.1094212*(E2/E5)^(1/3)+295.6641318*(E3/E5)^(1/3)+1.*h4*(E4/E5)^(1/3)))/(3240000+(146.0507832*(E1/E5)^(1/3)+197.1094212*(E2/E5)^(1/3)+295.6641318*(E3/E5)^(1/3)+1.*h4*(E4/E5)^(1/3))^2)^(3/2)))/E5-0.2430555555e-3/E5+(3937.500000*(.2/(3240000+(650+h4)^2)+(1800*(650+h4))/(3240000+(650+h4)^2)^(3/2)))/E5 = 0.3670000000e-1}, {E1, E2, E3, E4, E5, h4}, {E1 = 1000 .. 2000, E2 = 0 .. 2000, E3 = 0 .. 2000, E4 = 0 .. 2000, E5 = 0 .. 2000, h4 = 100 .. 400})

I have an equation,

     y = 2x2+7,

Plotting y against x will give a quadratic graph, but plotting y against x2 should give a straight line. The problem is I cannot do

  plot([y], x2 = -5 .. 5);

It gives me an error

    Error, (in plot) unexpected option: x^2 = -5 .. 5

I have searched online but no links. Can anyone help with this.

 

Thanks

Fairoz

 

 

Hi, I'm new to Maple and was trying to use it to solve 3 equations with 3 unknowns, in terms of another 2 parameters. This is what I put in, and the error that came up:

solve({(1-x)/(1+b) = b*y, (1-y)(1-a)/(-a*b+1) = b*z, (1-z)(1-b)/(-a*b+1) = a*x}, {x, y, z}); 

Error, (in SolveTools:-LinearSolvers:-Algebraic) unable to compute coeff

I want to solve the equations to get x,y,z in terms of a,b but I don't understand the error coming up - have I done something wrong or is it because not all the variables appear in all of the equations? As there are 3 equations and 3 unknowns there is a solution, but I want to check the answer I found on paper with something (the algebra got a bit messy!)

Any help greatly appreciated! :)

Is it possible to define and use an abstract linear operator L which is a function of y(x) and z(x)? 

For example, Maple should be able to recognise that L(y(x),z(x)) = L(z(x),y(x)) and simplify expressions such as 

L(c*y(x),d*z(x)) = c*d*L(y(x),z(x)) where c and d are constants or numbers.

It should also be able to expand things like L(c*y1(x)+d*y2(x),e*z1(x)+f*z2(x)).

Thanks.

I've got a function f(x_n) = (x_n-1)^3

and need to show that for the iterative method

x_(n+1)= x_n - f(x_n)/(sqrt(f'(x_n)^2-f(x_n)*f''(x_n), at a simple root we have cubic convergence while at a multiple root, it converges linearly.

I understand that the approach is to write either a recursive function or a sequence, but i'm confused about the structure since both x and n are being incremented

 

Hello guys

I have a coupled linear differentional equation which are in the 4th order. they are shown in the below:

P:=phi(x):
Q:=psi(x):

eq1:=a11*diff(P,x,x,x,x)+a22*diff(P,x,x)+a33*P+a44*diff(Q,x,x)+a55*Q:
eq2:=a44*diff(P,x,x)+a55*P+a66*diff(Q,x,x)+a77*Q:

eq1:=0:
eq2:=0:

The boundary values for this coupled equation are:
phi(a)=sigma1,phi(-a)=sigma1,diff(P,x)(a)=0,diff(P,x)(-a)=0,psi(a)=sigma2,psi(-a)=sigma2

Now consider:

a11:=6.36463*10^(-10):
a22:=-1.22734*10^(-9):
a33:=3.48604*10^(-10):
a44:=2.94881*10^(-11):
a55:=-5.24135*10^(-11):
a66:=-1.03829*10^(-9):
a77:=4.86344*10^(-10):
when I use dsolve for deriving a good answer in this equation, there are six real roots .How can I solve it with these boundary condition?

I need to extract phi(x) and psi(x) from this coupled equation.

Thanks

 

Hello guys

I have a linear differentional equation which is in the 4th order. It is shown in the below:

P:=phi(x):
eq:=a11*diff(p,x,x,x,x)+a22*diff(p,x,x)+a33*p:
eq:=0:
where a11 and a22 and a33 are constant coefficients. The boundary value for this equation is:

phi(a)=sigma1 , phi(-a)=sigma1 , diff(p,x)(a)=0 , diff(p,x)=0

Now consider :

a11:=2.731e-10:
a22:=-1.651e-9:
a33:=3.09027e-10:
a:=35.714:
sigma1:=200e6:

when I use dsolve for deriving a good answer in this equation. there are four real roots .How can I solve it with these boundary condition?

I need to extract phi(x) from this equation.

Thanks

Hello,

I need your help to display the matrix generated by a lenear system.

The linear system is okay, well written, but  there is an error in generate matrix, I get the wrong matrix.

Thank you very much for your help.

 

test.mw

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I have a set "EQ" containing N linear equations in N unknowns. The only symbolic variables in each "EQ[i]" are the unknowns. I want to write a procedure that derives the matrices "A" and "b" where A.x=b is the same linear system stored in "EQ". In other words, I want to write the linear system in the matrix form.

Can anyone guide me through writting such a procedure?
As an example do it with the system written in the following file.

Note: This procedure will be used for large linear systems (e.g. 2000 Equations, 2000 Unknowns) so it will be important that the procedure uses the least operations required.

LinearSystem.mw

Thanks in advance.

I have a large system of linear algebraic equations that I want to solve (2005 Unknowns, 2005 Equations). I was wondering that what are the proper commands to use in maple for solving the system as fast as possible. Take a look at the files in the download link if you want to see the system of linear algebraic equations.

http://pc.cd/h79

Please provide me any suggesitons that you may think will be helpful like using other sofwares that are good in doing this work such as MATLAB or something else.


Thanks in Advance




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