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Christian Wolinski

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20 years, 214 days
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These are answers submitted by Christian Wolinski

Option: power....

Christian Wolinski 1650
August 25 2016
1 3

The following is excessive, but it does achieve the goal:

 

e:= g^((2*(-sigma+k+1))/(-1+sigma))-tau^2;
`@`(factor, x -> combine(x, power), factor, expand)(e);

Coefficients of elements typed function....

Christian Wolinski 1650
August 13 2016
0 3

This appears to be a frequent question. This is a link to my previous response:
http://www.mapleprimes.com/questions/207267-Coefficients-Of-Differential-Polynomial#comment223370

 

function_coeffs := proc(A, v::set(name))
local S, T;
   S := indets(A, {function});
   S := select(has, S, v);
   T := {Non(map(identical, S))};
   frontend(proc(A, S) local V; [coeffs](collect(A, S, distributed), S, 'V'), [V] end proc, [A, S union v], [T, {}])
end proc;

fec:=(A,f,t)->frontend(function_coeffs,[A,f],[{Non(t)},{}]);

A:=diff(g(z),x)*g(z)^3+diff(g(z),z,z)*g(z)^4+diff(g(z),z,z,z)*g(z)^5+diff(g(z),z)/g(z)^2;
function_coeffs(A,{g});
fec(A,{g},specfunc(anything,diff));

Substitution....

Christian Wolinski 1650
August 13 2016
0 2

Try this substitution. Does it produce a different outcome?

 

restart;
assume(U,complex,V,complex,x,complex);
S := {b = U*(1+V^2)/V, a = U*(V-1)*(V+1)/V};
A:=diff(y(x),x)=a*cos(y(x))+b;
B:=subs(S,A);
dsolve(B,y(x));

Radical and multivariable algebraic statements are opportune to cause obstruction, so perhaps you should prepare your constants.

Some choices....

Christian Wolinski 1650
July 02 2016
0 0

I have aligned the grid with t and t+x. Perhaps this configuration is to your liking:

 

 

restart;
x:='x':y:=0:z:=0:
f:=unapply(abs(2*(-exp(-t-x-z)+exp(t+x+z))/(exp(-t-x-z)+tanh((1+I)*t+(1/2-1/2*I)*y+z)+exp(t+x+z))),t,x);
g:=(u,v)->(u,v-u);
(f@g)(u,v);
p1:=plot3d([g,f@g](u,v),u=-5..5,v=-10..10,numpoints=20000):
mp:=proc() global p1; plots[display](p1,'args'); end proc:
plots[display](
mp(style=contour,thickness=2,shading=XY,contours=[seq(i/4,i=0..12)]),
mp(style=point,symbol=POINT,color=blue),
mp(style=patchnogrid,shading=XYZ,lightmodel=light3),
scaling=constrained,orientation=[120,45],projection=0.1,axes=boxed,view=[-6..6,-6..6,0..3]);

-1/32*(-1)^(1/8)*2^(1/4)*(2*(-1)^(1/4)-3...

Christian Wolinski 1650
June 02 2016
0 30

Considering this is a residue of a rational polynomial with coefficients in Q, at a singular point.

Edited:
used evala@AFactor instead of split
corrected use of roots
replaced evala@residue with evala@coeff@series
added seq to RootOf definition

 

 

C0 := rationalize(expand(2^(1/4)*exp(3/8*I*Pi)));
P := z^2 / (z^4 + 2*z^2 + 2)^2;
Pf := numer(P) / (evala@AFactor)(denom(P),z);
Pfs := select(`@`(evalb, 0 = evalc, evala, Norm, `+`), map2(op, 1, (roots@denom)(Pf)), -C0);

 

QRatpolyResidue := proc(Pf, z, Pfs, C0)
local ANS, R, X, Y, Z, W, Wn, A, k, d, i;
    _EnvExplicit := false;
    A := NULL;
    for R in Pfs do
        #ANS := evala(residue(Pf, z = R));
        ANS := evala(coeff(series(Pf, z = R, 1 + degree(numer(Pf), z) + degree(denom(Pf), z)), z - R, -1));

