vv

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@pallav This is not possible as my answer shows. It would be, if e.g. A were symmetric in x and sigma, but then A1 is independent of x, sigma. Take a simpler A to see what happens, e.g. A = x + 2 sigma^2 + x^2*c.  

@pallav You may use e.g. coeff(coeff(A,x^2),sigma^4).

But what do you want to do with the relations  

sigma+x=a, sigma*x=b

?

@Carl Love I mean after substitution, i.e. A1 or A2.

IsSubgraphIsomorphic has the option isomorphism (in version 2022).  Why don't you use it?

@Carl Love  It would be interesting to find a composite (i.e. non-prime) number for which isprime gives true (there should exist such a number!). 

@Carl Love The standard method for "quasi-homogeneous" functions:
g:=eval(f, [x=X+x0,y=Y+x0]);
determine x0,y0 such that g be homogeneous in X,Y; then z = Y/X.

You should not do this. It makes no sense in maths and in Maple (where D[1] acts on a procedure rather than an expression).

@rlopez To increase the chances for a global minimum, use:

SOL:=[infinity]; 
to 100 do
  sol:=Optimization:-Minimize(f,{G=0},initialpoint={x1=rand(0. .. 1.)(), x2=rand(-1. .. 1.)()}, feasibilitytolerance=1e-8, iterationlimit=1000);
  if sol[1]<SOL[1] then SOL:=sol fi
od:
SOL;

 

@nm It is exactly maths which does not agree with your opinion. See e.g. Richardson's theorem - Wikipedia

@Carl Love Should be:

IsRowPerm:= (A::Matrix, B::Matrix)-> #Is A a row permutation of B?
    evalb(`=`((sort@convert)~([A,B], 'listlist')[])):

to avoid 
IsRowPerm( <1;2;2;1>, <1;1;1;2> )  = true

@nm Yes, simplification is sometimes "art and craft":

r := sqrt(-x + 1/2 - sqrt(-4*x + 1)/2) + 1/2 - sqrt(-4*x + 1)/2:
xx := solve(sqrt(-4*x + 1)=t, x):
simplify(eval(r,x=xx)) assuming t>1;  # 0
simplify(eval(r,x=xx)) assuming t>0;  # ((signum(-1+t) - 1)*(-1+t))/2

 

@rlopez Of course, but for the function defined implicitely by w=0, x = c is a (vertical) asymptote.

@Carl Love The fact that

sys:=[a*x1^3 + b*x1^2 + c*x1 + d, a*x2^3 + b*x2^2 + c*x2 + d, a*x3^3 + b*x3^2 + c*x3 + d, x1*x2*x3 = -d/a, x1 + x2 + x3 = -b/a, x1*x2 + x1*x3 + x2*x3 = c/a]:
solve(sys, {x1, x2, x3}, explicit);

does not work, is a weakness of solve.
It is easy to see (but not for a CAS) that sys is equivalent to sys[4..6], which works!

@Carl Love 

g := (x, T) -> T*x + x^2:  #just an example
dgdx1:=(x,T) -> D[1](g)(x,T);
dgdx2:=  D[1](g);
dgdx1(1, 2);  # 4
dgdx2(1, 2);  # 4

g := (x, T) -> T*x + 7*x^2:
dgdx1(1, 2); # 16
dgdx2(1, 2); # 4

For a fixed g, dgdx2 is of course more efficient, but I don't think this matters here.

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