dharr

Confirmed...

Answered: dharr 8205

March 12 2023

1 1

Yes, my Maple 2023.0 on Windows 10 shows this behaviour.

Try Insert->Table and choose 1 row, 1 column, Title: "Fig. 1" and tick Show caption. Then Insert->Image->From file and choose the file. After you have it, right-click (cmd-click on Mac) and choose Table -> Properties, Table size 50% and the image will reduce size 50% keeping the aspect ratio. That menu also has a "Caption" tab which gives several ways to specify the caption. As far as I know there isn't automatic renumbering.

(As you probably already found, you can insert images in a text area but resizing with the handles doesn't maintain the aspect ratio.)

another way...

Answered: dharr 8205

March 08 2023

1 0

(Notice that _B2 stands for a binary variable with values 0 or 1, but _Z1 and _Z2 are (different) integers.)

>

${alpha = Pi*_Z1, beta = (1/2)*Pi-Pi*_Z1-_B2*Pi+2*_B2*Pi*_Z1+2*Pi*_Z2, lambda = -arctan((2/3)*sin(Pi*_Z1)+(2/3)*cos((1/2)*Pi-Pi*_Z1-_B2*Pi+2*_B2*Pi*_Z1+2*Pi*_Z2))+Pi*_Z4}$

>	$undervars := indets(sol, suffixed({_B, _Z}))$

${_B2, _Z1, _Z2, _Z4}$

>

${alpha = n*Pi, beta = (1/2)*Pi+2*n^2*Pi, lambda = arctan((2/3)*sin(2*n^2*Pi))+Pi*_Z4}$

>

Download underline.mw

Unexpected list M...

Answered: dharr 8205

March 06 2023

2 0

Your code has a list M := [0, 0.5, 1.0, 1.5, 5] which you insert into other expressions as though it were a single number. So even for F[3] that isn't sensible. This ultimately leads to the error message.

one way...

Answered: dharr 8205

March 05 2023

1 0

mul(map(mul,ListTools:-Collect(A)))

one way...

Answered: dharr 8205

March 05 2023

0 0

CN and N1 are

CN := (G, u) -> add(map2(GraphTheory:-Degree, G, GraphTheory:-Neighborhood(G, u, 'closed')));
N1 := G -> add(map(e->(CN(G,e[1])+CN(G,e[2]))/2, convert(GraphTheory:-Edges(G),list)));

The others are straightforward modifications.

step by step...

Answered: dharr 8205

March 02 2023

3 1

This gives the answer in term of four algebraic numbers, each of which can be written in terms of the first root, which I think is more in the spirit of Galois theory than radicals. I also show the minimal polynomials that show the Galois group is order 72, But I'm not using the Galois group for much of this, as the OP asked for. Sorry if this is already obvious to the OP; perhaps we can have some clarification if the answers here are off track.

>

_local(gamma); poly := x^6-3*x+3

Doesn't factor in Q

>

Let alpha be the first root (index=1 by default)

>

alpha is the only root that can be written in terms of alpha. Notice that the rest factors, i.e. the quintic is reducible in Q(alpha)

>

Collect the factors

>

Let beta be the first root of the quadratic; the other can be expressed in terms of alpha and beta

>

(11/17)*alpha^5+(3/17)*alpha^4+(7/17)*alpha^3+(5/17)*alpha^2+(6/17)*alpha-36/17+((6/17)*alpha^5+(14/17)*alpha^4+(10/17)*alpha^3+(12/17)*alpha^2+(11/17)*alpha-15/17)*x+x^2

>

The cubic can't be done in terms of alpha and beta, so let gamma be its first root.

>

p3alpha; irreduc(p3alpha, {alpha, beta}); alias(gamma = RootOf(p3alpha, x))

x^3+((6/17)*alpha-(6/17)*alpha^5-(14/17)*alpha^4-(10/17)*alpha^3-(12/17)*alpha^2+15/17)*x^2+((6/17)*alpha^2-33/17+(3/17)*alpha^5+(7/17)*alpha^4+(5/17)*alpha^3-(3/17)*alpha)*x+(5/17)*alpha^3+18/17+(7/17)*alpha^4+(6/17)*alpha^2+(3/17)*alpha^5-(3/17)*alpha