        #print(R, Residue, ANS);
        X := R - k;
        Y := X;
        W := NULL;
        Wn := NULL;
        while hastype(Y, RootOf) do           
            Z := `evala/toprof`(Y);
            d := frontend(degree, [op(1, Z), _Z], [{Non}(function), {}]);
            W := W, seq((evala@Expand)(Y*Z^i), i = 0 .. d - 1);
            Wn := Wn, Z;
            Y := (evala@Norm)(Y, {Z}, indets(Y, RootOf) minus {Z})
        end do;
        ANS := frontend(factor@simplify, [ANS, [W], [Wn, k]],
            [({Non}@map)(identical, {Wn}), {}]);
        A := A, map(evalc@rationalize, subs(k = C0, ANS))
    end do;
    A
end proc:

 

ANS := {QRatpolyResidue}(Pf, z, Pfs, C0);
ANS2 := {QRatpolyResidue}(Pf, z, Pfs, k), k = C0;

 

returns:

C0 := 2^(1/4)*(-1)^(3/8)
P := z^2/(z^4+2*z^2+2)^2
Pf := z^2/(z+RootOf(RootOf(_Z^4+2*_Z^2+2)^2+_Z^2+2))^2/(RootOf(_Z^4+2*_Z^2+2)+
z)^2/(z-RootOf(RootOf(_Z^4+2*_Z^2+2)^2+_Z^2+2))^2/(z-RootOf(_Z^4+2*_Z^2+2))^2
Pfs := [RootOf(_Z^4+2*_Z^2+2), RootOf(RootOf(_Z^4+2*_Z^2+2)^2+_Z^2+2), -RootOf
(_Z^4+2*_Z^2+2), -RootOf(RootOf(_Z^4+2*_Z^2+2)^2+_Z^2+2)]

ANS := {(1/32-3/32*I)*2^(1/4)*(1/2*(2-2^(1/2))^(1/2)+1/2*I*(2+2^(1/2))^(1/2))}
ANS2 := {-1/32*k*(2+3*k^2)}, k = 2^(1/4)*(-1)^(3/8)

 

 

Simplify....

Christian Wolinski 1650
May 23 2016
0 0

   But for the RootOfs, simplify/mod2 would have been sufficient, so I suggest use frontend and simplify, which are meant for this task.

 

cmod2:=`@`(
f -> collect(f mod 2, RootOf),
(f, n) -> `if`(n = [], f, frontend(simplify, [f, map(x -> x^2 - x, n), n], [{Non}(function), {}])),
f -> (f, [op](select(type, frontend(indets, [f], [{Non}(function), {}]), name))),
f -> Expand(f) mod 2
):
alias(alpha = RootOf(x^4 + x + 1));
z := add(a[i] * alpha^i, i = 0 .. 3);
seq(cmod2(z^i), i = 0 .. 15);

Counterexamples ?

GB....

Christian Wolinski 1650
May 23 2016
2 4

@Carl Love There is no GB? I am certain there would be an equivalent. It is grobner basis modulo p.

 

 

with(share);

readshare(GB,'`mod`');
#readshare( `mod/GB` ):
L := 1+X+X^6+X^7+X^8, (X^3+X+1)*Q+Q^3+(X^7+X^6+X^4+1)*Q^2+X^7+X^6+X^4+X^3+X^2+X+1;
GB([L], [X, Q], plex) mod 2;

Results:

[X+Q^22+Q^21+Q^19+Q^17+Q^16+Q^15+Q^14+Q^9+Q^7+Q^6+Q^5+Q^2+Q, Q^19+1+Q^5+Q^21+Q^8+Q^10+Q^9+Q^12+Q^14+Q^24+Q^23+Q^13+Q^6+Q^16+Q^3+Q+Q^22]

Selector....

Christian Wolinski 1650
May 20 2016
0 3

The operator D is indexed.(and not parameterized) meaning it is a selector.

Consider:

 

D[1](f):=g;
D[1](g):=h;
D[1,1](f):=z;

and

 

(D[1]@@2=D[1,1])(f);

is a new statement altogether and not a simplification.

Expecting....

Christian Wolinski 1650
April 05 2016
0 0
 

A:=(-(1/2)*ib-(1/2)*ia+ic)*vc+(-(1/2)*ia-(1/2)*ic+ib)*vb+(-(1/2)*ic-(1/2)*ib+ia)*va;
B:=va+vb+vc:
C:=ia+ib+ic:
collect(A+1/2*B*C,indets(C));

Trial...

