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MaplePrimes Activity

These are replies submitted by vv

(for Example 2)

K := x -> piecewise(x<0.5, 1, 10):
pde := {diff(u(x,t),t)=diff(v(x,t),x),
bc:={ u(0,t)=0, u(1,t)=1, u(x,0)=2 }:
sol:=pdsolve(pde, bc, numeric, spacestep=1/200):
sol:-animate(u, t=0..0.15);


   local i; global A,B,F;
   if ev<>true then return 'last'(A,B)=F fi;
   F:=add(evalf(sin(a+b+i)), i=1..10^5)


@Carl Love 
In the 2019 help page:

This feature is currently not supported in 2-D math input

@Carl Love 

Maybe I am too conservative but I prefer the old style

ur:= proc(r) local i:=op(procname); A[i]*r + B[i]/r end;



Unfortunately you have not checked your statement.


Sorry, I don't know. But usually in such problems some heuristic is used.
BTW, I have corrected a typo in cond3.

Here is an idea.
Solve the problem in real x[i,j,k],  in the interval 0..1. Maple can do it quickly.
One obtains an "approx" solution (with some conditions violated). Maybe you can start from here.


fracsol:=Optimization:-LPSolve( add(q[i],i=1..20), {seq(cond||i, i=1..6), i01} );
sol01_approx:=map(t -> (lhs(t)=round(rhs(t))), fracsol[2]);


It is always a good idea to post the mathematical presentation of the problem. Then, an experienced user could come with a simpler/alternative solution.


Actually, it's the absolute error. For the relative error, use

numapprox:-infnorm((y-mExact)/mExact, 0..1);



I can easily give recurrences with "complex" sums which cannot be solved.
If you have a specific one, why don't you post it?

You have just produced a duplicate. You already asked this in your previous question and have received a partial answer.
Why don't you wait, maybe someone will come with the rest.
Note that usually such duplicates are simply deleted!


For a polyhedron the same method can be used but the formulae for  anew,bnew in doit are more complicated.
Actually even for a non-regular tetrahedron these formulae are much more complicated.


@Carl Love 

And it's not so?

@Carl Love 

You are right, LinearAlgebra:-Generic works with GF. I had an error when I have checked but probably I forgot a restart.
So, a conversion is not needed.


LinearAlgebra[Generic] seems to be incompatible with GF.
So, I made some conversions.

I have also changed your example, because it was already in a reduced form and had the rank=3.

G := GF(2, 4);
ex := G:-extension;
#  A := `~`[G:-input](Matrix(3, 6, [[1, 0, 0, 5, 0, 4], [0, 1, 0, 8, 7, 2], [0, 0, 1, 3, 0, 1]]));
A0 := <1, 0, 0, 5, 0, 4; 2, 0, 3, 10, 7, 8; 1, 0, 3, 5, 7, 4>;
A := G:-input~(A0);
GG[`0`] := 0:
GG[`1`] := 1:
GG[`=`] := `=`:
GG[`+`] := ()->modp(`+`(args),p):
GG[`-`] := (a,b)->`if`(nargs=1,modp(-a,p),modp(a-b,p)):
GG[`*`] := (a,b)->modp(Rem(a*b,poly,z),p):
GG[`/`] := proc(a,b) local i;
             if b=0 then error "division by zero"; end if;
             Gcdex(b,poly,z,'i') mod p;
             Rem(a*i,poly,z) mod p;
           end proc:



Yes, such an m exists. But unfortunately it depends on (a,b,c) which does not make the inequality very interesting.

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