vv

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8 years, 158 days

MaplePrimes Activity


These are answers submitted by vv

Proj:=proc(A::Matrix,b::Vector[column],v::Vector[column])
uses LinearAlgebra;
local u0,P,_xxx;
u0:=LinearSolve(A,b,free=_xxx);  #error if {x:Ax=b} is empty
u0:=eval(u0,_xxx=0);
P:=ProjectionMatrix(NullSpace(A));
#   u0+P.(v-u0)
u0+P.(Vector(v)-u0) # To work with Gradient without ConvertVector(...,free)
end;



Example

Proj(<<1|2|3>,<2|4|6>>, <1,2>, <0,0,0>);

G:=proc(k::posint, n::posint)
local f,g,x,y,u,v;
f:=(1/2)*(1-cos(k*Pi*x)*cos(n*Pi*y))/(sin((1/2)*Pi*x)^2+sin((1/2)*Pi*y)^2);
g:=simplify(subs([x=2/Pi*arctan(u),y=2/Pi*arctan(v)],f)*4/(Pi^2*(u^2+1)*(v^2+1)));
expand(g);
int(int(%,u=0..infinity),v=0..infinity)
end:

t:=time():
G(9,9);
time()-t;

   1.279

After some "reverse engineering", it is a group but for mod 70.

Simply construct the Cayley table:

G:=[5, 15, 25, 45, 55, 65];
Matrix(6, (i,j) -> G[i]*G[j] mod 70);

 

Alternatively, you may use Magma (see ?Magma) but the problem is too simple to need it..

 

phasor:=(r,a)->r*exp(I*a*Pi/180):   #simpler version
ec:=phasor(U__C, -90) = phasor(400, X)-phasor(150, 0)+phasor(1744/10, 90):
s:=solve(evalc([Re(ec),Im(ec)]),allsolutions);


uc:=eval(U__C,s):
[eval(uc,[_Z1=0,_B1=0]),eval(uc,[_Z1=0,_B1=1])];


evalf(%);


So, you will take the second (positive) solution.

Note that
fsolve(evalc([Re(ec),Im(ec)]), {U__C=0..1000, X=0..360});
does not provide a solution (it should!), but
fsolve(evalc([Re(ec),Im(ec)]), {U__C=200, X=270});



 

 

Here pi is actually numtheory[pi]; the package was loaded previously.

The function

y:=numtheory[pi](floor(x))/(x/ln(x))-1;

has indeed the graph:

Note that this function has an infinity of discontinuities and

limit(y, x=infinity) = 0.

(this is the celebrated  Prime number theorem; unfortunately Maple cannot compute the limit).

 

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 );

 

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