It factors into a quadratic and the linear term, so we need another algebraic number as the first root of the quadratic

>

(x^2+(-(6/17)*alpha^5-(14/17)*alpha^4-(10/17)*alpha^3-(12/17)*alpha^2+(6/17)*alpha+15/17+gamma)*x+(3/17)*alpha^5+(7/17)*alpha^4+(5/17)*alpha^3+(6/17)*alpha^2-(3/17)*alpha-33/17-(6/17)*gamma*alpha^5-(14/17)*gamma*alpha^4-(10/17)*gamma*alpha^3-(12/17)*gamma*alpha^2+(6/17)*gamma*alpha+(15/17)*gamma+gamma^2)*(x-gamma)

>

x^2+(-(6/17)*alpha^5-(14/17)*alpha^4-(10/17)*alpha^3-(12/17)*alpha^2+(6/17)*alpha+15/17+gamma)*x+(3/17)*alpha^5+(7/17)*alpha^4+(5/17)*alpha^3+(6/17)*alpha^2-(3/17)*alpha-33/17-(6/17)*gamma*alpha^5-(14/17)*gamma*alpha^4-(10/17)*gamma*alpha^3-(12/17)*gamma*alpha^2+(6/17)*gamma*alpha+(15/17)*gamma+gamma^2, x-gamma

>

Can put it all together

>

(x-delta)*(x+(6/17)*alpha^5+(14/17)*alpha^4+(10/17)*alpha^3+(12/17)*alpha^2+(11/17)*alpha-15/17+beta)*(x-beta)*(x-alpha)*(x-gamma)*(x-(6/17)*alpha^5-(14/17)*alpha^4-(10/17)*alpha^3-(12/17)*alpha^2+(6/17)*alpha+15/17+gamma+delta)

This can be done all in one step with PolynomialTools:-Split - the splitting field K with four algebraic numbers is returned (but figuring out the nested RootOf's is painful).

>

(x-delta)*(x+(6/17)*alpha^5+(14/17)*alpha^4+(10/17)*alpha^3+(12/17)*alpha^2+(11/17)*alpha-15/17+beta)*(x-beta)*(x-alpha)*(x-gamma)*(x-(6/17)*alpha^5-(14/17)*alpha^4-(10/17)*alpha^3-(12/17)*alpha^2+(6/17)*alpha+15/17+gamma+delta)

>

x^2+(6/17)*x*alpha^5+(14/17)*alpha^4*x+(10/17)*alpha^3*x+(12/17)*alpha^2*x+(11/17)*x*alpha-(15/17)*x+(11/17)*alpha^5+(3/17)*alpha^4+(7/17)*alpha^3+(5/17)*alpha^2+(6/17)*alpha-36/17

>

x^3-(6/17)*alpha^5*x^2-(14/17)*alpha^4*x^2-(10/17)*alpha^3*x^2-(12/17)*alpha^2*x^2+(6/17)*alpha*x^2+(15/17)*x^2+(3/17)*x*alpha^5+(7/17)*alpha^4*x+(5/17)*alpha^3*x+(6/17)*alpha^2*x-(3/17)*x*alpha-(33/17)*x+(3/17)*alpha^5+(7/17)*alpha^4+(5/17)*alpha^3+(6/17)*alpha^2-(3/17)*alpha+18/17

>

x^2-(6/17)*x*alpha^5-(14/17)*alpha^4*x-(10/17)*alpha^3*x-(12/17)*alpha^2*x+(6/17)*x*alpha+(15/17)*x+x*gamma+(3/17)*alpha^5+(7/17)*alpha^4+(5/17)*alpha^3+(6/17)*alpha^2-(3/17)*alpha-33/17-(6/17)*gamma*alpha^5-(14/17)*gamma*alpha^4-(10/17)*gamma*alpha^3-(12/17)*gamma*alpha^2+(6/17)*gamma*alpha+(15/17)*gamma+gamma^2

Order of Galois group.

>

Download Splits.mw

precision and algorithms...

Answered: dharr 8205

March 01 2023

3 2

That is a surprising result, and sufficiently far enough from the correct answer that I would call it a bug, even though it is just a numerical issue. Certainly since the matrix is "nice" (symmetric, diagonally dominant). As @vv says it depends on the precision, but certainly you don't need to go to 40 to get OK results. At Digits=14, M_MP.M_MP - C is recognizably close to zero and at Digits=16, the entries are all about 1e-6 or 1e-7.

