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These are answers submitted by vv

co:=coeffs(p1-p2, [ln(E),ln(K),ln(L)]);

seq(u_=0, u_=[co]);

D(F) is the derivative function of F. For example, for F = sin, D(F) equals cos.

If you define an explicit function such as

F := sin


F:= x -> x^10;

and execute again your definition, the derivatives will be computed effectively.

BTW, subs is superfluous.

for m from 0 to n do
end do;

(I would not write such formulas in 2D math, but maybe I am too old).


This is THE ART GALLERY problem invented by the well known mathematician Victor Klee in 1973:

Find the fewest number of (stationary) guards needed to protect a polygonal gallery.

The following theorem is known.

To cover a polygon with n vertices, floor(n/3) guards are needed for some polygons, and sufficient for all of them.

This particular configuration could be solved with some combinatorics and a series of simplex-es.
If I remember correctly, a general algorithm for the minimal number of guards does not exist for the moment.

[attachment deleted]

It seems that there is a bug in plottools:-`transform/object`

I have included the altered procedure [one line only] in the startup code of the attached worksheet

(Maple 2015).


Edit: it's line 20, from

     outdim := nops(t);


     if [t]<>[] then outdim := nops(t) end if;

Why don't you use theta0(z)  ( or theta[0](z) ) instead of theta(z,0) ?

ode := diff(theta0(z), z, z)+theta0(z):
ibc:=theta0(0)=beta, D[1](theta0)(0)=0:
                    theta0(z) = beta cos(z)


The result of fsolve could be NULL (if no solution found).

You should test this first, otherwise  NULL < 0   will produce the error.


res:=seq(x-pts[i][1] <=  a||i,i=1..nops([pts])),
     seq(x-pts[i][1] >= -a||i,i=1..nops([pts])),
     seq(y-pts[i][2] <=  b||i,i=1..nops([pts])),
     seq(y-pts[i][2] >= -b||i,i=1..nops([pts])):
                 ["x" = 5, "y" = 5, "min" = 51]


I always prefer a homemade version:




What is degrevlex?

Use e.g.

G:=Basis(K, 'tord', variables={r,u,v,w}); tord;

You could define the polynomial functions and use the composition operator @.

Or, better, use subs.
Notice first that the degrees of a and b must be 1.

                           a1 x + a0
                           b1 x + b0
                      b1 (a1 x + a0) + b0

# So, the general form of a,b is

{'a','b'} = eval({a,b},op(sol)); 

           {a,b} = {x/b1+a0, -a0*b1+b1*x}

# here a0,b1 are arbitrary constants, b1<>0.

Edit: actually c computes b o a, but the final result is of course the same. 

> min(extrema(x,{x^2 + x*y + y^2 = 2}));

It's a bug.

It seems that Apolonius works correctly only if all the 8 circles exist.

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