vv

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8 years, 304 days

MaplePrimes Activity


These are answers submitted by vv

restart;
with(plottools): with(plots):
dis3d:=proc(p1,p2)
  local A,B;
  A:=getdata(p1)[3];
  B:=getdata(p2)[3];
  min( seq(seq(seq(seq(add( (A[i,j]-B[ii,jj])^2 ), i=op(2,A)[1]),j=op(2,A)[2]),ii=op(2,B)[1]),jj=op(2,B)[2]));
  sqrt(%)
end:
num:=50: np:=9:tp:=9: eps:=3:
#  increasing  np,tp  too much will slow down the program


cen:=rand(0.0..100.0):
dir:=2*rand(0..1)-1:
h:=rand(1.0..60.0):
rr:=rand(0.25..5.0):
t[0]:= tubeplot([dir()*h()*t, dir()*h()*t, dir()*h()*t], t = -1 .. 1,
        radius = rr(),style=surface,numpoints=np,tubepoints=tp):
for n to num do
  print("n=",n);
  ok:=false;
  while not(ok) do
     t[n]:= tubeplot([cen()+dir()*h()*t, cen()+dir()*h()*t, cen()+dir()*h()*t],
            t = -1 .. 1, radius = rr(), style=surface,numpoints=np,tubepoints=tp);
     for k from 0 to n-1 do
       ok := dis3d(t[k],t[n]) >eps;
       if not(ok) then break fi
     od;
  od;
od:
display( seq( t[n],n=0..num), axes=none,scaling=constrained);


 Edit. dir() was not initially used; it is intended for "more random" directions of the cylinders.
dis3d(p1,p2)  computes the distance between two 3d plots.
A new cylinder is added only if the distance to the previous ones is > eps.


Use a procedure.

f:=unapply(Determinant(M2),omega):
fsolve(f, complex);

It seems that there is no solution.

N:=Matrix( 400,400, (i,j)-> abs(f(-10+i/20 (-10+j/20)*I)) );
min(N),max(N);

So, your function is almost constant and very close to 0. You can't hope to solve it numerically; even if you find a program providing a solution, I don't think you should trust it. You should re-think the problem i.e, try another approach to solve it if you are sure that a solution (of the problem, not of this equation) exists.

P.S. I don't understand the reason for evalf[4].

Use Digits:=10000     (!!!)

and it works.

What do you want to do with this? It is quasi-nonsense to have floats such as  -9.8455282400 (with 11 significant digits) and to be forced to set Digits to such a huge number).

You have an error in the algorithm. The corrected iterate is:

But there is also a more subtle error.

XX:=x;

must be replaced by

XX:=Vector(x);

(or XX:=LinearAlgebra[Copy](x)). Otherwise x and XX will share the same memory, creating a mess in this case.

The program will work after that.

sort(collect(%, Nu));

 

Pascal := proc (n::posint)
local x,i,k,dd;
k:=length(binomial(n,floor(n/2)));
dd:=cat("%",2*k,"d");
for i from 0 to n do
  printf(cat(" "$(n-i)*k,dd$(i+1),"\n"), coeffs(expand((x+1)^i)));
end do end proc:
Pascal(12);

plot( [2,3,5]*~x^2, x=-2..2);

h:=(n,x)->1/2+1/2*tanh(n*x):
f:=proc(a,b,c,d,n)
   proc(x) c+ (d-c)*h(n,x-a)*h(n,b-x) end
   end:

The following function will look the same as the first of OP's function

plot( f(-80,40,11/9,19,50), -200..200);


 

 

Of course it is not constant:

f:=1+18*(sinh(9*x-9)-sinh(3*x-477))^2/(9*cosh(9*x-9)+cosh(3*x-477))^2;
limit(f,x=infinity),limit(f,x=-infinity);

eval(f,x=0);evalf(%);

     19.000000

To be able to plot, Digits must be increased.

Digits:=20;  # this is enough

plot(f,x=-200..200); # almost an "impulse"

 

                                             
              

You must use the proper syntax for Surface. E.g.

cone:= Surface(<r*cos(theta),r*sin(theta),2*a-r>,r=0..2*a, theta=0..2*Pi):

Flux( VectorField(<x, y, z>, cartesian[x, y, z]),  cone );

 

It should be not too difficult, following the next steps.

1. Find the 11 regions (connected components) R_1,...,R_11
and fix a point P_k in each of them
(each region is determined by 2, 3 or 4 inequalities of the form lines[i]<0 or lines[i]>0)
2.  OK:={}
3. For each S subset {P_1,...,P_11} having 4 elements do
    If S is admissible (i.e. in each halfspace there are exactly 2 points of S), then OK:=OK union {S}.

The set OK will contain all the desired configurations.

Here is a DirectSearch approch for finding a configuration. It is possible to find several configurations if we repeat the call.

lines := [-(1/2)*x+y, (1/2)*x+y, 1-x+(1/2)*y, 3-x-(1/2)*y]:
pl:=(i,j) -> eval(lines[i],[x=x[j],y=y[j]]):
cond1:=seq(seq(pl(i,j)^2>eps,i=1..4),j=1..4):
cond:= seq( op( [
      pl(i,1)*pl(i,2)*pl(i,3)*pl(i,4)>eps,
      min(pl(i,1),pl(i,2),pl(i,3),pl(i,4))<-eps,
      max(pl(i,1),pl(i,2),pl(i,3),pl(i,4))>eps  ])  ,
      i=1..4):
bounds:=seq(x[i]=-10..10,i=1..4), seq(y[i]=-10..10,i=1..4),  eps=0..10:
vars:=[seq(x[i],i=1..4),seq(y[i],i=1..4),eps]:
with(DirectSearch):
ds := Search(-eps, [cond, cond1,bounds], variables=vars):
pts:= [seq( [ds[2](i),ds[2](i+4)],i=1..4 )];

plot1:=plots[pointplot](pts,color=red,symbol=solidcircle,symbolsize=8):
plot2:=plots[implicitplot](lines,x=-10..10,y=-10..10):
plots[display](plot1,plot2,axes=none);

 

 

 

 

 

   
 

You will have to use IntegrationTools (or op()) for this.

J:=int( int( a(x)*b(y),x=-infinity..infinity), y=-infinity..infinity):

JJ:=expand(J);

with(IntegrationTools):

J1:=op(1,JJ); J2:=op(2,JJ):
int(int(GetIntegrand(J1)*GetIntegrand(J2),GetVariable(J1)=GetRange(J1)),GetVariable(J2)=GetRange(J2));

 

You should show how to reproduce this.

Otherwise, see my worksheet:

12x.mw

I get correct results.

Why should  int( fD(x)*fA(x), x=a..b ) be between 0 and 1? If you integrate the product of two densities, the result could be even infinity!

This is the first problem of the "Hundred-dollar, Hundred-digit Challenge problems":

https://en.wikipedia.org/wiki/Hundred-dollar,_Hundred-digit_Challenge_problems

The Maple solution was obtained by Robert Israel:

https://www.math.ubc.ca/~israel/challenge/challenge1.html

 

Edit: Mathematica seems to have better algorithms for computing oscillatory integrals.

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