a, b = var('a,b')
((a + b)^2).expand()
How can I compute/expand the last expression assuming that $ab=0$? I.e. I would like the result to be $a^2+b^2$. I tried with assuming(..) but that doesn't have any effect.
It's easy for polynomial (rather than symbolic) variables:
K.<a,b> = QQ[]
J = K.ideal( [a*b] )
((a+b)^2).reduce(J)
Thanks, makes sense. What if the base field is more complicated, though? E.g. E.<w> = CyclotomicField(3) and I would like to assume $a + b + c = 0$ in the expression $(a + bw + cw^2)^3$ and would like to group the monomials by the powers of w instead of the monomials in a,b,c.
Something like this will do the job:
K.<w,a,b,c> = QQ[]
J = K.ideal( [cyclotomic_polynomial(3,w), a+b+c] )
f = ((a+b*w+c*w^2)^3).reduce(J)
Then you can extract the coefficients of powers of w as f.coefficient({w:0}) and f.coefficient({w:1}).
That's great! The only downside to this approach is the false symmetry between w,a,b,c in the poly ring definition. This doesn't correspond to reality and brings additional problems. For instance, I want to express various combinations of the w-coefficients as elementary symmetric polynomials. Is there a better way than this?
ABC.<a,b,c> = QQ[]
fABC = ABC(f.coefficient({w:0}))
SymK = SymmetricFunctions(QQ).e()
SymK.from_polynomial(fABC)
Plugging in f to SymK directly doesn't work because f.coefficient({w:0}) is of course not symmetric in K.<w,a,b,c>.
Instead of ABC(f.coefficient({w:0})) you can use f.coefficient({w:0}).specialization({w:0}). In fact for w^0 you can have simply f.specialization({w:0}) but for w^1 it has to be f.coefficient({w:1}).specialization({w:0}).
For an alternative approach, see my answer in https://ask.sagemath.org/question/75522/
Asked: 2024-01-17 23:26:43 +0100
Seen: 221 times
Last updated: Jan 18
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