Adri van der Meer

Adri vanderMeer

1355 Reputation

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15 years, 172 days
University of Twente (retired)
Enschede, Netherlands

My "website" consists of a Maple Manual in Dutch

MaplePrimes Activity

These are answers submitted by Adri van der Meer

Ck := [ seq( RandomMatrix(4,4), i=1..6 ) ];

makes a list of six 4×4-matrices, Ck[1], ... Ck[6].

If you have a similar list P of matrices, a matrix product is

Ck[i] . P[j];

(not Ck[i] * P[j] )

Your program is calculating several values for the variable NewLambda, and in each step is is overwritten by the new value.
Perhaps you mean a Matrix?

NewLambda := Matrix(18,18):
for i from 1 to 17 do
for j from 2 to 18 do
 NewLambda[i,j] := (Lambda[i,i]-Lambda[j,j])/2;
end do
end do;


alias( f = f(x+y, z+y), 
   f[xx] = D[1, 1](f)(x+y, z+y),
   f[xy] = D[1, 2](f)(x+y, z+y),
   f[yy] = D[2,2](f)(x+y, z+y) ,
   f[x] = D[1](f)(x+y, z+y),
   f[y] = D[2](f)(x+y, z+y) );

Perhaps your TeX-machine can't find the stylefile maplestd2e.sty?

format := "%+03.1f %+04.2Zf %+04.2f \n":

In the first equation you have the conflicting ICs (D(x))(0) = 0 and (D(x))(0) = v0
The second must be (D(y))(0) = v0, I suppose

  • Write the differential equation in the right way:
      diff(phi(zeta),zeta) = psi(zeta), 
    diff(psi(zeta),zeta) = phi(zeta)/alpha + beta*phi(zeta)^3/(3*alpha*c)
    indicating that φ amd ψ are the dependent variables and ζ the independent variable
    (Choose Tools → Options → Display → Input Display: Maple Notation to get full control over your input)
  • Assign a numerical value to the parameters α β and c
  • Give some initial points in the phase plane
  • Perhaps ?DEplot is better in this instance

With no more information on the nature of Record(...) I can only refer to ?fprintf

plot( [[x,2*x^2,x=-1/2..0], [x,1/2-x,x=-1/2..0],
      [x,1/2-2*x^2,x=0..1/2], [x,x^2-1,x=0..1/2]] );

see ?plot,parametric

Format list for each column of the matrix:

formats := "%+03.1f %+5.3e %+07.4f \n":

(mind the newline character at the end)

B := convert(A,listlist):
for i to nops(B) do fprintf( terminal, formats, op(B[i]) ) end do:

First possibility: you want to plot the derivative of x'(t), that is to say: the derivative of a numeric procedure. I suppose that odeplot is not designed for that. So use the option output=listprocedure:

sol3 := dsolve({DE1, DE2, ICs}, numeric, output=listprocedure );
v := subs(sol3,diff(x(t),t));
st := time(): plot(D(v),0..3); T := time() - st;

Faster is to calculate timeseries for x(t) and v(t) and use a(t) from the differential equation:

solve( DE1, diff(diff(x(t), t), t) ): subs( {m(t)=m, x(t)=x, diff(x(t),t)=v}, % ):
a := unapply( %, t,m,x,v );     #acceleration as a function of t, m, x and x'
sol2 := dsolve({DE1, DE2, ICs}, numeric, output=Array([seq(i,i=0..3,0.001)]) );
   # sol2 has two parts: a list with the names of the variables and a matrix of calculated values
plot( <sol2[2,1][1..-1,1]|a~(seq(sol2[2,1][1..-1,i], i=1..4))> );

You have two times " = 0 ", so


If you want to use the option coords = polar, then you must use the "normal" plot-command, to plot r(theta).

q1 := dsolve( {sys1, theta(0)=Pi, r(0)=2}, numeric, output=listprocedure);

This gives you numeric procedures for r(t) and theta(t).

Theta := subs(q1,theta(t)): R := subs(q1,r(t)):
plot( [R(t),Theta(t),t=0..10], coords=polar, axiscoordinates=polar );

The second argument of writedata must be of type {set, hfarray, list, matrix, array(1), list(list)}, and L is a table. But the result of evaluating a table is its name, and not its content (to get this, you must do: eval(L);) and that is wthe error message seems to say that L has no value.

I think the type "table" is not what you want, because a table consists of a set of indices and a set of corresponding entries, and that is probably not what you want to store.

To make L a list (for example):

L := [seq(i^2, i=1..3)];

If you really want to write the table, that is: both the indices and the corresponding entries, do:

LL := [seq( [i,L[i]], i in op~( [indices(L)] ) )];

Mind the tilde in the op-command.

Use the option gridrefine=6 or so in,implicitplot, and be patient

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