In the general case, the algorithm for matrix functions is nontrivial, but Maple often has special cases for shape=symmetric or shape = hermitian, so then I think just applying the function to the eigenvalues as you did is straightforward.

@vv's comment about the determinant might be true (I agree), but it seems to imply that the reason that <0,1;0,0> doesn't have a square root is because its determinant is zero. That has more to do with the fact that it is non-diagonalizable; for example <1,0;0,0> (positive semidefinite) and <1,1;0,0> both have square roots and have determinant zero.

Detecting edge cuts...

Answered: dharr 8205

February 25 2023

2 5

For the edge separator, Maple's IsCutSet can be used - it just checks for an increase in the number of components. Checking for a Cut is more complicated since just checking for 2 components can fail to detect extra removed edges that are in one of the components. Here is one way to do it.

[Edit - this actually tests for a cutset, which is a minimal Cut, so I've renamed it IsMinCut; see below for the test for a Cut]

>

restart;

IsMinCut returns true if the edges are a minimal cut (cutset), i.e., removal of the edges cuts the graph into two components, and no edge can be added back without reconnecting the graph.

This version doesn't handle loops for efficiency, but could be modified to use the underlying graph and ignore loops (provided edges doesn't contain a loop).

>

IsMinCut:=proc(G::GRAPHLN,edges::set)
       local partition,G1,G2;
       uses GraphTheory;
       if not IsConnected(G) or IsDirected(G) then error "Graph must be connected and undirected" end if;
       partition:=ConnectedComponents(DeleteEdge(G,edges,'inplace'=false));
       if nops(partition) <> 2 then return false end if;
       G1:=InducedSubgraph(G,partition[1]);
       G2:=InducedSubgraph(G,partition[2]);
       if (Edges(G) minus Edges(G1) minus Edges(G2)) = edges then
          true
       else
          false
       end if;
end proc:

>	with(GraphTheory): with(combinat): G:=CompleteGraph(3,3): edge:=choose(Edges(G), 7): `and`(seq(IsCutSet(G,s),s in edge)); # any 7 edges cuts the graph into 2 or more components

>	`or`(seq(IsMinCut(G,s),s in edge)); # no set of 7 edges is a cut

>	DrawGraph(G,size=[250,250]);

>	edges:={{1,4},{1,5},{1,6},{3,6}}; # {3,6} edge is extra IsMinCut(G,edges);

>	edges:={{1,4},{1,5},{1,6}}; IsMinCut(G,edges);

>	edges:={{1,4},{1,5},{1,6},{3,6},{3,5},{3,4}}; #3 components after cut IsMinCut(G,edges);

>

Download IsMinCut.mw

a guess...

Answered: dharr 8205

February 24 2023

1 0

It looks as though when there are more unkowns than equations that Mathematica sets the extra ones first in order to be zero. So something like this achieves that result.

Download solve.mw

ListTools:-Occurrences...

Answered: dharr 8205

February 24 2023

0 5

There are many ways; here is one. (Note that CycleBasis doesn't give all cycles in the graph)

Download countcycles.mw

variation...

Answered: dharr 8205

February 23 2023

2 2

A variation on @acer 's and @Kitonum's responses that bypasses radicals or floating points.

>

restart;

>	local gamma: alias(alpha=RootOf(_Z^3-4_Z^2+_Z+1,index=1),beta=RootOf(_Z^3-4_Z^2+_Z+1,index=2),gamma=RootOf(_Z^3-4*_Z^2+_Z+1,index=3));

>

eqs:=eval~(k[1]*x^3+k[2]*x^2*y+k[5]*x^2*z+k[3]*x*y^2+k[6]*x*y*z+k[8]*x*z^2+k[4]*y^3+k[7]*y^2*z+k[9]*y*z^2+k[10]*z^3,{{x=1,y=1,z=1},{x=RootOf(_Z^3-4*_Z^2+_Z+1,index=1),y=RootOf(_Z^3-4*_Z^2+_Z+1,index=2),z=RootOf(_Z^3-4*_Z^2+_Z+1,index=3)},{x=RootOf(_Z^3-4*_Z^2+_Z+1,index=2),y=RootOf(_Z^3-4*_Z^2+_Z+1,index=3),z=RootOf(_Z^3-4*_Z^2+_Z+1,index=1)},{x=RootOf(_Z^3-4*_Z^2+_Z+1,index=3),y=RootOf(_Z^3-4*_Z^2+_Z+1,index=1),z=RootOf(_Z^3-4*_Z^2+_Z+1,index=2)}});