Christian Wolinski 1650
April 04 2016
0 5

I've just inputted this command and the return was NULL response:

 

isolve({x>3/2,x<5/2},{x});

Reading from the help file "The procedure isolve solves the equations in eqns over the integers. It solves for all of the indeterminates occurring in the equations."

It has no zeros....

Christian Wolinski 1650
March 26 2016
0 0

Try the code below and consider what it implies:

 

A := -144*z-44+12*sqrt(-12*z^3+96*z^2+24*z-15);;
B := 1728*z^2+2304*z+1024+432*z^3;


rationalize(B/A);

q := rationalize(B/A)*A-B;
evalb(rationalize(q)=0);

 

member(0,map(`@`(radnormal,unapply(A,z)),{solve}(B)));

Solve attempt....

Christian Wolinski 1650
March 24 2016
2 12

Is the following at all close to the anticipated solution?Result of solve.

Recurse with index....

Christian Wolinski 1650
March 22 2016
0 1

Try this code:

 

OperandsTable := proc()
local F;
    F := proc(A)
        local i, j, B;
            j := args[2 .. -1];
            if nops(A) = 1 then
                B := op(1, 'A');
                if type(B, '{indexed, list, set, `*`, `+`, `^`, function, relation, boolean, `::`, `..`, `.`, uneval}') then
                    j = eval(B, 1), procname([op(0, B)], j, 0), seq(procname([op(i, B)], j, i), i = 1 .. nops(B))
                else j = eval(B, 1)
                end if
            else j = op('A'), seq(procname([op(i, 'A')], j, i), i = 1 .. nops(A))
            end if
        end proc;
    table([F(['args'])])
end proc;

An example:

 

K:=OperandsTable('map((x->x)=SetPartitions,[op](5..6,remove(has,combinat[partition](10),1)))'):
eval(K[],1);
leafsindices:=remove(proc(L,S) hastype(S,[op(L),anything]) end,{indices}(K)$2):
leafs:=map((S,T)->eval(T[op(S)],1),leafsindices,K);
leafs_parameters:=map((S,T)->`if`( op(-1,S)=0 , NULL, eval(T[op(S)],1)),leafsindices,K);
leafs_operand0:=map((S,T)->`if`( op(-1,S)=0 , eval(T[op(S)],1), NULL),leafsindices,K);
all_operand0indices:=select(S->evalb(nops(S)>0 and op(-1,S)=0),{indices}(K)):
all_operand0:=map((S,T)->eval(T[op(S)],1),all_operand0indices,K);

Let me know if there is anything amiss.

 

A tiny bit of programming....

Christian Wolinski 1650
March 21 2016
1 1

Two versions of same but with small distinctions:

 

F := proc(N::nonnegint, lo::nonnegint, hi::nonnegint) local q, A, i; A := 'irem(q, 2, 'q')'; irem(N, 2^lo, 'q'), [seq(eval(A), i = lo .. hi)], q end proc;

or this:

 

F  := proc(N::nonnegint, range::(nonnegint .. nonnegint))
local q, A, i;
    A := unapply('irem(q, 2, 'q'), i'); irem(N, 2^lhs(range), 'q'), map(A, [`$`(range)]), q
end proc;


So F(8,0..7)[2]; gives [0, 0, 0, 1, 0, 0, 0, 0]

and F(11+5*256+7*256*256,8..15); gives 11, [1, 0, 1, 0, 0, 0, 0, 0], 7

Lookup ?listlist...

Christian Wolinski 1650
March 07 2016
0 0

Lookup ?member. This function is intended to compare expressions. In your example you are looking for matrices. The difference here is a matrx is a storage and identified by a reference and not its contents. On the other hand type listlist is an expression.

This code will match matrices only if the expressions of the elements are a match. This maybe sufficient for matrices of constants. Otherwise lookup ?iszero, ?Normalizer and test the difference.

fl:=a->convert(a,listlist):Jfl:=map(fl,J):
for i to 1 do for j to 2 do a := J[i].J[j]; member(fl(a), Jfl, 'k'); print(i, j, k, a, whattype(a)) end do end do;

 

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