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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.

 

 

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.

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