${k[1]+k[2]+k[3]+k[4]+k[5]+k[6]+k[7]+k[8]+k[9]+k[10], k[1]*alpha^3+k[2]*alpha^2*beta+k[5]*alpha^2*gamma+k[3]*alpha*beta^2+k[6]*alpha*beta*gamma+k[8]*alpha*gamma^2+k[4]*beta^3+k[7]*beta^2*gamma+k[9]*beta*gamma^2+k[10]*gamma^3, k[1]*beta^3+k[2]*beta^2*gamma+k[5]*beta^2*alpha+k[3]*beta*gamma^2+k[6]*alpha*beta*gamma+k[8]*beta*alpha^2+k[4]*gamma^3+k[7]*gamma^2*alpha+k[9]*gamma*alpha^2+k[10]*alpha^3, k[1]*gamma^3+k[2]*gamma^2*alpha+k[5]*gamma^2*beta+k[3]*gamma*alpha^2+k[6]*alpha*beta*gamma+k[8]*gamma*beta^2+k[4]*alpha^3+k[7]*alpha^2*beta+k[9]*alpha*beta^2+k[10]*beta^3}$

>	eqs2:=solve(evala(eqs));

>	isolve(eqs2);

${k[1] = _Z1, k[2] = 10*_Z1-_Z2+2*_Z3+_Z4-_Z5-5*_Z6, k[3] = -4*_Z1-4*_Z2-_Z3-_Z5-4*_Z6, k[4] = _Z2, k[5] = _Z3, k[6] = -4*_Z1+3*_Z2-_Z3-3*_Z4+4*_Z5+22*_Z6, k[7] = -3*_Z1+_Z2-_Z3+_Z4-3*_Z5-14*_Z6, k[8] = _Z4, k[9] = _Z5, k[10] = _Z6}$

>

Download algeqns.mw

Condition for stable roots...

Answered: dharr 8205

February 20 2023

1 2

You can use PolynomialTools:-Hurwitz to find the conditions for stable roots, i.e for roots with negative real parts.

Download Hurwitz.mw

not easy to do...

Answered: dharr 8205

February 20 2023

3 2

I also would like an easy way to do this. I think evala(Reduce()) and evala(Normal()) want to have the the answer in the same algebraic field extension so don't do this. It can be done somewhat awkwardly through Minpoly; perhaps you had another objection to this.

>

q := -4*RootOf(_Z^3-3*_Z^2-10*_Z-1)^2*(1/5)+19*RootOf(_Z^3-3*_Z^2-10*_Z-1)*(1/5)+3/5; simplify(fnormal(evalf([allvalues(q)])))

-(4/5)*RootOf(_Z^3-3*_Z^2-10*_Z-1)^2+(19/5)*RootOf(_Z^3-3*_Z^2-10*_Z-1)+3/5

>

q2:=-45658*RootOf(37*_Z^6 - 382*_Z^5 + 1388*_Z^4 - 2188*_Z^3 + 1475*_Z^2 - 406*_Z + 37, index = 6)^5 + 417257*RootOf(37*_Z^6 - 382*_Z^5 + 1388*_Z^4 - 2188*_Z^3 + 1475*_Z^2 - 406*_Z + 37, index = 6)^4 - 1252087*RootOf(37*_Z^6 - 382*_Z^5 + 1388*_Z^4 - 2188*_Z^3 + 1475*_Z^2 - 406*_Z + 37, index = 6)^3 + 1463384*RootOf(37*_Z^6 - 382*_Z^5 + 1388*_Z^4 - 2188*_Z^3 + 1475*_Z^2 - 406*_Z + 37, index = 6)^2 - 558475*RootOf(37*_Z^6 - 382*_Z^5 + 1388*_Z^4 - 2188*_Z^3 + 1475*_Z^2 - 406*_Z + 37, index = 6) + 69230;